问:
How to compute the infinite product?
Given the following infinite product:
$$\lim_{n\to \infty} \frac{1}{2}\frac{3}{4}\cdots \frac{2^n-1}{2^n}.$$

It is easy to see that above infinite product is convergent by the convergence of the following series:
$$\sum_{n=1}^{\infty}\ln \left(1-\frac{1}{2^n}\right)$$.

But I do not know how to calculate the infinite product. I would appreciate it if someone can give any suggestions and comments.

Title: How to compute the infinite product?

Body:
For the infinite product
$$P=\frac{\prod_{i=0}^\infty (2^i-1) } {\prod_{i=0}^\infty 2^i }=\frac{315}{1024}\frac{\prod_{i=5}^\infty (2^i-1) } {\prod_{i=5}^\infty 2^i }$$

$$A=\log \left(\frac{\prod_{i=5}^\infty (2^i-1) } {\prod_{i=5}^\infty 2^i } \right)=\sum_{i=5}^\infty\log \left(1-\frac1 {2^i}\right)=-\sum_{i=5}^\infty \Big[\frac1 {2^i}+ \frac1 {2^{2i}}+\frac1 {2^{3i}}+\cdots\Big]$$ and we face geometric progressions. So, summing up to infinty, we have
$$A=-\Big[\frac{1}{16}+\frac{1}{768} +\frac{1}{28672}+\frac{1}{983040}+\frac{1}{32505856}+\frac{1}{1056964608}+\cdots\Big]$$ Using only these numbers
$$A \sim -\frac{10458550091}{163829514240}$$ Now using the approximation
$$e^A=\frac{1+\frac{A}{2}+\frac{A^2}{12}}{1-\frac{A}{2}+\frac{A^2}{12}}=\frac{311910183016998882664441}{332471213189757736214521}$$
$$P=\frac{98251707650354648039298915}{340450522306311921883669504}$$ which, converted to decimals is
$P=0.28859$ while the "exact" value is $P=0.28879$


问:
Closed form for $I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$
Is there a closed form for this integral?

$$I= \int_{-\infty}^{\infty} e^{-e^x-e^{-x}+x}~ dx$$

I tried $u=e^x$ but it didn't simplify things much.

I think the closed form is a Bessel function, but I can't verify this directly.

I'd like to see the steps involved to reach the closed form.

答:
The substitution actually helps quite a bit.
$$I=\int_{-\infty}^\infty \exp(-e^x-e^{-x}+x)\mathrm{d}x$$
Using $u=e^x$,
$$I=\int_0^\infty \exp\left(-u-\frac{1}{u}+\ln u\right)\frac{\mathrm{d}u}{u}=\int_0^\infty\exp\left(-u-\frac{1}{u}\right)\mathrm{d}u$$
Using the integral identity I referenced in your previous question this is
$$I=2K_{-1}(2)=2K_1(2)$$

ContentLicense: CC BY-SA 4.0


问:
$\int_X |f|^p d\mu < \infty$ $\leftrightarrow$ $\sum_{n=-\infty}^{\infty} a^n \mu$ ({$(x\in X:a^n \leq |f|^p (x) < a^{n+1}$}) < $\infty $
Given a measurable function: $f:X\rightarrow C $ on a measured space (X, A, $\mu$)

Show that $\int_X |f|^p d\mu < \infty$ $\leftrightarrow$ $\sum_{n=-\infty}^{\infty} a^n \mu$ ({$(x\in X:a^n \leq |f|^p (x) < a^{n+1}$}) < $\infty $, where $p\in [1,\infty[$, $a\in ]1,\infty[$, by using disjunct decomposition.

I have absolutely no idea, how to approach this problem. For example for the proof of "$\rightarrow$", I don't know, from where I get that given sum..

How do I approach this problem?

答:
HINT For fixed $a>1$, $f\ne0\iff |f|^p>0\iff \exists n\in\Bbb Z$ such that $a^n\le |f|^p< a^{n+1}$.

Ovbserve that $(0,+\infty)=\bigcup_{n\in\Bbb Z} [a^{n},a^{n+1})$ and the union is disjoint.


问:
Calculating the limit of a series
We have two series: $$x+\frac{x^3}{3}-\frac{x^4}{2}+\frac{x^5}{5}+\frac{x^7}{7}-\frac{x^8}{4}++-\ldots\tag{1}$$
and$$x+\frac{x^3}{3}-\frac{x^2}{2}+\frac{x^5}{5}+\frac{x^7}{7}-\frac{x^4}{4}++-\ldots\tag{2}$$

(the leading coefficients are taken from the alternating harmonic series, by a rearrangement, where we take two positive terms followed by a negative one)

EDIT: I think my book intented the following series:
for example, the first series can be written in compact form as $\sum_{n=0}^{\infty}\left(\frac{x^{4n+1}}{4n+1}+\frac{x^{4n+3}}{4n+3}-\frac{x^{4n+4}}{2n+2}\right)$. I'm not definitely sure, because the author provides those expressions, but it seems reasonable to use this one. For the second, I think $4n+4$ as exponent should be replaced by $2n+2$.

We want to study the behavior of them as $x\to1^-$. Clearly the first is a power series, and since it converges for $x=1$ to the value $3/2\log(2)$ (as well as the second which however isn't a power series, rather a rearrangement), by Abel's theorem we conclude that $1)$ tends to the same limit as $x\to 1^-$. What can we say about $2)$? I'm quite sure it doesn't converge uniformly in $(a, 1)$, where $a < 1$. Is it continuous at $x=1$? Is it bounded? For me it's clear it converges pointwise for every $x$ in $[0, 1]$.

A similar question can be raised for the series $$1+\frac{1}{3^x}-\frac{1}{2^x}+\frac{1}{5^x}+\frac{1}{7^x}-\frac{1}{4^x}++-\ldots\tag{3}$$ which is not a Dirichlet series, and hence Abel's theorem does not apply. What is its limit as $x\to1^+$?
How can I approach these problems? For $2)$ I thought I could study the derivative series, but it turns out it's not uniformly convergent.

答:
For $(2)$ the series $$\sum_{n=1}^\infty(-1)^n\frac {x^n}n=\log(1+x)$$ converges absolutely on $(-1,1)$ and so may be rearranged without altering the sum.  The lit as $x\to1^-$ is then $\log2$.  It seems to me that the convergence is uniform on closed intervals in $(-1,1)$.  If we write a partial sum as the sum of the odd terms plus the sum of the even terms, we have the sum of two partial sums of power series, both of which converge uniformly.

I don't know how to deal with the Dirichlet seies.

Title: Calculating the limit of a series


问:
I want to prove that for all normal number (M ) Takes place of GCD(m!+1,(m+1)!+1)=1 is true
I want to prove that for all normal number (M ) Takes place of
(m!+1,(m+1)!+1)=1 is true</p>

答:
I think (but am not sure) that you are asking:  for $n\in \mathbb N$ show that $\gcd(n!+1,(n+1)!+1)=1$

To see that, suppose we had a counterexample $n$. Then let $p$ be a prime which divided $\gcd(n!+1,(n+1)!+1)$.  It would then follow that $p$ divided the difference $(n+1)!+1-n!-1=(n+1)!-n!=n!\times n$.  But if $p\,|\,n!\times n$ then $p\,|\,n!$ in which case we certainly have $p\,\nmid (n!+1)$ so we have a contradiction, and we are done.


问:
Separate incomplete elliptic integral into real and imaginary parts
I am working in a problem that involves Incomplete Elliptic Integrals of the First and Second kind of the form $F(\sin^{-1}x~|~m)$ and $E(\sin^{-1}x~|~m)$ where the parameters $m$, $x$ are real numbers in the range $m>1$ and $1/\sqrt{m} \le x \le 1$.

($x$ and $m$ are related to the commonly used argument $\phi$ and modulus $m$ by $x \equiv \sin \phi$ and $m \equiv k^2$)

As can be seen by plotting them, in this range the real part of the integrals is independent of $x$ while the imaginary part isn't. As an example, see this plot for F and this other for E for a value $m=5$.

What I would like to do is to separate the real and imaginary part of this integrals, at least in this particular range. In other words, finding the real valued functions $f_{re} (m)$, $g_{re} (m)$, $f_{im} (x,m)$ and $g_{im} (x,m)$ that satisfy:

$$
F(\sin^{-1}x~|~m) \equiv f_{re} (m) + \text{i} f_{im} (x,m)
$$
$$
E(\sin^{-1}x~|~m) \equiv g_{re} (m) + \text{i} g_{im} (x,m)
$$
in the range $m>1$ and $1/\sqrt{m} \le x \le 1$.

By using the Reciprocal Modulus Transformations (see DLMF section 19.7) and taking the limit $x\rightarrow 1/\sqrt{m}$, I have found the real parts to be:

$$
f_{re}(m) \equiv \frac{1}{\sqrt{m}} K\left(\frac{1}{m}\right)
$$
$$
g_{re}(m) \equiv \sqrt{m} \left[ E \left( \frac{1}{m} \right) - K \left( \frac{1}{m} \right) \right] + \frac{1}{\sqrt{m}} K\left(\frac{1}{m}\right)
$$

However, the imaginary parts $f_{im} (x,m)$, $g_{im} (x,m)$ escape me. I reckon there should be a way of expressing them in terms of incomplete elliptic integrals with parameters in the real valued range.

If I use the reciprocal modulus transformations I will bring the parameter inside the range $0<m<1$ but the argument will now be complex as I will have $x>1$. I have looked everywhere in the literature but I can't seem to find any identity that solves the problem. I could perhaps do something if there was a way of expressing elliptic integrals of complex argument as a combination of elliptic integrals of real argument and imaginary pure argument, but I don't know how it can be done.

Does someone have any insight on how those imaginary parts could be found?

答:
The reciprocal modulus transformation is Byrd and Friedman 114.01:
$$F(\sin^{-1}x,m)=\frac1{\sqrt m}F(\sin^{-1}x\sqrt m,1/m)$$
$$E(\sin^{-1}x,m)=\frac1{\sqrt m}(mE(\sin^{-1}x\sqrt m,1/m)+(1-m)F(\sin^{-1}x\sqrt m,1/m))$$
Now the parameter is in $[0,1]$ but the sine-amplitude is a real number in $[1,\sqrt m]$. Thus B&F 115.02 applies:
$$F(\sin^{-1}x\sqrt m,1/m)=K(1/m)-iF(A,1-1/m)$$
$$E(\sin^{-1}x\sqrt m,1/m)=E(1/m)-i\left(F(A,1-1/m)-E(A,1-1/m)+\frac{(1-1/m)\sin A\cos A}{\sqrt{1-(1-1/m)\sin^2A}}\right)$$
where
$$A=\sin^{-1}\frac{\sqrt{mx^2-1}}{x\sqrt{m-1}}$$
Note that I have flipped the signs of the imaginary parts from B&F to match the values of $E(\cdot)$ and $F(\cdot)$ as calculated by Mathematica and mpmath. Finally we get
$$F(\sin^{-1}x,m)=\frac1{\sqrt m}(K(1/m)-iF(A,1-1/m))$$

$$E(\sin^{-1}x,m)=\frac1{\sqrt m}\left[mE(1/m)+(1-m)K(1/m)\\
-i\left(F(A,1-1/m)-mE(A,1-1/m)+\frac{(m-1)\sin A\cos A}{\sqrt{1-(1-1/m)\sin^2A}}\right)\right]$$

ContentLicense: CC BY-SA 3.0


问:
Evaluate $\int_0^3\sqrt\frac{x^3}{3-x}dx$.

Evaluate $\int_0^3\sqrt\frac{x^3}{3-x}dx$.

We can solve it by putting $x=3\sin^2\theta$. But it's not really an intuitive method.

Is there any other way to solve it?

Or, is there some pattern to this question that tells us instinctively that the substitution is $x=3\sin^2\theta$?

答: You can do the substitution $t^2=\frac{x}{3-x}$, since $$\sqrt{\frac{x^3}{3-x}}=x\sqrt{\frac{x}{3-x}}\qquad\text{for }x\ge0.$$


问:
Continuous functions that coincide in dense set are equal
Any corrections or feedback to the proof below will be highly appreciated.

Proposition. We have that $f\colon\mathbb{R}\to\mathbb{R}$ e $g\colon \mathbb{R}\to\mathbb{R}$ are continuous functions. Let $X_\alpha$ be a dense set in $\mathbb{R}$. If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f=g$.

Proof. As $f$ and $g$ are continuous functions, we have for every point $x_0\in\mathbb{R}$ such that $\lim\limits_{n\to \infty}{f(x_n)}=f(x_0)$ for every sequence of points $(x_n)_n \in \Bbb R$ with $\lim\limits_{n\to\infty} x_n = x_0$, and analogous definition is valid for $g$.

Because $X_\alpha$ is dense, then $\overline{X_\alpha}=\mathbb{R}$, that is, we can find for every $z\in\mathbb{R}$ a sequence $(z_n)_n$ in $X_a$ such that $z_n\to z$.

Then for every $z\in\Bbb R$ we do exactly that, defining a sequence $(z_n)_n$ in $X_\alpha$ such that $z_n\to z$. And as $f$ and $g$ are both continuous, we say that we have $f(z_n)\to f(z)$ as well as $g(z_n)\to g(z)$.

And if $\lim\limits_{n\to\infty}{f(z_n)}=f(z)$ e $\lim\limits_{n\to\infty}{g(z_n)}=g(z)$, then we have that

$$\forall \epsilon_1>0\ \exists n_f\in\mathbb{N}:\ n>n_f\implies \left|f(z_n)-f(z)\right|<\epsilon_1$$

and also

$$\forall \epsilon_2>0\ \exists n_g\in\mathbb{N}:\ n>n_g\implies \left|g(z_n)-g(z)\right|<\epsilon_2$$

We take $n_0=\min{(n_f,n_g)}$, and we have

$$n>n_0\implies \left|f(z_n)-f(z)\right|<\epsilon_1\ \text{e}\ \left|g(z_n)-g(z)\right|<\epsilon_2$$

We define $\varepsilon=\epsilon_1+\epsilon_2$, and therefore

$$\left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|<\varepsilon$$

If $f(x)=g(x)$ for all $x\in X_\alpha$, then $f(z_n)=g(z_n)$ for all natural $n$, because $z_n\in X_\alpha$. And by triangular inequality we have

$$\left|f(z)-g(z)\right|\leq \left|f(z_n)-f(z)\right| + \left| g(z_n)-g(z)\right|$$

But $\left|f(z)-g(z)\right|$ is absolute value, therefore

$$0\leq \left|f(z)-g(z)\right|<\varepsilon;\ \text{for all $\varepsilon>0$}$$

We claim that $0= \left|f(z)-g(z)\right|$, because if we assume the opposite, we would have $\left|f(z)-g(z)\right|= a >0$, then we could define $\varepsilon=\frac{a}{2}$ and find $a=\left|f(z)-g(z)\right|<\frac{a}{2}$, contradiction.

Hence,

$$0= \left|f(z)-g(z)\right|\implies f(z)=g(z)\ \text{for all $z\in\mathbb{R}$}$$

Ergo $f=g$. $\tag*{$\blacksquare$}$

$\rule{10cm}{0.4pt}$

What if $X_\alpha$ is not a dense set?

If $X_\alpha$ is not dense set, that is, $\overline{X_\alpha} \subsetneq \mathbb{R}$ is a proper subset. Well, this means that there is some $z\in\mathbb{R}$ such that $z\not\in\overline{X_\alpha}$, or yet, $z$ is not a limit point of $X_\alpha$, then there is no sequence $(z_n)_n$ of $X_\alpha$ such that $z_n\to z$.

答:
In fact you can simplify your proof a lot by denoting $h=f-g$. $h$ is continuous as the difference of two continuous maps. You then just have to prove that if a continuous map $h$ vanishes on a dense set $X_\alpha$, then $h$ vanishes on $\mathbb R$. Which is almost immediate. Take $a \in \mathbb R$. As $X_\alpha$ is dense, it exists a sequence $\{a_n\}$ of $X_\alpha$ such that $\lim\limits_{n \to \infty} a_n = a$ and then

$$f(a) = \lim\limits_{n \to \infty} f(a_n) = 0.$$


问:
$θ_2$ is better than $θ_1$ to estimate $μ$?
We offer two estimators for the average concentration $μ$ of lead in the atmosphere of a region of Quebec where factories manufacturing dyes are located. The first estimator $θ_1$ has a bias equal to $0.2$ and a variance of $0.02$. The second estimator $θ_2$ is unbiased and has a variance equal to $0.06$.

Which one is the best estimator?

I think $θ_2$ is better than $θ_1$ to estimate $μ$, but I am not sure.

EDIT

A PhD student in statistics explained to me that if $MSE(\theta_1) = MSE(\theta_2)$, then we cannot conclude. In other words, $\theta_2$ is not preferred over $\theta_1$ or inversely. I am not sure about that.

答:
In classical statistics, an estimator is better than another if its MSE is lower.

In this case

$$MSE(\theta_1)=0.02+0.2^2=0.06$$

$$MSE(\theta_2)=\mathbb{V}[\theta_2]=0.06$$

Being

$$MSE(\theta_1)=MSE(\theta_2)$$

$\theta_2$ is preferred as it is unbiased


问:
How to explain why $(1-i)^{2n+1} = a_n - ib_n$
I've just proved by induction that $(1+i)^{2n+1} = a_n + ib_n$ where $a_n$ and $b_n$ are $2^n$. And now I need to explain why $(1-i)^{2n+1} = a_n - ib_n$ and deduce from it why $|P_n(1)=2^n|$ where $P_n(X) = \frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ I have no ideas how to proceed.

答:
If you prove that $(1+i)^{2n+1}=a_n+ib_n$, for real $a_n,b_n$, then taking the conjugates shows that
$$
(1-i)^{2n+1}=a_n-ib_n
$$
There's no other computation to do.

Let's try to see what $a_n$ and $b_n$ should be. For $n=0$, we have $a_n=b_n=1$. Then
$$
a_{n+1}+ib_{n+1}=(1+i)^{2n+3}=(a_n+ib_n)(1+i)^2=2i(a_n+ib_n)=-2b_n+2ia_n
$$
Hence you have $b_{n+1}=2a_n$ and $a_{n+1}=-2b_n$.

However, there's a straighter path: $(1+i)^2=2i$, so you have
$$
(1+i)^{2n+1}=2^{n+1}i^n(1+i)
$$
and it should now easy to compute $a_n$ and $b_n$.


问:
can graph $K_5$ be embedded in a Möbius band?
I know that $K_5$ can be embedded in torus, but I am not sure if it is possible in Möbius band.
Please, can someone explain or show me this?

Thanks.

答:
The answer is yes.

Here a more general question was asked about the criteria for embedding a graph in the Möbius strip. There it was pointed out that graph embedding in a Möbius strip is equivalent to embedding a graph in a projective plane, because cutting out a disk from the latter gives you the former, or equivalently, gluing the former to a disk along their boundaries gives you the latter.

Now since the projective plane can be thought of as a disk with opposite points on the boundary identified, we can actually use the same idea for the torus embedding to construct a projective plane embedding.

In other words, in the above picture, instead of identifying opposite edges to form a torus, we identify points along the edge that are centrally symmetric to each other. You will notice that the ends of the "broken" edges are still identified to each other, so this picture also illustrates an embedding of $K_5$ into the projective plane.


答:
Or, is there some pattern to this question that tells us instinctively that the substitution is $x=3\sin^2\theta$?

To the familiar eye, yes, there is.

Note powers of $x,\,3-x$ are important here. Whenever two quantities in the integral you should care about have a nonzero constant sum, it's worth leveraging $\sin^2\theta+\cos^2\theta=1$, or possibly another sum-of-squares identity such as $\tanh^2\phi+\operatorname{sech}^2\phi=1$, but with the right transformation these are basically the same idea.

Similarly, if two quantities you have to care about have a nonzero constant difference, exploit $\sec^2\theta-\tan^2\theta=1$ or, by the same logic, maybe $\cosh^2\phi-\sinh^2\phi=1$.

(Mind you, out of circular vs hyperbolic you often find one more directly helpful with whichever problem is at hand, e.g. I'd recommend the circular option here.)

Once you've switched to trigonometry, further tips often write the rest of the calculation for you. Actually, in this case the latter brings you back round to no trigonometry at all; the real trick is $x=3y$. But diving into trigonometry often helps you spot a useful rational transformation, which may well be much less trivial in other cases.

Body:
Evaluate $\int_0^3\sqrt\frac{x^3}{3-x}dx$.

We can solve it by putting $x=3\sin^2\theta$. But it's not really an intuitive method.

Is there any other way to solve it?

Or, is there some pattern to this question that tells us instinctively that the substitution is $x=3\sin^2\theta$?


问:
Geometry - Equilateral triangle covered with five circles
I have to cover an equilateral triangle (whose sides are 1m long) with 5 identical circles: what's the minimum radius of the circles?</p>

答:
As a lower bound, consider circles that perfectly cover the triangle (which is impossible) and thus have the same combined area as the triangle. The area of the triangle is $\frac {\sqrt{3}}{4}$, so each circle would need to have an area of $\frac {\sqrt{3}}{20}$, or a radius of $\sqrt \frac {\sqrt{3}}{20 \pi} \approx 0.166$.

As an upper bound, consider three copies of the inscribed circle, each moved halfway to each corner of the triangle (the placement of the fourth and fifth circles is unimportant in this solution). The radius of those circles is $\frac {\sqrt{3}}{6}\approx 0.289$.

Thus the minimum radius to solve your problem is $0.166 \lesssim r \lesssim 0.289$.


问:
$|| P ||\leq 1$
Can somebody explain the final step why $ || P ||\leq 1$ here:
https://projecteuclid.org/download/pdf_1/euclid.pcma/1416340413

$P$ is a Projector and it self-adjoint.

Title: $|| P ||\leq 1$

Body: $$\|Px\|^2\le \|Px\|^2+\|Px-x\|^2=\|x\|^2$$ so $\|Px\|\le\|x\|$ which implies that $\|P\|\le 1$.


问:
Absolute Convergence of Improper Integral of sinx/(e^x-e^(-x))
I want to prove absolute convergence of improper integral ;

\begin{equation}
\int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx.
\end{equation}

But I couldn't prove this. (I don't have to calcurate the value. I only have to prove absolute convergence.)

My attempt is as follows.

$
\left|
\dfrac{\text{sin}x}{e^x-e^{-x}}
\right|
\leqq
\dfrac{1}{e^x-e^{-x}} \\
\int_{\epsilon}^N \dfrac{1}{e^x-e^{-x}} dx
=_{(t=e^x)}
\int_{e^{\epsilon}}^{e^N} \dfrac{1}{t^2-1} dt
=
\dfrac{1}{2}
\left[
\text{ln} \left| \dfrac{t-1}{t+1} \right|
\right]^{e^{N}}_{e^{\epsilon}}
$

I want to let $\epsilon \rightarrow 0$ and $N \rightarrow \infty.$

However, if I let $\epsilon \rightarrow 0$,
$
\text{ln} \left| \dfrac{e^{\epsilon}-1}{e^{\epsilon}+1} \right| \rightarrow - \infty
$

I would like you to give me any ideas or hints to prove absolute convergence of

\begin{equation}
\int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx.
\end{equation}

答:
The integrand is $$\frac 12 \sin (x) \text{csch}(x)$$ Composing Taylor series, you have
$$\frac 12 \sin (x) \text{csch}(x)=\frac{1}{2}-\frac{x^2}{6}+O\left(x^4\right)$$ so, no problem around $x=0$


问:
Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$
We have to prove the convergence and find the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$. I tried to find $a_{n+1}$ in terms of $a_{n}$ and got $a_{n+1}=\frac{n+2}{2(n+1)}\left(a_n+1\right)$ and could not proceed further.Any help is appreciated.

Edit : Following K.defaoite's answer limit appears to be 1 . But we still need to prove convergence.

答:
OK, I see what needs to be done here. Denote the sum in brackets by $S_n$

$$S_n = \left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$$

First we prove by induction (for $n \ge 4$) the following:

$$S_n \gt \frac{ 2^{n+1} (n+2)}{(n+1)n} \tag{1}$$

Then using (1) it is easy to prove by induction that:

(2) the sequence is decreasing for $n\ge4$ i.e. $a_n \gt a_{n+1}$

(3) $a_n \ge 1$ for $n\ge4$ (that is even true for every $n$ but for us it's enough that it's true for $n\ge4$)

Once these two statements are proved, the convergence follows.

And then you take limits (that's the easy part as K.defaoite showed) and you're done.

Here is how using (1) we can prove the sequence is decreasing: we have $a_n=\dfrac{n+1}{2^{n+1}}S_n$ and $a_{n+1}=\dfrac{n+2}{2^{n+2}}S_{n+1}$ $=\dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$. Now, the condition that $a_n\gt a_{n+1}$ is that $\dfrac{n+1}{2^{n+1}}S_n\gt \dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$; this is equivalent to $2(n+1)S_n\gt (n+2)(S_n+\dfrac{2^{n+1}}{n+1})$, which is equivalent to $nS_n\gt \dfrac{(n+2)2^{n+1}}{n+1}$ (subtract $(n+2)S_n$ from either side), and dividing this by $n$ gives (1).


问:
Determining the largest interval of definition of a Cauchy problem
I have some problems trying to solve this exercise:

Consider the following Cauchy problem: $\left\{\begin{matrix}x'=t+\frac{y}{1+x^2}
\\ y'=txe^{-ty^2}
\\ x(0)=y(0)=1
\end{matrix}\right.$

Discuss the existence of a solution and its uniqeness. Then I'm asked to prove or to disprove the fact that the supremum of the max interval on which the solution is definted is $[0,+\propto)$.

I've defined a function as following $F:\mathbb{R}^3 \rightarrow \mathbb{R}^2$, $F(t,**z**=(x,y))=\begin{bmatrix}t+\frac{y}{1+x^2}
\\ txe^{-ty^2}
\end{bmatrix}$

Since $F$ is infinitely differentiable (which implies it's continuous and locally lipschitz) on $\mathbb{R}^3$, we can conclude that $\exists \delta>0, \ \exists \ \phi:[-\delta,\delta] \rightarrow \mathbb{R}^2$ which is a unique solution of the Cauchy problem on that interval. I have some troubles while trying to compute the max interval of definition of the solution. Can someone suggest me how to procede?

Thanks in andvance for the help.

答:
If the function is sub-linear in state-space direction,
$$
\|F(t,u)\|\le a(t)\|u\|+b(t)
$$
then the solution is bounded per Grönwall lemma by the solution of the linear ODE
$$
v'(t)=a(t)v(t)+b(t),~~ v(0)=\|u(0)\|.
$$
As solutions of linear DE exist wherever the coefficients are continuous, the bound and thus the solution exist wherever the sub-linearity can be established.

Sublinearity is easy to see for $t>0$, for $t<0$ the second equation is not sub-linear due to the exponential term.


问:
How to calc this series?
I am supposed to calculate this series $$\sum_{n = 1}^{\infty} \frac{2}{(4n²+8n+3)}$$
I couldn't do it exactly, actually. I saw that:
$$\sum_{n = 1}^{\infty} \frac{1/2}{(n+1)^2} < \sum_{n = 1}^{\infty} \frac{2}{(4n²+8n+3)} < \sum_{n = 1}^{\infty} \frac{1/2}{(n^2)}$$

And i know that $$\sum_{n = 1}^{\infty} \frac{1}{(n^2)} = \pi/6$$

So that, the only answer between the alternatives (It was a question with alternatives) that satisfy this condition is $1/3$, which after i put on the wolframalpha it is really the result. But i would like to know how to calc the answer exactly.

Title: How to calc this series?

Body:
$$\frac{2}{4n^2+8n+3}=\frac{1}{2n+1}-\frac{1}{2n+3}.$$
The sum should telescope.


问:
If $\sum_{n=1}^\infty a_n$ converges with $a_n \geq 0$, then $\sum_{n=1}^\infty n a_n^2$ also converges or diverges?
Given $\sum_{n=1}^\infty a_n$ converges, with $a_n \geq 0$ $\forall n \in \bf{N}$, I want to know whether $\sum_{n=1}^\infty n a_n^2$ also converges.

Given the condition that $S_n =\sum_{n=1}^\infty a_n$ converges with $a_n \geq 0$ $\forall n$, then $\sum_{k=1}^\infty a_n^2$ also converges (proved result).

In addition, it is not always the case that $\sum_{n=1}^\infty n a_n$ converges (counterexample: let $a_n = \frac{1}{n^2}$), although $\lim_{n \to \infty} n a_n$ converges.

I am trying to see if $\sum_{n=1}^\infty n a_n^2$ would converge given the previous information as well as the results that I have obtained above and given to me by theorems. I cannot seem to figure out any counterexamples (if at all). Note that the sum goes from $n = 1 \to \infty$, so no series of the form $a_n = \frac{1}{n^p(\ln{n})^q}$ cannot be used (as the case $n=1$ is left out).

I don't see a good comparison test nor other convergence tests to use, along with monotone-bounded or Cauchy series principles. I am also at a loss of certain theorems that may be handy.

答:
Let
$$ a_n=\begin{cases}\frac1{k^2}&\text{if }n=k^4\\0&\text{otherwise}\end{cases}$$


问:
Show that if a quadratic has rational coefficients and $x + y \sqrt{z}$ is a root, then $x - y \sqrt{z}$ must also be a root.
Given $x, y$ and $z$ are rational and $\sqrt{z}$ is not, show that if $x + y\sqrt{z}$ is a root of a quadratic with rational coefficients, then $x - y\sqrt{z}$ is also a root.

In the situation the discriminant of the equation is positive. I think I have an idea how to start but haven't been able to go any further.

\begin{equation}
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = x \pm y \sqrt{z}
\end{equation}

Is my setup incorrect? How can I approach this problem?

答:
Maybe try using the following relationships of the roots and coefficients of a quadratic polynomial:
$\text{Let } a, b, c \in \mathbb{Q}.$

$$ax^2 + bx + c = a(x-x_1)(x-x_2) = a(x^2 -(x_1 + x_2) + x_1 \cdot x_2) = ax^2 - a(x_1 + x_2)x +a \cdot x_1 \cdot x_2.$$
From here we get the following system of equalities:
\begin{cases} x_1 + x_2 = -\frac{b}{a} \\\ x_1 \cdot x_2 = \frac{c}{a} \end{cases}

Notice that if $x_1 = x + \sqrt{z}$, then $x_1 + x_2 = x + \sqrt{z} + x_2 = -\frac{b}{a} \in \mathbb{Q}$, then we can solve for $x_2$:
$$x_2 = (-\frac{b}{a} - x) - \sqrt{z} = m - \sqrt{z}, \text{ where } m \in \mathbb{Q}$$
Now let's use the second relationship we proved:
$$x_1 \cdot x_2 = (x+\sqrt{z})(m-\sqrt{z}) = (xm - z) + (m  - x)\sqrt{z} = \frac{c}{a} \in \mathbb{Q}$$
If $m - x \neq 0$ we could solve for $\sqrt{z}$ to get that it must be rational, but this contradicts our assumption that it is irrational, hence $m = x$, which implies $x_2 = m -\sqrt{z} = x - \sqrt{z}$.


问:
Is $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$ if $g$ is a Sobolev function on the real line?
If we are given a function $g\in W_2^k(\mathbb{R})$ (even consider $k=1$ for simplicity), then is it true or not that $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$?  That is, do we have $$\underset{n\in\mathbb{Z}}\sum|g(n)|^2<\infty$$
?

At first I thought that this must be false, but after trying for a while to construct a counterexample, I found it a little more difficult than I thought.  The conclusion is certainly not true for functions only in $L_2(\mathbb{R})=W_2^0(\mathbb{R})$, because consider the function $g$ so that $g^2$ forms triangles or tent functions at each integer of height 1 and width $2^{-n}$.  The area is square summable, but the above sum is infinite.  However when we have to have control over even the first derivative it becomes a little harder, because such a tent function no longer works.

My idea was to try to set $g(n)=n^{-1/2}$ for an infinite subset of the natural numbers (consider $g(x)=0$ for $x\leq 0$), and try to ensure that the functions derivative was small enough to be square summable, but this didn't seem to work.

Any suggestions or comments would be helpful.  I am sure somebody knows this even though I have never seen it before.

答:
Theorem

Let $(x_n)_{n\in\mathbb{Z}}$ be a set of separated points in $\mathbb{R}$ such that $\underset{n\in\mathbb{Z}}\inf x_{n+1}-x_n>c$.
If $g\in W_p^k(\mathbb{R})$, $1\leq p\leq\infty$, $k\in\mathbb{N}$, then $(g(x_n))\in\ell_p$.

Proof:

First, in the $p=\infty$ case, if $g\in W_\infty^k(\mathbb{R})$, then $g$ is Lipschitz, and so the result is obvious. (That $W_\infty^1(\mathbb{R})=Lip(\mathbb{R})$ is a nontrivial result).

So we tackle the case $1\leq p<\infty$.  Let $$I_n:=\left[x_n-\frac{c}{3},x_n+\frac{c}{3}\right],\quad n\in\mathbb{N}$$
We define them this way so that they are disjoint, and also $|I_n|=\frac{2}{3}c$ for all $n$. By disjointness, we have that
$$\underset{n\in\mathbb{Z}}\sum\int_{I_n}|g|^p\leq\int_\mathbb{R}|g|^p<\infty$$
On the other hand, if $y_n:=\underset{x\in I_n}{argmin}|g(x)|^p$, then we have that
$$|g(y_n)|^p\frac{2}{3}c\leq\int_{I_n}|g|^p$$
and consequently, $(g(y_n))\in\ell_p$.

The following Lemma will be very useful presently.

Lemma: [A First Course in Sobolev Spaces, Leoni, Thm 7.13, p.222]

Let $\Omega\subset\mathbb{R}$ be open and $u:\Omega\to\mathbb{R}$.  Let $1\leq p<\infty$.  Then $u\in W_p^1(\mathbb{R})$ if and only if it admits an absolutely continuous representative, $\tilde{u}:\Omega\to\mathbb{R}$ such that $\tilde{u}$ and its classical derivative $\tilde{u}'$ belong to $L_p(\mathbb{R})$.  Moreover, if $p>1$, then $\tilde{u}$ is Holder continuous of exponent $1/p'$.

So since $W_p^k(\mathbb{R})\subset W_p^1(\mathbb{R})$, taking $\Omega=\mathbb{R}$ in the Lemma allows us without loss of generality to assume that $g$ is absolutely continuous.  Therefore by the Fundamental Theorem of Calculus,
$$g(x_n)=g(y_n)+\int_{y_n}^{x_n}g(t)dt$$
Using the inequality $|a+b|^p\leq 2^p(|a|^p+|b|^p)$, we have
$$|g(x_n)|^p\leq 2^p\left[|g(y_n)|^p+\left|\int_{y_n}^{x_n}g(t)dt\right|^p\right]\leq 2^p\left[|g(y_n)|^p+\left(\frac{2c}{3}\right)^\frac{p}{p'}\int_{I_n}|g(t)|^pdt\right]$$
The second inequality follows by Holder's Inequality:
$$\left|\int_{y_n}^{x_n}g(t)dt\right|^p\leq\left(\int_{y_n}^{x_n}|g(t)|dt\right)^p\leq |I_n|^\frac{p}{p'}\int_{I_n}|g(t)|^pdt$$

We conclude that
$$\underset{n\in\mathbb{Z}}\sum|g(x_n)|^p\leq C\left[\underset{n\in\mathbb{Z}}\sum|g(y_n)|^p+\underset{n\in\mathbb{Z}}\sum\int_{I_n}|g|^p\right]<\infty$$
and the conclusion follows.


问:
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$?
What is the maximum value of $\frac{x-x^3}{x^4+2x^2+1}$  ?

I've got the hint that this problem can be solved by replacing $x$ with tangent $K$.

So it became $\frac{\tan K(1-\tan^2K)}{(\tan^2K+1)^2}=\frac{\tan K}{\tan^2K+1}\cdot\frac{1-\tan^2 K}{\tan^2K+1}=\frac{\sin(2K)}{2(1-\sin^2K)}$
then I have no idea what to do next.

Any help is appreciated, thank you.^^

答:
$$\frac{x-x^3}{\left(x^2+1\right)^2}$$
substitute $x=\tan k$ and get
$$\frac{\tan k-\tan ^3 k}{\left(1+\tan ^2 k\right)^2}=\frac{\tan k}{1+\tan^2 k} \cdot \frac{1-\tan^2 k}{1+\tan^2 k}=\frac{1}{2}\sin 2k\cos 2k=\frac{1}{4}\sin 4k $$
as $\sin 4k$ has a maximum of $1$, the expression has a maximum of $\frac{1}{4}$.


问:
Integer solutions to Polynomial
Let $f(x) = x^3-px^2+qx$ and $g(x) = 3x^2-2px+q$ where $p,q \in \mathbb{Z}^+$. Show that if $f=0$ has three distinct integer solutions and $g = 0$ has 2 distinct integers solutions then $p$ is a multiple of 3 and $q$ is a multiple of 9.

I've been trying to do this problem for some time and I keep getting back to the discriminant,and I know it must be a perfect square, so I must have that $p^2-4q = n^2$ and $p^2-3q = m^2$ and then I thought to subtract them:
$$m^2 - n^2 = p^2 -3q - p^2 +4q = q$$

But then that got me nowhere too.. I feel like there is something basic here I am missing.

答:
Using Vieta, as the sum of the roots of $g(x)$ is $\dfrac{2p}3$, for integer roots, we must have $3\mid p$.  Similarly, as the product of the roots is $\dfrac{q}3$, we must have $3\mid q$.

Further, if $f(x)=x\,h(x)$, we need $h(x) = x^2-px+q$ to have two integer roots, and as $3\mid q$ the product of roots, one of the roots is divisible by $3$, and as $3\mid p$ the sum also, both the roots must be multiples of $3$, hence $9 \mid q$.  (Essentially Eisenstein's criterion, if you're familiar with it.)

--

P.S. The distinct root condition seems unnecessary here.

Title: Integer solutions to Polynomial


问:
singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$
How to show that all possible collection of analytic continuations of $\displaystyle f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} $ has singular point at $z = 1$. I know that $f(z)$ converges for $|z| \le 1$. Also is there a theorem that relates the singularity of analytic continuation with circle of convergence?</p>

答:
This is one of the polylogarithm, also $s=2$ case is called dilogarithm as GEdgar pointed out.
The definition is
$$
\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,
$$
Your case it is $s=2$, so $\operatorname{Li}_2(z)$.

According to http://en.wikipedia.org/wiki/Polylogarithm, we have a integral representation of $\operatorname{Li}_s(z)$:

$$
\operatorname{Li}_{s}(z) = {1 \over \Gamma(s)}
\int_0^\infty {t^{s-1} \over e^t/z-1} \,\mathrm{d}t \,.
$$

So in your case
$$
\operatorname{Li}_{2}(z) =
\int_0^\infty {zt  \over e^t  - z} \,\mathrm{d}t \,.$$

This allows an analytic continuation to $z\in \mathbb{C}-[1,\infty)$.


问:
Proving that the the sequence $x_n = \frac{n^{100}}{2^n}$ converges using only elementary methods
I realise that there are already questions on the site about this sequence or related sequences, but I'm looking for a completely elementary method that doesn't use:

L'Hopital's Rule

Tests like the comparison test or ratio test

Logarithms

Statements like "exponentials grow faster than polynomials"

I'm trying to show that the sequence $$x_n = \frac{n^{100}}{2^n}$$ converges as $n \to \infty$, where $n$ is a positive integer. I haven't been able to find a way to do this directly using the epsilon definition for convergence. Is there a completely elementary way to show that this converges (not necessarily needing to show what it converges to) without using any of the things mentioned above?

答:
I think this is the most elementary we can get. Look at the ratio between two consecutive terms:
$$
\frac{x_{n+1}}{x_n}=\frac{(n+1)^{100}/2^{n+1}}{n^{100}/2^n}\\
=\frac{(n+1)^{100}}{2n^{100}}\\
=\frac12\left(1+\frac1n\right)^{100}
$$
Now note that from some point on, $1+\frac1n<\sqrt[100]2$. This means that from that point on, we get $\frac{x_{n+1}}{x_n}<1$, which means that the sequence is monotonically decreasing. At the same time, it is trivial that $x_n>0$ for all $n$. Thus $x_n$ must converge.


Title: Proving that the the sequence $x_n = \frac{n^{100}}{2^n}$ converges using only elementary methods

Body:
I realise that there are already questions on the site about this sequence or related sequences, but I'm looking for a completely elementary method that doesn't use:

L'Hopital's Rule

Tests like the comparison test or ratio test

Logarithms

Statements like "exponentials grow faster than polynomials"

I'm trying to show that the sequence $$x_n = \frac{n^{100}}{2^n}$$ converges as $n \to \infty$, where $n$ is a positive integer. I haven't been able to find a way to do this directly using the epsilon definition for convergence. Is there a completely elementary way to show that this converges (not necessarily needing to show what it converges to) without using any of the things mentioned above?

Body:
Let $h=\sqrt[200]2-1$ and note that $h>0$.
Then using Bernoulli's inequality
$$ 2^n=(2^{n/200})^{200}=((1+h)^n)^{200}\ge (1+hn)^{200}>h^{200}n^{200}$$
so that
$$0<\frac{n^{100}}{2^n}<\frac{1}{h^{200}n^{100}}<\epsilon $$
as soon as $n>{\frac1{h^2\sqrt[100]\epsilon}}$.


问:
Intersection of collection of sets not in the collection of sets?
Hello I was doing a practice problem and I came across something that kind of tripped me up. The question says:

Suppose the Collection $F$ is given by $F=\{ [1, 1+ 1/n] : n\in \mathbb{N}\}$

Find the union of this collection of sets and find the intersection of these sets (I'm struggling to type this).

I looked at the solution and I don't understand why the intersection is $\{1\}$. This element is not actually contained in any of the sets since all the elements of the set are intervals. Can anyone please explain why this is? Intuitively on some level it makes sense since $1$ is the only common "thing" for lack a of a better, word but isn't the intersection of a collection of sets be the elements that are common to all the sets?

答:
The element $1$ is contained in all of the sets of the collection; the set $\{1\}$ is not a member of the collection, but it is a subset of each member of the collection. (It actually is an interval, by the way: $\{1\}=[1,1]$, a degenerate closed interval. By definition $[a,b]=\{x:a\le x\le b\}$, and if $a=b$ that set contains just the one element $a=b$.)

For each $n\in\Bbb Z^+$ let $F_n=\left[1,1+\frac1n\right]$, and let $\mathscr{F}=\{F_n:n\in\Bbb Z^+\}$. ($\mathscr{F}$ is the collection called $F$ in your question; I’ve changed the name to $\mathscr{F}$ to distinguish it more clearly from the names that I’ve given to the individual sets in the collection.) You want

$$\begin{align*}
\bigcap\mathscr{F}&=\bigcap_{n\in\Bbb Z^+}F_n\\\\
&=\bigcap_{n\ge 1}\left[1,1+\frac1n\right]\\\\
&=[1,2]\cap\left[1,\frac32\right]\cap\left[1,\frac43\right]\cap\left[1,\frac54\right]\cap\ldots\cap\left[1,\frac{n+1}n\right]\cap\ldots\;.
\end{align*}$$

This is the intersection of a bunch of sets of real numbers, so it will itself be some set of real numbers. Specifically, it’s the set of all real numbers that are in all of the sets $F_n$:

$$\begin{align*}\bigcap\mathscr{F}&=\{x\in\Bbb R:x\in F_n\text{ for all }n\in\Bbb Z^+\}\\\\
&=\left\{x\in\Bbb R:1\le x\le 1+\frac1n\text{ for all }n\ge 1\right\}\;.
\end{align*}$$

If $n$ is any positive integer, it’s certainly true that $1\le 1\le 1+\frac1n$, so $1\in F_n$. Thus, $1\in\bigcap\mathscr{F}$. Suppose that $x\ne 1$. If $x<1$, then obviously $x$ does not belong to any of the intervals $F_n$. If $x>1$, then $x-1>0$, and there is a positive integer $n$ large enough so that $\frac1n<x-1$. But then $x>1+\frac1n$, so $x\notin\left[1,1+\frac1n\right]=F_n$. That is, if $x$ is any real number other than $1$, we can find an $n\in\Bbb Z^+$ such that $x\notin F_n$, and therefore $x\notin\bigcap\mathscr{F}$. Thus, $1$ is the only real number that belongs to all of the intervals $F_n$, so it’s the only thing in the intersection of all those intervals:

$$\bigcap\mathscr{F}=\{1\}\;.$$


答:
Attacking discriminants is the natural try, but is the wrong approach for this particular problem.  The problem can be solved via
vieta's formulas.

Since $3x^2 - 2px + q = 0$ has two roots,
$r_1, r_2$, then these roots also satisfy:

$$x^2 - \frac{2}{3}px + \frac{1}{3}q = 0.$$

Therefore, since $r_1$ and $r_2$ are integers, where $r_1 + r_2 = \frac{2}{3}p,$ then $3$ must divide $p$.

Further, since $(r_1 \times r_2) = \frac{1}{3}q,$ then $q$ must be a multiple of $3$.

Now transfer attention to the other equation.

Since $x^2 - px + q = 0$ has integer solutions $s_1, s_2$ with $s_1 \times s_2 = q$, at least one of $s_1, s_2$ is a multiple of $3$.

WLOG, $s_1$ is a multiple of 3.

Then clearly, $(s_1)^2 -ps_1$ is a multiple of $9.$

Therefore, $q$ must be a multiple of $9$.

Title: Integer solutions to Polynomial

Body:
Let $f(x) = x^3-px^2+qx$ and $g(x) = 3x^2-2px+q$ where $p,q \in \mathbb{Z}^+$. Show that if $f=0$ has three distinct integer solutions and $g = 0$ has 2 distinct integers solutions then $p$ is a multiple of 3 and $q$ is a multiple of 9.

I've been trying to do this problem for some time and I keep getting back to the discriminant,and I know it must be a perfect square, so I must have that $p^2-4q = n^2$ and $p^2-3q = m^2$ and then I thought to subtract them:
$$m^2 - n^2 = p^2 -3q - p^2 +4q = q$$

But then that got me nowhere too.. I feel like there is something basic here I am missing.


问:
Is there a relation between the ZFC set axioms and our intuitive notion of sets?
A question about history.

When I took my first course in set theory, I had the perception that ZFC axioms weren't the most intuitive thing in relation to what common people would usually call a set. So I have this curiosity: is the intention at using them to formalise our intuitive notion of sets (in the sense of 'a collection of things' but avoiding paradoxes) or rather are they just used because all mathematics (all the stuff that is done in analysis, algebra, etc.) can be built upon them?

If the first is true, how is that correspondence between ZFC sets and the ordinary use of 'set'?

If the second is true, what is mathematics about? I have the idea (please correct me if I'm wrong) that mathematics is generally based upon sets, specifically ZFC sets. But if a set is just 'a structure upon which mathematics can be done', what are then the foundations of mathematics?

Thanks!

答:
History, unfortunately, is more complicated than your choices permit.

Presumably, Zermelo had proposed his 1908 axioms in support of his well ordering theorem. In so far as his axioms might relate to intuitive notions of set, his original axioms admitted urelements and relied on singletons to substantiate the truth of equality statements. The idea that a theory of pure sets suffice for mathematics is attributed to Skolem as far as I know. With a little further consideration, this may be seen as assuming the arithmetization of mathematics.  As such, one must recognize the chicken and egg problem that arises if linguistic symbols that are not pure sets become conflated with some Goedellian encoding for the sake of claiming that everything is a set.

Zermelo's reliance on singletons reflects the relationship of his system with Frege's logic. Category theorists claim that Zermelo had to make s choice between differing views expressed by Frege and Cantor. Since the relevant text is not translated into English, I cannot judge this claim. Nevertheless, Frege speaks of object identity and concept identity in a circular fashion that arises from Leibniz' principle of the identity of indiscernibles. Zermelo's use of singletons ought to be understood in this context. Foundationalism in the sense of Russell's logical atomism stipulates that object identity is fundamental. Hence, one has "first-order logic" with the proviso that "identity" is a "second-order" concept.

Since modern accounts of Zermelo-Fraenkel set theory are now based upon the first-order paradigm and the presumption that a theory of pure sets suffice for mathematics, urelements are treated in a derivative theory often called ZFA where the 'A' stands for (logical) atoms. The theory assumes that the urelements form a set (in conformance with the "everything is a set" presupposition). And, it is an important theory because of the models that arise from permuting the urelements.

It is, however, entirely plausible to understand urelements with respect to an antichain relation in a different signature.
Among other things to be found in Russell's writings, is the question of how a set is comparable to an object. While no one would ask that particular question today, the fact that urelements are not sets means that an antichain whose denial yields an order based on membership is entirely conceivable.

Such a theory would permit one to assert how non-sets satisfy the real number axioms, for example.

Understanding the distinction between urelements and sets in this manner permits one to understand how Cohen's forcing can extend a given model with "ideal subsets." The axiom of extension is contingent upon the denial of an antichain.

But, since such a view is not of interest to established mathematicians, students of set theory need to focus on the affirmative claims of first-order logic and set theory without concern for "intuition."

With regard to your second question, let me suggest that you look at the papers by Augustus de Morgan found in Ewald's two-volume anthology, "From Kant to Hilbert. "  This will provide you with a good foundation for understanding what the assertion, "mathematics is formal," really means.
The received view corresponding with first-order model theory reduces mathematics to language signatures, stipulation of relations and operations involving the symbols of such signatures, and a specification of the arities for each symbol in the signature. These constructs are often called "similarity types."

Of course, the "first-order" designation goes back to the foundationalist stipulation that object identity is fundamental.  So, language signatures will contain a symbol denoting an extensional domain of discourse. Extension is important to mathematical practice so that linguistic expressions intended to have "the same meaning" denote "singularly."

It had been Skolem's criticism of Zermelo's theory that gave impetus to this use of language signatures in mathematics over the notion that "mathematics is extensional." This particular claim arises from a different history. It is, of course, important for model theory. But, in the model theory of set theory it yields the "unfolding" and "geology" of possible models studied by set theorists using Cohen's forcing. What Skolem observed about Zermelo's axioms is that they could not be "categorical" in the sense of specifying a single, unique interpretation of the language symbols.


问:
Lie algebra of the automorphism group of an arbitrary Lie Algebra?
I'm trying to prove the Lie algebra of $Aut(\mathfrak{g})$ is $Der(\mathfrak{g})$, the set of all derivations over $\mathfrak{g}$. Where $\mathfrak{g}$ is an arbitrary finite lie algebra.

Due to $\mathfrak{g}$ is finite Lie algebra, I am using that $Lie(Aut(\mathfrak{g}))=\{X \in M(n, \mathbb{C}) / e^{tX}\in Aut(\mathfrak{g})\}$.

How can I prove if $\delta \in Der(\mathfrak{g})$ then $e^{t\delta}([X,Y])=[e^{t\delta}(X),e^{t\delta}(Y)]$?

And if $\delta \in Lie(Aut(\mathfrak{g}))$ then $\delta \in Der(\mathfrak{g})$?

答:
Let $\varphi(t)$ be a smooth path of automorphisms with $\varphi(0)=\mathrm{id}$. Then for $X,Y\in\mathfrak g$ you've got:

$$0=\frac d{dt}[X,Y]\lvert_{t=0}= \frac d{dt}[\varphi(t) X,\varphi(t)Y]\lvert_{t=0}=[\varphi'(0)X, Y]+[X,\varphi'(0)Y]$$
and $\varphi'(0)$ is a derivation. Hence the elements of Lie algebra of the automorphism group are derivations.

And the other direction? Let $\delta$ be a derivation, then:

$$[\delta^n X,\delta^m Y]= (-1)^m[\delta^{n+m}X,Y]$$

and

$$[\sum_n\frac{(t\delta)^n}{n!}X, \sum_m \frac{(t\delta)^m}{m!}Y]=\sum_{n,m}\frac{(-1)^mt^{n+m}}{n!\,m!}[\delta^{n+m}X,Y]$$
do a substitution of the sums, let $k=n+m$ and multiply by $\frac{k!}{k!}$ and $1^{k-m}$. Then the sum becomes:

$$\sum_k\frac{t^k}{k!}\sum_{m≤k}\frac{(-1)^m 1^{k-m}k!}{m!\,(k-m)!} [\delta^{k}X,Y] = \sum_k \frac{t^k}{k!}\sum_{m≤k}(1+(-1))^k[\delta^kX,Y]$$

Now only the $k=0$ term survives, so you are left with $[X,Y]$.


问:
substituting algebraic numbers for their conjugates in rational functions
If I have two algebraic numbers $\alpha$ and $\beta$ and a rational function $w$ with rational coefficients (a function that's the ratio of two rational polynomials)  that relates the two $\alpha = w(\beta)$ if I were to substitute $\beta$ with one of it's algebraic cojugates $\beta^{'}$ in the rational function will I get a conjugate of $\alpha$ or rather is $w(\beta^{'})$ an algebraic conjugate of alpha? I feel like there is a very obvious counter example but I've been struggling to find one.

答:
Let $f,F\in\Bbb Z[X]$ be irreducible. Let $g,h\in \Bbb Z[X]$ with $h\ne 0$. Let $\beta\in\Bbb C$ with $f(\beta)=0$ and $F(\frac{g(\beta)}{h(\beta)})=0$ (and in particular, $h(\beta)\ne 0$).
Note that by cancellation of $h$'s, $$p(X):=h(X)^{\deg F}F(\tfrac{g(X)}{h(X)})$$ is actually $\in \Bbb Z[X]$. As $p(\beta)=0$, we conclude that $\beta$'s minimal polynomial $f$ divides $p$. On the other hand, $h(\beta)\ne 0$ implies that $f$ does not divide (or even has a common factor with)  $p$.
But that means that for every $\beta'$ with $f(\beta')=0$, we have $p(\beta')=0$ and $h(\beta')\ne 0$, hence $$F(\tfrac{g(\beta')}{h(\beta')})=0,$$
i.e., $\tfrac{g(\beta')}{h(\beta')}$ is a conjugate of $\tfrac{g(\beta)}{h(\beta)}$.


问:
How to solve $z^3+z+1=0$?
I have troubles with solving an equation
$$z^3+z+1=0$$
in complex numbers. I know that in real numbers we can use Cardano's formula and it's clear to me. But what about the situation when $z \in \mathbb{C}$?

答:
$$ z = re^{i \theta}$$

Or,

$$ r^3 \left[  e^{i3 \theta} \right] + re^{i \theta} + 1 =0$$

$$ r^3 \left[ \sum_{k=even}^{2}\binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta+ \sum_{k=odd}^{3} \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta \right] + r e^{i \theta} + 1 =0$$

Equating real and imaginary part to zero,

$$ r^3 \left[ \sum_{k=odd}^3 \binom{3}{k} (i)^{3-k} \sin^{3-k} \theta \cos^k \theta  \right] + r \cos \theta + 1 = 0$$

Or,

$$ r^3 \left[ \sum_{k=0}^1 \binom{3}{2k+1}  (-1)^{1+k} \sin^{2-2k} \theta \cos^{2k+1} \theta  \right] + r \cos \theta + 1 = 0 \tag{1}$$

And,

$$  r^3 \left[\sum_{k=even}^2 \binom{3}{k} \cos^k \theta (i)^{3-k} \sin^{3-k} \theta  \right] + r i \sin \theta = 0$$

Or,

$$  r^2 \left[\sum_{k=0}^1 \binom{3}{2k} \cos^{2k} \theta (-1)^{k+1}  \sin^{3-2k} \theta  \right] +  \sin \theta = 0 \tag{2}$$

I'll leave you with solving the trignometric equation of (1) an d (2)

Title: How to solve $z^3+z+1=0$?


问:
Show that $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even..., using the binomial coefficients formula
In the previous questions i've proved that $(1+i)^{2n+1}=a_n+ib_n$ where $a_n$ and $b_n$ are $\pm 2^n$ and that $(1-i)^{2n+1}=a_n - ib_n$ and that $|P_n(1)|=2^n$ using the previous statements. But now I need to show that the polynomial $P_n(X)=\frac{(X+i)^{2n+1} - (X - i)^{2n+1}}{2i}$ is of degree $2n$, even (the coefficient in front of $X^n$ is non-zero only when $n$ is even, with real coefficients, and with $2n+1$ as a dominant coefficient, using the binomial coefficient formula.

I also need to show that the coefficient $X^{2n-2}$ in $P_n$  is $\frac{n(1-2n)(1+2n)}{3}$.

I don't really know how to use the binomial coefficient formula to prove those claims, any help would be greatly appreciated!

答:
Hint: Use
$$ (a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k, i^{2k}=(-1)^k, i^{2k-1}=(-1)^{k+1}i $$
and you will get the answer.


问:
Finding N in binomial question
A normal die with probability 1/6.

Question
The number of times you have to throw a cubic die so that we have 80percent change of getting at least one 1.

答:
Setting $X$ the rv that counts how many times "1 is drawn", say $X\sim Bin(n;\frac{1}{6})$

the probability you are looking for is

$$\mathbb{P}[X=0]\leq 0.2$$

$$\Bigg(\frac{5}{6}\Bigg)^n\leq 0.2$$

$$n\geq 9$$

Title: Finding N in binomial question


问:
Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$
We have to prove the convergence and find the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$. I tried to find $a_{n+1}$ in terms of $a_{n}$ and got $a_{n+1}=\frac{n+2}{2(n+1)}\left(a_n+1\right)$ and could not proceed further.Any help is appreciated.

Edit : Following K.defaoite's answer limit appears to be 1 . But we still need to prove convergence.

答: By the Stolz–Cesàro theorem (the equivalent for sequences of L'Hôpital's rule),$$\lim_{n\to\infty}\tfrac{\sum_{k=1}^n\frac{2^k}{k}}{\tfrac{2^{n+1}}{n+1}}=\lim_{n\to\infty}\frac{\tfrac{2^{n+1}}{n+1}}{\tfrac{n2^{n+1}}{(n+1)(n+2)}}=\lim_{n\to\infty}(1+\tfrac2n)=1.$$


问:
Is my proof of the equation $d=dx^i\wedge \partial_i$ correct?
Let $M$ be some manifold, $x$ a chart on $U\subset M$ and $\omega$ a $p$-form such that
\begin{equation}
    \omega=\omega_I\wedge dx{}^I:=\sum_{1\leq i_1<\ldots<i_p\leq n}\omega_{i_1,\ldots,i_p}dx^{i_1}\wedge\cdots\wedge dx^{i_p}
\end{equation}
on $U$. From The Geometry of Physics, page $141$:

In any manifold the operation of exterior differentiation [...] can be written symbolically as $d = dx^j\wedge\partial/\partial x^j$; the operator $\partial/\partial x^j$ acts only on the
coefficients.

Here is my attempt of a proof: We have
\begin{align}
&d\omega=d(\omega_Idx^I)=d(\omega_I\wedge dx^I)=d\omega_I\wedge dx^I+\omega_I\wedge ddx^I\\
&=d\omega_I\wedge dx^I+0=d\omega_I\wedge dx^I=d\omega_I\wedge dx^I=\partial_j\omega_Idx{}^j\wedge dx^I\\
&=dx{}^j\wedge \partial_j\omega_Idx^I=dx^j\wedge\partial_j\omega
\end{align}
where we have defined $\partial_j(\omega_Idx^I):=(\partial_j\omega_I)dx^I$.

Is this correct? In particular, is it allowed to write $\omega_I dx^I=\omega_I\wedge dx^I$?

Comment: The definition of $\partial _i\omega$ might seem a bit weird, but I noticed that it also appears in other parts of the book:

From the overview before the first chapter:

The electric field intensity is a $1$-form $\mathscr E$ and the magnetic field intensity is a $2$-form $\mathscr B$ and Faraday’s law says
\begin{equation}
d\mathscr E = −\frac{\partial \mathscr B}{\partial t}
\end{equation}
where $\partial \mathscr B/\partial t$ means take the time derivative of the components of $\mathscr B$.

From page $119$:

We define the time derivative of an exterior form by differentiating each component\begin{equation}
\frac{\partial}{\partial t}\omega:=\frac{\partial\omega_I}{\partial t}dx^I
\end{equation}

Thus I thought that maybe we can define $X\omega$ - where $X$ is a vector field and $\omega$ is a $k$-form - to be the $k$-form $X\omega:=X(\omega_I)dx^I$. If we can prove that the definition doesn't depend on the coordinate system, this would be well defined...

答:
Your $X\omega$ is not well defined. It's coordinate dependent. Consider the coordinate systems $x$ and $y$ in which $y = f(x)$ for some non-constant increasing and nonvanishing function (the manifold is $\mathbb{R}$ here). Consider the 1 form $\omega := dy = f'(x) dx$. Let $X = \frac{\partial}{\partial y} = \frac{1}{f'(x)} \frac{\partial}{\partial x}$.

Then $X(dy) = \frac{\partial}{\partial y} (1) dy = 0$

And $X(dy) = \frac{1}{f'(x)} \frac{\partial}{\partial x} f'(x) dx = \frac{f''(x)}{f'(x)} dx \neq 0 $.

There is indeed an invariant formula for the exterior derivative. If $\omega$ was a 2-form: then

$$d\omega (X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])$$
for any vector fields $X,Y$.
The general invariant formula looks a little more complicated. Check it out on Wikipedia:

https://en.wikipedia.org/wiki/Exterior_derivative#In_terms_of_invariant_formula.

EDIT: Your formula for $d\omega$ is correct. In local coordinates,
$$d\omega = dx^i \wedge \frac{\partial}{\partial x^i} \omega_I dx^I$$
which is indeed coordinate independent. That is how many authors define the exterior derivative actually. They define it locally in that way and show that this expression is coordinate independent, which shows that $d\omega$ is well defined globally. However, $\frac{\partial}{\partial x^i} \omega_I dx^I$ is not coordinate independent and so is not a well defined global 1-form as I show above.


问:
Absolute Convergence of Improper Integral of sinx/(e^x-e^(-x))
I want to prove absolute convergence of improper integral ;

\begin{equation}
\int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx.
\end{equation}

But I couldn't prove this. (I don't have to calcurate the value. I only have to prove absolute convergence.)

My attempt is as follows.

$
\left|
\dfrac{\text{sin}x}{e^x-e^{-x}}
\right|
\leqq
\dfrac{1}{e^x-e^{-x}} \\
\int_{\epsilon}^N \dfrac{1}{e^x-e^{-x}} dx
=_{(t=e^x)}
\int_{e^{\epsilon}}^{e^N} \dfrac{1}{t^2-1} dt
=
\dfrac{1}{2}
\left[
\text{ln} \left| \dfrac{t-1}{t+1} \right|
\right]^{e^{N}}_{e^{\epsilon}}
$

I want to let $\epsilon \rightarrow 0$ and $N \rightarrow \infty.$

However, if I let $\epsilon \rightarrow 0$,
$
\text{ln} \left| \dfrac{e^{\epsilon}-1}{e^{\epsilon}+1} \right| \rightarrow - \infty
$

I would like you to give me any ideas or hints to prove absolute convergence of

\begin{equation}
\int_0^{\infty} \dfrac{\text{sin}x}{e^x-e^{-x}} dx.
\end{equation}

答:
For $x>1$ we have
$$\left|\frac{\sin x}{e^x-e^{-x}}\right|\leq \frac{1}{e^{x/2}}$$

because for $x>1$
$$e^x-e^{-x}>e^{x/2}$$

thus
$$\int_1^{\infty} \frac{\sin x}{e^x-e^{-x}} \,dx\le \int_1^{\infty} \frac{dx}{e^{x/2}} $$
as the last integral converges, the given integral converges at $+\infty$.

It converges at $x=0$ as proved above by Mr.Claude Leibovici:

as $$\frac{\sin (x)}{e^x-e^{-x}}=\frac{1}{2}-\frac{x^2}{6}+O\left(x^3\right)$$
the integral converges at $x=0$ either.

Its value is $\frac{1}{4} \pi  \tanh \left(\frac{\pi }{2}\right)$

Thanks to contributor Gary for the useful collaboration.


问:
How come such different methods result in the same number, $e$?
I guess the proof of the identity
$$
\sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x
$$
explains the connection between such different calculations. How is it done?</p>

答:
This identity, in isolation, is certainly confusing. It becomes less so once you know that it is just a special case of a more general identity

$$\sum_{n \ge 0} \frac{x^n}{n!} = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n.$$

Why does this hold? Well, let's call the function on the left $e^x$. Then by inspection, $\frac{d}{dx} e^x = e^x$. General theory tells you that the solution to $y' = y$ is unique provided you fix the value of $y(0)$ (and this is also physically intuitive), and here $e^0 = 1$ by inspection, so that tells you that $e^x$ is the unique function satisfying $e^0 = 1$ and $\frac{d}{dx} e^x = e^x$.

Chandru1's deleted answer reminded me that the proof of this is actually completely elementary. Suppose $f$ is a function which satisfies $f'(x) = f(x)$. Then

$$\frac{d}{dx} \left( e^{-x} f(x) \right) = e^{-x} f'(x) - e^{-x} f(x) = 0$$

hence $e^{-x} f(x) = C_f$ for some constant $C_f$. Substituting $f(x) = e^x$ gives $e^{-x} e^x = C$, and substituting $x = 0$ gives $e^{-x} e^x = 1$. Now it follows that $f(x) = C_f e^x$.

Why does the function on the RHS also satisfy this property? Let's call this function $\exp(x)$. Then

$$\exp(cx) = \lim_{n \to \infty} \left( 1 + \frac{cx}{n} \right)^n = \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^{mc} = \exp(x)^c$$

where $m = \frac{n}{c}$. Setting $c = 1 + \frac{y}{x}$, this gives

$$\exp(x + y) = \exp(x)^{1 + \frac{y}{x}} = \exp(x) \exp(x)^{ \frac{y}{x} } = \exp(x) \exp(y)$$

from which it follows that

$$\frac{d}{dx} \exp(x) = \lim_{h \to 0} \frac{\exp(x) \exp(h) - \exp(x)}{h} = \exp(x) \lim_{h \to 0} \frac{\exp(h) - 1}{h}$$

so it follows that $\exp(x)$ satisfies $\exp(0) = 1$ and $\frac{d}{dx} \exp(x) = \exp(x)$, which means it must equal $e^x$ - at least, as soon as we know that

$$\lim_{h \to 0} \frac{\exp(h) - 1}{h} = 1.$$

But

$$\exp(h) = \lim_{n \to \infty} \left( 1 + \frac{h}{n} \right)^n = \lim_{n \to \infty} \left( 1 + {n \choose 1} \frac{h}{n} + {n \choose 2} \frac{h^2}{n^2} + ... \right) = \lim_{n \to \infty} \left( 1 + h + R(h, n) \right)$$

where $|R(h, n)| \le \sum_{k \ge 2} |h|^k = O(|h|^2)$ by the ratio test, from which the conclusion follows.

That last step, by the way, is an echo of the standard proof, where one expands

$$\left( 1 + \frac{x}{n} \right)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$

and uses the fact that $\frac{1}{n^k} {n \choose k} \to \frac{1}{k!}$ as $n \to \infty$. While this is fairly hands-on, in my opinion it obscures an underlying principle behind this problem, which is that $e^x$ is a very special function and all of the equivalent ways of defining it are based on one of its special properties.

One really nice way to interpret the above identity is that it is precisely what you get by applying Euler's method to the ODE $y' = y$ with initial conditions $y(0) = 1$.


问:
Definition of Induced Module - Typo in Corps Locaux?
This is from the beginning of the section on group cohomology in Corps Locaux (English Edition).

Serre states that $A$ is an induced $G$-module if

(1) $A\cong A\otimes_\mathbb{Z}X$ for an abelian group $X$,

or, equivalently,

(2) $A=\bigoplus_{s\in G}s\cdot X$.

Is (1) a typo? This is a very strict condition not just on $A$, but also on $X$. It seems to me that the correct version of the first definition above should read $A\cong A\otimes_{\mathbb{Z}[X]}X$, which is in line with the usual corresponding notion for group representatinos over fields, but perhaps I'm just missing something simple.

答:
Condition (1) says nothing about $G$. It should be something like

(1) $A \cong X \otimes_{\Bbb Z} \Bbb Z[G]$ for some abelian group $X$.

We assume that the $G$-action on $X \otimes_{\Bbb Z} \Bbb Z[G]$ acts only on the $\Bbb Z[G]$ factor.


问:
Define the Event A Given B?
Let ($\Omega , T , P$) be a measured space , P being a probability measure.

Let A,B$\in T$.
A usual question is to calculate P(A|B) (Probability of A Given B);
I want to demonstrate the expression has a sense showing A|B is in T.

What is the definition of the Event A|B ?
In most courses , i can find the expression of the probability of the event, but i can't find anybody giving a definition of the set it defines.

答:
There is no such thing as an event $A|B$. The point is, given $B \in T$, we can define a new probability measure $P_B$ in $T$, defininig $P_B(A)=P(A|B)$.

Note that this probability is "associated" with the event $A$. There is no need to define a new "event" $A|B$. The events are the same, what changes is the probability measure.

Title: Define the Event A Given B?


问:
Combinatorics problem - choosing $6$ cards out of $32$-card deck so that there are exactly three different suits (Inclusion-Exclusion)
Suppose we have a deck of $32$ cards with $8$ cards of each of the four suits. In how ways we can choose six cards such that there are cards of exactly three different suits among the chosen cards?

I believe inclusion-exclusion principle is the way to solve it, where we first count the number of total ways to choose $6$ cards out of $32$ (which is $\binom{32}{6}$), then exclude the number of combinations where exactly two of the suits are missing (which is $\binom{4}{2}\binom{16}{6}$) and then by inclusion-exclusion formula add the combinations where all three suits are missing (which is $\binom{4}{3}\binom{8}{6}$). The number of combinations of all $4$ suits missing is, of course, zero.

My question is - where is my logic incorrect? I know it is, but can't seem to spot the error.

答:
There are two issues here. The first is that the standard inclusion-exclusion formula assumes you start by subtracting events where at least one thing is missing, so that everything where two are missing is counted twice, three missing counted three times, etc, and that is what means you only have to add or subtract once, and which you do alternates.

Here you start by subtracting things where two are missing. This means that you have subtracted everything with three missing three times (from three suits you can choose a pair in three ways), and so you need to add on twice the number of ways to have three suits missing. (If it was a possible situation, you'd then need to subtract three times the number of ways to have four missing: you would have so far subtracted these configurations six times and added them eight times.)

The second problem is that you haven't accounted for all the situations where all four suits are present. So after making the change from the previous paragraph, what you have calculated is the number of ways to have at least three suits among the chosen cards.


问:
Evaluating limits using a series
I'm trying to use a Taylor series centered at $0$ to evaluate this limit:

$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$

I rewrote the function as its Maclaurin series:

$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$

In expanded form:

$$\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}=4x^{3-\frac{2}{x^3}}+\frac{4x^{3-\frac{4}{x^3}}}{2!}+\frac{4x^{3-\frac{6}{x^3}}}{3!}+...$$

As $x$ goes to $\infty$, $\frac{2}{x^3}$ goes to $0$.  Thus, the limit of the first term is simply the limit of $4x^3$, which is $\infty$.  Based on this fact alone, I would assume, the limit of the entire series is $\infty$, but apparently the answer is $-8$.  What did I do wrong?

答:
As $x\to +\infty$, it should be
$$4x^3\left(e^{\color{blue}{\frac{-2}{x^3}}}-1\right)=4x^3\sum_{k=1}^{\infty}\frac{(\color{blue}{\frac{-2}{x^3}})^k}{k!}=
4x^3\left(-\frac{2}{x^3}+o(1/x^3)\right)=-8+o(1).$$
Then what is the limit?

Title: Evaluating limits using a series


答:
$$e^{-\dfrac2{x^3}}=1-\dfrac2{x^3}+\dfrac{\left(-\dfrac2{x^3}\right)^3}{2!}+\cdots=1-\dfrac2{x^3}+O\left(\dfrac1{x^6}\right)$$

Alternatively, Set $-\dfrac2{x^3}=h$ to find

$$=-4\cdot2\lim_{h\to0^-}\dfrac{e^h-1}h$$

Title: Evaluating limits using a series

Body:
I'm trying to use a Taylor series centered at $0$ to evaluate this limit:

$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$

I rewrote the function as its Maclaurin series:

$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$

In expanded form:

$$\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}=4x^{3-\frac{2}{x^3}}+\frac{4x^{3-\frac{4}{x^3}}}{2!}+\frac{4x^{3-\frac{6}{x^3}}}{3!}+...$$

As $x$ goes to $\infty$, $\frac{2}{x^3}$ goes to $0$.  Thus, the limit of the first term is simply the limit of $4x^3$, which is $\infty$.  Based on this fact alone, I would assume, the limit of the entire series is $\infty$, but apparently the answer is $-8$.  What did I do wrong?


问:
Existence of a homeomorphism on ambient space mapping submanifolds
Let $M,N\subseteq \mathbb{R}^n$ be smooth sub-manifolds both of which are diffeomorphic to $\mathbb{R}^d$ for some $0\leq d<n$.  Let $\gamma:=n-d$.  For what values of $d$ does there necessarily exist a homeomorphism $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ for which
$$
\phi(M)= N?
$$

In other words:  What is the minimal co-dimension of the ambient space for which any two topologies copies of $\mathbb{R}^n$ can be mapped into one another by a homomorphism?

答:
Assuming I'm interpreting your question correctly, the answer is basically never.  Consider \begin{align*}
M &= (-\infty, \infty) \subseteq \mathbb{R} \\
N &= (-1,1) \subseteq \mathbb{R}
\end{align*}

Then $M$ and $N$ are both diffeomorphic to $\mathbb{R}$, but there is no ambient homeomorphism $\phi$ of $\mathbb{R}$ carrying $M$ to $N$, since $M$ is closed in $\mathbb{R}$ and $N$ is not closed in $\mathbb{R}$.

For other dimensions, take
\begin{align*}
\tilde{M} = M \times \mathbb{R}^{d-1} \subseteq \mathbb{R} \times \mathbb{R}^{n-1} \cong \mathbb{R}^n \\
\tilde{N} = N \times \mathbb{R}^{d-1} \subseteq \mathbb{R} \times \mathbb{R}^{n-1} \cong \mathbb{R}^n
\end{align*}
where $\mathbb{R}^{d-1} \subseteq \mathbb{R}^{n-1}$ is a standard linear embedding.

Then $\tilde{M}$ and $\tilde{N}$ are both diffeomorphic to $\mathbb{R}^d$, but $\tilde{M}$ is closed in $\mathbb{R}^n$ and $\tilde{N}$ is not, so no ambient homeomorphism taking $M$ to $N$ exists.

So it seems the only time such a homeomorphism exists is if $d = 0$.


问:
Bifurcation value and description
Find the bifurcation of $a$ and describe the bifurcation that take place at each value

$\displaystyle dy/dt=e^{-y^2}+a$

I let $\displaystyle e^{-y^2}+a=0$ then solve for y. I got $y^2=-\ln(a)$ What do I do next to find $a$?

答:
You started well to let $e^{-y^2} + a =0$ and then solve for $y$. This will allow us to determine the equilibrium points. You will arrive at $$y^2 = -\ln(-a).$$ (I believe you misplaced a negative sign.) From this we see that first of all, we must have $a < 0$. In order to find bifurcation points, we need to consider what values of $a$ will yield a change in the nature of the equilibrium points. That is, what value of $a$ will cause a change in the number or behavior of the equilibrium points? Is there a value of $a$ that will cause no solution to the equation?

I leave this to you, but please ask a question if you get stuck.

Title: Bifurcation value and description


问:
10 ball as put in 7 boxes randomly. What is the probability that at least one box is empty?
My solution:
First, put one ball at each box and then distribute the remaining 3 balls in 7 boxes.

$p=1-\frac{\binom{9} {3} }{\binom{16}{10}}$

Is this correct?

Or?

$p=1-\frac{\binom{10} {7} 7!7^3}{7^{10} }$

答:
Your first approach uses stars and bars method and you should not be using for computing probability as the events are not uniformly distributed.

So your second approach is what you need to take (considering each ball distinct) -

Say the event of distributing $10$ balls in $7$ boxes without restriction is $A$.

Say the event of distributing $10$ balls in $7$ boxes such that no boxes are empty is $B$.

Number of ways for $A = 7^{10}$

Number of ways for $B = 7! \times {10 \brace 7} =  29635200$ where ${n \brace k}$ is the Stirling Number of the second kind. It places $n$ distinguishable objects into $k$ indistinguishable heaps.

So desired probability where at least one box is empty $= 1 - \frac{29635200}{7^{10}} \approx 0.895 $

Instead of using Stirling Number of the second kind, you could also use principle of inclusion exclusion.

Number of ways for $B = \sum \limits_{i = 0}^{7} (-1)^i {7 \choose i} (7-i)^{10} = 29635200$


问:
An essential formula for reconstructing the Poisson bivector from a Poisson bracket
This was essentially addressed in this question. Nick's comment suggests to use that $\left\{f,g \right\}=\sum_{i,j} \frac{\partial f}{\partial x^i} \frac{\partial g}{\partial x^j} \left\{ x_i, x_j \right\}$ on some coordinate neighborhood, but I can't figure out why this is.

I know that $\left\{ f, \cdot \right\}$ and $\left\{ \cdot, g \right\}$ are both derivations, so $X_f(g) = X_f(x_j) \frac{\partial g}{\partial x^j}$. So I'd like to show that $X_f(\cdot) = \left\{ \frac{\partial f}{\partial x^i} x_i, \cdot \right\}$. I get that, if $df=df_1$ on some neighborhood, then $f-f_1$ is a constant on that neighborhood and since $\left\{\cdot , g \right\}$ is a derivation for any $g$, we have that $\left\{f-f_1, g \right\} = 0$ for any $g$ on the coordinate neighborhood. I'd like to use this fact, but I don't see why we necessarily have that $df = d \left( \frac{\partial f}{\partial x^i} x_i \right)$, because the right hand side is $ df + d \left( \frac{\partial f}{\partial x^i} \right)x_i$.

答:
As you remark $df$ and $d(\sum_i \partial_i f x_i)$ are not the same. Indeed you have in general
$$\{f,g\}_x\neq \sum_i \{\partial_i f \,x_i,g\}_x$$
Remember also that you cannot pull out the $\partial_if$ without getting a $x_i\{\partial_i f,g\}_x$ term since this is a derivation.

What you do have is:
$$\{f,g\}_x= df(\{\cdot ,g\}_x)=\sum_i \partial_i f\ dx_i(\{\cdot ,g\}_x) = \sum_i \partial_if\  \{x_i,g\}_x$$


问:
Proving that the the sequence $x_n = \frac{n^{100}}{2^n}$ converges using only elementary methods
I realise that there are already questions on the site about this sequence or related sequences, but I'm looking for a completely elementary method that doesn't use:

L'Hopital's Rule

Tests like the comparison test or ratio test

Logarithms

Statements like "exponentials grow faster than polynomials"

I'm trying to show that the sequence $$x_n = \frac{n^{100}}{2^n}$$ converges as $n \to \infty$, where $n$ is a positive integer. I haven't been able to find a way to do this directly using the epsilon definition for convergence. Is there a completely elementary way to show that this converges (not necessarily needing to show what it converges to) without using any of the things mentioned above?

Body:
$$\frac{n^{100}}{2^n} =2^{100} \frac{n^{100}}{2^{n+100}}$$
Now $$2^{n+100}> \binom{n+100}{101} = \frac{(n+100)(n+99) \cdots (n+1)n}{101 !}> \frac{n^{101}}{101!}$$

Therefore
$$\frac{n^{100}}{2^n} < 2^{100}\cdot  101! \cdot \frac{1}{n}$$


问:
Why is it important to know a family of functions has compact closure?
In Brezis' book on functionnal analysis, the author sais that

It is important to be able to decide whether a family of functions in
$L^p (\mathbb{R})$ has compact closure in $L^p (\mathbb{R})$ (for the strong topology).

Also, I guess that this is important as well in $C^0(I)$ since it is the subject of Arzelà-Ascoli theorem.

But why is this important intuitively(or not)?

答: Because then any sequence of functions of the family will have a convergent subsequence. So, if $(f_n)_{n\in\Bbb N}$ is a famliy of approximate solutions of a problem, in which the aproximations keep getting better, and if $(f_{n_k})_{k\in\Bbb N}$ is a subsequence converging to $f$, then $f$ might be a solution of that problem.


问:
Computing homology groups of glued hexagon
I've just started studying about homology groups, and I got an exercise in computing the homology groups of the following hexagon, with all edges identified as in the picture:

The first thing I did is representing it as a $\Delta$-complex as follows:

I then proceed in computing the chain groups:
$$C_0=\langle v\rangle =\mathbb{Z}$$
$$C_1=\langle a,b,c,d\rangle =\mathbb{Z}^4$$
$$C_2=\langle M,N,P,Q\rangle =\mathbb{Z}^4$$
$$C_n=\{0\}, n\geq3$$

Since there is only one simplex of dimension $0$ ($v$), we have that $\partial_1=0$. Moreover:
$$\partial_2(M)=b-2a$$
$$\partial_2(N)=c-a-b$$
$$\partial_2(P)=d-a-c$$
$$\partial_2(Q)=-2a-d$$
$$Ker(\partial_2)=\langle -2M+N+P+Q \rangle = \mathbb{Z}$$
So:
$$H_0=\mathbb{Z} $$
$$H_1=\langle a,b,c,d|b-2a=c-a-b=d-a-c=0 \rangle = \langle a|6a=0 \rangle = \mathbb{Z}_6$$
$$H_2= \mathbb{Z}$$
$$H_n=\{0\}, n\geq 3$$

Does this seem okay? This is basically the first time I'm computing the homology group of an unknown space on my own, so I'm a bit unsure about my work. My main concern is about $H_1$ - it turned out to be pretty "simple", while I expected a more complicated group, which makes me think that maybe I've made a mistake.

Anyway, I'd be happy to hear your thoughts about my solution, and generally about how you suggest approaching such problems (what to expect and such).

Thanks a lot in advance!

答:
If you know about cellular homology, then your space has a CW structure with just one 0-cell, one 1-cell, and one 2-cell, i.e., $C_0 \cong C_1 \cong C_2 \cong \mathbb{Z}$.  The boundary maps are $\partial_1 = 0$ and $\partial_2 = \pm 6$, giving
\begin{align}
H_0 &\cong \mathbb{Z} \\
H_1 &\cong \mathbb{Z}_6 \\
H_i &= 0, \quad i \geq 2.
\end{align}

So I would double-check your calculation of $\ker \partial_2$.


问:
Limits calculation approaches not given the same result
Friends, I was given a limit to calculate but it is not intuitive for me, if I solve it algebraically, it gives me 0, but if I take an analytical approach it gives infinity. Can you guys tell me how I can solve this on a algebraic matter?

The limit goes like:

The f(x) graph plotted:

答:
Is that the limit as x goes to 1 from the left? The limit does not exist. I think your graph is wrong. We get the graph going to positive infinity at the positive root of $x^2+4x-3$ which $\sqrt{7}-2$.

The function is undefined over $[\sqrt{7}-2,1)$ (over the reals) because we have square root of a negative. So we have an undefined limit from 1-. The limit isn't positive or negative $\infty$ either.

We can't plug in $x=1$ because the function is not continuous at $x=1$.


问:
$\lim_{n \to +\infty}$ with $\sqrt[n]{\ln(n)}$ (in the example: $n(1 + \sqrt[n]{\ln(n)})$
So, I found lots of exercises for $\lim_{n \to +\infty}$ with $\sqrt[n]{\ln(n)}$ in my calculus book and I didn't know how to solve those. The example is $\lim_{n \to +\infty} n(1 + \sqrt[n]{\ln(n)})$.

I actually know the limit from this post: Limit:$ \lim\limits_{n\rightarrow\infty}\left ( n\bigl(1-\sqrt[n]{\ln(n)} \bigr) \right )$. The problem is, all answers use "Taylor, h'Hospital or d'Alembert" and I didn't learn about those theorems yet. So my question is - is there any way to deal with $\lim_{n \to +\infty} \sqrt[n]{\ln(n)}$ (taking $\lim_{n \to +\infty} n(1 + \sqrt[n]{\ln(n)})$ as an example) using most basic calculus?

By most basic calculus I mean not using Taylor, h'Hospital nor d'Alembert.

答:
Here is a hint:

For $n\geq e^2$

$$
e^{1/n}<(\log n)^{1/n}\leq n^{1/n}
$$

There are several "elementary" methods to show that

$\lim_{n\rightarrow\infty}p^{1/n}=1$ for $p>0$ constant and

$\lim_{n\rightarrow\infty}n^{1/n}=1$
See for instance Rudin's  Principle's of mathematical analysis, p. 57.

The conclusion follows from the squeeze lemma.

Proofs for 1 and 2  type of methods also have been explained on this site.


问:
Can all linear operators be represented as matrices? (In infinite dimension.)
Old title: Components of unbounded endomorphism wrt Schauder basis? Changed for discoverarbility.

In finite-dimensional linear algebra, given an endomorphism $T$ and a basis $\boldsymbol{e}_i$, we can easily find a matrix representation of $T$:
$$
 T(v)
 = T(\sum_i v^i \, \boldsymbol{e}_i)
 \overset{*}{=} \sum_i T(v^i \, \boldsymbol{e}_i)
 = \sum_i v^i T(\boldsymbol{e}_i)
 =: \sum_{i,j} v^i \, T_{ij} \, \boldsymbol{e}_j
$$

Let us now assume that $X$ is a separable Banach space with a Schauder basis $\{\boldsymbol{e_i}\}_{i \in \mathbb{N}}$ and $T: X \to X$ is an unbounded linear operator. If we try to replicate the previous equation, the sum will become infinite, and since for unbounded operators $T(\lim_n v_n) = \lim_n T(v_n)$ doesn't hold, the equality $(\overset{*}{=})$ above doesn't hold.

This would seem to indicate that unbounded operators don't have a coordinate representation in a Schauder basis, at least not in the same sense as bounded operators do. Or is there a simple generalization that I just don't see?

If we now consider $X$ to be a separable Hilbert space and $\boldsymbol{e}_i$ an orthonormal basis, we know that it's isomorphic to $\ell_2$ with basis:
$$
  \begin{pmatrix}1\\0\\0\\\vdots\end{pmatrix},
  \begin{pmatrix}0\\1\\0\\\vdots\end{pmatrix},
  \begin{pmatrix}0\\0\\1\\\vdots\end{pmatrix},
  \;\dots
$$
This means that either the endomorphisms on the space of „infinite column vectors“ aren't always „infinite matrices“, or that my previous conclusion about the „non-representability“ of unbounded operators was wrong. Which of these is true?

In summary, these are my questions:

Is it possible to represent an unbounded operator using its coordinates wrt. a Schauder basis of a separable Banach/Hilbert space?

If yes, how? If no, are there other practical ways to represent it? (Eg. using Hamel basis? Or as an integral kernel on $L^p$?)

Are all operators on $\ell_2$ „infinite matrices“?

答:
The whole field of Analysis is highly dependent on the fact that we may approximate complicated things (think a vector $x$ in your space $X$) by simpler things (think truncated sums of the representation of $x$ in your Schauder basis).  These approximations are only useful because the observations we want to make (e.g. $T(x)$) respect the approximations we made (i.e., are continuous functions).

Analysis is therefore (almost) completely useless if the quantities we want to measure do not respect approximations.

Therefore, in order to study discontinuous linear transformations you might as well ignore the norm structure of $X$ and hence your Schauder basis will be devoid of any interesting properties, beyond being any old linearly independent set.  In this case, as you noted, you will need a Hamel basis to describe what you see!

Disclaimer. Many discontinuous linear transformtions may be successfully studied with analytic methods because they sometimes possess hidden continuity properties, such as closed operators.


问:
singularity of analytic continuation of $f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$
How to show that all possible collection of analytic continuations of $\displaystyle f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2} $ has singular point at $z = 1$. I know that $f(z)$ converges for $|z| \le 1$. Also is there a theorem that relates the singularity of analytic continuation with circle of convergence?</p>

答:
Note that when $|z|<1$,
$$f'(z)=\sum_{n=1}^\infty\frac{z^{n-1}}{n}=-\frac{\log(1-z)}{z}.\tag{1}$$
Since the right hand side of $(1)$ has a unique singularity at $z=1$, it implies that on the unit circle, the analytic continuation of $f$ has a unique singularity at $z=1$.

For the general situation, note that for one complex variable, any (nonempty) open set in $\mathbb{C}$ is a domain of holomorphy, which implies that for any closed subset $S$ of the unit circle, there exists a holomorphic function $f$ defined on the unit disk, such that the collection of sigularities of the analytic continuation of $f$ on the unit circle is precisely $S$.

Remark: Just in case, $(1)$ follows from integrating
$$(zf'(z))'=\sum_{n=0}^\infty z^n=\frac{1}{1-z}.$$


问:
How to check whether Mersenne number is a prime or composite by using theorem $q = 2kp + 1$?
How to know that whether any Mersenne Number, $M_{m}$ whether is a prime, or composite?

I have learnt that $M_{m}=2^m-1$ , for $m$ is any positive interger.

And there is a theorem, says that :

if $p$ is odd prime, then any divisor of Mersenne number is of the form
$2kp + 1$, where $k$ is a positive integer.

Then I have carried out an example by plugging $m=37$,

$M_{37}=2^{37}-1=137438953471$ (although I know that it is not a prime), and

we have $q=2kp+1=2k(37)+1 =74k+1$

For

$k=2,q=149\longrightarrow$ Since $149$ is not the divisor for $M_{37}$, so this implies that $M_{37}$ is currently a prime.

$k=3,q=223\longrightarrow$ Since $223$ is the divisor for $M_{37}$, so this implies that $M_{37}$ is immediately not a prime.

$\therefore M_{37}$ is not a prime.

Do my reasoning is correct? Can anyone give me a comment about it? Also, can I start with using $k=1$ ?

Will be very thankful !

答:
Yes, it is correct, but you should no say that “$M_{37}$ is currently a prime”; at this moment, you simply do not know whether or not it is a prime number.

And, yes, you can start with $k=1$. Take $M_{11}(=2047)$, for instance. If $k=1$, then $q=23$. And, in fact, $23\mid2047$.


问:
Coloring 8x8 square board?
A square board with side-length of 8 cm is divided
into 64 squares with side-length of 1 cm each. Each square can be painted black or
white. Find the total number of ways to color the board so that every square with
side-length of 2 cm formed with 4 small squares with a common vertex has two
black squares and two white squares.
After taking rhe hint ''Try to solve with columns''I did it by coloring first column in any way there are 2 ways for each square in a column thus $2^8$ ways then all other colorings have to bi fixed and this is where I am wrong? Please help

答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Consider the first column, with $2^8$ ways of coloring the squares (as you did).

Hint:  We can always extend the coloring to another column.

However, the extension need not be unique. When is it not unique? Find a necessary and sufficient condition.

 If (any) 2 consecutive squares are the same color, then it forces their neighbors in the next column to be of the opposite color. This then forces the rest of the column.

 The row that the 2 consecutive squares have uniquely determined colors, which then forces the rest of the columns. Hence, there is a unique way to extend the coloring.

 Conversely, if no 2 conseuctive squares are of the same color, meaning the column is either BWBWBWBW or WBWBWBWB, then the next column is either BWBWBWBW or WBWBWBWB (only).

So, in the unique cases, each original way of coloring the column leads to (how many?) chessboards?

 1

In the non-unique case, each original way of coloring the column leads to (how many?) chessboards.

 $2^7$.

Hence, the total number of ways is

 $(2^8 - 2) \times 1 + 2 \times 2^7 = 510.$

Title: Coloring 8x8 square board?


问:
Sequent calculus: $0<1$
Considering the axioms of real closed ordered fields, I'm having trouble proving $0<1$ using Gentzen sequent calculus enriched with the cut rule, i.e, $$\text{Ax}_{\text{RCOF}}\vdash 0<1.$$
Using the trichotomy axiom I was able to obtain the following sequents $$0<1,..., R\rightarrow 0<1$$
$$1<0,..., R\rightarrow 0<1$$
$$0\cong 1,..., R\rightarrow 0<1,$$ where $R$ is some subset of the set of axioms of RCOF and "$...$" are consequences of applying the rules $L\forall$ and $L\vee$. Now, the first sequent is an axiom and. Knowing that $\neg\, 0\cong 1$ belongs to $R$, I apply the rule $L\neg$ to it and I derive the third sequent.

My question: I've tried a couple different approaches but wasn't able to derive the second. Any suggestions? Many thanks in advance.

答:
Wou have Trichotomy: exactly one of $1 < 0$ or $1=0$ or $0 < 1$ holds.

If we use the Law of Order: if $1 < 0$ and $z > 0$, then $1z <0z$, we have $z < 0$, contradicting Trichotomy.

Thus, we have $\lnot (1 < 0)$.

And we have the Non-triviality axiom: $\lnot (1=0)$.

In terms of sequents, we have:

$\text {Ax} \to \lnot (1=0)$, and

$\text {Ax} \to \lnot (1 < 0)$.

Here is the derivation:

$(1 < 0) \to (1 < 0)$

$[(1=0) \lor (0 < 1)] \to [(1=0) \lor (0 < 1)]$

$(1 < 0) \lor [(1=0) \lor (0 < 1)] \to (1 < 0), [(1=0) \lor (0 < 1)]$ --- from 1) and 2) by $(\lor \text L)$, that is:

$\text {Ax}\to (1 < 0), [(1=0) \lor (0 < 1)]$

$\text {Ax} \to \lnot (1 < 0)$ --- see above

$\text {Ax}, (1 < 0) \to$

The next step is to apply Cut to 3) and 5):

$\text {Ax} \to (1=0) \lor (0 < 1)$.

Now repeat:

$(1=0) \to (1=0)$

$(0<1) \to (0<1)$

$(1=0) \lor (0<1) \to (1=0),(0<1)$

Cut on 6) and 9):

$\text {Ax},\to (1=0),(0<1)$.

$\text {Ax}, \lnot (1=0) \to (0<1)$

$\text {Ax} \to \lnot (1=0)$ --- see above

$\text {Ax} \to (0<1)$ --- from 11) and 12) by Cut.

Title: Sequent calculus: $0<1$


问:
Any reference for induces exact sequence $0 \to Ker(f) \to Ker(gf) \to Ker(g) \xrightarrow{\delta} Coker(f) \to Coker(gf) \to Coker(g) \to 0.$
Let $\mathscr{C}$ be an abelian category and $f:A \to B$ and $g:B \to C$ morphisms in $\mathscr{C}$, then we have the following exact sequence

$$0 \to \operatorname{Ker}(f) \to \operatorname{Ker}(gf) \to \operatorname{Ker}(g) \xrightarrow{\delta} \operatorname{Coker}(f) \to \operatorname{Coker}(gf) \to \operatorname{Coker}(g) \to 0.$$

I need to use this result which I know its true since I almost got it, but instead of looking help ending the proof. I'm wondering if someone knows a book or text where I can reference this result? This one looks pretty much like the Snake Lemma but I cannot find them in Literature. Or if someone help me see this as an application of a the Snake Lemma it would also be helpful. Thanks

答:
It is a consequence of the snake lemma applied to $$\array{
  \operatorname{Ker}(f) &\to & A &\stackrel{f}{\to} & B &\to& \operatorname{Coker}(f) &\to & 0
    \\
    && \downarrow^{gf} && \downarrow^{g} && \downarrow
    \\
    0 &\to& C &\to & C &\to& 0
  }
$$
and you get
$$ 0 \to \operatorname{Ker}(f)  \to \operatorname{Ker}(gf)  \to \operatorname{Ker}(g)  \to \operatorname{Coker}(f)  \to \operatorname{Coker}(gf)  \to \operatorname{Coker}(g) \to 0$$


问:
How should I show that all natural numbers bigger than 17 can be shown as 3m +10K?
Suppose that a bank machine can dispense money in either 3\$ or 10\$ bills. Show that any amount over 17\$ could be dispensed with combinations of only the 3\$ or the 10\$ bills</p>

答:
For any $a,b$, such that $g.c.d(a,b)=1$, all numbers greater than $ab-a-b$ can be written in the form $na+mb$. This is is the Fröbenius problem, you can find more information here:
The Frobenius Coin Problem


问:
How is it possible that $d(\ln V) = \frac1VdV$?
This is just confusing me. I have encountered in thermodynamics the following:

$$d(\ln V) = \frac1VdV$$

Which I think comes from knowing that:

$$\frac d{dV} (\ln V) = \frac1V$$

But I never thought you could just rearrange derivatives in that manner.

I would appreciate any help! Thanks!

答:
It depends how formal you want to make things, but usually the differential $df$ of a function is defined as
$$df = f'(x)\,dx,$$
so in your case,
$$d(\ln V) = (\ln V)'\,dV = \tfrac1V\,dV.$$
There are various other questions on this site related to yours.


问:
Prove $f$ is closed in compact spaces
I'm dealing with a problem of identifying a quotient space and find out if this map $f$ is closed (or open) would be really helpfull.

Let $f:(z,w)\in S^1\times S^1 \rightarrow f(z,w)=(z^2,w)\in S^1\times S^1$

with

$S^1=\left \{ z \in \mathbb{C}  : \left | z \right |=1\right \}$

I guess it's closed because it's a continuous map defined in compact spaces, but how can I show that more detailed?

答: Every closed subset of a compact metric space is compact. And a continuous map maps compact sets onto compact sets. And, finally, every compact subset of a metric space is closed. So, yes, your map is compact.</p>

Title: Prove $f$ is closed in compact spaces


答: Indeed, $f$ is a closed map. Take a closed subset $F\subseteq \mathbb S^1\times\mathbb S^1$. Since $\mathbb S^1\times \mathbb S^1$ is compact, it follows that $F$ is compact. Then, by continuity, $f(F)$ is compact. Finally, since $f(F)$ is compact in a Hausdorff space (because $\mathbb S^1\times\mathbb S^1$ is a Hausdorff space),  $f(F)$ is closed.

Title: Prove $f$ is closed in compact spaces

Body:
I'm dealing with a problem of identifying a quotient space and find out if this map $f$ is closed (or open) would be really helpfull.

Let $f:(z,w)\in S^1\times S^1 \rightarrow f(z,w)=(z^2,w)\in S^1\times S^1$

with

$S^1=\left \{ z \in \mathbb{C}  : \left | z \right |=1\right \}$

I guess it's closed because it's a continuous map defined in compact spaces, but how can I show that more detailed?


问:
Find $f(x)$, given $f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2$
For a differentiable function $f(x)$, $$f(x+y) - 2f(x-y) + f(x) - 2f(y) = y - 2.$$ Then find $f(x)$.

I am not able to think of any method which is working. I tried replacing $x$ and $y$, putting $x$ or $y = 0$, putting $y = x$, etc.
Can someone help me

Answer given at back is $f(x)=x+1$

答:
Another trick for solving. As PNDas has already mentioned $f(0) = 1$. Now substitute $y = h$ and do some grouping:

$$(f(x+h) - f(x)) - 2(f(x +(-h)) - f(x)) = h + 2(f(0 + h) - f(0))$$

Divide by $h$

$$\frac{f(x+h) - f(x)}h +2\frac{f(x +(-h)) - f(x)}{-h} = 1 +2\frac{f(0 + h) - f(0)}h$$
And take the limit as $h \to 0$ to get
$$3f'(x) = 1 + 2f'(0)$$
which by the assumption of differentiability is guaranteed to converge. Set $x = 0$ to find $f'(0)$, then it is a simple integration to solve.


问:
$\{s_n\}$ is sequence of arithmetic means of $\{a_n\}$ Prove that $\liminf a_n \leq \liminf s_n \leq \limsup s_n \leq \limsup a_n$
I thought that it would be useful to prove all these inequalities separately. The second inequality is obvious. Now we have to understand why the first and the third inequations are true either. The idea was to suppose inversed and come to contradiction. But I haven't managed to finish proof using this approach.

Thank you for help in advance!

答: If $a_n<q$ for all $n>N$ and $a_n<M$ for all $n$, show that $$s_n<\frac{MN+(n-N)q}{n}=q+\frac{NMq}n$$ for all $n>N$ and conclude.


问:
How to solve Probability questions?
I am preparing GRE Test which has Math Part.

How to solve probability questions in a real quick ?

One marble is randomly selected from a bag that contains only 4 black marbles, 3 red marbles, 5 yellow marbles, and 4 green marbles.

Quantity a) The probability of selecting either a black marble or a red marble
Quantity b) The probability of selecting either a yellow marble or a green marble

Whether Quantity a will be greater ?
Whether Quantity b will be greater ?
Whether two Quantities are equal ?
Relationship cannot be determined?

答:
There's no algorithm (ie solution machine) that you can use to solve probability questions.
You have to think about each question.

Most GRE questions are about equally likely outcomes.

The probability is the number of outcomes favourable to the question, divided by the total number of possible outcomes.

In your question: there are 18 marbles, so there are 18 possible outcomes.
For (a) Favourable outcomes are black or red.  There are 7 such outcomes, so probability is 7/18
For (b) Favourable outcomes are yellow or green.  There are 9 such outcomes, so probability is 9/18

cheers

Title: How to solve Probability questions?


问:
Dirichlet convolution of the small prime omega function and the Mobius function
I have seen that: $$(\omega\star\mu)(n)=\sum_{d\vert n}\mu(d)\omega\left(\frac{n}{d}\right)=\begin{cases}1 & n\ \text{is prime}\\ 0 &\text{otherwise} \end{cases}$$ where $\mu(n)=\delta_{\omega(n)}^{\Omega(n)} \lambda(n)$ is the Mobius function, $\omega(n)$ is the small prime omega (counting prime factors without multiplicity) and $\star$ is their Dirichlet convolution.

I was able to show that if $n$ is a prime then this is equal to 1, and if $n$ is any power $\ge2$ of a prime this is 0. However, I'm kinda confused about what to do with other cases.

I tend to think one should look at 2 other cases:first, when $n$ is the product of more than one prime and $n$ is square free, and the other case is when $n=p_1^{a_1}\cdot..\cdotp_k^{a_k}$ and $a_1,..,a_k\ge2$. However, I suspect I'm wrong and in fact, the entire statement can be proven in one fell swoop.

So I'd really like a proof that this convolution is indeed the indicator function for the primes, and I'll appreciate any comments on my attempt.

答:
Consider the Dirichlet convolution of the prime indicator function, $1_{p}=\begin{cases}1& p\ \text{is prime}\\ 0& \text{otherwise}\end{cases}$ with $1_{\mathbb{N}}$, which is 1 for every $n$, i.e.;
$$(1_{p}\star 1_{\mathbb{N}})(n)=\sum_{d\vert n}1_{p}(d)$$
this sum is over all prime divisors of $n$, since they are the only nonzero terms, so we have:
$$\sum_{d\vert n}1_{p}(d)=\sum_{p\vert n}1=\omega(n)$$ by definition, since $\omega(n)$ counts the number of prime factors without multiplicity. This shows that:
$$(1_{p}\star1_{\mathbb{N}})(n)=\omega(n)$$ and hence by The Mobius inversion theorem:
$$(\omega\star\mu)(n)=1_{p}.$$


问:
Proof of a compound proposition being a tautology
The question states the following compound proposition is a tautology, which I have to prove.

$$(p \rightarrow q) \wedge (q \rightarrow r) \rightarrow (p \rightarrow r)$$

My attempt is as follows:

$$\equiv (\neg p \vee q) \wedge (\neg q \vee r) \rightarrow (\neg p \vee r)$$
$$\equiv \neg((\neg p \vee q) \wedge (\neg q \vee r)) \vee (\neg p \vee r)$$

$$\equiv \neg[((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee (q \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee F )\vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv \neg[(\neg p \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$
$$\equiv [((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [(F \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Negation law]}$$

This goes on for quite a while, but I don't get the desired $T$ result. I'm guessing I've done something horribly wrong in the above calculation. I'll be so grateful for your help here. Anything regarding this text, from calculation error to using some incorrect terms. Thank yoou.

答:
I don't know which rules of inference you are using. The easiest would be the following using the rules that are in Understanding symbolic logic of Virginia Klenk:

$1)(p\rightarrow q)\wedge(q\rightarrow r)$ (Conditional Assumption)

$2) p $ (Conditional Assumption)

$3) p\rightarrow q$ (simplification 1)

$4) q$ (Modus Ponens 2,3)

$5) q\rightarrow r$ (Simplification 1)

$6) r$ (Modus Ponens  4,5)

$7) p\rightarrow r$ (Conditional Proof 2-6)

$8) (p\rightarrow q)\wedge(q\rightarrow r)\rightarrow(p\rightarrow r)$ (Conditional Proof 1-7)

Otherwise you can always use truth tables
(Note Virginia Klenk also allows the use of the hypothetical syllogism which can reduce this proof to just a few steps)


答:
Given a number $n \in \mathbb{N}$, then one of the following can happen:

$n \equiv 0 \ (mod 3)$

$n \equiv 1 \ (mod 3)$

$n \equiv 2 \ (mod 3)$

i.e. $n$ can be of the form $3k$ or $3k+1$ or $3k + 2$.

If $n$ is a multiple of $3$ then you can write $$n = 3k + 10 \cdot 0$$
If $n = 3k + 1$ then
$$n = 3k +1 = 3k + 10-9 = 3(k-3) + 10$$
Notice that $k$ is at least $6$ since $n >17$.

If $n = 3k +2$ then
$$n = 3k+2 = 3k+20-18 = 3(k-6) + 10 \cdot 2$$
Notice that $k-6$ is non-negative since $n > 17$.

Title: How should I show that all natural numbers bigger than 17 can be shown as 3m +10K?

Body: Suppose that a bank machine can dispense money in either 3\$ or 10\$ bills. Show that any amount over 17\$ could be dispensed with combinations of only the 3\$ or the 10\$ bills</p>


问:
Chebyshev's inequality assumptions
Why does Chebyshev's inequality demand that $\mathbb{E(}X^2) < \infty$?

答: Otherwise $\mathbb{E(}X) $ may not exist.

Title: Chebyshev's inequality assumptions


答: A common strategy you might not have seen: labelling false as $0$ and true as $1$, $\to$ becomes $\le$ and is therefore transitive.

Title: Proof of a compound proposition being a tautology

Body:
The question states the following compound proposition is a tautology, which I have to prove.

$$(p \rightarrow q) \wedge (q \rightarrow r) \rightarrow (p \rightarrow r)$$

My attempt is as follows:

$$\equiv (\neg p \vee q) \wedge (\neg q \vee r) \rightarrow (\neg p \vee r)$$
$$\equiv \neg((\neg p \vee q) \wedge (\neg q \vee r)) \vee (\neg p \vee r)$$

$$\equiv \neg[((\neg p \vee q) \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee (q \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Distribution of conjunction over disjunction]}$$
$$\equiv \neg[((\neg p \wedge \neg q) \vee F )\vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv \neg[(\neg p \wedge \neg q) \vee ((\neg p \vee q) \wedge r)] \vee (\neg p \vee r)\text{ [Idempotent law]}$$
$$\equiv [(p \vee q) \wedge ((p \wedge \neg q) \vee \neg r)] \vee (\neg p \vee r)\text{ [De Morgan's law]}$$
$$\equiv (p \vee q) \wedge ((p \wedge \neg q) \vee \neg r) \vee (\neg p \vee r)$$
$$\equiv (p \vee q) \wedge (\neg p \vee r) \vee ((p \wedge \neg q) \vee \neg r)\text{ [Commutative law]}$$
$$\equiv [((p \vee q) \wedge \neg p) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [((p \wedge \neg p) \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Conjunction over disjunction]}$$
$$\equiv [(F \vee (q \wedge \neg p)) \vee ((p \vee q) \wedge r)] \vee ((p \wedge \neg q) \vee \neg r)\text{ [Negation law]}$$

This goes on for quite a while, but I don't get the desired $T$ result. I'm guessing I've done something horribly wrong in the above calculation. I'll be so grateful for your help here. Anything regarding this text, from calculation error to using some incorrect terms. Thank yoou.