问:
Do there exist planar curves that can be constructed from geometric figures beyond conic sections?
Conic sections are algebraic curves that can be constructed from different cross-sections of a cone in $\mathbb{R}^3$. Do there exist other geometric shapes in $\mathbb{R}^3$ besides the cone whos cross-sections can yield algebraic curves analogous to conic sections? Are there higher dimensional equivalents for 3D curves or sheets in 3D that can be defined in by some geometric figure like a cone (or potential alternative geometric figures) in $\mathbb{R}^4$?

Maybe I can be a bit more concrete. Conic sections are an example of a two-variable second degree polynomial of the form
$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.
$$
I can construct a similar two-variable higher-degree polynomials as such
$$
\text{Curve}_n(x,y) =
A_0x^n + A_2x^{n-1}y + \ldots + A_{n-1}xy^{n-1} + A_ny^n + \text{Constant} = 0
$$
Does there exist something analogous to a "cone" for such a polynomial? What if I add another variable $z$?

答:
A 3D generalization of the conic curves are the quadric surfaces, of general equation

$$Ax^2+By^2+Cz^2+2Dxy+2Eyz+2Fzx+2Gx+2Hy+2Iz+J=0.$$

There are nine different types, not counting the degenerate ones (I would not count the sphere as different of the ellipsoid).

They can be seen as sections of an hypercone by an hyperplane,

$$\begin{cases}Ax^2+By^2+Cz^2+Jw^2+2Dxy+2Eyz+2Fzx+2Gxw+2Hyw+2Izw=0,\\w=1.\end{cases}$$

A cross-section of a quadric by a plane is a conic. Nothing new.

The intersection of two quadrics is a 3D quartic curve, forming a hair-raisingly complex family.


问:
Prove that a recurrent sequence $u_n$ is convergent if $\lim \limits_{n \to \infty}(u_{n+1}-u_n)=0$
Let $f:[a;b] \to [a;b]$ be a continuous function ($a$ and $b$ real numbers) and $(u_n)$ a sequence defined by  $u_0 \in [a;b]$ and $u_{n+1}=f(u_n)$

Prove that if $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$ then $(u_n)$ converges

I tried using subsequences and the bolzano weierstrass theorem but couldn't even get close.

Any help would be welcome.

答:
Lemma : If $(u_n)$ is a sequence such that $u_{n+1} - u_n \rightarrow 0$, then the set of all the adherence value (the word in french is "valeur d'adhérence") is an interval.

Proof : Let $x$ and $y$ two different adherence value (if they exist). Let $c\in ]x,y[$. To show it's an adherence value of $u$, we just need to show that $$\forall \varepsilon >0, \forall N\in \mathbb{N}, \exists n\geqslant N, |u_n -c| \leqslant \varepsilon$$

We take $\varepsilon >0$. Since $u_{n+1} - u_n \rightarrow 0$, there exists $N$ such that for all $n\geqslant N$, $|u_{n+1} - u_n|\leqslant \varepsilon$. Since $x$ is an adherence value of $u$, with $x<c$, there exists a rank $p\geqslant N$ such that $u_p <c$. Since $y$ is also one with $y>c$, there exists $q>p$ such that $u_q >c$. Let's consider the smallest $n>p$ such that $u_n >c$. Then we have $$u_n >c, u_{n-1}\leqslant c \text{ and } |u_n-u_{n-1}|\leqslant \varepsilon$$

So $u_n$ is in $]c, c+\varepsilon]$, which gives the result

Proof of the theorem :

We first note that all adherence values are fixed points of $f$.

From the Lemma, we get that if we have $2$ adherence values, for instance $x$ and $y$, then $[x,y]$ is only consituted of adherence values, and so of fixed points.
That means since we have to go from $x$ to $y$ with jumps that tend to $0$, when the length of the jumps becomes inferior strictly to $|y-x|/2$, then the sequence take a value in $[x,y]$, so takes a fixed point as value, so that the sequence gets constant after this rank.

That contradicts the fact that $x$ and $y$ are adherence values.

So $(u_n)$ converges.

ContentLicense: CC BY-SA 4.0


问:
Non-existence of this stochastic process?
I am intrigued by the following statement in Oksendal , Stochastic differential equations, Chapter 3:

(A)

There does not exist a 'reasonable' stochastic process satisfying the following conditions:

(i) $t_1 \ne t_2 \implies$ $W_{t_1}$ and $W_{t_2}$ are independent

(ii) $W_t$ is stationary ie the joint distribution of $\{W_{{t_1}+t}...W_{{t_k}+t}\}$ does not depend on t

(B)

Moreover if we require $E[W_t^2]=1$ then the function $(t,\omega)\rightarrow W_t(\omega) $ cannot be measurable with respect to the $\sigma$-algebra $B \text{ x } F$ where $B$ is the Borel $\sigma$-algebra on $[0,\infty[ $ (See Kallianpur 1980 p10)

I do not have access to that reference. Is there another one that someone could suggest to explain these statements?

答:
Here is the desired result, still looking for an answer worth a bounty as this answer could use additional explanation. For instance:

After the part about applying Fubini's theorem,

It says $\int_I X_t(\omega)dt = 0$ for $\omega \notin N_I$ where $P(N_I) = 0$, then they construct this union $N = \cup_I N_I$ such that $P(N) = 0$-- it is not entirely clear why the introduction of $N$.

Nonetheless here is the result

ContentLicense: CC BY-SA 3.0


问:
Example of a map which is not closed but every closed ball maps into closed set
I'm trying to find a counter-example for the following statement:
"Consider $X$ and $Y$ be any two metric spaces. If $f: X\rightarrow Y $ is a map such that $f$ maps closed balls in $X$ to closed Sets in $Y$ then $f$ is a closed map "
I thought about inclusion maps , projection maps, constant maps, but I was unable to think counter example from there. Please help me on how to think of any counter example? I am a beginner in topology and metric spaces.

答: why are you saying that projection maps would not work? What about the projection of the plane onto the real line? The graph of the hyperbola $h(x)=1/x$ is closed in the plane, but its projection is the real line minus the origin, which is not closed.


问:
A matrix w/integer eigenvalues and trigonometric identity
Any intuition and/or rigorous arguments on the proofs of the following statements would be appreciated:

Let $n$ be a natural number.
(a) Consider the following Toeplitz/circulant symmetric matrix:

$$
\Lambda_{kl} = \begin{cases}\frac{2n(n+1)}{3}, & k = l, \\[6pt]
\frac{-1}{1-\cos\frac{2\pi(k-l)}{2n+1}}, & k \ne l,\end{cases}
$$

where $k,l = 1,2, \ldots, 2n+1$.

Prove that its eigenvalues are natural numbers(!)

$$2n, 4n-2, 6n-6, \ldots, n(n+1)$$

with multiplicity $2$ and $0$ w/multiplicity $1$.

(b) Find a formula for the right signs in the following trigonometric identity

$$
\tan\left(\frac{l\pi}{2n+1}\right) = 2\sum_{k=1}^n (\pm)\sin\left(\frac{k\pi}{2n+1}\right),\qquad l = 1,2,\ldots, n.
$$

and show that the signs are uniquely determined in the case of prime $2n+1$.

The problems (that are formulated to be self-contained) arise from discrete approximation of continuous inverse boundary problem (see http://en.wikibooks.org/wiki/On_2D_Inverse_Problems for background) and the statements were formulated w/computer help.

答:
Since (b) has been largely answered in the comments, I'll deal with (a) here.
It is the case here that there is a very simple (and otherwise useful and well-known) basis of eigenvectors. Let $\zeta$ be a primitive $2n+1$-th root of unity. Our eigenvectors for $\Lambda$ will be the “cyclic” vectors, whose coordinates are successive powers of a root of unity :

$$
C_s=(\zeta^s,\zeta^{2s},\zeta^{3s}, \ldots ,\zeta^{(2n)s},1) \ (0 \leq s \leq 2n)
\tag{1}
$$

Let us compute all the coordinates of $\Lambda C_s$ ; in other words, we must
compute the sum

$$
f(i,s)=\sum_{j=1}^{2n+1} \Lambda_{ij} \zeta^{sj} \tag{2}
$$

We have $f(i,s)=\frac{2}{3}n(n+1)\zeta^{si}+g(i,s)$, where

$$
g(i,s)=\sum_{j\neq i} \frac{\zeta^{sj}}{\frac{\zeta^{i-j}+\zeta^{j-i}}{2}-1}=
\sum_{t\neq 0} \frac{\zeta^{s(i+t)}}{\frac{\zeta^{-t}+\zeta^{t}}{2}-1}=
2\zeta^{si}\sum_{t=1}^{2n} \frac{(\zeta^t)^{s+1}}{(\zeta^{t}-1)^2}=
2\zeta^{si}\sum_{\eta\in U, \eta\neq 1} \frac{\eta^{s+1}}{(\eta-1)^2}
\tag{3}
$$

where $U$ is the set of all $(2n+1)$-th roots of unity. There are classical techniques to compute the RHS of (3) above ; see the answer to this question, where the final
formula obtained is

$$
\sum_{\eta\in U, \eta\neq 1} \frac{\eta^{s+1}}{(\eta-1)^2}
=\frac{6(2n+1-s)s-((2n+1)^2-1)}{12}=
\frac{s(2n+1-s)}{2}-\frac{n^2+n}{3}
 \tag{4}
$$

Combining (3) with (4), the terms in $\frac{n^2+n}{3}$ cancel out and we thus obtain

$$
\sum_{j=1}^{2n+1} \Lambda_{ij} \zeta^{sj}=s(2n+1-s)\zeta^{si} \tag{5}
$$

Or, to express things vectorially,

$$
\Lambda C_s=s(2n+1-s) C_s \tag{6}
$$

So $C_s$ is an eigenvector associated to the eigenvalue $s(2n+1-s)$ as wished.


答:
$\{u_n\}$ is a sequence taking values in the compact segment $[a,b]$. Hence it has a convergent subsequence $\{u_{\varphi(n)}\}$ where $\varphi : \mathbb N \to \mathbb N$ is strictly increasing. Let say that $\lim\limits_{n \to \infty} u_{\varphi(n)} = l \in [a, b]$.

Suppose that $\{u_n\}$ has a second limit point $L \neq l$. We can find a subsequence $\{u_{\psi(n)}\}$ converging to $L$. Without loss of generality, we can suppose $l \lt L$. As $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$, any $ x \in [l , L]$ is a limit point of $\{u_n\}$.

Now as $f$ is supposed to be continuous, we get that $f(x) = x$ for all $x \in [l,L]$. But if $u_N \in (l,L)$ for $N \in \mathbb N$, we get the contradiction $u_p = u_N$ for all $p \ge N$ and therefore that $u_N$ is the only limit point of $\{u_n\}$.

Conclusion: $\{u_n\}$ has a single limit point and therefore converges.

Title: Prove that a recurrent sequence $u_n$ is convergent if $\lim \limits_{n \to \infty}(u_{n+1}-u_n)=0$

Body:
Let $f:[a;b] \to [a;b]$ be a continuous function ($a$ and $b$ real numbers) and $(u_n)$ a sequence defined by  $u_0 \in [a;b]$ and $u_{n+1}=f(u_n)$

Prove that if $\lim \limits_{n \to \infty}\left(u_{n+1}-u_n\right)=0$ then $(u_n)$ converges

I tried using subsequences and the bolzano weierstrass theorem but couldn't even get close.

Any help would be welcome.


问:
Proving that $TS^n \oplus (S^n \times \mathbb{R}) \cong S^n \times \mathbb{R}^{n+1}$ as vector bundles.
I was going through some exercise to prepare for an upcoming exam and I stumbled upon the following one, I am not fully able to solve.

Exercise 6. Consider the tangent bundle $TS^n$ of the $n$-dimensional sphere. Show that the direct sum of $TS^n$ with the trivial bundle of rank one is isomorphic to the trivial vector bundle of rank $n+1$. Note: This does not imply that $TS^n$ is isomorphic to a trivial bundle.

My Own Attempts

Little remark before I begin:

In the manual I am using, we call two vector bundles $E$ and $F$ isomorphic if there exists a collection of maps $\{u_x \colon E_x \to F_x\}_{x \in M}$　that are isomorphisms as vector spaces such that the induced map $U \colon E \to F$ is smooth.

Using this definition I proved that a vector bundle is trivial if and only if you can find a global frame. So my goal was to find a global frame of $TS^n \oplus (S^n \oplus \mathbb{R})$.
I had a bit of trouble doing this, so I decided to look at the fibres:
$$
(TS^n \oplus (S^n \oplus \mathbb{R}))_x \cong T_x S^n \oplus \mathbb{R}, \quad (S^n \oplus \mathbb{R}^{n+1})_{x} \cong \mathbb{R}^{n+1}.
$$
Without proving it, I sort of intuitively thought that
$$
T_x S^n = \{v \in \mathbb{R}^{n+1} \mid \langle v, x \rangle = 0\},
$$
where $\langle \cdot, \cdot \rangle$ denotes the inner product on the Euclidean space. I wasn't directly able to prove it, but I found another stackexchange post confirming this train of thought. I hoped I was able to then to be able to construct a global section using this description, but I was unfortunately not able to do so.

Other stackexchange posts

After these attempts, I tried to find out whether I could gather some hints by looking at some other stackexchange posts. First I came across this post, in which it is stated "the sum of the tangent bundle and normal bundle of the sphere is trivial." I have personally never worked with normal bundles, but I think that the proof I have to deliver is a simpler version of this proof. I went on looking, but I found other posts that seemed to concern that exact same point (e.g. this one, and this one).

Concluding

I hope that I have provided enough resources for you to identify where I stand with this problem. If it isn't too much trouble, I would like some tips on how to further tackle this exercise.

Thanks a lot in advance!

P.S. If I need to specify more definitions, don't be afraid to ask!

答:
Thanks to Tom Ariel's sharp insight, I was set on the right path again.
The question is flawed, we should actually want to prove that
$$
TS^n \oplus \mathbb{R} \cong S^{n} \oplus \mathbb{R}^{n+1}.
$$
See the comments under the original question to see why I made this initial mistake.

The strategy I used now is to explicitly write out what both sides look like.
We can view
$$
TS^n = \{(x,v) \mid \langle v, x \rangle = 0, x \in S^n,  v \in \mathbb{R}^{n+1} \},
$$
Now using that any vector $w \in R^{n+1}$ can be uniquely written as a sum of vectors tangent to and normal the sphere (see comments in this post), we explicitly write down the map
$$
S^n \times \mathbb{R}^{n+1} \to TS^n \times \mathbb{R}: (x,v) = (x, v - \langle v,x \rangle x, \langle x, v \rangle).
$$
You can easily check that this map is well-defined:
$$
\langle x, v - \langle v,x \rangle x \rangle = \langle x, v \rangle - \langle v,x \rangle \langle x, x \rangle = 0,
$$
where in the last step we used that $x \in S^n$. It is (I think) relatively easy to show that this map is bijective by using the fact that any vector $w \in R^{n+1}$ can be uniquely written as a sum of vectors tangent to and normal the sphere. I also think we can claim that this map is smooth as the inner-product is a smooth function and the total function is thus a composition of smooth maps. It is also easy to see that this map is compatible with the fibers of both vector spaces (as it preserves the base point $x$). Using this, we can safely assume that the two spaces are isomorphic as vector bundles.


问:
Why are math texts usually so compact?
I taught a couple years in a secondary school, where teachers are supposed to explain every little step in the simplest but accurate way.

In high school things don't change much. Perhaps the pace is a little faster but still the goal is to be as clear as possible. Teachers are supposed to use pictures, animations and justify every step when solving problems.

In the university things are a lot different. We get exercises every week and solutions are extremely compact. I don't mean to show every little step when computing a determinant of a $3\times 3$ matrix but often one reads "one implication is trivial, one can (easily) see that, it is clear that..." and so on, which makes me personally feel incompetent because most of the times I don't easily see what I'm supposed to. At least in secondary school, it is mandatory to avoid using words such as "trivial, easy" and so on because of that reason. Still, some teachers do it and some believe this to be one of the causes of the spread of anxiety for mathematics, which certainly hinders the learning process even more.

This applies to this platform as well. The style completely changes.
Why is that so? Are compact solutions more elegant? Are mathematicians just lazy?

答:
In school, mathematics lessons largely focus on teaching you how to follow certain algorithms, so it's important to spell out every step well enough each student will get the hang of it. But from the undergraduate level onwards, you're being trained to write proofs for mathematicians. There are two key challenges here: inculcating the right vocabulary, and ensuring a proof's author and reader see the forest for the trees. As I've discussed here, the latter goal often requires concision, some of which makes the reader write half the proof in their head.

Some concision techniques serve both goals simultaneously. For example, "it's trivial" not only keeps the write-up short (partly by leveraging the reader as aforesaid), but also trains the reader to do the same when they write a proof. For example, if I have to use a fact of the form $\sum_{k=1}^na_k=b_n$ in a proof, I'm not going to write out a proof by induction before I move on. What would be the point? If anything, the reader might get lost. "Wait, why do we care again? Wait, what were we working on? Wait, what was it we proved before they're using now?"

Teachers at every level train students to write what they're expected to. In an exam, you show your work so if you make a mistake you can get partial credit. (That's what I was already told, anyway; how you're supposed to work out the answer without some work you might as well write down is a separate issue.) In a paper you're preparing for publication, you keep it short because they don't want to allow you many pages, nor do readers want you to waste their time. If I saw someone waste four lines on a trivial induction proof in a 7-page paper that's trying to teach the mathematics community something new, I'd be gobsmacked. That's four lines that could have gone towards one of the less trivial things that made the paper worthwhile.

Much of this comes down to chunking. Once you've learned how to do something that has several parts, it can become one part in learning to do something else etc. This is as true of writing proofs as it is of executing algorithms. Before you know it, 1,000 pages of secondary school textbooks will feel like the contents of a post card in your mind. So don't sweat what you're struggling with now; you'll be surprised how quickly it all becomes second nature.

Edit at @BCLC's suggestion: the concision I described can be preferable to relegating details to appendices because, whereas certain steps can break the flow even if brief, appendices are reserved for examples that nevertheless cannot be omitted due to being non-obvious.

Title: Why are math texts usually so compact?


问:
When does a nonhomogeneous system of equations have a unique solution for every right-hand side of the equation?
I was solving the following question from my textbook but couldn't conclude an answer.

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for every right-hand side?

$Ax=b\\$, where A is a 7 x 6 matrix. Since n = 6, $Rank(A)\leq6$

Because there are 7 equations, $m=7$, thus $Col(A)$ is a subspace of $\mathbb{R^7}$. Since $Rank(A)$ is at most 6, $Col(A)$ is at most a 6-dimensional subspace of $\mathbb{R^7}$.

At this point, I'm not sure what to conclude from this reasoning.

答: That's is: since $\operatorname{col}A$ has dimension $\le6$, it isn't the whole $\Bbb R^7$, and therefore there is a vector $b\in\Bbb R^7\setminus\operatorname{col}A$. Any such vector satisfies $[\forall x\in\Bbb R^6, Ax\ne b]$.


问:
Prove that $z_1z_2z_3 = 1$ for complex numbers $z_1$, $z_2$, $z_3$.
Let $z_1$, $z_2$, $z_3$ be distinct complex numbers such that $|z_1| = |z_2| = |z_3| > 0$. If $z_1 + z_2z_3$, $z_2 + z_3z_1$, and $z_3 + z_1z_2$ are real numbers, prove that $z_1z_2z_3 = 1$.

I believe that this problem comes from 1979 Romanian Math Olympiad (State Competition, 10th grade).

I tried several approaches: The first was to substitute $a + ib$, $c + id$, and $f + ig$ for $z_1$, $z_2$, and $z_3$, respectively, and bash. I used $(1+z_1)(1+z_2)(1+z_3)$, which didn't really get me anywhere.

After a couple more attempts at substitutions and such, I hit a wall and didn't really know what to do.

答:
Gosh... well, I just looked at the three-way symmetry, noted all points must lie on a circle around the origin, and deduced $z_1 = 1$, $z_2 = e^{2 i \pi/3}$ and $z_3 = e^{-2 i \pi/3}$, and confirmed these worked.

I suppose one could work through the algebra more rigorously, but on an exam... I think this is fine!


问:
Fermat factorisation in one step?
Apparently if $N$ has a factor within $\sqrt[4] N$ of $\sqrt N$ then Fermat factorisation works in one step.

Specifically this would mean that for $r = \lfloor \sqrt N \rfloor +1$ we have
$$r^2 - N = s^2$$
for some integer $s$.

I've tried bounding $r^2 -N$ in-between $s^2\pm1$ for some $s$ but haven't got anywhere, and I certainly don't know how to make use of that factor close to $\sqrt N$.

Can someone prove this?

答:
A small correction: The algorithm starts with $r = \bigl\lceil \sqrt{N}\,\bigr\rceil$.

Usually, that's the same as $\lfloor \sqrt{N}\rfloor + 1$, but when $N$ is a square it's a difference that matters. A square $N$ of course has a factor within $\sqrt[4]{N}$ of $\sqrt{N}$, but $(\lfloor \sqrt{N}\rfloor + 1)^2 - N = 2\sqrt{N} + 1$ then rarely is a square too, so that we need to start with $r = \bigl\lceil \sqrt{N}\,\bigr\rceil$ to be sure to find a factor in the first step.

And of course, $N$ needs to be odd or divisible by $4$, since Fermat's method cannot be applied to $N \equiv 2 \pmod{4}$.

Having this out of the way, we can assume that $N$ is not a square and $N \not\equiv 2 \pmod{4}$.

Let $N = pq$ with $p < \sqrt{N} < q$ such that $q - \sqrt{N} < \sqrt[4]{N}$ or $\sqrt{N} - p < \sqrt[4]{N}$. If $q < \sqrt{N} + \sqrt[4]{N}$, then
$$p = \frac{N}{q} > \frac{N}{\sqrt{N} + \sqrt[4]{N}} = \frac{N - \sqrt{N}}{\sqrt{N} + \sqrt[4]{N}} + \frac{\sqrt{N}}{\sqrt{N} + \sqrt[4]{N}} > \sqrt{N} - \sqrt[4]{N}\,,$$
from which we obtain $q - p < 2\sqrt[4]{N}$ and hence
$$\biggl(\frac{q+p}{2}\biggr)^2 = N + \biggl(\frac{q-p}{2}\biggr)^2 < N + \sqrt{N}\,.$$
If $p > \sqrt{N} - \sqrt[4]{N}$, then $q$ may be slightly larger than $\sqrt{N} + \sqrt[4]{N}$:
$$q = \frac{N}{p} < \frac{N}{\sqrt{N} - \sqrt[4]{N}} = \sqrt{N} + \sqrt[4]{N} + 1 + \frac{1}{\sqrt[4]{N} - 1}\,.$$
For $N > 16$ we thus have $q < \sqrt{N} + \sqrt[4]{N} + 2$, whence $q - p < 2\sqrt[4]{N} + 2$ and
$$\biggl(\frac{q+p}{2}\biggr)^2 = N + \biggl(\frac{q-p}{2}\biggr)^2 < N + \sqrt{N} + 2\sqrt[4]{N} + 1\,.$$
However, since $N < (\lfloor \sqrt{N}\rfloor + 1)^2$ we have
$$(\lfloor \sqrt{N}\rfloor + 2)^2 = (\lfloor \sqrt{N}\rfloor + 1)^2 + 2(\lfloor \sqrt{N}\rfloor + 1) + 1 > N + 2\sqrt{N} + 1 > N + \sqrt{N} + 2\sqrt[4]{N} + 1$$
for $N > 16$, as then $2\sqrt[4]{N} < \sqrt{N}$.

This implies that for $N > 16$ not a square and not $\equiv 2 \pmod{4}$, if $N = pq$ with one of $p,q$ within $\sqrt[4]{N}$ of $\sqrt{N}$, then
$$\frac{q+p}{2} = \lfloor \sqrt{N}\rfloor + 1\,,$$
which shows that Fermat's method succeeds at the first step.

The cases $N = 3,5,7,8,11,12,13,15$ are dealt with by inspection. The equations
$$3 = 2^2 - 1^2, \quad 5 = 3^2 - 2^2, \quad 8 = 3^2 - 1^2, \quad 12 = 4^2 - 2^2, \quad 15 = 4^2 - 1^2$$
show that the method succeeds at the first step for $N \in \{3,5,8,12,15\}$, and the primes $7,11,13$ have no factor within $\sqrt[4]{N}$ of $\sqrt{N}$ (which is a close call for $N = 7$, as $\sqrt{7} - \sqrt[4]{7} - 1 \approx 0.01917475$).

Altogether, when Fermat's method is applicable (i.e. $N \not \equiv 2 \pmod{4}$) and $N$ has a factor within $\sqrt[4]{N}$ of $\sqrt{N}$, then the method succeeds at the first step, $\bigl\lceil \sqrt{N}\,\bigr\rceil^2 - N$ is then a perfect square.

Title: Fermat factorisation in one step?


问:
Simple question about difference in percentiles -- what's wrong with my solution?

An insurer's annual weather-related loss, X, is a random variable with density function
$$f(x) = \frac{2.5(200)^{2.5}}{x^{3.5}} \qquad \text{for } x > 200$$
and $f(x) = 0$ otherwise. Calculate the difference between the $25$th and $75$th percentile.

My approach:
$$F(X) = \int_{200}^x2.5(200)^{2.5}(t^{-3.5}) \; dt = 1-\left(\frac{200}{x}\right)^{2.5}$$

So, the desired difference is $$F(0.75)-F(0.25) = 1-\left(\frac{200}{0.75}\right)^{2.5} - 1+\left(\frac{200}{0.25}\right)^{2.5} = \left(\frac{200}{0.25}\right)^{2.5} - \left(\frac{200}{0.75}\right)^{2.5}$$ which gives $16940694$ (approximated) while the correct answer is $124$. Where have I gone wrong?

Body: If $$F(x) = 1-\left(\frac{200}{x}\right)^{2.5}$$ then $$F^{-1}(y) = \frac{200}{(1-y)^{0.4}}$$  and $$F^{-1}(0.75)-F^{-1}(0.25) \approx 348.22-224.39 = 123.83$$


问:
Questions on Nilpotent Operator
I've some questions from the following theorem:

Let $T:V\to V$ be a nilpotent linear operator with index of nilpotency $k.$ Then $T$ can be expressed as a block diagonal matrix representation where each block $N$ is a Jordan nilpotent matrix. Moreover there is at least one such $N$ of order $k$ and all other $N$ are of order $\le k.$ Moreover the no of $N$ of order $i(\le k)$ is uniquely determined by $T$ and total no of $N$ of all order is $\ker T.$

$$Now~my~questions~are: $$

Does such representation of $T$ look like this?

What is meant by "the no of $N$ of order $i(\le k)$ is uniquely determined by $T$"? Does it mean if $T$ is represented by two different such (Jordan nilpotent) block diagonal matrices then the no of block matrices of order $i$ remain same in those two?

答:
Yes, the matrix looks like that. Note that at least one of the blocks is notjust $\le k\times\le k$ but in fact $k\times k$.

And Yes, two different such block matrices do have the "same" blocks, but possibly permuted.

The claims can be easily verified by looking at the numbers $\dim \ker T^i$ for $i=1, \ldots, k$.

Title: Questions on Nilpotent Operator


答:
Levels of clarity/terseness and the accompanying motivations behind them will vary a lot from context to context.

For terse teacher-written solutions to homework, it may just be due to time constraints, with the expectation that if something is unclear you would go ask for help/clarification.

Sometimes books are written very compactly partly for elegance, but also partly because they are aimed at an audience who are able to fill in the details. (For example, imagine reading a physics textbook that wrote out the details of long division every time they divided two numbers.) For this reason, it can be difficult for newcomers to learn effectively purely from reading a book since they are still learning how to fill in the gaps

Teachers also have the opportunity to be more clear in person/lecture (faster mode of communication) than in writing (more time/effort needed to write things down formally), and you have the opportunity to directly ask questions when things are not clear. Take advantage of having an expert available by asking them questions even if you think it is a "stupid question."

Even at high levels of math, everyone appreciates having concrete examples to make abstract concepts more tangible, especially when communicating about research in a niche area to a broader audience. Terseness is not "more acceptable" as math gets more advanced, but the level expected of the audience may be higher.

The use of "trivial" in pedagogy is not helpful in my opinion. But more generally, it can be a helpful indication that the justification for some claim follows from some simple definition rather than some complicated argument.

Title: Why are math texts usually so compact?

Body:
I taught a couple years in a secondary school, where teachers are supposed to explain every little step in the simplest but accurate way.

In high school things don't change much. Perhaps the pace is a little faster but still the goal is to be as clear as possible. Teachers are supposed to use pictures, animations and justify every step when solving problems.

In the university things are a lot different. We get exercises every week and solutions are extremely compact. I don't mean to show every little step when computing a determinant of a $3\times 3$ matrix but often one reads "one implication is trivial, one can (easily) see that, it is clear that..." and so on, which makes me personally feel incompetent because most of the times I don't easily see what I'm supposed to. At least in secondary school, it is mandatory to avoid using words such as "trivial, easy" and so on because of that reason. Still, some teachers do it and some believe this to be one of the causes of the spread of anxiety for mathematics, which certainly hinders the learning process even more.

This applies to this platform as well. The style completely changes.
Why is that so? Are compact solutions more elegant? Are mathematicians just lazy?


答:
Short answer: Your $F(0.75) - F(0.25)$ is computed incorrectly and doesn't mean what you think it means.

Long answer: recall that
$$F(x) = 1 - \left(\dfrac{200}{x} \right)^{2.5}$$
for $x > 200$ (this is key) and $F(x) = 0$ for $x \leq 200$.

Thus, $F(0.75) - F(0.5) = 0 - 0 = 0$ (because both $0.75 \leq 200$ and $0.5 \leq 200$). This would be the probability $$F(0.75) - F(0.5) = \mathbb{P}(X \leq 0.75) - \mathbb{P}(X \leq 0.5) = \mathbb{P}(0.5 < X \leq 0.75) = 0\text{.} $$
Obviously, this isn't what you're looking for. What you're actually looking for are the percentiles $p_{0.25}$ and $p_{0.75}$. These are the values satisfying
$$F(p_{0.25}) = 0.25 = 1 - \left(\dfrac{200}{p_{0.25}} \right)^{2.5}$$
$$F(p_{0.75}) = 0.75 = 1 - \left(\dfrac{200}{p_{0.75}} \right)^{2.5}$$
Use algebra to solve for $p_{0.25}$ and $p_{0.75}$, and then compute the difference $p_{0.75} - p_{0.25}$.

Title: Simple question about difference in percentiles -- what's wrong with my solution?

Body:
An insurer's annual weather-related loss, X, is a random variable with density function
$$f(x) = \frac{2.5(200)^{2.5}}{x^{3.5}} \qquad \text{for } x > 200$$
and $f(x) = 0$ otherwise. Calculate the difference between the $25$th and $75$th percentile.

My approach:
$$F(X) = \int_{200}^x2.5(200)^{2.5}(t^{-3.5}) \; dt = 1-\left(\frac{200}{x}\right)^{2.5}$$

So, the desired difference is $$F(0.75)-F(0.25) = 1-\left(\frac{200}{0.75}\right)^{2.5} - 1+\left(\frac{200}{0.25}\right)^{2.5} = \left(\frac{200}{0.25}\right)^{2.5} - \left(\frac{200}{0.75}\right)^{2.5}$$ which gives $16940694$ (approximated) while the correct answer is $124$. Where have I gone wrong?


问:
All vectors $(a, b, c, d)\in K^4$ are described by a 2×2-matrix
I am currently taking a course in linear algebra and for the first time in my life I will probably fail a maths test. Since there are about 100 theorems in my linear algebra book and each theorem has an average of 10 rules and some theorems use each other, it gets pretty tough. I decided to try to understand the material instead of memorizing every rule.

When watching 3blue1brown's video entitled "matrix multiplication  as composition | Essence of linear algebra, chapter 4" He describes the following matrix multiplication

$$
\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}e&f\\g&h\end{bmatrix} = \begin{bmatrix}ae+bg&af+bh\\ce+dg&cf+dh\end{bmatrix}
$$

I understand the rule and can memorize it... however it makes no sense to me. How my professor and my book describes matrices is, not exactly but basically, as follows:

Matrices (plural of matrix) are made up of equations. For example:
$$
2x+3y+6z\\
4x+1y+5z
$$

which is the same as:

$$
\begin{bmatrix}2&3&6\\4&1&5\end{bmatrix}
$$

Meaning the first column is the x value for each equation, the second column contains all the y values, and so on.

I would think that only the x values of a matrix could affect the x values of another matrix. However in the first example the first element in the matrix $ae+bg$ is a contradiction to my assumption in that $bg$ is the y value of one equation times the x value of another equation.

I can memorize the rule but I have no way of visualizing it (however 3blue1brown's video did a great job of visualizing what he described it makes not sense. Is the first elements of the first column, going doing the rows, x and y? I thought that the first row was an equation and that x and y is the two elements of the row.) nor a logical explanation as to why we derived this method. Can anyone guide me further?

P.S
English isn't my first language so if I drifted of in a nonsensical way somewhere, please comment or propose an edit and I'll change it as soon as I can!

Edit:
I am getting down votes, how can I improve my question? The ways I can improve it by myself is limited because I have limited knowledge.

答:
I find it useful to keep the  variables around when setting up the matrix multiplication. For for example I would write $$\begin{pmatrix}2 & 3 &6\\4 & 1 & 5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2x+3y+6z\\
4x+1y+5z\end{pmatrix}$$ and all we've done here is add some braces around the equations but otherwise this is the same. Now lets say I have some other linear equations that define the system given by $$\begin{pmatrix} -1 & 1\\-2 & 1\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix} = \begin{pmatrix}-u + v\\-2u + v\end{pmatrix}$$

and say we want to compose these two maps by first applying one then the other. We could do this by substitution using that $u=2x+3y+6z$ and $v=4x+y + 5z$ so apply both maps using the variables $x,y$ and $z$ directly. After doing this several times you quickly begin to see some patterns. One of them is that you're working with the coefficients, not the variables. Secondly, the dot product is used over and over again to calculate each new coeffecient. This is the motivation for defining matrix multiplication as it is. We can set it up as $$\begin{pmatrix}-1&1\\-2&1\end{pmatrix}\begin{pmatrix}2 & 3 &6\\4 & 1 & 5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}$$

and we can use matrix multiplication to calculate that $$\begin{pmatrix}-1&1\\-2&1\end{pmatrix}\begin{pmatrix}2 & 3 &6\\4 & 1 & 5\end{pmatrix}=\begin{pmatrix}2 &-2& -1 \\ 0 & -5& -7 \end{pmatrix}$$

This is exactly the same result as the substitution method would yield.


问:
The solution of the heat equation $u_t=\alpha u_{xx}$ is continuous on time $[0,\infty)$
Consider the heat equation with Dirichlet boundary condition on a metal rod of length $l$
\begin{equation}
\frac{\partial}{\partial t}u(t,x)=\alpha\frac{\partial^{2}}{\partial x^{2}}u(t,x)\label{eq:edp}
\end{equation}
\begin{equation}
u(t,0)=0\,\,\,\,\,\text{et}\,\,\,\,\,u(t,l)=0\label{eq:bords}
\end{equation}
\begin{equation}
u(0,x)=f(x)\label{eq:condition initiale}
\end{equation}
Using separation of variable method we can solve this system and we get $$u(t,x)=\sum_{n=1}^{\infty}b_{n}e^{-\alpha\left(\frac{\pi n}{l}\right)^{2}t}\sin\left(\frac{\pi n}{l}x\right),$$where $$b_n=\frac{2}{l}\int_0^lf(y)\sin\left(\frac{\pi n}{l}y\right)dy.$$

Using the M-test of uniform convergence, we can prove that for each $x\in[0,l]$ the time function $t\mapsto u(t,x)$ is continuous on all intervals $[\epsilon,\infty)$ thus continuous on $(0,\infty)$. Moreover, we can prove that for each $t>0$ the space function $x\mapsto u(t,x)$ is continuous on $[0,l]$ no matter how the function $f$ is ($f(x)$ is hidden inside the $b_n$ in the series formula and does not have a role in the $x$  variable )

I am just curious about the continuity of the time function $t\mapsto u(t,x)$ on $[0,\infty)$. For an initial date such as $f(x)=\sin\left(\frac{\pi }{l}x\right)$, we have this continuity since in this case we avoid the mystery of infinities and we have simply $$u(t,x)=be^{-\alpha\left(\frac{\pi}{l}\right)^{2}t}\sin\left(\frac{\pi }{l}x\right).$$
What can we say for some general initial data $f(x)$?

答:
You have shown that the series converges uniformly for $(t,x) \in [\delta,\infty)\times[0,l]$ where $\delta > 0$ and, thus,  $t \mapsto u(t,x)$ is continuous for all $t > 0$.  If we can extend the uniform convergence for $t \in [0,\infty)$, then continuity at $t=0$ would follow, i.e., $\lim_{t \to 0+}u(t,x) = u(0,x) = f(x)$ for all $x \in [0,l]$.

Given  that the Fourier series $$f(x) = u(0,x)= \sum_{n=1}^\infty b_n \sin \frac{\pi n x}{l}$$ is  convergent, it is uniformly convergent, trivially, for $t \in [0,\infty)$. We also have that $e^{-\alpha n^2\pi^2t/l^2}$ is monotone decreasing  with respect to $n$, and, uniformly bounded for all $t\in [0,\infty)$ since $|e^{-\alpha n^2\pi^2t/l^2}| \leqslant 1$.  By Abel's test for uniform convergence of series, it follows that (with $x$ fixed) the series

$$u(t,x)= \sum_{n=1}^\infty b_n e^{-\alpha n^2\pi^2t/l^2}\sin \frac{\pi n x}{l}$$

is uniformly convergent for all $t \in [0,\infty)$, and, therefore, $t \mapsto u(t,x)$ is continuous at $t = 0+$.


问:
Showing that three statements are equivalent (Topology)
$(a)$ Assume that $X$ is a compact metrizable space. Prove that the following conditions are equivalent:

$(i)$ $T$ is transitive.

$(ii)$ The set of transitive points for $T$ is dense in $X$.

$(iii)$ The set of transitive points for $T$ is non-empty.

Hint: Recall that every compact metrizable space has a countable basis (why?).

$(b)$ Does the equivalence in part $(a)$ still hold if instead of being compact and metrizable, the space $X$ is separable and completely metrizable?

A discrete-time (topological) dynamical system can be identified by a continuous map $T : X → X$ on a topological space $X$. A map $T$ is said to be transitive if for every pair of non-empty open sets $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$.

The (positive) orbit of a point $x ∈ X$ under $T$ is the set $O^+_T(x) :=$ {$T(x), T2(x), . . .$} .
A point $x ∈ X$ is said to be transitive for the map $T$ if its orbit $O^+_T(x)$ is dense in $X$.

My attempt:

$(i) => (ii):$ let $T: X→X$ be transitive, then for all $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$. But $O^+_T(x)$ $=$ {$T^n(U)$}$_{n>0}$ $=$ {$T(x), T^2(x),...$}, for $x \in X$, and $n>0$.

Hence $O^+_T(x) \cap V ≠ ∅$ and so $O^+_T(x)$ is dense.

$(ii) => (i):$ $O^+_T(x)$ is dense in $X$, so for all non empty $V ⊆ X$, $V \cap O^+_T(x)≠∅$. But $O^+_T(x)$ $=$ {$T^n(U)$}$_{n>0}$, for $x \in X$.

Hence there exists $n>0$ such that $V \cap T^n(U) ≠∅$.

$(i) => (iii):$ $T$ transitive implies that $U, V ⊆ X$, there exists an integer $n > 0$ such that $T^n(U) ∩ V ≠ ∅$. To get a contradiction, suppose $O^+_T(x) = \phi$, therefore, {$T^n(U)$}$_{n>0} = O^+_T(x) =\phi$. Then, $T^n(U) ∩ V = ∅$. Contradiction. So $O^+_T(x)≠ ∅$.

$(iii) =>(i):$ let $O^+_T(x)≠ ∅$. And let $V⊆ X$ be non empty. Thus, $O^+_T(x) \cap V≠ ∅$. Therefore, $T$ is transitive.

$(ii) => (iii):$ $O^+_T(x)$ is dense in $X$, so for all non empty $V⊆ X$, $O^+_T(x) \cap V ≠ ∅$

Hence $O^+_T(x)$ is non empty.

$(iii) =>(ii):$ $O^+_T(x)≠ ∅$. And let $V⊆ X$ be non empty. Thus, $O^+_T(x) \cap V≠ ∅$.

Hence, $O^+_T(x)$ is dense in $X$.

Wondering whether my attempt is correct or not, because I didn't use the given hint nor the fact that $X$ is compact metrizable. And what about part $(b)$, how to solve it? Thank you.

答:
I really can’t make sense of most of your argument. In most parts of it, for instance, you make assertions about $O_T^+(x)$ without ever specifying what $x$ is. In several places you seem to have confused $O_T^+(x)$ with the set of transitive points of $T$. There just isn’t enough substance there for me to turn it into a proof, so I’m simply going to show you one. I will show that (i), (ii), and (iii) are equivalent by showing that (i) implies (iii), (iii) implies (ii), and (ii) implies (i).

If $X$ is finite, it must be a discrete space, and it’s easy to check that each of the three conditions is equivalent to saying that $T$ is a permutation of $X$. Henceforth I will assume that $X$ is infinite.

(i) implies (iii):

Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable base for $X$, and let $T$ be transitive. Let $V_0=B_0$. Given a non-empty open set $V_n$ for some $n\in\Bbb Z^+$, there is a $k_{n+1}\in\Bbb Z^+$ such that $T^{k_{n+1}}[V_n]\cap B_{n+1}\ne\varnothing$. Then $V_n\cap\left(T^{k_{n+1}}\right)^{-1}[B_{n+1}]$ is a non-empty open set, so there is a $V_{n+1}\in\mathscr{B}$ such that $\operatorname{cl}V_{n+1}\subseteq V_n\cap\left(T^{k_{n+1}}\right)^{-1}[B_{n+1}]$ and hence $T^{k_{n+1}}[\operatorname{cl}V_{n+1}]\subseteq B_{n+1}$.

In this way we can produce open sets $V_n$ for $n\in\Bbb N$ such that for each $n\in\Bbb N$ we have $\operatorname{cl}V_{n+1}\subseteq V_n$ and $T^{k_{n+1}}[\operatorname{cl}V_{n+1}]\subseteq B_{n+1}$. Let $K=\bigcap_{n\ge 1}\operatorname{cl}V_n$; $X$ is compact, so $K\ne\varnothing$. Let $x\in K$; then for each $n\in\Bbb Z^+$ we have $x\in\operatorname{cl}V_n$, so $T^{k_n}(x)\in B_n$. And $x\in B_0$, so $x$ is a transitive point of $T$.

(iii) implies (ii):

Let $x\in X$ be a transitive point of $T$. Suppose that $x$ is an isolated point of $X$. Then $\{x\}$ is open, so there is an $n\in\Bbb Z^+$ such that $T^n(x)=x$. But then $\{T^k(x):k=1,\ldots,n\}$ is dense in $X$, so $X$ is finite, contradicting our assumption that $X$ is infinite. Thus, $x$ is not isolated.

Now suppose that $T^n(x)$ is isolated for some $n\in\Bbb Z^+$. $O_T^+(x)$ is dense in $X$, so there is a strictly increasing sequence $\langle k_i:i\in\Bbb Z^+\rangle$ in $\Bbb Z^+$ such that $\langle T^{k_i}:i\in\Bbb Z^+\rangle$ converges to $x$. It follows that $\langle T^{k_i+n}:i\in\Bbb Z^+\rangle$ converges to $T^n(x)$, and since $T^n(x)$ is isolated, this implies that $T^{k+n}(x)=T^n(x)$ for some $k\in\Bbb Z^+$. But then, just as in the preceding paragraph, $X$ is finite, contrary to our assumption, and we conclude that no point of $O_T^+(x)$ is isolated.

Let $n\in\Bbb Z^+$ be arbitrary, and let $y=T^n(x)$. Let $U$ be any non-empty open set in $X$. Then $V=U\setminus\{T^k(x):k=1,\ldots,n\}$ is a non-empty open set, so there is a $k\in\Bbb Z^+$ such that $T^k(x)\in V$. Clearly $k>n$, so $T^k(x)=T^{k-n}(y)\in O_T^+(y)$, and $y$ is a transitive point of $T$. Thus, every point of the dense set $O_T^+(x)$ is a transitive point of $T$.

(ii) implies (i):

Let $D\subseteq X$ be a dense set of transitive points of $T$, and let $U$ and $V$ be non-empty open sets in $X$. $U\cap D\ne\varnothing$, so let $x\in U\cap D$. Then $T^n(x)\in T^n[U]\cap V$ for some $n\in\Bbb Z^+$, so $T$ is transitive.


问:
A map is homogenous iff $r\mathbb{B}(1) = \mathbb{B}(|r|)$
On Planet Math it is stated that th homogenity of a semi-norm is equivalent to the fact  that $r\mathbb{B}(1) = \mathbb{B}(|r|).$ I interpret this that for any real-valued map on a vector space for which homogenity is equivalent to this unit-ball property. This would give a nice geometric characterisation of homogenity.

However I can only show $p(r v) \leq |r| p(v),$ the other direction eludes me. Is this claim correct?

答:
Suppose we know that $p(rv) \leq |r|p(v)$ whenever $v \in V$ and $r \in K$.  We then know that
\begin{align*}
p(v) = |r|^{-1} |r| p(r^{-1} \cdot r v) \leq |r|^{-1} p(rv)
\end{align*}
which gives $|r| p(v) \leq p(rv)$ and hence equality.


问:
Isomorphism between the Tangent Space and Euclidean Space
I attempt to understand the invariance of dimension of topological manifolds from Loring Tu's An Introduction to Manifolds (Second Edition, page no. 89). The corresponding section of the book is given below.

My Question

In the proof of Corollary 8.7, it is claimed that the tangent space $T_p(U)$ at point $p \in U \subset \mathbb{R}^n$ is isomorphic to the Euclidean $n$-space $\mathbb{R}^n$. How do we know that this is the case? In other words, what is the proof of this claim?

答:
In Tu's book, it is considered proved at page 10, section 2.1, where $T_{p}\mathbb{R}^n$ is "defined" as the vector space of all arrows emanating from a point $p \in \mathbb{R}^n$.

Besides the fact there is a misprint in the book, (should be $T_{F(p)}V$ isomorphic to $\mathbb{R}^m$ in the last line) if you have shown it is an $n$-dimensional vector space, then this is equivalent to say it is isomorphic to $\mathbb{R}^n$. If otherwise, you have not proved this, then your line of proof is to build a linear injective and surjective map from $\mathbb{R}^n$ to $T_{p}\mathbb{R}^n$.

In general, for a smooth $n$-dimensional manifold, the tangent space at a point of the manifold will be a vector space isomorphic to $\mathbb{R}^n$. Proving this may be more or less difficult, depending on which of the many (mostly equivalent) definitions of manifold (and tangent space) you're using.

If you go on to Tu's chapter 8 section 4 you find "Bases for tangent spaces at a point". In this section, you prove formally that there is an isomorphism" in the general case of manifolds.


问:
Distribution of a division of two absolutely continuous random variables
Assume $Z,Y $ are independent standard normal random variables.

Find the distribution of $ Z/Y $.

The answer is that $ Z/Y $ is absolutely continuous random variable with density

$$ f_{\frac{Z}{Y}}(x)=\frac{1}{\pi(1+x^2)} $$

Here's what I don't understand:

I tried to write:

\begin{align}
F_{\frac{Z}{Y}}(t) & =\mathbb{P}\left(\frac{Z}{Y}\leq t\right)=\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y<0\right)+\mathbb{P}\left(\frac{Z}{Y}\leq t\cap Y>0\right) \\[8pt]
& =\mathbb{P} (Z\geq tY\cap Y<0)+\mathbb{P}\left(Z\leq tY\cap Y>0\right) \\[8pt]
& =\mathbb{P}(Z\geq tY) \mathbb{P}(Y<0)+\mathbb{P}(Z\leq tY) \mathbb{P}(Y>0) \\[8pt]
& =\frac{1}{2}\mathbb{P}(Z\geq tY)+\frac{1}{2}\mathbb{P} (Z\leq tY)
\end{align}

And then to calculate each term. for example :

$$ \mathbb{P}(Z\leq tY)=\mathbb{P} ((Z,Y)\in \{ (z,y):z\leq ty \} ) = \intop_{-\infty}^\infty \intop_{-\infty}^{ty} \frac{1}{2\pi} e^{-\frac{z^2+y^2}{2}} \, dz\, dy $$

But this leads me to a wrong answer, and I'm pretty sure that the problem is somewhere in what I wrote before.

Any help would be appreciated, thanks in advance.

答:
You seem to have concluded that $\Pr (Z\geq tY\cap Y>0) = \Pr(Z\geq tY) \Pr(Y>0).$

But these events are not independent, as may be seen by drawing a graph. Try it, for example, with $t=3.$ Look at the region of the $y,z$-plane where $y>0$ and $z>3y$ and the region where $y<0$ and $z>3y.$ These will get two very different probabilities.

You have $\Pr(Y>0) = 1/2,$ so $\Pr(Z\ge tY\mid  Y>0) = \dfrac{\Pr(Y\ge tY \cap Y>0)}{1/2}.$

This is $\displaystyle 2\int_0^{+\infty} \int_{-\infty}^{ty} f(y,z)\,dz\,dy,$ where $f$ is the joint density.

PS: If you want to use polar coordinates here, notice that $\big[y>0 \text{ and } z>ty\big]$ amounts to $\big[ \arctan t < \theta < \pi/2 \big],$ and from there to the bottom line is a very short route.


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
Another approach that is slightly more involved, but comes at the problem from a different angle and treats both $A_n$ and $B_n$ as special cases of a more general formula. We will actually compute the inverse matrices and relate them to graph laplacians, then use Kirchoff's theorem to eveluate the determinants.

Let $x$ be a variable, and define the $n\times n$ matrix
$$L_n(x)=\left(\begin{array}{ccccc}
 -1 & 1 \\
 1 & -2 & 1\\ \\ &\ddots& \ddots &\ddots\\ && 1 &-2 & 1\\ &&& 1 &-x-1 \end{array}\right)$$

The connection with graph theory is that if we consider the linear graph on $n+1$ nodes, with the edge weight between the last two nodes being equal to x, and the other edge weights equal to 1, then $L_n$ is equal to the corresponding graph Laplacian with the last row and column removed (and opposite the usual sign convention). As a consequence, we get the following:

Lemma $det(L_n(x))=(-1)^nx$

Proof
Follows from Kirchoff's theorem about spanning trees. Since the graph is linear, it has only a single spanning tree, and the weighting is $(1)^nx$.   The factor of $(-1)^n$ arises because we are using the opposite of the usual sign convention.

Now let us compute $L_n^{-1}$. We use the block decomposition
$$L_n=\left(\begin{array}{cc} L_{n-1}(1) &  e_{n-1} \\  e_{n-1}^t & -x-1\end{array}\right):=\left(\begin{array}{cc} A & B \\ C  & D\end{array}\right)$$
where $e_{n-1}$ denotes the vector of length $n-1$ with a 1 in the last entry and 0 elsewhere.

Now apply the formula for the inverse of a block matrix. The main observation is that $A\textbf{1}=-e_{n-1}$, so that $A^{-1}e_{n-1}=-\textbf{1}$. The remaining calculations are routine, with the final result being

$$L_n^{-1}(x)=\left(\begin{array}{cc} L_{n-1}^{-1}(1) -\textbf{1}\textbf{1}^t/x& -\textbf{1}/x \\ -\textbf{1}/x & -1/x\end{array}\right)=\left(\begin{array}{cc} L_{n-1}(1)^{-1} & 0 \\ 0 & 0\end{array}\right)-\textbf{1}\textbf{1}^t/x$$

If we set $x=1$ then we get a recurrence for $L_{n}^{-1}(1)$, and it easily follows by induction that the solution is $(L_{n}^{-1}(1))_{ij}=-\min(n-i+1,n-j+1)$.

Computing $\det(A_n):$

By the previous discussion, we can write $A_n=-P^t(L_n^{-1}(1))P$ where $P$ is the permutation matrix the reverses the order of the rows. Conjugating does not change the determinant, so applying the Lemma, $det(A_n)=(-1)^n det(L_n(1))^{-1}=(-1)^n(-1)^n=1$.

Computing $\det(B_n):$

For $1\leq i,j\leq n$, observe that
$max(i,j)=-min(n-i,n-j)+n=((L_{n-1}(1))^{-1})_{ij}+n$. Thus
$$B_n=\left(\begin{array}{cc} L_{n-1}(1)^{-1} & 0 \\ 0 & 0\end{array}\right)+n\textbf{1}\textbf{1}^t$$

and therefore $B_n=L_n(-1/n)^{-1}$. By the lemma, $det(B_n)=1/det(L_n(-1/n))=n(-1)^{n+1}$.


答:
This is more of a comment than an answer and might be seen as some sort of addendum to J.G.‘s answer.

It is completely correct that in school one is trained to follow algorithms, while in university the focus switches to writing correct algorithms (=proofs). Moreover it is very tempting (if not necessary in the case of research papers for example) to gloss over standard details to get to more interesting stuff.

Yet I feel like using words like trivial, obvious or easy is very bad practice, especially in the context of mathematical education. It unnecessarily makes students feel bad, if they don’t get the detail, sometimes even after hours of staring at it. It always is a triviality at the end, but most of the time it needs that one standard trick, which just isn’t memorized yet.

I don’t understand, why it isn’t more common to give concise sketches of the idea required to prove a result:

by definition says plugging in the definitions will make the result obvious

by iteratively using lemma X on Y is a pretty concrete recipe

by strong induction on $n$ gives information about which type of induction hypothesis is needed

et cetera, et cetera. I don’t think this requires that more time and space and most importantly it gives some sort of reality check: If you cannot summarize the triviality in like one sentence, it most likely is not so trivial at all...

Title: Why are math texts usually so compact?


问:
Working out expected values of coin flipping games when starting with an initial value
Suppose I start with $\$50$ and I flip a fair coin, where flipping heads rewards me with $\$2$ and flipping tails loses me $\$1$, and I either flip until I run out of money, or I flip 100 times otherwise. What is the expected value of this game, considering that my profits do not include the initial $\$50$?

Clearly, if I consider the more basic game of starting with $\$0$ and flip 100 times, then the expected value can be calculated to be $\$50$. Moreover, the case of running out of money yields an expected value of $0$, but I don't know how to put this all together

答:
(3rd edit...)

You either finish 100 rounds, or you don't. If you finish 100 rounds, you end up with an expected profit of 50\$, but if you don't, it means you have lost your initial 50\$, i.e., -50\$ profit. You lose if you flip 50 tails in a row, or 52 tails and one head, or 54 tails and 2 heads, etc. The only condition in each case is that your last three outcomes should be tails, which makes all losing scenarios disjoint. The probability of losing being:

$P_L = \sum\limits_{k=0}^{16} {47+3k \choose k} \left(\frac{1}{2}\right)^{47+3k} \approx 1\times 10^{-10}$

So, keep playing! You will end up with +50\$.


问:
Parametric plot of a cycle - transition between two lines - dealing with a jump
I have the following problem. I am modelling a cycle that consist of "loading" and "unloading" parts.

Firstly, the material is being loaded from the origin on the "blue line", that is the simple linear function ( $y = 2x$ ), and then I would like to perform transition to the bottom orange line that represents "unloading" of the material (now it is the linear function $y = x$ ):

I created the following solution based just on the ratios of the linear coefficients,

but it is not accurate enough. I would like to have steeper transition between the blue and orange line. I tried to use $tanh$ function as suggested here but in my situation there is a jump between linear functions and I do not know how to overcome it.

Please, do you have any advices?

Thank you very much in advance!

EDIT:

I would like to get a curve, that comes from the origin to the point $[1;2]$, then I need some transition from the upper line to the bottom line (that I need to be steep enough, using hyperbolic tangens?) and from the point that is as close as possible to the point $[1;1]$ on the bottom line go back on the curve to the origin.

Actually, I have a solution (it is my black curve), but it is not precise enought. I would like to kindly ask if anybody knows some improvement of the transition...

答:
Consider a function like

$$f(x) = \frac{e^{kx}-1}{e^k-1}$$

for some $k \in \Bbb R$. We have $f(0)=0$ and $f(1)=1$, and as $k$ increases, the curve flattens toward $y=0$, making an ever more abrupt vertical transition near $x=1$. It looks like this for $k=30$:

$\hskip 4cm$

Let's call your two bounding line segments $L$ and $M$, so $L(x) = 2x$ and $M(x) = x$. You can use a function like $f(x)$ to transition between $L$ and $M$ like so:

$$g(x) = f(x)L(x) + (1-f(x))M(x);$$

in other words, this is just a weighted average of $L$ and $M$, with the weight almost entirely on $M$ until we get close to $x=1$. Again with $k=30$, it looks like so:

$\hskip 4cm$


问:
a linear combination of polynomials is a homogeneous polynomials implies that each polynomial is homogeneous
Let $F$ and $G$ be relatively prime homogeneous polynomials in $k[X,Y,Z]$ and $A,B\in k[X,Y,Z]$ such that $AF+BG$ is a  homogeneous polynomial. How to show that $A$ and $B$ are homogeneous polynomials ?

I am interested in this question since in the proof of Bezout Theorem of Fulton's book

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

page 58, it is asserted the following.

Let $\pi:k[X,Y,Z]_d\longrightarrow k[X,Y,Z]_d/(F,G)$ be the canonical map where $k[X,Y,Z]_d$ is the set of homogeneous polynomials of degree $d$. Let $\varphi:k[X,Y,Z]_{d-m}\times k[X,Y,Z]_{d-n} \longrightarrow k[X,Y,Z]_d$ defined by $\varphi(A,B)=AF+BG$ where $m=deg(F)$ and $n=deg(G)$. Fulton asserts that $Im(\varphi)=Ker(\pi)$.

However, $Ker(\pi)$ is the set of $AF+BG$ such that $AF+BG$ is homogeneous of degree $d$ with $A$, $B$  in $k[X,Y,Z]$. In order to show that it is $Im(\varphi)$, one has to show that $A$ and $B$ are homogeneous. The counter-example in the previous comment of K.Reiser seems to show that the assertion of Fulton is wrong ????

答:
It is not true that if $F,G$ are relatively prime homogeneous polynomials in $k[x,y,z]$ and $A,B\in k[x,y,z]$ are such that $AF+BG$ is homogeneous, then $A,B$ are homogeneous: consider $F=x$, $G=y$, $A=y+1$, $B=-x+1$.

This is fine in your situation, though: the goal is to say that if $P$ is a polynomial in $k[x,y,z]_d$ which lies in $(F,G)$, then there exist $A\in k[x,y,z]_{d-m}$ and $B\in k[x,y,z]_{d-n}$ so that $P=AF+BG$. (Note that this is not what you've written!) We can start by using the fact that $P\in (F,G)$ to write $P=CF+DG$ for some polynomials $C,D\in k[x,y,z]$. Now write $C=\sum C_i$ and $D=\sum D_i$ where $C_i,D_i$ are homogeneous of degree $i$. This gives that $$P= \left(\sum C_iF\right)+\left(\sum D_iG\right).$$ Therefore the degree-$e$ homogeneous part of $P$ is equal to $C_{e-m}F+D_{e-n}G$, and by the fact that $P$ is homogeneous of degree $d$, we have that $C_{e-m}F+D_{e-n}G=0$ when $e\neq d$ and $C_{e-m}F+D_{e-n}G=P$ when $e=d$.

So $P=C_{e-m}F+D_{e-n}G=P$ and therefore if $P\in \ker(\pi)$, $P$ is in the image of $\varphi$ as $\varphi(C_{e-m},D_{e-n})$ and the claim is proven.


问:
Showing that degree of polynomial reduced modulo $p$ divides degree of splitting field
Let $f$ be a polynomial of degree $n$ with integer coefficients and $p$ a prime for which $f$, considered modulo $p$, is a degree-$k$ irreducible polynomial over $\mathbb{F}_p$. Show that $k$ divides the degree of the splitting field of $f$ over $\mathbb{Q}$.

(this is Miklos Schweitzer 2020, Problem 10)

My miscellaneous thoughts: We know that for a polynomial of degree $n$, the degree of its splitting field is at most $n!$ from induction on the Ring Tower Theorem. Additionally, since $f(x)$ is irreducible over $\mathbb{F}_p$, then the field $\mathbb{F}_p[x]/(f(x))$ has a root of $f(x)$. Now suppose $k = 2$. Then $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in \mathbb{Z}[x]$ reduces to $f(x) \equiv b_2 x^2 + b_1x + b_0 \in \mathbb{F}_p [x]$ where $b_i \equiv a_i \pmod{p}$ where applicable. I'm really not sure if these things are particularly useful or how to combine them.

This is really hard!

答:
$f\in \Bbb{Z}[x]$ is such that $f\bmod p$ is irreducible and non-constant. Factorize $f=\prod_j f_j\in \Bbb{Z}[x]$, then one of the $f_j$ is such that $f_j=c f\bmod p$ with $c\in \Bbb{F}_p^\times$.

Let $L$ be the splitting field of $f$ so that $f_j$ has a root $a/b\in Frac(O_L)$.

Note that $p$ is not a unit of $\Bbb{Z}[a/b]\cong\Bbb{Z}[x]/(f_j)$. Since $O_L[a/b]$ is integral over $\Bbb{Z}[a/b]$ it means that $O_L[a/b]$ contains a maximal ideal $P$ above $p$.

Being finite fields $O_L[a/b]/P$ contains and is integral over $O_L/(P\cap O_L)$, thus $a/b\in (O_L)_{(P\cap O_L)}$ and $O_L/(P\cap O_L)\cong O_L[a/b]/P$.

So $O_L/(P\cap O_L)$ contains $\Bbb{Z}[a/b]/(p) \cong \Bbb{F}_p[x]/(f)$.

Thus $\deg(f\bmod p)$ divides $[O_L/(P\cap O_L):\Bbb{F}_p]$.

The factorization $pO_L = \prod_j Q_j^e$ with $Q_1=P\cap O_L$ and $N(Q_j)=N(Q_1)=p^{[O_L/(P\cap O_L):\Bbb{F}_p]}, N(pO_L)=p^{[L:\Bbb{Q}]}$ gives that $\deg(f\bmod p)$ divides $[L:\Bbb{Q}]$.


问:
confusion in finding $I=\frac{1}{2\pi j} \oint_c Re\{ z\} \, dz$

If C denotes the counterclockwise unit circle, then the value of the contour integral
$$\frac{1}{2\pi j} \oint_c Re\{ z\} \, dz \quad \text{is _____ }?$$

Method $1$:

$$Re\{ z\} = \frac{z + \bar{z}}{2} \quad \dots (1)$$
$$\implies \frac{1}{2\pi j} \oint_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint_c \frac{z + \bar{z}}{2} \, dz$$
$$=\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint_c \frac{\bar{z}}{2} \, dz \quad \dots (2)$$
Now using Cauchy-Integral theorem, we get:
$$\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz=0 \quad \dots (3)$$
as $f(z)=z$ is differentiable at every point inside C ($|z|=1$)

and $\bar{z}=\frac{1}{z}$ for C( $|z|=1$),

so, $eq(2)$ can be rewritten as:
$$\frac{1}{2\pi j} \oint_c \frac{z }{2} \, dz + \frac{1}{2\pi j} \oint_c \frac{\bar{z}}{2} \, dz = 0 + \frac{1}{2}\frac{1}{2\pi j} \oint_c \frac{1}{z} \, dz$$
$$\frac{1}{2} .1=\frac{1}{2}$$

Method $2$:

Let $z=re^{j \theta}$

$$\implies \frac{1}{2\pi j} \oint_c Re\{ z\} \, dz = \frac{1}{2\pi j} \oint_c r cos (\theta) \, d(re^{j \theta}) \quad \dots (4)$$
Now, for C : $r=1 $ and $  \theta \to 0 $ to $2\pi $
$$=\frac{1}{2\pi j} \int_0^{2\pi} cos (\theta) e^{j \theta}j \, d\theta$$
$$=\frac{1}{2\pi } \int_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta \quad \dots (5)$$
Now,
$$\int e^{ax}cos(bx) dx = \frac{e^{ax}}{a^2 +b^2}\{ a cos(bx) + b sin (bx)\} +C \quad \dots  (6)$$
$$\implies \int_0^{2\pi} cos (\theta) e^{j \theta} \, d\theta = [\frac{e^{jx}}{j^2 +1^2}\{ j cos(x) +  sin (x)\}]_0^{2\pi}$$
Here, denominator is coming to $0$ , so we can't use the formula described in $eq(6)$;

Hence how to evaluate $eq(5)$ ?

Any help or suggestions please...

答: As has been noted, method 2 should be$$\frac{1}{2\pi j}\oint_cr\cos\theta d(re^{j\theta})=\frac{1}{2\pi}\int_0^{2\pi}r^2\cos\theta e^{j\theta}d\theta=\frac{r^2}{2\pi}\int_0^{2\pi}\cos^2\theta d\theta=\frac12r^2=\frac12$$(although I'd rather insert $r=1$ at the start), as $\int_0^{2\pi}\cos\theta\sin\theta d\theta=0$.


问:
Integral representation of the Dirac Delta function
Someone told me, without so much details, that I can use the following representation of $\delta$
$$
\delta(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{g(-i\partial_x)}\tilde{g}(x),
$$
for a suitable function $g$ and where $\tilde{g}(x)=1/\sqrt{2\pi}\int g(y)e^{ixy}dy$ is the Fourier transform of $g$. How to show this? I believe one can start from the $\delta$ definition
\begin{align}
\delta(x)&=\frac{1}{2\pi}\int e^{ixy}dy\\
&=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}}\int \frac{1}{g(y)}g(y)e^{ixy}dy\\
\end{align}
and an integration by parts, but what boundary conditions or what properties the function $g$ must have (clearly a series decomposition)?

答:
We start from
$$
\delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} \, d\xi
.
$$

If $g$ is analytic then
$$
g(x) = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}x^k
$$
and we define
$$
g(-i\partial_x) = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\partial_x)^k
.
$$

Note now that
$$
g(-i\partial_x) e^{ix\xi}
= \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\partial_x)^k e^{ix\xi}
= \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\cdot i\xi)^k e^{ix\xi}
= \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}\xi^k e^{ix\xi}
= g(\xi) e^{ix\xi}
.
$$

Therefore, applying $g(-i\partial_x)$ on $\delta$ gives
$$
g(-i\partial_x) \delta(x)
= \frac{1}{2\pi} \int_{-\infty}^{\infty} g(-i\partial_x) e^{ix\xi} \, d\xi
= \frac{1}{2\pi} \int_{-\infty}^{\infty} g(\xi) e^{ix\xi} \, d\xi
= \frac{1}{\sqrt{2\pi}} \tilde{g}(x)
$$
from which we conclude
$$
\delta(x) = \frac{1}{\sqrt{2\pi}} g(-i\partial_x)^{-1} \tilde{g}(x)
.
$$


问:
Finding an expression for the probability that one random variable is less than another, given a condition.
Let $X$ and $Y$ be two independent random variables, who's supports are $[0,\infty]$. We can express $\mathbb{P}[X<Y]$ as:

$$\mathbb{P}[X < Y] = \int_{y=0}^{\infty}\int_{x=0}^{y}P_{X}(x)P_{Y}(y)dxdy.$$

Can we find a similar expression for:

$$\mathbb{P}[X < Y|Y<k],\; \mathrm{given}\;k\in[0,\infty)?$$

答:
You can find an expression for the desired conditional probability. Assume for simplicity that $X$ and $Y$ have respectively density functions $f_X(x)$ and $f_Y(y)$. Then
$$\Pr(X\lt Y|Y\lt k)=\frac{\Pr((X\lt Y)\cap (Y\lt k)}{\Pr(Y\lt k))}.$$
Both numerator and denominator can be expressed as integrals. For the numerator, we want $\int_{y=0}^k\int_{x=0}^y f_X(x)f_Y(y)\,dx\,dy$. For the denominator, it is much the same, except that $x$ goes from $0$ to $\infty$.

Remarks: $1.$ The independence does not play a large role here, apart from (in concrete cases) making the integrations easier. For joint density functions $f_{X,Y}(x,y)$ the expression for the conditional probability is
$$\frac{\int_{-\infty}^k\int_{-\infty}^y f_{X,Y}\,dx\,dy}{\int_{-\infty}^k\int_{-\infty}^\infty f_{X,Y}\,dx\,dy}.$$

$2.$ Note that as pointed out by @Jon Claus, the denominator is just the probability that $Y\lt k$, so it can be expressed in the simple form $\int_0^k f_Y(y)\,dy$.


问:
$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle
The question contains a hint: at the appropriate point use the result $a^{3} - b^{3} =
\left(a - b\right)\left(a^{2} + b^{2} + ab\right)$.
$$
\frac{{\rm d}}{{\rm d}x}\sin\left(x^{1/3}\,\right)
$$

My attempt:

I did not use the hint as it was not immediately obvious to me what the simplification was. Instead I wrote with binomial expansion that $(x+h)^{\frac{1}{3}}=x^{\frac{1}{3}}(1+\frac{h}{x})^{\frac{1}{3}} \approx x^{\frac{1}{3}}(1+(\frac{1}{3})\frac{h}{x})$, used $\sin(()+())=\sin()\cos()+\sin()\cos()$ and applied $\cos() \approx 1$, $\sin() \approx ()$ for small $h$ which gave the correct answer.

But I do not see the point of using the cubes difference formula given in the question? Could this be a typo because it would make more sense to apply the identity to something like e.g. $\sin({x})^{\frac{1}{3}} (\text{not }\sin(x^{\frac{1}{3}}))\to \sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}} \iff (\sin(x+h)-\sin(x)=(\sin(x+h)^{\frac{1}{3}}-\sin(x)^{\frac{1}{3}})(\sin(x+h)^2+\sin(x)^2+\sin(x+h)\sin(x))$

Or am I missing something here?

Body:
hint

Replace in

$$\frac{a-b}{a^3-b^3}=\frac{1}{a^2+ab+b^2}$$

$ a \;\; \text{by} \;\; (x+h)^\frac 13 $

and

$ b\;\; \text{by}\;\; x^\frac 13$.


问:
Method to solve this indefinite integral $ \int \frac{x^{4}}{\sqrt{x^{2}+9}}dx $
I'm not able to solve this integral. Can anybody help me?
I tried doing it by parts noticing that
$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = \int x^{3} \cdot \frac{x}{\sqrt{x^{2}+9}} .
$$
where we know that $$ D[\frac{x}{\sqrt{x^{2}+9}}] = \sqrt{x^{2}+9} $$

so that we can view the integral as

$$
\int \frac{x^{4}}{\sqrt{x^{2}+9}} \, dx \, = x^{3} \cdot \sqrt{x^{2}+9} - \int 3 \cdot x^{2} \cdot \sqrt{x^{2}+9} \, dx  .
$$
which led me in another integration by parts and in another one again, so i realized that maybe this wasn't the right way to do it

答:
Since other answerers have already explained how to solve the question at hand, I'll focus on a more general problem solving strategy. The integral in question,
$$
\int \frac{x^4}{\sqrt{x^2+9}} \, dx \, ,
$$
has an expression involving $\sqrt{x^2+9}$. This should make it obvious that a trigonometric substitution is the way to go, and I'll explain why. The identities
\begin{align}
\sin(\theta)^2+\cos(\theta)^2&=1 \tag{1}\label{1}\\
\tan(\theta)^2+1&=\sec(\theta)^2 \tag{2}\label{2}
\end{align}
come in handy here. $\eqref{1}$ can be rearranged to
$$
\cos(\theta)=\sqrt{1-\sin(\theta)^2} \, .
$$
(The reason that we don't bother writing $\cos(\theta)=\pm\sqrt{1-\sin(\theta)^2}$ is quite subtle; I'll leave it as a footnote if you're interested.)

These identities imply that any integral of the form
$$
\int f\left(\sqrt{1-x^2}\right) \, dx
$$
can be simplified if we let $x=\sin(\theta)$. Since $dx=\cos(\theta)d\theta$, we obtain
$$
\int f(\cos(\theta))\cos(\theta) \, d\theta \, ,
$$
meaning we can avoid the pesky square root sign. For much the same reason,
$$
\int f(x^2+1) \, dx
$$
can often be simplified if we write $x=\tan(\theta)$. Your problem was almost the same—it is just that we had to substitute $x=3\tan(\theta)$ because we were dealing with $\sqrt{x^2+9}$ rather than $\sqrt{x^2+1}$. Integration by parts doesn't offer an obvious way of simplifying the integral, unlike this method.

To fully appreciate why we don't write $\cos(\theta)=\sqrt{1-\sin(\theta)^2}$ one must be familiar with why integration by substitution works in the first place:

Integration by substitution

Integration by substitution comes from reversing the chain rule. Recall that if $y=f(g(x))$, then
$$
\frac{dy}{dx}=f'(g(x))g'(x) \, .
$$
Thus,
$$
\int f'(g(x))g'(x) \, dx = f(g(x))+C \, .
$$
On the other hand, if we make the substitutions $u=g(x)$ and $du = g'(x) \, dx$, then  the integral becomes
$$
\int f'(u) \, du=f(u)+C=f(g(x))+C \, .
$$
It just so happens that working out $du/dx$ and then 'multiplying by $du$' means that you will make the correct substitutions. The formula can be summarised as
$$
\int f(g(x))g'(x) \, dx = \int f'(u) \, du \quad \text{where $u=g(x)$} \, .
$$
In Leibnizian notation, it reads particularly well
$$
\int \frac{dy}{du} \cdot \frac{du}{dx} \, dx = \int \frac{dy}{du} \, du \, .
$$
Again, the $dx$'s 'cancel' in a similar fashion to the chain rule.

More Sophisticated Substitutions

A more sophisticated way of making substitutions is given in Michael Spivak's Calculus. Integrals of the form
$$
\int f'(g(x))g'(x) \, dx
$$
can be easily solved by substituting $u=g(x)$. In fact, many of these integrals are so simple that you can do them in your head. However, the substitution $u=g(x)$ can still be made even if the factor $g'(x)$ does not appear. In general,
$$
\int f(g(x)) \, dx \tag{*}\label{*}
$$
can be solved in the following way, provided that $g$ is one-to-one:
\begin{align}
u &= g(x) \\
x &= g^{-1}(u) \\
\frac{dx}{du} &= (g^{-1})'(u) \\
dx &= (g^{-1})'(u)du
\end{align}
This means that \eqref{*} can be transformed to
$$
\int f(u)(g^{-1})'(u) \, du \, ,
$$
which in practice often makes the integral simpler to solve than before. Hence,
$$
\int f(g(x)) \, dx = \int f(u)(g^{-1})'(u) \, du \, .
$$
The validity of this approach can be demonstrated by differentiating $\int f(u)(g^{-1})'(u)$ with respect to $x$:
\begin{align}
\frac{d}{dx}\int f(u)(g^{-1})'(u) \, du &= f(u)(g^{-1})'(u)\frac{du}{dx} \\
&= f(u) \frac{dx}{du} \frac{du}{dx} \\
&= f(g(x)) \, .
\end{align}

Why we ignore the $\pm$ signs

This means that we can make a substitution of the form $x=\ldots$, what we are really doing is making a substitution of the form $u=\ldots$ and then taking the inverse. Say we are solving
$$
\int \sqrt{1-x^2} \, dx \, .
$$
When we write
$$
\text{let $x=\sin(\theta)$} \, ,
$$
what we really mean is let $\theta=\arcsin(x)$, meaning that $x=\sin(\theta)$ with $-\pi/2 \leq \theta \leq \pi/2$. Notice that $\arcsin$ is a one-to-one function, meaning that the substitution is valid. Then, since $-\pi/2 < \theta < \pi/2$, we know that $\cos(\theta)$ is positive, so we can drop the $\pm$ sign.


问:
Given that a variable's probability density function is $p(t)=kf(t)$, how to find $k$?
The variable t is defined in the domain $t>0$. At first, I thought $k$ could be calculated by normalizing $p(t)$ to $1$. So $$k = \frac{1}{F(\infty) - F(0)}$$ where $F(t)$ is the primitive function of $f(t)$.

However, I encountered a problem with this type of function, (I applied specific values to give a concrete example):

\begin{equation}
p(t) = k(1+\sin(20t))e^{\frac{-t}{1.5}}
\end{equation}

when using the method above, I found $$F(\infty) - F(0) \approx 1.55$$ But, when plotted, $f(t)$ looks like this:

enter image description here

So for some $t\to 0$, where $f(t)$ peaks, the probability $p(t)$ will be larger than $1$ because $1/k$ is smaller than $f(t)$ there. In this case, how do I find $k$? What constraints should I put in my calculation to make sure that $p(t)$ is always smaller than $1$.

答: It's the integral of density over its domain that needs to be $1,$ not the density itself, which in some cases can exceed $1$ at some points. For example in a uniform density defined over $[0,1/10]$ the density will be $10$ between $0$ and $1/10.$


问:
Non-linear coupled differential equations
I was trying to solve these coupled differential equations but can´t quite get to the solution. The differential equations are:

$$ H^2= \frac{1}{3}\left[\frac{1}{2} \dot\phi^2+V(\phi)\right] \space\space\space\space\space(1)$$

$$  \ddot\phi+3H\dot\phi-\lambda V_0e^{-\lambda \phi}=0  \space\space\space\space\space\space (2)$$

where $ H(t)=\frac{\dot a(t)}{a(t)} $ , $\lambda$ and $V_0$ are constants and the dot notation represents $\space \dot a= \frac{da}{dt}$. The respective solutions of $a(t)$ and $\phi(t)$ are the following:

$$ a(t)=a_0 t^{P} \space \space \space, \space\space\space P=\frac{2}{\lambda^2}\space\space\space\space\space\space (3)$$

$$ \phi(t) =\phi_0 + \frac{2}{\lambda}\ln \left( \frac{t}{t_0}\right)\space\space\space\space\space\space\space\space (4)$$

I was told to do a power law solution for $a(t)$ and $\psi(t)$, given that $\psi(t)=e^{- \lambda \phi (t)}$ , but making that substitution $ \psi(t)$ I got an equation that seems even worse:

$$ -\ddot \psi \psi^2 \left(\frac{1}{\lambda} \right) + \left( \frac{\dot \psi}{\lambda}- \frac{3H \psi}{\lambda}  \right) \dot \psi = \lambda V_0 \psi^3$$

I tried using aproximations too but to no luck. Any help would be appreciated.

答:
This question is not answerable without knowing $V(\phi)$. In your similar post on Physics SE, you mentioned that $V(\phi)=V_0e^{-\lambda\phi}$ so I will assume that.

The two equations to be solved are

$$\left(\frac{\dot a}{a}\right)^2=\frac13\left(\frac12\dot\phi^2+V_0e^{-\lambda\phi}\right)\tag{1a}$$

and

$$\ddot\phi+3\left(\frac{\dot a}{a}\right)\dot\phi-\lambda V_0e^{-\lambda\phi}=0.\tag{1b}$$

The presence of the exponential $e^{-\lambda\phi}$ complicates these equations so change variables from $\phi$ to

$$\psi\equiv e^{-\lambda\phi}.\tag2$$

In terms of $\psi$ we have

$$\phi=-\frac{1}{\lambda}\ln\psi\tag{3a}$$
$$\dot\phi=-\frac{1}{\lambda}\frac{\dot\psi}{\psi}\tag{3b}$$
$$\ddot\phi=-\frac{1}{\lambda}\left[\frac{\ddot\psi}{\psi}-\left(\frac{\dot\psi}{\psi}\right)^2\right]\tag{3c}$$

so equations (1) become

$$3\left(\frac{\dot a}{a}\right)^2=\frac{1}{2\lambda^2}\left(\frac{\dot\psi}{\psi}\right)^2+V_0\psi\tag{4a}$$

and

$$-\frac{1}{\lambda}\left[\frac{\ddot\psi}{\psi}-\left(\frac{\dot\psi}{\psi}\right)^2\right]-\frac{3}{\lambda}\left(\frac{\dot a}{a}\right)\left(\frac{\dot\psi}{\psi}\right)-\lambda V_0\psi=0.\tag{4b}$$

These look complicated, but let's see if they have a simple power-law solution of the form

$$a=a_0t^p\tag{5a}$$

$$\psi=\psi_0t^q\tag{5b}.$$

It seems reasonable to try this because all the terms will be just powers of $t$ and maybe we can make those powers work out to satisfy the equations. We have

$$\frac{\dot a}{a}=\frac{p}{t}\tag{6a}$$

$$\frac{\dot\psi}{\psi}=\frac{q}{t}\tag{6b}$$

$$\frac{\ddot\psi}{\psi}=\frac{q(q-1)}{t^2}\tag{6c}$$

so equations (4) become

$$\frac{3p^2}{t^2}-\frac{q^2}{2\lambda^2t^2}=V_0\psi_0t^q\tag{7a}$$

$$\frac{q}{\lambda t^2}-\frac{3pq}{\lambda t^2}=\lambda V_0\psi_0t^q\tag{7b}.$$

Now look at the dependence of each term on $t$. These equations can only be satisfied if

$$q=-2\tag{8}$$

and, with this value of $q$, equations (7) become

$$3p^2-\frac{2}{\lambda^2}=V_0\psi_0\tag{9a}$$

$$-\frac{2}{\lambda}+\frac{6p}{\lambda}=\lambda V_0\psi_0\tag{9b}.$$

We can eliminate $V_0\psi_0$ by taking (9a) and subtracting (9b) divided by $\lambda$ from it. This gives

$$3p^2-\frac{6p}{\lambda^2}=0\tag{10a}$$

or

$$p=\frac{2}{\lambda^2}.\tag{10b}$$

So we have our solution!

$$a(t)=a_0t^{2/\lambda^2}\tag{11a}$$

$$\phi(t)=-\frac{1}{\lambda}\ln(\psi_0t^{-2})=-\frac{1}{\lambda}\ln\psi_0+\frac{2}{\lambda}\ln t\tag{11b}.$$

This is equivalent to the form you gave,

$$\phi=\phi_0+\frac{2}{\lambda}\ln\frac{t}{t_0}\tag{12}$$

if we take

$$\phi_0=\frac{2}{\lambda}\ln{t_0}-\frac{1}{\lambda}\ln{\psi_0}.\tag{13}$$


问:
Proof of a new identity for the finite sum $\sum _{k=0}^{n-1} (-1)^k \sec \left(\frac{\pi\, k}{n}+\frac{\pi }{2 \ n}\right)$
Problem statement:

Prove the following identity

\begin{align}
\frac{\pi}{2\,\mathscr{n}} \sum _{k=0}^{\mathscr{n}-1} (-1)^k \sec \left(\frac{\pi\, k} {\mathscr{n}}\,+\,\tag{1}\label{eq:1}
\frac{\pi }{2\,\mathscr{n}}\right)+\frac{\pi}{2} \sec \left(\frac{\pi\, \mathscr{n}}{2}\right)\,
=
\end{align}

\begin{align*}
=\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, \mathcal{G_\texttt{1}(\mathscr{n})} \, + \,
 \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,\mathcal{G_\texttt{2}(\mathscr{n})}
\end{align*}

\begin{align*}
=\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, %
G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert
\begin{array}{c}
\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\
0,0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}%
\end{array}%
\right. \right)\,+
\end{align*}

\begin{align*}
 +\, \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,%
G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert
\begin{array}{c}
\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\
0,0,\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 2}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n}{\mathscr{n}}%
\end{array}%
\right. \right)
\end{align*}

by serial representations of the two Meijer G-functions on the right hand side. The finite sum on the left hand side converges for any even number $\mathscr{n}$.

The identity is a special case of a more general solution of the integral
Vladimir Reshetnikov
and can be derived starting from stocha - Summary (1. - 3.) - and solve the integral  with the help of the Plancherel theorem, similar to Sasha.
The more challenge is to prove the identity by a serial representation of the right hand side of \eqref{eq:1}. As a result new series representations of each Meijer G function follow, respectively.

Introduction:

For a more general discussion the representation of the Meijer G-functions through their generalizations - the Fox H - Functions - Mathai are needed:

\begin{align*}
\mathcal{H_\texttt{1}(\mathscr{n})}\, = \mathcal{G_\texttt{1}(\mathscr{n})}\, =  H_{3,3}^{3,2}\left( \tag{2}\label{eq:2}1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right)
\end{array}\right. \right)
\,\text{and}\qquad
\end{align*}
\begin{align*}
\mathcal{H_\texttt{2}(\mathscr{n})}\, = \mathcal{G_\texttt{2}(\mathscr{n})}\, =  H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right)
\end{array}\right. \right)\text{,}
\end{align*}

where we used the following coefficients and indices:

\begin{align*}
\left(a_1,A_1\right) = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right),\, \left(a_n,A_n\right) = \left(a_2,A_2\right) = \left(0,1\right)
\end{align*}
\begin{align*}
\left(a_p,A_p\right) = \left(a_{n+1},A_{n+1}\right) = \left(a_ 3,A_3\right) = \left(1,1\right)
\end{align*}

\begin{align*}
\left(b_1,B_1\right) = \left(0,1\right),\, \left(b_2,B_2\right) = \left(0,1\right)
\end{align*}
\begin{align*}
\left(b_q,B_q\right) = \left(b_m,B_m\right) = \left(b_3,B_3\right) = \left(a_1,A_1\right)  = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right)
\end{align*}

\begin{align*}
m = 3,\, n = 2,\, p = 3,\, q = 3
\end{align*}

Motivation:

The identity is a special case of a more general solution of the integral Vladimir Reshetnikov

\begin{align}
\int_{0}^\infty\operatorname{sech}(x)\operatorname{sech}(a\, x)\ dx\,=\,\left( 1-\frac{1}{\mathscr{n}}\right)\,\mathcal{H_\texttt{1}(\mathscr{a})}\,+\, \frac{2}{\mathscr{n}}\,\mathcal{H_\texttt{2}(\mathscr{a})}\tag{3}\label{eq:3}\\
\end{align}

\begin{align*}
=\left( 1-\frac{1}{\mathscr{n}}\right)\,H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right)
\end{array}\right. \right)\,+
\end{align*}

\begin{align*}
+\,\frac{2}{\mathscr{n}}\,
H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right)
\end{array}\right. \right)
\end{align*}

Besides of the identity \eqref{eq:1}, which is subject of this post, other interesting unknown identities arise from the discussion in stocha and especially the interesting answer Somos. An asymptotic expansion of $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ may also give more insight in the approximations of Claude Leibovici.

Moreover the study of the Mellin transformation of  $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ is a valuable
replenishment to the paper "Evaluation of a class of definite integrals" by M.L. Glasser (Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz.76,498-No.541(1975), 49-50.)

Since the argument of the Fox H-functions are equal to 1, we can use Mathai and write $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ in different forms:

\begin{align*}
\mathcal{H_\texttt{1}(\mathscr{n})} = H_{3,3}^{3,2}\left( 1\left\vert \tag{4}\label{eq:4}
\begin{array}{c}
\left(a_p,A_p\right)
\\
\left(b_q,B_q\right)
\end{array}\right. \right)
= H_{3,3}^{2,3}\left( 1\left\vert
\begin{array}{c}
\left(1-b_q,B_q\right)
\\
\left(1-a_p,A_p\right)
\end{array}\right. \right)
= \frac{1}{k} \,H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_p,\frac{A_p}{k}\right)
\\
\left(b_q,\frac{B_q}{k}\right)
\end{array}\right. \right)\,\text{and}\qquad
\end{align*}

\begin{align*}
\mathcal{H_\texttt{2}(\mathscr{n})} = H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_p,A_p\right)
\\
\left(b_q + 1,B_q\right)
\end{array}\right. \right)
= H_{3,3}^{2,3}\left( 1\left\vert
\begin{array}{c}
\left(1-b_q + 1,B_q\right)
\\
\left(1-a_p,A_p\right)
\end{array}\right. \right)
\end{align*}

\begin{align*}
= \frac{1}{k} \,H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_p,\frac{A_p}{k}\right)
\\
\left(b_q + 1,\frac{B_q}{k}\right)
\end{array}\right. \right)
\end{align*}

where $k > 0$

Attempt at a solution:

For the prove we followed two ways: The first way is to express the Fox H - Functions $\mathcal{H_\texttt{1}(\mathscr{n})}$ and $\mathcal{H_\texttt{2}(\mathscr{n})}$ in serial representation  Mathai, take advantage of the fact, that the poles of  $\Gamma \left(1-a_j+s A_j\right)$ at least for $\mathcal{H_\texttt{1}(\mathscr{n})}$ are simple and receive an infinite sum:

\begin{align}
\mathcal{H_\texttt{1}(\mathscr{n})} = (-1)^{\frac{\mathscr{n}}{2}} \,\pi \,\log\left(2\right) + \frac{1}{\mathscr{n}}\,\sum _{\nu =0}^{\infty } (-1)^{\nu } \,\frac{\Gamma\left(\frac{\mathscr{n} - 2\,\nu -1}{2\,\mathscr{n} }\right) \Gamma\left(1 + \nu\right) \Gamma\left(\frac{1 + \mathscr{n} + 2\,\nu }{2\, \mathscr{n} }\right)^2}{\nu ! \,\Gamma\left(\frac{1 + 3 \,\mathscr{n} + 2 \,\nu }{2\, \mathscr{n} }\right)}  \tag{5}\label{eq:5}
 \end{align}

\begin{align*}
= (-1)^{\frac{\mathscr{n}}{2}} \,\pi \,\log\left(2\right) + 2 \pi \, \sum _{\nu =0}^{\infty } \frac{(-1)^{\nu }\, \sec\left(\pi\,\frac{1+2\,\nu }{2\, \mathscr{n} }\right)}{1 + \mathscr{n} + 2 \,\nu }
 \end{align*}

It turned out, that the equivalent sum for $\mathcal{H_\texttt{2}(\mathscr{n})}$ in contrast do not converge. Therefore we use the series representations of the Meijer G function Mathematical Functions Site, which do converge:

\begin{align}
\mathcal{G_\texttt{1}(\mathscr{n})} = \sum _{k=1}^{\mathscr{n}+1} \frac{\left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}  \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}+2} \Gamma\left(1+b_j-a_k\right)\right)}{\Gamma\left(1+a_{\mathscr{n}+2}-a_k\right)}\,_{\mathscr{n}+2}F_{\mathscr{n}+1}(\left\{1+b_{1}-a_k,\ldots ,1+b_{\mathscr{n}+2}-a_k\right\},  \tag{6}\label{eq:6}
\end{align}
\begin{align*}
\left\{1+a_1-a_k,\ldots ,1+a_{k-1}-a_k,1+a_{k+1}-a_k,\ldots,1+a_{\mathscr{n}+2}-a_k\right\},-1)\\
\end{align*}

Note that $b_{1} = b_{2}=0$, $a_{\mathscr{n}+1} = 0$, $a_{\mathscr{n}+2} = 1$, $a_j = b_{j+2},\,j = 1,\ldots,n$
and

\begin{align}
\frac{{\Gamma\left(1-a_{\mathscr{n}+1}\right)}^2}{\Gamma\left(2-a_{\mathscr{n}+1}\right)} = \frac{\Gamma\left(2-a_{\mathscr{n}+1}\right)}{{\left(a_{\mathscr{n}+1}-1\right)}^2}\tag{7}\label{eq:7}\\
\end{align}

\begin{align}
\mathcal{G_\texttt{1}(\mathscr{n})} =  \sum _{k=1}^{\mathscr{n}+1}  \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}  \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{{\left(a_{k}-1\right)}^2} \tag{8}\label{eq:8}\\
\end{align}
\begin{align*}
 \,_{\mathscr{n}+2}F_{\mathscr{n}+1}(\left\{1-a_k,1-a_k,1+a_{1}-a_k,\ldots ,1+a_{\mathscr{n}}-a_k\right\},
\end{align*}
\begin{align*}
 \left\{1+a_1-a_k,\ldots ,1+a_{k-1}-a_k,1+a_{k+1}-a_k,\ldots,1+a_{\mathscr{n}+2}-a_k\right\},-1)\\
\end{align*}

\begin{align}
\mathcal{G_\texttt{1}(\mathscr{n})} = \sum _{k=1}^{\mathscr{n}+1}  \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}  \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{{\left(a_{k}-1\right)}^2} \tag{9}\label{eq:9}
\end{align}
\begin{align*}
 \,_{3}F_{2}(\left\{1,1-a_k,1-a_k\right\},\left\{1-a_k,2-a_k\right\},-1)\\
\end{align*}

\begin{align}
\mathcal{G_\texttt{1}(\mathscr{n})} = \frac{1}{2}\,\sum _{k=1}^{\mathscr{n}+1}  \left(\left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}  \Gamma \left(a_k-a_j\right)\right) \prod _{j=1}^{\mathscr{n}} \Gamma\left(1+a_j-a_k\right)\right)\frac{\Gamma\left(2-a_{k}\right)}{\left(a_{k}-1\right)}\,\left(\psi\left(\frac{1}{2}-\frac{a_k}{2}\right)-\psi\left(1-\frac{a_k}{2}\right)\right) \tag{10}\label{eq:10}\\
\end{align}

Note that $a_{\mathscr{n}+1} = 0$. For further simplifications we have to express the coefficients $a_k$ and $a_j$ as function of $\mathscr{n}$

\begin{align}
a_j\left(\mathscr{n}\right) = \frac{1-\mathscr{n}}{2 \mathscr{n}}+\frac{j-1}{\mathscr{n}}\quad a_k\left(\mathscr{n}\right) = \frac{1-\mathscr{n}}{2 \mathscr{n}}+\frac{k-1}{\mathscr{n}} \tag{11}\label{eq:11}
\end{align}

The second product in \eqref{eq:6} is done with the duplication formula Wikipedia

\begin{align}
\prod _{j=1}^\mathscr{n} \Gamma\left(1+a_j-a_k\right)=\prod_{j=0}^{\mathscr{n}-1} \Gamma\left(1 + \frac{k-1}{\mathscr{n}} + \frac{j}{\mathscr{n}}\right)=(2 \pi )^{\frac{\mathscr{n}-1}{2}}\mathscr{n}^{\frac{1}{2}-(\mathscr{n}-(k-1))} \,\Gamma\left(\mathscr{n}-(k-1)\right) \tag{12}\label{eq:12}
\end{align}

The simplification of the first product in \eqref{eq:6} seems to be very difficult, since the sign in the gamma function for some coefficients is negative.  The same derivations have to be done similar for $\mathcal{G_\texttt{2}(\mathscr{n})}$. We find a definite sum over $\mathscr{n}$ as expected, but make no progress at this point at all.

EDIT:

In the meantime, progress is done with the sum of the two Meijer - G- Functions \eqref{eq:1}:

\begin{align}
\frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\tag{13}\label{eq:13}
G_{\mathscr{n} + 1,\mathscr{n} + 1}^{\mathscr{n} + 1,\mathscr{n} + 1}\left( 1\left\vert
\begin{array}{c}
0,\frac{ \frac{1-\mathscr{n}} {2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}\\
0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}%
\end{array}%
\right. \right) =
\end{align}

\begin{align*}
=\, \frac{\left(1-\frac{1}{\mathscr{n}}\right)}{(2 \pi )^{\mathscr{n}-1}}\,\, %
G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert
\begin{array}{c}
\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\
0,0,\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}}%
\end{array}%
\right. \right)\,+
\end{align*}

\begin{align*}
 +\, \frac{2}{(2 \pi )^{\mathscr{n}-1}}\,\,%
G_{\mathscr{n} + 2,\mathscr{n} + 2}^{\mathscr{n} + 2,\mathscr{n} + 1}\left( 1\left\vert
\begin{array}{c}
\frac{\frac{1-\mathscr{n}}{2}}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n - 1}{\mathscr{n}},0,1\\
0,0,\frac{\frac{1-\mathscr{n}}{2} + 1}{\mathscr{n}},\frac{\frac{1-\mathscr{n}}{2} + 2}{\mathscr{n}},...,\frac{\frac{1-\mathscr{n}}{2} + n}{\mathscr{n}}%
\end{array}%
\right. \right)
\end{align*}

This identity can be generalized for the Fox H-Function, too:

\begin{align*}
2\,H_{2,2}^{2,2}\left( 1\left\vert
\begin{array}{c}
\left(c_1,C_1\right) ,\left(c_2,C_2\right)
\\
\left(d_1,D_1\right) ,\left(d_2,D_2\right)
\end{array}\right. \right)
\end{align*}

\begin{align*}
=\left( 1-\frac{1}{\mathscr{n}}\right)\,H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3,B_3\right)
\end{array}\right. \right)\,+
\end{align*}

\begin{align*}
+\,\frac{2}{\mathscr{n}}\,
H_{3,3}^{3,2}\left( 1\left\vert
\begin{array}{c}
\left(a_1,A_1\right) ,\left(a_2,A_2\right) ,\left(a_3,A_3\right)
\\
\left(b_1,B_1\right) ,\left(b_2,B_2\right) ,\left(b_3 + 1,B_3\right)
\end{array}\right. \right)
\end{align*}

where we used the following coefficients:

\begin{align*}
\left(c_1,C_1\right) =\left(d_1,D_1\right) = \left(0,1\right),\,
\left(c_2,C_2\right) =\left(d_2,D_2\right) = \left(\frac{1-\mathscr{n}}{2},\mathscr{n}\right)
\end{align*}

With the new relationship \eqref{eq:13}, considering the appropriate series representation, we only have to show that:

\begin{align}
\frac{\sum _ {k=1}^{\mathscr{n}+1} \left(\prod _{j=1,j\neq k}^{\mathscr{n}+1}\Gamma\left(c_k-c_j \right)\right) \prod_{j=1}^{\mathscr{n}+1}\Gamma\left(1+c_j-c_k \right)}{(2\pi )^{\mathscr{n}-1}} =
\end{align}

\begin{align}
\frac{\pi}{2\,\mathscr{n}} \sum _{k=0}^{\mathscr{n}-1} (-1)^k \sec \left(\frac{\pi\, k} {\mathscr{n}}\,+\,
\frac{\pi }{2\,\mathscr{n}}\right)+\frac{\pi}{2} \sec \left(\frac{\pi\, \mathscr{n}}{2}\right)\,
\end{align}
We may express again the coefficients $c_k$ and $c_j$ as function of $\mathscr{n}$

\begin{align}
c_1=0, \quad c_{j/k}=\frac{\left(\frac{1-\mathscr{n}}{2}+\left(j/k-2\right)\right)}{\mathscr{n}} \quad j,k\geq 2
\end{align}

Question:

How can we proof the new identity by a series representation of the two Meijer G-functions?

What are the appropriate series representation of each of the Meijer G-functions and how can the finite sum over the secant - function be splitted?

Is there a closed - form expression for the product $\prod_{j=1,j\neq k}^{\mathscr{n}+1}\Gamma\left(a_k-a_j\right)$ ?

Final Question - Final question remains after progress

The proof is done, if we show that:

\begin{align}
\prod _{j=0,j\neq k}^{\mathscr{n}-1} \Gamma\left(\frac{k}{\mathscr{n}}-\frac{j}{\mathscr{n}}\right)=\frac{(-1)^{1+k} \,\mathscr{n}^{\mathscr{n}-k-\frac{3}{2}} \,\,(2 \pi )^{\frac{\mathscr{n}-1}{2}}}{\Gamma\left(\mathscr{n}-k\right)}
\end{align}

答:
To obtain the last identity we use the Gauss's multiplication formula:
\begin{equation}
 \Gamma\left(nz\right)=(2\pi)^{(1-n)/2}n^{nz-(1/2)}\prod_{r=0}^{n-1}\Gamma\left(z+\frac{r}{n}\right)
\end{equation}
where $nz\ne 0,\pm1,\pm2\ldots$.
Some modifications to the product to be evaluated are to be done. First, by changing $j=n-1-s$, we have
\begin{align}
 P&=\prod _{j=0,j\neq k}^{n-1} \Gamma\left(\frac{k}{n}-\frac{j}{n}\right)\\
 &=\prod _{s=0,s\neq n-k-1}^{n-1} \Gamma\left(\frac{k-n+1}{n}+\frac{s}{n}\right)
\end{align}
Now, as $k-n+1$ is a non-positive integer, we shift the arguments of 1, using the recurrence formula for the Gamma function:
\begin{equation}
 \Gamma\left(\frac{k-n+1}{n}+\frac{s}{n}\right)=\frac{\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\frac{k-n+1}{n}+\frac{s}{n}}
\end{equation}
Then,
\begin{align}
 P&=n^{n-1}\frac{\prod_{s=0,s\neq n-k-1}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\prod_{s=0,s\neq n-k-1}^{n-1}(k-n+1+s)}
\end{align}
In the numerator, we can add the factor corresponding to $s=n-k-1$ which is $\Gamma(1)=1$ to express
\begin{equation}
 P=n^{n-1}\frac{\prod_{s=0}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)}{\prod_{s=0,s\neq n-k-1}^{n-1}(k-n+1+s)}
\end{equation}
The numerator can now be evaluated with the multiplication formula with $z=(k+1)/n$:
\begin{equation}
 \prod_{s=0,s\neq n-k-1}^{n-1}\Gamma\left(\frac{k+1}{n}+\frac{s}{n}\right)=(2\pi)^{(n-1)/2}n^{-k-1/2}\Gamma(k+1)
\end{equation}
The factors of the denominator $D$ are
\begin{equation}
 -(n-k-1),-(n-k-2),\ldots,-2,-1,1,2,\ldots,k-1,k
\end{equation}
then,
\begin{equation}
 D=(-1)^{n-k-1}\Gamma(n-k)\Gamma(k+1)
\end{equation}
Finally
\begin{align}
 P&=n^{n-1}\frac{(2\pi)^{(n-1)/2}n^{-k-1/2}\Gamma(k+1)}{(-1)^{n-k-1}\Gamma(n-k)\Gamma(k+1)}\\
 &=(-1)^{n-k+1}(2\pi)^{\frac{n-1}{2}}\frac{n^{n-k-3/2}}{\Gamma(n-k)}
\end{align}
Which is the expected result (up to a different sign rule. There may be a typo in the OP, as the present result seems to be numercally correct).


问:
Eigenfunction of $T^n$ vs. eigenfunction of $T$ when $T$ is ergodic
Let $(X, \mathcal{B}, \mu, T)$ be an ergodic system, and suppose $f \in L^2(X, \mathcal{B}, \mu)$ is an eigenfunction of $T^n$ for some integer $n > 1$. Is it possible to write $f$ as a linear combination (finite or infinite) of eigenfunctions of $T$?

It seems like a fairly easy question (and I believe the answer is yes), but I've been stuck with this for a couple of days. Any thoughts would be appreciated.

Thanks!

答:
You do not need $T$ to be ergodic for this to hold. Measure-preserving is enough. And you get finite sums.

We are interested in the unitary operator
$$
u_T:f\longmapsto f\circ T
$$
on $L^2$ via the so-called Koopman representation.

By the spectral mapping, the spectrum of $u_T^n=u_{T^n}$ is $\sigma(u_T^n)=\{\lambda^n\;;\;\lambda\in \sigma(u_T)\}$. In particular, if $\mu=\lambda^n$ is an eigenvalue in the latter, we have
$$
\{0\}\neq \mbox{Ker}(u_T^n-\mu Id)=\bigoplus_{\lambda^n=\mu}\mbox{Ker}(u_T-\lambda Id).
$$
So any eigenfunction of $u_T^n$ is a finite sum of eigenfunctions of $u_T$. Not every summand of the rhs decomposition must be nonzero, but at least one must be.

Note: the above really is for the case $\mu\neq 0$ and is due to the fact that $X^n-\mu$ has $n$ distinct roots. Since $u_T$ is unitary, so is $u_T^n$ and their spectra are both contained in the unit circle. So the case $\mu=0$ never occurs. The lemma we used is, for an operator $u$, and a polynomial $p(X)=\prod_{j=1}^np_j(X)$ whose factors are pairwise coprime, the one that says: $$\mbox{Ker}\,p(u)=\bigoplus_{j=1}^n\mbox{Ker}\,p_j(u).$$


问:
Give a counterexample of the compactness property
I want to refute that if $\Omega$ is complete metric space, $F$ is Borel $\sigma$-algebra $\Omega$, then for any positive $\varepsilon$ there is a compact set $K$ such that $\mathbb{P}(K)>1−\varepsilon$.

I know that for this limitation to be true, I need need separability of $\Omega$ or at least separable support of $\mathbb P$. Therefore, I need to give a counterexample, but unfortunately I cannot come up with one. I would be very grateful for help

答:
In addition to spaceisdarkgreen's answer:

In  Convergence of probability measures, 1999, by Billigsley, p.12, there's the next place:

(in other words): Let $S$ be uncountable discrete space. Then it is complete and not separable. If $S$ has the power of continuum, if we assume the axiom of choice and the continuum hypothesis, it may be shown that all probability measures are tight, i.e. for all $\epsilon$ there exists compact $K$ such that $P(K) > 1- \epsilon$.

The idea of the proof: if there is a not tight measure on such $S$ then it's easy to get a noatomic probability measure on $S$ and then work with uncountable measurable cardinal, see also  math.stackexchange.com/questions/2511273

It also shows that your question refers to deep problems of set theory, because even in "not very complicated" cases like $[0,1]$ with metric $d(x,y) = \mathbf{1}_{\{ x \ne y \}}$ the question about existence of not tight measures implies reference to continuum hypothesis.


问:
Why a graph generated from a graphon is said to be dense?
Now, I'm studying about a Graphon proposed by Lovasz.
Why a graph generated by a graphon is said to be dense?
I cannot image this.</p>

答:
An $n$-vertex graph is dense in this sense if it has $m = \Theta(n^2)$ edges: if its density $m/\binom n2$ is positive. Such a graph has a constant fraction of all the positive edges it could have.

Suppose we have a graphon $W : [0,1]^2 \to [0,1]$, and let's exclude examples which are zero almost everywhere (such graphons will actually produce empty graphs almost surely). Then there is an $\epsilon>0$ such that the set $S = \{(x,y) \in [0,1]^2 : W(x,y) \ge \epsilon\}$ has positive measure $\mu(S)$.

We generate an $n$-vertex graph from this graphon by choosing values $u_1, u_2, \dots, u_n \in [0,1]$ independently and uniformly at random, and then adding edge $(i,j)$ with probability $W(u_i, u_j)$ (also independently for each $(i,j)$.) We have
$$
   \Pr[(i,j) \in E] = \Pr[(i,j) \in E \mid (u_i, u_j) \in S] \cdot \Pr[(u_i, u_j) \in S] \ge \epsilon \cdot \mu(S).
$$
Therefore the expected number of edges in the graph is at least $\epsilon \mu(S) \binom n2$.

In fact the number of edges is tightly concentrated around the mean. We can prove this, for example, with McDiarmid's inequality: changing each $u_i$ can change the number of edges by at most $c_i = n-1$, so
$$
   \Pr\left[|E| \le (\epsilon \mu(S)- \delta) \binom n2 \right] \le \exp\left(\frac{2 (\delta \binom n2)^2}{\sum_{i=1}^n (n-1)^2}\right) = e^{-\delta^2 n/2}.
$$
Therefore the graph has density at least $\epsilon \mu(S)-\delta$ with probability $1 - e^{-\delta n^2/2}$, which goes to $1$ as $n \to \infty$. This proves that the graph we get is dense asymptotically almost surely.