问:
$dX(t) = a(t)dt +b(t)dW(t)$ and $Y(t)=exp(X(t)$ find $dY(t)$
Having trouble with the following

$$dX(t) = a(t) + b(t)dW(t)$$
$$Y(t) =exp(X(t))$$

find $dY(t)$

答:
What is your trouble then?

If you are to exercise your understanding on Ito formula, everything is direct
.
$dY(t)= \exp ( X(t))\left[ dX_t+\frac{1}{2}d \langle X \rangle_t \right]=Y_t \left[ dX_t+ \frac{1}{2}b(t)^2dt \right]  $


问:
Conditional Probability with Binomial and Beta r.vs
$X$ is a Beta random variable with parameters $\alpha$, $\beta$. $Y$ is a binomial random variable with parameters $n$ and $p=x$ conditional on $X$. Find $P(Y = k)$.

I currently have $$P(Y = k) = \int_0^1 P(Y=k | X = x)f_X{(x)}\,dx = \int_{0}^{1}\binom{n}{k}x^{k}\left(1-x\right)^{n-k}\cdot\frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha) \Gamma(\beta)}x^{\alpha - 1}(1-x)^{\beta - 1}dx.$$ I simplify this expression down to just $1$, using the fact that $\int_{0}^{1}x^{\alpha-1}\left(1-x\right)^{\beta-1}dx\ = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)}$. Am I correct?

答:
Yes, the resulting integrand is still the kernel of a Beta thus

$$\mathbb{P}[Y=k]=\binom{n}{k}\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}\underbrace{\int_0^1x^{a+k-1}(1-x)^{n+b-k-1}dx}_{=\frac{\Gamma(a+k)\Gamma(n+b-k)}{\Gamma(a+b+n)}}$$

All the $\Gamma$ function can be simplified but I do not know if we can arrive at a nice formula...

I get

$$\binom{n}{k}\frac{\overbrace{a\cdot(a+1)\cdot(a+2)\cdot\dots\cdot(a+k-1)}^{\text{k-terms}}\cdot \overbrace{b\cdot(b+1)\cdot(b+2)\cdot\dots\cdot(b+n-k-1)}^{\text{(n-k) terms}}}{\underbrace{(a+b)\cdot(a+b+1)\cdot\dots\cdot(a+b+n-1)}_{\text{n-terms}}}=$$

$$=\binom{n}{k}\frac{\prod_{i=0}^{k-1}(a+i)\prod_{j=0}^{n-k+1}(b+j)}{\prod_{l=0}^{n-1}(a+b+l)}$$

I do not see how to further simplify it using binomial coefficients...perhaps it can be done but at the moment I am stuck

ContentLicense: CC BY-SA 4.0


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$</p>

答:
The beautiful idea of Raymond Manzoni can actually be made rigorous. Consider a finite product $\prod_{n=1}^{L}$ and take its logarithm. After using duplication formula for the gamma function and telescoping, it simplifies to the following:
$$\sum_{n=1}^{L}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\left(1-2^{-L}\right)\ln\left(2\sqrt{\pi}\right)-2L\ln2+2\cdot\frac{1}{2^{L+1}}\ln\Gamma(2^{L+1}).$$
This is an exact relation, valid for any $L$. Now it suffices to use Stirling,
$$\frac{1}{N}\ln\Gamma(N)=\ln N-1+O\left(\frac{\ln N}{N}\right)\qquad \mathrm{as}\; N\rightarrow\infty$$
to get
$$\sum_{n=1}^{\infty}\frac{1}{2^n}\ln\frac{\Gamma(2^n+\frac12)}{\Gamma(2^n)}=\ln\left(2\sqrt{\pi}\right)+2\left(\ln 2-1\right)=\ln\frac{8\sqrt{\pi}}{e^2}.$$
So the answer is indeed $\displaystyle\frac{8\sqrt{\pi}}{e^2}$.

ContentLicense: CC BY-SA 3.0


问:
Definite integral: exponential times non-integer powers
I would like to calculate:
$$
\int_0^1 \sqrt{1+ a x^2} e^{-a x^2/c} \, \mathrm{d}x
$$
for any positive constants $a,c>0$. I am happy with any closed expression involving any kind of special functions.

Failed attempts so far:

standard changes of variables seem not to effectively simplify this integral. Indeed one can change variables (e.g. $ax^2/c\to y$) to simplify the dependence on parameters, but I am not able to reduce the indefinite integration to a known one.

Mathematica fails

I am not able to find this integral (or a similar one, or those integrals that I get by change of variables) on the standard references, e.g. Gradshteyn–Ryzhik.

Contour integration: this I did not try extensively by lack of a reasonable contour. Anyway, the integrand has singularities $\pm \tfrac{i}{\sqrt{a}}$ (choose the standard branch cut of $\sqrt{.}$ on the negative real semi-axis) and $\infty$.

Exploiting the symmetries of the integrand also does not seem to be of use.

Body:
$\int_0^1\sqrt{1+ax^2}~e^{-\frac{ax^2}{c}}~dx$

$=\int_0^1\sqrt{1+ax}~e^{-\frac{ax}{c}}~d(\sqrt{x})$

$=\dfrac{1}{2}\int_0^1x^{-\frac{1}{2}}(1+ax)^\frac{1}{2}e^{-\frac{ax}{c}}~dx$

$=\Phi_{1}\left(\dfrac{1}{2},-\dfrac{1}{2},\dfrac{3}{2};-a,-\dfrac{a}{c}\right)$ (according to https://en.wikipedia.org/wiki/Humbert_series)


问:
Predictive distribution in Bayesian Analysis
How do I find the posterior predictive distribution in Bayesian Analysis? I'm not looking for a specific case, I would like the general solution.</p>

答:
Say $\Theta$ is distributed as $p(\theta)\, d\theta,$ i.e.
$$
\Pr(\Theta\in S) = \int\limits_S p(\theta)\,d\theta
$$
and
$$
X_1,\ldots,X_n \mid \Theta=\theta\sim\text{i.i.d. } f_\theta(x)\,dx
$$
i.e.
$$
\Pr(X_i\in A\mid \Theta=\theta) = \int\limits_A f_\theta(x)\,dx
$$
and $X_1,\ldots,X_n$ are conditionally independent given $\Theta=\theta.$

Then the posterior distribution, i.e. the conditional distribution of $\Theta$ given $X_i=x_i$ for $i=1,\ldots,n,$ is
$$
q(\theta)\, d\theta = \text{constant} \times p(\theta) f_\theta(x_1)\cdots f_\theta(x_n)\, d\theta
$$
where the constant is so chosen that $\displaystyle \int q(\theta)\,d\theta$ (the integral being taken over the whole parameter space) is $1.$

Then the predictive distribution is given by
$$
\Pr(X_{n+1}\in A \mid X_1=x_1,\ldots,X_n=x_n) = \int \left( \, \int\limits_A f_\theta(x)\,dx\right) q(\theta) \, d\theta,
$$
the outer integral again being taken over the whole parameter space.


问:
Space $X$ so that $\pi_{1}(X)=\mathbb{Z}_{3}$
Given a group $G=\langle S|R \rangle$ we can construct a 2-dim complex $K$ s.t  $\pi_{1}(K)=G$. I get that the main idea is to construct 2-cells that "read" the relators and glue them to the wedge that represents the free group on $S$. With that in mind I can see for example that if $G=\mathbb{Z}\times\mathbb{Z}$ we glue the rectangle $aba^{-1}b^{-1}$ in the graph $R_{2}$ (rose graph) that represents the free group on $a,b$ and after that glueing we get  a torus. WIth the same procedure we can easily see that if $G=\mathbb{Z}_{2}$ we get the projective plane. But what about all the other cyclic groups? For example which space has fundamental group $\mathbb{Z}_{3}$? I am having trouble to see if this is a familiar space.

答:
The real projective plane has fundamental group $\mathbb Z_2$, not $\mathbb Z_3$. To obtain a space with fundamental group $\mathbb Z_n$, $n\geq 2$, you can form the quotient of the 2-disk $D^2$ identifying each point $x\in \partial D^2 = S^1$ with its counterclockwise rotation by an angle of $\frac{2\pi}{n}$.

The projective plane is the case $n=2$ in this construction.


问:
Manifolds with Betti numbers like a projective plane
Is there a classification of integers $k>0$ such that there can be a simply connected closed $4k$-manifold $M$ with $b_0(M)=b_{2k}(M)=b_{4k}(M)=1$ and all other betti numbers $0$? (I don't mind if there is some torsion in some of the integral homology)

答: These are known as rational projective planes. Su has studied these and found many dimensions in which they exist or don't exist. See https://arxiv.org/pdf/1702.07892.pdf for one of her papers on the subject. Also this talk by Zagier https://www.youtube.com/watch?v=lLEWKDtCPTU announces the dimensions less than a million in which these things can exist.</p>


问:
I have difficulty counting
Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?

答:
We want
\begin{align}
\color{red}{S = 84 + 87 + 90 + 93 + \cdots + 180 + 183}
\end{align}
We also have
\begin{align}
\color{blue}{S = 183 + 180 + 177 + 174 + \cdots + 87 + 84}
\end{align}
This gives us
\begin{align}
\color{green}{2S} & \color{green}{= (84+183) + (87+180) + (90+177) + \cdots + (180+87) + (183+84)}\\
& \color{green}{= (267) + (267) + (267) + \cdots + (267) + (267)}\\
\end{align}
Let us count how many such $267$'s are there: We have $183 = 84 + 3k$. Note that $k+1$ gives the number of terms in the summation. We get $k = \dfrac{183-84}3 = 33$. Hence, there are $34$ terms. We hence get that
$$\color{green}{2S = 34 \times 267 \implies S = 17 \times 267 = 4539}$$</p>

Title: I have difficulty counting


问:
For any $X\subset \mathbb{R}$, is $|X|<|\mathbb{R}|, |X|\leq|\mathbb{R}|,$ or $|X|=|\mathbb{R}|$?
So take for example $(20,35) \subset \mathbb{R}$

$(20,35)$ has an infinite number of elements and is uncountable

$\mathbb{R}$ has an infinite number of elements and is uncountable

Can we find a function $f:(20,35)\rightarrow \mathbb{R}$ that maps every element of the domain to the codomain?

Or is the infinity posed by $|\mathbb{R}|$ bigger than the infinity posed by $|(20,35)|$?

Like, for example, we know that $|\mathbb{N}|<|\mathbb{R}|$ since for any $n \in \mathbb{N}$, we can find an $x \in \mathbb{R}:n<x<n+1$, thus $|\mathbb{N}|<|\mathbb{R}|$

I would like to think that since $(20,35)\subset\mathbb{R}$, that by design, $|(20,35)|< |\mathbb{R}|$ but the uncountability of the reals makes this dangerous to assume.

答:
By definition, a set is infinite if and only if it contains a proper subset with the same cardinality as itself. Put another way:

A set $K$ is infinite $\iff \exists L \subset M, L \neq M : |L|=|M|$

We can see examples of this with $2\mathbb{Z} \subset \mathbb{Z}$

We can find a bijective function $f: 2\mathbb{Z} \rightarrow \mathbb{Z}$ so that every element of $2\mathbb{Z}$ gets mapped to an element in $\mathbb{Z}$

In the same manner, with the uncountable reals,

For any nonempty proper subset $X \subset \mathbb{R},$ where $X$ is not countable,

Then $|X|=|\mathbb{R}|$

Here, also we can find a bijective function $f: X \rightarrow \mathbb{R}$ so that every element of $X$ gets mapped to an element in $\mathbb{R}$


问:
another definition of Lebesgue functions
Assume $U(I)$ is all the functions $f:[a,b]\to \mathbb{R}$ such that there exists an increasing sequence $(\phi_n)_n$ of simple functions such that  $ \lim\int_{a}^{b}\phi_n<\infty $ and  $\phi_n\to f$ (almost everywhere). And assume  $U_0(I)$ is all the functions $f:[a,b]\to \mathbb{R}$ such that there exists an increasing sequence $(\phi_n)_n$ of simple functions such that  $ \lim\int_{a}^{b}\phi_n<\infty $ and  $\phi_n\to f$.

Is it true that every Upper function(= every element of U(I))  is almost everywhere equal to an element of $U_0(I)$ ?

答: If $\phi_n\to f$ almost everywhere this mean that $\mathbf{1}_{A}\phi_n\to \mathbf{1}_{A}f$ everywhere for some measurable set $A\subset \mathbb{R}$ such that $\mu(A^\complement )=0$ (here $\mu$ is the measure defined in $[a,b]$, it can be the Lebesgue measure or any other). Now note that $\mathbf{1}_{A}f=f$ almost everywhere.


答: When you count $1$ through $10$ you have $10$ elements, but $10-1=9$. The extra one comes because you are counting both ends, while taking the difference counts the gaps between them. You have to add $1$ to account for that.</p>

Title: I have difficulty counting

Body:
Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?


问:
Closure of a compact subset of an open set in a T3 space.
Near the top of page 25 of Steen & Seebach's "Counterexamples in Topology" (2nd edition) it says that if A is a compact subset of a T3 space, and U is an open set containing A, then there is an open set V such that $A\subset V\subset\bar{V}\subset U$ (where $\bar{V}$ is the closure of V).
I can see that we can get an open set V from the intersection of U with the finite open cover of A, but I've been banging my head against the wall trying to figure out why $\bar{V}\subset U$.
Can anyone enlighten me, or it is a mistake in the book?

答: It is easy to show this to be true when $A$ is a point, so in general there is a cover $\{V_x\}_{x\in A}$ of $A$ such that each $\bar V_x$ is contained in $U$. By compactness of $A$ there is a finite subcover $\{V_i\}_1^n$. Set $V = \cup_1^n V_i$, then $$A\subset V \subset \bar V = \cup_1^n \bar V_i \subset U.$$


问:
Integrate counterclockwise around unit circle. Indicate whether Cauchy Riemann theorem applies. Show details
$$\frac{1}{2z-1} = f(z)$$

So I don't think CR theoreum applies. So if the denominator is 0, then it's not defined. If z = 1/2, then it is not defined at 0. So f(z) it is not analytic even within C right? So there does not exist a simply connected domain D on which f is analytic and such that D contains C. So cannot apply Cauchy's Riemann equation right?

So we have to integrate another way.

So unit circle could be expressed as $e^{it} = z(t)$ where t is between 0 and $2\pi$

Furthermore $dz/dt = ie^{it}$

$$\int_0^{2\pi} \frac{1}{2e^{it} - 1} \cdot ie^{it}$$

Am I on the right track? If so, can someone help with the integral?

答:
METHODOLOGY $1$:  CONTOUR DEFORMATION

We can deform the contour $|z|=1$ and apply Cauchy's Integral Theorem to assert that for $0<r<1/2$

$$\begin{align}
\oint_{|z|=1}\frac1{2z-1}\,dz&=\oint_{|z-1/2|=r}\frac1{2z-1}\,dz\\\\
&=\int_0^{2\pi}\frac1{2re^{it}}ire^{it}\,dt\\\\
&=i\pi
\end{align}$$

METHODOLOGY $2$:  PARAMETERIZATION

Alternatively, we can parameterize the contour $|z|=1$ and write

$$\begin{align}
\oint_{|z|=1}\frac1{2z-1}\,dz&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{ie^{it}}{2e^{it}-1}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\int_{\varepsilon}^{2\pi-\varepsilon}\frac{\sin(t)+i2(1-\cos(t))}{5-4\cos(t)}\,dt\\\\
&=\lim_{\varepsilon\to 0^+}\left.\left(\frac14\log(5-4\cos(t))+i2\left(\frac t4-\frac16 \arctan\left(3\tan(t/2)\right)\right)\right) \right|_{\varepsilon}^{2\pi -\varepsilon}\\\\
&=i\pi
\end{align}$$

as expected!

METHODOLOGY $3$:  COMPLEX LOGARITHM

As a third approach, we cut the plane so that the natural logarithm is defined as $\log(z)=\text{Log}(|z|)+i\arg(z)$, where $0\le \arg(z)<2\pi$ we have

$$\oint_{|z|=1}\frac1{2z-1}\,dz=\lim_{\varepsilon\to 0^+}\left.\left(\frac12\log(2z-1)\right)\right|_{z=1+i\varepsilon}^{z=1-i\varepsilon}=i\pi$$


问:
Generalization of natural density on $\omega_1$
The natural density (or asymptotic density) of a subset $A$ of $\mathbb{N}$ is defined as $$d(A) = \lim_{n \rightarrow +\infty} \frac{|A\cap \{0,1,\dots, n-1\}|}{n}$$ when such limit exists.

I was wandering if there exists a generalization of such concept that captures a sort of asymptotic density of subsets of larger cardinals. For example a measure (not necessarily in the measure theoretic sense) that extends the natural density and that discriminates between cofinal and bounded subsets of $\omega_1$. Is there? Any idea?

Thanks!

EDIT: I'll try to frame the question in a more specific and different way. What I'm looking for is an additive measure $\mu: \Sigma\subseteq\mathcal{P}(\omega_1) \rightarrow X$ with $(X,\le,+,0)$ being a ordered divisible group, and $\Sigma$ a "reasonable" algebra of the power set (i.e. it includes non-pathological subsets of $\omega_1$, like in the $\omega$ case when it included sets having a definite asymptotic density)  s.t.

$\mu(\omega_1) = 1$, where $1$ is an element of the group strictly greater than $0$

$\mu(\emptyset) = 0$

$A\subseteq B \Rightarrow \mu(A) \le \mu(B)$

$A \cap B = \emptyset \Rightarrow \mu(A \cup B) = \mu(A)+\mu(B)$

$\mu(n\omega_1) = \frac{1}{n}$, where $n\omega_1$ is the set of all the ordinals less than $\omega_1$ having finite part divisible by $n$

Given $A,B \subseteq \omega_1$ with $A$ bounded and $B$ cofinal, then $\mu(A) < \mu(B)$

答:
This isn't going to be a complete answer, but just some thoughts that may help you go in the right direction.
First, here are two small points about your question:

Like bof mentioned, natural density on $\omega$ allows cofinal subsets to have density $0$.
So, it might be more reasonable to change your condition (6) to $\mu(A) = 0$ for any bounded $A \subseteq \omega_1$.

You said $\Sigma \subseteq \mathcal{P}(\omega_1)$ should be an algebra, but in fact this is not even true for natural density.
It's a fun exercise to try to find $A, B \subseteq \omega$, both having natural density, for which neither $A \cup B$ nor $A \cap B$ have natural density.

With that out of the way, the more important point here is that if you want to analogize with natural density as closely as possible you probably want to think about Følner sequences.
If $G$ is a countable left-cancellative semigroup, a (left) Følner sequence of $G$ is a sequence $\Phi := (\Phi_n)_{n \in \mathbb{N}}$ of finite subsets of $G$ satisfying
$$
\lim_{n \to \infty} \frac{|\Phi_n \cap g\Phi_n|}{|\Phi_n|} = 1
$$
for every $g \in G$.
Then, given $A \subseteq G$, we define the density of $A$ with respect to $\Phi$ as
$$
\operatorname{d}_\Phi(A) := \lim_{n \to \infty} \frac{|A \cap \Phi_n|}{|\Phi_n|}.
$$
Then many of the familiar properties of natural density hold for $\operatorname{d}_\Phi$.
In particular, your properties (1)-(4) are immediate from the definition, and you get some other nice properties as well, such as left-translation invariance.

Of course, in your case, the semigroup $(\omega_1, +)$ is not countable, which means that one has to be a bit more careful.
It is fortunately left-cancellative.
Because of this, it may be useful to first try finding “natural” Følner sequences for countable ordinals, such as $(\omega^2, +)$.
This doesn't seem too difficult to do, and might give some ideas for how to approach the uncountable case.

In the case of $(\omega_1, +)$, it seems that no Følner sequence, under the definition I gave, can exist.
Indeed, if $(F_n)$ is any sequence of finite subsets of $\omega_1$, then $\cup F_n$ is finite and so there is some $\alpha \in \omega_1$ larger than any element of this set.
From there it is immediate that the Følner condition will fail with $g = \alpha$.
An idea which might be worth pursuing would be to consider Følner sequences of uncountable length or even Følner nets.
A good reference is probably the paper “Density in arbitrary semigroups” by Neil Hindman and Dona Strauss.


答:
It's one of those typical "off by one" errors. The best way to see how to count the elements in such a sequence is to start with few elements:

If you have $84+87$ you have $(87-84)/3=1$ and there are $2$ elements in the sum

If you have $84+87+90$ you have $(90-84)/3=2$ and there are $3$ elements in the sum

If you have $84+87+90+93$ you have $(93-84)/3=3$ and there are $4$ elements in the sum

I guess by now you see the pattern ...

If you have $84+87+90+\cdots183$ you have $(183-84)/3=33$ and there are $34$ elements in the sum

Title: I have difficulty counting


问:
How to apply a change of variables in $\int_0^1\int_0^1 \frac{\mathrm{d}x\mathrm{d}y}{1-xy}$?
Denote $\mathcal{D}=\{(x,y)\in \mathbb{R}^2: 0\le x\le 1 ;\ 0 \le y\le 1   \}$.

I have to use : $x=\cos{\alpha} - t$ and $y=\cos{\alpha}+ t$.

I noticed that $\frac{x+y}{2}=\cos{\alpha}$ and $\frac{y-x}{2}=t$. So I can take $\alpha \in [0, \pi/2]$ and $t\in[-\min\{\ \cos(\alpha) \ ; 1-\cos(\alpha)\ \}, +\min\{\ \cos(\alpha)\ ; 1-\cos(\alpha)\ \}]$.

And now I do not know how to continue. I'm maybe wrong for the $\mathcal{C}^{1}$-diffeomorphism...

Thanks in advance !

答:
This doesn't directly answer your question, but I would solve the integral as follows:

$$\int_0^1\int_0^1 \frac{\mathrm{d}x\mathrm{d}y}{1-xy} = \int_0^1 \int_0^1 \sum_{n=0}^{\infty}(xy)^n~\mathrm{d}x\mathrm{d}y =  \sum_{n=0}^{\infty}\int_{0}^{1}\frac{y^n}{n+1}~dy = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$$


问:
Linear Algebra Textbook for Reference?
Just want a good linear algebra textbook for reference.
I know that there is a lot of good ones, but I am not a mathematician and I don't want anything way too abstract like Axler, Curtis, Hoffman&Kunze, etc.

However, I want the text to cover "advanced" things like inner product, Jordan forms, Bilinear forms, etc, and at least prove some of the theorems.

I am currently trying Blyth& Robertson. Though I prefer a book with examples and exercises more interesting. Also LADW is good, though I prefer paperback.

Any other suggestions?

Edit: Not that I cannot understand Axler...

Due my limited exposure to algebra and laziness, I find it way harder to see the connection between LADR and first-year linear algebra than between, say, Rudin and calculus (You just want to replace "compact set" with "closed intervals" most of the time...right?). That's why I think LADW is better for me, even if many people think it is harder than LADR.

答:
This book is great : Linear Algebra and Its Applications, David C. Lay, University of Maryland. https://www.pearson.com/us/higher-education/product/Lay-Linear-Algebra-and-Its-Applications-4th-Edition/9780321385178.html

Many examples and exercices (even corrected). I know some others, but these are in french.

Title: Linear Algebra Textbook for Reference?


答:
You have to go back to the definition of cardinality to clear up your misunderstandings about the sizes of sets.

Here are some definitions and provable statements that you should understand first.

Two sets have the same cardinality when there is a bijection between
them.

The size of a subset of a set can be no larger than the size of the
set, because there is an injection from the subset to the set.

A subset of a set may have the same size as the whole set: there can
be a bijection even if the natural inclusion injection isn't a
bijection. The standard first example of this is the argument that
there are just as many even integers as integers.

A set is infinite precisely when it is the same size as one of its
subsets. That tells you there are infinitely many integers.

There can be infinite sets of different sizes. The class argument due
to Cantor shows that the size of the set of rational numbers is
strictly smaller than the size of the reals.

So the answer to the question in the title is the weak inequality: $|X| \le \mathbb{R}$.

Finally this

Like, for example, we know that |N|<|R| since for any n∈N, we can
find an x∈R:n<x<n+1, thus |N|<|R|

is a correct conclusion from a false argument. The argument must be wrong since it would apply to show the false result that the size of set of irrational numbers is strictly smaller that that of the reals, since you can always find a rational between any two irrationals.

Title: For any $X\subset \mathbb{R}$, is $|X|<|\mathbb{R}|, |X|\leq|\mathbb{R}|,$ or $|X|=|\mathbb{R}|$?

Body:
So take for example $(20,35) \subset \mathbb{R}$

$(20,35)$ has an infinite number of elements and is uncountable

$\mathbb{R}$ has an infinite number of elements and is uncountable

Can we find a function $f:(20,35)\rightarrow \mathbb{R}$ that maps every element of the domain to the codomain?

Or is the infinity posed by $|\mathbb{R}|$ bigger than the infinity posed by $|(20,35)|$?

Like, for example, we know that $|\mathbb{N}|<|\mathbb{R}|$ since for any $n \in \mathbb{N}$, we can find an $x \in \mathbb{R}:n<x<n+1$, thus $|\mathbb{N}|<|\mathbb{R}|$

I would like to think that since $(20,35)\subset\mathbb{R}$, that by design, $|(20,35)|< |\mathbb{R}|$ but the uncountability of the reals makes this dangerous to assume.


问:
set of all possible $n$-tuple combinations
I am dealing with the really simple problem of generating a pin code programmatically. Like "A1B" And now I would like to write it down mathematically. To show how much unique variations of pin codes exists for a specific $n$.

What I have already:

A code contains of $n$ symbols from the set $\text{C}$

where

$n \in \mathbb{N}; n \ge 1$

$\text{A} = \{ ``A",``B",``C"\}$

$\text{B} = \{ ``1",``2",``3"\}$

$\text{C} = \text{A} \cup \text{B}$

The set $\text{D}_n$ of all possible combinations of a $n$-long code is described as

$\text{D}_n =$ ???

So that
$\text{D}_3 = \{ ``AAA", ``AAB", ``AAC", ..., ``AA1", ``AA2", ..., ``333"\}$

So the number of all possible variations
$\vert \text{D}_n \vert = \vert \text{C} \vert ^n$

EDIT:
Better Question:

How can I write that the elements of $\text{C}$ come togehter as an Tuple (or whatever) and how to place $n$ in this form?

$\text{D}_n =$ ???

答:
After more research I think this could be a solution.

Now I describe a code as a $n$-tupel, eg a triple ("A", "2", "B"):

$\text{D}_n = \{ \space (\text{x}_1, \dots, \text{x}_n ) \space | \space \text{x}_1, \dots, \text{x}_n \in \text{C} \space \}$


问:
Linear shooting method for Second Order BVP
How can we use the linear shooting method to solve the boundary value problem $$y'' = 2y' - y, ~y(0) = 1, \text{and} ~~y(1) = 2?$$

I tried to convert it to a first order system, but didn't get what I needed.

答:
Characteristic equation of
$$
y''-2y+y = 0
$$
is
$$
\lambda^2 - 2\lambda+1 = 0 \\
\lambda_{1,2} = 1
$$
so general solution can be found as
$$
y = e^x(C_1 + C_2 x)
$$
To find unknown coefficients one can use boundary conditions
$$
y(0) = C_1 = 1 \\
y(1) = e(1+C_2) = 2 \\
C_2 = \frac 2e-1
$$
And finally
$$
y(x) = e^x \left [1 + x \left( \frac 2e-1\right) \right ]
$$</p>


问:
Arrangements of Solutions to $n$ Linear Equations with $n$ Unknowns
In a system of two linear equations with two unknowns, there are three "arrangements" that we can see when we graph the two lines in the $xy$ plane:

The two lines intersect at a single point (one solution to the system).

The two lines are parallel and never intersect (no solution to the system).

The two equations describe the same line (infinitely many solutions to the system).

When we move up to a system of three linear equations with three unknowns, now we have three planes in space, and there are eight distinct arrangements of the three planes:

If we consider one linear equation with one unknown, I suppose it makes sense to say that there is one arrangement, which is a single point on the real number line.

So, for one, two and three unknowns, we have the start of a sequence: $1, 3, 8, ...$

I am interested in how this sequence continues. I have searched in vain on the OEIS. Alas, there are many sequences that have $1, 3, 8,...$ and I'm not sure which, if any, are the right one.

This one: https://oeis.org/A001792 looks like it might be it, because the comments say that sequence is related to matrices in certain ways. Also I expect a formula for this sequence should involve powers of 2. But, I wouldn't bet any money on it.

Is there a sequence with a simple formula to work out the number of arrangements?

答:
More than an answer, the following represent some considerations on how the problem could be tackle.

Start and consider just two planes (in $3$-D) with the relevant
matrix of coefficients $\bf C$ and the augmented matrix (coefficients plus known terms) $\bf A$
$$
\underbrace {\left( {\overbrace {\matrix{
   {a_1 } & {b_1 } & {c_1 }  \cr
   {a_2 } & {b_2 } & {c_2 }  \cr
 } }^{\bf C}\;
\left| {\;\matrix{
   {d_1 }  \cr
   {d_2 }  \cr
 } } \right.} \right)}_{\bf A}\;
$$

Now, if the matrix $\bf C$

has full rank $2$, then the two normal vectors are independent, i.e.
the planes are incident;

has rank $1$, then the two vectors are dependent, i.e. parallel;

the rank $0$ corresponds to having all the coefficients null, which is a
degenerated situation.

The rank of $\bf A$ (call it $R$) cannot be less than that of $\bf C$ ($r$), nor it can be grater than $r+1$.

When $r=0$, $R$ can be $0$ or $1$ and it will be in any case a degenerated situation.

When $r=1$, $R$ can be $1$ (planes coincident), or $2$ (planes parallel).

When $r=2$ also $R=2$ and the planes are incident.

Passing to consider three $3$-D planes, your scheme of distinction
is unfortunately wider than that provided by considering  the ranks of the $3 \times 4$ matrices.

For the eight cases that you sketched in fact we have
$$
\begin{array}{c|cccccccc}
   {} &  &  {\left( 1 \right)} & {\left( 2 \right)} & {\left( 3 \right)}
 & {\left( 4 \right)} & {\left( 5 \right)} & {\left( 6 \right)}
 & {\left( 7 \right)} & {\left( 8 \right)}  \\
\hline
   r &  &  1 & 2 & 1 & 2 & 1 & 3 & 2 & 2  \\
   R &  &  1 & 2 & 2 & 3 & 2 & 3 & 2 & 3  \\
\end{array}
$$
We have that the three couples $[(2),\, (7)] , \; [(3),\, (5)], \; [(4),\, (8)]$ cannot be distinguished.

To be able to distinguish between them, for $r < 3$,  we could introduce a fourth plane, with coefficients
independent from the previous three, which means that it increases $r$ to $r+1$, and count
the crosses we are going to have.

Equivalently, we can consider both the ranks of the first two planes and how they change by
adding the third.

However this way would soon become impractical for higher dimensions.


问:
How to set bounds in cylindrical coordinates?
I'm trying to evaluate the following integral in cylindrical coordinates

$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_x^{\sqrt{1-x^2}}e^{-x^2-y^2} \, dy \, dx \, dz$$

After attempting to set the bounds in cylindrical coordinates, I got
$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_{\rho \cos\varphi }^{\sqrt{1-\rho^2 \cos^2\varphi }}e^{-\rho ^2}d\varphi \rho \, d\rho \, dz$$
But I know this doesn't make sense. Can someone explain how to switch the bounds analytically? I don't understand how to transform the bounds.

答:
Analytically and without drawing: in short, solving inequalities. You have set
$$\left\lbrace \begin{array}{a}
0 \leqslant y \leqslant 6 \\
0 \leqslant x \leqslant \frac{1}{\sqrt{2}} \\
x \leqslant z \leqslant \sqrt{1-x^2}
\end{array} \right\rbrace$$
taking cylindrical coordinates
$$\begin{cases}x=\rho \cos \phi \\
y= \rho \sin \phi \\
z=z'\end{cases}$$
we obtain restrictions for Oxy
$$\left\lbrace \begin{array}{a}
0 \leqslant \rho \sin \phi \leqslant 6 \\
0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}}
\end{array} \right\rbrace$$
Solving this inequalities we obtain $0 \leqslant \rho \leqslant \min \left\lbrace \frac{1}{\sqrt{2}\cos \phi }, \frac{6}{\sin \phi} \right\rbrace$. From here we obtain angle $\tan\phi_1 = 6\sqrt{2}$. So integral will be

$$\int\limits_{0}^{\phi_1}\int\limits_{0}^{\frac{1}{\sqrt{2}\cos \phi }}\int\limits_{\rho\cos \phi}^{\sqrt{1-(\rho \cos \phi)^2}}+\int\limits_{\phi_1}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{6}{\sin \phi }}\int\limits_{\rho\cos \phi}^{\sqrt{1-(\rho \cos \phi)^2}}$$

Addition:
As pointed in comments below I proceed from that sequence of limits in OP integral matches integration variables. Taking reverse direction we obtain set
$$\left\lbrace \begin{array}{a}
0 \leqslant z \leqslant 6 \\
0 \leqslant x \leqslant \frac{1}{\sqrt{2}} \\
x \leqslant y \leqslant \sqrt{1-x^2}
\end{array} \right\rbrace$$
now we have restrictions
$$\left\lbrace \begin{array}{a}
0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}} \\
\rho \cos \phi \leqslant \rho \sin \phi \leqslant \sqrt{1-(\rho \cos \phi)^2} \\
\end{array} \right\rbrace$$
first inequality in second line gives restriction $\tan \phi \geqslant 1$, which together with $\cos \phi \geqslant 0, \sin \phi \geqslant 0$ from first line gives $\phi \in \left[\frac{\pi}{4},\frac{\pi}{2}\right]$. At last second inequality from second line gives $\rho \leqslant 1$, so integral have limits
$$\int\limits_{0}^{6}\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{1}$$

At end let me say, that although drawing is most often  recommended solution for multi variable integrals, but it is not exact formal proof. We can consider it as good tool for creating intuition about solution, but correct proof should come from solving inequalities system.


问:
Example of Diverging functions with converging integral
{Hey thanks for sparing some time I would very much appreciate your help on this problem :}

Give a example of a continuous function $f$ such that $f(x)\geq0$
$(1\leq{x}\leq\infty)$
and such that $$\int_1^\infty{f(x)dx}$$ converges but $$\sum_1^\infty{f(n)}$$  diverges.

I have just started studying analysis from R.R.Goldberg and find few of the problem a bit tricky  Will very much appreciate your help.

答: Take the always vanishing piecewise linear map $f$ except that $f(n-1/2n^2)=f(n+1/2n^2)=0$ and $f(n)=1$ for all $n \in \mathbb N$.


答:
"Can we find a function f:(20,35)→R that maps every element of the domain to the codomain?" Yes.

If $g(x) = \frac {x-20}{15}$ then $g:(20,35)\to (0,1)$ is a bijjection.

And if $h(x)=\pi x - \frac \pi 2$ then $h:(0,1)\to (-\frac \pi 2, \frac \pi 2)$ is bijection.

And $\tan:(-\frac \pi 2,\frac \pi 2) \to (-\infty, \infty) = \mathbb R$ is a bijection.

So if $f= tan(\frac{x-20}{15}\cdot \pi - \frac \pi 2)=\tan(h(g(x))) $ the $f:(20,35)\to \mathbb R$ is a bijection.

SO $|(20,35)| = |\mathbb R|$.


答:
Using residue theorem we easily get
$$\oint\limits_C {\frac{{dz}}{{2z - 1}} = 2\pi i}\;\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right) =\pi i$$

Because
$$\text{Res}\left(\frac{1}{2 z-1},\left\{z,\frac{1}{2}\right\}\right)=\underset{z\to \frac{1}{2}}{\text{lim}}\frac{z-\frac{1}{2}}{2 z-1}=\frac{1}{2}$$

Title: Integrate counterclockwise around unit circle. Indicate whether Cauchy Riemann theorem applies. Show details

Body:
$$\frac{1}{2z-1} = f(z)$$

So I don't think CR theoreum applies. So if the denominator is 0, then it's not defined. If z = 1/2, then it is not defined at 0. So f(z) it is not analytic even within C right? So there does not exist a simply connected domain D on which f is analytic and such that D contains C. So cannot apply Cauchy's Riemann equation right?

So we have to integrate another way.

So unit circle could be expressed as $e^{it} = z(t)$ where t is between 0 and $2\pi$

Furthermore $dz/dt = ie^{it}$

$$\int_0^{2\pi} \frac{1}{2e^{it} - 1} \cdot ie^{it}$$

Am I on the right track? If so, can someone help with the integral?


问:
Conceptual difference between Poisson and uniform distribution
I feel very stupid asking this question, because they're obviously different concepts, but I can't understand why. Every textbook I read has them both thoroughly explained, but at some point I can't grasp WHY they're different. Where do they defer?

Basically, what I can see is that in any Poisson process, if $N(t)$ is the amount of successes in a interval $[0,t)$, then $N(t_i)$ is independent of $N(t_j)$, $t_i$ and $t_j$ being sections of the interval, having the same length.

Besides, successes are distributed uniformly. The probability of a success does not depend on its position in the interval, but only its size.

I don't understand why those characteristics are not describing a Uniform distribution too. I mean, where is the line drawn?

Thank you very much for reading, and I'm sorry if you didn't understand something i said: English is not my prime language, and I acknowledge I have some problems using it.

答:
Not a stupid question at all!

For a Poisson process, if one and only one event occurs in the interval between 0 and $t$, then the timing of when the event occurs is uniform between 0 and t. The total number of occurrences $N(t)$ is a Poisson random variable. It's when we "zoom in" and look at a single occurrence that we observe a uniform distribution (or an exponential distribution if we're interested in the waiting time instead of the time of the occurrence).


问:
Are Transition Maps Implied within an Atlas?
From my understanding of (smooth) manifolds, all you need is an atlas to describe a manifold. However, if you have some atlas ={($U_n$,$\phi_n$)} with $n$ charts, we still haven't defined our transition maps. My questions are:

Are the transition maps implied within the atlas (i.e. you can derive all transition maps from a given atlas) or do we have to store our transition maps along with our atlas to prove we have a smooth atlas?

If you have $n$ charts within an atlas, does that mean you are going to have something like $n!$ (maybe its a bit more complex than that) transition maps? For example, if $n=3$ and a chart $c\in A$, wouldn't you need a transition map from $c_1 -> c_2$, $c_1 -> c_3$, $c_2 -> c_3$ plus all of the inverses (that are implied)? When don't you need a transition map between two charts in the same atlas?

答: Once you have the charts $\phi_n$, the transition maps are determined, as $\phi_m\circ\phi_n^{-1}$.  (That uses my favorite convention for the direction of these maps; you might need to move the "inverse" if your convention is different.)


问:
Graph Theory Software with simple GUI
To the best of my knowledge I cannot find, on this site, any graph theory program resources.

I am looking for a program where I can draw nodes and edges and most importantly drag and drop vertices while keeping edges in tact.  Does any such program exist?  Can you provide me with resources to find such a program?

I want such a program, where I can drag and drop vertices, so that I may better visualize different isomorphisms of the same graph.  I have a tough time seeing these sorts of things in my minds eye (although I'm trying to get better at it).

Although this appears as though it may be a duplicate, I want to stress that I am looking for a program with a simple GUI, where I can drag and drop vertices while keeping edges in tact.  The "duplicate" question's answers provide an extensive of list of programs that can return information on a graph and allow you to programmatically generate graphs.  I am not interested in this at the moment.  I want an incredibly simple program.

Please only post answers with these conditions in mind.

答:
It sounds like Geogebra Geometry is exactly what you are looking for, at least for casual use on a small number of vertices.  It lets you drop down points, connect segments between them, and move the points around freely (bringing the lines with them).  It does it in a simple web-based interface, with no software installation needed.

I think it might require some work to automatically import a graph specified in a text file, as you'd want to do if you were working with something of moderate size.   But for gathering intuition about small graphs (as the original question describes), Geogebra is pretty well-suited.

Title: Graph Theory Software with simple GUI


问:
Why are the functions $H_t$ in a homotopy continuous for all $0 \le t \le 1$?
If $f,g: X \to Y$ are two continuous functions, the definition of a homotopy between them that is a continuous function $H:I \times X \to Y$ where $H(0,x) = f(x)$ and $H(1,x) = g(x)$ for all $x \in X$.

In Introduction to Topological Manifolds it then states that the functions $H_t:X \to Y$ defined by $H_t(x) = H(t,x)$ are a family of continuous functions. I'm confused about how to deduce that the functions $H_t$ are continuous given that $H$ is continuous.

I thought about looking at the universal property of product spaces but this gives a statement about continuous maps into $I \times X$ as opposed to continuous maps out of $I \times X$.

答: For a specific $t$, Try showing that the function $f:X \to I \times X$, defined by $f(x) = (t,x)$ is continuous. Then deduce that $H_t$ is continuous, using composition of functions.


答:
You can simply define the transition maps, once the atlas is given.

There is a transition map which I shall denote $\psi_{m,n}$ for every pair of indices $m,n$ having the property that $U_m \cap U_n \ne \emptyset$.

The domain of $\psi_{m,n}$ is the set $\phi_m(U_m \cap U_n) \subset \mathbb R^k$ (I'm assuming implicitly that $k$ is the dimension of the manifold).

The range (or codomain) of $\psi_{m,n}$ is the set $\phi_n(U_m \cap U_n) \subset \mathbb R^k$.

And the formula for $\psi_{m,n} : \phi_m(U_m \cap U_n) \to \phi_n(U_m \cap U_n)$ is
$$\psi_{m,n}(p) = \phi_n(\phi^{-1}_m(p)), \quad p \in \phi_m(U_m \cap U_n)
$$

Also, once all of this is written down, one can use the definition of a manifold together with the Invariance of Domain Theorem to prove that the domain and range of $\phi_{m,n}$ are both open subsets of $\mathbb R^k$, and one can show that $\psi_{n,m}$ is an inverse map of $\psi_{m,n}$, hence each transition map is a homeomorphism from its domain to its range.

And once that is done, you can now ask yourself questions that are aimed at determining whether your manifold is a $C^\infty$ manifold, or a $C^2$ manifold, or a $C^1$ manifold or whatever smoothness property you want. Namely: Are the functions $\{\psi_{m,n}\}$ all $C^\infty$? or are they all $C^2$? or $C^1$?

Title: Are Transition Maps Implied within an Atlas?

Body:
From my understanding of (smooth) manifolds, all you need is an atlas to describe a manifold. However, if you have some atlas ={($U_n$,$\phi_n$)} with $n$ charts, we still haven't defined our transition maps. My questions are:

Are the transition maps implied within the atlas (i.e. you can derive all transition maps from a given atlas) or do we have to store our transition maps along with our atlas to prove we have a smooth atlas?

If you have $n$ charts within an atlas, does that mean you are going to have something like $n!$ (maybe its a bit more complex than that) transition maps? For example, if $n=3$ and a chart $c\in A$, wouldn't you need a transition map from $c_1 -> c_2$, $c_1 -> c_3$, $c_2 -> c_3$ plus all of the inverses (that are implied)? When don't you need a transition map between two charts in the same atlas?


答:
As indicated by @HansLundmark, drawing a picture clarifies. The $2$-d section is

The limits indicate $y$ varies from $x$ to $\sqrt{1-x^2}$. This means region of interest is between the curves $y=x$ and $y=+\sqrt{1-x^2}$. $y=+\sqrt{1-x^2}$ is just part of unit circle $x^2+y^2=1$ in first quadrant.

As $x$ varies from $0$ to $1/\sqrt{2}$, we obtain the shaded area in the diagram. From this we see, $0 \le \rho \le 1$ and $\pi/4 \le \varphi \le \pi/2$ and $z$ is independent.

Hence our integral is separable as

$$ \int_0^6 dz \int_{\pi/4}^{\pi/2} d \varphi  \int_0^{1} \rho e^{-\rho ^2}d\rho $$

which is easy to evaluate.

Title: How to set bounds in cylindrical coordinates?

Body:
I'm trying to evaluate the following integral in cylindrical coordinates

$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_x^{\sqrt{1-x^2}}e^{-x^2-y^2} \, dy \, dx \, dz$$

After attempting to set the bounds in cylindrical coordinates, I got
$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_{\rho \cos\varphi }^{\sqrt{1-\rho^2 \cos^2\varphi }}e^{-\rho ^2}d\varphi \rho \, d\rho \, dz$$
But I know this doesn't make sense. Can someone explain how to switch the bounds analytically? I don't understand how to transform the bounds.


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答: Here's an algebraic version of Alon's comment, basically your initial argument hopefully written more-intuitively: the $n$ roots are $1, \zeta, \zeta^2, \ldots, \zeta^{n-1}$ where $\zeta$ is a primitive $n$th root ($\zeta^n=1$).  Let $S$ be the sum of all of these: $S=1+\zeta+\zeta^2+\ldots = \Sigma_{i=0}^{n-1} \zeta^i$.  Then $\zeta S = \zeta+\zeta^2+\ldots+\zeta^n = \zeta+\zeta^2+\ldots+\zeta^{n-1}+1 = S$; rotating so that the zero'th root becomes the first, the first becomes the second, etc. - that is, multiplaying by $\zeta$ - can't change the sum.</p>


问:
How to find the first three positive values of $\theta=\arctan(3)$?
How would I find the first three positive values of that? All I know is that the equation can also be written as $\theta=\arctan(3)$</p>

答:
Note that the tangent function is postive in Quadrant I and III. Now every trigonometric have only one, unique value in each quadrant, which means only one angle in quadrant I satisfy your equation, also it's case in the third.

In the first quadrant the angle $\theta$ that satisfy your equtaion is: 71,56 degrees. This is the first one. Now we know that if we add 360 degrees we'll get the same tangent value for the angle so the general equation for the angles in Quad. I is

$$\theta = 360k + 71.56\text {; k=1,2,3...}$$

Know to get the value for Quad. III we add 180 to the value from the Quad. I and the general equation is:

$$\theta = 360k + 251.56\text {; k=1,2,3...}$$

And from these equation we get that angles that satisfy your equation can be written as:

$$\theta = 180k + 251.56\text {; k=1,2,3...}$$

$$\text{ or in radians}$$
$$\theta = k\pi + 1.25\text {; k=1,2,3...}$$

And if you are interested in generating trigonometric function I recommend you using the Taylor-Maclaurin Series.I use it and it's one of the easiest way to generate values for trigonometric function


问:
How can I prove that a function $f: \mathbb{R} \rightarrow \mathbb{N}$ is continuous only if $f$ is constant?
I have to find all functions of the type:

$$f: \mathbb{R} \rightarrow \mathbb{N}$$

that are continuous. The way I see it, such a function can only be continuous if it is a constant function like so

$$f(x) = n$$

for any $x \in \mathbb{R}$ and for some $n \in \mathbb{N}$. If this condition wouldn't be satisfied then the point where the value of the function changes to a different natural number (say that is changes from $n$ to another natural number $m$) would break continuity, because at that point we would have either a jump discontinuity or a removable discontinuity, so obviously the function wouldn't be continuous. If, on the other hand the function if constant, then it's natural to think that it is continuous (duh).

But how can I formalize my intuition? I was taught the definition of continuity in terms of neighborhoods. It goes something like this:

$$\forall V \in \mathcal{V}(f(c)), \exists U \in \mathcal{V}(c) \text{ such that } \forall x \in U \cap A \text{ we have } f(x) \in V$$

But I don't see how I could use this definition to prove my point (that is, if my intuition is correct!). So my question is this: How can I formalize my intuition and come up with a real, mathematical solution to this problem? I don't see how to manipulate the above definition in order to prove my point.

答: The key insight is to note, for any $n\in\Bbb N$, that $V=\{n\}$ really is a neighborhood of $n$ in $\Bbb N$.


问:
differential equation of first order - omitting absolute value operator
i have the equation:
$$y'+2y\:=1$$

and i solve it the regular way for first order differential equation:
$$y'\:=1-2y$$
$$\frac{dy}{dx}=1-2y$$
$$\int \:\frac{1}{1-2y}dy=\int \:1dx$$
$$-\frac{1}{2}\int \:-\frac{2}{1-2y}dy=\int \:1dx$$
and using the integral formula:

$$-\frac{1}{2}\ln\left(\left|1-2y\right|\right)=x+\ln\left(c\right)$$
Why Symbolab omits the absolute value operator and writes:

答:
Since
$$\int\frac{1}{1-2y}dy=\int 1dx, \text{ or }-\int\frac{1}{2y-1}dy=\int1dx $$
one has
$$ -\frac12\ln|2y-1|=x+C $$
or
$$ \ln|2y-1|=-2x-2C$$
So
$$ 2y-1=\pm e^{-2C}e^{-2x}$$
Let $k=\pm e^{-2C}$. Then
$$ y=\frac12+\frac12ke^{-2x}. $$
Since $k$ absorbs the signs, it does not matter if you have absolute value for $2y-1$ or not.


答: A continuous function preserves connectedness, $\mathbb{R}$ is connected and since $\lbrace n\rbrace$ is open in the discrete topology on $\mathbb{N}$, the image can't have more than a point.

Title: How can I prove that a function $f: \mathbb{R} \rightarrow \mathbb{N}$ is continuous only if $f$ is constant?

Body:
I have to find all functions of the type:

$$f: \mathbb{R} \rightarrow \mathbb{N}$$

that are continuous. The way I see it, such a function can only be continuous if it is a constant function like so

$$f(x) = n$$

for any $x \in \mathbb{R}$ and for some $n \in \mathbb{N}$. If this condition wouldn't be satisfied then the point where the value of the function changes to a different natural number (say that is changes from $n$ to another natural number $m$) would break continuity, because at that point we would have either a jump discontinuity or a removable discontinuity, so obviously the function wouldn't be continuous. If, on the other hand the function if constant, then it's natural to think that it is continuous (duh).

But how can I formalize my intuition? I was taught the definition of continuity in terms of neighborhoods. It goes something like this:

$$\forall V \in \mathcal{V}(f(c)), \exists U \in \mathcal{V}(c) \text{ such that } \forall x \in U \cap A \text{ we have } f(x) \in V$$

But I don't see how I could use this definition to prove my point (that is, if my intuition is correct!). So my question is this: How can I formalize my intuition and come up with a real, mathematical solution to this problem? I don't see how to manipulate the above definition in order to prove my point.


问:
Characteristic of a field to make projective points the same
In the resolution of a linear algebra exercise, they tell me that given a field $\Bbb K$ and the projective space $\Bbb P^n(\Bbb K)$, the projective points $(1:-1:2)$ and $(2:1:1)$ are the same if and only if $\text{char}(\Bbb K)=3$.

I understand the right to left implication, since if $\text{char}(\Bbb K)=3$ then $3=0$, hence $-1=-1+0=-1+3=2$ and $4=3+1=1$. Therefore, $(1:-1:2)=(1:2:2)=(2:4:4)=(2:1:1)$.

However, I have trouble understanding the left to right implication, I am not really sure of why any other value of the characteristic of the field would result in those two points being different.

Thanks in advance!

答:
For $P=(1:-1:2)$ to be the same as $Q=(2:1:1)$ we need the existence of some $\lambda\neq0$ such that $\lambda(1,-1,2)=(2,1,1)$. From the first component we get that $\lambda=2$, which means $P=Q$ if and only if $(2,-2,4)=(2,1,1)$ if and only if $-2=1\land 4=1$.

Both $-2=1$ and $4=1$ are true if and only if $3=0$ (add $2$ to the first equality and subtract $1$ to the first one), so we can conclude $P=Q$ if and only if $\text{char}(\Bbb K)=3$.

Another way to see this is thinking that you need $(1,-1,2)$ and $(2,1,1)$ to lie in the same vector line, since then they're projected onto the same projective point. This is equivalent to $(1,-1,2)$ and $(2,1,1)$ being linearly dependent,

which can be checked looking at the range of
$\begin{pmatrix}
         1 &      2  \\
          \llap{\text{-}}1&1\\
         2 & 1
    \end{pmatrix}$

Clearly, this matrix will have range greater or equal to 1, since it has entries with $1$ and the multiplicative identity of a field can't be zero (by convention). We need the range to be 1 in order to have the linear dependence, so we need to make sure the minors of order two are all 0:

$\left|\begin{matrix}
         \ 1 &      2  \\
          \ \llap{\text{-}}1&1
    \end{matrix}\right|=3,\quad\left|\begin{matrix}
         1 &      2  \\
         2 & 1
    \end{matrix}\right|=-3,\quad\left|\begin{matrix}
          \ \llap{\text{-}}1&1\\
         \ 2 & 1
    \end{matrix}\right|=-3$

and these are 0 if and only if $3=0$ if and only if $\text{char}(\Bbb K)=3$.


问:
Integrate and determine if cauchy integral theorem applies
$$\int_c \frac{dz}{z-3i}$$

$$f(z) = \frac{1}{z-3i}$$ where C the circle |z|=π counterclockwise.

So since f(z) is not analytic when z = 3i, we have to integrate another way since we cannot apply cauchy theorem of integrals.

$$\int_c \frac{1}{z - 3i}dz = \int_c (z - 3i)^{-1}$$

Where do I go from here?

答:
Using residue theorem, as $z=3i$ is internal to the circle $|z|=\pi$
$$\oint_{|z|=\pi}\frac{dz}{z-3i}=2\pi i\text{Res}\left[\frac{1}{z-3 i},\{z,3 i\}\right]=2\pi i$$
because
$$\text{Res}\left[\frac{1}{z-3 i},\{z,3 i\}\right]=\underset{z\to 3 i}{\text{lim}}\frac{z-3 i}{z-3 i}=1$$


答:
The simplest way is to rewrite $(1: -1: 2)$ as $(-1, 1, -2)$.

If we have $(-1, 1, -2) = \lambda(2, 1, 1)$, then by comparing the second coordinate, it's clear that $\lambda = 1$ and hence $-1 = 2$ and $-2 = 1$. This of course implies $3 = 0$.

Title: Characteristic of a field to make projective points the same

Body:
In the resolution of a linear algebra exercise, they tell me that given a field $\Bbb K$ and the projective space $\Bbb P^n(\Bbb K)$, the projective points $(1:-1:2)$ and $(2:1:1)$ are the same if and only if $\text{char}(\Bbb K)=3$.

I understand the right to left implication, since if $\text{char}(\Bbb K)=3$ then $3=0$, hence $-1=-1+0=-1+3=2$ and $4=3+1=1$. Therefore, $(1:-1:2)=(1:2:2)=(2:4:4)=(2:1:1)$.

However, I have trouble understanding the left to right implication, I am not really sure of why any other value of the characteristic of the field would result in those two points being different.

Thanks in advance!


问:
How to prove $\min(\delta a_k)\le\delta S\le \max(\delta a_k)$?
We have sum of exact numbers: $$S=a_1+a_2+a_3+\cdots+a_n$$
And sum of their corresponding approximate numbers:

$$S_0=a_{0_1}+a_{0_2}+a_{0_3}+\cdots +a_{0_n}$$
Our lecturer wrote the following inequality without any proof: $$\min(\delta a_k)\le\delta S\le \max(\delta a_k)$$

Where $\delta_{a_k}$ is relative error of $a_k$.

I am very curious to know why this inequality works.

I know $\delta S=\cfrac{\Delta a_1}{|S|}+\cfrac{\Delta a_2}{|S|}+\cfrac{\Delta a_3}{|S|}+\cdots+\cfrac{\Delta a_n}{|S|}\quad\quad,\quad \delta a_k=\cfrac{\Delta a_k}{|a_k|}$ If I'm not mistaking

Therefor :
$$\min(\cfrac{\Delta a_k}{|a_k|})\le\ \cfrac{\Delta a_1}{|S|}+\cfrac{\Delta a_2}{|S|}+\cfrac{\Delta a_3}{|S|}+\cdots+\cfrac{\Delta a_n}{|S|}\le \max(\cfrac{\Delta a_k}{|a_k|})$$
But no idea comes to my mind how to prove this inequality works.

答:
For simplicity, denote $m=\min \left(\frac{\Delta a_k}{|a_k|}\right)$ and $M=\max \left(\frac{\Delta a_k}{|a_k|}\right).$

Then for each $k$ $$m \cdot|a_k|\le \frac{\Delta a_k}{|a_k|}|a_k|\le M\cdot {|a_k|} \tag 1$$
Rewrite
$$\ \cfrac{\Delta a_1}{|S|}+\cfrac{\Delta a_2}{|S|}+\cfrac{\Delta a_3}{|S|}+\cdots+\cfrac{\Delta a_n}{|S|}=\color{blue}{\frac{1}{|S|}\left(\frac{\Delta a_1}{|a_1|}|a_1|+\dots +\frac{\Delta a_n}{|a_n|}|a_n|\right)}.$$

With the use of $(1)$ we get
$$\frac{1}{|S|}\left(m|a_1|+\dots + m|a_n|\right)\le \color{blue}{\frac{1}{|S|}\left(\frac{\Delta a_1}{|a_1|}|a_1|+\dots +\frac{\Delta a_n}{|a_n|}|a_n|\right)}\le \frac{1}{|S|}\left(M|a_1|+\dots + M|a_n|\right)$$
Left inequality (that you want to prove) is true as $|S|\le|a_1|+\dots + |a_n|,$ the right one holds provided $a_k$ have equal signs.


问:
Working with the Indegrees and Outdegrees of a Tournament
When going through an Indian math magazine, I found this cool problem, but I don't really get how to solve it:

Given a tournament $T$ on $m$ vertices with the vertex set $V(T)$, show that $n=2$ is the only solution where $n>1$ to the following equation: $$\sum_{v\in V(T)} \bigl((\operatorname{indeg} v)^n - (\operatorname{outdeg}v)^n\bigr) = 0.$$

I know how to show that $n=2$ works. For all $n \ge 3$, I tried using an induction argument but it didn't really lead me anywhere. Does anyone have any ideas for how to prove this?

答:
I think the problem is sloppily worded.

In a tournament on $2k+1$ vertices where each vertex has in-degree $k$ and out-degree $k$, the formula holds for any value of $n$, so the problem must mean that there is no value of $n>2$ such that the formula holds for every tournament.  This is not what it seems to say.  I read it as asserting that there is no value of $n>2$ such that the formula holds for any tournament, but we've just seen that this is false.

To prove what I think is the intended statement, it's enough to consider a tournament on $4$ vertices, one of which has out-degree $3$, and the others all with out-degree $1$.  If the formula holds, then $$3^n+3\cdot1= 3\cdot2^n+1\cdot0^n$$or
$$3^{n-1}+1=2^n$$ and it's easy to show that this is false for $n>2$.


问:
I need help to evaluate this integral with a constant $c$
Find the value of the constant $c$ for which the integral converges, and evaluate the integral:

$$\int_0^\infty \left(\frac{9x}{x^2+1}-\frac{9c}{2x+1}\right)dx$$

答:
The integrand can be simplified to
$$9\frac{2x^2+x-cx^2 -c}{(x^2+1)(2x+1)}.$$
It $c\ne 2$, then for large $x$ the absolute value of the numerator is dominated by the $9(|2-c|x^2)$ term. it follows that for large $x$, the integrand behaves like $\frac{9(|2-c|)}{2x}$, so the integral does not converge. </p>

<p>Thus for convergence we need $c=2$. The calculation of the integral for $c=2$ has been described by user17762. </p>

Body:
The integrand can be simplified to
$$9\frac{2x^2+x-cx^2 -c}{(x^2+1)(2x+1)}.$$
It $c\ne 2$, then for large $x$ the absolute value of the numerator is dominated by the $9(|2-c|x^2)$ term. it follows that for large $x$, the integrand behaves like $\frac{9(|2-c|)}{2x}$, so the integral does not converge.

Thus for convergence we need $c=2$. The calculation of the integral for $c=2$ has been described by user17762.


问:
How do I find $m$ for which $-2x^2 + (3m-1)x - 1=0$ would have no real roots?

Find $m$ for which $-2x^2 + (3m-1)x - 1=0$ would have no real roots

I'm not certain on how to complete this question. I'm aware that the discriminant has to be less than $0$ in order for there to only be complex roots, and I can solve the question (I think) up to $$3m-1 < 2\sqrt 2$$ but I'm not sure on how to continue.

答:
You obtained

$$ (3m-1)^2 < 8 $$
$$ \Leftrightarrow (3m-1) \in (-2\sqrt2, 2\sqrt2)$$
$$ \Leftrightarrow m \in \Big(\dfrac{1-2\sqrt2}{3}, \dfrac{1+2\sqrt2)}{3} \Big)$$


问:
Change of Basis Matrix and repeated composition.

Hi I was wondering if anyone could help me with part d) e) and f) of this linear algebra question.

So far I have found:

a) $T(x^2 - 2x + 1)= 3x^2 - 5x - 1$ using linear map definitions.

b) the associated matrix $A$ is:

$A=
    \begin{pmatrix}
    -1 & 0 & 0\\
    -2 & 2 & 1 \\
    2 & 0 & 1 \\
    \end{pmatrix}
$

c) used the fact that the dimension is equal to 3 and have proven they are linearly independent by showing that $C_o=C_1=C_2=0$.

And here is where i get stuck. Any help would be appreciated.

Body:
d)

$$\mathcal{C}=\left(
\begin{array}{ccc}
 0 & -1 & 1 \\
 0 & 1 & 0 \\
 1 & 1 & -1 \\
\end{array}
\right)$$
$$\mathcal{C}\cdot \mathcal{B}=(x^2-x,x,-x^2+x+1)$$
$$\mathcal{C}^{-1}=\left(
\begin{array}{ccc}
 1 & 0 & 1 \\
 0 & 1 & 0 \\
 1 & 1 & 0 \\
\end{array}
\right)$$
$$\mathcal{C}^{-1}\cdot (x^2-x,x,-x^2+x+1)=(1,x,x^2)$$


问:
Existence of proper self-map of Euclidean space whose image has non-zero co-dimension
I'm trying to show that, for every $d> 2$, there exists a proper continuous self-map $f$ on $\mathbb{R}^d$ such that $\partial f(\mathbb{R}^d)$ is a sub-manifold of $\mathbb{R}^d$ of positive co-dimension.  I feel that the construction should be obvious, but I haven't managed to demonstrate its existence...

Am I wrong in assuming that such a map exists?

What I have in mind:
I'm trying to build a map from $\mathbb{R}^d$ to $\mathbb{R}^d - \left\{(x_1,\dots,x_d):\, x_1=\dots = x_{d-1}=0,\, x_d \geq 0 \right\}$ but that's all I've got so far.

答:
Here is an example if you are talking about the topological boundary: define
\begin{align}
f : \mathbb{R}^n &\longrightarrow \mathbb{R}^n \\
\left(x^1,\ldots,x^n\right) & \longmapsto \left(|x^1|,x^2,\ldots,x^n\right)
\end{align}
This is a continuous map. One can easily show that it is proper. As a topological statement, $\partial f(\mathbb{R}^n)$ is the subspace $\left\{\left(0,x^2,\ldots,x^n\right) ~|~ \left(x^2,\ldots,x^n \right)\in \mathbb{R}^{n-1}\right\}$ and then is a submanifold of codimension $1$.

If you want $f$ to be injective, and smooth, you can define $f(x^1,x^2,\ldots,x^n) = \left(e^{x^1},x^2,\ldots,x^n\right)$, but $f$ would not be proper.

Notice that, if you require for $f$ to be injective, and for $\partial f(\mathbb{R}^n)$ as the boundary of the manifold $f(\mathbb{R}^n)$, then obviously, the boundary will be empty as $f$ will be an open map, and its image will be boundaryless as a manifold.


问:
Why does $(1)$ imply $(2)$?
Let $\mathcal{V}=\mathcal{V}(S,T)$ be the class of functions $f(t,\omega):[0,\infty)\times\Omega\to\mathbb{R}$ such that $(t,\omega)\to f(t,\omega)$ is $\mathcal{B}\times\mathcal{F}$-measurable (where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $[0,\infty)$), $f(t,\omega)$ is $\mathcal{F}_t$-adapted and $\mathbb{E}\bigg[\int_{S}^T f(t,\omega)^2 dt\bigg]<\infty$.

If $f\in\mathcal{V}$ one can show that it is possible to choose elementary functions $\phi_n\in\mathcal{V}$ such that:
$$\mathbb{E}\left[\int_S^T|f-\phi_n|^2 dt\right]\to0\tag{1}$$
I know that $(1)$ implies that:
$$\mathbb{E}\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]\to0\tag{2}$$

Could you please help me understand why $(1)$ implies $(2)$?

答:
\begin{align}
E\left[\int_S^T|\phi_n(t)-\phi_m(t)|^2 dt\right]&=E\left[\int_S^T|\phi_n(t)-f(t)+f(t)-\phi_m(t)|^2 dt\right] \\
&\leq E\left[\int_S^T 2|\phi_n(t)-f(t)|^2+2|f(t)-\phi_m(t)|^2 dt\right]  \\
&= 2E\left[\int_S^T |\phi_n(t)-f(t)|^2\right]+2E\left[\int_S^T|f(t)-\phi_m(t)|^2 dt\right] \rightarrow 0.
\end{align}
The '$\leq$' follows from the elementary inequality $(a+b)^2 \leq 2a^2 + 2b^2$ and the convergence to zero from (1).

Title: Why does $(1)$ imply $(2)$?


答: Take $c\in\Bbb R$. Then $\{f(c)\}$ is both a closed and an open subset of $\Bbb N$. So, $f^{-1}\bigl(\{f(c)\}\bigr)$ is both a closed and an open subset of $\Bbb R$. It is also non-empty, since it contains $c$. But, since $\Bbb R$ is connected, its only subset which is open, closed and non-empty is $\Bbb R$ itself. Therefore, for each $x\in\Bbb R$, $f(x)=f(c)$.


问:
Does $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ converge?
Let's consider the series: $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$.

I would suggest that it doesn't converge. One can see it as follows:

Let be $S_{2n}:=\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n}$ and $S_{2n+1}:=\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}$ two subsequences (subseries?) of $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$. Both converge due to alternating series test (Leibniz criterion). However, if I take a look at $|S_{2n}-S_{2n+1}|$, I notice:

$$|S_{2n}-S_{2n+1}|=\Big|\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}\Big|\\=\Big|2\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\frac{1}{2n+1}\Big|=2\sum\limits_{k=1}^{2n}\frac{(-1)^{k-1}}{k} +\frac{1}{2n+1}>1.
$$
Hence, $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ violates the Cauchy-criterion, or in other words I can always find two partial sums which are not arbitrarily close to each other.

Is this correct? Is there a faster or more elegant way two show this result?

答:
The sequence $$u_n =\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}$$

converges to $-\ln(2) < 0$. Therefore the sequence $$\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n=(-1)^n u_n$$

does not converge.


问:
If $z \in\mathbb { C } $ is an algebraic integer then $\Pi_z \in\mathbb {Z} [X] $.
A complex number $z $ is called an algebraic number if there is $P \in\mathbb {Q }[X]\setminus\{0\} $  such that $P (z) = 0$.

We say that $x\in\mathbb {C} $  is an algebraic integer  algebraic integer if there exists a monic polynomial $P\in\mathbb {Z} [X] $  unitary such that  $P (x) = 0$.

we fix an algebraic number $z $. The set
$$
 I (z) = \{P \in\mathbb {Q}[X]\ :P (z) = 0\}
$$
is an ideal of $\mathbb {Q}[X]$. There is therefore a unique monic polynomial $\Pi_z \in\mathbb {Q}[X]$, called minimal polynomial of $z$, such that
$$I (z) = \{\Pi_z Q :Q  \in\mathbb {Q}[X]\}.$$

We admit the following results:

(1) The set of algebraic integers is a subring of $\mathbb {C} $.

(2) If $x \in\mathbb {Q}$ is an algebraic integer, then $x\in\mathbb {Z}$.

Problem

Show that if $z\in\mathbb { C } $ is an algebraic integer then $\Pi_z \in\mathbb {Z} [X] $.

An idea please

答:
Fix a nonzero algebraic integer $k$.

Show that there is some monic $p\in \mathbf Z[x]$ which is irreducible (in $\mathbf Z[x]$) and such that $p(k)=0$ (to see this, notice that a proper factor of a monic polynomial over $\mathbf Z[x]$ is also monic, up to sign, of lower degree, and use induction with respect to the degree).

Then by Gauss, it follows that $p$ is irreducible in $\mathbf Q[x]$, so $p=\Pi_k$.


答:
There’s actually a simple proof not involving some nontrivial considerations on $\mathbb{Z}[X]$ given what you admit.

The roots of $\Pi_z$ are all algebraic integers (because $\Pi_z$ divides a monic polynomial with integral coefficients). So the coefficients of $\Pi_z$ (by the Vieta relations) are algebraic integers as well, and rational numbers. So they’re integers.

Title: If $z \in\mathbb { C } $ is an algebraic integer then $\Pi_z \in\mathbb {Z} [X] $.

Body:
A complex number $z $ is called an algebraic number if there is $P \in\mathbb {Q }[X]\setminus\{0\} $  such that $P (z) = 0$.

We say that $x\in\mathbb {C} $  is an algebraic integer  algebraic integer if there exists a monic polynomial $P\in\mathbb {Z} [X] $  unitary such that  $P (x) = 0$.

we fix an algebraic number $z $. The set
$$
 I (z) = \{P \in\mathbb {Q}[X]\ :P (z) = 0\}
$$
is an ideal of $\mathbb {Q}[X]$. There is therefore a unique monic polynomial $\Pi_z \in\mathbb {Q}[X]$, called minimal polynomial of $z$, such that
$$I (z) = \{\Pi_z Q :Q  \in\mathbb {Q}[X]\}.$$

We admit the following results:

(1) The set of algebraic integers is a subring of $\mathbb {C} $.

(2) If $x \in\mathbb {Q}$ is an algebraic integer, then $x\in\mathbb {Z}$.

Problem

Show that if $z\in\mathbb { C } $ is an algebraic integer then $\Pi_z \in\mathbb {Z} [X] $.

An idea please


问:
Alternate to Vieta's formula to find $a,b,c,d$ in this question.
Question :

Roots of the equation $$x^{4}+2 x^{3}-5x^{2}+7x+10=0$$ are $\alpha, \beta, \gamma, \delta$ and that of $x^{4}+a x^{3}+b x^{2}+c x+d=0$ be
$\alpha+\beta+\gamma, \alpha+\beta+\delta, \alpha+\gamma+\delta ; \beta+\gamma+\delta,$ then find the value of
$a+b-c-d$.

What I tried:

By applying Vieta's formula(In 2nd Equation):
$$\alpha+\beta+\gamma+\alpha+\beta+\delta+\alpha+\gamma+\delta+\beta+\gamma+\delta=-a$$
$$=3(\alpha+\beta+\gamma+\delta)=-a \tag1\label{eq1} $$
and By applying in Vieta's formula in 1st equation we get:
$$\alpha+\beta+\gamma+\delta=-2 \tag2\label{eq2}$$
From $\eqref{eq1}$ and $\eqref{eq2}$:

$$a=6$$
In similar way, I would find $b,c,d $ which is a very tedious and time consuming method(and it's not guaranteed that answer will come).

Is there any other way?

Hint

(Not complete answer)

Thanks

答:
Idea: Perform a change of roots.

More specifically, if $x_i$ are the roots to the original equation, we want $y_i = f(x_i)$ to be the roots to the new equation. What is $f(x)$?

 Hint:  Like you stated, $\alpha + \beta + \gamma + \delta = -2$

 $f(x) = -2 - x$

$$x^4 + ax^3 + bx^2 + cd + d = f(x) ^ 4 + 2f(x) ^3  - 5f(x)^2 + 7 f(x) + 10  .$$

Using the transformation, expanding the expression gives us

 $ x^4 + 6x^3 + 7x^2 - 19x - 24 $


问:
Find the length of rope when horse grazes double area

A field of grass  $12$ ft by $20$ ft has a horse tied to a corner of the field. The initial length of the rope was $10$ ft but when the horse needed twice as much area over which to graze, the rope was adjusted accordingly.

Now how can I calculate the length of the rope? Do I have to use the Pythagoras theorem to calculate the diagonal of 12 by 20 rectangle? Will  that be the length of the rope? Waiting for your answer :).

答:
Horse is tied to corner $B$. Initially the horse could reach circular sector BEF of radius $10$. Now it needs double the area so we have a circular sector ($BHG$) and a triangle ($\triangle BCG$).

Add area of both triangle and circular sector. It is double of $25\pi$ which was initial area.

Say new radius is $r$ and $\angle GBH = \theta,$ then $\angle CGB = \theta$.

$\sin \theta = \displaystyle \frac{BC}{BG} = \frac{12}{r}$

Total area (new) $ \displaystyle = \frac{\theta}{2}r^2 + \frac{1}{2} BC \times CG = \frac{\pi}{4} \times 10^2 \times 2$

$\displaystyle \theta \times r^2 + 12 \times \sqrt{r^2-144} = 100 \pi$

$\displaystyle \arcsin (\frac{12}{r}) \times r^2 + 12 \times \sqrt{r^2-144} = 100 \pi$

Of course this is not trivial to solve. One way is to first approximate using triangle $BGH$.

WolframAplha gives $r \approx 14.911$ ($15$ ft would surely work).


问:
What happens to the singular values if we take absolute values of all matrix entries?
$\newcommand{\norm}[1]{\left\lVert#1\right\rVert}$
Consider $|A|:=(|a_{ij}|)$. I have to prove or disprove that $\norm{|A|}_2=\norm{A}_2$, where $\norm{A}_2=\sqrt{\rho(A^*A)}$ and $\rho(M)$ denotes the spectral radius of a matrix $M$.

First I tried to find counterexpmples using any random matrix I could think of. Than I constructed some, where taking the abs of every entry causes maximal changes to the matrix. But nothing worked.

Then I tried to proof it. First I defined $A:=\pmatrix{a & b \\ c & d}$ and formally calculated all the other matrix, hoping something nice will show up. Then I tried to do some algebra, using defintions and norm properties. But I always stuck at $|A|^*|A|$ since I do not think I can give any statements about its eigenvalues depending from $A$.

Can anyone tell me if this claim is true or not and give me a little hint how to solve it?

答:
The claim is false. The simplest counterexample I believe is $A=\pmatrix{1&-1\\ 1&1}$. It is $\sqrt{2}$ times an orthogonal matrix. Hence $\|A\|_2=\sqrt{2}$. However, as $|A|=ee^T$ where $e^T=(1,1)$, we have $\big\||A|\big\|_2=\|e\|_2^2=2$.

In general, by triangle inequality, we have $\|Ax\|_2\le\big\||A||x|\big\|_2$. Hence it is always true that $\|A\|_2\le\big\||A|\big\|_2$. Equality holds if all entries on each column of $A$ happen to have the same phases, i.e. if $A=BD$ for some entrywise nonnegative matrix $B$ and some unitary diagonal matrix $D$, but this condition is not necessary. E.g. when $n\ge2$ and
$$
A=\pmatrix{1&0\\ 0&C}\in M_{n+1}(\mathbb C),
$$
we have $\|A\|_2=\big\||A|\big\|_2$ for any matrix $C\in M_n(\mathbb C)$ whose entries are smaller than $\frac1n$ in moduli.

However, when equality holds and $A$ has an entrywise nonzero singular vector $x$ corresponding to its largest singular value, the ties in the triangle inequality $\|Ax\|_2\le\big\||A||x|\big\|_2$ will force $A$ to take the form of $BD$.


问:
Interesting question about Borel-sigma algebra and Lebesgue measure.
Let $a , b\in \Bbb R$ , $a>0$

$T_{a,b}: \Bbb R\to \Bbb R$ defined as:

$T_{a,b}(x)=ax+b$.

Show that

$T_{a,b}(B_R)=B_R$

2.$\forall A\in{B_R}$ $m(T_{a,b}(A))=am(A)$

Hints:

Notice that $T_{a,b}$ is bijection and its inverse is $T_{\frac{1}{a},\frac{-b}{a}}$.
First prove that $T_{a,b}$ is a sigma-algebra that contains all the open finite intervals and then look at:

$s: B_R\to [0,\infty]$

$s(A)=am(T^{-1}_{a,b}(A))$
$m$ is Lebesgue measure.

I tries to write:

$T_{a,b}(B_R)={aI+b : I\in B_R}$.

We know that the borel sigma algebra is generated by the collection of open intervals, closed intervals..
It's obvious that $T_{a,b}(B_R)\subset B_R$ as $I\in T_{a,b}(B_R)$ and $I=T_{1,0}$.
Then I have to show that $T_{a,b}(B_R)$ is a sigma-algebra that contain all of the finite segments so by the fact that $B_R$ is the minimal sigma algebra that contains all the finite segments, we get that $B_R\subset T_{a,b}(R)$.
So, we must prove two axioms, T is closed under complements and countable unions.
I tried to use the hint mentioned, but it's not obvious for me how to do it.

Thanks for any help.

答:
For the first question, you can prove something more general( also more clean so see)

Lemma : For any function between two space $f: A \rightarrow B$, we have: $ f^{-1}( \sigma(S))= \sigma (f^{-1}(S))$ , where $S$ is any family of subsets of $B$.

The first question is the direct consequence of the above lemma by using $f(x)= \frac{x-b}{a}$

For the second question, we denote the righthand side measure as $\mu_2$, the lefthand side measure as $\mu_1$, they clearly equal on any elements in the semiring $ S=\{ (a,b] : a<b\}$, hence by the unicity of Caratheodory's extension theorem (+ the $\sigma$-finiteness), we get the equality.

Hints for lemma:
You can firstly prove the $LHS$ is a $\sigma$-algebra, then the inclusion $ RHS \subset LHS$ is clear.

For the other direction, you define $P:= \{ A | f^{-1}(A) \in \sigma( f^{-1}(S))\}$ , that is the familiy of set $A$ such that $f^{-1}(A) \in \sigma( f^{-1}(S))$. So what is left to do is proving that $ \sigma(S) \subset P$.  So do this, you prove that $S \subset P$ and $P$ is a sigma algebra.

P/s: I assume that you are familiar with monotone class argument.


问:
Differential Geometry - Computation Help
I'm trying to learn differential geometry through one of MIT's online courses (lecture notes found here:  http://ocw.mit.edu/courses/mathematics/18-950-differential-geometry-fall-2008/lecture-notes/ch1_revised.pdf) and am stuck with what should be an easy question.  The question asks to show the following:  Suppose $c(s)$ is a regular curve ($c'(s) \ne 0$) in the plane with $|c(s)| \le 1$ and suppose there is a point t with $|c(t)| = 1$.  Then $|\kappa(t)| \ge 1$.  Here, $\kappa$ is defined as

$\frac{det(c', c'')}{||c'||^3}.$

It's should be straightforward if we write $c$ as the graph of a function and rotate so that $c(t) = (0,1)$.  However, I feel as if there should be a more elegant solution using only what's found in the first day's notes (found in the link).  I just don't see it.

答:
Geometrically, you should understand that if the curve $c$ stays inside the unit circle and touches at a point, it must be at least as curved as the circle at that point.

Here's a hint: Use the fact that $t$ is a local maximum of the function $|c(s)|^2$, and put your calculus to work :)


问:
Splitting of primes in finite extensions of Noetherian normal rings
Let $A/B$ be a finite extension between noetherian normal rings. Let ${\frak p}$ be a prime of $B$ and ${\frak P}$ be the unique prime of $A$ lying over ${\frak p}$.

Let $e \colon= [k(A) \colon k(B)]$, where we denote by $k(A)$ (resp. $k(B)$) the fractional field of $A$ (resp. $B$) be the degree of the extension.

Q. Does the following equality hold as the ideal of $A$?$\colon$

\begin{equation*}
{\frak P}^e = {\frak p},
\end{equation*}

where ${\frak P}^e$ is defined to be the ideal generated by $g_{i_1} \cdots g_{i_e}$'s with each $g_{i_j}$ being a member of generators of ${\frak P}$.

答:
Note that by "$\mathfrak{P}$ the unique prime of A lying above $\mathfrak{p}$" you are actually choosing special ideals $\mathfrak{p}$, because it can happen that $\mathfrak{p}$ factors as a product of multiple distinct prime ideals lying above it.

The answer to your question is: No.

As an example consider integer ring $A$ of a number field as extensions of $\mathbb{Z}=:B$: $A$ is a Dedekind domain (Noetherian normal integral domain of Krull dimension $\leq 1$) as required.

We call a prime $\mathfrak{p}$ of $B$ completely ramified if it exhibits the property you describe above, i.e. $\mathfrak{p}A=\mathfrak{P}^e$. But there are other behaviours a prime of $B$ can exhibit: If we assume that there is just one prime $\mathfrak{P}$ lying above $\mathfrak{p}$, that could also be the ideal $\mathfrak{p}A$, in which case we would call $\mathfrak{p}$ inert.

For an example in the Gaussian integers, you can check out the Wikipedia article about the splitting of primes in Galois extensions


问:
Dynamical system and attractors
Let $X$ be a compact metrizable space and $T : X → X$ a continuous map. We think of $T$
as the transition map of a discrete-time dynamical system with state space $X$. Namely, starting from a state $x$,
the system evolves by first going to state $T(x)$, then to state $T^2(x)$, and so on and so forth.
A closed subset $U ⊆ X$ is said to be inward if $T(U) ⊆ int(U)$. Observe that if $U$ is inward, then $U ⊇ T(U) ⊇ T^2(U) ⊇ · · ·$ . A set $A ⊆ X$ is called an attractor if $A =\cap_{n=0}^∞T^n(U)$ for some inward $U ⊆ X$.

$(a)$ Consider the map $T : [0, 1] → [0, 1]$ defined by $T(x) := x^2$

Show that $A_1 :=$ {$0$} and $A_2 := [0, 1]$ are
attractors but $A_3 :=$ {$1$} is not.

$(b)$ Show that every attractor is non-empty.

$(c)$ Let $U$ be an inward set and $A$ the associated attractor.

Show that for every open set $B ⊇ A$, there exists
an $n ∈ \mathbb N$ such that $T^n(U) ⊆ B$

My attempts:

$(b)$ I'll prove this hy contradiction. Suppose that every attractor is empty, i.e. for $A ⊆ X$, we have $A =\cap_{n=0}^∞T^n(U) = \phi$ for some inward $U ⊆ X$. But then this means that $A = U \cap T(U) \cap T^2(U) \cap ... \cap T^n(U) = \phi$, and this means that the $T^n(U)$'s for $n \in \mathbb N$ don't have any elements in common. However, $U ⊇ T(U) ⊇ T^2(U) ⊇ · · ·$ so they indeed have elements in common. Therefore, every attractor is non-empty.

$(c)$ We are given an open set $B ⊇ A$, thus $A =\cap_{n=0}^∞T^n(U) ⊆ B$, and this means that $A = U \cap T(U) \cap T^2(U) \cap ... \cap T^n(U) ⊆ B$. Let $x \in U \cap T(U) \cap T^2(U) \cap ... \cap T^n(U) ⊆ B$, thus $x \in U$, and $x \in T(U)$, and $x \in T^2(U)$,..., and $x \in T^n(U)$, also $x \in B$. Thus, there exists $n \in \mathbb N$ such that $T^n(U) ⊆ B$

Are parts $(b)$ and $(c)$ correct?

$(a)$ I don't know how to solve it, any help please?

答:
For $a)$, see that
$$A_1 = \bigcap_{n=0}^{\infty} T^n\left(\left[0,\frac{1}{2}\right]\right)$$
$$A_2 = \bigcap_{n=0}^{\infty} T^n\left(\left[0,1\right]\right)$$

but $A_3=\lbrace 1 \rbrace$ cannot be written like this. Indeed, the only subset $U$ of $[0,1]$ such that $A_3 = \bigcap_{n=0}^{\infty} T^n(U)$ is $\lbrace 1 \rbrace$, but $\lbrace 1 \rbrace$ is not inward (since it has empty interior).

Title: Dynamical system and attractors


问:
Space $X$ so that $\pi_{1}(X)=\mathbb{Z}_{3}$
Given a group $G=\langle S|R \rangle$ we can construct a 2-dim complex $K$ s.t  $\pi_{1}(K)=G$. I get that the main idea is to construct 2-cells that "read" the relators and glue them to the wedge that represents the free group on $S$. With that in mind I can see for example that if $G=\mathbb{Z}\times\mathbb{Z}$ we glue the rectangle $aba^{-1}b^{-1}$ in the graph $R_{2}$ (rose graph) that represents the free group on $a,b$ and after that glueing we get  a torus. WIth the same procedure we can easily see that if $G=\mathbb{Z}_{2}$ we get the projective plane. But what about all the other cyclic groups? For example which space has fundamental group $\mathbb{Z}_{3}$? I am having trouble to see if this is a familiar space.

答:
Here's a way to get adequate spaces, but not using the procedure you describe (MathsIsFun described what this procedure looks like in the case of a cyclic group).

Suppose you have a (discrete) group $G$ acting very nicely (there are, of course, precise hypotheses) on a simply-connected space $X$.

Then covering theory tells us that the quotient $X/G$ has $\pi_1\cong G$.

The case $G=\mathbb Z, X= \mathbb R$ with action by translation gives $S^1$ with the correct $\pi_1$, and the case $G=\mathbb Z/2, X= S^n$ with the antipodal action gives $\mathbb RP^n$, with the correct $\pi_1$ as well.

For cyclic groups $G$, note that $G$ can be seen as a subgroup of the multiplicative group of nonzero complex numbers. Unfortunately, $\mathbb C^*$ isn't simply connected, but fortunately for us, it embeds as a group into a simply-connected group : the unit group of the quaternions $\mathbb H$. If we look at the elements of norm $1$, we thus get a group embedding $G\to S^3$, and thus, an action by translation produces a quotient $S^3/G$ with $\pi_1\cong G$

(you may be interested in looking up "lens spaces")

In particular, every finite subgroup of $S^3$ (with the "quaternionic" multiplication) can be realized as the fundamental group of a quotient of $S^3$.


问:
Finding Blowup of an ODE with maple
I've been interested in finite blowups of odes lately. I'm looking for a good way to calculate blowup times in maple. Right now all I do is to use dsolve, plot the ode, and then look where the solution appears to blowup. Unfortunately, this is rather cumbersome for my computer to handle for the odes I'm interested in. Just wondering if someone knew of another way to calculate the blowup times. I can use other math programs as well, I'm just most comfortable with maple. Thanks in advance for your help.</p>

答:
At least in Maple 16, dsolve tells me about singularity explicitly, without me having to look at the graph. For example:

```<code>IVP:=[diff(w(t),t$4)+w(t)^3=0, w(0)=0, D(w)(0)=0, (D@@2)(w)(0)=0, (D@@3)(w)(0)=1];
dsolve(IVP, w(t), numeric,  range=0..6);
</code>```

Result:

Warning, cannot evaluate the solution further right of 5.5472807, probably a singularity.

Or, for the equation you were originally interested in:

```<code>IVP:=[diff(w(t),t$4)+diff(w(t),t$2)+w(t)^3=0, w(0)=0, D(w)(0)=0, (D@@2)(w)(0)=0, (D@@3)(w)(0)=1];
dsolve(IVP, w(t), numeric,  range=0..7);
</code>```

Result:

Warning, cannot evaluate the solution further right of 6.2805222, probably a singularity

Hm, that looks suspiciously close to $2\pi$...

Title: Finding Blowup of an ODE with maple

Body:
At least in Maple 16, dsolve tells me about singularity explicitly, without me having to look at the graph. For example:

IVP:=[diff(w(t),t$4)+w(t)^3=0, w(0)=0, D(w)(0)=0, (D@@2)(w)(0)=0, (D@@3)(w)(0)=1];
dsolve(IVP, w(t), numeric,  range=0..6);

Result:

Warning, cannot evaluate the solution further right of 5.5472807, probably a singularity.

Or, for the equation you were originally interested in:

IVP:=[diff(w(t),t$4)+diff(w(t),t$2)+w(t)^3=0, w(0)=0, D(w)(0)=0, (D@@2)(w)(0)=0, (D@@3)(w)(0)=1];
dsolve(IVP, w(t), numeric,  range=0..7);

Result:

Warning, cannot evaluate the solution further right of 6.2805222, probably a singularity

Hm, that looks suspiciously close to $2\pi$...


问:
What are the pros and cons of natural deduction relative to Hilbert-style systems?
What are the pros and cons of natural deduction relative to Hilbert-style systems?

From Wikipedia, I get the impression that natural deduction proofs tend to be shorter and closer to how humans do it. If so, why do some proof-checkers, like Metamath, use Hilbert-style systems?

答:
It is hard to give a comprehensive answer, but the Metamath Proof Explorer(MPE) has some relevant things to say as to why they chose to go with a Hilbert-style system in terms of its strengths and lack of downsides. I will quote them at length so that this answer is more self-contained.

The appendix on Traditional Textbook Axioms of Predicate Calculus explains:

Both our system and the traditional system are called Hilbert-style systems. Two other approaches, called natural deduction and Gentzen-style systems, are closely related to each other and embed the deduction (meta)theorem into its axiom system.

The "Weak Deduction Theorem" section of the page on the "Weak Deduction Theorem for Classical Logic explains why they do not (directly) use the standard deduction theorem from natural deduction/Gentzen-style systems:

One of the goals of Metamath is to let you plainly see, with as few underlying concepts as possible, how mathematics can be derived directly from the axioms, and not indirectly according to some hidden rules buried inside a program or understood only by logicians. If we added the Standard Deduction Theorem to the language and proof verifier, that would greatly complicate both and largely defeat Metamath's goal of simplicity. In principle, we could show direct proofs by expanding out the proof steps generated by the algorithm of the Standard Deduction Theorem, but it is not feasible in practice because the number of proof steps quickly becomes huge, even astronomical. Since the algorithm is driven by proof of the deduction, we would have to go through that proof all over again—starting from axioms—in order to obtain the theorem form. In terms of proof length, there would be no savings over just proving the theorem directly instead of first proving the deduction form.

It is notable that the ease of constructing proofs in natural deduction can be recovered essentially completely in a Hilbert-style system like MPE's. Basically, just write the hypotheses and conclusion in "deduction form" where there is an extra "$P\to$" in front of everything (a trick that may have been first pointed out by Mario Carneiro). This allows you to use Hilbert-style translations of the rules of natural deduction, such as the ones listed in MPE's "natded" list of translations. For a lot more commentary on this approach, see MPE's page "Deduction Form and Natural Deduction". Most of that page might be helpful, but I want to highlight a bit from the section "Strengths of the current approach":

As far as we know there is nothing in the literature like either the weak deduction theorem or Mario Carneiro's natural deduction method (Mario Carneiro's method is presented in "Natural Deductions in the Metamath Proof Language" by Mario Carneiro, 2014). In order to transform a hypothesis into an antecedent, the literature's standard "Deduction Theorem" requires metalogic outside of the notions provided by the axiom system. We instead generally prefer to use Mario Carneiro's natural deduction method, then use the weak deduction theorem in cases where that is difficult to apply, and only then use the full standard deduction theorem as a last resort.


答:
In fact, if $T:X\to X$ is any homeomorphism, then $T$ carries Borel sets bijectively to Borel sets. Prove this in the usual way, letting $B=\{T(E):E\in \mathscr B(X)\}$ and showing that $B$ is a $\sigma$-algebra that contains the open sets in $X$:

If $T(E_i)\in B$ then $\bigcup T(E_i)=T\left(\bigcup E_i\right)\in B$ because $\bigcup E_i\in \mathscr B(X).$ A similar argument works for complements.

Now, if $U$ is open in $X$, then $T^{-1}(X)$ is open in $X$ and therefore $T^{-1}(U)\in \mathscr B(X).$ But then $U=T(T^{-1}(U))\in B$ by defnition.

So, $B$ is a $\sigma$-algebra on $X$ that contains the open sets. That is, $\mathscr B(X)\subseteq B$. Replacing $T$ by $T^{-1}$ gives the reverse inclusion, so $\mathscr B(X)=B.$

Title: Interesting question about Borel-sigma algebra and Lebesgue measure.

Body:
Let $a , b\in \Bbb R$ , $a>0$

$T_{a,b}: \Bbb R\to \Bbb R$ defined as:

$T_{a,b}(x)=ax+b$.

Show that

$T_{a,b}(B_R)=B_R$

2.$\forall A\in{B_R}$ $m(T_{a,b}(A))=am(A)$

Hints:

Notice that $T_{a,b}$ is bijection and its inverse is $T_{\frac{1}{a},\frac{-b}{a}}$.
First prove that $T_{a,b}$ is a sigma-algebra that contains all the open finite intervals and then look at:

$s: B_R\to [0,\infty]$

$s(A)=am(T^{-1}_{a,b}(A))$
$m$ is Lebesgue measure.

I tries to write:

$T_{a,b}(B_R)={aI+b : I\in B_R}$.

We know that the borel sigma algebra is generated by the collection of open intervals, closed intervals..
It's obvious that $T_{a,b}(B_R)\subset B_R$ as $I\in T_{a,b}(B_R)$ and $I=T_{1,0}$.
Then I have to show that $T_{a,b}(B_R)$ is a sigma-algebra that contain all of the finite segments so by the fact that $B_R$ is the minimal sigma algebra that contains all the finite segments, we get that $B_R\subset T_{a,b}(R)$.
So, we must prove two axioms, T is closed under complements and countable unions.
I tried to use the hint mentioned, but it's not obvious for me how to do it.

Thanks for any help.


答:
Your proof is correct. $|S_{2n}-S_{2n+1}| > 1$ for all $n$ implies that $(S_n)$ is not a Cauchy sequence and therefore not convergent.

You also showed that $(S_{2n})$ and $(S_{2n+1})$ are convergent, but this is not needed for the proof.

Title: Does $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ converge?

Body:
Let's consider the series: $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$.

I would suggest that it doesn't converge. One can see it as follows:

Let be $S_{2n}:=\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n}$ and $S_{2n+1}:=\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}$ two subsequences (subseries?) of $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$. Both converge due to alternating series test (Leibniz criterion). However, if I take a look at $|S_{2n}-S_{2n+1}|$, I notice:

$$|S_{2n}-S_{2n+1}|=\Big|\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\sum\limits_{k=1}^{2n+1}\frac{(-1)^k}{k}(-1)^{2n+1}\Big|\\=\Big|2\sum\limits_{k=1}^{2n}\frac{(-1)^k}{k}(-1)^{2n} -\frac{1}{2n+1}\Big|=2\sum\limits_{k=1}^{2n}\frac{(-1)^{k-1}}{k} +\frac{1}{2n+1}>1.
$$
Hence, $\sum\limits_{k=1}^{n}\frac{(-1)^k}{k}(-1)^n$ violates the Cauchy-criterion, or in other words I can always find two partial sums which are not arbitrarily close to each other.

Is this correct? Is there a faster or more elegant way two show this result?


问:
Details of Hecke Summation for weight 2 Eisenstein Series
In reference to https://math.stackexchange.com/a/51467/356443

I don't understand the details of the Hecke summation trick in order to define an Eisenstein Series of weight $2$.

For convergence issues, we can't just define
$$ E(z) = \sum_{c, d}'' \dfrac{1}{(cz+d)^2}. \qquad (z \in \mathbb{H}) $$
The trick is to consider
$$ E(z,s) = \sum_{c, d}'' \dfrac{1}{(cz+d)^2 |cz+d|^{2s}}, \qquad (z\in\mathbb{H}; \Re(s) > 0) $$
and try to find an analytic continuation for $s$.
This justifies setting $s=0$ to get a holomorphic function that should satisfy the right transformation rule.

It is noted that it is not always possible to continue $E(z,s)$ analytically.
In the cases that such an analytic continuation does exist, what is the strategy for the computation to construct such a continuation?

答:
So far as I can tell, the original "Hecke summation", meromorphic continuation in an auxiliary variable "$s$" and then evaluation at a special point (often normalized to $s=0$) occurs in

E. Hecke, Theorie der Eisensteinschen Reihen
h"oheren Stufe und ihre Anwendung auf Funktiontheorie und
Arithmetik, Abh. Math. Sem. Hamburg 5 (1927), 199-224;
collected works, 461-486.

Papers of G. Shimura that use this and related ideas are

On some arithmetic properties of modular forms of one and several variables, Ann. Math. 102 (1975) no. 3, 491-515.

The special values of the zeta functions associated with cusp forms, Comm. Pure Appl. Math. 29 (1976) no. 6, 783-804.

Looking superficially at those two papers, Shimura does not redo Hecke's determination that for weight 2 a non-trivial condition must be met so that the meromorphically-continued-and-evaluated-at-$s=0$ thing is holomorphic in the upper-half-plane coordinate. This fails for weight $k=2$ (over $\mathbb Q$) as we know. Hecke computes the Fourier expansion fairly directly to see this: the constant term has a leftover not-holomorphic summand.

I first saw this in lectures of Shimura about such things in 1975-6, where he did prove such things, using Fourier expansions for a proof of meromorphic continuation, and also to see what happens at the special point. I don't know whether he later put this into his books. Similar, but more complicated, computations make sense to attempt for higher-rank groups admitting holomorphic principal series.

My old book on Hilbert modular forms reproduces essentially Hecke's argument, with supplements from Shimura's lectures referring to Hilbert modular forms as well.

The germ of a higher-level explanation for the phenomenon is already hinted-at in some remarks of Shimura's in those papers, concerning the differential operators often called "Maass-Shimura" operators. Namely, the holomorphic discrete series repns of weight $2k$ occur as subreps of principal series with natural parameter $s=k$. Shimura's differential operators are "raising" operators inside the family of principal series, and increase the "weight" by $2$. There are also "lowering" operators acting, and the "holomorphic" vector in the weight-$2k$ discrete series is annihilated by the lowering operator. Shimura does not speak in these terms.

In particular (because formation of Eisenstein series is an intertwining operator) the lowering operator maps $E_{2k,s}$ to $(s-k)\cdot E_{2k-2,s}$. (The normalization here is more natural, but less elementary: the "holomorphic point" is at $s=k$, not $s=0$.) For $2k>2$ the Eisenstein series $E_{2k-2,s}$ is holomorphic at the special point, so in that case the lowering operator annihilates $E_{2k,k}$. As it happens (though there are reasons...), the lowering operator is essentially the Cauchy-Riemann operator, so this gives holomorphy.

For $2k=2$, the lowering operator maps $E_{2,s}$ to $(s-1)E_{0,s}$, and the special point is $s=1$. In this case, the weight-zero Eisenstein series has a pole at $s=1$. This is cancelled by the factor $s-1$, picking up the residue, which is a constant. Thus, the lowering operator does not quite annihilate $E_{2,1}$.

Analogous, more complicated, things happen on higher-rank groups admitting holomorphic discrete series.


问:
Complete Directed Graph Indegree and Outdegree summations
Let the indegree of a vertex $v$ be $i(v)$ and the outdegree be $o(v)$. Consider a single tournament (a directed graph obtained by assigning a direction for each edge in an undirected complete graph) occurring. For $a\in\mathbb{Z}^+$, $$s_a=\sum_v (i(v))^a$$ and $$t_a=\sum_v (o(v))^a.$$

For which $x$ does $s_x=t_x$ hold, regardless of the tournament?

Does $s_3\ge t_3$ imply $s_4\ge t_4$ ?

Does $s_4\ge t_4$ imply $s_5\ge t_5$ ?

Progress: So clearly $s_1 = t_1$ because of # of edges being the same, but I also noticed $s_2 = t_2$ for the cases I drew out and I'm not sure how to prove this yet.

答:
Please do not answer this question. This problem is pretty much copy-pasted from this year's MIT PRIMES Entrance Test: https://math.mit.edu/research/highschool/primes/materials/2021/entpro21.pdf

Check Problem M3. It is literally the same.

In the MIT PRIMES rules, it explicitly says "Note that posting these problems on problem-solving websites before the application deadline is strictly forbidden! Applicants who do so will be disqualified, and their parents and recommenders will be notified."

Solutions are due on December 1 - this is a clear violation of the rules.


问:
What does mean for all $t\in [0,1]$, $Z_t^n\to Z_t$ a.s.?
What does mean : for all $t\in [0,1]$, $Z_t^n\to Z_t$ a.s. ?

Does it mean that

1) $$\mathbb P\left(\forall t\in [0,1], \lim_{n\to \infty }Z_t^n=Z_t\right)=1$$

2) or $$\forall t\in [0,1], \ \ \mathbb P\left(\lim_{n\to \infty }Z_t^n=Z_t\right)=1.$$

Obviously, if 1) hold, then so does 2). But the converse is wrong. But I think that 2) should be written as $Z_t^n\to Z_t$ a.s. for all $t\in [0,1]$. But after all, "for all $t\in [0,1]$, $Z_t^n\to Z_t$ a.s." could also mean that if you fix $t\in [0,1]$ then $Z_t^n\to Z_t$ a.s. (i.e. 2) holds).

I'm a bit confuse, because I often see that in some article/book, and I never know which one it is. Is there any rule ?

答:
I don't think the countability matters a lot here. Any set can be defined as measurable.

Anyway, back to your question, well, as you pointed out, two conclusions are different and your statement can be any one of those. So you should look up at your textbook again to find out what it is.

Either way, as you may know, if you provide some uniform-continuity (independent of n) for the sequence of processes $Z^n_t$, the two notions are equivalent.


问:
How can I prove that a function $f: \mathbb{R} \rightarrow \mathbb{N}$ is continuous only if $f$ is constant?
I have to find all functions of the type:

$$f: \mathbb{R} \rightarrow \mathbb{N}$$

that are continuous. The way I see it, such a function can only be continuous if it is a constant function like so

$$f(x) = n$$

for any $x \in \mathbb{R}$ and for some $n \in \mathbb{N}$. If this condition wouldn't be satisfied then the point where the value of the function changes to a different natural number (say that is changes from $n$ to another natural number $m$) would break continuity, because at that point we would have either a jump discontinuity or a removable discontinuity, so obviously the function wouldn't be continuous. If, on the other hand the function if constant, then it's natural to think that it is continuous (duh).

But how can I formalize my intuition? I was taught the definition of continuity in terms of neighborhoods. It goes something like this:

$$\forall V \in \mathcal{V}(f(c)), \exists U \in \mathcal{V}(c) \text{ such that } \forall x \in U \cap A \text{ we have } f(x) \in V$$

But I don't see how I could use this definition to prove my point (that is, if my intuition is correct!). So my question is this: How can I formalize my intuition and come up with a real, mathematical solution to this problem? I don't see how to manipulate the above definition in order to prove my point.

答: The image under a continuous real map $f$ of $\mathbb R$ is an interval. As the only intervals of $\mathbb N$ are singletons, $f$ has to be constant.


问:
Conditional expectation (uniform distribution)
We have $X_1, X_2, \ldots$ - independent random variables with uniform distribution on interval $[0,1]$. $N$ - random variable independent of $X_1,X_2,\ldots$ with Poisson distribution with $\lambda=2$. I have to calculate $\operatorname{cov}({V_N, N})$ where $V_N=\frac{X_1}{S_N}$ where $S_N=X_1+\cdots+X_{N+1}$.
$$\operatorname{cov}(V_N,N)=E(V_NN)-E(V_N)E(N)$$

and now I have problem with calculation of $E(V_NN)$ and $E(V_N)$

I write $$E(V_N)=E(E(V_N\mid N=n))$$
but then I dont know how to calculate $E(V_N\mid N=n)=E\left(\frac{X_1}{X_1+\cdots+X_{n+1}}\right)$

答:
There are surely typos in your post as you have the same notation for both uniform rvs and Poisson rvs. Either way, the last expectation is not that hard.

You see, $X_1,...,X_n$ are iids, hence .
$ \mathbb{E}\left( \frac{X_1}{X_1+...+X_n} \right) = \mathbb{E}\left( \frac{X_2}{X_1+...+X_n} \right)=etc$

As their sum is clearly equation to 1, the desired expectation is equal to $\frac{1}{n+1}$


答:
Let $\angle HBG = x = \angle BGC$. $BG = 12 \csc x$

We have

$$ 2\cdot 25\pi = \dfrac{x}{2\pi}\cdot \pi (12\csc x)^2 + \dfrac{1}{2}\cdot 12 \cdot 12\cot x$$

$$ \Rightarrow x(\csc x)^2 + \cot x = \dfrac{25}{36}\pi$$

WA gives acute $x \approx 0.9353$

Length of rope found by WA is $$ 12\csc x \approx 14.91 \, \text{ft}$$

Title: Find the length of rope when horse grazes double area

Body:
A field of grass  $12$ ft by $20$ ft has a horse tied to a corner of the field. The initial length of the rope was $10$ ft but when the horse needed twice as much area over which to graze, the rope was adjusted accordingly.

Now how can I calculate the length of the rope? Do I have to use the Pythagoras theorem to calculate the diagonal of 12 by 20 rectangle? Will  that be the length of the rope? Waiting for your answer :).