问:
Lagrange Multiplier In Proving AM-GM inequality
I am dealing with a question in Stewart Calculus Book which states the following:

Using Lagrange multiplier, find the maximum of the function $$f(x_1,x_2,...,x_n)=(x_1x_2...x_n)^{1/n}$$subject to the constraint$$\sum_{i=1}^{n} x_i=c$$,where $x_i$'s are all positive, and use this to prove the AM-GM inequality. Actually in AM-GM inequality, it holds for all nonnegative numbers, but I think the author wants to get rid of the case where $x_i$'s are all zero, where Lagrange multiplier can no longer be applied.

And the result I obtained is: $f$ will have its maximum/minimum (the arithmetic mean), or even saddle point at $(\frac{c}{n},\frac{c}{n},...,\frac{c}{n})$, since Lagrange method gives no definiteness on the behaviour of the critical points found (whether max, min, or saddle).

And my question is: In this case, how do we show that the point I found gives the maximum value of the function? My teacher gave me an argument that said "the minimum value would be some place very near to $0$, so the arithmetic mean will not be the minimum value". But I think the argument is not persuasive enough to explain why the arithmetic mean is the maximum value. So, I hope that there will someone who are willing to give their opinions on this. Thanks.

答:
Let $A$ be the set of $(x_1, x_2, ..., x_n)$ such that
$$\sum x_i = c$$
and $x_i \geq 0$ for all $i$. On a point $x$ of the boundary of $A$, there is always a component $x_i$ of $x$ which is equal to zero, hence $f = 0$ on the boundary. As $f$ is continuous and $A$ is compact, $f$ has a global maximum on $A$. Denote by $x^\ast$ one such maximum. The fact that $f = 0$ on the boundary of $A$ implies that $x^\ast$ is in the interior of $A$. Then what you have done with Lagrange multipliers shows that
$$x^\ast = (\frac{x_1}{c}, ..., \frac{x_n}{c})$$
as it is the only extremum of $f$ in the interior of $A$.

EDIT : Let me clarify my notations. Write
$$\mathbb{R}^n_+ = \{ x = (x_1, ..., x_n), \ x_i \geq 0 \text{ for } i = 1, 2, ... n \}$$
$${\mathbb{R}^n_+}^\ast = \{ x = (x_1, ..., x_n), \ x_i > 0 \text{ for } i = 1, 2, ... n \}$$
and
$$h(x) = \sum x_i - c.$$
Then $f$ has a maximum $x^\ast$ on the compact
$$\{x \in \mathbb{R}^n_+, \ h(x) = 0 \}$$
and in fact
$$ x^\ast \in \{x \in {\mathbb{R}^n_+}^\ast, \ h(x) = 0 \}$$
One can conclude using Lagange multipliers in this last set.


问:
proof limits with $\epsilon$ and $\delta$

proof that $\lim\limits_{x \to 1} x \neq 2$ with $\epsilon$ and $\delta$ (with definition)

let $\lim\limits_{x \to c}f(x)=L$ so we have for every $\epsilon$ there is a $\delta$ such that $0<|x-c|<\delta => |f(x)-L|< \epsilon$.

Can we imply that $0<|x-c|<\delta_1 => |f(x)-L|< \epsilon$ which $\delta_1 < \delta$. (why?)

Can we show that $0<|x-c|<\delta => |f(x)-L|< {\epsilon\over a}$. (what can we do with $\epsilon$ ? Can we write $\sqrt{\epsilon}$? why?)

答:
If $\epsilon=\dfrac12$,

$$\exists\,\delta:\forall\,0<|x-1|<\delta\implies |x-2|<\dfrac12$$

is false.

ContentLicense: CC BY-SA 4.0


问:
How to show that if $p(A) = 0 \implies p(\lambda_0)=0$?
Let $V$ be a finite dimensional vector space and $V \ne \{ 0 \}, A\in L(V), \lambda_0 \in \sigma(A)$. If $p(\lambda)$ is an arbitrary polynomial for which the following applies: $p(A) = 0  $,  prove that $p(\lambda_0)=0$

$L(V)$ = The set of all linear mappings (linear operators) from V to V

$\lambda_0$ = eigenvalue

$\sigma(A)$ = spectrum of A

If I am correct this would involve the Hamilton-Cayley theorem but I don't know how to apply it. I know that $\lambda_0$ is a zero of the polynomial $det(A-\lambda I)$ but I don't even know where to start.

Edit (hint from mrf):

Since $V$ is finite, we can mark the degree of the polynomial with n $\implies p(\lambda) = \alpha_n\lambda^n + \alpha_{n-1}\lambda^{n-1} + ...  +  \alpha_2\lambda^2  + \alpha_1\lambda + \alpha_0$

So $p(A) = \alpha_nA^n + \alpha_{n-1}A^{n-1} + ... + \alpha_2A^2 + \alpha_1A + \alpha_0I$

Let $v$ be an corresponding eigenvector to the eigenvalue $\lambda_0 \implies Av = \lambda_0v$

$p(A)v = (\alpha_nA^n + \alpha_{n-1}A^{n-1} + ... + \alpha_2A^2 + \alpha_1A + \alpha_0I)v $

$p(A)v = \alpha_nA^nv + \alpha_{n-1}A^{n-1}v + ... + \alpha_2A^2v + \alpha_1Av + \alpha_0Iv$

If $\lambda \in \sigma(A) \implies \lambda^k \in \sigma(A^k)$

Proof:

$\lambda \in \sigma(A) \implies \exists v \in V$ so that $Av = \lambda v$

$Av = \lambda v \hspace{10px}| A$

$A^2v = A(Av) = A(\lambda v) =$ A is linear $ =\lambda(Av) = \lambda^2v$

$A^2v = \lambda^2v \hspace{10px}| A$

$A^3v = A(A^2v) = A(\lambda^2 v) =$ A is linear $ =\lambda^2(Av) = \lambda^3v$

And by induction we get: $A^kv = \lambda^k v$

$\implies$

$p(A)v = \alpha_n\lambda_0^nv + \alpha_{n-1}\lambda_0^{n-1}v + ... + \alpha_2\lambda_0^2v + \alpha_1\lambda_0v + \alpha_0v \hspace{10px}| :v$

$p(A) = \alpha_n\lambda_0^n + \alpha_{n-1}\lambda_0^{n-1} + ... + \alpha_2\lambda_0^2 + \alpha_1\lambda_0 + \alpha_0$

$0 = \alpha_n\lambda_0^n + \alpha_{n-1}\lambda_0^{n-1} + ... + \alpha_2\lambda_0^2 + \alpha_1\lambda_0 + \alpha_0$

答: Hint: Let $v$ be a corresponding eigenvector. Compute $p(A)v$.</p>

ContentLicense: CC BY-SA 3.0


问:
Calculation of the integral $\int_0^\infty d x\, x^{2n+1} \tanh(\pi x)\, \log (1 - e^{-2\pi x})$ for non-negative integer $n$
In some physics problem I ran into the integral with a non-negative integer $n$ of the form:
\begin{align}
I(n) \equiv \int_0^\infty d x\, x^{2n+1} \tanh(\pi x)\, \log (1 - e^{-2\pi x}) \ ,
\end{align}
which I cannot perform even with Mathematica for general $n$, but for small $n$ the values are given analytically (e.g. in Eq. (3.10) and (3.23) of https://arxiv.org/pdf/1708.00305.pdf):
\begin{align}
\begin{aligned}
I(0) &= - \frac{1}{8}\,\log 2 + \frac{9}{16\pi^2}\,\zeta(3) \ ,\\
I(1) &= - \frac{1}{64}\,\log 2 - \frac{3}{32\pi^2}\,\zeta(3) + \frac{225}{128\pi^4}\,\zeta(5) \ ,
\end{aligned}
\end{align}

It is likely that the integral generally takes the form
\begin{align}
I(n) = c_0\,\log 2 + \sum_{i=1}^{n+1}\,c_i\,\zeta(2i+1)\ .
\end{align}
Can we fix the values of the coefficients $c_i~(i=0,1,\cdots,n+1)$ for integer $n\ge 0$?

Thank you for your help in advance.

答:
Yes, every $I(n)$ can be expressed in $\zeta$.

Note that
$$I(n) = \frac{-1}{(2\pi)^{2n+2}}\int_0^1 \left(\frac{1}{u}-\frac{2}{u+1}\right)\log^{2n+1} u \log (1-u) du$$

Such kind of integral is well-known on this site, for example, see links under this question. We have
$$\int_0^1 \frac{\log^{2n+1} u \log (1-u)}{u} du = (2n+1)! \zeta(2n+3)$$
and in term of multiple-zeta function $$\int_0^1 \frac{\log^{2n+1} u \log (1-u)}{1+u} du = -(2n+1)! \zeta(\overline{2n+2},\overline{-1})$$
The follow formula exists:
$$\zeta(\overline{2n+2},\overline{1}) = \sum _{k=0}^{n+1} \zeta (2 k) \bar{\zeta}(-2 k+2 n+3)-\frac{\zeta (2 n+3)}{2}+\bar{\zeta}(1) \bar{\zeta}(2 n+2)-n \bar{\zeta}(2 n+3)-\bar{\zeta}(2 n+3)$$
where $\bar{\zeta}(s) = \left(1-2^{1-s}\right) \zeta (s)$


问:
Using $\int_{0}^{1} x^n dx =\frac{1}{n+1}$ find the sums of the series

$$1-\frac{1}{4}+\cdots +\frac{(-1)^{n+1}}{3n-2}+\cdots$$

Sorry for being silly. I thought this
$\frac{1}{3n-2}=\frac{1}{n+1}$
so  if $n=3/2$.

Then
$$\sum_{n=1}^{\infty }  (-1)^{n+1} \int_{0}^{1} \frac{1}{2}x^{\frac{3}{2}n} dx=\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{3n-2}$$

But, I don't know if they are asking me this and I don't know how to fix it.

答:
Transform
$$\frac{1}{3n-2} = \int_0^1 x^{3n-3} \mathrm d x$$
and
$$(-1)^{n+1}=(-1)^{3n-3}$$
Thus
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{3n-2} =
 \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 x^{3n-3} \mathrm d x =
\int_0^1 \sum_{n=1}^{\infty} (-1)^{n+1} x^{3n-3} \mathrm d x =$$
$$=\int_0^1 \sum_{n=1}^{\infty} (-x)^{3n-3} \mathrm d x = \int_0^1 \frac{1}{1+x^3}\mathrm d x$$
note that we can swap sum and integral because the geometric series is uniformly convergent on compact sets inside $(-1; 1)$.

Now, you have to compute this integral, which is not that easy.
WA says it evaluates $$\frac{\pi}{3 \sqrt 3} + \frac{\log 2}{3} \approx 0.835$$


问:
Questions using Iterated Integrals.
so Ive just been introduced to the idea of iterated integrals and im finding it hard to work out how to complete questions on this subject and was wondering if anyone could help.

So if I have an intial question, say:

$\int_{0}^{1}\int_{0}^{y^5}(\pi sin(5\pi x)/(1-x^{1/5}) dxdy$

I know that I have to intially look at the inequalities of the limits which I think is:

$0\leq y\leq1$ and $0\leq x\leq y^5$

And I believe that I have to draw the region defined by the inequalities but I'm not entirely sure how to do that.

From there I think I have to described the drawn region by the new inequalities but with the variables in the opposite direction. And once I have defined them, put the new inequalities back onto the integral and do the process of double integration.

So hence I think I know what I am doing with the beginning and the latter but I am stuck on the process in the middle.

Any help would be greately appreciated.

答:
If you change the order of integrations you get

$$\int_{0}^{1}\int_{0}^{y^5}(\pi \sin(5\pi x)/(1-x^{1/5}) dxdy=\int_{0}^{1}\int_{x^{1/5}}^{1}(\pi \sin(5\pi x)/(1-x^{1/5}) dydx.$$

This gives the integral

$$\int_{0}^{1}(\pi \sin(5\pi x) dx.$$

Title: Questions using Iterated Integrals.


问:
Set that is not well-ordered
I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.

Any hint?

答:
(I argue on the basis of $\sf{ZF}$).

Let $X$ be a set such that every nonempty subset has an $\in$-maximal element (or more precisely: greatest element).
Assume that $\in$ is not a wellorder on $X$.
As $a,b\in X$ implies that $\{a,b\}$ has an $\in$-maximal element, exactly one of $a\in b$, $a=b$, $a\ni b$ holds (the Axiom of Foundations forbids that two of these hold concurrently).
For $a,b,c\in X$ with $a\in b$ and $b\in c$, it is again the Axiom of Foundation that forbids $a=c$ and $c\in a$, hence $a\in c$. We conclude that $\in$ is a linear order on $X$.
Then the inverse relation $\ni$ is a well-order, hence there is an ordinal $\alpha$ and a bijection $f\colon \alpha\to X$ that inverts the order, i.e. $a\in b$ implies  $f(b)\in f(a)$.
Then a nonempty subset of $X$ without minimal element corresponds to a nonempty subset of $\alpha$ without maximal element.
For such a subset to exist, we need $\alpha\ge\omega$.
The Axiom of Foundation demands that $S:=\{f(n)\mid n\in\omega\}$ has an element $f(n)$ with $f(n)\cap S=\emptyset$. As $f(n+1)\in f(n)\cap S$, we get a contradiction.

No such $X$ can exist (and we didn't even use transitivity).

Title: Set that is not well-ordered


答:
For (1): if you take $\epsilon = 1/2$

$$0<|x-1|<\delta_0 => |f(x)-2|< \frac{1}{2}$$

does this $\delta_0$ exist? no becouse we could still take $x=1+\frac{1}{n}$ for some $n$ such that $1/n < \delta_0$, so we have that $|f(1+\frac{1}{n})-2|=|1+\frac{1}{n}-2|=|\frac{1}{n}-1|\ge \frac{1}{2}$ for n large enough.

For (2): if we have that
$$\forall \epsilon >0 \mbox{, }\exists \delta>0 \mbox{ s.t. }0< |x-c|<\delta => |f(x)-L|< \epsilon$$
then taking $0<\delta_1<\delta$, $|x-c|<\delta_1< \delta$. so the definition written above holds for all $x$ such that $|x-c|<\delta_1$.

Finally, for (3): since by definition it holds for all $\epsilon>0$, you can take $\epsilon$ to be whaterer you want.

so if we take $\epsilon_1=\frac{\epsilon}{a}$ for some positive $a$, $\epsilon_1>0$, so the definition holds:

$$0< |x-c|<\delta_1 => |f(x)-L|< \epsilon_1=\frac{\epsilon}{a}$$
taking for example $\delta_1=\frac{\delta}{a}$.

Also, if we select $\epsilon_2=\sqrt{\epsilon}$ for some $\epsilon>0$, $\epsilon_2>0$ again, so the definition still holds:

$$0 < |x-c|<\delta_2 => |f(x)-L|< \epsilon_2=\epsilon_2=\sqrt{\epsilon}$$
taking again the right $\delta_2$.

Title: proof limits with $\epsilon$ and $\delta$

Body:
proof that $\lim\limits_{x \to 1} x \neq 2$ with $\epsilon$ and $\delta$ (with definition)

let $\lim\limits_{x \to c}f(x)=L$ so we have for every $\epsilon$ there is a $\delta$ such that $0<|x-c|<\delta => |f(x)-L|< \epsilon$.

Can we imply that $0<|x-c|<\delta_1 => |f(x)-L|< \epsilon$ which $\delta_1 < \delta$. (why?)

Can we show that $0<|x-c|<\delta => |f(x)-L|< {\epsilon\over a}$. (what can we do with $\epsilon$ ? Can we write $\sqrt{\epsilon}$? why?)


问:
Example for a sequence of operators converging pointwise, but not with respect to the operator norm
I am trying to understand the following example.

Define $$T_n: l^2 \rightarrow l^2$$
$$T_n(x)=(0, ..., 0, x_{n+1}, ...).$$

It's rather clear that $T_n(x)$ converges for $0$ for every $x \in l^2$. However, the script says that $(T_n)_{n \in \mathbb{N}}$ does not converge with respect to the operator norm.

The last part is difficult to understand for me. So I thought maybe my understanding of the first statement is wrong?

If we arbitrarily choose $x \in l^2$, we have $\sum_{i=0}^\infty |x_i|^2 < \infty$. We also have that $\left|| T_n(x) \right||= \left( \sum_{i=n+1}^\infty |x_i|^2 \right)^{1/2}$ is decreasing for increasing $n$. If look at the subsequence of all $x_i \neq 0$, it becomes strictly decreasing and by assumption bounded. So $\left|| T_n(x) \right||$ should converge to $0$.

Is that reasoning flawed?

For the operator norm, I have to look at the limit of $$\left|| T_n \right|| = \sup \left\{ \left|| T_n(x) \right|| : x=1 \right\}.$$ Intuitively, it makes sense that the limit is not $0$, since infinitely many elements of my sequence always remain and thus the supremum over all $x \in l^2$ shouldn't be $0$.

Yet, I have some trouble with the coincidence of limit and supremum here. E.g., how could I calculate the supremum for a fixed $n \in \mathbb{N}$? Does the sequence $\left|| T_n \right||$ converge to a value that is not zero or not at all?

Clearly, I wouldn't know how to properly examine this problem. Occurrence of suprema in general has always provided difficulties for me.

答:
First of all, if $x = (a_k) \in \ell^2$ is the sequence with $a_{n+1} = 1$ and $a_k = 0$ otherwise, we have that
$$\|T_n x\|_{\ell^2} = \|(0, \dots, 0, 1, 0, \dots)\|_{\ell^2} = 1.$$
Hence $\|T_n\|_{\text{op}} \geq 1$.

On the other hand, note that for any $y = (b_k) \in \ell^2$,
$$\|T_n y\|_{\ell^2} = \sum_{k = n+1}^\infty |b_k|^2 \leq \sum_{k = 1}^\infty = \|y\|_{\ell^2}.$$
Therefore
$$\|T_n\|_{\text{op}} = \sup \{ \|T_n y\|_{\ell^2} : \|y\|_{\ell^2} = 1\} \leq 1.$$

Since $\|T_n\|_{\text{op}} \geq 1$ and $\|T_n\|_{\text{op}} \leq 1$, it must be that
$$\|T_n\|_{\text{op}} = 1.$$


问:
For any prime p>5 proving the existence of consecutive quadratic residues and consecutive quadratic non residues
This question was asked in an assignment which I am solving and I couldn't solve it.

Prove that for any prime $p>5$ there exist integers $1\leq a,b \leq  p-1$ for which $\binom{a}{p}=\binom{a+1}{p}=1$ and $\binom{b}{p}=\binom{b+1}{p} =-1$.

Quadratic residues and non-residues are equal in number , if I prove that any 2 residues are consecutive then non-residues will be automatically proved consecutive.

But I am unable to prove and 2 residues to be consecutive. For theory, I am studying David M Burton's book.

Can you please help?

答:
Let $p>5$. We always have $\left(\frac{1}{p}\right)=1$ and $\left(\frac{4}{p}\right)=1$. Then there are 2 cases:

Case 1: $\left(\frac{2}{p}\right)=1$ or $\left(\frac{3}{p}\right)=1$. In this case there are at least 3 quadratic residues in $\{1,2,3,4\}$. Hence there are at least $\frac{p-1}{2}-1$ quadratic non-residues and at most $\frac{p-1}{2}-3$ quadratic residues in the set $\{5,\dots,p-1\}$ (because the number of quadratic residues is equal to the number of quadratic non-residues). Thus by the pigeon hole principle, there have to be consecutive quadratic non-residues.

Case 2: $\left(\frac{2}{p}\right)=\left(\frac{3}{p}\right)=-1$. We know that $p>7$ in this case, because $3^2\equiv 2\; (7)$, but $2$ is not a quadratic residue. Thus we may look at the residue classes of $\{4,5,6,7,8,9,10\}$. If it contains consecutive quadratic residues, we are done. Otherwise, because $4$ and $9$ are squares, $\{5,6,7,8,10\}$ contains at least $4$ quadratic non-residues. This means that the set $\{1,2,3,4,5,6,7,8,9,10\}$ contains at least $6$ quadratic non-residues and at most $4$ quadratic residues. We then conclude similarly to Case 1.


问:
Understanding The Automorphisms to make a Semidirect Product
I have used semi-direct products for a while now, and frankly, they still don't make much sense to me. Especially where we define the homomorphism $\phi = K \to \text{Aut}(H)$. The trivial homomorphism makes sense to me, but non-trivial homomorphisms don't. I understand the definition,  but it seems very hard to work with. How do you actually construct the non-trivial homomorphisms, and just as importantly, how do you actually write them? Most of the examples I have seen are written similarly to this: $H \rtimes_{\phi} K$ with $\phi(x) = x^{m}$, but this makes no sense to me.

When are the different homomorphisms in the semidirect products isomorphic, and when are they not?

Once you do construct a non-trivial homomorphism, how do you find the presentation of the resulting semidirect product?

Thank you!

答:
Take for example $H=C_4$ and $K=C_2$, you can think of $C_4$ as $\langle i\rangle$, where $i^2=-1$. I like to think of the homomorphism $K\rightarrow Aut(H)$ as an action of $K$ on $H$. In this case the action of $C_2=\{1,\sigma\}$ on $C_4=\{1,i,-1,-i\}$ is given my multiplication by $-1$, i.e. $\varphi(\sigma): i\mapsto -i$, $1\mapsto -1$. This is a nontrivial homomorphism and you can easily check that  $C_4\rtimes_{\varphi} C_2\simeq Q_8$ with this action. Now write $C_4=\{1,\rho,\rho^2,\rho^3\}$, and consider the action $\psi(\sigma):1\mapsto 1, \rho\mapsto \rho^3,\rho^2\mapsto \rho^2$. Now you can check that $C_4\rtimes_{\psi}C_2\simeq D_8$.

I also like to think of the semidirect product as a split solution to the extension problem
$$ 1 \rightarrow K \rightarrow G \rightarrow H\rightarrow 1,$$
where the above is a short exact sequence of groups, and split means that there is a section $H\rightarrow G$ for the map $G\rightarrow H$. As far as I know the solutions to this problem are captured by the Ext functor and the classes of nonisomorphic solutions are given by $Ext^1$.

As to how to construct nontrivial homomorphisms, I suppose that it depends much on the particular groups at hand. But hopefully seeing them as permutation representations of groups might make things more clear.


问:
Are all sets of zero measure (not necessarily Lebesgue measure) measurable?
I know that all sets with Lebesgue outer measure equal to zero are measurable.  This is not hard to prove using the definition of a Lebesgue measurable set, E $\subseteq X$, that I was given : $$m^*(A)= m^*(A\cup E) + m^*(A\cup E^C) ,\forall A\subseteq X$$

But what about the case in a general measure space, $(X,\mathfrak{X} ,\mu)$?

Say $\mu (E)= 0$. Does this imply that $E$ is measurable?

All I have about general measurable sets is that they belong to $\mathfrak{X}$, where $\mathfrak{X}$ is a subset of the power set of $X$ that is closed under compliments and unions and contains $X$ and $\emptyset$.  All I have about the measure, $\mu$, is that it maps from $\mathfrak{X}$ to $[0,\infty]$,
$\mu(\emptyset)=0$ and we have disjoint additivity.

Does the very fact that $\mu(E)$ is defined imply that E is measurable?  This would seem, perhaps, obvious but I am not confident with Measure Theory.  I find it best not to make assumptions.

答:
Well, $\mu$ is a map $\mathfrak X\longrightarrow[0,\infty]$, so the fact that $\mu(E)$ is defined in the first place makes $E$ an element of $\mathfrak X$, so measurable.

If you're asking about outer measures (which measure spaces don't need!), then your question makes more sense and the answer is still yes, sets of measure $0$ are measurable. Let $\mu^\ast:\mathcal P(X)\longrightarrow [0,\infty]$ be an outer measure, $E\subseteq X$ such that $\mu^\ast(E)=0$. $E$ is measurable by definition if for all $A\subseteq X$ we have $\mu^\ast(A)=\mu^\ast(E\cap A)+\mu^\ast(E^c\cap A)$. Since outer measures are $\sigma$-subadditive this is equivalent to $\mu^\ast(A)\geq \mu^\ast(E\cap A)+\mu^\ast(E^c\cap A)$. Because of monotonicity we have $\mu^\ast(E\cap A)\leq\mu^\ast(E)=0$, so $\mu^\ast(E\cap A)=0$. So we only need to show $\mu^\ast(A)\geq \mu^\ast(E^c\cap A)$. But that's just monotonicity of the outer measure.
So $E$ is measurable.


问:
Question about the thought process at multivariable limit calculation
Is the following calculation correct, using the substitution method?

$$\lim_{(x,y)\to(0,0)}\frac{{x^2 - y^2}}{{2x + y}} = \lim_{(x,y)\to(0,0)}\frac{{x(1 - y/x)}}{{2 + y/x}} = \lim_{(x,y) \to(0,0)}\frac{{1 - (y/x)^2}}{{2 + y/x}} \cdot \lim_{x\to0} x,$$ where $u = y/x \to 1/2 \cdot 0 = 0$.

I'd like to know, firstly if this is correct, and after that if there is other way to solve problems like this, thank you.

答:
Since you are asking how to solve problems like this, I add this answer.

Look carefully at the denominator: For $y=-2x$ it is not defined while for $x\neq 0$ the numerator $x^2-y^2 = x^2-(-2x)^2 = -3x^2\neq 0$.

This suggests that if you move to $(0,0)$ along a curve close t0 $y=-2x$, you may push the value of the expression $\frac{{x^2 - y^2}}{{2x + y}}$ to $+\infty$ or $-\infty$.

Now you choose curves close to $y=-2x$. Often the following trick works quite well: Add a power $x^t$ for $t>0$. With $t$ you can control how fast you get closer to $y=-2x$ while approaching $(0,0)$:

$$y=-2x+x^t \Rightarrow \frac{{x^2 - y^2}}{{2x + y}} = \frac{x^2-(-2x+x^t)^2}{x^t}\stackrel{expanding}{=} -3\frac{x^2}{x^t}-\underbrace{x^t+4x}_{\stackrel{x\to 0}{\longrightarrow}0}$$

Now, you see if you choose, for example, $t=4$, the limit for $x\to 0$ would be $-\infty$. If you choose, for example, $t=2$, the limit would be $-3$.

So, the limit you are looking for does not exist.


问:
Fundamental group of Hawaiian earring is uncountable
I have to show fundamental group of Hawaiian earring($H=\cup^{\infty}_{n=1}K_{n}$, where $K_{n}$ is the circle centered at $\frac{1}{n}$ with radius $\frac{1}{n}$) is uncountable, without using Seifert-van Kampen theorem. So I have come up two ideas of proof:

1.Denote $[n]_{m}$ be the loop that travels counter-clockwise n times in $K_{m}$. Then $\{[n_{1}]_{1}[n_{2}]_{2}...|n_{i}\in\mathbb{Z}, i\in\mathbb{N}\}$ is uncountable, since every element in this set belongs to  $\pi_{1}(H,0)$, the fundamental group is thus uncountable.

2.Using the same notation above, the set $\{[1]_{f(1)}[1]_{f(2)}...|f $ is any bijective map from  $\mathbb{N} $ to itself$\}$ is uncountable, since $f$ is a reordering of natural numbers and there are uncountable many reordering exist. Thus this set as a subset of fundamental group, the group itself is uncountable.

Are these valid idea of proof?

答:
Your ideas are correct, but you have to make explicit how you regard an element of $\mathbb Z^{\mathbb N}$ as an element of $\pi_1(H)$ and that the resulting function $\phi : \mathbb Z^{\mathbb N} \to \pi_1(H)$ is injective. Let us elaborate 1.

Let us write $l_n^m : [0,1] \to K_n$ for the loop based at $0$ that travels counter-clockwise $m$  times around $K_n$. Explicitly, $l_n^m(t) = \frac{1}{n}(1- e^{2m\pi i t})$. Define

$$\psi((m_n)) : [0,1] \to H, \psi((m_n))(t) = \begin{cases}l_n^{m_n} (n(n+1)t - n) & t \in [\frac{1}{n+1},\frac{1}{n}] \\
0 & t = 0 \end{cases}$$
This is a well-defined continous map (since each neigborhood of $0$ contains all but finitely many $K_n$). Let $\phi((m_n)) = [\psi((m_n))]$, where $[-]$ denotes homotopy class of paths.

Let us show that $\phi$ is injective. There is a  retraction $r_n : H \to K_n$  which maps all $K_r$, $r \ne n$, to $0$. Let $i_n : K_n \to H$ denote inclusion.  The map $F_n = (r_n)_* \circ \phi$ has the property that the sequence $(m_n)$ is sent to the homotopy class of the path given by $l_n^{m_n} (n(n+1)t - n)$ for $t \in [\frac{1}{n+1},\frac{1}{n}]$ and maps all other $t$ to $0$. This path is clearly homotopic to $l_n^{m_n}$. Thus, if $\phi((m_n)) = \phi((m'_n))$, then $F_n((m_n)) = F_n((m'_n))$ for all $n$, i.e. $[l_n^{m_n}] = [l_n^{m'_n}]$ for all $n$. But this implies $m_n = m'_n$ for all $n$.

Identifying $\pi_1(K_n)$ with $\mathbb Z$ via the isomorphism $\iota_n : \mathbb Z \to \pi_1(K_n), \iota_n(m) = [l_n^m]$, we can express this alternatively as follows: The homomorphism
$$R : \pi_1(H) \to \mathbb Z^{\mathbb N}, R(u) = ((\iota_n)^{-1}(r_n)_*(u))$$
has the property $R \circ \phi = id$.

Also have a look at Fundamental group of mapping cone of quotient map from suspension to reduced suspension .


问:
Third Vertex of a Triangle
If I have two vertices of a triangle $(x_1,y_1)$ & $(x_2,y_2)$ and I know the lengths of three sides of the triangle $a,b,c$ , how can I find the third vertex ? Is there any derived formula?

答:
The third vertex will lie at the intersection point(s) of two circles centered at the two given vertices and radii equal to corresponding side lengths.

Clearly there'll be two possible positions for third vertex.

So one can write equations of two circles $S_1=0$ and $S_2=0$, equate them to get equation of common chord $L=0$ and further solve for intersection of $L$ with $S_i$'s.

Here is a diagram explaining the situation.

Title: Third Vertex of a Triangle


问:
Checking independence of random variables.
I'm revisiting the coupon collector's problem and I'm not sure how to prove that my variables are independent. Here's what I have:

Let $X$ denote the number of tries required to collect all the coupons, and $X_i$ denote the number of tries required to collect a new coupon, after having collected $i-1$ coupons. Then $X_i \sim Geom(p_i)$ with $p_i= \frac{n-i+1}{n}$.

Intuitively they should be independent but how would one go about proving this? Would calculating $E[X_i X_j]$ and showing it equal to $E[X_i]E[X_j]$ be enough? Is this the right way to proceed?

答:
From Mood, Graybill and Boes, Introduction to the theory of statistics, Pp. 150, Third Ed., 1974:

``...supose $X$ and $Y$ are two independent random variables, then $f_{X,Y}(x,y)=f_{x}(x)f_{y}(y)$ by definition of independence; however $f_{X,Y}(x,y)=f_{Y|X}(y|x)f_{X}(X)$ by definition of conditional density, which implies that $f_{Y|X}(y|x)=f_{Y}(y)$; that is, the conditional density of $Y$ given $x$ is the conditional density of $Y$''.

In your example, find the conditional density function for $p_i$ and $p_j$. If conditional density of $p_i$ given $p_j$ equals marginal density of $p_i$, then they are stochastically independent.

If you have the conditional density $f_{X,Y}(x,y)$, another way to show $X$ and $Y$ are independet is to rewrite $f_{X,Y}(x,y)=g(x)h(y)$, if you can find both g(x) and h(y), then $X$ and $Y$ are independent.


问:
Maxima and minima of function of two variables.
If $$f(x,y)=x^{2}+y^{3}+\frac{768}{x+y}$$
then find the points where it attains its extrema.

For this I have done the following:
$$f_x=2x-\frac{768}{(x+y)^{2}}$$
$$f_y=3y^{2}-\frac{768}{(x+y)^{2}}$$
Putting $f_x=0$ and $f_y=0$,
$$2x=\frac{768}{(x+y)^{2}}$$
$$3y^{2}=\frac{768}{(x+y)^{2}}$$

$$\implies\frac{2x}{3y^{2}}=1$$

$$\implies\frac{x}{y^{2}}=\frac{6}{4}$$

So, $x=6, y=2$ is a stationary point.

But in the book, it is given that $(5.99,-2.45)$ is another stationary point which satisfies the parabola $2x=3y^{2}$. But I can not find the way to proceed. Any help would be appreciated!

答:
Starting from @WA Don's answer, wa have
$$9y^6+12y^5+4y^4-1024=(y-2) \left(3 y^2+8 y+16\right) \left(3 y^3+2 y^2+32\right)$$
$(3 y^2+8 y+16)$ does not show any real root nad we are left with the solutions of
$$3 y^3+2 y^2+32=0$$ whidh has only one real root since $\Delta=-249856$.

Using the hyperbolic method, the real root is given by
$$y=-\frac{2}{9} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}(487)\right)\right)\approx -2.44743$$ and, from $2x=3y^2$,
$$x=\frac{2}{27} \left(1+2 \cosh \left(\frac{1}{3} \cosh ^{-1}(487)\right)\right)^2\approx 8.98489$$

All of this leads to a minimum value $\approx 183.545$.


问:
Computing an iterated integral $\int_{D}(2x+3y) \ dx \ dy$

Compute the iterated integral $$\int_{D}(2x+3y) \ dx \ dy, \text{ when } D= \{(x,y) \in \mathbb{R}^2 \mid 1\leqslant x^2+y^2 \leqslant 4, x\leqslant0, y \geqslant0 \}$$

Plotting the inequality $1\leqslant x^2+y^2 \leqslant 4$ it seems to be a circle not including the middle part and $2x+3y$ is a plane cutting through the middle of the circle. Also, the solutions for $\mid 1\leqslant x^2+y^2 \leqslant 4$ seems to be $x = \pm 2$ and $y=0$. So the bounds for $x$ would be $[-2, 2]$ right? For $y$ I'm not sure what should I do... Any hints would be appreciated.

答:
The domain $D= \{(x,y) \in \mathbb{R}^2 \mid 1\leqslant x^2+y^2 \leqslant 4, x\leqslant0, y \geqslant0 \}$ can be easily written with polar coordinates as follows.

Let $x=r\cos\theta$ and $y=r\sin\theta$. Since $1\leq x^2+y^2\leq 4$, you get that $r\in [1,2]$. Also, $x\leq 0$ and $y\geq 0$, which means we only have the part of the annulus in the second quadrant, that is, the angle goes from $\pi/2$ to $\pi$. In other words, we can reparametrize $D$ as
\begin{equation}
D=\lbrace (r,\theta)\in\mathbb{R}^2\mid r\in [1,2],\theta\in [\frac{\pi}{2},\pi]\rbrace.
\end{equation}
Hence, the integral
\begin{equation}
\int_{D}(2x+3y) \ dx \ dy=\int_1^2\int_{\pi/2}^\pi\,\left(2r\cos\theta+3r\sin\theta\right)r\,d\theta\,dr,
\end{equation}
which can be solved with elementary methods.


问:
Lie derivative of r-form
I actually try to read the Geometry, Topology and Physics by M.Nakahara and I'm stuck on the Lie derivative of a r-form and the demonstration of  $i_{[X,Y]}\omega = [L_X,i_Y]\omega $ . I found some proof of this relation like in this pdf https://www.mathi.uni-heidelberg.de/~lee/Xiaoman_Wu.pdf . But this use a different definition of the Lie derivative than the Nakahara :

 $L_X \omega =\frac{1}{r!} X^\nu\partial_\nu\omega_{\mu_1 \dots\mu_r}dx^{\mu_1}\wedge \dots \wedge dx^{\mu_r} + \frac{1}{r!}\sum_{s=1}^r \partial_{\mu_s}X^\nu\omega_{\mu_1 \dots\nu\dots\mu_r}dx^{\mu_1}\wedge \dots \wedge dx^{\mu_r}$

in Nakahara become :

$(L_X \omega)(X_1,\dots,X_r) = X[\omega(X_1,\dots,X_r)] -\sum_{i=1}^r \omega(X_1,\dots,[X,X_i],\dots,X_r) $

in the pdf.

I can't find how to connect this 2 definitions, thanks for help

答:
I think I've fund the solution :
$$\begin{align} (L_X w)(X_1\dots X_r)&=\frac{1}{r!}X^\nu \partial_\nu w_{\mu_1 \dots \mu_r}dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} (X_1 \dots X_r) + \frac{1}{r!}\sum_{s=1}^r \partial_{\mu_s} X^\nu w_{\mu_1\dots\nu\dots\mu_r}dx^{\mu_1}\wedge \dots\wedge dx^{\mu_s}\wedge\dots \wedge dx^{\mu_r} (X_1 \dots X_r)\\
&=\frac{1}{r!}X^\nu \partial_\nu w_{\mu_1 \dots \mu_r}\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(r)}^{\mu_r} + \frac{1}{r!}\sum_{s=1}^r \partial_{\mu_s} X^\nu w_{\mu_1\dots\nu\dots\mu_r}\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(s)}^{\mu_s} \dots X_{\sigma(r)}^{\mu_r}\\
&=\frac{1}{r!}X^\nu \partial_\nu \left[\sum_{\sigma \in P}w_{\mu_1 \dots \mu_r} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(r)}^{\mu_r}\right] - \frac{1}{r!}X^\nu w_{\mu_1 \dots \mu_r} \partial_\nu \left[\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(r)}^{\mu_r}\right]  + \frac{1}{r!}\sum_{s=1}^r \partial_{\mu_s} X^\nu w_{\mu_1\dots\nu\dots\mu_r}\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(s)}^{\mu_s} \dots X_{\sigma(r)}^{\mu_r}\\
&=X\left[w(X_1,\dots, X_r)\right] - \frac{1}{r!}X^\nu w_{\mu_1 \dots\mu_s\dots \mu_r} \sum_{\sigma \in P} \text{sgn}(\sigma)\sum_{s=1}^r X_{\sigma(1)}^{\mu_1} \dots\partial_\nu X_{\sigma(s)}^{\mu_s}\dots X_{\sigma(r)}^{\mu_r} + \frac{1}{r!}\sum_{s=1}^r \partial_{\mu_s} X^\nu w_{\mu_1\dots\nu\dots\mu_r}\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(s)}^{\mu_s} \dots X_{\sigma(r)}^{\mu_r}\\
&=X\left[w(X_1,\dots, X_r)\right] - \frac{1}{r!}X^\nu w_{\mu_1 \dots\mu\dots \mu_r} \sum_{\sigma \in P} \text{sgn}(\sigma)\sum_{s=1}^r X_{\sigma(1)}^{\mu_1} \dots\partial_\nu X_{\sigma(s)}^{\mu_s}\dots X_{\sigma(r)}^{\mu_r} + \frac{1}{r!}\sum_{s=1}^r \partial_{\nu} X^{\mu_s} w_{\mu_1\dots\mu_s\dots\mu_r}\sum_{\sigma \in P} \text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1} \dots X_{\sigma(s)}^{\nu} \dots X_{\sigma(r)}^{\mu_r}\\
&= X\left[w(X_1,\dots, X_r)\right] - \frac{1}{r!} w_{\mu_1\dots\mu_s\dots\mu_r}\sum_{s=1}^r\sum_{\sigma \in P}\text{sgn}(\sigma)X_{\sigma(1)}^{\mu_1}\dots \left(X^\nu\partial_\nu X_{\sigma(s)}^{\mu_s}-X_{\sigma(s)}^{\nu}\partial_{\nu}X^{\mu_s}\right) \dots X_{\sigma(r)}^{\mu_r}\\
&= X\left[w(X_1\dots X_r)\right] - \sum_{s=1}^r w(X_1,\dots, [X,X_s], \dots,X_r)
\end{align}$$

I hope there aren't too many mistakes

Title: Lie derivative of r-form


问:
Test the convergence of $\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$

Test the convergence of $$\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$$

My Attempt: Using the root or ratio test would be too inconvenient here. Looking at the denominator, I used the Cauchy Condensation Test, i.e. if the condensed sequence $\sum_{k=1}^{\infty} 2^k a_{2^k}$ converges, then $\sum a_n$ converges. By this test, we have
$$\sum_{k=1}^{\infty} 2^{k} \frac{(2^k)^{\frac{1}{2^k}} - 1}{2^k} = \sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1$$

Consider the term $2^{\frac{k}{2^k}} - 1$. If we rewrite 2 as a binomial $1+1$, we get
$$(1+1)^{\frac{k}{2^k}} - 1 \lt \frac{k}{2^k}$$
Summing both sides,
$$\sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1 < \sum_{k=1}^{\infty} \frac{k}{2^k}$$
For the latter sum, it is trivial to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{2} < 1$, and the series converges by ratio test. Hence $\sum_{k=1}^{\infty} 2^{\frac{k}{2^k}} - 1$ converges and therefore, $\sum_{n=1}^\infty \frac{n^{1/n} - 1}{n}$ is convergent. $\; \blacksquare$

Is this proof valid? Please point out the errors and suggest improvements as well. I think there may be a less complicated way to solve this, but I'm not able to put my finger on it.

答: Another way: $\frac {e^{x}-1} x \to 1$ as $x \to 0$ so $e^{x}-1 <2x$ for $x>0$ close enough to $0$. So for $n$ sufficiently large we get $n^{1/n}-1 =e^{\frac1 n \ln n} -1< \frac {2 \ln n} n$. Now $\ln n=\ 2 \ln (n^{\frac 1  2})<2n^{1/2}$, Now you can compare the given series with $\sum \frac 1 {n^{3/2}}$ which is convergent.


问:
Expand $5e^{2x}$ using the power series?
I have scoured the internet for solutions or workthroughs to a similar question to no avail.

I am assuming I need to use either the Taylor or Maclaurin Series but I have no idea where to start.

Any help would be greatly appreciated :)

答:
$e^x$ is the function with the "nicest" power series. Most operations on finite polynomials work the same with $e^x$:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
$$e^{2x}=\sum_{n=0}^\infty\frac{(2x)^n}{n!}$$
$$5e^{2x}=\sum_{n=0}^\infty\frac{5\cdot2^n}{n!}x^n$$


问:
Signed measure of subsets of zero-measured sets?
Hey, I'm reading the proof of proposition 6.8 in Big Rudin, pg. 120. I'll just mention where my confusion lies: Suppose K is an arbitrary measure (complex, real, whatever). In the proof of property (f), Rudin says that if K(A) = 0, and F is a subset of A, then K(F) = 0. Rudin has forced that by construction in chapter 1 for positive measures, but why should it work for complex or signed measures? Does there exist two sets A,B such that A is a subset of B, measure of B is 0, but measure of A is non-zero (negative?), for some arbitrary measure?

EDIT: Rudin's prop 6.8, part (f) says: given two arbitrary measures $\lambda_1$, $\lambda_2$, a positive measure $\mu$. If $\lambda_1 \ll \mu$, and $\lambda_2 \perp \mu$, then $\lambda_1 \perp \lambda_2$.

Also, $\ll$ and $\perp$ denote absolute continuity and mutual singularity, respectively.

答: Observe that $\mu$ is positive. Since $\mu$ is positive and $\mu \perp \lambda_{2}$, the latter is concentrated on a $\mu$-null set $A$. But as $\lambda_{1} \ll \mu$ and $\mu$ is positive, we conclude for all measurable $E \subset A$ that $\mu(E) = 0$, hence $\lambda_{1}(E) = 0$ by absolute continuity, and thus $\lambda_{1}$ is concentrated on the complement of $A$. In other words, $\lambda_{1} \perp \lambda_{2}$.</p>


问:
Problem with proving Brownian Motion
Hey i want to prove that $Y_t= \frac{1}{\sqrt{2}}M_t $ is a B.M, while
$$
M_t = X_t-X_0 + \int_{0}^{t} X_s ds
$$
and $$
X_t=e^{-t} B_{e^{2t}}, $$
where $(B_t)_{t \geq 0}$ is a standart B.M..

In order to show this i wanted to show that $<Y>_t=t$.

We have that $dM_t=dX_t+X_tdt.   $ Hence $$ d<Y>_t=dY_t * dY_t = \frac{1}{2}dM_t*dM_t = \frac{1}{2} (dX_t+X_tdt)(dX_t+X_tdt)=(1).$$
Since $dX_t= d(e^{-t}B_{e^{2t}}) = -e^{-t} B_{e^{2t}} dt + e^{-t} dB_{e^{2t}}, $

we get $$ (1)= \frac{1}{2} e^{-2t}dB_{e^{2t}} dB_{e^{2t}} = \frac{1}{2} e^{-2t}e^{2t} dt = \frac{1}{2} dt.  $$
Hence $<Y>_t \neq t.$ So $Y_t$ is not a Brownian Motion? So what am i missing?
Furthermore i am quiet sure that $Y_t$ is not a martingale.
Any help would be appreciate.

答:
Definition of Standard BM:

$(i) B(0):=0$

$(ii)$ The increments $B(t) - B(s)$ are independent for all $0\leq s < t$

$(iii)$ The increments $B(t) - B(s) \sim N(0, (t-s))$

$(iv)$ The function $t \rightarrow B(t)$ is almost surely continuous.

We need to show that the above hold for $Y_t$. I will start by trying to show that $Var(Y_t)=t$:

$Var(\frac{1}{\sqrt{2}}M_t)=\frac{1}{2}Var(M_t)=\\=0.5\mathbb{E}[M_t^2]-0.5\mathbb{E}[M_t]^2$

First note that $X_0=1B(1) \sim N(0,1)$

Now taking expectation of $M_t$:

$\mathbb{E}[M_t]=\mathbb{E}[X_t]-\mathbb{E}[X_0]+\int_0^t \mathbb{E}[X_s]ds=\\=0-0+0$

Now:

$M_t^2=\left(X_t-X_0+\int_{s=0}^{t}X_s ds\right)^2=\\=X_t^2-2X_tX_0+2X_t\int_{s=0}^{t}X_s ds-X_0^2+2X_0\int_{s=0}^{t}X_s ds+\int_{s=0}^{t}X_s ds\int_{h=0}^{t}X_h dh$

To work out the expectation over $M_t^2$, I will do the following:

$\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)\sim N(0,e^{2t})$

So that:

$X_t=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}B(t)$

And:

$X_t^2=e^{-2t}\frac{e^{2t}}{t}B(t)^2$

Taking the expectation:

$$\mathbb{E}[M_t^2]=\mathbb{E}[X_t^2]-2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]+2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]-\mathbb{E}[X_0^2]+2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]+\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]$$

Now, the following terms are straight forward:

$$\mathbb{E}[X_t^2]=e^{-2t}\frac{e^{2t}}{t}\mathbb{E}[B(t)^2]=e^{-2t}e^{2t}=1$$

$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_tB_1]=min(1,t)2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$$

$$\mathbb{E}[X_0^2]=1$$

To workout the other three expectations, I will try to derive the process for $X_t$:

$\frac{\partial X}{\partial t}=\frac{2}{\sqrt{t}}-0.5e^{-0.5t}$

$\frac{\partial X}{\partial B_t}=e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}$

$\frac{\partial^2 X}{\partial B_t^2}=0$

By Ito's Lemma:

$$X(B_t,t)= X(B_0,t_0)_{=0} + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial t}+\frac{\partial X}{\partial B_h}a(B_h,h)_{=0}+\frac{1}{2}\frac{\partial^2 X}{\partial B^2}_{=0}b(B_h,h)^2_{=1}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial X}{\partial B}b(X_h,h)_{=1}\right)dB_h=\\=\int_{h=0}^{h=t}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=t}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h$$

So that:

$$2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}[B_t \int_{s=0}^{t}X_s ds]=\\=2e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\mathbb{E}\left[B_t\int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right\} ds\right]$$

And:

$$2\mathbb{E}\left[X_0 \int_{s=0}^{t}X_s ds \right]=\\=2\mathbb{E}\left[X_0 \int_{s=0}^{t}\left\{\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h \right\} ds \right]$$

And:

$$\mathbb{E}\left[\int_{s=0}^{t}X_s ds\int_{u=0}^{t}X_u du\right]=\\=\mathbb{E}\left[\int_{s=0}^{t}\left(\int_{h=0}^{h=s}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=s}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) ds\int_{u=0}^{t}\left(\int_{h=0}^{h=u}\left(\frac{2}{\sqrt{t}}-0.5e^{-0.5t}\right)dh + \int_{h=0}^{h=u}\left(e^{-t}\frac{\sqrt{e^{2t}}}{\sqrt{t}}\right)dB_h\right) du\right]$$

I will have a go at working out these later.

Ps: I am not 100% sure that $M_t$ is an Ito process (which would of course immediately imply that $Y_t$ cannot be a Brownian motion).

Title: Problem with proving Brownian Motion


问:
Fair coins or not
I am reading Introduction to Probability, 2nd Edition, and at the end of chapter one the following problem appears:

We are given three coins: one has 'heads' in both faces, the second has
tails in both faces, and the third has a 'head' in one face and a tail in the other. We
choose a coin at random, toss it, and the result is 'heads'. What is the probability that
the opposite face is tails?

I solve it assuming the general case where the coin might not be fair so I assume that the probability of having a 'head' in the second coin is p.

And the other two coins would not be affected by the nature of the coin.

But the solution manual which was written by the same author assumes that the coins are fair.

So what's is the right thing to assume?

答: I don't think there is 'right' or 'wrong' here as the exercise was poorly worded. I personally would interpret the third coin as 50:50 for the lack of information, but your interpretation is equally reasonable. For what it's worth, you solved a generalization of the exercise and therefore would receive full credit in case this was an exam (at least if it was one of mine).</p>

Title: Fair coins or not


问:
Choose N numbers in order to produce maximum equidistant numbers by summing the chosen numbers

Preliminaries

I have a device, an attenuator which is made up of resistive cells. To set the desired level of attenuation there are 7 switches on its front panel with the following values of attenuation (in dBs):
$$
{0.5, 1, 2, 3, 6, 12, 20}
$$
These switches allow to choose the attenuation between 0 and 44.5 dB with the step of 0.5 dB.

I was wondering how the designers of this device have chosen those values. I was thinking that maybe they wanted to produce the maximum possible attenuation range with a step of 0.5 dB by means of 7 resistive cells. But then I realized that it was not the case because the following values produces a wider range:
$$
{0.5, 1, 2, 4, 8, 16, 32}
$$
The range in this case is from 0 to 63.5 dB (and the step is still 0.5dB). There the problem came.

Problem

Let's consider an $N$-tuple
$$
a(N) \in \mathbb{N}^N
$$
and a set
$$
S(a(N)) = \bigg\{\sum_{i\in p,\, p \in \mathcal{P}(\overline{[1,N]})} a_i \bigg | a_i \text{ is the $i$th coordinate of $a(N)$}\bigg\},
$$
where
$\mathcal{P}(A)$ denotes the power set of $A$ and $\overline{[1,N]}$  is $\{i|0 < i < N, i\in\mathbb{N}\}$
So $S(a(N))$ is the set of all possible sums of coordinates of $a(N)$ (including the empty sum).
Let's call an $N$-tuple $a(N)$ good if
$$S(a(N)) = \overline{[1,M]},\text{ for some } M\in\mathbb{N}$$
Let's denote the set of all good $N$-tuples as $G(N)$.
For example, the tuple $a=(1,2,3)$ is good because $S(a) = \{1, 2,3,4,5,6\}$. On the other hand the tuple $b=(1,2,5)$ is not good because $S(b) = \{1,2,3,5,6,7,8\}$ and 4 is missing.

Given $N \in \mathbb{N}$ prove that
$$\bigg [a(N)\in G(N) \text{ and } \sum_i a_i  = \max_{a(N)\in G(N)}({\max_{s_i \in S(a(N))}s_i})\bigg] \iff \bigg[a(N) = (1, 2, 4, ..., 2^{N-1})\bigg]$$
In other words, a good $N$-tuple with the maximum possible sum of its coordinates is $(1,2,...,2^{N-1})$.

Proof

I will prove the statement by induction. Without loss of generality, it is assumed that the coordinates of tuples is sorted in the ascending order.
The base case $N=1$ is obvious.
Let's suppose that the statement is true for all $k\leq N$. Then consider the case for $N+1$. Suppose that in this case the desired tuple $a\neq (1,2,4,...,2^{N})$. Then take the tuple $a' = (a_1, a_2, ..., a_N)$ i.e. $a$ without the last coordinate. There are two possibilities here:

Case $a' = (1,2,4, ..., 2^{N-1})$. In this case $S(a)$ looks like this:
$$
S(a) = \{ \color{red}{0, 1, 2, 3, ..., (2^N-1)}, \color{blue}{(a_{N+1} + 0), (a_{N+1} + 1), (a_{N+1} + 2), ..., (a_{N+1} + 2^N - 1)}\},
$$
where red elements are those obtained only from $a'$ and blue - are those obtained from the rest ($a_{N+1}$ and $a'$). Due to our convention we know that $a_{N+1} \geq a_{N}$. Therefore, taking into account the fact that $a$ is a good tuple, we conclude that $\color{blue}{a_{N+1}+0}$ must be equal to $\color{red}{2^N-1} + 1$. Of course, the reds and blues might intersect but in this case the sum of all coordinates wouldn't be the maximum sum. I.e.
$$
a_{N+1} = 2^N
$$
This contradicts the initial suggestion that $a$ is not $(1,2,4, ..., 2^N)$.

Case $a'\neq (1,2,4, ..., 2^{N-1})$. In this case $S(a)$ is the following (if $a\in G(N+1)$ then $a'\in G(N)$, see Update):
$$
S(a) = \{\color{red}{0, 1, 2, 3, ..., M}, \color{blue}{(a_{N+1} + 0), (a_{N+1} + 1), ..., (a_{N+1} + M)}\},
$$
where the colors have the same meaning as in the previous case and
$$
M < 2^N - 1\quad \text{by induction}
$$
Using the same reasoning as in the previous case we conclude that
$$
a_{N+1} = M + 1
$$
So
$$
\sum_i a_i = 2M + 1 < 2^{N+1} - 1
$$

Update

If $a\in G(N+1)$ then $a'\in G(N)$. I thought it was true but now I cannot prove it :). But this doesn't really violate the proof I provided: suppose that $a'\notin G(N)$ then there is a gap between 0 and $M$ (red numbers), and therefore at least one of the blue sums must fill this gap. So in this case the maximum would be even smaller.

Is my proof a valid one? I am asking because I feel a bit uncomfortable that I don't proof the $\Leftarrow$ and $\Rightarrow$ separately.
I feel that this "fact" has a more elegant proof. Could you provide your own?

答:
Firstly, the proof I provided in the post is not correct.

The conventions and notations are the same as in the post. Also let's denote the set of $N$-tuples which coordinates are distinct powers of two as $B(N)$ so
$$
B(N) := \{a\in \mathbb{N} | a_i = 2^{i-1} \}
$$

It is obvious that
$$
|\mathcal{P}(\overline{[1,N]})| = 2^N
$$
So the maximum possible number in $S(a):a\in G(N)$ is $2^N -1$. This is because there are at most $2^N$ numbers in $S(a)$ for any $a\in \mathbb{N}^N$. But for good $N$-tuples these numbers are consequent from 0 to $M$. So the maximum possible $M$ will be for $a\in G(N)$ for which $S(a)$ contains exactly $2^N$ elements (i.e. there are no duplicates).

So we prove
$$
\big[ a\in G(N)\, \wedge\, \max S(a) = 2^N - 1\big] \iff a\in B(N)
$$

($\Leftarrow$). Obviously, the tuple $a = (1, 2, 4, ..., 2^{N-1})$ doesn't duplicate elements in $S(a)$. And
$$\forall n\in \mathbb{N}, n<2^{N} \quad\exists s\in S(a) : s = n$$
So this $N$-tuple possesses the desired feature.

($\Rightarrow$). Suppose that there exists an $N$-tuple $b\notin B(N)$ such that $$\max_{s_i\in S(b)}s_i = 2^{N} - 1$$
Without loss of generality we assume that
$$b_N \neq 2^{N-1}\tag{1}\label{one}$$
Indeed, otherwise we throw away $b_N$ and consider $b' = (b_1, b_2, b_3, ..., b_{N-1})$ such that its maximum is $2^{N-1}-1$ and the problem stands the same.

Let's consider the tuple $b'=(b_1, b_2, ..., b_{N-1})$. We know that $|S(b')| = 2^{N-1}$. What is the maximum of $S(b')$.

Case 1. $\max S(b') = 2^{N-1} -1$. Then we obtain $b_N = 2^{N-1}$. Contradiction with \ref{one}.

Case 2. $\max S(b') < 2^{N-1} - 1$. Then $b_N > 2^{N-1}$. But then gaps are guaranteed (e.g. the number $2^{N-1} - 1$ would not appear in $S(b)$).$\tag{*}\label{aster}$

Case 3. $\max S(b') > 2^{N-1} - 1$. Then $b_N < 2^{N-1}$. We notice that $S(b)$ divides into two disjoint subsets
$$
S(b) = S(b') \cup S(b)\setminus S(b')\tag{**}\label{twoaster}
$$
In $S(b')$ are the sums which are obtained only from $b'$ and in $S(b)\setminus S(b')$ are the sums which are of the form $b_{N} + X, X\in S(b')$ so, obviously,
$$
\overline{[0, b_{N} - 1]} \subset S(b')\tag{2}\label{two}
$$
But then
$$
\overline{[b_{N}, 2b_{N} -1]} \subset S(b)\setminus S(b')\tag{3}\label{three}
$$
Those numbers may fill the entire set $S(b)$ but then $2b_{N} - 1 = 2^{N} - 1$ so $b_{N} = 2^{N-1}$. Contradiction with \ref{one}. Therefore there are more numbers in $S(b)$. The next one is $2b_{N}$ and it is, obviously, lies in $S(b')$. And then $3b_{N}$ is also in $S(b)$ and lies in $S(b)\setminus S(b')$. But then all the numbers between $2b_{N}$ and $3b_{N}$ are also in $S(b)$:
$$
\overline{[2b_N, 3b_N-1]}\subset S(b')\tag{4}\label{four}
$$
But then
$$
\overline{[3b_N, 4b_N-1]}\subset S(b)\setminus S(b')\tag{5}\label{five}
$$
Again it may be the entire $S(b)$. Then $b_N = 2^{N-2}$. Acting in the same manner one can prove that
$$
b_N = 2^K, K < N-1\tag{6}\label{six}
$$
I feel that there should be a shorter proof using \ref{six} but I couldn't find it.

We see that the last element $b_{N}$ divides the entire range into equal intervals of length of $2^{K}$.
Let's look at the first such interval $I_1 := [0, b_{N}-1]$ together with the tuple $b'' = (b_1', b_2', ..., b_{N-2}')$ i.e. $b'$ without its last element. Obviously, $S(b'')\subset S(b')$ and $S(b')$ itself can be represented as a union of two disjoint subsets (cf.\ref{twoaster}):
$$
S(b') = S(b'') \cup S(b')\setminus S(b'')
$$
But more important, $I_1\subset S(b')$ is also divided by $b_{N-1}$ into two subsets
$$
\overline{[0, b_{N}-1]} = A \cup B, A\subset S(b''), B\subset S(b')\setminus S(b'')
$$
The first we can say is that $B\neq \emptyset$. Indeed if it was so it would mean that the tuple $S(b'')$ fills the entire interval including the number $b_{N}-1$ but then adding $b_{N-1} < b_{N}$ to it would give the number between $b_{N}-1$ and $2b_{N}$ which is in contradiction with \ref{three}. So we know that

$$
\max_{s_i''\in S(b''),s_i''\in I_1} s_i'' < b_{N} -1
$$

More over, it must be greater then or equal to $b_{N}/2 - 1$. Otherwise, $S(b')$ wouldn't fill the entire interval (cf.\ref{aster}):
$$
b_{N}/2 - 1 \leq \max_{s_i''\in S(b''),s_i''\in I_1} s_i'' < b_{N}-1
$$
Using the same reasoning as in the first part of this proof one can show that

$$
b_{N-1} = 2^{K}, K\leq N-3
$$
And eventually
$$
b_{i} = 2^{K_i}, b_{N}  < 2^{N-1}, i < j \Rightarrow b_{i} < b_{j}\tag{7}\label{seven}
$$
But this makes no sense.


问:
If the set of all linear maps is itself a vector space, then what is its basis and dimension?
In linear algebra, we learn about linear mappings and it is an easy proof that the set of all linear maps is itself a vector space over the same field. But if it's a vector space how would one go about figuring out what a basis for the set of linear maps are? To be more specific we can take the set of all linear map from R^3 to R^2 over the field R.</p>

答:
A bit more basic: Let $V,W$ be vector spaces over a field $F$. Let $B_V$ and $B_W$ be bases of these vector spaces. Then each linear map $L:V\longrightarrow W$ can be uniquely written as
$$\begin{align}L\left(\sum_{x\in B_V} a_x x\right)&=\sum_{x\in B_V}a_x L(x)\\
&=\sum_{x\in B_V}a_x\sum_{y\in B_W}b_y y\\
&=\sum_{x\in B_V}\sum_{y\in B_W}a_xb_y y\end{align}$$
The coefficients $c_{xy}:=a_x b_y$ uniquely determine the function. That's the idea of the matrix representation of a linear map. Those coefficients are the matrix entries in the finite dimensional case. Now let's define the linear maps $L_{xy}:V\longrightarrow W$ which map $x\in B_V$ to $y\in B_W$ and map every other element of $B_V$ to $0$. We can see that using the coefficients from above, we can write $L=\sum_{x\in B_V}\sum_{y\in B_w} c_{xy}L_{xy}$., and there is only one way to do so. That means, the maps $L_{xy}$ are a basis of the space of linear maps $V\longrightarrow W$. There is exactly one such map for each combination of vectors in $B_V$ and $B_W$, that is, one for each element of $B_V\times B_W$. So the dimension is the cardinality of this product. In the finite dimensional case, that's the product of the cardinalities of $B_V$ and $B_W$, so the product of the dimensions of $V$ and $W$.

So that's the takeaway: if $\dim V=n,\dim W=m$, then $\dim\operatorname{Hom}(V,W)=nm$.


问:
Solve $ h(t) = \int_{0}^{t} e^{ - (\eta + \mu) s } \{\mu + \eta \, \, h(t-s)^2 \} ds. $
Suppose that $h(t) : [0, \infty) \rightarrow [0, \infty)$ satisfies,
$$
 h(t) = \int_{0}^{t} e^{  - (\eta + \mu) s   } \{\mu + \eta  \, \, h(t-s)^2 \} ds,
$$
where $\eta, \mu \geq 0$.
I need to prove  that, if  $\mu \neq \eta$, then,
$$
 h(t) = \frac{ \eta e^{\eta t}    - \eta e^{  \mu t }  }{    \eta e^{\eta t} -   \mu e^{\mu t} }.
 $$
Would you have some hint?

答: This is hint in the form of a suggestion, as one way to approach the problem. Since you don't have to derive the expression for $h(t)$, you could first check that it satisfies the integral equation, and second show that the integral equation has a unique solution. One way to show uniqueness would be to derive from the integral equation, an ordinary differential equation satisfied by $h$, since we have a standard uniqueness theorem for ode's.


问:
A question on how to prove a local martingale of continuous time
I have been reading a paper on the stochastic epidemic model. Paper deals with the following model.

$dS(t) = (\Lambda– μS(t) – βS(t)I(t)/(1+aI(t)) + r_1I(t) + δR(t)) dt – σS(t)I(t)/(1+aI(t)) dB(t)$,

$dI(t) = (βS(t)I(t)/(1+aI(t)) – (μ + r1 + r2 + α)I(t)) dt + σS(t)I(t)/(1+aI(t)) dB(t),$

$dR(t) = (r_2I(t) – (μ + δ)R(t)) dt.$

I have encountered a place where it is assumed
$M_1(t)$ = $\int_{0}^{t}$ σS(τ)/(1+aI(τ)) dB(τ) and $M_1(t)$ is the local continuous martingale with $M_1(0)=0$. I completely stuck how to prove $M_1 (t)$ is a martingale. Please give any hint. Thanks a lot. This article can be found here: https://advancesindifferenceequations.springeropen.com/articles/10.1186/s13662-018-1759-8

答:
user6247850 is right in stating that, in general, this is a continuous local martingale by the construction of the stochastic integral. If you want to prove that this is a true martingale, then it suffices that the following criterion is satisfied: $\mathbb{E} \left[ \langle M_{1} \rangle_t \right] < \infty$ for all $t>0$. The proof of this result is short but non-trivial:  it involves reducing the continuous local martingale to obtain $L^2$ bounded martingales, then using Vitali's convergence theorem and strong $L^1$ convergence.

In this case, by properties of quadratic variation and Fubini's theorem:

\begin{equation}
\mathbb{E} \left[ \langle M_{1} \rangle_t \right] = \int_{0}^{t} \mathbb{E} \left[  \dfrac{\sigma^2 S_{u}^{2}}{(1 + \alpha I_u)^2} \right] du.
\end{equation}

Assuming you are working with a closed population (so that $S+I+R  = const.$) and that $\alpha > 0$, the expectation inside the integral is clearly bounded above, so the integral always converges, the criterion holds and this is a true martingale.


问:
Solving Differential Equation $y'=\cos(xy)$
My question is about solving differential equation $y'=\cos(xy)$.
I tried to changing variable $u=xy$
\begin{align*}
u &= xy \\
\frac{du}{dx} &= y+ x\frac{dy}{dx}\\
\frac{dy}{dx} &= \frac{1}{x}\frac{du}{dx}-\frac{y}{x}\\
\end{align*}
then since $u=xy\to \frac{y}{x}=\frac{u}{x^2}$
\begin{align*}
\frac{1}{x}\frac{du}{dx}-\frac{u}{x^2}-\cos(u)=\;0\\
\end{align*}
finally
\begin{align*}
x\,du-(u+x^2\cos(u))\,dx=0\\
\end{align*}
since this is not an Exact Differential Equation$($Because of the existence of $\cos(u))$, what is multiplicative "integrating" factor? Is this equation solvable or not?

答:
Hint:

Let $u=xy$ ,

Then $y=\dfrac{u}{x}$

$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$

$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\cos u$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\cos u+u}{x^2}$

$(x^2\cos u+u)\dfrac{dx}{du}=x$

Let $v=x^2$ ,

$\dfrac{dv}{du}=2x\dfrac{dx}{du}$

$\therefore\dfrac{(x^2\cos u+u)}{2x}\dfrac{dv}{du}=x$

$(x^2\cos u+u)\dfrac{dv}{du}=2x^2$

$(v\cos u+u)\dfrac{dv}{du}=2v$

This belongs to an Abel equation of the second kind.

Let $w=v+u\sec u$ ,

Then $v=w-u\sec u$

$\dfrac{dv}{du}=\dfrac{dw}{du}-(u\tan u+1)\sec u$

$\therefore(\cos u)w\left(\dfrac{dw}{du}-(u\tan u+1)\sec u\right)=2(w-u\sec u)$

$(\cos u)w\dfrac{dw}{du}-(u\tan u+1)w=2w-2u\sec u$

$(\cos u)w\dfrac{dw}{du}=(u\tan u+3)w-2u\sec u$

$w\dfrac{dw}{du}=(u\tan u\sec u+3\sec u)w-2u\sec^2u$


问:
The properties of $A^{1 /2} B^{1 / 2 } - B^{1 / 2} A^{1 / 2}$.
Suppose $A$ in $\mathbb{R}^{p \times p}$ is a strictly positive defined symmetric matrix. Now suppose there exists a weakly consistent estimator for the matrix $A$, say $B_n$, which satisfies $\| B_n - A\|_{\infty} \rightarrow_p 0$.

Does any know the properties of
$$A^{1 /2} B_n^{1 / 2 } - B_n^{1 / 2} A^{1 / 2}.$$
Thanks so much!

PS: the property I would like to obtain is something like
$$\frac{\|A^{1 /2} B_n^{1 / 2 } - B_n^{1 / 2} A^{1 / 2}\|}{\|A - B_n\|} \, \rightarrow_p 0$$
under some proper norm $\|\cdot\|$ as $n$ goes to infinity. Does this possible?

答:
Let $C:=A^{1/2}$ and $D:=B^{1/2}$. Then it is not necessarily the case that $$\frac{\|CD-DC\|}{\|C^2-D^2\|}\to0$$ as $D\to C$.

Here is a counterexample. Let $C:=\begin{pmatrix}2&0\\0&1\end{pmatrix}$, $D:=\begin{pmatrix}2&\epsilon\\\epsilon&1\end{pmatrix}$, with $\epsilon\to0$, and $A=C^2$, $B=D^2$. Then $$\|CD-DC\|_F=\left\|\begin{pmatrix}0&\epsilon\\-\epsilon&0\end{pmatrix}\right\|_F=\sqrt2|\epsilon|$$
$$\|A-B\|_F=\|C^2-D^2\|_F=\left\|\begin{pmatrix}\epsilon^2 & 3 \epsilon \\
 3 \epsilon &\epsilon^2\end{pmatrix}\right\|_F=\sqrt2|\epsilon|\sqrt{9+\epsilon^2}$$ Hence the ratio $\frac{\|CD-DC\|_F}{\|A-B\|_F}\to\frac{1}{3}$ as $\epsilon\to0$.

N.B. All norms on matrices are equivalent, so it does not matter what specific norm one picks.


问:
How do I prove the formula of the general term(with induction) of this sequence?
$a_{n+2}=\frac{a_{n+1}+a_{n}+6}{2} ,a_{1}=1 , a_{2}=3$

That is the formula for the recursive form of the sequence,I found the general term formula for the sequence(the sequence was $1,3,5,7,....$) ,the formula is $a_{n}=2n-1$.So the statement for the induction is $P_{(n)}: a_{n+2}=\frac{a_{n+1}+a_{n}+6}{2}=2(n+2)-1$ so I tried to do with $P_{(k)}$ and with $P_{(k+1)}$ but i did not get anywere so I tried to say that for any even number the statement is true,so from that I said that $P_{(2k)}$ is true ,from it I got that $a_{(2k+1)}=4(2k+4)-2-a_{2k}-6$ knowing that for any even number the statement is true we can calculate $a_{(2k)}=2(2k+2)-1=4k+3$ ,from that I plugged it in $a_{2k+1}$ and got $a_{2k+1}=4k+5$ so I did the formula for $a_{2k+1}$ and got it true,so the formula is good.Am I wrong?(was I supposed to do it for $P_{(2k+1)}$,as well?)

答:
The base case is clear.

Now assume that $n \in \mathbb N$ and

$(*) \quad a_k=2k-1$ for $k=1,2,...,n.$

It follows that $a_{n+1}= \frac{a_n+a_{n-1}+6}{2}.$

From $(*)$ we get $a_n=2n-1$ and $a_{n-1}=2n-3.$

Now show that $a_{n+1}= \frac{a_n+a_{n-1}+6}{2}=2(n+1)-1.$


问:
Variational Calculus Optimization Problem with x-independent Lagrangian Solution via Beltrami Identity

Optimize $$\int_0^1 y^2 (y')^2 dx$$
subject to $y(0)=0, y(1)=1$

For x-independent Lagrangians it is easier to use Beltrami Identity: $$ F-y'{\frac{\partial F}{\partial y'}}=C $$

答:
According to Beltrami's formula

$$
y^2(y')^2-y'(y^2 2 y')= -y^2(y')^2= C
$$

then

$$
\left(\frac 12 \frac{d}{dx}y^2\right)^2= C_1
$$

or

$$
\frac 12 \frac{d}{dx}y^2 = C_2
$$

and

$$
\frac 13 y^3=C_3x+C_4
$$

and with the boundary conditions

$$
y = \sqrt[3]{x}
$$


问:
Find conditions on two parameters $c,d>0$ such that the asymptotics of an integral diverges
Let us start from defining:
$$\underset{c,d}{g(x)}=\displaystyle{\int_{\frac{1}{2}}^{x}}z^{c-1}(1-z)^{d-1}dz\tag{1}$$
and:
$$m(y)=\frac{2}{l^2}\displaystyle{\int_{\frac{1}{2}}^{y}}\frac{\underset{c,d}{g(x)}}{x^c(1-x)^d}dx\tag{2}$$
with $c,d>0$.

I would like to find conditions on $c,d>0$ such that:
$$\lim\limits_{y\to0^+}m(y)=\lim\limits_{y\to1^{-}}m(y)=\infty$$

Could you please help me understand how to proceed^?

答:
This is merely an attempt to inspire, i was not able to conclude this way.

So i think you can procede in this way.

$$g_{c,d}(x)=\int_{\frac{1}{2}}^xz^{c-1}\sum_{i=0}^{d-1}{{d-1}\choose{i}}(-z)^{d-i-1}dz=$$
$$=\sum_{i=0}^{d-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c-1}(-z)^{d-i-1}dz=\sum_{i=0}^{d-1}(-1)^{d-i-1}{{d-1}\choose{i}}\int_{\frac{1}{2}}^xz^{c+d-i-2}dz$$
where $i \le d-1 \Rightarrow$ $(c+d-i-2)\ge (c-1)>-1$, so there is no "log-case" for the intrgral:
$$\int_{\frac{1}{2}}^xz^{c+d-i-2}dz=\frac{1}{(c+d-i-1)}\biggl[ z^{(c+d-i-1)} \biggr]_{\frac{1}{2}}^x=\frac{x^{(c+d-i-1)}-\frac{1}{2^{(c+d-i-1)}}}{(c+d-i-1)}$$
so you would have
$$m(y)=\sum_{i=0}^{d-1}\frac{(-1)^{d-i-1}{{d-1}\choose{i}}}{(c+d-i-1)}\frac{2}{l^2}\int_{\frac{1}{2}}^y\frac{x^{(c+d-i-1)}-\frac{1}{2^{(c+d-i-1)}}}{x^c(1-x)^d}dx$$

Looking at this sum we can arrive at least at a partial conclusion if we consider the sign of the "relevant" term, noting that:

if $y\to 1^-$, $d\ge 1$ gives that the "relevant" integral diverges;

if $y\to 0^+$, $c \ge 1$ gives that the "relevant" integral diverges.

Hope this helps, and also that I made no mistakes.


问:
How is a matrix connected to a grid?
I have a hard time finding information about and understanding how a matrix (adjacency matrix) is connected to a grid used in numerical analysis. What would the nodes be and are the matrix weighted or unweighted? I understand that it may very depending on what the grid is used for, but if anyone could give me an example i would appreciate it.</p>

答:
For a graph of N nodes, the adjacency matrix will always be a square matrix of size N x N, and it will generally be symmetric too (though there might be exceptions to this). The nodes correspond to whatever data you are working with: they could be cities, pixels within an image, positions within the brain etc. Thats determined by the problem you are working on, but from the point of view of the adjacency matrix it doesn't matter what the underlying data represents.

The simplest form of adjacency matrix just contains 1 and 0 values denoting connections between nodes. For example, if on row A there is a 1 in column B, then node number A is connected to node number B. The symmetry means this works the other way round: column B will also contain a 1 in row A because node B must also be connected to node A. In this form, the adjacency matrix is basically a boolean structure denoting 'connection' or 'no connection'. More info here

More sophisticated adjacency matrices may have values reflecting the 'strength' of connection between two nodes. For example, if both B and C are connected to A, but C is much further away than B, then the weight of the connection may be smaller. That is, M(A,C) < M(A,B), where M is the adjacency matrix.

Finally, various forms of (discrete) Laplacian are closely related to the adjacency matrix, though they represent very different things. The Laplacian operator is a measure of curvature of a function (ie, second derivative), and in order to calculate this quantity, you need to know the neighbours for each given node (you can't calculate rates of change without knowing how the data varies from one node to the next). This is why the Laplacian is closely related to the adjacency matrix: calculating it requires adjacency information.

The 'graph' or 'mesh' Laplacian is the simplest one and is constructed by taking the normal adjacency matrix with 1/0 values and then modifying the diagonal values. In the standard adjacency matrix, diagonal elements are zero (because nodes being 'connected to themselves' doesn't count as a connection); in the mesh Laplacian, the diagonal is set to the negative sum of how many connections that node has. So, if a node is connected to 4 other nodes, then the diagonal element for that node will be -4. This means that every row or column of the Laplacian then sums to zero, which is beneficial in many use cases. This definition also extends to cases where the weights are not 1/0 values. This article explains graph Laplacians in more detail 2.

Finally, the concept of adjacency isn't just related to graphs. For example, in image processing, many operations are performed on neighbourhoods of pixels/voxels, so an adjacency matrix is useful to quickly identify the appropriate neighbourhoods.

A note on implementation: adjacency matrices are by nature very sparse (almost all elements are zero) and very large (size N x N). Hence, if doing any serious computing with them you'll need to use some kind of 'sparse data structure' to speed things up.

Title: How is a matrix connected to a grid?


问:
Conditional expectation is square-integrable
I am given the following definition:

Let $(G_i:i\in I )$  be a countable family of disjoint events, whose union is the probability space $\Omega$. Let the $\sigma$-algebra generated by these events be $\mathcal{G}$. Let $X$ be an integrable random variable, that is $E|X|<\infty$. The conditional expectation of X given $\mathcal{G}$ is given by $Y=\sum_iE(X1_{G_i})1_{G_i}/P(G_i)$.

I am told that $Y$ is square integrable, that is $E(Y^2)<\infty$. $1_A$ denotes the indicator function for the event $A$.

But I could not prove that $Y$ is square integrable. Can anyone offer a proof? Thanks.

答:
If $X$ is $\mathcal G$-measurable, then $Y=X$. Thus, with the only assumption that $X$ is integrable, there is no way to deduce that $Y$ is square integrable.

Counterexample: $X=\sum\limits_ix_i\mathbf 1_{G_i}$ with $\sum\limits_ix_iP[G_i]$ finite and $\sum\limits_ix_i^2P[G_i]$ infinite.

On the other hand, in the general case, if $X$ is square integrable then $E[Y^2]\leqslant E[X^2]$ hence $Y$ is square integrable.


问:
Starting position in rule of product problem-solving
Consider the following basic question:

With the numbers $0,1,2,3,4,5$, how many odd 3-digit numbers can we
write without repeating any of the numbers.

Albeit we can start from any position, starting from the left side will force us to branch our choices and use rule of sum to find the final answer. But starting from the right and then going left enables us to calculate the solution by merely doing $4\times4\times3$.

Now my question is that whether there is a concise way to find that "optimal path" in similar questions.

答:
You have to use commonsense.

In fact, to avoid branching for this particular problem, choosing from right to left won't eliminate branching,

you need to choose in the sequence $last\, digit,\,first\, digit\, and \,middle\, digit$,

thus writing in chosen order, $3\times4\times4$


问:
How to prove $m(A)＝0$
Assume $A\subset\mathbb{R}$ and is Lebesgue measurable. For $d$ arbitrary small, we have $d+A=A$. (here $d+A=\{d+a:a\in A\}$) Prove $m(A)＝0$ or $m(A^C)＝0$.

I assume $m(A^C)＞0$ and try to prove $m(A)=0$. But I can't figure out what to do next. It seems that I should construst some $f(x)$ corresponding to $A+x$ and then integral it by Fubini Thm to have $m(A^C)$.

答:
First assume that there is a dense sequence $(x_n)$ such that $x_n+A=A$ for all $n$. If $m(A)>0$ and $m(A^{c})>0$ then $\exists $ $a<b$ such that $(a,b)\subset
A-A^{c}.$

[Sketch of proof: the convolution $I_{A}\ast I_{A^{c}}$ is continuous. If it is
identically $0$ we get a contradiction by noting that its integral is $%
m(A)m(A^{c})>0$. Hence it is positive on some interval $(a,b)$].

Also there is an interger $n$ such that $-x_{n}\in (a,b).$ But then $-x_{n}=x-y$ with $%
x\in A,y\in A^{c}$ so $y=x+x_{n}\in x_{n}+A=A,$ a contradiction.

Now for the general case just observe that $\{nx_m:  n \in \mathbb Z, m \geq 1\}$ is dense in $\mathbb R$ where $(x_m)$ is sequence of distinct numbers tending to $0$ such that $x_m+A=A$ for all $m$.

Title: How to prove $m(A)＝0$


问:
Upper bound on iterations of a decreasing function
I have the following decreasing function:

$$n_0 =n$$
$$n_{i+1} = n_i \left ( 1 - \frac{4}{\ln n_i + 2} \right)$$

Given $1 < k \ll n$ what is a good upper bound for the number of iterations needed to reach a value $n_i < k$ ?

答:
Your sequence is decreasing, so as long as $\ln n_i > 2$:
$$
\begin{align*}
n_i &\le n_0 \\
\frac{4}{\ln n_i + 2} &\ge \frac{4}{\ln n_0 + 2} \\
1 - \frac{4}{\ln n_i + 2} &\le 1 - \frac{4}{\ln n_0 + 2} \\
n_i &\le n_0 \left( 1 - \frac{4}{\ln n_0 + 2} \right)^i
\end{align*}
$$
So the (easy) solution for $i$ to:
$$
\frac{k}{n_0} = \left( 1 - \frac{4}{\ln n_0 + 2} \right)^i
$$
gives an upper bound. Dunno if "good"...</p>


问:
Would this be the correct approach for solving y'' - 4y = 4H(t-2)
If not, what would be a better way?

Please see attached imageenter image description here

答:
You get from the general setup of such tasks $y(t)=0$ for $t<0$, which includes $y'(0)=0$. In consequence also $y(t)=0$, $y'(t)=0$ for $t<2$. For $t>2$ the equation reads
$$
y''-4y=4,~~y(2)=0,~~y'(2)=0
$$
with a solution
$$
y(t)+1=A\sinh(2(t-2))+B\cosh(2(t-2)),
$$
so that $B=1$ and $A=0$. In total
$$
y(t)=H(t-2)(\cosh(2(t-2))-1)=2H(t-2)\sinh^2(t-2).
$$

Please explain, in typed text, not another image, where the contemplation of the Laplace transform of $tf(t)$ for the treatment of the right side comes from.


答: The second coin, which has head on one side and tail on the other is assumed to be fair in the sense that both outcomes are equally likely. And we only need to consider the HH coin and HT coin for computing the indicated Pr because the TT coin can't yield H

Title: Fair coins or not

Body:
I am reading Introduction to Probability, 2nd Edition, and at the end of chapter one the following problem appears:

We are given three coins: one has 'heads' in both faces, the second has
tails in both faces, and the third has a 'head' in one face and a tail in the other. We
choose a coin at random, toss it, and the result is 'heads'. What is the probability that
the opposite face is tails?

I solve it assuming the general case where the coin might not be fair so I assume that the probability of having a 'head' in the second coin is p.

And the other two coins would not be affected by the nature of the coin.

But the solution manual which was written by the same author assumes that the coins are fair.

So what's is the right thing to assume?


问:
An ideal generated by a polynomial
What does it mean if an ideal is generated by a polynomial where the ring is Q[X]? e.g if the polynomial that generates the ideal is $x^2-5x+6$ does this mean that its scalar multiples, $nx^2-5nx+6n$ for $n \in \mathbb{Q}$ are also generators? And also are its factors i.e. $(x-3)(x-2)$ in the ideal too? So would the ideal also be generated by $(x-2)$?

答:
The ideal $I= (x^2-5x+6)$ generated by the polynomial $p(x)= x^2-5x+6$ is the subset

$$I= \{(x^2-5x+6) q(x) \mid q(x) \in \mathbb Q[x]\}.$$

In particular neither $x-2$ nor $x-3$ belong to $I$.

Title: An ideal generated by a polynomial


问:
Inner product of unit vector with a vector uniformly distributed on a hypersphere
Let $v$ be some vector in $\mathbb R ^n$ and $a$ a random vector uniformly distributed on the sphere in $n$ dimensions with a radius of $\sqrt n$. Show that:

$$\mathbb P(|\langle a, v \rangle| \geq 1/2 \|v\|_2) \geq 1/2$$

What I've tried

Since $v$ is arbitrary, by symmetry we can take the first coordinate axis unit vector ($e_1$) instead.

$$\mathbb P(|\langle a, v \rangle| \geq 1/2 \|v\|_2) = \mathbb P(|\langle a, e_1 \rangle| \geq 1/2)$$

Now since $\|a\|_2 = \sqrt n$ then:

$$\mathbb P(|\langle a, e_1 \rangle| \geq 1/2) = \mathbb P\left(|\langle s, e_1 \rangle| \geq \frac{1}{2\sqrt n}\right)$$

where $s$ is uniformly distributed on the surface of the unit sphere. I now need to lower bound this expression. Since $\langle s, e_1 \rangle$ is distributed as $|Z_1| / \|Z\|_2$ for $Z \sim N(0, 1)$ and since $\|Z\|_2$ concentrates around $\sqrt n$ then we can argue that the probability in question is at least $1/2$, but I need a non-asymptotic result (for all $n$, not just large $n$).

My more formal attempt is to use the Paley-Zygmund inequality, but that might be overkill for this application. I'm still struggling to evaluate (or get an upper bound for) the expectation of $X = |Z_1| / \|Z\|_2$. The expectation of $X^2$ is easy by linearity of expectation: $1/n$

答:
We shall follow a geometric approach.

Let $S\subset\Bbb R^n$ be a sphere of radius $r=\sqrt{n}$ centered at the origin $0$ of $\Bbb R^n$. If $v=0$ then $|\langle a, v \rangle|\ge 0$ for any $a\in S^{n-1}$. If $v\ne 0$ then it is easy to check that a set $\{a\in S:|\langle a, v \rangle| \geq 1/2 \|v\|_2\}$ consists of two spherical caps, cut from $S$ by hyperplanes orthogonal to $v$ and placed at a distance $1/2$ from the origin. It follows that each cap is spanned by an angle $2\psi$, where $\cos\psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}$.

When $n=1$ then $S$ consists of two points covered by the caps. When $n=2$ then $S$ is a circle, $\cos\psi=\tfrac 1{2\sqrt{2}}<\tfrac 1{1\sqrt{2}}=\cos\tfrac{\pi}4$, so $\psi>\tfrac{\pi}4$  and the caps cut more than half of the circle length.
So from now we assume $n\ge 3$.

We have that ($n-1$) dimensional area of $S$ equals $$A_n=r^{n-1}\frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}.$$
According to [Li], an area $C_n$ of the spherical cap equals $$r^{n-1}\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_0^\phi\sin^{n-2} \theta d\theta,$$ where, as I understood, $2\phi$ is the angle spanned by the cap.

Thus we have to show that the area remaining after cutting the caps is at most half  of the sphere area, that is

$\frac{2\pi^{(n-1)/2}}{\Gamma\left(\tfrac {n-1}2\right)}\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac 14 \frac{2\pi^{n/2}}{\Gamma\left(\tfrac n2\right)}$

$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}.$

Since $n\ge 3$,
$$\int_\psi^{\pi/2}\sin^{n-2} \theta d\theta\le \int_\psi^{\pi/2}\sin \theta d\theta=-\cos\frac \pi{2}+\cos \psi=\tfrac 1{2r}=\tfrac 1{2\sqrt{n}}.$$

Now we estimate the right-hand side of the inequality.

By Legendre’s formula, $$\Gamma\left(\tfrac {n-1}2\right)\Gamma\left(\tfrac n2\right)=\frac{\sqrt{\pi}}{2^{n-2}}
\Gamma(n-1)=\frac{(n-2)!\sqrt{\pi}}{2^{n-2}}.$$

1)) If $n$ is even then $\Gamma\left(\tfrac n2\right)=\left(\tfrac n2-1\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}= \frac{(n-2)!\sqrt{\pi}}{2^{n-2}\left(\tfrac n2-1\right)!^2}=\frac{\sqrt{\pi}}{2^{n-2}}{n-2\choose \tfrac {n-2}2}.$$

Robbins’ bounds imply that  for any positive integer $m$
$$\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m-1}\right)<{2m\choose m}<\frac {4^{m}}{\sqrt{\pi m}}\exp\left(-\frac {1}{8m+1}\right).$$

So
$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=\frac{\pi}{2^n}{n-2\choose \tfrac {n-2}2}>
\frac{\pi}{2^n} \frac {2^{n-2}}{\sqrt{\pi (n-2)/2}}\exp\left(-\frac {1}{4n-9}\right)=$$
$$\sqrt{\frac{\pi}{8n-16}}\exp\left(-\frac {1}{4n-9}\right)>\sqrt{ \frac{\pi}{8n}}\exp\left(-\frac {1}{7}\right)
>\frac{0.54}{\sqrt{n}}>\frac 1{2\sqrt{n}}.$$

2)) If $n$ is odd then $\Gamma\left(\tfrac {n-1}2\right)= \left(\tfrac {n-3}2\right)!$, so
$$\frac{\Gamma\left(\tfrac {n-1}2\right)}{\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-2}\left(\tfrac {n-3}2\right)!^2}{(n-2)!\sqrt{\pi}}=\frac{2^{n-2}}{(n-2)\sqrt{\pi}}{n-3\choose \tfrac {n-3}2}^{-1}.$$

If $n=3$ then the latter expression equals $\tfrac {2}{\sqrt\pi}>\tfrac 1{2\sqrt{n}}$. If $n>3$ then Robbins’ bounds imply that

$$\frac{\sqrt{\pi}\Gamma\left(\tfrac {n-1}2\right)}{4\Gamma\left(\tfrac n2\right)}=
\frac{2^{n-4}}{n-2}{n-3\choose \tfrac {n-3}2}^{-1}>\frac{2^{n-4}}{n-2}\cdot \frac {\sqrt{\pi (n-3)/2}}{2^{n-3}}\exp\left(\frac {1}{4n-11}\right)>\frac {\sqrt{\pi (n-3)/2}}{2(n-2)}> \frac 1{2\sqrt{n}}.$$

References

[Li] S. Li, Concise Formulas for the Area and Volume of a Hyperspherical Cap, Asian Journal of Mathematics and Statistics 4:1 (2011) 66–70.


问:
Evaluating the integral: $\lim_{R \to \infty} \int_0^R \frac{dx}{x^2+x+2}$
Please help me in this integral:

$$\lim_{R \to \infty} \int_0^R \frac{dx}{x^2+x+2}$$
  I've tried as usually, but it seems tricky. Do You have an idea?

Thanks in advance!

答:
$$\dfrac1{x^2+x+2} = \dfrac1{\left(x+\dfrac12 \right)^2 + \left(\dfrac{\sqrt{7}}2 \right)^2}$$
Recall that
$$\int_a^b \dfrac{dx}{(x+c)^2 + d^2} = \dfrac1d \left.\left(\arctan\left(\dfrac{x+c}d\right)\right)\right \vert_{a}^b$$
I trust you can finish it from here. You will also need to use the fact that
$$\lim_{y \to \infty} \arctan(y) = \dfrac{\pi}2$$</p>


问:
proof that a stopped martingale is a martingale?
Definition.

$\mathcal{F}_{\tau}=\{F\subset \Omega:\; \forall n \in N \cup \{\infty\},\; F\cap(\tau\leq n)\in \mathcal{F}_{n}$} is a sigma-algebra.

Definition.

$\forall \omega \in \Omega:\; M_{n}^{\tau}(\omega) \;=\; M_{\min(\tau(\omega), n)}(\omega).$

Question.

Prove that this collection of functions is also a martingale with respect to the given filtration.

(First try to prove that the following is true: $\,M_{n}^{\tau} \;=\; M_{n-1}^{\tau}+1_{\tau \le n}(M_{n}-M_{n-1}).\,$)

My Attempt.

First of all, I am not able to prove that $M_{n}^{\tau}=M_{n-1}^{\tau}+1_{\tau \le n}(M_{n}-M_{n-1})$. It would be great if someone could help me with that!

Secondly, if I assume that the hinted expression for $M_{n}^{\tau}$ is correct, I continued as follows:

We want to show that $E(M_{n}^{\tau}\mid \mathcal{F}_{n-1})=M_{n-1}^{\tau}$.

By assumption (now)

$$E(M_{n}^{\tau}\mid \mathcal{F}_{n-1}) \\=\, E\left(M_{n-1}^{\tau}+1_{\tau \leq n}(M_{n}-M_{n-1})\mid \mathcal{F}_{n-1}\right) \\=\, M_{n-1}^{\tau}+E\big(1_{\tau \leq n}(M_{n}-M_{n-1})\mid \mathcal{F}_{n-1}\big) \\=^{?}\, M_{n-1}^{\tau}.$$

Can someone help me to continue and complete this proof?

Body:
A bit late to the party, but for completeness's sake, in order to show that $M^{\tau}_n = M^{\tau}_{n-1} + 1_{\tau \geq n} (M_n - M_{n-1})$, just note that, for any $k$,

$$ M^{\tau}_k = \sum_{j=0}^{k-1} M_j 1_{\tau = j} + M_k 1_{\tau \geq k}. $$

Thus,

$$ M^{\tau}_n = \sum_{j=0}^{n-1} M_j 1_{\tau = j} + M_n 1_{\tau \geq n} = \sum_{j=0}^{n-1} M_j 1_{\tau = j} + M_n 1_{\tau \geq n} + M_{n-1} 1_{\tau \geq n} - M_{n-1} 1_{\tau \geq n}$$

and since $M_{n-1} 1_{\tau \geq n} + M_{n-1} 1_{\tau = n-1} = M_{n-1} 1_{\tau \geq n-1}$ we get

$$  M^{\tau}_n = \sum_{j=0}^{n-2} M_j 1_{\tau = j} + M_{n-1} 1_{\tau \geq n-1} + (M_n - M_{n-1}) 1_{\tau \geq n} = M^{\tau}_{n-1} + 1_{\tau \geq n} (M_n - M_{n-1}). $$


问:
Why do we represent complex numbers as the sum of real and imaginary parts?
We know complex numbers have a real and an imaginary part. We can visualize the imaginary part by adding another axis, hence a plane for both parts.

My question is really simple but may be unusual.

Why is the plus sign used to represent a complex number? $z=a+bi$. Is it just historically common to use the plus sign, or does one actually do arithmetic by using the sign? Why can't it be represented by a comma, for example, like a vector?

$z=a+bi$
versus
$z=a,bi$

I'm really curious to know if $a+bi$ is a way of telling if the real and imaginary parts are connected or if it is a mathematical expression.

答:
A natural way to think of complex numbers is to represent a typical complex number as a $2\times2$ matrix

$$\begin{bmatrix}
    x & y\\
    -y & x \\
    \end{bmatrix},$$
where $x$ and $y$ are real. This matrix may be decomposed as $$\begin{bmatrix}
    x & 0\\
    0 & x \\
    \end{bmatrix}+\begin{bmatrix}
    0 & y\\
    -y & 0 \\
    \end{bmatrix}=x\begin{bmatrix}
    1 & 0\\
    0 & 1 \\
    \end{bmatrix}+y\begin{bmatrix}
    0 & 1\\
    -1 & 0 \\
    \end{bmatrix}.$$We may identify the first matrix as the real part and the second as the imaginary part of the complex number, with $x$ and $y$ respectively denoting the real and imaginary coefficients. Considered geometrically, addition of such a matrix represents a shift in the plane, while multiplication by the above matrix represents the combination of scaling by a factor $r:=\surd(x^2+y^2)$ and an anticlockwise rotation through the angle $\theta$ defined by $$x=r\cos\theta\quad\text{and}\quad y=r\sin\theta.$$
The algebraic operations (addition, multiplication, etc.) and properties of complex numbers then arise entirely from the corresponding operations and properties in the algebra of $2\times2$ matrices.


问:
Evaluate $\lim_ {n \to\infty} n (n^ {\frac {1} {n}}-2^ {\frac {1} {n}}) ^a$ as $a$ varies in the reals
How would you evaluate $\lim_ {n \to\infty} n (n^ {\frac {1} {n}}-2^ {\frac {1} {n}}) ^a$ with $n$ an integer and $a$ a real parameter? I tried to apply the most common criteria and to compare this with some other easier functions but without success.
Thank you

答: Now I wasn't able to figure out a closed form but maybe it has something to do with the idea that I will present here. We will try different values for the parameter a. If a is equal to $0$ then obviously you just have the limit $\lim_{n\to\infty}n$, which diverges to $\infty$. If you let a be in the interval $(0;1]$ then it will diverge to $\infty$ once more. The same goes for the interval $(-\infty;0)$. The only place where this limit would converge is for a in the interval $(1;\infty)$, where the resulting limit will attain a value of $0$. Hopefully this helps, although someone more profficient in maths will surelly offer you a better advice.


答:
When thinking about ideals keep in mind the analogy between Vector space and subspace. If $V$ is a vector space over a field $F$ and $W$ is a subspace of $V$ means if $a,b \in W$ then $a+b \in W$ and $\lambda a\in W,\: \lambda \in F$.

Same way an ideal is defined as a subspace is closed under addition and the scalars are coming from a ring rather than a field.

Here the ring is $\mathbb{Q}[X]$. So the ideal generated by the "vector" $x^2 - 5x+6$ would be, by definition must contain all the "scalar" multiples which will be $p(x) \in \mathbb{Q}[X]$. Hence the ideal is $\{p(x)(x^2-5x+6) : p(x) \in \mathbb{Q}[X]\}$

As for the next question, in the ring $\mathbb{Q}[X]$ there are some elements called the units. They are nothing but elements whose multiplicative inverse also exist in that ring. In $\mathbb{Q}[X]$ the units are precisely the elements of $\mathbb{Q}$. An ideal doesn't change as long its generators are multiplied by units. Hence $n(x^2-5x+6)$ are also generators of the ideal for every $n\in \mathbb{Q}$.

The ideal generated by $x-2$ is "bigger" than the ideal generator $x^2 - 5x +6$ because the former contains $x-2$ but no matter with which polynomial you multiply $x^2 - 5x +6$ you'll never get $x-2$ because the degree always increases when you multiply by a non zero polynomial

Title: An ideal generated by a polynomial

Body: What does it mean if an ideal is generated by a polynomial where the ring is Q[X]? e.g if the polynomial that generates the ideal is $x^2-5x+6$ does this mean that its scalar multiples, $nx^2-5nx+6n$ for $n \in \mathbb{Q}$ are also generators? And also are its factors i.e. $(x-3)(x-2)$ in the ideal too? So would the ideal also be generated by $(x-2)$?


问:
Confusion in conditional convergence
As we know if a series is conditionally converges to a some finite value L , we can change this value(L) of convergence to some other other value just by changing the order of series .

But, is it possible to change the order in such a way that the new series becomes divergence?

and if it is not possible then how to prove it ?

答:
Hint

Yes this is possible. The idea is that as $\sum \vert a_n \vert$ diverges, you can add enough positive terms in order for $\sum a_{\sigma(n)} $ to be larger than any $N \in \mathbb N$, bring only a negative term and so on.

Title: Confusion in conditional convergence


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答:
Great video lecture series by Benedict Gross at Harvard:

http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra

Outstanding short pieces on many facets of algebra by Keith Conrad at UConn. Very clear exposition and lots of intuitive discussion any plenty of warnings - teaching at its best(even one on Rubik"s cube):

http://www.math.uconn.edu/~kconrad/blurbs/


问:
Can an Integral Domain be considered a Vector Space?
If $D$ is an integral domain that contains a field $F$. Can $D$ be considered a Vector Space over $F$?

答: For a vector space $(V,+)$ should be an abelian group. Your integral domain is a ring. With respect to the first operation $+$ if that is abelian then you can certainly form a vector space


问:
If $\;\lim\limits_{x\rightarrow 10}g(x)=a\;$ and if$\;\lim\limits_{x \rightarrow10}\dfrac{f(x)}{g(x)}=b$, find $\lim\limits_{x\rightarrow10}f(x).$
If $\;\lim\limits_{x \rightarrow10}g(x)=a\;$ and if $\;\lim\limits_{x\rightarrow10}\dfrac{f(x)}{g(x)}=b\;,\;$ where $\;a,b\neq 0\;,\;$ what's $\;\lim\limits_{x\rightarrow10}f(x)\;?$

The answer is insufficient information to determine. But I thought there was a "limit multiplication rule" I could use to determine it's $ab$? Can someone give an example when this doesn't hold?

[EDIT] based on below discussion, I guess the question boils down to, can we conclude $\lim\limits_{x\rightarrow10}f(x)$ exists based on above conditions?

I am also curious if the answer changes if $b=0$ or $a=0$?

答:
$f(x)= \frac{f(x)}{g(x)} \cdot g(x)$ for $x $ in a neighborhood of $10$, since $a \ne 0.$

Hence

$$f(x)= \frac{f(x)}{g(x)} \cdot g(x) \to b \cdot a$$

as $x \to 10.$


答:
Here:
https://en.wikipedia.org/wiki/Riemann_series_theorem

you can find a proof for the existence of a rearrangement that diverges to $ \infty.$

Title: Confusion in conditional convergence

Body:
As we know if a series is conditionally converges to a some finite value L , we can change this value(L) of convergence to some other other value just by changing the order of series .

But, is it possible to change the order in such a way that the new series becomes divergence?

and if it is not possible then how to prove it ?


问:
An estimate on number of solutions $e^z=a$ in $|z|\leq r.$
Let $n(r,a)$ be the number of solutions of $e^z=a$ in $|z|\leq r, $ where $a\in\mathbb{C}\setminus\left\{0\right\}.$

I have to estimate this number.

Attempt: Writing $a=|a|e^{i\theta}.$ Then $e^{z}=a$ gives $$z=\log|a| + i\left(\theta + 2k\pi\right),$$ where $k\in\mathbb{Z}.$

For these solutions to lie inside $|z|\leq r,$ we must  have
$$|\log|a| + i\left(\theta + 2k\pi\right)|\leq r$$ which means that $$(\log|a|)^2 + (\theta + 2k\pi)^2\leq r^2$$
This finally gives $$k\leq\frac{\sqrt{r^2-(\log|a|)^2}}{2\pi}-\theta/2\pi.$$
It has been proved that $$n(r,a)=\frac{\sqrt{r^2-(\log|a|)^2}}{\pi} + O(1).$$ However, I am unable to understand how to arrive at this estimate.

答:
Your calculation is correct up to
$$
(\log|a|)^2 + (\theta + 2k\pi)^2\leq r^2 \, .
$$
That is equivalent to
$$
 |\theta + 2k\pi| \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \sqrt{r^2 - (\log|a|)^2} \le \theta + 2k\pi \le \sqrt{r^2 - (\log|a|)^2} \\
\iff - \frac{\sqrt{r^2 - (\log|a|)^2}}{2 \pi} - \frac{\theta}{2 \pi}
\le k \le \frac{\sqrt{r^2 - (\log|a|)^2}}{2 \pi} - \frac{\theta}{2 \pi} \, .
$$
So $k$ lies in an interval of length $\frac{\sqrt{r^2 - (\log|a|)^2}}{\pi}$. It follows that
$$
\frac{\sqrt{r^2 - (\log|a|)^2}}{\pi} \le n(r, a) \le \frac{\sqrt{r^2 - (\log|a|)^2}}{\pi} + 1
$$
because a closed interval of length $L$ contains between $L$ and $L+1$ integral points.


问:
Marginalization of conditional probabilities
An exam question asked to select all expressions that are equal to 1, given no independence assumptions.
The solution stated that the following expression was not equal to 1, but I don't understand why.
Is it possible this is a mistake?

$\sum_{a}\sum_{b}P(A=a | B=b)$

答:
For a fixed $b$, the sum is $1$ since you go over all possibilities of $A$. But then you also sum over all values of $b$, which means you sum over many $1$s.

For example, suppose you throw a dice. $A$ is the value of it, $B=1$ is it is even, $0$ if odd. Then this sum should be $2$, as given even, the sum of all options of $A$ is $1$ and given odd the sum of all options of $A$ is $1$.


问:
If $A$ is a subgroup of $B$, just it hold that $\text{Aut}(A)$ is a subgroup of $\text{Aut}(B)$?
Let $\text{Aut}(G)$ denote the group of automorphisms of $G$ and let $A\subseteq B$ denote $A$ is a subgroup of $B$. Does the following hold:

$$A\subseteq B\implies \text{Aut}(A)\subseteq \text{Aut}(B)$$

If not, is there a necessary and sufficient condition for this to hold?

答:
It is certainly not true in general.

For a counterexample take $A:=S_6$ whose automorphism group is well-known to have order $2\cdot 6!$; and take $B:=S_7$ whose automorphism group is well-known to have order $7!=7\cdot 6!$.


问:
Line integral on piece of helix

Calculate $\int_\gamma fdx$ for $f(x)=(x_1^2+5x_2+3x_2x_3,5x_1+3x_1x_3-2,3x_1x_2-4x_3)$ and $\gamma$ is the helix with radius $1$ and pitch $1$ from $(0, 1, 0)$ to $(0, 1, 2\pi)$

Now, the coordinates of a circular helix of radius $a$ and pitch $2\pi b$ are:
$(x(t),y(t),z(t))=(a\cos{t},a\sin{t},bt)=(\cos{t},\sin{t},t)$ since $a = 1$ and $b=1$

My definition for calculating the line integral would be:
$\int_\gamma fdx=\int(f(\gamma(t)))·\gamma'$

My doubt here is the parametrization of $\gamma(t)$, I don't know for which values of $t$ :

Let $\gamma(t)=(\cos{t},\sin{t},t)$, for $t\in[0,\pi]$? For $t\in[0,2\pi]$? For $t\in$ something else?

答:
As $\gamma$ starts at $(0,1,0)$, and not at $(1,0,0)$, you have
$$\gamma:\quad t\mapsto \gamma(t)=(-\sin t,\cos t, t)\qquad(0\leq t\leq2\pi)\ ,$$
and then have to calculate
$$\int_\gamma f\cdot dx=\int_0^{2\pi}f\bigl(\gamma(t)\bigr)\cdot\gamma'(t)\>dt\ .$$
This example has a special flavor: The force field $f$ is in fact the gradient of some scalar function $F$, i.e., $$f(x_1,x_2,x_3)=\nabla F(x_1,x_2,x_3)\ .$$
If you can guess $F$ the value of the integral is then given by
$$\int_\gamma f\cdot dx=F(0,1,2\pi)-F(0,1,0)\ .$$
In fact $F(x_1,x_2,x_3)={1\over3}x_1^3+5x_1x_2+3x_1x_2x_3-2x_2-2x_3^2$.

Title: Line integral on piece of helix


问:
Differentiable on open unit disc
I have come across the following questions on a past exam paper, but I'm not sure how to go about answering the,. Any help would be greatly appreciated....

(i) Does there exist a function $f$ differentiable on the open unit disc $\{z\in \mathbb{C}: |z|<1\}$ such that $f(1/n)=f(-1/n)=1/n^2$ for all $n=1,2,3,\ldots$?

(ii) Does there exist a function $g$ differentiable on the open unit disc $\{z\in \mathbb{C}: |z|<1\}$ such that $g(1/n)=g(-1/n)=1/n^3$ for all $n=1,2,3,\ldots$?

答:
By differentiable, I assume you mean complex differentiable, i.e. holomorphic.

(a) How about $f(z) = z^2$?

(b) Compare the presumptive $g$ to $z^3$ and $-z^3$ and use the identity theorem for holomorphic functions.

Title: Differentiable on open unit disc


问:
$X$ and $Y$ are $\text{unif}(0,1)$. Find the Joint Distribution of $S$ and $U$, where $S=X+Y$, $U=\min(X,Y)$
$X$ and $Y$ are $\text{unif}(0,1)$. Find the Joint Distribution of $S$ and $U$, where $S=X+Y$, $U=\min(X,Y)$.

I know the density of $S$ is $f(s)=s$ for $0<s<1$, $f(s)=2-s$ for $1<s<2$, and $0$ otherwise.
The density of $U$ is $f(u)=2-2u$, for $0<u<1$.
How to find the joint distribution $f(s,u)$?

答:
As I stated in my comment, without assuming independence between $X$ and $Y$ the problem cannot be solved.

Assuming independence and using the jacobian method, easy find that

$$f_{US}(u,s)=2\cdot \mathbb{1}_{(0;1)}(u)\cdot \mathbb{1}_{(2u;1+u)}(s)$$

that is defined in the following purple area

Just for a double check, you can derive the marginals

$$f_U(u)=2\int_{2u}^{u+1}ds=2(1-u)$$

$$f_S(s)=2\int_{0}^{\frac{s}{2}}du=s$$

for $0<s<1$

and

$$f_S(s)=2\int_{s-1}^{\frac{s}{2}}du=2-s$$

for $1<s<2$

as already known

The easiest way to derive the joint distribution of $(S,U)$ is the following.

Let's avoid to assume independence between $X,Y$ but state that $X<Y$ now $f(x,y)=2$, the jacobian is 1 and the $f(s,u)$ is again uniform in the support I showed. Note that the support area is 0.5, so perfectly coherent with the joint density =2.


问:
Filling a piecewise continuous linear shape with a constant volume of liquid
We have a piecewise continuous linear function (representing topography). The shape is to be filled with a constant volume of liquid (representing an ocean). How can we find the 'sea level', and where are the boundaries of this ocean?

As an example, here is a region defined by four points: Filling problem example.

The forward problem of finding the volume $V$ of the liquid given a known liquid level $l$ is a quadratic:

$$ V = -\frac{1}{2}(l-y_1)^2\frac{x_1-x_0}{y_1-y_0} + \frac{(2l-y_1-y_2)(x_2-x_1)}{2} + \frac{1}{2}(l-y_2)^2\frac{x_3-x_2}{y_3-y_2} = 1.625 $$

This quadratic can be easily simplified and solved for $l$ if $V$ is known, and for the example problem where $l=2.5$ we obtain $V = 1.625$.

However, this method requires prior knowledge of which points the boundary of the liquid will lie between, and so does not generalise. Nonlinear optimisation tools can provide a numerical answer, but I am ideally looking for solutions that might cast the problem in terms of a linear or quadratic programming problem, or some other efficient method of solution that might ideally be able to be generalised to the 2D case.

答:
The good news is that the volume is a monotone function of the depth. So you can simply do binary search on the depth. And if you've worked out that
$$
V(a) < T
$$
and
$$
V(b) > T
$$
where $T$ is the target volume, and $V(d)$ denotes the volume of water when the depth is $d$, then in computing $V(\frac{a+b}{2})$, it's possible that you can simply compute the volume between depth $b$ and depth $\frac{a+b}{2}$, re-using much of your prior work. But that'd take some effort to get right.

You could also do a little preprocessing, looking at each "cell" (presumably something like a quad in $xy$, divided by a well-chosen diagonal), and identifying the minimum $z$ coordinate (for depths below that, this cell can be ignored), the max $z$ coordinate (for depth $d$ above this level $s$, the cell contributes $\Delta d \cdot A$ when the depth increases by $\Delta d$, where $A$ is the area of each $xy$-rectangle), and the volume within the cell when the depth is exactly $s$. Those numbers let you rapidly estimate an upper and lower bound on the volume at some depth, so that your search-region gets small.

Anyhow, assuming an $n \times n$ grid of data, there are $n^2$ depth values, and the right answer lies between some two of these. Identifying which two involves $\log (n^2) = 2 \log n$ evaluations of volume, each of which takes time proportional to $n^2$ unless you do something clever with clustering your depths into larger blocks in some spatial data structure, so this gives about $n^2 \log n$ runtime. That's not too bad, when you realize that $n^2$ is the size of the input data rather than $n$.

By the way, you should expect this problem to be hard. Imagine a (1D) region of rough bottom topography, more or less sawtooth, and a target depth that's about half way up the "tooth". In this situation, every single cell in the domain is involved in the depth computation at the critical depth.

If, on the other hand, your ocean bottom has a gradual rise towards the shore, then at the critical level, exactly one cell will be involved in the computation, so some instances of the problem should be reasonably straightforward.

Wait...it's easier than that. Still not pretty, but...that's OK.

Let's assume the grid-squares are unit squares, i.e., the domain grid is something like $\{1, \ldots,  n\} \times \{1, \ldots, n \}.$

Let's look at one "cell", the rectangular column of liquid above a unit square. To each corner of this cell is associated a height, which I'll call a $z$-value. Let's call them $z_{00}, z_{01}, z_{10}, z_{11}$. You say that the depth is piecewise linear, so lets make a triangulation: in the domain square $0 \le x,y \le 1$, we draw a diagonal from $(0,0)$ to $(1,1)$. Now instead of a rectangular column of liquid, we have two triangular columns, each with base-area $\frac12$. Let's take one, with heights I'll call $z_A, z_B, z_C$; to make things simple, let's assume $z_A \le z_B \le z_C.$ If we plot the volume of water in this triangular cylinder when the water-depth is $d$, and vary $d$, we'll get an increasing function $V(d)$.

For $d < \min(z_A, z_B, z_C)$, we'll have $V(d) = 0.$ For $z_A \le z \le z_B$, we'll have $V(d)$ is some cubic function of $d$, which can be expressed in terms of $z_A, z_B, z_C$ by doing some geometry which I'm not prepared to write out, because...it's just basic geometry. For $z_B \le z \le z_C$, we have a similar result. But for $z > z_C$, $V(d)$ has the particularly nice form
$$
V(d) = K + \frac{1}{2} (z - z_C).
$$

Each of the four "pieces" of this function $V$ is a cubic whose coefficients depend only on $v_A, v_B, v_C$. So you can represent it by four numbers (the coefficients $p, q, r, s$ of $d^3, d^2, d,$ and $1$).

So let's represent each of our $2n^2$ "half-cells" (triangular prisms) by a list of four quintuples, each of which looks like this:
$$
(e; p, q, r, s)
$$
indicating that starting at depth $e$, the next "segment" of the function is represented by the quadratic $d \mapsto pd^2 + qd + r$. So if the depth $z_A, z_B, z_C$ are $1, 2, 5.3$, then our four quadruples would look look
$$
(0; 0,0,0, 0)\\
(1; *, *, *, *) \\
(2; *, *, *, *) \\
(5.3; 0, 0, \frac12, *)
$$

Now suppose that we want to compute volume as a function of $d$. We have $n^2$ depth-values at the grid cells, and between any two of these, the depth is a sum of $n^2$ cubics, one contributed by each grid cell. But a sum of cubics is still a cubs, so from the $10n^2$ data items you've got, you can build a table, sorted by depth, that looks just like the list of 4 quintuples above: each item is a depth, followed by the coefficients of the cubic volume-function starting at that depth and in use until the next depth.

It might appear that at each of the $n^2$ depth values, you'd need to sum up all the $n^2$ cubics, but ... assuming all depths are distinct for the moment ... at each depth value, at most four cells change their cubics, so you need only deal with a constant amount of work.

In short, in $n^2$ time you can build a table describing the entire piecewise-cubic depth function. And then you can do binary search to find the two $z$-depths between which your goal-volume occurs, and then you just solve the cubic for the depth. (You could use Cardano's formula, but numerical methods will work just fine here.)

So...building the table takes time proportional to your input size $n^2$. Binary search on that takes time $\log (4n^2) \approx \log n$. And the you solve one cubic. All in all: linear in the input size, which is pretty good!