问:
Reference for Exercise 5.3.1 in [Probability: Theory and Examples, 4th edition, Durrett, 2010]
Exercise 5.3.1 on page 205 of [Probability: Theory and Examples, 4th edition, Durrett, 2010] reads:

Exercise 5.3.1. Let $X_n$, $n \ge 0$, be a submartingale with $\sup X_n < \infty$. Let $\xi_n = X_n − X_{n−1}$ and suppose $\mathbb{E}(\sup \xi_n^+) < \infty$. Show that $X_n$ converges a.s.

In [Probability: Theory and Examples, 5th edition, Durrett, 2019], the same question appears as Exercise 4.2.4. on page 223.

I would like to cite this result, but I hesitate to cite an exercise. Could anyone kindly provide a reference for this conclusion, or for a variant similar to the following?

Theorem. Let $X_n$ be a submartingale with  $\mathbb{E}(\sup (X_n-X_{n-1})^+) < \infty$ (or more strongly, $\sup|X_n-X_{n-1}|\le c <\infty$, which is adequate for my application).
Then
$$
\mathbb{P}\left(\{X_n \text{ converges}\}\bigcup \{\sup X_n = \infty\}\right) = 1
$$

I would be surprised if this kind of result has never been mentioned in any other monographs/textbooks, although I have found nothing after searching.

Any comments or criticism will be appreciated. Thank you.
(I am looking for a reference. Nevertheless, if you would like to post a proof, it is also very welcome.)

答: If you are okay with the stronger statement(uniform bounded increments), then you can cite Theorem 4.3.1 on Durrett's PTE(5th ed.) directly, just several pages after the exercise. It works for submartingales as well, as suggested by @Jason, you just need to change $D$ to be $\{\limsup X_n=+\infty\}$.


问:
How close do multiples of 3 different unit fractions sum to 1?
I am wondering how to solve or make advance in the following problem.
Given 3 different unit fractions, i.e. $\frac1{n_i}$ for $n_i\in\mathbb{N}$, and gcd($n_i,n_j$)=1. Hence no multiples of two or three of these numbers sum to 1.

I want to add multiples of at least two of these three numbers such that their sum is less than 1 but as close as possible. So I want $$\frac1{n_1} k_1 + \frac1{n_2} k_2 + \frac1{n_3} k_3\text{  for }k_i<n_i,\ k_i\in\mathbb{N}$$ to be as close as possible to 1 and I want to know how close it can get.

Ist there any way to determine a upper bound on how close they sum to 1?

I am sure this is possible by enumerating all possibilities and always solving a linear diophantine equation for 3 numbers but this is inefficient and I was more thinking about properties on the size of the smallest $\frac1{n_i}$.
I was thinking that one of the numbers must be very small in order for a possible sum to be very close to 1. Is there any way to make this precise?

Any ideas?

Thanks

答:
There are infinitely many cases where Jyrki's lower bound $1/(n_1 n_2 n_3)$ is achieved.  Consider for example

$$ \frac{m-1}{2m-1} + \frac{1}{2m} + \frac{m}{2m+1} = 1 - \frac{1}{(2m-1)(2m)(2m+1)}$$

EDIT: Another example:

$$ {\frac {m}{8\,m+1}}+{\frac {2\,m+1}{8\,m+3}}+{\frac {5\,m+3}{8\,m+5}}=
1-{\frac {1}{ \left( 8\,m+1 \right)  \left( 8\,m+3 \right)  \left( 8\,
m+5 \right) }}
$$

ContentLicense: CC BY-SA 3.0


问:
Isomorphic as $R$-modules v.s. Isomorphic as abelian groups
Let $(R,\mathfrak m, \mathbb Q)$ be a Noetherian local ring. Let $M$ be a finitely generated $R$-module such that for some integer $n\ge 0$, there is an isomorphism of abelian groups $M \cong \mathbb Q^{\oplus n}$.

My question is: Is $M$ Isomorphic, as an $R$-module, to a direct sum of copies of $\mathbb Q$? i.e., is it true that $\mathfrak m M=0$?

答: No : take $R=\mathbb Q[\epsilon] = \mathbb Q[X]/(X^2)$; then $R$ itself satisfies your hypotheses.

ContentLicense: CC BY-SA 4.0


问:
Convergence of Isoclines
Suppose we have a differential equation $y' = f(x,y)$, then I know isoclines are defined as :

$S_c = \{(x,y)|f(x,y) = c\}$ where $c$ is any real number.

Now, consider $y'= x-y$, for this equation isoclines are given by :

$x-y =c$, Now , graphically I can see that all the isoclines asymptotically converge to $x = y+1$

But, how do I show this fact analytically ?

Any hints will be helpful.

答:
Let us solve the linear differential equation
$$y'+y=x$$

Without RHS
$$y_h'=-y_h \implies y_h=\lambda e^{-x}$$

A particular solution $y_p=\lambda(x) e^{-x} $ satisfies

$$\lambda'(x)=xe^x$$
or
$$\lambda(x)=e^x(x-1)$$

So, the general solution will be
$$y=y_h+y_p=\lambda e^{-x}+x-1$$

When $ x\to +\infty\;  $, $ \;\lambda e^{-x} \to 0 $ and

$ \; f(x,y)= x-y\to 1$.

Title: Convergence of Isoclines


问:
$n > 1$, consecutive integers $a, a+ 1, . . . , a+ (n−1)$. Prove that the product thereom is always a multiple of $n!$
Consider $n > 1$ consecutive integers $a, a+ 1, . . . , a+ (n−1)$. Prove that the product thereof is always
a multiple of $n!. (Think about a number of the form C
y
x
.)

答:
(Not a native English speaker, so please pardon my mistakes)

Sorry, but I can't think of anything using C y x. But the proof below may be the most common one.

Consider a mod n, (a + 1) mod n, ... (a + (n - 1) ) mod n. (1)

Obviously, these values are all different. There are n values.

On the other hand, the possible results of (a number mod n) include 0, 1, ..., n-1. There are also n values.

So we can say that the values in (1) take up 0, 1, ..., n-1. Therefore, one of them must be 0, so there must be a number in a, a+1, ..., a+n-1 that is a multiple of n. So the product must be a multiple of n.


问:
Chance for two numbers to be same if first k digits matches
Two people pick two numbers (same number of digits say n). What is the chance that the number will be same number if their first k digits matches.</p>

答: If they both pick all the digits randomly and independently from each other, then the probability in the question is the same as the probability that the pick the remaining $n-k$ digits the same. This is $0.1^{n-k}$ since whatever $n-k$ remaining digits the first person chose, the second person will have that given probability of choosing the same digits.


问:
Name for this map of power sets associated with a function?
Given a function $f: X \to Y$, there are two associated maps of power sets:  the direct image $f^{\to}: \mathcal{P}(X) \to \mathcal{P}(Y)$ and the inverse image $f^{\leftarrow}: \mathcal{P}(Y) \to \mathcal{P}(X)$.

Consider the map

$$
f^?: \mathcal{P}(X) \to \mathcal{P}(Y)
$$

defined by

$$
f^?(A) = \{y \in Y: f^\leftarrow(\{y\}) \subset A\}
$$

In words, $f^?(A)$ is the set of all $y \in Y$ so that every $x$ which maps to $y$ is in $A$.

Does this map have a standard name and notation?

Note:  I became interested in this while reading this nCategory Cafe post.  Knowing the context is not needed to answer the question, but $f^?$ is associated with universal quantification in the sense of the post.

答:
(Adapted from an email I sent the OP, who was also asking about connections with "the six operations" of Grothendieck.)

I guess it might depend which community you're talking to. I believe the notation $\exists_f$ (for direct image along $f$) and $\forall_f$ notation (for the thing OP is asking about) is somewhat standard for those working in topos theory and categorical logic. For example, you see both notations in Mac Lane-Moerdijk, Sheaves in Geometry and Logic, Proposition IV.9.3 and just before. In the same place you will see the parallel notation for functors between slice toposes, $\Sigma_f: E/X \to E/Y$ and $\Pi_f: E/X \to E/Y$.

(These notations make sense when you consider that

$$f^\to(A) = \exists_f(A) = \{y \in Y: \exists_{x \in X}\; y = f(x) \wedge x \in A\}$$

$$f^?(A) = \forall_f(A) = \{y \in Y: \forall_{x \in X}\; y = f(x) \Rightarrow x \in A\}$$

These notations may have begun with Lawvere, who uses it e.g. in his 1970 paper Quantifiers and Sheaves.)

For some reason, $\forall_f$ seems to be the "weak sister", much less well-known than her brother the direct image along $f$. I am not aware of any standard name or notation for it in the traditional logical literature. And yet, it plays a role here and there. For example, if X is a compact space with a countable base $B$ and $f: X \to Y$ is a continuous surjection to a Hausdorff space, then $Y$ also has a countable base; the most natural choice for that base is $\{\forall_f(U): U \in B\}$. Another example: it's more direct to prove that if $X$ is compact and $Y$ is a space, and $p: X \times Y \to Y$ is projection, then the operator $\forall_p$ takes open sets to open sets, than it is to prove that the direct image $\exists_p$ takes closed sets to closed sets (which is the De Morgan dual statement).

In representation theory, the corresponding operations are induction and coinduction (of a $G$-representation along a group homomorphism $f: G \to H$) that are left and right adjoint to the restriction functor along $f$. These can also be described by (enriched) Kan extensions. So $Res_f$ would be like pullback $f^\ast = f^\leftarrow$, $Ind_f$ would be like $\exists_f$, and $Coind_f$ would be like $\forall_f$.

Yes, morally this is related to "the six operations" although I don't know much about that either. In particular, what I know as "Frobenius reciprocity", whose logical form is $\exists_f(S) \wedge T = \exists_f(S \wedge f^\ast(T)$), is formally like the projection formula in algebraic geometry (and there's a similar-looking one for Chow rings). The name "Frobenius reciprocity" here was introduced by Lawvere (naturally), borrowing from terminology of representation theorists, who often refer just to the adjunction $Ind_f \dashv Res_f$ as "Frobenius reciprocity", but who sometimes include as part of this the analogous isomorphism

$$Ind_f(V) \otimes W \cong Ind_f(V \otimes Res_f(W)).$$

In the language of hyperdoctrines, these reciprocity isomorphisms arise from the fact that pulling back or restriction along $f$ preserves the internal hom (see again Mac Lane & Moerdijk, IV.9, although they don't mention the word "hyperdoctrine").

Sorry that I don't have a better answer to your question!


问:
Partial fraction decomposition help
In a text that I am reading, they state that the following partial fraction ($r$ fixed) expansion is "readily computed":

$$f(z) = \frac{z^r}{(1-z)(1-2z)(1-3z)\cdots (1-rz)} = \frac{1}{r!} \sum_{j=0}^r \binom rj \frac{(-1)^{r-j}}{1-jz}$$

I know how to do partial fractions, or at least I thought I did. I tried to set it up with something like
$$\frac{A_1}{1-z} + \frac{A_2}{1-2z} + \cdots + \frac{A_r}{1-rz}$$
but it got messy. Also, the answer given above has a constant term $\frac{(-1)^{r}}{r!}$ when $j=0$, which shouldn't show up when doing this method normally.

Could someone point me in the right direction? Thanks!

I now see why the constant term is necessary; doing a "long division" would result in the $\frac{(-1)^r}{r!}$ term. The remainder would be something like
$$\frac{z^r - \frac{1}{r!}(1-z)(1-2z)\cdots(1-rz)}{(1-z)(1-2z)\cdots(1-rz)}$$
which technically could be solved by partial fractions...

Induction also looks promising, as suggested by Maesumi.

答:
Multiply by $\frac{1}{1-jz}$ and plug in $\frac{1}{j}$ to get
\begin{align*}
  A_j&=\frac{1/j^r}{\left( 1-\frac{1}{j} \right)\left( 1-\frac{2}{j} \right)\cdots \left( 1-\frac{j-1}{j} \right)\left( 1-\frac{j+1}{j} \right)\cdots \left( 1-\frac{r}{j} \right)} = \\
  &= \frac{1}{j\left( j-1 \right)\left( j-2 \right)\cdots \left( j-j+1 \right)\left( j-j-1 \right)\cdots \left( j-r \right)}= \\
  &= \frac{1}{\underbrace{j\left( j-1 \right)\left( j-2 \right)\cdots 1}_{j!} \cdot \underbrace{(-1)(-2) \cdots (-(r-j))}_{(-1)^{r-j}(r-j)!}}= \\
  &= \Big\{\binom{r}{j} = \frac{r!}{j!(r-j)!}\Big\}= \frac{1}{r!}\binom{r}{j}\left( -1 \right)^{r-j}
\end{align*}

By polynomial division, we can write
\begin{equation}
  \frac{z^r}{(1-z)(1-2z)\cdots (1-rz)}=q+\frac{c}{(1-z)(1-2z)\cdots (1-rz)}\tag{1}
\end{equation}
where $q, c\in \mathbb{R}$. By theory of partial fractions, we can have at most $r$ fractions, so by the above, we get
\begin{equation*}
  \frac{c}{(1-z)(1-2z)\cdots (1-rz)}=\sum_{j=1}^{r}\frac{A_j}{1-jz}
\end{equation*}
Plugging in $0$ in (1), we thus obtain
\begin{align*}
  q=A_0&=-\frac{1}{r!}\sum_{j=1}^{r}\binom{r}{j} \frac{(-1)^{r-j}}{1-0}= \\
  &=-\frac{1}{r!}\Big[\underbrace{\sum_{j=0}^{r}\binom{r}{j}(-1)^{r-j}}_{(1-1)^r=0}-\binom{r}{0}(-1)^r \Big]=\frac{(-1)^r}{r!}
\end{align*}

Title: Partial fraction decomposition help


答:
We have two numbers with $n$ digits having first $k$ digits same. In order that the numbers are same, the remaining $n-k$ digits must be same as well.

Now the number of ways of choosing the remaining $n-k$ digits of the first number are $10^{n-k}$. The two numbers are only equal when the remaining $n-k$ digits are same for second number as well. Thus there are $10^{n-k}$ cases where the two numbers are same given that first $k$ digits are same.

Now the total cases can be $10^{2(n-k)}$, since each of the two numbers have $10^{n-k}$ ways to choose remaining digits.

$\therefore P(\text{Two numbers are same})=\dfrac{10^{n-k}}{10^{2(n-k)}}=\dfrac{1}{10^{n-k}}$

Title: Chance for two numbers to be same if first k digits matches

Body: Two people pick two numbers (same number of digits say n). What is the chance that the number will be same number if their first k digits matches.</p>


问:
Shape operator of a strictly convex surface
Define a strictly convex surface in $\mathbb{R}^3$ to be one that bounds a strictly convex closed region. Is it true that the Gauss map on this surface has invertible derivative everywhere? Clearly the second fundamental form is non-negative definite, but must it be positive definite?

答:
Not necessarily. Such a surface must be "positively curved" in the sense that it contacts each tangent plane at exactly one point, but the second fundamental form only measures 2nd order effects, and this may only happen at higher order. With this in mind, consider the following surface:
$$
\{(x,y,z)\in\mathbb{R}^3:x^4+y^4+z^4=1\}
$$
One can show that this surface is strictly convex, but its second fundamental form vanishes at $(0,0,1)$.


问:
Proof without using truth tables
I need some further explanation, as I feel I am just missing some critical piece of information. I have shown my work thus far (I excluded the truth tables as they take up so much space but included the pertinent values). Can someone shed some light on what I'm overlooking here?

Which of the following formulas are equivalent? Without showing truth tables, explain why which of the following five statements are equivalent. Hint – analyze the possible values of P, Q, and R that make each of the statements false.

a. P → (Q → R).

b. Q → (P → R).

c. (P → Q) ∧ (P → R).

d. (P ∧ Q) → R.

e. P → (Q ∧ R).

My work:

So using truth tables I was able to determine that a,b,d are all equivalent to each other AND c and e are equivalent to each other.
A, B, D are all false ONLY when p and q are true and r is false, as that is the  only time that the conditional statement becomes true implies false.
C and E are false for several instances, when p=T q=T r=F; p=T q=F r=T; p=T q=F r=F.

What has me stumped is how to write a proof from that information. I have tried using the logical equivalence identities, but am unable to  come up with the correct answer, so I feel like that is not the proper way to do it. So I'm wondering, is there somethin I'm missing? Is this supposed to be a proof by implication, or and indirect proof, and if so, how am I supposed to start that? I understand that these are conditional statements, but again, I feel as if I am missing something.

答:
If you are not allowed to use truth-tables to prove equivalence (even though that's a perfectly acceptable method), but you also don't want to rely on laws like DeMorgan, you could do something like this:

Since any conditional $P \to Q$ is False if and only if $P$ is True and $Q$ is False, we have:

$P \to (Q \to R)$ is False iff

$P$ is True and $Q \to R$ is False iff

$P$ is True and $Q$ is True and $R$ is False iff

$Q$ is True and $P$ is True and $R$ is False iff

$Q$ is True and $P \to R$ is False iff

$Q \to (P \to R)$ is False

... which of course also means that $P \to (Q \to R)$ is True iff  $Q \to (P \to R)$ is True ... and hence a) and b) are equivalent

You should be able to do something similar for the other equivalences. Indeed, note how the equivalence of d) to these two statements can be established right after the third line from the above demonstration (which is called a proof by semantics by the way: a proof that simply applies the basic truth-conditions of the operators involved)

Good luck with your presentation!

Title: Proof without using truth tables


问:
In Gauss integer integral domain, $p$ is prime iff the equation $x^2+y^2=p$ has no solution
In Gauss integer integral domain $G= \mathbb Z[i]$, if p is prime integer then $p$ is prime in $\mathbb Z[i]$ iff the equation $x^2+y^2=p$ has no solution in $\mathbb Z$.

the first side is easy, if $p$ is prime, suppose $x^2+y^2=p$ has a solution then

$(x+iy)(x-iy)=x^2+y^2=p$ where either $x+iy$ is a unit or $x-iy$ is a unit

$\Rightarrow$ either      $ x=\pm1$  and $  y=0,$ or $x=0$ and $y=\pm1$

This makes the equation $p=x^2+y^2=1$ and that's a contradiction.

the other side is harder. I thought of using contrapositive ( if $p$ is composite then $x^2+y^2=p$ has a solution in $\mathbb Z$)  but I couldn't reach something.

答:
Using the contrapositve.
Let $p$ be prime integer but not prime in $\mathbb Z[i]$

so that $p$ can be written as product of 2 non invertible elements in $\mathbb Z[i]$ i.e

$p =\alpha \beta$

taking the norm for both sides

$N(p) = N(\alpha) N(\beta) = p^2$

Now since the norms are integers and neither $\alpha $ nor $\beta$ are units then $N(\alpha) =p$

Let $\alpha = x+iy$ , for some $x,y \in \mathbb Z$

$N(\alpha) = x^2+y^2= p$

so the equation $x^2+y^2= p$ has a solution


问:
Properties of Partial Derivatives?
Do Partial Derivatives have the property, for some general set of variables x,y,z,
$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=\frac{\partial x}{\partial z}$$
like exact derivatives do, i.e., $$\frac{d x}{d y}\frac{d y}{d z}=\frac{d x}{d z}$$

答:
Partial differentials don't have properties like you are wanting, because partial differentials aren't, generally speaking, unique.

The $\partial y$ in $\frac{\partial y}{\partial x}$ is a different variable than the $\partial y$ in $\frac{\partial y}{\partial z}$.  This is why the partials are never split up - part of their identity is based on the lower part of the fraction.

There are some notational options you can use to keep track of which quantity is which, and then you can treat them algebraically as above.  One option is to mark which variable is allowed to freely vary.

For instance, if I had $z = y^2 + x^3$, then I could mark the partial of $z$ when $y$ was allowed to vary freely as $\partial_y z$.  So, $\partial_y z = 2y\,dy$.  Or, $\frac{\partial_y z}{dy} = 2y$.

If you keep track of which is which in a fashion like this, then normal algebraic rules apply.

Title: Properties of Partial Derivatives?


问:
Homogenous Systems Question
$A, B \in \mathbb{C}^{n\times n}$

If $I - B$ is singular, then $\exists x$ from $\mathbb{C}^n$  so that $(I - B)x = 0$.

How do I show that $|| B || \geq 1$ and if $|| A || \leq 1$ then $I-A$ isn't singular?

It's not really clear for me either, but I think the norm I'm supposed to use is $ || A || = \max \{ | a_{ij} | \}$

答:
The correct statement should be:

if $\|A\|\color{red}{<}1$, then $I-A$ is nonsingular.

The usual proof is to show that the power series $I+A+A^2+\ldots$ converges (we need $\|A\|<1$ here) and it is the inverse of $I-A$. Which norm you are using is irrelevant, as long as the norm is a submultiplicative matrix norm.

Your statement for $B$ is just the contrapositive of the statement in the grey box.

Title: Homogenous Systems Question


问:
If $\int_{0}^{\infty} f(x)dx$ converges, $f(x)$ is differentiable on $[0,\infty)$, does $\int_{0}^{\infty} f'(x)dx$ also converge?
If $\int_{0}^{\infty} f(x)dx$ converges, and $f(x)$ is differentiable on $[0,\infty)$, does $\int_{0}^{\infty} f'(x)dx$ also converge?

答:
The result is not true. Let $\varphi \in {\mathcal C}^1({\mathbb R})$ be such that

$\varphi(x)\ge 0$ for all $x$

$\varphi(x) = 0$ when $x\not\in [0, 1]$

$\int_0^1\varphi(x) d x = 1$

Let $f: [0, \infty)\to {\mathbb R}$ be defined by $f(x) = \varphi(k^2(x-k))$ for $x\in [k, k+1)$.

Then $\int_0^\infty f(x) d x  = \sum_{k=1}^\infty \frac{1}{k^2} <\infty$ but $\int_0^\infty f'(x) d x$ cannot converge because $f$ reaches the maximum and the minimum of $\varphi$ on every interval $[k, k+1)$ for $k\ge 1$.


问:
confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test
I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.

$$\frac{n(n+1)}{(n+3)^3}$$

Using the ratio test, then simplifying in stages:

$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$

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

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

$$\frac{(n+3)^3}{(n+4)^22n}$$

The concluding remark:

$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$

Hence the series is divergent, unless I went wrong somewhere?

答:
In the vein of @lulu 's comment, the fast way uses asymptotic equivalence: as a polynomial is equivalent to its leading term, we have
$$\frac{n(n+1)}{(n+3)^3}\sim_\infty\frac{n^2}{n^3}=\frac 1n,$$
which is a divergent series.


问:
sequence limits - correctness of the adjustment
I would like to ask why the following modification of the example is not allowed $$\displaystyle\lim_{n\rightarrow \infty} \frac{1+2+ \cdots + 2^n}{1+5+ \cdots + 5^n} = \lim_{n\to \infty}\left(\frac{2^n}{5^n}\right)\cdot \frac{\left(\frac{1}{2^n}+\frac{2}{2^n}+.....+1\right)}{\left(\frac{1}{5^n}+\frac{5}{5^n}+.....+1\right)}$$
and similarly

$$\displaystyle\lim _{n\rightarrow \infty} \left(\frac{3n+1}{4n+5}\right) ^n = \lim_{n \to \infty}
\left(\frac{3}{4}\right)^n\cdot\left(\frac{1 + \frac{1}{3n}}{1 + \frac{5}{4n}}\right)^n$$

For these two examples, I am not concerned with the calculation, but rather with describing why this procedure is wrong.

答:
To actually evaluate the limits I'd go as follows:

$$\frac{1+2+\ldots+2^n}{1+5+\ldots+5^n}=\frac{\frac{1-2^{n+1}}{1-2}}{\frac{1-5^{n+1}}{1-5}}=4\frac{1-2^{n+1}}{1-5^{n+1}}=4\left(\frac25\right)^{n+1}\;\frac{\frac1{2^{n+1}}-1}{\frac1{5^{n+1}}-1}\xrightarrow[n\to\infty]{}4\cdot0\cdot1=0$$

As for the second one:

$$\left(\frac{3n+1}{4n+5}\right)^n=\frac1{\left(\frac43\right)^n}\left(\frac{3n+1}{3n+\frac{15}4}\right)^n=\frac1{\left(\frac43\right)^n}\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^n=$$

$$=\frac1{\left(\frac43\right)^n}\left[\left(1-\frac{\frac{11}4}{3n+\frac{15}4}\right)^{3n+\frac{15}4}\right]^{-2n-\frac{15}4}\xrightarrow[n\to\infty]{}0\cdot0=0$$


答:
Ratio test is not effective because
$$\lim_{n\to \infty }\frac{(n+2) (n+3)^3}{n (n+4)^3}=1$$
You can use comparison test as
$$\frac{n (n+1)}{(n+3)^3}\sim \frac1n ;\text{ as }n\to\infty$$
and $\sum\frac1n $ diverges

Title: confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test

Body:
I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.

$$\frac{n(n+1)}{(n+3)^3}$$

Using the ratio test, then simplifying in stages:

$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$

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

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

$$\frac{(n+3)^3}{(n+4)^22n}$$

The concluding remark:

$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$

Hence the series is divergent, unless I went wrong somewhere?


问:
Properties of a dense set in a metric space
Let $(M,d)$ be a metric space and $X \subset M$, with $X$ being a dense set in $M$.

a) Show that $d(p,X)=0$, $\forall p \in M$

b) Show that if $M$ is a limited set, then $\operatorname{diam}(M)=\operatorname{diam}(X)$

c) Let $(x_n) \subset M$ be a sequence. Proof that if $\exists p \in M$ that $\lim\limits_{n \to \infty} d(x_n,x)=d(p,x)$, $\forall x \in X$, then $x_n \to p$.

My thoughts for a):

Since $\forall p \in M$ and $\forall \varepsilon > 0$, we have $B(p,\varepsilon)\cap X \not= \emptyset$, let $\varepsilon = \frac{1}{n}, n \in \mathbb{N}$.

It follows that for any $p \in M$, we have an $x_n \in X$ that $d(p,x_n)< \frac{1}{n}$.

So $d(p,X) = \inf\bigl\{\frac{1}{n}, n \in \mathbb{N}\bigr\} = 0, \forall p \in M$. (right?)

For b) and c) I can't think anything to proceed. Any leads?

答:
$diam(M)=max_{x,y\in M}d(x,y)$. For every $n>0$, there exists $x_n,y_n\in M$ such that $d(x_n,y_n)>diam(M)-{1\over{2n}}$. Since $X$ is dense, there exists $a_n,b_n\in X$ such that $d(a_n,x_n)<{1\over{4n}}$ and $d(b_n,y_n)<{1\over{4n}}$.

We have: $diam(M)-{1\over{2n}}<d(x_n,y_n)<d(x_n,a_n)+d(a_n,b_n)+d(b_n,y_n)<d(a_n,b_n)+{1\over{2n}}$, we deduce that:

$d(a_n,b_n)>diam(M)-{1\over n}$ and $diam(X)\geq diam(M)$.

For $c)$.

Let $x_n$ be a sequence, suppose that there exists $p$ such that $d(x_n,x)=d(x,p)$ for every $x\in X$.
Since $X$ is dense, for every $m>0$, there exists $p_m\in X$ such that $d(p,p_m)<{1\over m}$. We have $d(x_n,p)\leq d(x_n,p_m)+d(p_m,p)=2d(p_m,p)<{2\over n}$ implies $x_n=p$. The sequence is a constant.


问:
To which group is the $\mathbb Z_{20}^*$ isomorphic?
I have a question saying that to which group is $\mathbb{Z}_{20}^{*}$ is isomorphic, where $\mathbb{Z}_{20}^{*}$ is the set of the not zero divisors of $\mathbb{Z}_{20}$.

Here is what i think: $\mathbb{Z}_{20}^{*}=\{1,3,7,9,11,13,17,19\}$. It has 8 elements. Then it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ or $\mathbb{Z}_{8}$. But how can I decide which of them? And is this group a multiplicative group or an additive group? How can I know?

Thank you

答:
To hint your last question first, if $a,b$ are each relatively prime to $20$, then $ab$ is also relatively prime to $20$.  $a+b$ need not be relatively prime to $20$.

To hint your first question, $\mathbb{Z}_8$ has an element of order 8,  $\mathbb{Z}_2\times\mathbb{Z}_4$ has an element of order 4 but no element of order $8$, while in $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ all elements but the identity have order 2.


答:
$$\int_0^\infty \sin(e^x)\:dx < \int_{-\infty}^\infty \sin(e^x)\:dx = \frac{\pi}{2}$$

But the integral of its derivative $e^x\cos(e^x)$ does not converge.

Title: If $\int_{0}^{\infty} f(x)dx$ converges, $f(x)$ is differentiable on $[0,\infty)$, does $\int_{0}^{\infty} f'(x)dx$ also converge?

Body: If $\int_{0}^{\infty} f(x)dx$ converges, and $f(x)$ is differentiable on $[0,\infty)$, does $\int_{0}^{\infty} f'(x)dx$ also converge?


问:
Euler-Mascheroni and von Mangoldt function
I know that by their respective definitions:

$$\gamma_0 = \sum_{n=1}^{\infty}{\frac{1}{n}-\ln(\frac{n+1}{n})}$$

$$\ln(n)=\sum_{d|n}{\Lambda(d)}$$

and I want to get there:

$$\gamma_0 = \sum_{n=1}^{\infty}{\frac{1-\Lambda(n)}{2n}}$$

I suspect that a number theoretic aproximation must be made regarding to the sum over divisors of $n$ and divisors of $n+1$, but no idea from where can it come.

答: As others have noted, it's hard to prove more than$$\lim_{s\to1^+}\sum_{n\ge1}\frac{1-\Lambda(n)}{n^s}=\gamma_0,$$which I'll focus on. (In particular, partial summation of $\sum_{n\ge1}\frac{1-\Lambda(n)}{n}=\gamma_0$ implies the prime number theorem viz. $\sum_{n=1}^N(\Lambda(n)-1)\in o(N)$.) In terms of Dirichlet convolutions $\ln=\Lambda\ast1\implies\Lambda=\ln\ast\mu$. We want to evaluate at $s=1$ the Dirichlet series of $\tfrac{1-\Lambda}{2}$. Since $1,\,\ln,\,\mu$ have respective Dirichlet series $\zeta,\,-\zeta^\prime,\,1/\zeta$, we want$$\lim_{s\to1^+}\frac{\zeta(s)+\zeta^\prime(s)/\zeta(s)}{2}=\gamma_0.$$Now use the $0<s-1\ll1$ results$$\begin{align}\zeta(s)&\in-\tfrac{1}{1-s}+\gamma_0+\gamma_1(1-s)+O((1-s)^2),\\\zeta^\prime(s)&\in-\tfrac{1}{(1-s)^2}-\gamma_1+O(1-s),\\\tfrac{\zeta^\prime(s)}{\zeta(s)}&\in\tfrac{1}{1-s}+\gamma_0+O(1-s),\\\zeta(s)+\tfrac{\zeta^\prime(s)}{\zeta(s)}&\in2\gamma_0+O(1-s).\end{align}$$


问:
Finding, writing down and visualizing images of functions
I apologize in advance if this is a duplicate (I would expect to see something like this on here), but I haven't found anything that matches my question.

I have this general difficulty (perhaps, to lack of practical experience) of finding and writing down images of functions given a domain. For example, I have recently seen this problem in a book:

Given $f(x,y) = \left(\begin{array}{c}u(x,y)\\v(x,y)\end{array}\right) = \left(\begin{array}{c}x^2-y^2\\2xy\end{array}\right)$, find image of $f$ when $x>0$, $y>0$, $x^2+y^2<1$

Basically, it's asking for a set of inequalities for what $u$ and $v$ trace out in the $u$-$v$ plane. How do I go about solving such problems for a general case of $f : \mathbb{R}^m \to \mathbb{R}^n$? For example, in the above problem, the domain of $f$ is a quarter of a circle of radius 1 in the first quadrant of the $x$-$y$ plane. How do I find what $f$ maps this to in the $u$-$v$ plane? Where do I start?

Update:
So apparently (by empirically constructing a table of values), $f$ maps the quarter of a circle to a half-circle ($v$>0, $u^2+v^2<1$). Is there a way to see this intuitively? Most importantly, how can I get these two inequalities algebraically?

答: If you want an algebraic solution, follow Qiaochu Yuan and parametrize the region in polar coordinates by $(x,y)=(r \cos t, r \sin t)$, for $r\in [0,1)$ and $t \in (0,\pi/2)$. Then $f(x,y)=r^2(\cos^2 t - \sin^2 t, 2 \sin t \cos t) = r^2(\cos 2t, \sin 2t)$. So, the image is the (open) half-disk.</p>


答:
Hints:

Is $1 + 3$ in $\Bbb Z^*_{20}$? On the other hand, one can show that it's closed under multiplication.

Is $\Bbb Z^*_{20}$ cyclic? What's the order of $3$? Answer both questions and one possibility will remain.

To elaborate on the second hint. We have:
\begin{align}
3^1 &= 3 &\equiv 3 \pmod{20} \\
3^2 &= 9 &\equiv 9 \pmod{20} \\
3^3 &= 27 &\equiv 7 \pmod{20} \\
3^4 &= 81 &\equiv 1 \pmod{20} \\
\end{align}

Therefore the order of $3$ is $4$. This eliminates the possibility $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. With similar computations, you can see that none of the elements of $\Bbb Z^*_{20}$ generates the group. Hence, it's not cyclic. One possibility remains and it's $\Bbb Z_4 \times \Bbb Z_2$.

To show that all elements of $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ have order $1$ or $2$, consider $(a, b, c) \in \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. We have:
$$
(a, b, c)^n = (a^n, b^n, c^n)
$$

Since each of $a$, $b$, $c$ is a member of $\Bbb Z_2$, it has order $1$ or $2$. This forces:
$$
(a, b, c)^2 = (a^2, b^2, c^2) = (1, 1, 1) = 1
$$

Title: To which group is the $\mathbb Z_{20}^*$ isomorphic?

Body:
I have a question saying that to which group is $\mathbb{Z}_{20}^{*}$ is isomorphic, where $\mathbb{Z}_{20}^{*}$ is the set of the not zero divisors of $\mathbb{Z}_{20}$.

Here is what i think: $\mathbb{Z}_{20}^{*}=\{1,3,7,9,11,13,17,19\}$. It has 8 elements. Then it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ or $\mathbb{Z}_{8}$. But how can I decide which of them? And is this group a multiplicative group or an additive group? How can I know?

Thank you


问:
Finding the function for a curve given its length and the length of its velocity?
I am working on an assignment and am trying to find a function for $\vec{r}(t)$ with constant length $x$, whose velocity function also has a constant length $y$.

I'm not sure where to begin with this. I know that, say, $\langle \cos(t),\sin(t),0\rangle$ has a constant length of 1, but its velocity curve also has a constant length of 1. How can I scale the velocity without scaling the original function?

Thanks!

答:
So what you want to achieve is that the path of the curve stays the same, but the velocity changes.

To try to provide some intuition, suppose you are going to the grocery store. There is a fixed path for you to travel, how do you get there faster? You just speed up.
To put a finer point on this, if you plot the curve of your walk to the grocery store, you start at your house $H$ at time $0$, and end up at the grocery store $G$ at time $t_G$.
So $$\vec{r}(0)=H\text{ and }\vec{r}(t_G)=G$$. Now you construct a faster path $\vec{s}$ such that your velocity is twice as fast, so $$\vec{s}(0)=H\text{ and }\vec{s}(\frac{t_G}{2})=\vec{r}(t_G)=G$$.

To summarize for any arbitrary $t$ we have $\vec{s}(t) = \vec{r}(2t)$. And of course the $2$ is arbitrary as well.


问:
Find the maximum likelihood estimate of probabilities of ordered pairs of random variables
$P((X_k,Y_k)=(i,j))=\pi_{ij}$ where $\sum \pi_{ij}=1$ and the frequency of $(i,j)$ is $n_{ij}$.

I'm now asked to find the MLE of $\pi_{ij}$ if $X$ and $Y$ are independent and again if they are not.

I found that the likelihood funciton is:
$L(\pi;x)= \prod_{ij} \pi_{ij}^{n_{ij}}$

If $X$ and $Y$ are independent then we can say $\pi_{ij}=\alpha_i \beta_j$ with constraints $\sum \alpha_i = 1$ and $\sum \beta_j =1$ and $\sum \sum \pi_{ij}=1$

Using Lagrange multipliers I find the MLE of $\alpha_i$ and $\beta_j$ are $\frac {2}{\sum_j n_{ij}}$ and $\frac {2}{\sum_i n_{ij}}$

Then the MLE of $\pi_{ij}$ is $\frac {4}{\sum_j n_{ij} \sum_i n_{ij}}$ which doesn't make logical sense to me... Is there some explanation as to why this answer makes sense?

Many thanks for your help!

答:
No, it doesn't make sense. That could even be greater than 1.

Let $N=\sum\sum n_{i,j}$. If they are independent, the mle of $\alpha_i$ is $\frac{\sum_{j}n_{i,j}}{N}$ and the mle of $\beta_j$ is $\frac{\sum_{i}n_{i,j}}{N}$.

Just re-write the likelihood in terms of the $\alpha_i$ and $\beta_j$ to derive those.

$$\prod_{i,j}(\alpha_i \beta_j)^{n_{i,j}}$$
Try to maximize this wrt $\alpha_1$ for example.

It's easier to see if you maximize the log-likelihood.


答:
$\int_{0}^{\infty} f'(x)dx = \displaystyle\lim_{x \to \infty} \left(f(x)\right) - f(0).$

The answer to the question is no. Let's start with a "bar graph"- looking function:

$$g(x)=\begin{cases}2^{n} & n \leq x < n+\frac{1}{4^n},\quad n\in\mathbb{N}\cup\{0\} \\   0 & \text{otherwise}.  \end{cases}$$

The area under each "bar" is $\ 1,\ \frac{1}{2},\ \frac{1}{4},\ \frac{1}{8}, ...,\ \frac{}{} $ so the total area under the graph is just $2.$ More importantly, it converges as $n \to \infty.$

Now all we need to do is make each ends of the "bar" differentiable. One way to do this is to make each "bar" into a sine curve, for example:

$$f(x)=\begin{cases}\frac{1}{2}(\sin\left((2\pi x)4^n - \frac{\pi}{2}\right) + 1)2^n & n \leq x < n+\frac{1}{4^n},\quad n\in\mathbb{N}\cup\{0\} \\   0 & \text{otherwise}.  \end{cases}$$


问:
Mathematical constant
Does there exists any important complex mathematical constant ?

It is known that there exist a large number of mathematical constants that are real numbers, like:

$$\pi, e, \phi, ...$$

Here is the link for much more of them:

https://en.wikipedia.org/wiki/Mathematical_constant

I have never seen and could not find any constant written as:
$$c = \alpha + \beta i$$
$$i = \sqrt{-1}$$
$$\alpha, \beta \in \Bbb{R}$$
Why ?

What I mean is that $\alpha$ and $\beta$ can't be also a combination of any other well known constants, so $\alpha$ and $\beta$ should be in some way unique like $\pi$ is unique for themselves and is strongly connected with circle, or like Feigenbaum constants that are strongly connected with bidfurcation and can't be rewriten as combination of any other constatns.

答:
You could argue the most important number in complex analysis is $2\pi i$, because of its appearance in the residue theorem.

Complex roots of unity, i.e. $\{e^{2\pi iq}|q\in\Bbb Q\}$, are also very important, partly because of their role in the theory of polynomials, partly because of complex analysis again. In particular, you often see results of the form $\oint_Cf(z)dz=(1-e^{2\pi is})\int_0^\infty f(x)dx$, which allow us to calculate some integrals of the form $\int_0^\infty f(x)dx$ with a keyhole contour. (If the upper limit is instead finite, doghole contours may encounter the same phenomenon.) You may wish to argue, then, that $1-e^{2\pi is}$ or its reciprocal is therein of more fundamental importance, for suitable $s$, than is $e^{2\pi is}$.

Title: Mathematical constant


问:
What does $\sum_{n=0}^\infty z^{n(n+1)/2}$, $|z|<1$ converge to?
Does anyone know what the series
$$
S(z) = \sum_{n=0}^{\infty} z^\frac{n(n+1)}{2}
$$
converges to for $|z|<1$? This came up in an application where $z$ is a probability and $S(z)$ an expected value. The exponent comes from the fact that $\sum_{k=0}^nk=\frac{n(n+1)}{2}$. I think $S(z)$ should definitely converge because the coefficients are nonnegative and form a subsequence of the well-known $\sum_{n=0}^\infty z^n=\frac{1}{1-z}$. Is there a closed-form representation of $S(z)$?

答:
Using Jacobi Elliptic theta function
$$\vartheta _2(u,q)=2 \sqrt[4]{q} \sum _{n=0}^{\infty } q^{n(n+1)} \cos ((2 n+1) u)$$
we can conclude that
$$\sum _{n=0}^{\infty } z^{\frac{1}{2} n (n+1)}=\frac{\vartheta _2\left(0,\sqrt{z}\right)}{2 \sqrt[8]{z}}$$
To satisfy my curiosity I plotted (for real $z$ with $|z|<1$) the graph of such an "exotic" function, below.


答:
The Speed is the magnitude of the velocity vector, in your example $\mathbf g(t)= (\cos(t),\sin(t),0)$ has as its velocity vector $\mathbf g'(t) = (-\sin(t),\cos(t),0)$, and thus the speed is going to be $||\mathbf g'(t)||=1$.

To increase the speed we can make use of the chain rule, in fact $\frac {d}{dt} (\mathbf g(2t))= 2\mathbf g'(2t).$ and thus the speed will be greater. Test this!

Title: Finding the function for a curve given its length and the length of its velocity?

Body:
I am working on an assignment and am trying to find a function for $\vec{r}(t)$ with constant length $x$, whose velocity function also has a constant length $y$.

I'm not sure where to begin with this. I know that, say, $\langle \cos(t),\sin(t),0\rangle$ has a constant length of 1, but its velocity curve also has a constant length of 1. How can I scale the velocity without scaling the original function?

Thanks!


问:
Given $P(x)=x^{4}-4x^{3}+12x^{2}-24x+24,$ then $P(x)=|P(x)|$ for all real $x$
$$P(x)=x^{4}-4x^{3}+12x^{2}-24x+24$$

I'd like to prove that $P(x)=|P(x)|$. I don't know where to begin. What would be the first step?

答:
$$x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 = \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix}^\top \begin{bmatrix} 1 && -2 && t \\ -2 && 12 - 2t && -12 \\ t && -12 && 24 \end{bmatrix} \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix}$$

where $t \in \mathbb R$. Using Sylvester's criterion, after some work, we learn that the (symmetric) matrix above is positive semidefinite when

$$t \in \left[ 2 - 4 \cos{\left(\frac{\pi}{9} \right)}, 2 +  4 \sin{\left(\frac{\pi}{18} \right)} \right]$$

Computing the Cholesky decomposition of the matrix for an admissible value of $t$, we obtain a sum of squares (SOS) decomposition, thereby proving that the given quartic polynomial is (globally) nonnegative.

$\boxed{t = 1}$

$$\begin{bmatrix} 1 && -2 && 1 \\ -2 && 10 && -12 \\ 1 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{\sqrt{6}} & 0\\ \color{blue}{1}  & \color{magenta}{- \frac{5 \sqrt{6}}{3}} & \color{red}{\frac{\sqrt{57}}{3}} \end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x + 1} \right)^2 + \left( \color{magenta}{\sqrt{6}\, x - \frac{5 \sqrt{6}}{3}} \right)^2 + \left( \color{red}{\frac{\sqrt{57}}{3}} \right)^2 \\ &= \left( x - 1\right)^4 + 6 \left( x - \frac{5}{3} \right)^2 + \frac{19}{3}\end{aligned}$$

which is the same sum of squares (SOS) decomposition in this answer.

$\boxed{t = 0}$

$$\begin{bmatrix} 1 && -2 && 0 \\ -2 && 12 && -12 \\ 0 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{2 \sqrt{2}} & 0\\ 0 & \color{magenta}{- 3 \sqrt{2}} & \color{red}{\sqrt{6}} \end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x} \right)^2 + \left( \color{magenta}{2 \sqrt{2} \, x - 3 \sqrt{2} } \right)^2 + \left( \color{red}{\sqrt{6}} \right)^2 \\ &= x^2 \left( x - 2\right)^2 + 2 \left( 2 x - 3 \right)^2 + 6\end{aligned}$$

$\boxed{t = 2}$

$$\begin{bmatrix} 1 && -2 && 2 \\ -2 && 8 && -12 \\ 2 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{2} & 0\\ \color{blue}{2} & \color{magenta}{-4} & \color{red}{2}\end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x + 2} \right)^2 + \left( \color{magenta}{2 x - 4} \right)^2 + \color{red}{2}^2 \end{aligned}$$

SymPy code

```<code>>>> from sympy import *
>>> t = symbols('t', real=True)
>>> Q = Matrix( [[  1,       -2,   t],
                 [ -2, 12 - 2*t, -12],
                 [  t,      -12,  24]] )
</code>```

For $t=1$,

```<code>>>> L = Q.subs(t,1).cholesky()
>>> L
Matrix([
[ 1,            0,          0],
[-2,      sqrt(6),          0],
[ 1, -5*sqrt(6)/3, sqrt(57)/3]])
</code>```

For $t=0$,

```<code>>>> L = Q.subs(t,0).cholesky()
>>> L
Matrix([
[ 1,          0,       0],
[-2,  2*sqrt(2),       0],
[ 0, -3*sqrt(2), sqrt(6)]])
</code>```

For $t = 2 - 4 \cos{\left(\frac{\pi}{9} \right)}$,

```<code>>>> L = simplify(Q.subs(t,2 - 4 * cos(pi/9)).cholesky())
>>> L
Matrix([
[              1,                                        0, 0],
[             -2,                  2*sqrt(1 + 2*cos(pi/9)), 0],
[2 - 4*cos(pi/9), -(4*cos(pi/9) + 4)/sqrt(1 + 2*cos(pi/9)), 0]])
</code>```

For $t = 2 + 4 \sin{\left(\frac{\pi}{18} \right)}$,

```<code>>>> L = simplify(Q.subs(t,2 + 4 * sin(pi/18)).cholesky())
>>> L
Matrix([
[               1,                                          0, 0],
[              -2,                   2*sqrt(1 - 2*sin(pi/18)), 0],
[4*sin(pi/18) + 2, 4*(-1 + sin(pi/18))/sqrt(1 - 2*sin(pi/18)), 0]])
</code>```

Note that the latter two $\rm L$ matrices are rank-$2$. Thus, the corresponding SOS decompositions are the sums of just $2$ squares.

For general $t$,

```<code>>>> L = simplify(Q.cholesky())
>>> L
Matrix([
[ 1,                           0,                                                                    0],
[-2,               sqrt(8 - 2*t),                                                                    0],
[ t, sqrt(2)*(t - 6)/sqrt(4 - t), sqrt(-t**2 + 24 - 2*(t - 6)**2*conjugate(1/sqrt(4 - t))/sqrt(4 - t))]])
</code>```

which suggests that the case $t = 2$ is interesting. Is it?

```<code>>>> L = Q.subs(t,2).cholesky()
>>> L
Matrix([
[ 1,  0, 0],
[-2,  2, 0],
[ 2, -4, 2]])
</code>```

Interesting, indeed.

Related:

How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?

polynomials sum-of-squares-method matrices matrix-decomposition cholesky-decomposition

Body:
$$x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 = \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix}^\top \begin{bmatrix} 1 && -2 && t \\ -2 && 12 - 2t && -12 \\ t && -12 && 24 \end{bmatrix} \begin{bmatrix} x^2 \\ x \\ 1 \end{bmatrix}$$

where $t \in \mathbb R$. Using Sylvester's criterion, after some work, we learn that the (symmetric) matrix above is positive semidefinite when

$$t \in \left[ 2 - 4 \cos{\left(\frac{\pi}{9} \right)}, 2 +  4 \sin{\left(\frac{\pi}{18} \right)} \right]$$

Computing the Cholesky decomposition of the matrix for an admissible value of $t$, we obtain a sum of squares (SOS) decomposition, thereby proving that the given quartic polynomial is (globally) nonnegative.

$\boxed{t = 1}$

$$\begin{bmatrix} 1 && -2 && 1 \\ -2 && 10 && -12 \\ 1 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{\sqrt{6}} & 0\\ \color{blue}{1}  & \color{magenta}{- \frac{5 \sqrt{6}}{3}} & \color{red}{\frac{\sqrt{57}}{3}} \end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x + 1} \right)^2 + \left( \color{magenta}{\sqrt{6}\, x - \frac{5 \sqrt{6}}{3}} \right)^2 + \left( \color{red}{\frac{\sqrt{57}}{3}} \right)^2 \\ &= \left( x - 1\right)^4 + 6 \left( x - \frac{5}{3} \right)^2 + \frac{19}{3}\end{aligned}$$

which is the same sum of squares (SOS) decomposition in this answer.

$\boxed{t = 0}$

$$\begin{bmatrix} 1 && -2 && 0 \\ -2 && 12 && -12 \\ 0 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{2 \sqrt{2}} & 0\\ 0 & \color{magenta}{- 3 \sqrt{2}} & \color{red}{\sqrt{6}} \end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x} \right)^2 + \left( \color{magenta}{2 \sqrt{2} \, x - 3 \sqrt{2} } \right)^2 + \left( \color{red}{\sqrt{6}} \right)^2 \\ &= x^2 \left( x - 2\right)^2 + 2 \left( 2 x - 3 \right)^2 + 6\end{aligned}$$

$\boxed{t = 2}$

$$\begin{bmatrix} 1 && -2 && 2 \\ -2 && 8 && -12 \\ 2 && -12 && 24 \end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ \color{blue}{-2} & \color{magenta}{2} & 0\\ \color{blue}{2} & \color{magenta}{-4} & \color{red}{2}\end{bmatrix}$$

and, thus,

$$\begin{aligned} x^{4} - 4 x^{3} + 12 x^{2} - 24 x + 24 &= \left( \color{blue}{x^2 - 2 x + 2} \right)^2 + \left( \color{magenta}{2 x - 4} \right)^2 + \color{red}{2}^2 \end{aligned}$$

SymPy code

>>> from sympy import *
>>> t = symbols('t', real=True)
>>> Q = Matrix( [[  1,       -2,   t],
                 [ -2, 12 - 2*t, -12],
                 [  t,      -12,  24]] )

For $t=1$,

>>> L = Q.subs(t,1).cholesky()
>>> L
Matrix([
[ 1,            0,          0],
[-2,      sqrt(6),          0],
[ 1, -5*sqrt(6)/3, sqrt(57)/3]])

For $t=0$,

>>> L = Q.subs(t,0).cholesky()
>>> L
Matrix([
[ 1,          0,       0],
[-2,  2*sqrt(2),       0],
[ 0, -3*sqrt(2), sqrt(6)]])

For $t = 2 - 4 \cos{\left(\frac{\pi}{9} \right)}$,

>>> L = simplify(Q.subs(t,2 - 4 * cos(pi/9)).cholesky())
>>> L
Matrix([
[              1,                                        0, 0],
[             -2,                  2*sqrt(1 + 2*cos(pi/9)), 0],
[2 - 4*cos(pi/9), -(4*cos(pi/9) + 4)/sqrt(1 + 2*cos(pi/9)), 0]])

For $t = 2 + 4 \sin{\left(\frac{\pi}{18} \right)}$,

>>> L = simplify(Q.subs(t,2 + 4 * sin(pi/18)).cholesky())
>>> L
Matrix([
[               1,                                          0, 0],
[              -2,                   2*sqrt(1 - 2*sin(pi/18)), 0],
[4*sin(pi/18) + 2, 4*(-1 + sin(pi/18))/sqrt(1 - 2*sin(pi/18)), 0]])

Note that the latter two $\rm L$ matrices are rank-$2$. Thus, the corresponding SOS decompositions are the sums of just $2$ squares.

For general $t$,

>>> L = simplify(Q.cholesky())
>>> L
Matrix([
[ 1,                           0,                                                                    0],
[-2,               sqrt(8 - 2*t),                                                                    0],
[ t, sqrt(2)*(t - 6)/sqrt(4 - t), sqrt(-t**2 + 24 - 2*(t - 6)**2*conjugate(1/sqrt(4 - t))/sqrt(4 - t))]])

which suggests that the case $t = 2$ is interesting. Is it?

>>> L = Q.subs(t,2).cholesky()
>>> L
Matrix([
[ 1,  0, 0],
[-2,  2, 0],
[ 2, -4, 2]])

Interesting, indeed.

Related:

How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?

polynomials sum-of-squares-method matrices matrix-decomposition cholesky-decomposition


问:
Real non algebraic numbers
I know that there are countably infinite many algebraic numbers and uncountably infinite many real numbers. Therefore there exist real numbers that are not algebraic numbers. Are there any simple proofs that show certain real numbers are not algebraic?</p>

答: Real non-algebraic numbers are called transcendental numbers. It's usually a lot harder to proof that a number is  transcendental than proving that a number is algebraic. Here is an example for the number $\pi$.

Title: Real non algebraic numbers


问:
For a tree T, the tree of increasing sequences of elements of T does not embed in T
For a tree T, let σT be the set of all increasing sequences in T of elements of T, ordered by initial segment (s<t of s is an initial segment of t). Why isn't σT embeddable in T ? In the sense there is no f from σT to T such that s<t --> f(s)<f(t) ?

I saw in a paper of Todorcevic that it is because such an embedding implies a one-to-one function between T and Ord, but I don't see where is the contradiction..

答:
Assume that there is an increasing map $i:\sigma T\to T$. Now inductively define

$$X_\xi =\langle a_\eta\mid \eta<\xi\rangle, \quad a_\xi=i(X_\xi).$$

Then $\langle a_\xi\mid \xi\in\mathrm{Ord}\rangle$ is a strictly increasing sequence over $T$, a contradiction.


问:
Differentiating $ f(x)= \frac{x + \sin x}{x - \cos x}$
Can someone help me? I'm having some trouble with this:
how can I differentiate $$ f(x)= \frac{x  + \sin x}{x - \cos x} \quad ?$$

P.S. Is there any trick or something to derive this kind of limit?
Thanks!

答:
This is the kind of problem where you want to take advantage of the quotient rule:

$$\text{Given}\;\;f(x) = \dfrac{g(x)}{h(x)},\quad f'(x) = \frac{g(x)h(x) - g(x)h'(x)}{\left(g(x)\right)^2}$$

In your case, we have $$f(x)= \frac{x  + \sin x}{x - \cos x}$$

$$\mathrm{Letting}\;\; g(x) = x + \sin x, \;\;g'(x) = 1 +\cos x\,; \\ \quad\mathrm{and }\quad h(x) = x - \cos x,\;\;h'(x) = 1 + \sin x$$

$$f'(x) = \frac{\left[(1+\cos x)(x-\cos x)\right] - \left[(x + \sin x)(1+\sin x)\right]}{(x-\cos x)^{2}}$$

Of course, you'll want to simplify!


问:
Coefficient of integral dependence and prime ideals
Let $A \subseteq B$ be an integral extension of rings. Let $b \in B$, and let $b^n+a_1b^{n-1}+\dots+a_{n-1}b+a_n=0$ be an integral dependence relation for $b$, with $a_1,\dots,a_n\in A$.

Given a prime ideal $\mathfrak{p}$ in $B$, I am trying to prove that $b\in\mathfrak{p}$ if and only if $a_1,\dots,a_n\in \mathfrak{p}\cap A$. I have managed to prove the implication $\Leftarrow$, but I am stuck with the other: I am only able to show the claim for $a_n$.

In particular, I am interested in the case where $A$ is the subring of invariants under some group action on $B$, but since I proved one implication in this more general setting I suspect that the other direction may hold too under these assumptions.

答:
Here are my thoughts about your question. For $\Rightarrow$, I think you can manage to prove that all the $a_i$'s are in $\mathfrak{p}\cap A$ by induction. For simplicity set $\mathfrak{q}= \mathfrak{p}\cap A$, and note that it is a prime ideal of $A$.

In what follows I treat the case $b\in B\setminus A$.

First step: $a_n\in \mathfrak{q}$. This follows from looking at the polynomial mod $\mathfrak{p}$, which gives you $a_n\equiv 0 \mod \mathfrak{p}$.

Induction step: suppose $a_i\in \mathfrak{q}$ for $i\in \left\lbrace k+1, \dots ,n\right\rbrace$, let us show that $a_k\in\mathfrak{q}$.
You can see that:
$$-b^{n-k} \sum_{i=0}^k a_i b^{k-i} = \sum_{i=k+1}^n a_i b^{n-i} \in \mathfrak{q} .$$
But as $b\in B\setminus A$, then $b^{n-k}\notin \mathfrak{q}$, and the latter being a prime ideal then we have $\sum_{i=0}^k a_i b^{k-i}\in\mathfrak{q}\subset\mathfrak{p}$.
Now as $a_ib^{k-i}\in\mathfrak{p}$ for $0\leq i < k$ looking at the previous expression modulo $\mathfrak{p}$ yields $a_k\equiv 0\mod \mathfrak{p}$. Thus $a_k\in \mathfrak{p}\cap A = \mathfrak{q}$, and this concludes the induction.

Conclusion: we proved that each $a_i$ lies in $\mathfrak{q}$.

This is a "naive" proof, where we do not really understand what's really happening. It might be enlightening to look for a geometric proof using the language of schemes but it'd need some more thought!


问:
$(a\neq -1\ \wedge\ b\neq -1)\implies (a+b+ab\neq -1)$ Using proof by contrapositive
Let A,B be two real numbers, using proof by contrapositive, show the following implication:
$(a\neq -1\ \ and \ \ b\neq -1)\implies (a+b+ab\neq -1)$

I applied it's contrapositive

$(a+b+ab=-1)\implies (a=-1\ or\ b = -1$)

But I'm currently stuck in proving the implication because it has OR.

Body: 1+a+b+ab=(1+a)(1+b)=0 implies a=-1 or b=-1


答: I think you're meant to assume the antecedent in your question, deduce the consequent AND THEN apply the contrapositive.</p>

Title: $(a\neq -1\ \wedge\ b\neq -1)\implies (a+b+ab\neq -1)$ Using proof by contrapositive

Body:
Let A,B be two real numbers, using proof by contrapositive, show the following implication:
$(a\neq -1\ \ and \ \ b\neq -1)\implies (a+b+ab\neq -1)$

I applied it's contrapositive

$(a+b+ab=-1)\implies (a=-1\ or\ b = -1$)

But I'm currently stuck in proving the implication because it has OR.


问:
Deciding whether an atlas is smooth
Let us begin with some definitions regarding smooth atlases. They are quoted from the book by John M. Lee.

Let $M$ be a topological $n$-manifold. If $(U,\varphi)$, $(V,\psi)$ are two charts such that $U\cap V\neq\emptyset$, the composite map $\psi\circ\varphi^{-1}:\varphi(U\cap V)\to\psi(U\cap V)$ is called the transition map from $\varphi$ to $\psi$. It is a composition of homeomorphisms, and is therefore itself a homeomorphism. Two charts $(U,\varphi)$ and $(V,\psi)$ are said to be smoothly compatible if either $U\cap V=\emptyset$ or the transition map $\psi\circ\varphi^{-1}$ is a diffeomorphism. Since $\varphi(U\cap V)$ and $\psi(U\cap V)$ are open subsets of $\mathbb{R}^n$, smoothness of this map is to be interpreted in the ordinary sense of having continuous partial derivatives of all orders.

We define an atlas for $M$ to be a collection of charts whose domains cover $M$. An atlas $\mathcal A$ is called a smooth atlas if any two charts in $\mathcal A$ are smoothly compatible with each other.

To show that an atlas is smooth, we need only verify that each transition map $\psi\circ\varphi^{-1}$ is smooth whenever $(U,\varphi)$ and $(V,\psi)$ are charts in $\mathcal A$; once we have proved this, it follows that $\psi\circ\varphi^{-1}$ is a diffeomorphism because its inverse $(\psi\circ\varphi^{-1})^{-1}=\varphi\circ\psi^{-1}$ is one of the transition maps we have already shown to be smooth. $\color{red}{Alternatively}$, given two particular charts $(U,\varphi)$ and $(V,\psi)$, it is often easiest to show that they are smoothly compatible by verifying that $\psi\circ\varphi^{-1}$ is smooth and injective with nonsingular Jacobian at each point, and appealing to Corollary C.36.

I wonder why the alternative way is considered to be often easiest by the author. It seems like no work is saved. Worse, we take more steps by looking at injectivity, Jacobian, etc. Based on the discussion prior to the alternative method, if $\psi\circ\varphi^{-1}$ is shown to be smooth, we are done with checking that the given atlas is smooth. Did I miss anything?

答:
The paragraph is referring to two different (albeit closely related) procedures:

To show an atlas is smooth, it suffices to show that every transition map $\psi \circ \phi^{-1}$ is smooth.

To show $(U, \phi)$ and $(V, \psi)$ are smoothly compatible, i.e. that $\psi \circ \phi^{-1}$ is a diffeomorphism, it suffices to show that $\psi \circ \phi^{-1}$ is smooth + injective + nonsingular Jacobian. This saves us the trouble of verifying that the inverse map $\phi \circ \psi^{-1}$ is smooth.

Title: Deciding whether an atlas is smooth


问:
Determining the sign of a polynomial given its factorization
Is there a quick way of determining where a polynomial is positive/negative without actually plugging values?

Say you have a polynomial

$$1) f(x)=(x+a)(x+b)$$

or

$$2) f(x)=(x-a)(-x+b)(x-c)$$

How can one quickly know whether function $f(x)$ is positive or negative without plugging in random interval values between the roots?

答:
Looking at $(x+a)(x+b)$, if $x$ is "large" then the polynomial is positive.  Now imagine $x$ decreasing.  Provided $a\neq b$, once $x$ crosses the larger of $-a, -b$, $f(x)$ will change sign.  Once $x$ crosses the other one, it will change sign again.

This strategy needs to be modified if any of the roots are repeated.  For example, $f(x)=(x-2)^5(x-3)^4$.  This is positive for "large" $x$.  Crossing below $3$ the sign does not change, because $4$ is even, so $f(x)$ is positive in $(2,3)$.  Crossing below $2$ the sign does change, because $5$ is odd, so $f(x)$ is negative on $(-\infty, 2)$.


答: Hint: $a + b + ab = -1$ iff $a(1+b) = -(1+b)$. If $b \neq -1$, then you can divide this through by $1 + b$. (I.e., one strategy to prove the disjunction $a = -1$ or $b = -1$ is to assume on of the disjuncts is false and use that fact to prove the other disjunct.)


问:
Error propagation of a variable for an integral
I have an integral that depends on two parameters $a\pm\delta a$ and $b\pm \delta b$. I am doing this integral numerically and no python function can calculate the integral with uncertainties.

So I have calculated the integral for each min, max values of a and b.
As a result I have obtained 4 values, such that;

$$(a + \delta a, b + \delta b) = 13827.450210 \pm 0.000015~~(1)$$
$$(a + \delta a, b - \delta b) = 13827.354688 \pm 0.000015~~(2)$$
$$(a - \delta a, b + \delta b) = 13912.521548 \pm 0.000010~~(3)$$
$$(a - \delta a, b - \delta b) = 13912.425467 \pm 0.000010~~(4)$$

So it is clear that $(2)$ gives the min and $(3)$ gives the max. Let us show the result of the integral as $c \pm \delta c$. So my problem is what is $c$ and $\delta c$ here?

The integral is something like this

$$I(a,b,x) =C\int_0^b \frac{dx}{\sqrt{a(1+x)^3 + \eta(1+x)^4 + (\gamma^2 - a - \eta)}}$$

where $\eta$ and $\gamma$ are constant.

Note: You guys can also generalize it by taking $\eta \pm \delta \eta$ but it is not necessary for now.

I have to take derivatives or integrals numerically. There's no known analytical solution for the integral.

$\eta = 4.177 \times 10^{-5}$, $a = 0.1430 \pm 0.0011$, $b = 1089.92 \pm 0.25$, $\gamma = 0.6736 \pm 0.0054$, $C = 2997.92458$

答:
What is inside the square root is
$$\gamma ^2+ (3 a+4 \eta )x+ 3( a+2 \eta )x^2+ (a+4 \eta )+\eta    x^4\tag 1$$ Write it as
$$\eta\, (x-r_1 ) (x-r_2 ) (x-r_3 ) (x-r_4)$$ where the $r_i$ are the roots of the quartic polynomial given in $(1)$ .

So, we need to compute
$$I(a,b)=\frac C {\sqrt \eta}\,\int_0^b\frac{dx}{\sqrt{(x-r_1 ) (x-r_2) (x-r_3 ) (x-r_4 )}}$$ and we have an elliptic integral  of the first kind (have a look here).

So, now, we can compute all the partial derivatives with repect to $(\eta,r_1,r_2,r_3,r_4)$ and use the chain rule.

So , assuming no cross terms, thz final result write
$$I = I_0 +\frac {\partial I}{\partial a} ( a-a_0)+\frac {\partial I}{\partial b} (b-b_0)+\frac {\partial I}{\partial \eta} (\eta-\eta_0)+\frac {\partial I}{\partial \gamma}  (\gamma-\gamma_0)$$ with
$$I_0=13869.7187382790600280056975524$$
$$\frac {\partial I}{\partial a}=-38667.5002882782982646434723$$
$$\frac {\partial I}{\partial b}=0.1916010843310452774261082$$
$$\frac {\partial I}{\partial \eta}=-1517907.851327789447487779$$
$$\frac {\partial I}{\partial \gamma}=-3984.5811163972118547061439$$


问:
Normalized sum converges in distribution to $N(0,1)$
Let $X_i$ be a sequence of i.i.d. random variables with $E(X_i)=0$ and $Var(X_i)=\sigma^2>0$.  Prove that the distributions of $(\sum_{i=1}^{n} X_i)/ (\sqrt(\sum_{i=1}^{n}X_i^2))$ converge weakly to the $N(0,1)$ distribution.

I tried using the Lyapunov central limit theorem and Lindeberg but I couldn't get anywhere.  Any help would be greatly appreciated.

答:
By SLLN, we have
$$\frac{\sum\limits_{k=1}^n X_k^2}{n}\overset{a.s.}{\rightarrow}\mathbb{E}X_1^2=\sigma^2,$$

as $n\rightarrow\infty$. Together with the classical CLT:
$$\frac{\sum\limits_{k=1}^n X_k}{\sqrt{n\sigma^2}}\Rightarrow N(0,1)$$

gives the final result you want. You should note that a.s. convergence is stronger than convergence in distribution so we can insert the first convergence here.


问:
Does this series converge or diverge by the comparison test?
I have the following series:

$\frac{1}{2}+\frac{2}{7}+\frac{3}{11} . . . \frac{n}{4n-1}$

I rearranged it like-so:

$\frac{n}{4n-1} < \frac{n}{3n-1} < \frac{n}{2n-1} < \frac{2n}{2n-1}$

Although, it's comparatively similar to:

$\frac{1}{2} ...\frac{1}{n}$

So I've made the assumption the series diverges, however, I'm not sure if my reasoning can imply this.

答:
There is no need for the comparison test here. The necessary condition for the convergence of the series $\sum_{n}a_n$ is
$$
\lim_{n\to\infty}a_n=0,
$$

and in your case it is $\frac14$ instead.


答:
Sure. If you have those factorizations, then it's not so bad. You want to look at the sign of each individual term.

In the case of $(1)$, we know that $f(x)$ will be positive when $(x+a)$ and $(x+b)$ are the same sign, and negative when they are different. When is $x + a < 0$? Well, when $x < -a$. When is $x + b < 0$? Similarly, when $x < -b$. Without loss of generality, assume $a > b$.

Then when $x < -a$, we also have $x < -b$. So we would have $f(x) = (x+a)(x+b) = (-)(-) = (+)$.

How about when $-a < x < -b$? Then we have $f(x) = (-)(+) = (-)$.

And when $-a < -b < x$? Then we have $f(x) = (+)(+) = (+)$.

Of course, when you have either $x = -a$ or $x = -b$, then $f(x) = 0$. And if $a = b$, $f(x) = (x+a)^{2}$, so it's always nonnegative, and zero only when $x = -a$.

Using this kind of analysis, can you try your second example?

Title: Determining the sign of a polynomial given its factorization

Body:
Is there a quick way of determining where a polynomial is positive/negative without actually plugging values?

Say you have a polynomial

$$1) f(x)=(x+a)(x+b)$$

or

$$2) f(x)=(x-a)(-x+b)(x-c)$$

How can one quickly know whether function $f(x)$ is positive or negative without plugging in random interval values between the roots?


问:
Existance of multivariable limit
We are asked to find whether the limit
$$
\lim_{(x,y)\to(0,0)}\frac{x^3y^3}{x^8+y^4}
$$
exists. I tried to find through the epsilon-delta definition but wasn't able to apply it successfully. So I tried to come up with a convergent sequence $(x_n)$ such that it converges to $(0,0)$ but $f(x_n)$ does not converge to a single point, thus establishing non-existence of the limit. But I wasn't able to come up with any such sequence.
When I checked on Wolfram-Alpha it showed the limit to be non existent. Can someone provide a proof of this or give example of a sequence $(x_n)$ converging to $(0,0)$ such that such that $f(x_n)$ does not converge ?

答:
By AM-GM inequality we have that

$$\frac{|x^3y^3|}{x^8+\frac{y^4+y^4+y^4}{3}}\leq \frac{3^{\frac{3}{4}}|x^3y^3|}{4|x^2y^3|} \leq |x|$$

thus the limit exists by squeeze theorem.

Title: Existance of multivariable limit


问:
Why is this integral always non-negative?
I don't understand why the following integral is always non-negative.

$$
\int_{0}^{1}(f'(x))^2dx $$

Where $f(x)$ can be any real valued function on $x$ and $$f(0)=f(1)=0$$.

答: Since $f'(x)$ is a real-valued function, $(f'(x))^2$ is always positive. The area under a positve graph cannot be negative. Here a visualization:


问:
Decomposing a rectangle in $\mathbb{R}^d$
Let $K=[a_1,b_1]\times\dots \times [a_d,b_d]$ be a closed rectangle in $\mathbb{R}^d$ and suppose that $K\subset\cup_{i=1}^n R_i$, where each $R_i$ is an open rectangle. In his measure theory book Cohn writes:

$K$ can be decomposed into a finite collection $\{K_j\}$ of rectangles that
overlap only on their boundaries and are such that for each $j$ the interior of $K_j$ is
included in some $R_i$.

If $\overline{R}_i$ denote the closure of $R_i$, then $K\cap\overline{R}_i$ is a closed rectangle. For each $1\leq m \leq d$, let $P_m$ denotes all the left and right $m$th coordinates appearing among the $K\cap\overline{R}_i$ $ (1\leq i\leq n)$. Then $P_m\subset [a_m,b_m]$, and if we order the elements of $P_m$ from left to right we obtain a partition of $[a_m,b_m]$ into a finite collection if closed intervals that overlap only at the boundary. Let $\{K_j\}$ denote the finite collection of all closed rectangles in $\mathbb{R}^d$ which are formed by taking $d$-products of those closed intervals (by choosing one closed interval in the partition of $[a_m,b_m]$ for each $m$). These rectangles intersect only at their boundaries and

$$K=\cup K_j$$

Indeed if $x\in K$, then $x_m \in [a_m,b_m]$ for each $m$, and so $x\in K_j$ for some $j$. Conversely if $x\in K_j$ for some $j$, then $x_m \in [a_m,b_m]$ for each $m$ and so $x\in K$.

Am stuck at proving that the interior of each $K_j$ is included in some $R_i$. It would suffice to show that $K_j\subset K\cap\overline{R}_i$ for some $i$. But I don't see how to show it.

Any help is greatly appreciated.

答:
For completeness I will finish the approach I proposed originaly. This approach might be closer to what Cohn had in mind.

Assume $K^\circ\neq \emptyset$ and $K\cap R_i \neq \emptyset$ for each $i$. Write
$K\cap \bar{R}_i=[a^i_1,b^i_1]\times\dots\times[a^i_d,b^i_d]$. Note that $K=\cup_{i=1}^n K\cap \bar{R}_i$. For each $j=1,\dots,d$, let

$$a_j=c_j^1<c^2_j<\dots<c_j^{m_j}=b_j$$

be the list of all distinct numbers among $a_j^i$ and $b_{j}^i$ for $1\leq i\leq n$. Denote by $\mathcal{L}$ any $n$-tuple of the form $\mathcal{L}=(l_1,\dots,l_d)$ with $1\leq l_j\leq m_j$. Then denote

$$K^{\mathcal{L}}=[c_1^{l_1},c_1^{l_1+1}]\times\dots\times[c_d^{l_d},c_d^{l_d+1}]$$

Clearly $K=\cup\{K^{\mathcal{L}}\}$ and the $K^{\mathcal{L}}$ overlap only on their boundaries. Suppose $(K^{\mathcal{L}})^\circ \cap (K\cap \bar{R}_i) \neq \emptyset$ for some $\mathcal{L}$ and $i$. Then $c_j^{l_j}<x_j<c_j^{l_j+1}$ and $a^i_j\leq x_j\leq b^i_j$ for $1\leq j\leq d$. By the choice of the $c_j$'s we must have $a^i_j\leq c_j^{l_j}<x_j<c_j^{l_j+1}\leq b^i_j$. Thus $K^{\mathcal{L}}\subset K\cap \bar{R}_i\subset \bar{R}_i$ and hence $(K^{\mathcal{L}})^\circ\subset\bar{R}_i^\circ=R_i$. This shows that the interior of $K^{\mathcal{L}}$ is included in some $R_i$.


问:
Proving a function is continuous through differentiability
I have the following function $f_n (x) = \sqrt{x^2 + \frac{1}{n^2}}$. I want to prove this is continuously differentiable. The method I have used seems like a bit of a shortcut but I was wondering if it was still valid.

I just computed the derivative $f_n'(x)$ and then showed that this functions derivative is defined over all $x \in \mathbb{R}$, by computing the second derivative.

Is this valid? Or do I need to prove continuity of the first derivative through a more standard method?

答: A function which is differentiable is continuous, $f_n$ is differentiable (continuous) since it is a composition of differentiable (continuous) functions $g_n(x)=x^2+{1\over n^2}$ and $h(x)=\sqrt{x}$ (defined on $(0,+\infty)$.


答:
Yes, it is valid to prove that a function is continuous by actually proving that it is differentiable.

However, it seems more natural for me to see that$$f_n'(x)=\frac x{\sqrt{x^2+1/n^2}}$$and to say that this function is continuous since it is the quotient of two continuous functions.

Title: Proving a function is continuous through differentiability

Body:
I have the following function $f_n (x) = \sqrt{x^2 + \frac{1}{n^2}}$. I want to prove this is continuously differentiable. The method I have used seems like a bit of a shortcut but I was wondering if it was still valid.

I just computed the derivative $f_n'(x)$ and then showed that this functions derivative is defined over all $x \in \mathbb{R}$, by computing the second derivative.

Is this valid? Or do I need to prove continuity of the first derivative through a more standard method?


问:
Partial fraction decomposition help
In a text that I am reading, they state that the following partial fraction ($r$ fixed) expansion is "readily computed":

$$f(z) = \frac{z^r}{(1-z)(1-2z)(1-3z)\cdots (1-rz)} = \frac{1}{r!} \sum_{j=0}^r \binom rj \frac{(-1)^{r-j}}{1-jz}$$

I know how to do partial fractions, or at least I thought I did. I tried to set it up with something like
$$\frac{A_1}{1-z} + \frac{A_2}{1-2z} + \cdots + \frac{A_r}{1-rz}$$
but it got messy. Also, the answer given above has a constant term $\frac{(-1)^{r}}{r!}$ when $j=0$, which shouldn't show up when doing this method normally.

Could someone point me in the right direction? Thanks!

I now see why the constant term is necessary; doing a "long division" would result in the $\frac{(-1)^r}{r!}$ term. The remainder would be something like
$$\frac{z^r - \frac{1}{r!}(1-z)(1-2z)\cdots(1-rz)}{(1-z)(1-2z)\cdots(1-rz)}$$
which technically could be solved by partial fractions...

Induction also looks promising, as suggested by Maesumi.

Title: Partial fraction decomposition help

Body:
The coefficient of $\frac{1}{1-kz}$ is
$$\left[\frac{z^r}{(1-z)(1-2z)(1-3z)\cdots\widetilde{(1-kz)}\cdots (1-rz)}\right]_{r=\frac{1}{z}}=\frac{\left(\frac{1}{k}\right)^r}{(1-\frac{1}{k})(1-2\frac{1}{k})(1-3\frac{1}{k})\cdots\widetilde{(1-k\frac{1}{k})}\cdots (1-r\frac{1}{k})}=\frac{1}{k(k-1)(k-2)\cdots\widetilde{0}(-1)(-2)\cdots(k-r)}=\frac{1}{r!}  \binom rj (-1)^{r-j}$$


问:
confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test
I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.

$$\frac{n(n+1)}{(n+3)^3}$$

Using the ratio test, then simplifying in stages:

$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+1)}$$

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

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

$$\frac{(n+3)^3}{(n+4)^22n}$$

The concluding remark:

$$\frac{1}{2}\lim_{n \to \infty}\frac{(n+3)^3}{(n+4)^2n}$$

Hence the series is divergent, unless I went wrong somewhere?

答:
Without using asymptotic equivalence, an easy way is
\begin{equation}
\sum_{n=1}^\infty\left(\dfrac{n}{n+3}\times\dfrac{n+1}{n+3}\times\dfrac{1}{n+3}\right)\ge\sum_{n=1}^\infty\left(\dfrac14\times\dfrac24\times\dfrac{1}{n+3}\right)=\infty
\end{equation}

Another alternative solution is
\begin{equation}
\sum_{n=1}^\infty\dfrac{n(n+1)}{(n+3)^3}\ge\sum_{n=1}^\infty\dfrac{n^2}{(n+3n)^3}=\infty
\end{equation}


问:
Quotient ring calculation
I'd love to show $\mathbb{R}[s,u,t] / (su, t(s+u)-1) = \mathbb{R}[s,t]/(ts-1) \times \mathbb{R}[u,t]/(tu-1)$. Is there any suggestion for this calculation?

Edit: What I get is $(ts-1)(tu-1) = t^2(su) - [t(s+u)-1]$. Is this useful?

答:
''Edit: What I get is $(ts−1)(tu−1)=t^2(su)−[t(s+u)−1]$. Is this useful?''

This shows that $(ts-1)(tu-1)$ lies in the ideal $\langle su,t(s+u)-1\rangle$.

Title: Quotient ring calculation


问:
When does injective implies surjective for an $R$-module endomorphism?
Let $R$ be a commutative ring.

It is known that if $M$ is a finitely generated $R$-module and if $f \in End_R(M)$, then surjectivity of $f$ implies injectivity of $f$ (Vasconcelos).

I am looking for sufficient conditions on $R$ or $M$ to have a converse to this statement.

For instance, if $R$ is a field or if $R$ is a finite commutative ring, this will be true (in the former case, because $M$ is a finite dimensional vector space over $R$, and in the latter case, because $|M| < \infty$).

If $M$ is a free module of finite rank, say $M \cong R^n$, then $f$ is just a matrix $A$. Injectivity is equivalent to the columns of $A$ to be linearly equivalent over $R$. By Exercise 5.23B in Exercises in Modules and Rings by T. Y. Lam, this is equivalent to $\det(A) \in R$ not being a zero divisor. If all such elements of $R$ are invertible, then $A$ and thus $f$ are invertible as well. By this question, all artinian rings $R$ work (and when $M$ is free of finite rank).

What are some other conditions?

答:
If $R$ is Artinian and $M$ is finitely generated, then any injective $R$-linear map $M \rightarrow M$ is surjective. Vasconcelos has also shown that all finitely generated $R$-modules have this property if and only if $R$ has Krull dimension $0.$

You can prove a nice theorem for finite rank free modules: $R^n$ has the property that all injective endomorphisms are surjective if and only if $R$ is a quoring (that is, all regular elements are invertible). Seehere. You have part of the argument already in your question.

Modules $M$ with the property you mention are called co-Hopfian. There is an extensive literature on them, and Lam has some exercises around them. It is worth noting that you can work with $M$ that are not finitely generatd as well. For intance, the abelian group $\oplus \mathbb{Z}(p),$ the sum over the positive primes with $\mathbb{Z}(p)$ the cyclic group of order $p,$ has this property. The property follows since there are no nonzero homomorphism between $\mathbb{Z}(p)$ and $\mathbb{Z}(q)$ if $p \neq q.$


问:
Counterexample about weak convergence
Let $X$ be Banach space. Let $S_n, S, T_n, T \in \mathcal{B}(X,X).$ Show that $S_n \to S $(strongly) and $T_n \to T$ (weakly) doesn't necessarily imply that $S_nT_n \to ST$ (weakly).

I can't find the counterexample, can anyone help me find it?

答:
Here is a hint / counter-example:

Let $\mathcal{H}$ be an infinite dimensional separable Hilbert space with ONB $\{e_n\}_{n\geq1}$. For any $n\geq1$, set $$T_nx=\sum_{k=n+1}^\infty\langle x,e_k\rangle e_{k-n},\;\;\;\;x\in\mathcal{H}$$

Show that $(T_n)\subset\mathcal{B(H)}$.

Show that $T_n\to0$ strongly, so $T_n\to0$ weakly, too. Since involution is WOT-WOT continuous, we also have that $T_n^*\to0$ weakly.

Show that $T_nT_n^*\require{cancel}\cancel{\longrightarrow}0$ weakly by observing that $T_nT_n^*e_1=e_1$ for all $n\geq1$.

Title: Counterexample about weak convergence


问:
Classification of groups of order $p^2q$, when $q|p+1$
When considering non-abelian groups of order $p^2q$, where $p>q$ and $p,q>3$, two cases can be identified, when $q|p^2-1$.

Case 1: $q |p+1, q \nmid p-1$

Case 2: $q |p-1, q \nmid p+1$

Can someone guide me to prove that in Case 1, a unique semidirect product is present.

$p,q$ are distinct primes.

Thanks a lot in advance.

答:
Let $G$ be a group of order $p^2q$, with $q|p^2-1$ and $p>q$.

You know that $n_p\equiv_p1$ and $n_p|q\implies n_p\in\{1,q\}$. If $n_p=1\implies\exists !P\in Syl_p(G)$ such that $P\trianglelefteq G$.

If $n_p=q\implies p|q-1\implies p<q$, which is not possible because we are supposing $p>q$, hence $n_p=1$ and we have a unique normal p-group in $G$.

Similarly we know that $n_q\equiv_q 1$ and $n_q\in\{1,p,p^2\}$.

If $n_q=p^2$, then we would have $p^2-1$ elements of order $q$ because every q-Sylow has prime order, hence it's cyclic.

Therefore we would have $|G|-1-|Syl_q(G)|q=p^2q-1-(p^2-1)=p^2(q-1)$ elements of order $p$, but this is not possible because we've shown that there is only one p-Sylow subgroup in $G$.


答:
Let your group be $G$.

First of all, the Sylow $p$-subgroup $P$ will be a unique, as it has index $q < p$.

If $P$ is cyclic, then there is no (non-abelian) group $G$, as $q$ does not divide the order $p (p - 1)$ of the automorphism group of $P$.

Let then $P$ be elementary abelian, so that its automorphism group $A$ has order $$a = (p^{2} - 1) (p^{2} - p) = p (p - 1)^{2} (p + 1).$$

As $q > 2$, $q$ is the highest power of $q$ which divides $a$.

Let $Q$ be a Sylow $q$-subgroup. The groups $G$ are parameterised by the non-trivial homomorphisms $\phi : Q \to A$. The image of such a $\phi$ is thus a Sylow $q$-subgroup of $A$, and all these Sylow subgroups are conjugate.

This, and an easy additional argument with powers, show that all the groups arising from these $\phi$ are isomorphic.

Title: Classification of groups of order $p^2q$, when $q|p+1$

Body:
When considering non-abelian groups of order $p^2q$, where $p>q$ and $p,q>3$, two cases can be identified, when $q|p^2-1$.

Case 1: $q |p+1, q \nmid p-1$

Case 2: $q |p-1, q \nmid p+1$

Can someone guide me to prove that in Case 1, a unique semidirect product is present.

$p,q$ are distinct primes.

Thanks a lot in advance.


问:
Definite Integral from or to; plus or minus infinity
I always knew integrating in a common way:

$\int_{n}^{z}{x^y} \ dx=[\frac{z^{y+1}}{y+1}] - [\frac{n^{y+1}}{y+1}]$

All we do is integrate $x^y$ and then replace the $x$ with $z$ first, then subtract it with itself but instead of replacing $z$, we replace $n$. But then I have seen that:

$e^z = \int_{-\infty}^{z}{e^x} \ dx ≠ [e^z] - [e^{-\infty}]$

Because $e^{-\infty}$ is undefined.

Question is, do we use limits for the second term? Unless, then how do we calculate such an integral?

Thanks for the answers.

答:
An integral
$$
\int_{-\infty}^b f(x)dx
$$
is understood as
$$
\lim_{a\to-\infty}\int_{a}^b f(x)dx,
$$
if such limit exists.


问:
Features of a given scalar field
I have to create a visualization of a scalar field given by the formular:

$$f(x,y) = x^3 - 3xy^2$$

I have to represent some features of this scalar field.

I plotted the following scalar fieldbut can´t really say something about its local maximum and minimum.

Every input is appreciated. Thanks in advance.

I plotted the Contour lines of my scalar field but don´t really understand what they tell me about it. It seems to be a 2d view of the field from above.

答:
The jacobian matrix is $O$ when $\frac {\partial f}{\partial x}= 3x^2 -3y^2 =0$ and $\frac {\partial f}{\partial y}=-6xy=0$.

$-6xy=0$ when $x=0$ or $y=0$, and $3x^2 -3y^2 =0$ iff $x= \pm y$, so $x=y=0$.

Thus $(0,0)$ is a critical point. The Hessian doesn't give us any further information, but if you notice:
let's restrict $f$ to some lines through the origin, $f(t,t)= -2t^3$; so for $t \gt 0$ we have $f \lt 0$; and for $t \lt 0$ we have $f \gt 0$.

Therefore $(0,0)$ is the only critical point and is also a saddle (without defining any compact set).

Title: Features of a given scalar field


问:
Comparing two events
Let $X_n$ be a sequence of real random variables, $X$ a real random variable and $\varepsilon > 0$

Let $A_n$ be the event $ \{ | X_n - X | \geq \varepsilon \} $ and $B_n$ the event $ \{ \sup_{k \geq n} | X_k - X | \geq \varepsilon \} $

I was wondering if we can compare $B_n$ and $ \lim \sup A_n $

Thank you for your answer,

答:
$$\bigcup_{m\geq n}\{|X_m-X|\geq \varepsilon \}\subset B_n$$
therefore $\limsup_{n\to \infty }A_n\subset B_n.$

Title: Comparing two events


问:
Closed form for $\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx$
Consider the following integral:
$$\mathcal{I}(\mu,\nu)=\int_0^\infty\ln\frac{J_\mu(x)^2+Y_\mu(x)^2}{J_\nu(x)^2+Y_\nu(x)^2}\mathrm dx,$$
where $J_\mu(x)$ is the Bessel function of the first kind:
$$J_\mu(x)=\sum _{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\mu+1)}\left(\frac x2\right)^{2n+\mu}$$
and $Y_\mu(x)$ is the Bessel function of the second kind:
$$Y_\mu(x)=\frac{J_\mu(x)\cos(\mu\pi)-J_{-\mu}(x)}{\sin(\mu\pi)}.$$
I was not able to rigorously establish a closed form for $\mathcal{I}(\mu,\nu)$, but based on numerical integration I made a conjecture:
$$\forall\mu,\nu\in\mathbb{R},\hspace{1cm}\mathcal{I}(\mu,\nu)\stackrel{?}{=}\frac{\pi}{2}(\mu^2-\nu^2).$$
Could you please help me to find out if this conjecture is true?

If the conjecture is true, then taking the derivative with respect to $\mu$ at $\mu=1$ we get the following corollary:
$$\int_0^\infty\frac{J_0(x)J_1(x)+Y_0(x)Y_1(x)}{J_1(x)^2+Y_1(x)^2}x^{-1}\mathrm dx\stackrel{?}{=}\frac{\pi}{2}.$$

As pointed by O.L., the conjecture is equivalent to
$$\int_0^{\infty}\ln\frac{\pi\,x\,H^{(1)}_\nu(x)H^{(2)}_\nu(x)}{2}\mathrm dx\stackrel{?}{=}\frac{\pi\,(4\nu^2-1)}{8},$$
where $H^{(1)}_\nu(x)=J_\nu(x)+i\,Y_\nu(x)$ and $H^{(2)}_\nu(x)=J_\nu(x)-i\,Y_\nu(x)$ are the Hankel functions of the first and second kind.

答:
I will assume that $\nu$ is real, as in the formulation
of the question. Something similar may be true for
complex $\nu$ by a similar argument, but there would be an  extra complication (perhaps
only notational) of working with $\nu$ and
its conjugate $\overline{\nu}$ instead of just $\nu$.
The Hankel functions $H^{(i)}_{\nu}(z)$ are entire
except for a branch cut along the negative real axis.
Let
$$A^{(1)}_{\nu}(z) =  \frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)}, \qquad
A^{(2)}_{\nu}(z) =  \frac{H^{(2)}_{\nu-1}(z)}{ H^{(2)}_{\nu}(z)},$$
$$ B^{(1)}_{\nu}(z) =  \frac{H^{(1)}_{\nu+1}(z)}{ H^{(1)}_{\nu}(z)}, \qquad
B^{(2)}_{\nu}(z) =  \frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)}.$$
We introduce the following notation:
for a function $F(z)$ with a branch cut along the negative axis, we let $F(x^{+})$
and $F(x^{-})$ denote the limit of $F(z)$ as $z \rightarrow x$ from the region
$\mathrm{Im}(z) > 0$ and $\mathrm{Im}(z) < 0$ respectively.
There are identities as follows:
$$\text{Eq. 1:} \quad  A^{(1)}_{\nu}(x^{+}) = \overline{A^{(2)}_{\nu}(x^{-})} = -A^{(2)}_{\nu}(-x)
= - \overline{A^{(1)}_{\nu}(-x)},$$
$$
A^{(2)}_{\nu}(x^{+}) = \overline{A^{(1)}_{\nu}(x^{-})}.$$
and the same equations hold for $B$. Note that
it's not true that
$A^{(2)}_{\nu}(x^{+}) = -A^{(2)}_{\nu}(-x)$, the lack of symmetry
here is related to the branch cut. This is an important point.  The behavior of  $A^{(1)}_{\nu}(z)$ is bad in the region near $x^{-}$, and correspondingly
 $A^{(2)}_{\nu}(z)$ is bad near $x^{+}$.
The Hankel function has nice  asymptotic expansions for large $z$.
The ratio of such functions at arguments
$\nu$ differing by integers is particularly nice, because the complex phase cancels.
In particular, one has the following:
$$A^{(1)}_{\nu}(z) \sim
i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}),$$
$$A^{(2)}_{\nu}(z) \sim
-i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}),$$
$$  B^{(2)}_{\nu}(z)  \sim
i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}),$$
$$  B^{(1)}_{\nu}(z)  \sim
-i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}).$$
This holds outside the bad regions mentioned above. In particular,
it holds for $A^{(1)}_{\nu}(z)$ and $B^{(1)}_{\nu}(z)$ for $z$ with argument
in $[-\pi + \epsilon,\pi]$, and for $A^{(2)}_{\nu}(z)$ and $B^{(2)}_{\nu}(z)$ with argument
in $[-\pi,\pi - \epsilon]$.
Let $C_R$ be the semi-circle
with centre $0$ and radius $R$ in the upper half plane, oriented anti-clockwise, and thought of as lying
above the branch cut in $(-\infty,0])$. Note that this is contained within the range where the asymptotic holds for $A^{(1)}_{\nu}(z)$,
and hence
$$\lim_{R \rightarrow \infty} \int_{C_R} z  \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - i \right)
-  \frac{(2 \nu - 1)}{2} dz
=   \pi i \cdot \left( -i  \cdot \frac{(4 \nu^2 - 1)}{8} \right) =
\frac{\pi (4 \nu^2 - 1)}{8}  $$
The main term comes from the residue theorem
(applied to a half circle, hence the $\pi i$ rather than $2 \pi i$ factor), and
the error term comes from the fact that the integral of $O(z^{-2})$ over a
half-circle of radius
$R$ and circumference $\pi R$ is $O(R^{-1})$.

We now use the following fact: $H^{(1)}_{\nu}(z)$ has no zeros in the upper  half plane.
I say that it is a fact, but I couldn't find a reference (Edit: proof of this fact included at the end of this answer). I proved it rigorously by an explicit contour integral
computation for various ranges of values of $\nu$, however. (Certainly, by the asymptotic expansion which is valid in the entire upper half plane, it follows that any such zeros, if they exist, must be within some small radius, which one can eliminate by computing $(2 \pi i)^{-1} \oint d \log(f)$.)
By the residue theorem (taking $C$ above to be the circle in the upper half plane), we get,
for any holomorphic integrand,
$$0 = \oint  = \int_{C} + \int^{R}_{-R},$$
and hence
$$ \lim_{R \rightarrow \infty}  \int^{R}_{-R} z  \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - i \right)
-  \frac{(2 \nu - 1)}{2} dz
 = - \frac{\pi (4 \nu^2 - 1)}{8} .$$
Note that the integrand has order $O(1/z)$, so one really has to take the integrand
from $-R$ to $R$ and then take the limit for this to make sense.
One may apply the same analysis to $H^{(2)}_{\nu}$, except now the zero free region
of $H^{(2)}_{\nu}$
is in the lower half plane --- this follows by symmetry from the
identity $H^{(1)}_{\nu}(\overline{z}) = \overline{H^{(2)}_{\nu}(z)}$, noting
that we are once again in the correct region as far as asymptotics goes.
  Hence we deduce that
$$\lim_{R \rightarrow \infty} \int^{-R}_{R} z  \left(\frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} - i \right)
-  \frac{(2 \nu + 1)}{2} dz = - \frac{\pi (4 \nu^2 - 1)}{8} .$$
Note that the direction of the integral has changed, for orientation reasons.

Warning! There's also another difference between this and the previous integral. The first integral was above the branch cut and this integral is below
 the branch cut. However, in the first case,
we were integrating values of the form $A^{(1)}_{\nu}(x^+)$, which was the good value (in the
sense that it was related to three other values by symmetry in equation 1), and
here we are integrating $B^{(2)}_{\nu}(x^{-})$, which also is related to three
other values by the same equations.

Correcting
the order of the second integrand and then subtracting the results,  we get
$$\lim_{R \rightarrow \infty} \int^{R}_{-R} 1 + z
\left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} - \frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} \right)
 dz = - 2 \cdot  \frac{\pi (4 \nu^2 - 1)}{8}.$$
We now make two observations: the integrand is now $O(z^{-2})$ for large $z$, and hence
it exists as a definite integral. Moreover, the integrand is even. In light of the warning,
we should really specify that the
integrand for values $x \in (-\infty,0]$ is precisely:
$$1 + x \left(A^{(1)}_{\nu}(x^{+}) - B^{(2)}_{\nu}(x^{-})\right).$$
(One should check this is the correct function to make the integrand
even.) We deduce that
$$
- \int^{\infty}_{0} 1 +  z \left(\frac{H^{(1)}_{\nu-1}(z)}{ H^{(1)}_{\nu}(z)} -
\frac{H^{(2)}_{\nu+1}(z)}{ H^{(2)}_{\nu}(z)} \right) dz = \frac{\pi (4 \nu^2 - 1)}{8}.$$
Let
$$I(\nu) = \int^{\infty}_{0} \log \frac{\pi x H^{(1)}_{\nu}(x)  H^{(2)}_{\nu}(x)}{2}  \cdot dx.$$
Integrating by parts, and being a little bit careful about what happens at $0$, and
expressing the derivatives on Hankel functions in terms of Hankel  functions of other arguments, we find that
$$I(\nu) = -  \int^{\infty}_{0} 1 +  x \left(\frac{H^{(1)}_{\nu-1}(x)}{ H^{(1)}_{\nu}(x)} -
\frac{H^{(2)}_{\nu+1}(x)}{ H^{(2)}_{\nu}(x)} \right) dx = \frac{\pi (4 \nu^2 - 1)}{8}.$$
As noted in the comments, this was the identity to be proved.
Alternatively, integration by parts also shows that
$$\frac{\pi (\mu^2 - \nu^2)}{2} = I(\mu) - I(\nu)
=  \int^{\infty}_{0} \log \frac{H^{(1)}_{\mu}(x)  H^{(2)}_{\mu}(x)}{H^{(1)}_{\nu}(x)  H^{(2)}_{\nu}(x)} \cdot dx,$$
and hence
$$ \int^{\infty}_{0} \log \frac{J_{\mu}(x)^2 + Y_{\mu}(x)^2}{J_{\nu}(x)^2 + Y_{\nu}(x)^2} \cdot
dx = \frac{\pi (\mu^2 - \nu^2)}{2}.$$

Edit: Proof that $H^{(1)}_{\nu}(z)$ has no zeros in the upper half plane for real $\nu > 0$.

I realized that the proof can be completed in a similar manner. Let
$$G^{(1)}_{\nu}(z) =  d \log H^{(1)}_{\nu}(z)
=\frac{1}{2} \left(A^{(1)}_{\nu}(z) -  B^{(1)}_{\nu}(z)\right).$$
Then we have an asymptotic formula, as before:
$$A^{(1)}_{\nu}(z) \sim
i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu - 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}), $$
$$  B^{(1)}_{\nu}(z)  \sim
- i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right)
+ \frac{(2 \nu + 1)}{2} \cdot \frac{1}{z}  + O(z^{-3}), $$
and thus
$$G^{(1)}_{\nu}(z) \sim  i \left(1 - \frac{(4 \nu^2 - 1)}{8} \cdot \frac{1}{z^2} +  \ldots \right) - \frac{1}{2z} + O(z^{-3}),$$
This is valid for $z$ with argument in $[-\pi + \epsilon,\pi]$, so in particular is valid in the upper half plane.
If $C_R$ denotes the circle in the upper half plane, we find that:
$$\lim_{R \rightarrow \infty} \int_{C_R} (G^{(1)}_{\nu}(z) - i) dz = - \frac{\pi i}{2},$$
because it is $O(R^{-1})$ plus the contribution from the $1/(2z)$ term.
Let $\Omega_R$ denote the boundary of the  region bounded by by the interval $[-R,R]$ and $C_R$.
The function $H^{(1)}_{\nu}(z)$ and thus $H^{(1)}_{\nu}(z) e^{-iz}$ is holomorphic in $\Omega_R$
away from $z = 0$. At zero (and $\nu > 0$)  we have an asymptotic
$$H^{(1)}_{\nu}(z) \sim  - i \cdot \frac{\Gamma(\nu)}{\pi} \left(\frac{2}{z}\right)^{\nu}.$$

It follows that the number of zeros
in the upper half plane is given, accounting for the singularity at $0$ (modified by a factor
of two since this integral only accounts for half of the singularity) by
$$\frac{\nu}{2} + \lim_{R \rightarrow \infty} \frac{1}{2 \pi i} \oint_{\Omega_R} d  \log (H^{(1)}_{\nu}(z) e^{-iz}) dz$$
$$= \frac{\nu}{2} + \frac{1}{2 \pi i} \left( \lim_{R \rightarrow \infty} \int_{C_R}  (G^{(1)}_{\nu}(z) - i) dz +
\lim_{R \rightarrow \infty} \int_{-R}^{R} (G^{(1)}_{\nu}(z) - i) dz\right).$$
We computed the first integral above, hence it suffices to compute:
$$\frac{\nu}{2}    -\frac{1}{4} + \frac{1}{2 \pi i} \lim_{R \rightarrow \infty} \int_{-R}^{R} (G^{(1)}_{\nu}(z) - i) dz.$$
We may write this as the integral
$$\frac{\nu}{2}  -\frac{1}{4} + \frac{1}{2 \pi i} \int_{0}^{\infty} \left( G^{(1)}_{\nu}(z) + G^{(1)}_{\nu}(-z) - 2 i \right) dz.$$
This expression evaluates to an integer, which is the number of zeros of $H^{(1)}_{\nu}(z)$
in the upper half plane.
In particular, evaluating this integral numerically for a random value (say $\nu = \pi$) shows
that, for this value, it is equal to zero. Now suppose that we vary $\nu$. Since this integral
evaluates to an integer, to complete the proof, it suffices to show that it varies continuously.
For the integral over $[R,\infty)$ this is clear  for large enough $R$ from the asymptotic formula. For small values, it suffices to show that $H^{(1)}_{\nu}(z)$ doesn't have any zeros on
the real line $[0,R]$, since otherwise the integrand is continuous in $\nu$, and the continuity
of the integral is clear. Now on the real axis, we have, by definition,
$$H^{(1)}_{\nu}(z) = J_{\nu}(z) + i \cdot Y_{\nu}(z).$$
Since $\nu$ is real, it suffices to show that $J_{\nu}(z)$ and $Y_{\nu}(z)$
do not have any simultaneous zeros.  However, the zeros of these Bessel functions
are well known to interlace (see Watson, A treatise on the theory of Bessel functions),
and the result is established. Edit: Actually, there's a much easier proof that  $J_{\nu}(z)$ and $Y_{\nu}(z)$
do not have any common zeros --- they are linearly independent solutions to a second order ODE!


问:
Wronskian of $x^3$ and $x^2 |x|$
The question is fairly straight forward, given $2$ functions $y_1=x^3$ and $y_2=x^2|x|$ determine whether the functions are linearly dependent or independent.

I am conflicted because in my differential equations textbook it states that two functions are linearly independent if their quotient is a non-constant function for all $x$. Here it is evident that for $x<0$ the quotient $\frac{y1}{y2} = 1$ while for $x<0$, $\frac{y1}{y2}=-1$. This would lead to a conclusion that the functions are linearly independent.

However, in the textbook it is also stated that two linearly dependent functions will have a Wronskian of zero on the real number line.

So the Wronskian $W(y_1(x), y_2(x)) = \det \begin{vmatrix}y_1(x)& y_2(x)\\ y_1'(x)& y_2'(x) \end{vmatrix}$ and $y_1'(x) = 3x^2$

the derivative of $y_2'(x)$ is slightly more tricky but checking with wolfram alpha
$y_2'(x)=\frac{3x^3}{|x|}$

so the Wronskian is given by:

$$W=y_1(x)y_2'(x)-y_2(x)y_1'(x)=x^3\times\frac{3x^3}{|x|}-x^2|x|\times3x^2$$

Now multiplying the second term by $\frac{|x|}{|x|}$ in order to add the terms yields

$$W=\frac{3x^6}{|x|}-3x^4\times \frac{|x|^2}{|x|}$$

At this point I assume $|x|^2=x^2$

which yields:

$$W=\frac{3x^6}{|x|}-\frac{3x^6}{|x|}=0 $$

This would imply the functions are linearly dependent....

The answer provided by the textbook states that the functions are linearly independent. Does anyone have some words of wisdom for me :)

Cheers

答: If the Wronskian is not equal to zero for some x in I, then the functions are linearly independent on the interval. Also if the functions are linearly dependent on I, then the  Wronskian is equal to zero for all x in the interval I. But W=0 does not mean the functions are dependent.</p>

Title: Wronskian of $x^3$ and $x^2 |x|$


问:
Coefficient of integral dependence and prime ideals
Let $A \subseteq B$ be an integral extension of rings. Let $b \in B$, and let $b^n+a_1b^{n-1}+\dots+a_{n-1}b+a_n=0$ be an integral dependence relation for $b$, with $a_1,\dots,a_n\in A$.

Given a prime ideal $\mathfrak{p}$ in $B$, I am trying to prove that $b\in\mathfrak{p}$ if and only if $a_1,\dots,a_n\in \mathfrak{p}\cap A$. I have managed to prove the implication $\Leftarrow$, but I am stuck with the other: I am only able to show the claim for $a_n$.

In particular, I am interested in the case where $A$ is the subring of invariants under some group action on $B$, but since I proved one implication in this more general setting I suspect that the other direction may hold too under these assumptions.

答:
It is not true in general. Note that $b\in\mathfrak{p}\Leftrightarrow a_1,\dots,a_n\in\mathfrak{p}\cap A$ implies $$f\left(D(b)\right)=\bigcup_i D(a_i)$$ where $f:\operatorname{Spec} B\to\operatorname {Spec} A,\mathfrak{p}\mapsto A\cap\mathfrak{p}$ is the induced map on the spectra. In particular $f$ would be an open map, and hence going-down would hold for the extension $B/A$. But there are integral extensions for which going-down does not hold (see e.g. here or here).

I don't know about the general case for  the invariant ring under group actions but the following special case is true: If $G$ is a finite group that acts on $B$ and $A=B^G$ is the ring of invariants, then for $b\in B$ the claim (almost, see edit) holds for the polynomial $$\prod_{\sigma\in G}(X-\sigma(b))\in A[X]$$
(Since the coefficients of that polynomials are sums of products of elements of the form $\sigma(b)$)

Edit: It is also not true for the invariant ring, here is an easy example (much easier than the argument above): Let $B=\Bbb Z[i]$ be the ring of integers in $\Bbb Q(i)$ and $A=\Bbb Z$. The element $b=i+2$ satisfies $b^2-4b+5=0$ and we have $b\in \langle b\rangle =:\mathfrak{p}$, but $4\notin \mathfrak{p}\cap A=\langle 5\rangle$.

Note that $A$ is the ring of invariants under the Galois action on $B$.

What is true is the following (in the general setting again): If $b$ is contained in all prime ideals $\mathfrak{p}$ lying over the same prime $\mathfrak{q}$ of $A$ then also all $a_i$ lie in $\mathfrak{q}$. Here the $a_i$ are defined by $$\prod_{\sigma\in G}(X-\sigma(b))=X^n+a_1X^{n-1}+\dots+a_n$$
Since the elements $\sigma\in G$ permute the primes lying over $\mathfrak{q}$ all sums of products of elements of the form $\sigma(b)$ are also contained in all primes $\mathfrak{p}$ lying over $\mathfrak{q}$. Now the coefficients $a_i$ are exactly of this form and are furthermore contained in $A$, so we have $a_i\in \mathfrak{p}\cap A=\mathfrak{q}$


问:
Midpoints of diagonals
Let $ABCD$ be convex quadrilateral such that $AB=CD$. And $E\neq F$ where $E, F$ is midpoint of $AC, BD$ respectively. Then prove that angle between$(AB, EF)$ and $(CD, EF)$ are equal.

I'll prove $\vec{AB}\cdot\vec{EF}=\vec{CD}\cdot\vec{EF}$. Using $\vec{EF}=\frac{1}{4}(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})$, it's sufficient to prove that $$(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{AB}=(\vec{AB}+\vec{CB}+\vec{CD}+\vec{AD})\cdot\vec{CD}\iff (\vec{CB}+\vec{AD})\cdot\vec{AB}=(\vec{CB}+\vec{AD})\cdot\vec{CD}\iff BC\cdot CDcos\angle C+AD\cdot CDcos\angle D=BC\cdot ABcos\angle B+BA\cdot ADcos\angle A\iff BCcos\angle C+ADcos\angle D=CBcos\angle B+ADcos\angle A$$. I can't continue after this. Can anyone help me?

答:
Let the position vectors of $A,B,C$ and $D$ be $a,b,c$ and $d$ (not writing arrow above the symbols for ease in writing).
\begin{align}
(\vec {AB}-\vec {CD})\cdot\vec{EF}&=(b-a-(d-c))\cdot\left(\dfrac{b+d}{2}-\dfrac{a+c}{2}\right)\\
&=\dfrac12\times(b-a+c-d)\cdot(b-a-(c-d))\\
&=\dfrac12\times(|b-a|^2-|c-d|^2)\\
&=\dfrac12\times(AB^2-CD^2)\\
&=0
\end{align}

Title: Midpoints of diagonals


问:
Area of a triangle formed by mid-points of two sides and the other vertex
Let the medians $AK$ and $BM$ of a triangle $ABC$ intersect at $O$, $AB=13,BC=14,CA=15$. Find the area of $\triangle AOM$.

The lengths of the medians are given by:$$m_a=AK=\dfrac12\sqrt{2b^2+2c^2-a^2},\\m_b=BM=\dfrac12\sqrt{2a^2+2c^2-b^2}.$$ So $AM=\dfrac{15}{2}$ and we can use the fact that the centroid $O$ divides each median into two parts, which are always in the ratio $2:1$ to find $AO$ and $MO$. Then we can find the area by Heron's formula. I am not sure this is the easiest approach. What do you think? Thank you in advance!

答: Draw also the median from $C$. Observe that the area of all six triangles are equal. Find the total area using Heron's Formula and divide it by 6 to obtain the desired area.


答:
Since $O$ is the centroid, the area of $\triangle AOM$ is $\frac16$ of that of $\triangle ABC$.

The area of $\triangle ABC$ is easily found by Heron's Formula; hence the area of  $\triangle AOM$ is easily found as well.

Title: Area of a triangle formed by mid-points of two sides and the other vertex

Body:
Let the medians $AK$ and $BM$ of a triangle $ABC$ intersect at $O$, $AB=13,BC=14,CA=15$. Find the area of $\triangle AOM$.

The lengths of the medians are given by:$$m_a=AK=\dfrac12\sqrt{2b^2+2c^2-a^2},\\m_b=BM=\dfrac12\sqrt{2a^2+2c^2-b^2}.$$ So $AM=\dfrac{15}{2}$ and we can use the fact that the centroid $O$ divides each median into two parts, which are always in the ratio $2:1$ to find $AO$ and $MO$. Then we can find the area by Heron's formula. I am not sure this is the easiest approach. What do you think? Thank you in advance!


问:
least squares minimum
it's the age old least-squares problem:

Find the value of b for which the following sum has the minimum value (assuming that x and y are one-dimensional vectors of the same length):

$$ \sum_{i=1}^n (y_i - bx_i)^2 = min $$

This must have been asked before but I can't find it, would appreciate a redirect from somebody thanks

答: Hint: Differentiate with respect to $b$ and equate to $0$.

Title: least squares minimum


答:
Altitude from $M$ to $AB$ is $\frac{1}{2}$ of altitude from $C$ to $AB$.

So $\triangle AMB = \frac{1}{2} \triangle ABC$.

Now taking base as $AM$, altitude from $O$ to $AM$ is $\frac{1}{3}$ of altitude from $B$ to $AM$.

(as $OM$ is $\frac{1}{3}$ of $BM$).

So $\triangle AOM = \frac{1}{3} \triangle AMB = \frac{1}{6} \triangle ABC$.

You can find area of $\triangle ABC$ using its side lengths.


问:
To which group is the $\mathbb Z_{20}^*$ isomorphic?
I have a question saying that to which group is $\mathbb{Z}_{20}^{*}$ is isomorphic, where $\mathbb{Z}_{20}^{*}$ is the set of the not zero divisors of $\mathbb{Z}_{20}$.

Here is what i think: $\mathbb{Z}_{20}^{*}=\{1,3,7,9,11,13,17,19\}$. It has 8 elements. Then it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ or $\mathbb{Z}_{8}$. But how can I decide which of them? And is this group a multiplicative group or an additive group? How can I know?

Thank you

答:
Hint By the Chinese Remainder Theorem $Z_{20}$ is isomorphic to $Z_{4}\times Z_5$. Also by the Chinese Remainder Theorem $Z_{20}^*$ is isomorphic to $Z_{4}^*\times Z_5^*$ [If you know rings, you can use directly the CRT for rings, otherwise you can prove the second statement from the standard CRT].

Since $5$ is prime $Z_5^*$ is a cyclic group with $4$ elements. $Z_{4}^*$ has 2 elements, and there are not too many groups with 2 elements...