问:
Reconstructions of Groups From Category of $G-\mathbf{Sets}$; Construction of a Group Homomorphism
I try to come up with a proof of the following statement, but I find it a little difficult. I hope that I can get some help from someone on this site. I think this is what they give a proof of, on Ncatlab - Tannakian Duality (at the section $G-\mathbf{Sets}$). But I can't really follow that proof: https://ncatlab.org/nlab/show/Tannaka+duality#ForPermutationRepresentations.

Statement. Let $F:G-\mathbf{Sets}\to\mathbf{Sets}$ be the forgetful functor, where $G-\mathbf{Sets}$ is the category of sets equipped with a group action by the group $G$. I'm trying to understand the proof of the following fact
$$\operatorname{Aut}(F)\cong G.$$

What I have done

I have managed to construct a map
$$\varphi:G\to\operatorname{Aut}(F)$$
This was done by the following rule $\varphi(g)=\eta^g$, where $\eta_S^g:S\to S$ is defined by $\eta_S^g(s)=s\cdot g$. It is straightforward to check that this gives a natural transformation from $F$ to $F$ and that it also is a group homomorphism.

However, the other way is more problematic for me. I want to find a map
$$\psi:\operatorname{Aut}(F)\to G.$$
That is, given a natural transformation $\eta$, I want to assign it to a group element $g\in G$.

The natural transformation $\eta$ is defined by the following commutative diagram
$\require{AMScd}$
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
F(X) & \ra{\eta_X} & F(X) \\
\da{F(f)} & & \da{F(f)} \\
F(Y) & \ra{\eta_Y} & F(Y) & \\
\end{array}
$$
where $\eta_X$ is a morphism in $\mathbf{Sets}$ and $f:X \to Y$ is a morphism in the category $G-\mathbf{Sets}$. Since $F$ is just the forgetful functor, the above diagram reduces to
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
X & \ra{\eta_X} & X \\
\da{f} & & \da{f} \\
Y & \ra{\eta_Y} & Y & \\
\end{array}
$$

Concerns and Questions

In the definition of natural transformation - I have that - given any $G-\text{Set}$ $X$, $\eta_X:F(X)\to F(X)$ is a morphism. A natural $G-\text{Set}$ is simply to take $X=G$ and to let it act on itself through the group structure: $$\varphi: G\times G\to G \\ (g,s)\mapsto g\cdot s.$$
So the commutative diagram now becomes
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
G & \ra{\eta_G} & G \\
\da{f} & & \da{f} \\
Y & \ra{\eta_Y} & Y & \\
\end{array}
$$

Remark 1. I remember a professor told me that the morphism $\eta_G$ is totally understood by what it does to the identity element $e\in G$ (from which I should be able to understand how to construct the group homomorphism),
$$e\mapsto \eta_G(e).$$

I don't really understand what the above means. I think I have misunderstood something about the forgetful functor. When I think about the forgetful functor $F:A\to B$, I think that the functor forgets everything that is present in $A$, but isn't present in $B$. In our case, it forgets the structure of group actions. And so, in particular, I cannot use the property of being a $G$-equivariant map. Only the properties of being a set-theoretic map.

Question 1.

If $\eta_G(e)=s$, and if I would like to make sense of what the professor told me, I think I would reason something as follows
$$\eta_G(g)=\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)=s\eta_G(g).$$
where I in the second equality used the property of being a group homomorphism. But on the other hand, if I want to treat it as a group homomorphism, then I think I had to do it to start with. That is, $\eta_G$ must map identities to identities (in order to be consistent in my reasoning). So I think my argument fails.

My question is: What does he mean?

I don't think what I did above makes any sense. But I think I have seen others using the properties of the morphisms in the category $A$, after having applied the forgetful functor, hence my reasoning. Once again, I am not really sure what I am doing. So I may very well be wrong.

Question 2. How does this tell me where to map a natural transformation?

Given a $\eta\in\operatorname{Aut}(F)$, where do I map it? Do I map it as follows
$$\eta\mapsto \eta_G(e)?$$
Doing so, do I know that I have exhaustively told where to map every natural transformation?

Question 3. I guess I also, somehow, have to use the commutative diagram in the definition of the natural transformation when I construct the group homomorphism, which I haven't done? I guess my suggestion above is not the correct way to do it. Do you have any ideas how I can construct the map?

I would be really happy I could have any help from someone on this site to understand this better. Because I am really lost, and confused.

Best wishes,

Joel

答:
I am going to use left $G$-sets, not right.

Question 1 & 3.

You cannot write $\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)$, we are not assuming $\eta_G:G\to G$ is a group homomorphism, only that it is a morphism of $G$-sets. You can ue that to say $\eta_G(g\cdot e)=g\cdot\eta_G(e)$ though (which you'd reverse the order of if you insist on right group actions).

Consider your commutative diagram again:

$$\require{AMScd}
\begin{CD}
G @>{\eta_G}>> G \\
@VVV @VVV \\
Y @>{\eta_Y}>> Y
\end{CD}$$

Here, we can let the map $G\to Y$ be the evaluation-at-$y$ map $g\mapsto gy$ where $y\in Y$ is fixed (note the evaluation map is also useful in establishing the orbit-stabilizer theorem - its fibers are cosets of $y$'s stabilizer). Then we chase the diagram starting from $e\in G$ in the top left.

If we follow the upper-right path, we get $e\mapsto \eta_G(e)\mapsto \eta_G(e)y$. In the lower-left path, $e\mapsto y\mapsto \eta_Y(y)$. Therefore we may equate $\eta_Y(g):=\eta_G(e)y$. That is, every automorphism $\eta$ applied to a $G$-set $Y$ is just applying a particular group element $\eta_G(e)\in G$.

Qusetion 3.

Yes, $\eta\mapsto \eta_G(e)$. This applies for all $\eta\in\mathrm{Aut}\,F$.


问:
Deriving exponential generating function for central trinomial coefficients
A recent question (link) asked for a derivation of the (ordinary) generating function for the central trinomial coefficients $\{T_n\}$. But the OEIS page (A002426) also lists an exponential generating function

$$\sum_{n=0}^\infty T_n \frac{x^n}{n!}=e^x I_0(2x)$$ where $I_0(x)$ is the zeroth Bessel function. How is this derived? I'll take a stab myself at showing this using the tools of analytic combinatorics, but I wanted to open this up to more knowledgeable folks as well.

答: $$\sum_{n=0}^\infty T_n\frac{x^n}{n!}=\frac{1}{2\pi i}\sum_{n=0}^\infty\frac{x^n}{n!}\oint\frac{(1+z+z^2)^n}{z^{n+1}}\,dz=\frac{e^x}{2\pi i}\oint e^{x(z+1/z)}\frac{dz}{z}=e^x I_0(2x),$$ using $T_n=[z^n](1+z+z^2)^n$, then the exponential series, then the contour integral representation of $I_0$ based on the generating function $e^{z(t+1/t)/2}=\sum_{n\in\mathbb{Z}}I_n(z)t^n$.

ContentLicense: CC BY-SA 4.0


问:
Probability of a bold gambler doubling their capital in fair even odds game
A gambler plays an even odds game, where they either win or lose the amount they staked, each with probability $1/2$. The gambler starts with a capital of $x$ dollars, where $0<x<2A$, and adopts the following strategy:

If the gambler's capital is $y\leq A$, then they stake all their money.

If they have $y>A$, then they bet $2A-y$.

The gambler stops playing once they have exactly $2A$ dollars. What's the probability that they reach their aim without going bankrupt, as a function of $x$?

If $S_n$ denotes the fortune of the gambler after round $n$, then $S_n$ is martingale, and $\mathbb E[S_n]=x$. It can be seen that the stopping time $T$ for the end of the game satisfies the conditions for the optional stopping theorem, so $$x=\mathbb E[S_0]=\mathbb E[S_T]=2A\cdot\mathbb P(S_T=2A).$$
Hence the probability of winning is $\frac{x}{2A}$.

This problem feels too way too cute to be nuked by martingales. I'm interested in a more elementary solution. Here's some headway I've made.

Let the probability be $f(x)$. Then $f$ satisfies
$$f(x)=\begin{cases}\frac{1}{2}f(2x)&\text{if }x\in[0,A] \\ \frac{1}{2}[1+f(2x-2A)]&\text{if }x\in[A,2A],\end{cases}$$and of course $f(0)=0$, $f(2A)=1$. I'm not sure how to go from here to establishing that $f$ is linear.

答: I can already tell, before doing any calculus at all, since the probability of winning is 1/2 at every turn and the probability of losing it's the same 1/2 at every turn, at you can either lose or gain the same amount that you can bet the average expectancy, that is the probability of achieving 2A before going bankrupt is going to be. $$P(achieving \ 2A \ capital)=x/2A$$


问:
Do we count only distinct roots in Descartes' rule of signs?
Descartes' rule of signs states that numbet of positive roots of a polynomial $P(x) = a_nx^n + ... + a_1x + a_0$ is not greater than the number of sign changes of coefficients.

By Gauss we know that number of complex roots of $P(x)$ is $n$ and therefore, number of real (in this case positive) roots is certainly $\leq n$ and we can count same roots more than once if $P(x)$ indeed has multiple roots.

Okay, suppose that number of signs changes of $P(x)$ is $k$ and also suppose that we have set $S$ of all positive roots of $P(x), S = \{x_1, x_2, ..., x_m\}, m \leq n$ and some $x_i, x_j$ can be equal. Now, take that same set and eliminate those equal elements (eliminate all $x_i = x_j$ when $i \neq j$), that new set call $S'$ and let's say $|S'| = r.$

Does Descartes' rule of signs tells us that $m$ cannot be greater than $k$ or that $r$ cannot be greater than $k$?

Does this rule counts multiple roots more than once or it counts just disctint roots?
I am asking this because the geometric proof given here https://math.hmc.edu/funfacts/descartes-rule-of-signs/ has sense to me if we count only distinct roots (because that's number of intersections of polynomial with the $x$-axis) but no otherwise. Thanks

I saw in the comments that the rule counts multiple roots (say $x_0$) more than once. Since the proof given in the link is geometric, does polynomial bounces near the crossing point $(x_0, 0)$?

How exatcly the number of $x_0$'s we counted reflect on the behaviour of $P(x)$ near that point? ($)

My hypothesis is that if that number is even polynomial has parabolic shape near that point and if it is odd polynomial has shape simillar to the curve $x^3$ (and since the parity of sign changes and positive roots (counted with multiplicity) are equal, then this does not change number of ups and downs of our curve - now proof has sense but please answer #)

答:
If $X$ is a root with multiplicity $m$, then $f(x)= (x-X)^m$ divides $p(x)$, so $p(x)=f(x) q(x)$ for some polynomial $q(x)$ where $q(X)\ne0$. Note that because $q(X)$ is (continuous) polynomial with $P(X)\ne0$, there is some small interval $[X_-,X_+]$ not containing $0$ (where $X_-<X<X_+$) on which $q(x)\ne0$ and does not change sign.

If $m$ is even, $f(X_-)$ and $f(X_+)$ are both positive (they are both even powers of a nonzero number), and therefore $p(X_-)$ and $p(X_+)$ have the same sign as $q(X)$. If $m$ is odd, $f(X_-)>0$ and $f(X_+)<0$ so $p(X_-)$ and $p(X_+)$ have opposite signs, as they are odd powers of a negative number ($X_--X$) and a positive number ($X_+-X$), respectively.

Therefore when $X$ is a root of $p(x)$ with odd multiplicity, the graph of $p(x)$ crosses over the $x$-axis at $x=X$. When $X$ is a root of $p(x)$ with even multiplicity, the graph of $p(x)$ “bounces” on the $x$-axis at $x=X$.


问:
Two $R$-module homomorphisms induce a bilinear map
Say $A,B,C$ are three $R$-algebras. And given two homomorphisms of $R$-algebras, $f:A\to C$ and  $g:B\to C$. We could induce the "product homomorphism" $f\times g$ which is an $R$-bilinear map $A\times B\to C,(a,b)\mapsto f(a)g(b)$ by forgetting the algebra structure.

Is there a natural way to do this merely in the $R$-module category?

答:
No. For modules the best you get is that if $f : M \to N$ is a map of right $R$-modules and $g : P \to Q$ is a map of left $R$-modules then they induce a tensor product map

$$f \otimes g : M \otimes_R P \to N \otimes_R Q.$$

In the algebra case an $R$-algebra $C$ admits an $R$-bilinear map $C \otimes_R C \to C$, namely the multiplication, and you compose by this map to get the map you're looking at. If $C$ is just a module it is of course in no way equipped with such a map.


问:
Closed Image and finite cokernel $\Leftrightarrow$ Closed image of the transpose and finite dimensional kernel
Let $D:X\rightarrow Y$ be a bound linear operator.

$\newcommand{\im}{\operatorname{im}}$
The claim given is:

A closed image $D(X)$ and finite dimensional cokernel is equivalent to a closed image $D^*(Y^*)$ of the transpose and finite dimensional kernel.

I am stuck on my way to show the properties for the transposed function $D^*$. I have some notes from class, that is what I was able to complete. I showed that if an element $a$ of the kernel is $0$ for all basis vectors of the cokernel, then it is also $0$. Thus $\dim(\ker(D^*))\leq \dim(\operatorname{coker}(D))<\infty$.

Now let $\{\beta_n\}$ be a sequence such that $D^*\beta_n\rightarrow \alpha\in X^*$. I want to show that $\im(D^*)$ is closed so in this case that $\alpha\in \im(D^*)$. Im my notes there is just the remark that we can deduce the rest from the fact that $\im(D)$ is closed, but I don't see how.

答:
I'm assuming $X$ and $Y$ are normed vector spaces and that $X^*$ and $Y^*$ are equipped with the standard dual norm, i.e. the norm of $\phi$ in $X^*$ is $\underset{v \neq 0}{\mathrm{sup}} \frac{|\phi(v)|}{\|v\|_X}$. I haven't done functional analysis for a long time, so I might have over-detailed things, but better safe than sorry!

($\Rightarrow$) Suppose $D(X)$ is closed and $Y/D(X)$ is finite-dimensional. Let $\{\phi_n\}$ be a sequence in $Y^*$ such that $\{D^*(\phi_n)\}$ converges to $\psi$ in $X^*$.

We can re-write $D^*(\phi_n)(x)$ as $\phi_n(D(x))$, and since $Y/D(X)$ is finite-dimensional, we can find a basis of $Y/D(X)$, say $\{y_1+D(X), \cdots, y_n+D(X) \}$ such that any vector in $Y$ has a unique writing of the form $y = \sum_{i=1}^n a_i y_i + d$, and $y$ is expressed uniquely as a function of $a_1,\cdots,a_n$ in $\mathbb R$ (or $\mathbb C$, don't know which field you're working over) and $d \in D(X)$. For $\phi \in Y^*$, this means that $\phi$ is uniquely determined by $\phi(y_1),\cdots,\phi(y_n)$ and $\phi|_{D(X)}$. In particular, if $D^*(\phi)$ is zero, then $\phi|_{D(X)} = 0$, so that $\ker D^*$ is a vector space of dimension $n$ spanned by the $n$ functionals $\omega_1,\cdots,\omega_n$ such that $\omega_i(y_j) = \delta_{ij}$ and $\omega_i|_{D(X)} = 0$ for $i,j=1,\cdots,n$.

The information that $\{D^*(\phi_n)\}$ converges to $\psi \in X^*$ tells us that $\|\phi_n \circ D - \psi\|_{X^*} \to 0$, or in other words, that $\underset{x \neq 0}{\mathrm{sup}} \frac{|\phi_n(D(x)) - \psi(x)|}{\|x\|_X} \to 0$, and in particular $\phi_n(D(x)) \to \psi(x)$. It follows that for any $d \in D(X)$, $\phi_n(d)$ is a convergent sequence. Denote by $\phi(d)$ its limit, so we get a function $\phi: D(X) \to \mathbb R$, which we extend by zero to all of $Y$, i.e. $\phi\left(\sum_{i=1}^n a_i y_i + d\right) \overset{\mathrm{def}}{=} \phi(d)$.

We wish to show that $\phi \in Y^*$ and that $D^*(\phi) = \psi$. The latter is by construction. As to the former, the linearity of $\phi$ follows from the linearity of $\psi$. For continuity, suppose a sequence $d_k$ converges to $d \in D(X)$ (the general case will follow since if a sequence $y_k$ converges to $y$, then writing $Y = D(X) \oplus Y'$ where $Y'$ is finite-dimensional as above with the basis, we can write $y_k = y'_k + d_k$, $y = y' + d$ and $d_k \to d$, $\phi(y_k) = \phi(d_k)$, $\phi(y) = \phi(d)$). Since $\phi_n$ is continuous, $\phi_n(d_k) \to \phi_n(d)$. Pick a sequence $x_k$ with $D(x_k) = d_k$ and some $x \in X$ with $D(x) = d$. Then

$$
|\phi(d_k) - \phi(d)| = |\phi(d_k) - \psi(x)| \le |\phi(d_k) - \phi_n(d_k)| + |\phi_n(d_k) - \phi_n(d)| + |\phi_n(d) - \psi(x)|.
$$
The above inequality is true for all $n$, so we can take its $\mathrm{liminf}$ when $n$ goes to infinity to get an upper bound that depends only on $k$. The first term goes to $0$ with $n$ by definition of $\phi$, the second term goes to $0$ with $k$ by continuity of $\phi_n$, and the last term is $|\phi_n(D(x)) - \psi(x)|$, which also goes to $0$ with $n$ since $D^*\phi_n \to \psi$ in $X^*$. This means $\phi$ is also continuous and therefore $\phi \in Y^*$.

($\Leftarrow$) Suppose $D^*$ has finite dimensional kernel. Consider the vector space $Y/D(X)$. The universal property of the quotient map says that $(Y/D(X))^* \simeq \ker D^*$ via the dual of the quotient map $\pi : Y \to Y/D(X)$. It follows that $(Y/D(X))^*$ has dimension $n$. Therefore, $Y/D(X)$ also has dimension $n$, and it follows that $D(X)$ is a closed subspace of $Y$. (To see the latter, take a basis $\{\omega_1,\cdots,\omega_n\}$ of $(Y/D(X))^*$ and lift those functionals to $\{\tilde{\omega}_1,\cdots,\tilde{\omega_n}\}$ in $Y^*$. An elementary argument shows that $\bigcap_{i=1}^n \ker \omega_i = D(X)$, showing that $D(X)$ is closed; hint: consider $K = \bigcap_{i=1}^n \ker \omega_i$ and show that $(K/D(X))^* = 0$.)

Let me know what you think of this argument, I think it's still very elementary but I can add some details if you like.


问:
Indefinite Integral of $\dfrac {x^{p-1}}{x^{2m} - a^{2m}}$
I am looking at a bunch of related integrals found in Spiegel's "Mathematical Handbook of Formulas and Tables", (Schaum, 1968), item $14.336$.

It's a complicated and messy old beast:

For $m \in \mathbb Z$ such that $m \ge 1$:

$\displaystyle \int \dfrac {x^{p-1} \, \mathrm d x}{x^{2m} - a^{2m}} = \frac 1 {2 m a^{2 m - p}} \sum_{k=1}^m \cos \frac {k p \pi} m \ln \left({x^2 - 2 a x \cos \frac {k \pi} m + a^2}\right)$

$\displaystyle - \frac 1 {m a^{2 m - p}} \sum_{k=1}^m \sin \frac {k p \pi} m \arctan \left({\frac {x - a \cos \frac {k \pi} m}{a \sin \frac {k \pi} m}}\right)$

$\displaystyle + \frac 1 {2 m a^{2 m - p}} (\ln (x - a) + (-1)^p \ln (x + a)) + C$

for all $0 < p \le 2 m$.

(Note that he glosses over the sign of $x$, presenting the expression for positive $x$ only -- I will explore the negative $x$ case later.)

I thought of $2$ different approaches:

Factorise the denominator using the standard result:

$\displaystyle x^{2 n} - y^{2 n} = (x - y) (x + y) \prod_{k \mathop = 1}^{n - 1} \left({x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}\right)$

and then explore the possibility of doing a partial fraction expansion. However, this did not work out so well, as I was then unable to see how to simplify the expression on the top of the resulting terms in all but the $x - a$ and $x + a$ denominators, so I didn't proceed further with that.

Prove it by induction on $m$, which first concerns establishing the base case:

$\displaystyle \int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\dfrac {x - a} {x + a}}\right) + C$ (that is: $m = 1$, $p = 1$)

$\displaystyle \int \frac {x \, \mathrm d x} {x^2 - a^2} = \ln \left({x^2 - a^2}\right) + C$ (that is: $m = 1$, $p = 2$)

But when I put $m = 1$ into the above expression, I get:

$\dfrac 1 {2 a^{2 - p} } ( (-1)^p \ln (x^2 + 2 a x + a^2) ) + \dfrac 1 {2 a^{2 - p} } (\ln (x - a) + (-1)^p \ln (x + a) )$

But that pesky term on the left, in $(-1)^p \ln (x^2 + 2 a x + a^2)$ is clearly incorrect.

This arises from the $\cos \dfrac {k p \pi} m \ln \left({x^2 - 2 a x \cos \dfrac {k \pi} m + a^2}\right)$ expression in the given solution.

But for integer $p$ and $m = 1$, that $\cos \dfrac {k p \pi} m$ term, which I would hope to vanish, defiantly remains in place.

It doesn't help much changing it to $\cos \dfrac {k p \pi} {2 m}$, because this time it does not vanish for $p = 2$.

So either Spiegel has reported this result wrong, or I'm misunderstanding something.

Should that cosine in fact be a sine? Then it would vanish away like I want it to.

For $m = 2$ and $p = 1$ you get this:

$\dfrac 1 {4 a^3} \ln \left( {\dfrac {x - a} {x + a} }\right) - \dfrac 1 {2 a^3} \arctan \dfrac x a$

which is consistent with the result quoted. I have gone a little further, trying $p = 2, 3, 4$, but now it is looking as though the cosine term is completely erroneous and perhaps should not be there at all, because any contribution it makes is consistently absent.

I have gone no further with higher $m$ values.

答:
There is a typo in the formula

\begin{align}
I(m,p) =\int \dfrac {x^{p-1} \, \mathrm d x}{x^{2m} - a^{2m}}
& = \frac 1 {2 m a^{2 m - p}} \sum_{k=1}^{m-1} \cos \frac {k p \pi} m \ln \left({x^2 - 2 a x \cos \frac {k \pi} m + a^2}\right)\\
 & \>\>\>\>\>- \frac 1 {m a^{2 m - p}} \sum_{k=1}^{m-1}\sin \frac {k p \pi} m \arctan \left({\frac {x - a \cos \frac {k \pi} m}{a \sin \frac {k \pi} m}}\right)\\
 & \>\>\>\>\>+ \frac 1 {2 m a^{2 m - p}} (\ln (x - a) + (-1)^p \ln (x + a))
\end{align}
where the summation should run from $1$ to $m-1$, instead of $m$. Then, the followings check out
$$I(1,1)=\int \frac {\mathrm d x} {x^2 - a^2} = \frac 1 {2 a} \ln \left({\dfrac {x - a} {x + a}}\right)$$

$$I(1,2)=\int \frac {x \, \mathrm d x} {x^2 - a^2} = \ln \left({x^2 - a^2}\right)$$

$$I(2,1)=\int \frac {\, \mathrm d x} {x^4 - a^4}
= \dfrac 1 {4 a^3} \ln \left( {\dfrac {x - a} {x + a} }\right) - \dfrac 1 {2 a^3} \arctan \dfrac x a$$


问:
$y=\sin^2x+\sin x+1$. Find the range of $y$
The range of $\sin x$ is $[-1,1]$ for $x \in \mathbb{R}$.

答:
You may convert the problem into
\begin{equation}
y = x^{2} + x + 1, -1 \leq x \leq 1.
\end{equation}
This is the standard form of quadratic functions $y = ax^{2} + bx + c$. Since $a = 1 > 0$ in this case, the function is of a minimum value if the domain is $\mathbb{R}$. The minimum value of $y$ for $x \in \mathbb{R}$ is achieved at $-b/2a = -0.5$, which is within the range of $\left[-1,1\right]$. Then $y\left(-0.5\right) = 0.75$ is the minimum value. The maximum of $y$ is achieved at either $x = -1$ or $x = 1$. We have $y\left(-1\right) = 1$ and $y\left(1\right) = 3$. The function $y$ is continuous, and we have
\begin{equation}
0.75 \leq y \leq 3.
\end{equation}


问:
Does having multiple limit values at a point imply essential discontinuity?
In Complex Analysis, do "jump discontinuities" exist?

If I find that a function of $z$ approaches two different values as z is approached from two different directions, can I immediately conclude that the point is an essential singularity? Or do other possibilities exist?

答:
I'll give a proof of what I've written in the comment to the question (when I wrote that, I was assuming in my mind than the singularity was isolated one).

Suppose that we have an holomorphic function $f\colon U\setminus \{p\} \longrightarrow \mathbb{C}$, where $U\subseteq \mathbb{C}$ is an open subset and $p\in U$ a point. We want to prove that

$p$ is a removable singularity if and only if $\lim_{z\to p}f(z)=c\in \mathbb{C}$

$p$ is a pole if and only if $\lim_{z\to p}f(z)= \infty$

$p$ is an essential singularity if and only if $\lim_{z\to p}f(z)$ does not exists

We can suppose up to translation that $p=0$.

First, if $0$ is a removable singularity, then by definition there is an holomorphic function $g\colon U \longrightarrow \mathbb{C}$ that extends $f$, so that $\lim_{z\to 0 }f(z)=g(0)\in \mathbb{C}$. Conversely, if the limit $\lim_{z\to 0}f(z)=c$ exists and belongs to $\mathbb{C}$, then we can define a function $g\colon U \longrightarrow \mathbb{C}$ by
$$ g(z)=\begin{cases} f(z) & \text{ if } z\ne 0 \\ c & \text{ if } z=0 \end{cases} $$
It is clear that this function is continuous and holomorphic on $U\setminus \{ p\}$ so that it is holomorphic on the whole of $U$, by Morera's theorem (this can be found for example in Cartan's "Elementary Theory of Analytical Functions of One or Several Complex Variables").

Now, if $f$ has a pole in $0$: this means that the Laurent series of $f(z)$ in $0$ is of the form

$$ f(z)=\sum_{n\geq N}a_n z^n $$

for a certain $N\geq 1$. This implies that the function $g(z)=z^Nf(z)$ is holomorphic, so that $\lim_{z\to 0}z^Nf(z)=a\in \mathbb{C}$, but since $N\geq 1$, this implies that $\lim_{z\to 0}f(z)=\infty$.

Instead, if $f$ has an essential singularity, then Casorati-Weierstrass' Theorem or Picard's Theorem tell us that the image through $f$ of an arbitrarily small neighborhood of $0$ is dense in $\mathbb{C}$, so that the limit $\lim_{z\to 0}f(z)$ does not exists.

Conversely, if $\lim_{z\to 0}f(z)=\infty$,then $0$ cannot be neither a removable singularity nor an essential singularity, for what we have proven before, so that it must be a pole. Now a similar reasoning proves that, if the limit does not exists, then the point must be an essential singularity,

ContentLicense: CC BY-SA 3.0


答:
Using the fact that $I_0 (2x) = \sum_{j=0}^\infty \frac{x^{2j}}{(j!)^2}$

\begin{equation}
e^x I_{0}(2x) = \sum_{i = 0}^\infty \sum_{j=0}^\infty \frac{x^{i+2j}}{(i!)(j!)^2}
\end{equation}
We can group the factors with the same exponent $i+2j $, obtaining:
\begin{equation}
e^x I_{0}(2x) = \sum_{n = 0}^\infty \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \frac{x^n}{(k!)^2 (n-2k)!} =
\sum_{n=0}^\infty \frac{x^n}{n!}\sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} \binom{2k}{k}
= \sum_{n=0}^\infty \frac{x^n}{n!} T_n
\end{equation}

Title: Deriving exponential generating function for central trinomial coefficients

Body:
A recent question (link) asked for a derivation of the (ordinary) generating function for the central trinomial coefficients $\{T_n\}$. But the OEIS page (A002426) also lists an exponential generating function

$$\sum_{n=0}^\infty T_n \frac{x^n}{n!}=e^x I_0(2x)$$ where $I_0(x)$ is the zeroth Bessel function. How is this derived? I'll take a stab myself at showing this using the tools of analytic combinatorics, but I wanted to open this up to more knowledgeable folks as well.


问:
What is the density of set of natural numbers that can be written as at most five or six fifth powers?
The asymptotic density of a subset $A \subset\mathbb{N}$ is defined as $d(A) = \lim_{n\rightarrow\infty} \frac{A \cap \{1, 2, ... , n\}}{n}$.

Let $A_{i} = \{ k \in\mathbb{N} : k$ can be written as a sum of at most $i$ fifth powers of natural numbers  $\}$.

When $i < 5$ I suspect that $d(A_{i}) = 0$. There is at most $n$ fifth powers $\leq n^{5}$. Therefore, we see that there are at most

${n \choose 1} + ... + { n \choose i} $ ~ $O(n^{i}$) many elements in $A_{i}$. For $i < 5$, this can be used to show $d(A_{i})$ = 0.

When $i = 5 $ it appears that the quantity $d(A_{5})$ can at most be $\frac{1}{5!}$.

How do I even dare to do this when $i = 6$? Can I put at least an upper bound to the density of $A_{6}$?

答:
According to page 7 of https://www.personal.psu.edu/rcv4/Waring.pdf every sufficiently large number is conjectured to be expressible as a sum of 6 fifth powers, which would mean the natural density in the $i=6$ case is 1.

I don't know if further progress has been made since that survey was written in 2000, but it certainly means you shouldn't be able to upper bound it.


答:
Given $$\displaystyle y=\sin^2(x)+\sin(x)+1=\frac{1}{4}\bigg[4\sin^2(x)+4\sin(x)+4\bigg]$$

$$y=\frac{1}{4}\bigg[\bigg(2\sin(x)+1\bigg)^2+3\bigg]\geq \frac{3}{4}$$

Also $$-1\leq \sin(x)\leq 1\Longrightarrow -2 \leq 2\sin(x)\leq 2$$

$$-1\leq 2\sin(x)+1\leq 3\Longrightarrow 0\leq \bigg(2\sin(x)+1\bigg)^2\leq 3$$

So we get $$y=\frac{1}{4}\bigg[\bigg(2\sin(x)+1\bigg)^2+3\bigg]\leq 3$$

So we get range as $$y\in\bigg[\frac{3}{4}\;,\; 3\bigg]$$

Title: $y=\sin^2x+\sin x+1$. Find the range of $y$

Body: The range of $\sin x$ is $[-1,1]$ for $x \in \mathbb{R}$.


问:
Shell Method Versus Disc Method
I'm trying to find the volume of a solid formed by revolving the curve $y=x^2+2$ around the y axis between $y=2$ and $y=4$, but I'm getting a different answer using the shell method and the disc method.

Using Disc Method: $V=\pi\int_{2}^{4}(\sqrt{y-2})^2dy = 2\pi$

Using Shell method: $V=2\pi\int_{0}^{\sqrt{2}}\left(x^{2}+2\right)x\ dx = 6\pi$

The solution says the answer is $2\pi$, but WolframAlpha says the volume is $6\pi$. Any explainations for this discrepancy would be great.

答:
The solid whose volume you calculated using the disk method is not the solid whose volume you calculated using the shell method. If you used shells on the first solid, your shells would run from the curve $y=x^2+2$ up to $y=4$, and the height of the shell would be $4-(x^2+2)=2-x^2$; you, however, have taken the shells from the $x$-axis up to the curve. Had you computed that volume using the disk method, you’d have had annuli, not disks, with inner radius $\sqrt{y-2}$ and outer radius $\sqrt2$.

The two volumes combine to form a cylinder of radius $\sqrt2$ and height $4$, with volume $8\pi$, the sum of your two figures.

Title: Shell Method Versus Disc Method


问:
Confusion on notation of multivariable differentiation in n-dimensional space for matrices (gradients)
I am learning gradients on my own.

One of the rules given in the textbook I am reading states this (without proof):

$$∇_\mathbf{x}\mathbf{A}\mathbf{x} = \mathbf{A}^⊤$$

and then proceeds onwards as if it's trivial to see. I haven't seen this before so I need to go prove that it is correct.

I do not know if the above equation is equal to

$$∇f(\mathbf{x}) = \mathbf{A}^⊤$$

where

$$f(\mathbf{x}) = \mathbf{A}\mathbf{x}$$

I decided to go ahead with this anyways and see where it takes me. I figure doing this with a simple $\mathbb{R}^{2 \times 2}$ like $\mathbf{A} = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ and seeing where it takes me when $\mathbf{x} = \begin{bmatrix}i\\j\end{bmatrix}$.

This means $\mathbf{A}\mathbf{x}$ is:

$$\begin{bmatrix}ai + bj\\ci + dj\end{bmatrix}$$

The gradient would be (for $\mathbf{Ax}$ as I'm passing that in for the parameter):

$$∇f(\mathbf{Ax}) = \begin{bmatrix} \frac{\partial f(\mathbf{Ax})}{\partial i} \frac{\partial f(\mathbf{Ax})}{\partial j}\end{bmatrix}^⊤$$

Which means we want to evaluate $∇_\mathbf{x} \begin{bmatrix}ai + bj\\ci + dj\end{bmatrix}$. By the definition of a gradient, then we should get $∇f(\mathbf{Ax}) = \begin{bmatrix} \begin{bmatrix}a\\c\end{bmatrix} \\ \begin{bmatrix}b\\d\end{bmatrix} \end{bmatrix}$

This however does not look right, or is it? Is that equal to $\begin{bmatrix}a&c\\b&d\end{bmatrix}$?

I wrote it out that way because I suspect that when we ask for $a_{21}$ we are saying "get the 2nd row in the first column, and in a programming fashion, A[2][1] (or A[1][0] for most languages) would get $b$ from the result above. This would make sense because I'd get the transpose I was looking for in the primitive example. My linear algebra class only covered vectors, then jumped to matrices and showed us matrix multiplication without covering matrices as if they were a vector of vectors, so if there is a fundamental gap in my knowledge here and what I've described above is correct... then we've found an issue.

Moving to an n-dimensional proof after figuring out what is wrong (if anything) seems like it'd be straight forward with more rigorous variable and index naming.

答:
Suppose that your matrix A is $m \times n$. First of all, you need to view $Ax$ as a function $f: \mathbb{R}^{n} \mapsto \mathbb{R}^{m}$ such that
\begin{equation}
\forall x \in \mathbb{R}^{n}, f\left(x\right) = Ax.
\end{equation}
Basically, this function is a linear transformation of $\mathbb{R}^{n}$. Then you should understand how derivative is defined for functions from multiple inputs to multiple outputs. The definitions and theorems can typically be found in real analysis. A function is differentiable at $x_{0}$ if and only if there exists a matrix $M$ such that
\begin{equation}
\lim_{x \to x_{0}}\frac{\lVert f\left(x\right)-f\left(x_{0}\right)-M\left(x-x_{0}\right)\rVert}{\lVert x - x_{0}\rVert} = 0.
\end{equation}
Then we write
\begin{equation}
M = f^{\prime}\left(x_{0}\right).
\end{equation}

Here, you may understand $\nabla$ as the concept differential $D$. The derivative of $f$ at a point $x \in \mathbb{R}^{n}$ is described as follows:
\begin{equation}
f^{\prime}\left(x\right) =
\begin{pmatrix}
\frac{\partial f_{1}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{1}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{1}\left(x\right)}{\partial x_{n}}\\
\frac{\partial f_{2}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{2}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{2}\left(x\right)}{\partial x_{n}}\\
\vdots & \vdots & \ddots & \vdots\\
\frac{\partial f_{m}\left(x\right)}{\partial x_{1}} & \frac{\partial f_{m}\left(x\right)}{\partial x_{2}} & \dots & \frac{\partial f_{m}\left(x\right)}{\partial x_{n}}
\end{pmatrix},
\end{equation}
which is a $m \times n$ matrix. Sometimes, the derivative is defined with a transpose symbol, which is essentially your case. The differential at a point $x_{0}$ is of the following property:
\begin{equation}
D_{x_{0}}f\left(x\right) = f^{\prime}\left(x_{0}\right)x, x \in \mathbb{R}^{n}.
\end{equation}
That is to say, the differential is a function $D_{x_{0}}f: \mathbb{R}^{n} \mapsto \mathbb{R}^{m}$. Again, real analysis is the best reference for these questions. An open-source book I recommend is "Introduction to Real Analysis" by William F. Trench.


问:
Is there a $\triangle ABC$, with $D$ on $BC$, such that $AB$, $BC$, $CA$, $AD$, $BD$, $CD$ are all integers, but only one is even?
This is just a question I came up with while working through another problem.

If we have a triangle $ABC$ and we draw a cevian from $A$ to $D$ on side $BC$, is it possible to assign integer values to lengths $AB$, $BC$, $AC$, $AD$, $BD$ or $CD$ such that only one of the listed lengths has an even valued length? If it is possible, what are these lengths?

I have done some parts of this problem on my own. I figured out that this even length could only be $BC$, $CD$, or $BD$, with $BD$ and $CD$ being the same case.

With $BC$ being the only even length I computed for $BC$ using Stewart's Theorem and got
$$BC = \frac{(AC^2)(BD)+(AB^2)(CD)}{(BD)(CD)+(AD)^2}$$ where every length but $BC$ is odd. The equation seems to work but I don't know how to obtain its integer solutions.

Similarly with $BD$/$CD$ case, I applied Stewart's theorem and got $$BD=\frac{(AB^2)(CD)-(BC)(AD^2)}{(CD)(BC)-(AC)^2}$$
Once again this seems to work, but I don't know how to obtain its integer solutions.

Pretty much all I need help is verifying whether or not my work is right, and if it  is how to find the integer solutions of my provided equations. If there is a better way or my work has flaws please let me know.

答:
A simple solution found by brute force (by setting $AD=AB$):

and three more interesting ones found by exhaustive search:


问:
Showing that Frobenius Norm is Submultiplicative
Considering the Frobenius norm $\|.\|_{F}$, how can we prove that it's submultiplicative?

N.b: I noticed there is a user who asked a similar question but proof looked like directing away from the question. I hope anyone can help me.

答:
\begin{align*}\|AB\|_{F}^{2}=
\left\|\left(\begin{array}{c}
\widehat{a}_{0}^{H} \\
\widehat{a}_{1}^{H} \\
\vdots \\
\widehat{a}_{m-1}^{H}
\end{array}\right)\left(\begin{array}{llll}
b_{0} & b_{1} & \cdots & b_{n-1}
\end{array}\right)\right\|_{F}^{2}
&=\left\|\left(\begin{array}{c|c|c|c}
\widehat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\
\hline \hat{a}_{0}^{H} b_{0} & \widehat{a}_{0}^{H} b_{1} & \cdots & \widehat{a}_{0}^{H} b_{n-1} \\
\vdots & \vdots & & \vdots \\
\hline \widehat{a}_{m-1}^{H} b_{0} & \widehat{a}_{m-1}^{H} b_{1} & \cdots & \widehat{a}_{m-1}^{H} b_{n-1}
\end{array}\right)\right\|_{F}^{2}\\ \\&=\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} b_{j}\right|^{2} \\
&\leq \sum_{i} \sum_{j}\left\|\hat{a}_{i}^{H}\right\|_{2}^{2}\left\|b_{j}\right\|^{2} \quad \text { (Cauchy-Schwartz) } \\
&=\left(\sum_{i}\left\|\widehat{a}_{i}\right\|_{2}^{2}\right)\left(\sum_{j}\left\|b_{j}\right\|^{2}\right)\\ &=\left(\sum_{i} \widehat{a}_{i}^{H} \widehat{a}_{i}\right)\left(\sum_{j} b_{j}^{H} b_{j}\right) \\
&\leq\left(\sum_{i} \sum_{j}\left|\widehat{a}_{i}^{H} \widehat{a}_{j}\right|\right)\left(\sum_{i} \sum_{j}\left|b_{i}^{H} b_{j}\right|\right)\\&=\|A\|_{F}^{2}\|B\|_{F}^{2}
\end{align*}
Now, to conclude this proof, we shall take the square root of both sides of the inequality  :$$\|AB\|_{F}^{2}\leq\|A\|_{F}^{2}\|B\|_{F}^{2}$$
And hence, this proves that Frobenius norm is submultiplicative.


答:
The disc method is correct.  The shell method gives $2\pi\int_0^\sqrt2(2-x^2)x\rm dx=2\pi$.

What you inputted for Wolfram Alpha was different.  It was the volume (swept out by the region) below the curve (and inside $x=\sqrt2$).

Title: Shell Method Versus Disc Method

Body:
I'm trying to find the volume of a solid formed by revolving the curve $y=x^2+2$ around the y axis between $y=2$ and $y=4$, but I'm getting a different answer using the shell method and the disc method.

Using Disc Method: $V=\pi\int_{2}^{4}(\sqrt{y-2})^2dy = 2\pi$

Using Shell method: $V=2\pi\int_{0}^{\sqrt{2}}\left(x^{2}+2\right)x\ dx = 6\pi$

The solution says the answer is $2\pi$, but WolframAlpha says the volume is $6\pi$. Any explainations for this discrepancy would be great.


问:
Clarification on convergence almost surely
The three aspects within convergence in distributions include:

Convergence in distribution

Convergence in probability

Convergence almost surely

Convergence in distribution and in probability intuitively make sense to me, since you can quite easily visualise what limit you are trying to evaluate as ${n\to \infty} $. However, convergence almost surely isn't quite the case. When trying to prove convergence almost surely, which is if and only if $ \Bbb{P}( [X_n-X|>\epsilon] $ infinitely often)=$0$.
An event, say $A_n$, is said to happen infinitely often if $$\sum_{n=1}^\infty \Bbb{P}(A_n)< \infty.$$
However, what does it mean (intuitively) for a random variable to converge almost surely?
And, what does the set notation of {$A_n$} infinitely often or eventually mean, i.e {$A_n$ infinitely often} = $\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_m$ and {$A_n$ eventually} = $\bigcup_{n=1}^\infty \bigcap_{m=n}^\infty A_m$?

If there's a simple way of portraying this example (eg a coin toss, die being rolled etc), the help would be greatly appreciated.

答:
I don't know what you mean by intuitively, but the definition of almost sure convergence is very simple to read: it says that $\mathbb P(X_n \to X) = 1$. Let's say you have a sequence $\{X_n\}$ of random variables all defined on a probability space $(\Omega, \Sigma, \mathbb P)$. We say that $X_n \to X$ almost surely if the event
$$
\{ \omega \in \Omega \, | \, X_n(\omega) \to X(\omega)\}
$$
has probability $1$, which is what the notation $\mathbb P(X_n \to X) = 1$ suggests.

Now that you have this event, you can express it in a way that is more useful for computing bounds, such as expressing this event as a limit of other events. Note that $X_n(\omega) \to X(\omega)$ is defined as
$$
\forall \varepsilon > 0, \quad \exists N_{\varepsilon} \quad s.t. \quad \forall n > N_{\varepsilon}, \quad |X_n(\omega) - X(\omega)| \le \varepsilon.
$$
In other words, for all $\varepsilon > 0$, $|X_n(\omega) - X(\omega)| > \varepsilon$ happens only finitely many times. This can be used to translate the above event to
$$
\bigcap_{\varepsilon > 0} \left(\bigcup_{N \in \mathbb N} \bigcap_{n \ge N}  \{ \omega \in \Omega \, | \, |X_n(\omega) - X(\omega)| \le \varepsilon \} \right)
$$
Since the collections of events under the intersection over $\varepsilon > 0$ form a decreasing collection, it has probability $1$ if and only if for each $\varepsilon > 0$,
$$
\mathbb P \left( \bigcup_{N \in \mathbb N} \bigcap_{n \ge N}  \{ \omega \in \Omega \, | \, |X_n(\omega) - X(\omega)| \le \varepsilon \} \right) = \liminf_{n \to \infty} \mathbb P(|X_n-X| \le \varepsilon) = 1.
$$
Note that this $\liminf$ was written just for computational simplicity, but conceptually, if the $\liminf$ is $1$, since all the values are probabilities, they are $\le 1$, so the $\limsup$ is also $1$ and the $\liminf$ is a well-defined limit.

(Remark: when intersecting and taking unions of events over ranges like this, it's useful to read intersections as "for every (possible subscript)" and unions as "there exists (a subscript)". So

"for every $\varepsilon > 0$" $\to$ $\bigcap_{\varepsilon > 0}$

"there exists $N$" $\to$ $\bigcup_{N \in \mathbb N}$

etc.

Hope that helps,


问:
Fourier Series: integrating the square of a Fourier series between $[0,\pi]$
Question: Suppose $f(x)= a_0 + \sum_{k=1}^N a_k\cos(kx)$

Show that
$$ \int_0^\pi f^{2}(x) = \pi a_0^2 + \frac{\pi}{2} \sum_{k=1}^N a_k^2 $$

So far I have:

$$ \begin{equation}\int_0^\pi f^{2}(x)=\int_0^\pi a_0^2 \,dx + 2a_0 \int_0^\pi \sum_{k=1}^N a_k\cos(kx) \,dx + \int_0^\pi \left(\sum_{k=1}^N a_k\cos(kx)\right)^2 \,dx \tag{$\ast$}\end{equation}$$

The first term in $(\ast)$ gives the term $\pi a_0^2$ and using term by term integration. For the second term in $(\ast)$, we get a series of $\sin(kx)$ with coefficients $\frac{a_k}{k} $ being evaluated at $0$ and $\pi$ hence the second term goes to $0$.

I am having trouble with caculating the third term which is $\int_0^\pi \left(\sum_{k=1}^N a_k\cos(kx)\right)^2 \,dx$.

Is my expansion of $f^2(x)$ correct and is splitting up the integral the best method?

My biggest question is how to evaluate the third term in $(\ast)$? If anyone could give me some guidance on how to approach this, it would be greatly appreciated.

答:
If $k,j\in\mathbb{N}$ are different integers we have $\int_{0}^{\pi}\cos(kx)\cos(jx)\,dx = 0$, so

$$\begin{eqnarray*} \int_{0}^{\pi}f(x)^2\,dx &=& \int_{0}^{\pi}\left(a_0+\sum_{k\geq 1}a_k \cos(kx)\right)\left(a_0+\sum_{k\geq 1}a_k \cos(kx)\right)\,dx\\&=&\int_{0}^{\pi}a_0^2+\sum_{k\geq 1}a_k^2\cos^2(kx)\,dx\\&=&\pi a_0^2+\frac{\pi}{2}\sum_{k\geq 1}a_k^2. \end{eqnarray*}$$


问:
Is the shortest distance between a point and a curve always normal to the curve?
Suppose we have some point $P$ and some differentiable function $f$. Then let $P_C$ be a point on $f$ such that $|P-P_C|\le |P-f(t)|$ for any $t$. In other words, $P_C$ is one of the points closest to $P$ and lies on $f$. Intuitively, it seems to me that the line normal to $f$ at $P_C$ will always intersect $P$. However, I don't know of a proof for this. Is my intuition correct? Is there a proof of this statement? Is this true for all dimensions? If it's not true for all differentiable functions, is there a subset of functions for which it is true?

答:
Corner cases aside, the distance attains minimum/maximum if and only if the square function also attains minimum/maximum

$$g(t) = d^2(t) = \|P-f(t)\|^2$$

On the other hand,

$$\frac{d}{dt}g(t) = 2(f(t)-P)\cdot f'(t)$$

and the statement follows.


问:
Simultaneous Diagonalization of two bilinear forms
I need to diagonalize this two bilinear forms in the same basis (such that $f=I$ and $g$=diagonal matrix):

$f(x,y,z)=x^2+y^2+z^2+xy-yz $

$g(x,y,z)=y^2-4xy+8xz+4yz$

I know that it is possible because f is positive-definite, but I don't know how can I do it

答:
I wonder if the method the OP was attempting is simply Lagrange's method from multivariable calculus. Because a quadratic form takes its extreme values on the unit circle (or any circle) at eigenvectors, and because we can diagonalize a symmetric matrix using a basis of orthonormal eigenvectors, we look for the stationary points of $g$ with the constraint $f(x,y,z)=1$ (here $f$ will be our scalar product since, as the OP mentioned, it is positive definite and this is enough to guarantee a solution). I'll represent $f$ and $g$ by matrices $A$ and $B$ respectively to match the other answers, so we have
$$A=\begin{bmatrix}1&\frac{1}{2}&0\\\frac{1}{2}&1&-\frac{1}{2}\\0&-\frac{1}{2}&1\\\end{bmatrix} \qquad B=\begin{bmatrix}0&-2&4\\-2&1&2\\4&2&0\\\end{bmatrix}$$
Lagrange's method results in the system of linear equations
$$\begin{bmatrix}-\lambda&-2-\lambda/2&4\\-2-\lambda/2&1-\lambda&2+\lambda/2\\4&2+\lambda/2&-\lambda\\\end{bmatrix}\begin{bmatrix}x\\y\\z\\\end{bmatrix}=\begin{bmatrix}0\\0\\0\\\end{bmatrix}$$
where the $(i,j)$-th entry in the leftmost matrix is $b_{ij}-\lambda a_{ij}$. The system has a nontrivial solution when that matrix's determinant vanishes, which is when $\lambda=6$ or $\lambda=4$ or $\lambda=-4$. Now, we can plug in each of these values in turn to the system to get solutions, which will be eigenvectors forming a basis $\{b_1,b_2,b_3\}$ that makes both quadratic forms diagonal. I get
$$b_1=\begin{bmatrix}-1\\2\\1\\\end{bmatrix},\quad b_2=\begin{bmatrix}1\\0\\1\\\end{bmatrix},\quad b_3=\begin{bmatrix}-1\\0\\1\\\end{bmatrix}$$
The final step is to normalize these vectors so that $A$ is not only diagonal, but the identity matrix. We need to do this because the information that $f(x,y,z)=1$ was lost when taking partial derivatives; we could have started out with $f(x,y,z)=\pi^3$ and got the same system of equations, and furthermore I arbitrarily chose $b_1,b_2,b_3$ to have third coordinate $1$ so there's no reason to think they'd be normalized. Just keep in mind that our scalar product is $f$ and not the typical dot product, so in fact every vector in the above basis has magnitude $\sqrt{2}$. If we put all this together, our change of basis matrix is
$$R=\dfrac{1}{\sqrt{2}}\begin{bmatrix}-1&1&-1\\2&0&0\\1&1&1\\\end{bmatrix}$$
and the matrices $\widetilde{A}$, $\widetilde{B}$ of the quadratic forms in the new basis are
$$\widetilde{A}=R^TAR=I \qquad \widetilde{B}=R^TBR=\begin{bmatrix}6&0&0\\0&4&0\\0&0&-4\\\end{bmatrix}$$
Two remarks: This method is described in Shilov's Linear Algebra, section 10.32. Also, I've been referring to column matrices as vectors, which if we're being pedantic is tacitly assuming that we're starting in the standard basis. However, the method works no matter what basis we start with. If it's not the standard basis, then the columns of the change of basis matrix simply represent the coordinates of the new basis vectors in terms of the starting basis, instead of the vectors themselves.


问:
Row swap changing sign of determinant
I was wondering if someone could help me clarify something regarding the effect of swapping two rows on the sign of the determinant. I know that if $A$ is an $n\times n$ matrix and $B$ is an $n\times n$ matrix obtained from $A$ by swapping two rows, then

$$\det(B)=-\det(A)$$

but I don't know how to prove this.

I have been looking for proofs at internet, and read in both in textbooks and lectures notes that are available that this result is very hard to prove and most approaches rely on induction and so was wondering if there is something wrong with using that $\det(AB)=\det(A)\det(B)$ and then writing $B=EA$ where $E$ is an elementary matrix swapping two rows and using this result to get $\det(B)=\det(E)\det(A)=-\det(A)$ (since showing that $\det(E)=-1$ in this case is not that hard).

答:
It is a chicken and an egg kind of a problem if you think about it that way. All of the following ideas are connected to each other;

1- Swapping any 2 rows of a matrix, flips the sign of its determinant.

2- The determinant of product of 2 matrices is equal to the product of the determinants of the same 2 matrices.

3- The matrix determinant is invariant to elementary row operations.

4- Multiplying an entire row (or column) of a matrix by a constant, scales the determinant up by that constant.

If you assume any subset of these, the rest follow through.

I have used the elementary row operations and multiplying the entire row by a constant to show that the proof is quite straightforward.

Swapping 2 rows inverts the sign of the determinant.

For any square matrix you can generalize the proof of swapping two rows (or columns) being equivalent to swapping the sign of the determinant by using the axiom that the determinant is invariant under elementary row (or column) operations.

Consider a Matrix $\mathrm{A} \in \mathbb{F}^{n\times n} $ as shown below;

$$
\mathrm{A}=
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{bmatrix}
$$

Then the determinant of $\mathrm{A}$ is given by;

$$
\det{\mathrm{A}}
=
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
$$

Consider two separate rows that you want to swap and add one of them to the other. W.L.O.G here assume that the first row is added to the second row. This holds in general for any 2 rows (or columns).

$Row 2 := Row 2 + Row 1$

$$
\det{\mathrm{(A)}}
=
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21}+a_{11} & a_{22}+a_{12} & \cdots & a_{2n}+a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
$$

Next subtract the resulting row 2 from row 1.

$Row1 := Row1 - Row2$

$$
\det{\mathrm{(A)}}
=
\begin{vmatrix}
-a_{21} & -a_{22} & \cdots & -a_{2n}\\
a_{21}+a_{11} & a_{22}+a_{12} & \cdots & a_{2n}+a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
$$

Add the resulting Row 1 to Row 2.

$Row2 := Row2 + Row1$

$$
\det{\mathrm{(A)}}
=
\begin{vmatrix}
-a_{21} & -a_{22} & \cdots & -a_{2n}\\
a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
$$

Take $-1$ common from the first row, this falls out of the determinant.

$$
\det{\mathrm{A}}
=
(-1)
\begin{vmatrix}
a_{21} & a_{22} & \cdots & a_{2n}\\
a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
$$

Finally we have;

$$
\begin{vmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
=
(-1)
\begin{vmatrix}
a_{21} & a_{22} & \cdots & a_{2n}\\
a_{11} & a_{12} & \cdots & a_{1n}\\
\vdots & \vdots & \ddots & \vdots\\
a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{vmatrix}
\quad \blacksquare
$$

The same argument holds for swapping any two rows (or columns).

Title: Row swap changing sign of determinant


问:
Rotate the axis of rotation of a quaternion by another quaternion
I have two quaternions, $q_1$ and $q_2$. I want to rotate the axis of rotation of $q_1$ by $q_2$. Is there any way of doing this directly, without extracting the axis of rotation from $q_1$, rotating it and re-inserting it again?

Please note that I am not trying to perform a combined rotation by quaternion multiplication ($q' = q_2q_1$). Instead I want to rotate the axis of rotation of $q_1$.

答:
If $q_1$ can be written as a real $c$ and a vector $a$ (so that $q_1 = c + a$), then using rschwieb's method, you get

$$q_1' = c + q_2 a q_2^{-1} = q_2 (c + a) q_2^{-1} = q_2 q_1 q_2^{-1}$$

where only normalization of a quaternion has been used to manipulate the expression.  So what you end up doing is back-rotating the whole system by $q_2^{-1}$, rotating by $q_1$ as originally written, and then rotating again by $q_2$.


问:
If I'm given the distances of some point $p$ from $n$ coplanar points, can I tell whether $p$ is also coplanar?
I've been thinking for some time about this problem, however I'm not yet sure how to approach it.

There are a few ways to think about this, however I'll give it exactly as in the original applied problem I'm trying to solve. Consider a network of $n$ coplanar nodes in $2$-dimensions, with known coordinates, and a signal with velocity $v$, with arbitrary coordinates $(x,y)$ and emitted at an unknown time $t$.

I've written an algorithm which is able to reconstruct the coordinates of the signal source $(x,y)$ given only the signal velocity $v$ and the time of arrival at each node, assuming the source is also coplanar with the nodes.

The Problem:

Given the time at which the signal reached each node (or, if you like, all $\frac{n(n-1)}{2}$ time of arrival differences), how can we determine whether it may correspond to a source coplanar with the nodes (a source in $2$-dimensions). Or, if you like, how can we flag input as "invalid" before passing it to the algorithm?

My Thinking:

I know that the minimum possible time of arrival difference can be found by computing the centroid of the $n$ nodes, and taking a signal source at that location, then computing the time of arrival for each node from this point. I also know that the maximum possible time of arrival difference will be for the two furthest nodes (the "diameter" of the set, if you like, or the diameter of the minimum bounding sphere).

Will restricting input to lying between this minimum and maximum be sufficient enough to exclude input that corresponds to sources outside of the plane formed by the nodes?

There will be $\infty$ possible source locations in $2$-dimensions, but (I think) an even larger $\infty$ of locations in $3$-dimensions.

I could also phrase this as "given the distance of some point $p$ from $n$ coplanar points, how can I determine whether $p$ is also coplanar with the $n$ points by examining the distances?

I'm interested not only in solving this as an applied problem, but also understanding the deeper meaning that must exist behind this. How can I look at $n$ times of arrival, and tell whether it definitely corresponds to a source in $3$-dimensions? Or, at least, something close to this.

EDIT: this looks like it may be useful, though I don't quite follow yet (and it seems a bit overkill).

答:
This is a start, and I'm very open to alternative answers & modifications

According to:

If $A$, $B$, $C$, and $D$ are four points in a plane, then the Cayley-Menger determinant vanishes. That is,

$$\det\begin{pmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & |AB|^2 & |AC|^2 & |AD|^2\\
1 & |AB|^2 & 0 & |BC|^2 & |BD|^2\\
1 & |AD|^2 & |BD|^2 & |CD|^2 & 0
\end{pmatrix} =0$$


问:
What kind of ODE is this
What kind of ODE is the following one:

$y^{(n)}(t) = \dfrac{x^{(n)}(t)}{f(n)}$

I have never seen such type. Any references would be very much appreciated. Thanks

答:
If

$$ y^{(n)}(t) = \dfrac{x^{(n)}(t)}{f(n)} $$

then we have

$$ y(t) = \dfrac{x(t)}{f(0)},\quad y^\prime(t) = \dfrac{x^\prime(t)}{f(1)},\quad y^{\prime\prime}(t) = \dfrac{x^{\prime\prime}(t)}{f(2)},\text{ etc.} $$

from which it follows that $f(0)=f(1)=f(2),\cdots$.

So, essentially you have
$$ y(t)=c\cdot x(t) $$

Title: What kind of ODE is this


问:
Proof that each prime power of ${2n \choose n}$ is $\leq \log_p 2n$
I'm trying to work through this proof of the prime number theorem.

Def: Let $P_p(x)$ be the prime power of $p$ in the prime factorization of $x$. I.e. for any natural number $x$, $x = \prod_{p\in \text{primes}} p^{P_p(x)}$.

I am trying to prove the following lemma.

Lemma: $\forall p \in \text{primes}, \forall n\in \mathbb N: P_p\left({2n \choose n}\right) \leq \log_p(2n)$

My attempt at a proof: (tldr, I end up getting $P_p\left({2n \choose n} \right) \leq \log_p(n^2) + \frac{1}{p-1}$, how do I get the $2n$ in the log instead of the $n^2$?)

For any $a, b \in \mathbb N$, we have that $P_p(ab) = P_p(a) + P_p(b)$. Similarly, if $a/b \in \mathbb N$ then $P_p(a/b) = P_p(a) - P_p(b)$. Consider the $p=2$ case:

\begin{align*}
P_2(n!) &= P_2(1) + P_2(2) + P_2(3) + P_2(4) + \dots\\
&= P_2(2) + P_2(4) + \dots\\
&= P_2(2\cdot 1)+ P_2(2\cdot 2) + \dots\\
&= \lfloor n/2 \rfloor \cdot P_2(2) + P_2(1) + P_2(2) + \dots + P_2(\lfloor n/2 \rfloor)\\
&= \lfloor n/2 \rfloor + P_2(\lfloor n/2 \rfloor !)
\end{align*}

Similarly, for general $p$,
\begin{align*}
P_p(n!) &= \lfloor n/p \rfloor + P_p(\lfloor n/p \rfloor!)\\
&= \lfloor n/p \rfloor + \lfloor n/p^2 \rfloor + P_p(\lfloor n/p^2 \rfloor !)\\
&= \sum_{i=1}^{\log_p n} \lfloor n/p^i \rfloor
\end{align*}

Then we can bound it as $P_p(n!) \leq \sum_{i=1}^{\log_p n} n/p^i$ and $P_p(n!) > \sum_{i=1}^{\log_p n}\left(n/p^i-1 \right)$. So then $$\frac{n-1}{p-1} - \log_p n < P_p(n!) \leq \frac{n-1}{p-1}$$

Then we finally get that
\begin{align*}
P_p\left({2n \choose n} \right) &= P_p((2n)!) - 2P_p(n!)\\
&< \frac{2n-1}{p-1} - 2\frac{n-1}{p-1} + 2\log_p n\\
&= 2\log_p n + \frac{1}{p-1}
\end{align*}

How can I fix this so that I prove the desired lemma?

答:
Thanks to @Greg Martin for his answer. For completeness, I'll include the result here. As I have above, $P_p\left(n!\right) = \sum_{i=1}^{\lfloor \log_p n\rfloor } \lfloor n/p^i\rfloor$. So

\begin{align*}
P_p\left({2n \choose n}\right) &= P_p((2n)!) - 2P_p(n!)\\
&= \sum_{i=1}^{\lfloor\log_p2n\rfloor}\lfloor 2n/p^i\rfloor - 2\sum_{i=1}^{\lfloor\log_p n\rfloor}\lfloor n/p^i\rfloor\\
&= \sum_{i=1}^{\lfloor\log_p2n\rfloor} \left(\lfloor 2n/p^i\rfloor - 2\lfloor n/p^i\rfloor \right)\\
&\leq \sum_{i=1}^{\lfloor\log_p2n\rfloor}1\\
&\leq \log_p2n
\end{align*}


答:
As this is probably coming from a set of noisy data, a least-squares algorithm might work better. I would go with $p=\arg\! \min_{\tilde{p}\in\mathbb{R}^3} \sum_{i\neq j} (|\tilde{p}-p_i|-|\tilde{p}-p_j|-(t_i-t_j))^2$,  and then check if $p$ is in the plane containing $p_i$'s, assuming that you have a large enough number of known points.

If there are few points, then for each pair of points, you actually know that the source point cannot be far from the hyperboloid determined by the difference in time of arrival. The intersections of these hyperboloids should have a point close to the plane.

Title: If I'm given the distances of some point $p$ from $n$ coplanar points, can I tell whether $p$ is also coplanar?

Body:
I've been thinking for some time about this problem, however I'm not yet sure how to approach it.

There are a few ways to think about this, however I'll give it exactly as in the original applied problem I'm trying to solve. Consider a network of $n$ coplanar nodes in $2$-dimensions, with known coordinates, and a signal with velocity $v$, with arbitrary coordinates $(x,y)$ and emitted at an unknown time $t$.

I've written an algorithm which is able to reconstruct the coordinates of the signal source $(x,y)$ given only the signal velocity $v$ and the time of arrival at each node, assuming the source is also coplanar with the nodes.

The Problem:

Given the time at which the signal reached each node (or, if you like, all $\frac{n(n-1)}{2}$ time of arrival differences), how can we determine whether it may correspond to a source coplanar with the nodes (a source in $2$-dimensions). Or, if you like, how can we flag input as "invalid" before passing it to the algorithm?

My Thinking:

I know that the minimum possible time of arrival difference can be found by computing the centroid of the $n$ nodes, and taking a signal source at that location, then computing the time of arrival for each node from this point. I also know that the maximum possible time of arrival difference will be for the two furthest nodes (the "diameter" of the set, if you like, or the diameter of the minimum bounding sphere).

Will restricting input to lying between this minimum and maximum be sufficient enough to exclude input that corresponds to sources outside of the plane formed by the nodes?

There will be $\infty$ possible source locations in $2$-dimensions, but (I think) an even larger $\infty$ of locations in $3$-dimensions.

I could also phrase this as "given the distance of some point $p$ from $n$ coplanar points, how can I determine whether $p$ is also coplanar with the $n$ points by examining the distances?

I'm interested not only in solving this as an applied problem, but also understanding the deeper meaning that must exist behind this. How can I look at $n$ times of arrival, and tell whether it definitely corresponds to a source in $3$-dimensions? Or, at least, something close to this.

EDIT: this looks like it may be useful, though I don't quite follow yet (and it seems a bit overkill).


问:
two simple laplace transformation with simple functions

this is a simple laplace transformation can someone please help me solve this I am stuck

答:
Let's remind the definition of the Laplace transform:
$$
\mathcal L(f)(s) = \int_0^{\infty} f(t) e^{-st} \, dt
$$
So for example, we have with $f(t) = e^{-t} \cos(2t)$ that
$$
\mathcal L(f)(s) = \int_0^{\infty} \cos(2t) e^{-(s+1)t} \, dt = \mathcal L(g)(s+1))
$$
with $g(t) = \cos(2t)$. You can either compute it yourself or lookup a table, but
$$
\mathcal L(t \mapsto \cos(at))(s) = \frac s{s^2+a^2}
$$
so $\mathcal L(g)(s) = \frac{s}{s^2+4}$. It follows that
$$
\mathcal L(f)(s) = \mathcal L(g)(s+1) = \frac{s+1}{(s+1)^2 + 4} = \frac {s+1}{s^2 + 2s + 5}.
$$
Similarly, you expand $(4e^{2t} - 2)^3$ as a polynomial in $e^t$ and rely on tables to compute the Laplace transform of the individual terms, and then rely on the linearity of the Laplace transform.

Hope that helps,


问:
Quotient group of finitely generated group is finitely generated
I am studying elementary level of algebra, and I'm trying to prove that a quotient group of finitely generated group is finitely generated. It is intuitively true, but I can't get an idea to prove it.</p>

答:
Suppose that a group $G$ is finitely generated. Let $g_1,\cdots,g_n \in G$ be generators, and let $H \trianglelefteq G$ be a normal subgroup. Can you show that $g_1H,\cdots,g_nH$ generate $G/H$? Hint: write an arbitrary element of $G/H$ as $gH$ for some $g \in G$, and then use the fact that the $g_1,\cdots,g_n$ generate $G$.

Another more involved way to prove it which is categorically more intuitive (but less intuitive at the beginning) is that saying that a subset $S \subseteq G$ generates $G$ means that one can form the free product $K \overset{\mathrm{def}}{=} \mathbb Z * \cdots * \mathbb Z$ ($|S|$ copies of $\mathbb Z$) and the map $K \to G$ sends the word $\omega_i$ to $g_i$ where $S = \{g_1,\cdots,g_n\}$ is the generating set. In this setting, saying that $S$ generates $G$ is the same as saying that the induced map $K \to G$ defined by $S$ is surjective. But then clearly if $K \to G$ is surjective, so is the composition $K \to G \to G/H$.

You sort of already know this argument if you know real vector spaces and quotient spaces of them: if $S = \{v_1,\cdots,v_n\}$ is a subset of a vector space $V$, you can form the map $\mathbb R^n \to V$ where the unit basis vector $e_i$ of $\mathbb R^n$ is mapped to $v_i$.  A finite subset $S$ of $V$ generates $V$ if and only if the induced map $\mathbb R^n \to V$ is surjective. It is then clear that if $S$ generates $V$, then $S$ generates $V/W$ for any subspace $W$, because the composition $\mathbb R^n \to V \to V/W$ remains surjective. If you replace $\mathbb R$ by $\mathbb Z$, you obtain the equivalent argument for $\mathbb Z$-modules, also known as abelian groups (and $\mathbb R^n$ becomes $\mathbb Z^n$ in the argument). But for general groups, the free object is not as pretty as $\mathbb R^n$ or $\mathbb Z^n$, that's why this free product appears to get every possible product of elements of the group.

Hope that helps,


问:
Defect version of Fatou's Lemma (repost)
Someone recently posted the following question, but it was deleted.  Since it was an interesting question and I didn't know the answer, I'm reposting it here.

The question: is the following limit theorem true?

Suppose $(X,\mathcal{A},\mu)$ is a measure space, $f \in L^{1}(X)$, and $\{f_{n}\}_{n \in \mathbb{N}} \subseteq L^{1}(X)$ with $f_{n} \geq 0$ a.e. for all $n \in \mathbb{N}$ and $f_{n} \to f$ pointwise a.e.  If $\{f_{n}\}_{n \in \mathbb{N}}$ is bounded in $L^{1}(X)$, then $f_{n} \to f$ in $L^{1}(X)$ if and only if $\int_{X} f_{n} \, d \mu \to \int_{X} f \, d \mu$.

The answer turns out to be yes.  This isn't completely obvious, and it's interesting since I'm aware of at least one very closely related result in the literature.  (Namely, if $\{\mu_{n}\}_{n \in \mathbb{N}}$ are finite signed Borel measures on a compact metric space $X$ and $\mu_{n} \overset{*}{\rightharpoonup} \mu$, then $\|\mu_{n} - \mu\| \overset{*}{\rightharpoonup} 0$ if and only if $\|\mu_{n}\|(X) \to \|\mu\|(X)$.)

答:
The result you are asking for is just a special case of a more general result, known as Scheffe's lemma:

Suppose $(X,\mathcal{A},\mu)$ is a measure space. Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in L^1 (X)$ and $f_n \to f$ pointwise a.e. in $X$. Then $f_{n} \to f$ in $L^{1}(X)$ iff $\lim_{n \to \infty} \int_X |f_n| d\mu = \int_X |f| d\mu$.

It can be proved in a rather simple way by using Fatou's lemma. Let us prove it:

Proof: (=>) It is trivial, since, from Minkowski's inequality, we have

$$\left | \int_X|f_n|d\mu-\int_X|f|d\mu \right |\leqslant \int_X|\,|f_n| -|f|\,| d\mu\leqslant \int_X|f_n -f| d\mu$$

So, if $f_{n} \to f$ in $L^{1}(X)$ then $\lim_{n \to \infty} \int_X |f_n| d\mu = \int_X |f| d\mu$.

(<=) Note that $|f_n -f|\leqslant |f_n| +|f|$. So, for each $n$, the function $|f_n| +|f| - |f_n -f|$ is non-negative and using Fatou's Lemma, we have
\begin{align}
2 \int_X|f|d\mu
&=\int_X \lim\inf(|f_n| +|f| - |f_n -f|)d\mu \leqslant \lim\inf \int_X (|f_n| +|f| - |f_n -f|)d\mu = \\
&=\lim\inf \left (\int_X|f_n|d\mu +\int_X|f|d\mu - \int_X|f_n -f|d\mu \right) = \\
&= \left(\lim\inf\int_X|f_n|d\mu\right) +\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right)  = \\
&=2\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right)
\end{align}
So we have
$$2 \int_X|f|d\mu \leqslant 2\int_X|f|d\mu - \left(\lim\sup\int_X|f_n -f|d\mu\right) $$
Since $f \in L^1 (X)$ , we know that $\int_X|f|d\mu<+\infty$, and so we get
$$\lim\sup\int_X|f_n -f|d\mu \leqslant  0$$
So we can conclude that
$$\lim\int_X|f_n -f|d\mu =  0$$
So, $f_{n} \to f$ in $L^{1}(X)$.

Remark: Note that the hypothesis that $\{f_{n}\}_{n \in \mathbb{N}}$ is bounded in $L^{1}(X)$ is not needed. In fact, it is a consequence of the other hypothesis of the result.


问:
Suppose $|A|=n, f:A\to A$ is injective $\implies \exists k\in [n]:f^k(x)=x$

My Attempt

Case 1

Suppose $f:A \to A$ is an identity function. then $k=1\in [n]:f(x)=x$.

Case 2

Suppose $f:A \to A$ is an injective function and non identity function.

so, there exists $x\in A$ such that $f(x) \neq x.$
So, there are $n-1$ possibilities for $f(x).$
if $f(f(x))=x, $ then $k=2.$
so on. I am not able to complete the proof. Is my method correct?

答: Since A is finite and f is injective, f is a permutation (ker (f) = 0). Every permutation in $S_n$ can be writen as a product of disjoint cycles, that is unique up to the ordering of it's cycles and up to a cyclic permutation of it's elements within each cycle. There is a cycle in each orbit of every $a \in A$, and that cycle has at most n elements thus $\implies \forall a \in A,\exists k\in [n]:f^k(a)=a$.


问:
Graph theory, proving the existence of a vertex satisfying some conditions
I am new to graph theory. I am reading this. I am having trouble understanding the following proof in the handout. Here's the statement and the proof which is from the link above.

Let $G$ be a tree and $v$ be any vertex of $G$. Let $v_1,v_2,...,v_t$ be the vertices adjacent to $v$. Let $e_i$ be the edge joining $v$ and $v_i$. Let $T_i$ be the subtree containing $v_i$ after removing edge $e_i$. (Draw a diagram for this. It will help.) Let $f(v)=max^{t}_{i=1}|V(T_i)|$.

Let's use a little intution here. Since $\sum |V(T_i)|=n-1$, if $f(v)$ is "large", then the tree looks unbalanced. If $f(v) \approx (n-1)/t$, then the tree looks balanced. We want to find vertex $v$ that minimizes $f(v)$, the vertex that makes the tree the most balanced.

Let $G$ be a tree with $n$ vertices and $\Delta > 1$ be the maximum degree
amongst all vertices in $G$. Using the same function $f$ as defined before, prove that there exists
a vertex $v$ such that

$$\frac{1}{\Delta}(n-1)\leq f(v)\leq \frac{\Delta-1}{\Delta} (n-1)$$

Left inequality follows for all $v$ from pigeonhole principle. To prove the right
inequality, choose $v$ such that $f(v)$ is minimum. Suppose $f(v) \ge (\Delta−1)/(\Delta)\cdot(n−1)+ 1$. Let $v_i$
be the neighbour of $v$ with $|T_i
| \ge (\Delta-1)/\Delta(n−1)+1$. Let $v = w_1, w_2, · · · , w_\Delta$ be the neighbours
of $v_i$. Then since the tree containing $v$ after removing $v_iv$ contains at most $1/\Delta(n - 1)$ vertices,
then $f(v_i) \leq (\Delta − 1)/\Delta(n - 1) - 1 < f(v)$, contradicting the minimality of $f(v)$. (Draw a
diagram to understand this proof better.)

I understood the proof of the left inequality. I don't really understand the proof of the right inequality. Can anyone elaborate? Thanks!

答:
The bound you're quoting is a bad bound. We can prove a better result: there is a vertex $v$ such that $f(v) \le \frac12 n$. This is easy to see once we draw the right diagram:

Suppose that $v$ has neighbors $v_1, v_2, \dots, v_k$. The subtrees $T_1, T_2, \dots, T_k$, which the text defines as

Let $T_i$ be the subtree containing $v_i$ after removing $e_i$, the edge joining $v$ and $v_i$

are simply the connected components of $T - v$. These are shown in red in the diagram above.

Claim. If $f(v) > \frac12 n$, then there is another vertex $v^*$ with $f(v^*) < f(v)$.

Proof. Suppose that $T_i$ is the largest of $T_1, T_2, \dots, T_k$, so that $f(v) = |V(T_i)|$. Then let $v^* = v_i$. Call $v^*$'s neighbors $v_1^*, v_2^*, \dots, v_m^*$, where $v_1^* = v$. Deleting $v^*$ leaves components $T_1^*, T_2^*, \dots, T_m^*$, which are shown in blue in the diagram above.

Since $T_2^*, \dots, T_m^*$ are all pieces of $T_i$, leaving out $v_i$ itself, we have $|V(T_j^*)| \le |V(T_i)|-1 < f(v)$ for $j=2, \dots, m$. Only $T_1^*$ can possibly be bigger than $T_i$.

What's more, $T_1^*$ and $T_i$ are the two connected components we get after deleting edge $v v_i$, so $|V(T_1^*)| = n - |V(T_i)| = n - f(v)$. If $f(v) > \frac12n$, then we conclude $|V(T_1^*)| < f(v)$ - and we've already proven that $|V(T_2^*)|, \dots, |V(T_m^*)| < f(v)$. Therefore $f(v^*) < f(v)$.

From the claim, it follows that if we pick the vertex $v$ that minimizes $f(v)$, we must have $f(v) \le \frac12n$.

It might seem like the bound $f(v) \le \frac{\Delta-1}{\Delta}(n-1)$ is better when $\Delta=2$, in which case it gives $\frac12(n-1) < \frac12n$. But in this case, then $\frac12(n-1)$ bound is false. Consider a path $P_n$, when $n$ is even. The best thing we can do is pick one of the two middle vertices, which will give $|V(T_1)| = \frac n2-1$ and $|V(T_2)| = \frac n2$. We'll get $f(v) = \frac12n$, not $\frac12(n-1)$.

(The flaw in the proof that gives $\frac{\Delta-1}{\Delta}(n-1)$ is the assumption that if $f(v) > \frac{\Delta-1}{\Delta}(n-1)$, then we must have $f(v) \ge \frac{\Delta-1}{\Delta}(n-1) + 1$. This is false when $\frac{\Delta-1}{\Delta}(n-1)$ is not an integer.)


问:
Help solving a recurrence relation
I'm quite stuck on how to solve recurrence/difference equations.

For example I have the following linear inhomogeneous equation:
$a_n = 2a_{n-1} + 2^n n$, and $a_0 = 1/2$

I know that $2a_{n-1} = 1/2(2^n)$, and then we have $a_n = 1/2(2^n) + n(2^n)$. This is where I'm stuck though and the inhomogeneous part is throwing me off.

Am I in the right direction thinking that: $a_n = \frac12 \prod 2^n n$ ?

No answers please but some nudges in the right direction would be greatly appreciated. Thanks!

答:
Hint:

Set $b_n = \frac{a_n}{2^n}$. Then dividing your recurrence by $2^n$ gives

$$b_n = b_{n-1} + n.$$

Do you see how to find a closed form for $b_n$? Do you see how to turn this into a closed form for $a_n$?

As a general "life pro tip", when you see a recurrence with a geometric term (i.e. something that looks like $r^n$) that's otherwise linear, you should try dividing by it. Since $r^{n+1} = r \cdot r^n$, this procedure turns your geometric term into a constant one, which is easier to handle.

I hope this helps ^_^

Title: Help solving a recurrence relation


问:
Find $A$ such that $f(0) = A$ and there exists $f'(0)$
Let
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
A & :x=0\\
\end{cases}$$
I'm supposed to find $A$ such that there exists $f'(0)$. My attempt:

First, I found $A$ such that $f$ is continuous at $x=0$, using de l'Hospital's rule to calculate $\lim_{x \to 0^{+}}f(x)$ and $\lim_{x \to 0^{-}}f(x)$.
$$f(x)=
\begin{cases}
(e^{3x}-3e^x+2)/x^2 & :x \neq 0\\
3 & :x=0\\
\end{cases}$$
Then I calculated one-sided limits of the difference quotient at $x=0$ using de l'Hospital's rule (but I'm not sure if I'm allowed to do so - I've read that applying this rule to the definition of derivative may lead to some kind of circular reasoning).
$$\lim_{h \to 0^{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}=f'(0)=4$$
Is this solution correct? Wolfram Alpha and Desmos tell me that $f'(0)=0$ for $A=3$. However, when I set $A$ to a different value (meaning the function isn't continuous at $x=0$ and hence not differentiable) I still get $f'(0)=0$, which is quite confusing.

Body:
You have calculated $A$ correctly now for finding $f^{\prime}(0)$
$$\begin{align} f^{\prime}(0) &= \lim_{x \to 0}\dfrac{\dfrac{e^{3x}-3e^x+2}{x^2} - 3}{x}\\
&=\lim_{x \to 0}\dfrac{e^{3x}-3e^x+2-3x^2}{x^3}\\
&=\lim_{x \to 0}\dfrac{e^{3x}-e^x-2x}{x^2}\\
&= \lim_{x \to 0}\dfrac{3e^{3x}-e^x-2}{2x} \\
&= \frac{9}{2}\lim_{x \to 0} \dfrac{e^{3x}-1}{3x} - \frac{1}{2}\lim_{x \to 0} \dfrac{e^x -1}{x} \\
&= 4 \end{align}$$


问:
Research question - prove inequality for $n \log n$.
The following inequality has come up in some research and appears true when plotted, but I'm not sure how to to prove it. If $g(z) = z \log z$, then for $x>3$ ($x$ real),
$$
g(x) < g(x-1)\cdot \left(1 + \frac{g(x-1) - g(x-2)}{g(x-1)-1} \right).
$$

Body:
Alternative solution:

Fact 1: $\frac{2u}{2+u} \le \ln (1 + u) \le \frac{2u + u^2}{2 + 2u}$ for all $u \ge 0$.

(The proof is easy. Hint: Take derivative.)

Now, first, by Fact 1, we have
$$g(x) - x\ln (x-1) = x \ln(1 + \tfrac{1}{x-1})
\le x \cdot \frac{2x-1}{2x(x-1)} = \frac{2x-1}{2(x-1)}$$
which results in
$$g(x) \le x\ln(x-1) + \frac{2x-1}{2(x-1)}.$$

Second, by Fact 1, we have
$$(x-2)\ln(x-1) - g(x-2) = (x-2)\ln(1 + \tfrac{1}{x-2})
\ge (x-2) \cdot \frac{2}{2x-3} = \frac{2(x-2)}{2x-3}$$
which results in
$$g(x - 2) \le (x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}.$$

Third, it suffices to prove that (note: $g(x-1) - 1 > 0$)
$$x\ln(x-1) + \frac{2x-1}{2(x-1)}
< g(x-1) \cdot \left(1 + \frac{g(x-1) - [(x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}]}{g(x-1)-1} \right)$$
that is
$$\frac{1}{2(2x-3)[(x-1)\ln(x-1)-1]}\left(4(x-1) - \frac{1}{x-1} - \ln(x-1)\right) > 0$$
which is true since $x-1 \ge \ln(x-1)$ and $x - 1 \ge \frac{1}{x-1}$.

We are done.


问:
Estimation from noisy observation
Assume that you have access to a noisy observation (static vector) y, which can be expressed as $y=x+b$ where the static vector $x$ is unknown. The noise $b$ is an independent and identically distributed (i.i.d.) zero-mean noise distributed according to a Gaussian distribution with covariance $C$. The mean of the noise b (0) and its covariance $C$ are known, as well as the noisy measurement $y$. What is the best technique to estimate $x$ from $y$ please?

答: Since the noise term is iid Gaussian with zero mean and equal variances, the most likely value for each $x_i$ is just $y_i$. If you have some reason to believe the $x_i$ have some prior distribution then you can apply Bayesian inference and then choose the posterior mean vector as a point estimate.

Title: Estimation from noisy observation


问:
If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.
Let $(G,*)$ be a group. If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.

$a^2*b=b*a \implies ab=a^{-1}*b*a$.
I can't proceed further.

答:
$$a^2b=ba \tag{*}$$Left-multiply both sides of $(*)$ by $b^{-1}$
$$ a = b^{-1}a^2 b$$
Square both side to get $$a^2 = b^{-1}a^4b = e$$
Plug this into $(*)$ to get $$\begin{align} ba &= a^2b \\ &= eb \\ &=b \end{align}$$
Hence, we have $ba = b$ which gives $a=e$.


答:
You just have to combine your two equations!

If $a^4 = e$ and $a^2b = ba$, then
$$
b = a^4b = a^2(a^2b) = a^2(ba) = (a^2b)a = (ba)a = ba^2
$$
and therefore $a^2 = e$. But if $a^2 = e$, then $b = a^2b = ba$, so $a=e$.

Hope that helps,

Body:
Let $(G,*)$ be a group. If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.

$a^2*b=b*a \implies ab=a^{-1}*b*a$.
I can't proceed further.


问:
Is using the letters of English in math a universal practice?
In the Anglosphere (and in Russia, my home country) some of the symbols used in mathematics are the 26 letters of the English alphabet (52 actually, because there are two cases), and not any other alphabet based on the Latin script. I am aware that Greek and Hebrew, and even the Russian Ш (I call it Russian and not Cyrillic as it was introduced in honor of a Russian mathematician) letters are used in math as well.

My question is: Is this practice universal for all countries, even those that officially use an alphabet based on the Latin script (with letters not present in the English alphabet)? In other words, is there a country in which at least one Latin letter is used routinely in higher (this is important, I'm not asking about high school math) math that would not normally be considered an English letter?

I believe the answer is most likely "yes" (to the first formulation of the question) because the vast majority of mathematical papers and the leading mathematical literature are in English. Plus, English is the global lingua franca. However, I am not completely sure.

N.B. The second time derivative of $o$, i.e. $\ddot o$, is not the same as the German letter 'ö'. I know that this example is from physics, the example is just to get the point across.
Also: to those that believe French uses the English alphabet, the Wikipedia article about the English alphabet disagrees (on the right it says "Languages: English" with no French or any other language in sight).

答:
The answer is NO.

It is not quite correct to say that letters of English alphabet are used. Math uses the letters of Latin. The Latin is not changing since many hundreds of years.

English alphabet is based on Latin. English had letters that are partially different from Latin, see old English Latin alphabet. Later on English alphabet has changed and is currently the same as Latin. Thus, despite they look now the same, it is more correct to say that Latin letters are used, no English.

Besides Latin letters, also Greek letters are widely used in mathematics: In USA, UK, France, Germany, Russia, as well as many other countries.

Greek: Here is an example of internationally understandable notation of limit that uses Greek letters "sigma" and "delta".

German: For instance, for "power set" an old-German letter is used which is similar to Latin P. Here are example 1 and example 2.

Russian: The two widely used **Russian letters are "Ш" (sha-function, from Шафаревич), see example,  and "Ж" (one of notations for Lagrange related expressions), see example.

Here is a good article about the history of mathematical notation.


问:
Show that the only continuous solution of $g(s+t) = g(s)g(t)$ with $s,t \ge 0$ is $g(x) = e^{-\lambda x}$
If for any $s,t \ge 0$:

\begin{align*}
  g(s+t) &= g(s) g(t) \\
\end{align*}

Show that the only right-continuous solution of this is:

\begin{align*}
  g(x) &= e^{-\lambda x} \\
\end{align*}

My textbook claims this. Clearly the solution is continuous and satisfies the given condition. I'd like to see a basic sketch of a proof and I don't know how to approach this.

答:
If $n$ is an integer then
$$
g(ns)=g(s)^n.
$$
Let $c=g(1).$ Then if $a/b$ is any rational number,
$$
g(a/b)^b=g(a)=g(1)^a=c^a
$$
so $c\geq 0$ and $g(a/b)=c^{a/b}.$
Then you can use right-continuity to show that $g(x)=c^x$ holds for all other real numbers. Then if $c>0$, define $\lambda=-\ln c$.


答: As noted in a comment constant $0$ is a solution. Suppose $g$ is non-zero. Then, one can show that $g$ is positive everywhere. Once you take the $\log $ of both sides, you obtain the Cauchy functional equation. See here.

Title: Show that the only continuous solution of $g(s+t) = g(s)g(t)$ with $s,t \ge 0$ is $g(x) = e^{-\lambda x}$

Body:
If for any $s,t \ge 0$:

\begin{align*}
  g(s+t) &= g(s) g(t) \\
\end{align*}

Show that the only right-continuous solution of this is:

\begin{align*}
  g(x) &= e^{-\lambda x} \\
\end{align*}

My textbook claims this. Clearly the solution is continuous and satisfies the given condition. I'd like to see a basic sketch of a proof and I don't know how to approach this.


问:
Is every involution a palindrome of transpositions?
An involution is a permutation $P$ which is its own inverse: $P\cdot P = \text{id}$.

Every permutation can be written (in various ways) as a sequence of single-element swaps (transpositions). Sometimes these sequences are palindromes, in that the reversal of the sequence is the same sequence.

If $T$ is a sequence of transpositions for $P$, then the reversal of $T$ is a sequence of transpositions for $P^{-1}$. It follows that if the sequence $T$ is a palindrome, then $P$ is an involution.

I'm wondering whether the converse is true:

Conjecture: Any involution $P$ can be decomposed into a sequence of transpositions which is a palindrome.

— and if it's false, if there is any alternative way to characterize the involutions which can be decomposed in such a way.

答:
It's not possible in general to express an involution as a palindrome of transpositions. In fact, because of cancellation properties, the only involutions that can be expressed as palindromes are (1) a single transposition, and (2) the identity.

Indeed, any even-length palindrome of transpositions cancels itself out:
$$\pi_1\pi_2\ldots \pi_m \pi_m\ldots \pi_2\pi_1$$
Each consecutive pair of transpositions in the middle will cancel out successively until they're all gone.

On the other hand, an odd length palindrome has the form $$\pi_1\pi_2\ldots \pi_m \pi_{m+1}\pi_m\ldots \pi_2\pi_1$$. Grouping the other transpositions together, we see that it has the form $Q\cdot \pi_{m+1}\cdot Q^{-1}$ for some permutation $Q$. Hence the palindrome is conjugate to the transposition $\pi_{m+1}$ by similarity; it is therefore itself a single transposition.


问:
Generalization of Doob Dynkin for Stochastic processes
Let $\{X_t\}_{t\geq 0}$ be continuous time stochastic process and $\{\mathcal{F}_t^X\}_{t \geq 0}$ be the filtration generated by it. If the process $Y$ is $\{\mathcal{F}_t^X\}_{t \geq 0}$ adapted, is there any simple representation of the process $Y$ in terms of $X$. This will be a generalization of Doob Dynkin Lemma</p>

答: Apart from the trivial consequence of Doob-Dynkin's lemma that for each $t$ there exists some measurable function $u_t$ such that $Y_t=u_t(X_s;s\leqslant t)$... Since every $(Y_t)$ thus defined satisfies the hypothesis, I doubt one can go much further.</p>


问:
Finding a periodic bijection on the set of bitstrings
Let $S$ be the set of bitstrings of length $n$. For $x\in S$, let $W(x)$ be the Hamming weight of $x$ i.e. number of ones in $x$.

I am trying to find a bijection $f:S\to S$ which maps bitstrings in the following periodic manner. Bitstrings of low weight are mapped to bitstrings of medium weight. Then bitstrings of medium weight are mapped to bitstrings of high weight and finally the high weight ones are mapped back to low weight ones.

My attempt: Consider $x\in S$. If we take the $XOR$ of the $n-$bits of $x$, append the result to $x$ and pop the first bit of $x$ to get a bitstring say $y$. Then let $f(x)=y$ defines a bijection. But this bijection doesn't satisfy the above requirements because $W(f(x))=W(x)+1$ if $W(x)$ is odd and the first bit of $x$ is $0$ and $W(f(x))=W(x)-1$ if $W(x)$ is even and the first bit of $x$ is $1$.

So with this $f$, we move up/down a weight class depending on whether the current weight is odd/even and it's clearly not periodic. So instead of moving up/down, I would like a bijection that only goes up weight classes initially, then we reach medium weight at which point we need to go down weight classes as described above.

I've gotten very helpful answers on other types of bijections on $S$ here before. This is a part of a larger problem I'm working on: Random walk on the set of bitstrings and this $f$ helps me shuffle the bitstrings so that the random walk converges to its stationary distribution faster.

EDIT: Just a note on the comment below by Ross. If we map weight $1$ strings to weight $2$ strings and so on and keep the remaining strings mapped to themselves, then a whole lot of strings do not change weight because they are mapped to themselves. I am trying to avoid that. The strings don't have to change weight only by $1$. Also, a lot of strings need to change weight as opposed to the previous example.

答: You can start with an order that has the numbers sorted by number of bits, then lexicographically.  For four bits, that would give the order $0,1,2,4,8,3,5,6,9,10,12,7,11,13,14,15$.  Then you can write a function that steps you once along the order.  If the lowest bit is $1$ it looks at the string of $1$ bits from the bottom and shifts the top one into the place above.  If the lowest bit is $0$ it looks at the lowest $1$.  If it has a $0$ above it it shifts that $1$ by one place.  If it has a $1$ above it, it shifts the top of the string of $1$s up one place and all the lower $1$s into the lowest places.  If it carries out the top it just fills one more bit from the bottom than the previous word had.  You need special rules for $0 \to 1$ and all $1$s $\to 0$.  Now choose a number of steps to go each time.  For the four bit one, three steps is reasonable, giving an order of $0,4,5,10,11,15,2,3,9,12,13,1,8,6,12,13$  Maybe you would prefer five steps, giving an order of $0,3,12,15,8,10,14,4,9,13,2,6,11,1,5,7$.  Now only $3$ has the same weight as its successor.


问:
A question about a property of spectral density $f(\lambda)$, i.e. $f(\lambda) \geq 0$.
I'm reading Time Series analysis from the book Introduction to time series and forecasting by Peter J. Brockwell and Richard A. Davis. I'm going through the chapter of Spectral Analysis.

Where spectral density defined as
$$
f(\lambda) = \frac{1}{2\pi} \sum_{|h| < \infty} \exp{(-ih\lambda)} \gamma(h)
$$
Here $\{\gamma(h)\}_{|h|<\infty}$ is is auto covariance function of time series $\{X_t\}_{|t|<\infty}$.

I'm basically stuck in the proof of $f(\lambda)\geq 0 $. there is a step
$$
\frac{1}{2\pi N}\sum_{|r|<N}\sum_{|s|<N} \exp{(i\lambda(r-s))}\gamma(r-s)
$$
$$
= \frac{1}{2\pi}\sum_{|h|<N}\bigg(1 - \frac{|h|}{N}\bigg) \exp{(i\lambda h)}\gamma(h)
$$

How to derive the second line from the first line?

答:
I think this is just a counting argument, but somehow I don't get the result you get in your question, so I'll leave my argument here so maybe it is useful to you or someone can correct it (or you can find a flaw in your book). The summand in the double summation only depends on $r-s$, and when $|r|,|s| < N$, the difference $r-s$ also satisfies $|r-s| < N$. So set $h \overset{def}{=} r-s$ and count the number of pairs $(r,s)$ satisfying $|r| < N$, $|s| < N$ and $r-s = h$.

One easily found such pair is $r=h$ and $s=0$. Since we want $|r-s| = h$ and $|r|$, $|s| < N$, the pairs look like this for say $h > 0$:
$$
(-N+1+h, -N+1)), \cdots, (0, -h), \cdots, (h, 0), \cdots, (N-1, N-1-h)
$$
Of course, the case $h < 0$ will have the same count, so the answer depends only on $|h|$. You see that the answers $(0,-h), (1,-h+1), \cdots, (h-1,1),(h,0)$ account for $|h|+1$ answers, and for $r < 0$ you have $N-(|h|+1)$ answers, and similarly for the answers with $s > 0$. This gives you a total of $2(N-(|h|+1)) + |h|+1 = 2N-(|h|+1)$ such pairs. So you can re-write your sum as
$$
\frac 1{2\pi N} \sum_{|r|,|s| < N} \mathrm{exp}(i\lambda(r-s)) \gamma(r-s)
$$
by putting together the terms involving $(r,s)$ with $|r|,|s| < N$ and $r-s = h$ and just put your count in front of the term, which gives
$$
\frac 1{2\pi N} \sum_{|h| < N} (2N-(h+1)) \mathrm{exp}(i\lambda h) \gamma(h)
$$
By bringing the factor $\frac 1N$ in the summation, you get
$$
\frac 1{2\pi} \sum_{|h| < N} \left( 2-\frac{|h|+1}N \right) \mathrm{exp}(i\lambda h) \gamma(h)
$$
One possible reason that I see for the discrepancy in our counts is if you're really summing over $0 \le r,s < N$ instead of $|r|,|s| < N$, because then the count becomes $N-|h|$ instead of $2N-(|h|+1)$, and you get the correct factor. That may explain your mistake (or the author's).

Hope that helps,


问:
Calculate $\int_0^{1/2}\, x\sin(\pi x)\Gamma(\frac{d-1}{2} + x)\Gamma(\frac{d-1}{2} - x) \mathrm dx$ for even $d\ge 4$
I would like to evaluate the integral of form:
\begin{align}
I(d) \equiv \int_0^\frac{1}{2}\,d x\, x\,\sin(\pi\,x)\,\Gamma\left(\frac{d-1}{2} + x\right)\,\Gamma\left(\frac{d-1}{2} - x\right) \ ,
\end{align}
for even $d\ge 4$. ($I(2)$ diverges due to the pole of the gamma function.)

After some explicit calculation, I speculate it is given by a sum of the Riemann zeta functions:
\begin{align}
I(d) = \sum_{n=1}^{\frac{d}{2}-1}\,c_{n,d}\,\zeta(2n+1) \ .
\end{align}

Is it possible to determine the coefficients $c_{n,d}$?

答:
Interestingly the structure of $I(2M)$ is similar to this question that OP asked before. Indeed, using functional equation of $\Gamma$ one have
$$\small I(2M)=\int_0^{\frac{1}{2}} x \sin (\pi  x) \Gamma \left(M+x-\frac{1}{2}\right) \Gamma \left(M-x-\frac{1}{2}\right) \, dx=\int_0^{\frac{1}{2}} \pi  x \tan (\pi  x) \prod _{k=0}^{M-2} \left(\left(k+\frac{1}{2}\right)^2-x^2\right) \, dx$$
Where the primitive of RHS is evaluable through polylogarithms. The following command calculates $I(2M)$ for arbitrary $M$:

```<code>Table[Integrate[Pi x Tan[Pi x] Product[((k + 1/2)^2 - x^2), {k, 0, M - 2}], {x, 0,
1/2}], {M, 2, *Large number*}] // Expand
</code>```

For a explicit calculation, one have (try it)

$$\small I(2 M)=\pi  \sum _{j=0}^{M-2} \left(\sum_{1\leq a_1<\cdots<a_{M-2-j}\leq M-2} \prod_{k=1}^{M-2-j}(a_k+\frac12)^2\right) (-1)^j  F_1(j)$$
Where
$$\small F_1(j)=\sum _{k=0}^{2 j+1} (-1)^k F_2(k+1) \left(\frac{1}{2}\right)^{2 j-k+2} \left(\binom{2 j+1}{k}+\binom{2 j+2}{k}\right)+F_2(2 j+3)$$
Where
$$\small F_2(k)=-\frac{k \sum _{l=0}^{k-1} (-1)^l F_3(l) \left(\frac{\pi }{2}\right)^{k-l-1} \binom{k-1}{l}}{\pi ^k}$$
Where
$$\small F_3(l)=\Re\left(\frac{(-1)^{l-1} \left(1-2^{-l-1}\right) l! \zeta (l+2)+\sum _{m=0}^l \frac{(-1)^{m-1} l! (i \pi )^{l-m} \zeta (m+2)}{(l-m)!}}{(2 i)^{l+1}}\right)$$
Enjoy (though I think a simple expression of $c_{n,d}$ in OP's (correct) speculation may not exist, other than the trivial observation that $c_{n,d}\in \mathbb Q/\pi^{2n}$)!

Body:
Interestingly the structure of $I(2M)$ is similar to this question that OP asked before. Indeed, using functional equation of $\Gamma$ one have
$$\small I(2M)=\int_0^{\frac{1}{2}} x \sin (\pi  x) \Gamma \left(M+x-\frac{1}{2}\right) \Gamma \left(M-x-\frac{1}{2}\right) \, dx=\int_0^{\frac{1}{2}} \pi  x \tan (\pi  x) \prod _{k=0}^{M-2} \left(\left(k+\frac{1}{2}\right)^2-x^2\right) \, dx$$
Where the primitive of RHS is evaluable through polylogarithms. The following command calculates $I(2M)$ for arbitrary $M$:

Table[Integrate[Pi x Tan[Pi x] Product[((k + 1/2)^2 - x^2), {k, 0, M - 2}], {x, 0,
1/2}], {M, 2, *Large number*}] // Expand

For a explicit calculation, one have (try it)

$$\small I(2 M)=\pi  \sum _{j=0}^{M-2} \left(\sum_{1\leq a_1<\cdots<a_{M-2-j}\leq M-2} \prod_{k=1}^{M-2-j}(a_k+\frac12)^2\right) (-1)^j  F_1(j)$$
Where
$$\small F_1(j)=\sum _{k=0}^{2 j+1} (-1)^k F_2(k+1) \left(\frac{1}{2}\right)^{2 j-k+2} \left(\binom{2 j+1}{k}+\binom{2 j+2}{k}\right)+F_2(2 j+3)$$
Where
$$\small F_2(k)=-\frac{k \sum _{l=0}^{k-1} (-1)^l F_3(l) \left(\frac{\pi }{2}\right)^{k-l-1} \binom{k-1}{l}}{\pi ^k}$$
Where
$$\small F_3(l)=\Re\left(\frac{(-1)^{l-1} \left(1-2^{-l-1}\right) l! \zeta (l+2)+\sum _{m=0}^l \frac{(-1)^{m-1} l! (i \pi )^{l-m} \zeta (m+2)}{(l-m)!}}{(2 i)^{l+1}}\right)$$
Enjoy (though I think a simple expression of $c_{n,d}$ in OP's (correct) speculation may not exist, other than the trivial observation that $c_{n,d}\in \mathbb Q/\pi^{2n}$)!


问:
Suspension of a CW complex is a CW complex
How can I show that a suspension of a CW complex is a CW complex?
I tried to start with showing first that a product of CW complexes is a CW complex but I didn't know how to construct the attaching maps.</p>

答:
Hint: The product of two CW complexes can (often) be made into a CW complex.  Same for the quotient of one.

Apparently the result does not always hold.  See @Tyrone's comment below.

Correction to the correction:  according to @SteveD it is true.


答:
You didn't use the generating-functions tag, so here is more than a nudge for that approach.  Let $A(z)=\sum_{n=0}^\infty a_n z^n$ be the ordinary generating function for $(a_n)$.  Then the recurrence and initial condition imply that
\begin{align}
A(z)
&= \frac{1}{2} z^0 + \sum_{n=1}^\infty \left(2 a_{n-1} + 2^n n\right)z^n \\
&= \frac{1}{2} + 2z \sum_{n=1}^\infty a_{n-1} z^{n-1} + 2z\sum_{n=1}^\infty n (2z)^{n-1} \\
&= \frac{1}{2} + 2z A(z) + 2z \frac{1}{(1-2z)^2},
\end{align}
so
\begin{align}
A(z) &= \frac{1/2 + 2z /(1-2z)^2}{1-2z} \\
&= \frac{1+4z^2}{2(1-2z)^3} \\
&= \frac{1}{(1 - 2 z)^3} - \frac{1}{(1 - 2 z)^2} + \frac{1/2}{1 - 2 z} \\
&= \sum_{n=0}^\infty \binom{n+2}{2} (2 z)^n - \sum_{n=0}^\infty \binom{n+1}{1} (2 z)^n + \frac{1}{2}\sum_{n=0}^\infty \binom{n+0}{0} (2 z)^n \\
&= \sum_{n=0}^\infty \frac{(n+2)(n+1)-2(n+1)+1}{2} (2 z)^n \\
&= \sum_{n=0}^\infty (n^2+n+1) 2^{n-1} z^n,
\end{align}
which immediately implies that $a_n = (n^2+n+1) 2^{n-1}$.

Title: Help solving a recurrence relation

Body:
I'm quite stuck on how to solve recurrence/difference equations.

For example I have the following linear inhomogeneous equation:
$a_n = 2a_{n-1} + 2^n n$, and $a_0 = 1/2$

I know that $2a_{n-1} = 1/2(2^n)$, and then we have $a_n = 1/2(2^n) + n(2^n)$. This is where I'm stuck though and the inhomogeneous part is throwing me off.

Am I in the right direction thinking that: $a_n = \frac12 \prod 2^n n$ ?

No answers please but some nudges in the right direction would be greatly appreciated. Thanks!


问:
How many tangents does a fixed curve have to the curves in a pencil?
Let $X$ be a smooth projective surface over $\mathbb{C}$ equipped with ample line bundles $L, L'$. Consider a fixed smooth curve $P$ in the linear system $|L|$, and consider a pencil $Q_\lambda$ of curves in $|L'|$.

Assume that $P$ does not intersect any critical point of $Q_\lambda$ (i.e. the singular point of a singular fiber), and does not pass through the base locus of $Q_\lambda$. Under these assumptions, I'm looking for a formula that enumerates the number of tangent points between $P$ and curves in $Q_\lambda$; I expect there should be one involving the degrees of the line bundles $L, L'$ as well as the canonical bundle of $X$.

This is probably an easy exercise somewhere, but any assistance would be appreciated!

答:
The pencil $Q_\lambda$ defines a rational map $X \dashrightarrow \mathbb{P}^1$ which restricts to a regular morphism
$$
P \to \mathbb{P}^1
$$
of degree $P \cdot Q_\lambda = \mathcal{L} \cdot \mathcal{L'}$. The points of tangency become the ramification points for this covering, so you can count them via the Hurwitz formula.


问:
Power-mean Inequality: Prove that for positive reals $a$, $b$, $c$ we have $3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$.

Taiwan Quiz 2014:

Prove that for positive reals $a$, $b$, $c$ we have

$$3(a+b+c) \ge 8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}$$

Applying Power-mean inequality (weighted) in $$a_1={abc} , a_2={\frac{a^3+b^3+c^3}{3}}$$ with $r=1$, $s=\dfrac{1}{3}$ and weights $\dfrac{8}{9}, \dfrac{1}{9}$ respectively, we get

\begin{align}
  P(1) &\ge P\left(\frac{1}{3} \right) \\
 \frac{8}{9}\cdot {abc} +\frac{1}{9}\cdot {\frac{a^3+b^3+c^3}{3}} & \ge
 \left(
   \frac{8}{9}\sqrt[3]{abc}+
   \frac{1}{9}\sqrt[3]{\frac{a^3+b^3+c^3}{3}}
  \right)^3 \tag{1}
\end{align}

In the book it says this simplifies into

$$a^3 +b^3 +c^3 +24abc \le (a+b+c)^3$$

which is true by $AM \ge GM$, we get on $2$ terms

$$8abc\le(a+b)(b+c)(c+a) \tag{2}$$

Can someone explain me how the simplification happened from $(1)$ to $(2)$?

Body:
From (1), we have
$$27(24abc + a^3 + b^3 + c^3) \ge \left(8\sqrt[3]{abc} + \sqrt[3]{\frac{a^3+b^3+c^3}{3}}\right)^3.$$
Thus, it suffices to prove that
$$[3(a+b+c)]^3 \ge 27(24abc + a^3 + b^3 + c^3)$$
or
$$(a+b+c)^3 \ge 24abc + a^3 + b^3 + c^3.$$


答:
I think the short answer is yes.

Decorated letters from $a, \ldots, z$ are very often used, like $i'$, but diacritic marks as found in French or German writing would be pretty unusual to be used as variables or parameters in math.  While $\tilde{n}$ could be used, and it looks Spanish, it really isn't since you could just as easily find $\tilde{x}$ or $\tilde{m}$ in math and those don't correspond to any real language.

I suspect the habit goes back to the wide use of Latin until the 1800s for writing math. So I would say the trend is really the use of letters from Latin rather than from English (allowing that $j$, $u$, and $w$ came later), but I don't want to get sucked into whatever rabbit hole you've fallen into arguing with other people in the comments about what language the alphabet $a, \ldots, z$ belongs to.

The only Cyrillic letter used systematically in math outside of Russia is Ш, although you know that В, Р, and Г may be used by your own instructors if they talk about graph theory.

Title: Is using the letters of English in math a universal practice?

Body:
In the Anglosphere (and in Russia, my home country) some of the symbols used in mathematics are the 26 letters of the English alphabet (52 actually, because there are two cases), and not any other alphabet based on the Latin script. I am aware that Greek and Hebrew, and even the Russian Ш (I call it Russian and not Cyrillic as it was introduced in honor of a Russian mathematician) letters are used in math as well.

My question is: Is this practice universal for all countries, even those that officially use an alphabet based on the Latin script (with letters not present in the English alphabet)? In other words, is there a country in which at least one Latin letter is used routinely in higher (this is important, I'm not asking about high school math) math that would not normally be considered an English letter?

I believe the answer is most likely "yes" (to the first formulation of the question) because the vast majority of mathematical papers and the leading mathematical literature are in English. Plus, English is the global lingua franca. However, I am not completely sure.

N.B. The second time derivative of $o$, i.e. $\ddot o$, is not the same as the German letter 'ö'. I know that this example is from physics, the example is just to get the point across.
Also: to those that believe French uses the English alphabet, the Wikipedia article about the English alphabet disagrees (on the right it says "Languages: English" with no French or any other language in sight).