答:
Set
$$
z=y\mathrm{e}^{kx}
$$
Then
$$
z''=(y\mathrm{e}^{kx})''=(y'\mathrm{e}^{kx}+ky\mathrm{e}^{kx})'
=y''\mathrm{e}^{-kx}+2ky'\mathrm{e}^{-kx}+k^2y\mathrm{e}^{-kx}=(y''+2ky'+k^2y)\mathrm{e}^{-kx}=0
$$
and hence
$$
z''=0\quad\Longrightarrow\quad z=c_1+c_2x\quad\Longrightarrow\quad y=(c_1+c_2x)\mathrm{e}^{-kx}
$$

Title: Second order homogeneous differential equations: why do repeated roots modify the solution set?

Body:
Consider the following differential equation with constant coefficients:
$$
y'' + 2ky' + k^2y = 0 \, .
$$
The auxiliary equation is
\begin{align}
m^2 + 2km + k^2 &= 0 \\[4pt]
(m+k)^2 &= 0 \\[4pt]
m &= -k \, .
\end{align}
This means that the solutions of the differential equation are $Ae^{-kx}+B\color{red}{x}e^{-kx}$, where $A$ and $B$ are arbitrary constants. The $x$-term highlighted in red is what confuses me. Generally speaking, if the roots of the auxiliary equation are $\alpha$ and $\beta$, then the solutions to the differential equation are $Ae^{\alpha x} + Be^{\beta x}$. However, if $\alpha=\beta$, then the solutions are not $Ae^{\alpha x} + Be^{\alpha x}$—or at least, this solution is incomplete. I can verify that $y=xe^{-kx}$ is a solution of the differential equation by plugging it back in to the equation, but I'm looking for a deeper reason for why this occurs.


问:
Extention of an Amenable Group by an Amenable Group is Amenable
I would like to prove that if $H\subset G$ is a normal amenable subgroup such that $G/H$ is amenable, then $G$ is amenable. The definition of amenability that I'm using is the following:

A group $G$ is amenable if every action of $G$ by homeomorphisms of a compact metric space admits an invariant probability measure.

This definition can be found on Navas's "Groups of Circle Diffeomorphisms". I've tried a lot of different ways but i couldn't prove it, I know there are many equivallent definitions for amenability but I would like (if possible) a proof that only uses this definition.

Here is what I've done so far: If $G$ acts on $(M,d)$ then $G/H$ acts on $M/H$ (the quotient of $M$ by the orbits of $H$), the problem is that this group is not necessarily metric, whe could endow the quotient group with the pseudometric $d'$ given in wikipedia https://en.wikipedia.org/wiki/Metric_space#Quotient_metric_spaces (the topology could be weaker than the quotient topology), and then do another quotient $X=(M/H)/\sim$ where $[x]\sim [y]$ if $d'([x],[y])=0$. Here $X$ is a compact metric space and we could take the action of $G/H$ on $X$ given by ${[g]}({[[x]]})=[[y]]$ if $[[g(x)]]=[[y]]$, since $G/H$ is amenable there exists an invariant probability measure, namely $\nu$. Now the sets $A_{[[x]]}=\lbrace y\in M:[[y]]=[[x]]\rbrace$ are compact and invariant under the action of $H$, so each one has an invariant probability measure namely $\mu_{[[x]]}$ and we could define the probability measure on $M$ as
$$\mu(B)=\int_X \mu_{[[x]]}(B\cap A_{[[x]]})d\nu([[x]]).$$

I don't know if this works in general, i couldn't prove or disprove it, i suppose this doesn't work since there could be some internal shifting of the orbits of $H$ in the sets $A_{[[x]]}$, but I hope this gives you some insight of what I'm trying so far.

I hope I was clear, many thanks in advance.

Something that might help: The space of probability measures on a metric space is compact, so you could use convergence of probability neasures.

答:
Fix a compact metric space $M.$ Let $W(M)$ denote the Wasserstein space for $M$: the space of probability measures on $M,$ with the Wasserstein metric. The important property is that this metric gives the topology of weak convergence, making $W(M)$ a compact metric space.

Let $W(M)^H$ denote the subspace of $H$-invariant measures. This is closed, so it's also a compact metric space.

An action of $G$ on $M$ gives an action $(gp)(A)=p(g^{-1}A)$ on $W(M).$ Since $H$ is normal, $G$ preserves $W(M)^H$: if $p$ is $H$ invariant then $p(g^{-1}hA)=p((g^{-1}hg)g^{-1}A)=p(g^{-1}A).$ But $H$ acts trivially on $W(M)^H,$ so in fact $G/H$ acts on $W(M)^H.$ Since $G/H$ is amenable there's a $G$-invariant measure $\xi$ on $W(M)^H.$

This is a probability measure on a space of probability measures. To get a measure on the original space $M,$ we need integration of measures. Or in other words the multiplication of the Kantorovich monad. Define $E\xi\in W(M)$ by $(E\xi)(A)=\int p(A)d\xi(p)$ for each Borel $A.$ The $G$-invariance of $\xi$ implies the $G$-invariance of $E\xi$: $$(gE\xi)(A)=\int (gp)(A)d\xi(p)=(E\xi)(A).$$

Finally I would like to mention that the same argument works if you drop the metrizability condition everywhere. The existence of an invariant probability measure for every $G$-action on a compact Hausdorff space is one of the few definitions of amenability that generalizes usefully to non-locally compact groups.

ContentLicense: CC BY-SA 4.0


问:
Second order homogeneous differential equations: why do repeated roots modify the solution set?
Consider the following differential equation with constant coefficients:
$$
y'' + 2ky' + k^2y = 0 \, .
$$
The auxiliary equation is
\begin{align}
m^2 + 2km + k^2 &= 0 \\[4pt]
(m+k)^2 &= 0 \\[4pt]
m &= -k \, .
\end{align}
This means that the solutions of the differential equation are $Ae^{-kx}+B\color{red}{x}e^{-kx}$, where $A$ and $B$ are arbitrary constants. The $x$-term highlighted in red is what confuses me. Generally speaking, if the roots of the auxiliary equation are $\alpha$ and $\beta$, then the solutions to the differential equation are $Ae^{\alpha x} + Be^{\beta x}$. However, if $\alpha=\beta$, then the solutions are not $Ae^{\alpha x} + Be^{\alpha x}$—or at least, this solution is incomplete. I can verify that $y=xe^{-kx}$ is a solution of the differential equation by plugging it back in to the equation, but I'm looking for a deeper reason for why this occurs.

答:
$$y'' + 2ky' + k^2y = 0 \, .$$
$$(y' + ky)' + k(y'+ky) = 0 $$
$$((y' + ky)e^{kt})' = 0 $$
Is equivalent to:
$$(ye^{kt})''=0$$
Inetgrate twice and the solution will be obvious.


问:
detail in proof of the mean value theorem for integrals
I am trying to prove the following result

If $f$ is a continuous function on a closed interval $[a,b]$, then
there exists a $c \in (a,b)$ such that,

$$f(c)=\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x$$

I have worked out the following proof

Because $f$ is continuous on the interval, it reaches a minimum $m$, and a maximum $M$ so that we have
$$\begin{aligned} m&\leq f(x)\leq M \\ m(b-a)\leq  & \int_a^bf(x)\;\mathrm{d}x\leq M(b-a) \\  m\leq &\frac{1}{b-a}\int_a^bf(x)\;\mathrm{d}x\leq M\end{aligned}$$
and the IVT guarantees the exxistence of $c$ as above.

How do I prove that $c \neq a\ \& \ c\neq b$?

Thanks for the help !

答:
If $f$ is constant, there is nothing to do.

Otherwise, let $M=\max f$ and $m=\min f$ on $[a,b]$ with $m\neq M.$ Then, $m\le \frac{\int_a^bf(t)dt}{b-a}\le M$ so the Intermediate Value Theorem gives us a $c\in [a,b]$ such that $f(c)=\frac{\int_a^bf(t)dt}{b-a}.$

If $c\in (a,b),$ we are done.

If $c=a$ then $f(a)=\frac{\int_a^bf(t)dt}{b-a}.$ If $f(a)=m,$ then $\int_a^b(f(t)-m)dt=0$ and since $f-m\ge 0$ on $[a,b]$ and continuous there, $f-m=0$ and $f$ is constant. Similarly, if $f(a)=M$ we get $f=M$.

So there are $a<x_0\neq x_1<b$ such that $f(x_0)=m$ and $f(x_1)=M.$ Without loss of generality, $x_0<x_1.$

Then, $m<f(a)<M$ so appying the IVT again, we get a $x_0\le d\le x_1$ such that $f(d)=f(a)$ and the result follows.

An almost identical argument shows that the result also holds if $c=b.$


问:
Precalculus in reverse?
In real analysis, I am aware of James Propp's "Real Analysis in Reverse" which does "naive reverse mathematics", showing the equivalence of various theorems of analysis to the completeness property of the real line.

Recently I've been thinking about the various geometric results that are typically covered in a precalculus course, such as:

The Pythagorean theorem

The law of cosines

The equivalence between algebraic and geometric dot product ($u\cdot v = \|u\|\|v\|\cos \theta$)

The formula for projections: $\operatorname{proj}_u(v) = \frac{v\cdot u}{\|u\|^2}u$

The sine and cosine angle addition formulas

The geometric interpretation of complex multiplication ("add the angles and multiply the lengths")

Determinant of a $2\times 2$ matrix as the signed area of the parallelogram spanned by its columns

...and so on. It seems to me that these results are all saying "basically the same thing" (whatever that means); in particular it's often possible to prove one of them using some of the others.

I am wondering if there is a textbook or paper that does something like "naive reverse mathematics of precalculus". What would the "base axioms" be in this setting? What is the relationship between these theorems? Why are they so closely related? Are there other "equivalent" theorems that are missing from this list? Are there any keywords I should be using in my search?

答:
There are many algebraic statements like the ones you list that are equivalent to the Pythagorean theorem.
That is essentially because the Pythagorean theorem is exactly what you need to show that the Euclid's geometric plane can be modeled as $\mathbb{R}^2$.

Even more interesting is the fact that there are many geometric equivalences. Among those are

The parallel postulate.

The angles of a triangle sum to $2\pi$.

Similar triangles that are not congruent exist.

See

Is Pythagoras' Theorem a theorem?

which axiom(s) are behind the Pythagorean Theorem

https://www.cut-the-knot.org/triangle/pythpar/PTimpliesPP.shtml

Title: Precalculus in reverse?


问:
Approximation of subsets of $\mathbb{R}$ with Borel subsets?
Let $E\subseteq \mathbb{R}$ be a subset, we want to show that there exists a Borel set $E\subseteq B\subseteq \mathbb{R}$ such that each Lebesgue measurable subset $H\subseteq (B\setminus E)$ has measure zero.

Here's my proof attempt:

Let $E\subseteq \mathbb{R}$. We have
$$m^*(E) = \inf\left\{\sum_k(b_k-a_k):E\subseteq \bigcup_k(a_k,b_k]\right\}$$
where $m^*$ denotes the Lebesgue outer measure. Hence, for each $n\in \mathbb{N}$ we can get a sequence of sets $\left\{(a_k^{(n)},b_k^{(n)}]\right\}_k$ such that
$$E\subseteq \bigcup_k \left(a_k^{(n)},b_k^{(n)}\right]\text{ and }\sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n.$$
Now, set $B_n = \bigcup_k\left(a_k^{(n)},b_k^{(n)}\right]$ and $B=\bigcap_n B_n$. So, $B$ is a Borel set and $E\subseteq B$. It remains to show that, for all $H\subseteq B\setminus E$, where $H$ is Lebesgue measurable, we have $m(H) = 0$. We first define
$$A=\bigcap_{\substack{S\in\mathcal{L}\\ E\subseteq S}}S\text{ where }\mathcal{L}\text{ denotes the set of Lebesgue measurable sets}.$$
Then, we have $A\in \mathcal{L}$ and $E\subseteq A\subseteq B$.

Claim: Suppose that $H\subseteq B\setminus E$ is a Lebesgue measurable set, then $H\subseteq B\setminus A$.

Assuming the above claim, we can complete the proof. It suffices to show that $m(B\setminus A) = 0$. Notice that, for all $n\in\mathbb{N}$, we have
$$m(B\setminus A) \leq m(B_n\setminus A) = m(B_n)-m(A) \leq \sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n$$
if $m(B_n)<\infty$ (this allows us to make the subtraction in the complement step). However, I'm not sure how to prove the infinite case and how to rigorously prove the claim (even though it seems intuitive to me). Please let me know if I'm on the right track / provide some hints. Thanks!

答:
Upon further attempts on my own and some suggestions from the comments, here is my improved solution to this problem (note that the first part regarding the construction of $B$ is copied from the question itself):

First, supposed that $E\subseteq \mathbb{R}$ is bounded. We have
$$m^*(E) = \inf\left\{\sum_k(b_k-a_k):E\subseteq \bigcup_k(a_k,b_k]\right\}$$
where $m^*$ denotes the Lebesgue outer measure. Hence, for each $n\in \mathbb{N}$ we can get a sequence of sets $\left\{(a_k^{(n)},b_k^{(n)}]\right\}_k$ such that
$$E\subseteq \bigcup_k \left(a_k^{(n)},b_k^{(n)}\right]\text{ and }\sum_k(b_k^{(n)}-a_k^{(n)})-m^*(E) < \frac1n.$$
Now, set $B_n = \bigcup_k\left(a_k^{(n)},b_k^{(n)}\right]$ and $B=\bigcap_n B_n$. So, $B$ is a Borel set and $E\subseteq B$. It remains to show that, for all $H\subseteq B\setminus E$, where $H$ is Lebesgue measurable, we have $m(H) = 0$.

Let $H\subseteq B\setminus E$ and set $A = B\setminus H$, then we have $E\subseteq A$ and $H = B\setminus A$ where $A$ is Lebesgue measurable. We then have the following (for all $n\in\mathbb{N}$):
$$m(H) = m(B\setminus A) = m(B)-m(A) \leq m^*(B_n)-m^*(E) < \frac1n.$$
Hence, we have $m(H)=0$.

Now, suppose that $E$ is unbounded and we write
$$E = \bigcup_j E_j\text{ where } E_j = [-j,j]\cap E.$$
Since $E_j$ is bounded for each $j\in\mathbb{N}$, we then have, for all $j$, there exists some $B_j$ such that $E_j\subseteq B_j$ and $m(H_j)=0$ for all Lebesgue measurable $H_j\subseteq B_j\setminus E_j$. Now, let $B = \bigcup_j B_j$ and consider some Lebesgue measurable $H\subseteq B\setminus E$. Now, for each $j$, consider $H_j = B_j\cap H$ and notice that $H_j\subseteq B_j\setminus E_j$ for each $j$ is Lebesgue measurable and that $H = \bigcup_j H_j$. Hence, we have
$$m(H) \leq \sum_j m(H_j) = 0\implies m(H) = 0.$$
Thus, our proof is complete.


问:
Chromatic number in relation to the amount of vertices

We have a finite undirected graph $G := (V, E)$ and its complementary graph $\overline G := (\overline V, \overline E)$. How do we show that $\chi (G) \cdot \chi (\overline G) \ge |V|$?

We know that there is a subset $V'$ in $V$ with $|V'| \ge \frac{|V|}{\chi (G)}$ and it also applys to $\overline G$. I started with transforming the given inequality to $\chi(G)$, respectivly $\chi(\overline G)$, so that
$$
\chi(G) \ge \frac{|V|}{|V'|} ,$$
and $$ \chi(\overline G) \ge \frac{|V|}{|\overline V'|} .$$ And put both together: $$\chi(G) \cdot \chi(\overline G) \ge \frac{|V|^2}{|V|\cdot|\overline V'|} .$$ But how do I eleminate the $|V|\cdot|\overline V'|$, or show that $$ \frac{|V|^2}{|V|\cdot |\overline V'|} \ge |V|?$$

答:
$$
\chi(G)\chi(\overline{G})\ge\chi(G)w(\overline{G})\ge\chi(G)\alpha(G)\ge V(G)
$$
where $w(\overline{G})$ is a maximal complete subgraph of $\overline{G}$, and $\alpha(G)$ is a maximal stable subset of $G$.</p>

ContentLicense: CC BY-SA 3.0


问:
Spectrum and point spectrum of quantum harmonic oscillator
I have this question concerning the Hermite operator on $\mathbb{R}^n$ (defined by $Hf:=-\Delta f+|x|^2f$). We know that $H$ is a densely-defined unbounded operator on $L^2(\mathbb{R}^n)$ and that $\mathcal{E}=\{2k+n\}_{k\in\mathbb{N}_0}$ is the set of its eigenvalues (the corresponding eigenfunctions are the Hermite polynomials). In general, we know that the spectrum $\sigma(T)$ of some operator $T$ contains the set of the eigenvalues $\mathcal{E}_T$ of $T$, but how do I prove that they coincide for $T=H$ (or, there is any theorem that grants it)?

Moreover, how do I prove that the spectral decomposition of $H$ is given by $\sum_k(2k+n)P_k$ where $P_k$ is the projection on the $k$-th eigenspace? I guess I shall prove that the projection valued measure $E$ given by the spectral theorem coincides with the counting measure and, then, invoke the spectral theorem (for unbounded operators).

If you have any good reference, please help me.

Thank you in advance.

答: As you have mentioned you have an eigenbasis given be the Hermite polynomals. Since every basisvector is only an eigenvalue once, the basis spans the entire Hilbert space and the set of eigenvalues is discrete you obtain that the spectrum is the set of eigenvalues and the spectral decomposition follows.</p>


问:
$L^2(\Omega) \subset D'(\Omega)$ , can we logically give a sense to this?
How can "$L_2(\Omega) \subset D'(\Omega)$" not be an aberration in term of logical symbolism ? $*_1$

I get that , the idea is to identify a L_2 function to it's regular associated distribution and to continusously inject the latter into D' , but i'm also very "shocked" that the logical symbol of "$\subset$" has been defined (in most books i've read) by :
$B \subset A \Leftrightarrow(\forall x)(x \in B \rightarrow x\in A)$
I'd see something like , maybe it exists a terminology for the symbol "$\subset$" such as , for two sets of differents objects , it is defined by "it exists a continuous injection from the first set to the second";

$*_1$: D'($\Omega$) is a set of function of $D(\Omega)\rightarrow R$ ,$D(\Omega)$ being itself a set of function ,$L_2$($\Omega$) is a set of function of $\Omega\rightarrow R$ , so it defines two different mathematical objects in my opinion.

Edit : $L_2$($\Omega$) is a set of Class of functions,

答:
Do you think that $\mathbb{Q} \subseteq \mathbb{R}$? Technically, these are completely different beasts: $\mathbb{Q}$ consists of pairs of integers (the second component non-zero) subject to a certain equivalence relation. $\mathbb{R}$ consists (depending on your taste) Cauchy sequences subject to a certain equivalence relation or Dedekind cuts (additional possibilities exist). These are certainly different types of mathematical objects but I think most people would agree that having $\mathbb{Q} \subseteq \mathbb{R}$ is useful. For example, we can now take the square root of a (positive) rational number even though it might not be rational again.

So here’s what we do: We find a copy of $\mathbb{Q}$ within $\mathbb{R}$ (i.e. an injection $\mathbb{Q} \hookrightarrow \mathbb{R}$) and then essentially redefine $\mathbb{Q}$ to mean the image of this injection. This is what we mean when we say that we “identify” the two sets. (We keep the original definition around because it often makes thinking about these objects easier.) One thing that is important here is that the inclusion is canonical, so there is no confusion which real number I mean when I talk about the rational number $\frac{23}{42}$.

The inclusion $L_2(\Omega) \subseteq D'(\Omega)$ works the same way. Here the canonical inclusion is given by the regular associated distribution you mentioned. But sometimes it is more convenient to just think of $L_2$-functions as distributions so you can apply stuff to it that generally only works for distributions (like the square root above).

This is a powerful tool but one also needs to be a bit careful with this trick. You shouldn’t simply apply it to any injection you see. For example, if you have three sets $A$, $B$, $C$ and injections $i : A \hookrightarrow B$, $j : B \hookrightarrow C$ and $k : A \hookrightarrow C$, the composition $j \circ i$ need not agree with $k$. If you use all three to make identifications $A \subseteq B \subseteq C$ and $A \subseteq C$ you can now consider elements $a \in A$ as elements of $C$ in two different ways; if these don’t agree, things get confusing.


问:
Characteristics of Metric Dual to Killing Vector Field
For a Killing vector field on a compact manifold, can we say that the metric dual one-form is closed, co-closed or both without appealing to additional information?</p>

答:
$X$ is a Killing field on a Riemannian manifold $(M,g)$ iff

$$g(D_Y X, Z) + g(Y, D_Z X) = 0$$

for any vector fields $Y$ and $Z$. On the other hand, if $\xi = g(X,-)$ is the $g$-dual of $X$, then

$$ (d\xi)(Y,Z) = D_Y(g(X,Z)) - D_Z(g(X,Y)) - g(X,[Y,Z]) = g(D_YX,Z) - g(D_ZX,Y),$$

using that the Levi-Civita connection is torsion-free and compatible with $g$. We thus see that $d\xi = 0$, that is to say $\xi$ is closed, iff

$$ g(D_YX,Z) - g(D_ZX,Y) = 0.$$

The two equations are not equivalent.

Edit 1: here is an easy counterexample. Let $g = \frac{dx^2 + dy^2}{y^2}$. Then $X = \partial_x$ is a Killing field of $g$. Its $g$-dual is

$$\xi = \frac{dx}{y^2}$$

which is not closed. It is also not co-closed, since its Hodge star is simply

$$\frac{dy}{y^2}$$

which is not closed.

Edit 2: I just realized that you asked about compact manifolds. My example in edit 1 is non-compact, in fact it is the hyperbolic plane. Here is another example, which is compact.

Let $M = S^2$ with spherical coordinates $\phi$ (the longitude) and $\theta$ (the co-latitude). Its round metric $g$ is

$$g = \sin^2(\theta) d\phi^2 + d\theta^2.$$

Then $\partial_\phi$ is a Killing field of $M$. Its metric dual is

$$\xi = \sin^2(\theta) d\phi$$

which is not closed. However, in this example, $\xi$ is actually co-closed.

Edit 3: I should probably mention Bochner's result. First note that if $(M,g)$ is compact oriented, then a $1$-form is harmonic iff it is closed and co-closed. Bochner's theorem says that if, moreover, $(M,g)$ has nonnegative Ricci, then any harmonic $1$-form is actually parallel.


问:
Precalculus please help
What is the smallest positive integer $n$ such that all the roots of $z^4 + z^2 + 1 = 0$ are $n$th roots of unity?

答:
Interesting query re pre-calculus tag combined with a complex analysis problem.

Hint:

The $n$-th roots of unity are the roots of

$$0 = (z^n - 1) = (z - 1)\left[z^{(n-1)} + z^{(n-2)} + \cdots + z^2 + z + 1\right]. \tag1$$

Therefore, one approach to the challenge is to look for a way to convert

$$\left[z^4 + z^2 + 1\right]$$

into the rightmost factor of equation (1) above.

One obvious try is to look for a specific value of $n$ such that $$z^4 = z^1 \implies \{z^3 = 1 ~\text{or}~ z = 0\}.$$

Title: Precalculus please help


问:
If $M_t$ is a martingale, when $\frac{1}{M_t}$ be a sub-martingale?
I am wondering what are constrains for using the argument that "martingale + convex function -> sub-martingale".

The problem I have is:
if $M_t$ is within $]0, +\infty[$, then whether $(\frac{1}{M_t}, t\geq 0)$ a sub-martingale ?

Generally, let $f$ a convex function, then by Jensen's inequality, we have:
$$\forall t\geq 0, \mathbb{E}[f(M_t)] \geq f(\mathbb{E}[M_t])$$

Here for this problem, $f$ will be $f(x) = 1/x$ and since $M_t$ is positive, we have the convexity condition satisfied. And since $M_t$ cannot be $0$ or $+\infty$, we have $f(\mathbb{E}[M_t])$ well defined.

So am I right to write:
$$\mathbb{E}[\frac{1}{M_t}|\mathcal{F_s}] \geq \frac{1}{\mathbb{E}[M_t|\mathcal{F}_s]} = \frac{1}{M_s}, \quad 0\leq s \leq t$$

so that $(\frac{1}{M_t}, t\geq 0)$ is a sub-martingale ? Or I have overmitted some important points ?

Besides, If we discard the constrain that  $M_t$ is within $]0, +\infty[$, and take brownian motion $(B_t, t\geq 0)$ as an example, can I say that   $(\frac{1}{B_t}, t\geq 0)$ a sub-martingale ? If not, what's the nature of $(\frac{1}{B_t}, t\geq 0)$ ?

Thanks !

答:
The only condition you omitted is that $E \frac{1}{M_t}$ must be finite.

If we don't have a condition $M_t \in (0, \infty)$ we can't say anything.

For example, $\frac{1}{B_t}$ is not a martingale or submartingale or supermartigale, because $E |\frac{1}{B_t}|$ is not finite.

If $M_t \in (-\infty, 0)$ and $E |\frac{1}{M_t}| < \infty$ we can prove that $\frac{1}{M_t}$ is supermartingale.


问:
Given a complex Gaussian variable $Z$ and a real Gaussian variable $X$, what is the mean of $Z\,e^{b\,X}$?
Eq. 19 of Gershgorin et. al. (2010) states that given a complex Gaussian variable $Z$ and a real Gaussian variable $X$, the mean of $Z\,e^{b\,X}$ is given by

$$\left\langle Z\,e^{b\,X} \right\rangle = \left( \left\langle Z \right\rangle + b\,\text{Cov}\left( Z,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)}.$$

The paper mentions that this quantity can be computed using the characteristic function of a Gaussian distribution, but I'm not sure how one would go about this.

I understand that the characteristic function is the inverse Fourier transform of the probability distribution function, and that the characteristic function of a linear combination of probability distribution functions is a product of the individual characteristic functions, but I don't know how the characteristic functions of $Z$ and $X$ would relate to the characteristic function of $Z\,e^{b\,X}$.

EDIT: I've made some progress on this problem. If we assume that $X$ and $Z$ are independent, with probability density functions $f_{X}\left( x \right)$ and $f_{Z}\left( z \right)$, then we may use the Law of the Unconscious Statistician

$$E\left[ g\left( X \right) \right] = \int_{\mathbb{R}} g\left( x \right)\,f_{X}\left( x \right)\,dx$$

to evaluate

\begin{align*}
     E\left[ Z\,e^{b\,X} \right] &= \int_{\mathbb{C}} \int_{\mathbb{R}} z\,e^{b\,x}\,f_{X}\left( x \right)\,f_{Z}\left( z \right)\,dx\,dz \\
&= \int_{\mathbb{C}} z\,f_{Z}\left( z \right)\,dz\,\int_{\mathbb{R}} e^{b\,x}\,f_{X}\left( x \right)\,dx \\
&= \left\langle Z \right\rangle\,M_{X}\left( b \right)
\end{align*}

where $M_{X}\left( t \right)$ is the moment-generating function of $X$

$$M_{X}\left( t \right) = E\left[ e^{t\,X} \right].$$

Since $X$ is Gaussian, we have

$$M_{X}\left( b \right) = e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)},$$

which means for independent $X$ and $Z$ we have

$$\left\langle Z\,e^{b\,X} \right\rangle = \left\langle Z \right\rangle\,e^{b\,\left\langle X \right\rangle + \frac{b^2}{2}\,\text{Var}\left( X \right)}$$

which is almost there! I am guessing that once we drop the assumption that the desired identity falls out.

答:
Alright, I figured it out based on some of the references ([15, 16]) in that Gershgorin et. al. 2010 paper.

To begin, let's write $Z = U + i\,V$ so that we may discuss its real and complex parts, and define a vector $\overrightarrow{V} = \left[ U\  V\  X \right]^T$. Note that $\overrightarrow{V}$ is normally distributed since each of $U$, $V$, and $X$ are as well.

The quantity we would like to find $\left\langle Z\,e^{b\,X} \right\rangle$ will come from finding its real and complex parts $\left\langle U\,e^{b\,X} \right\rangle$ and $\left\langle V\,e^{b\,X} \right\rangle$.

To find $\left\langle U\,e^{b\,X} \right\rangle$, we begin by considering the partial derivatives of the moment-generating function $M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)$ for $\overrightarrow{V}$

$$M_{\overrightarrow{V}} \left( \overrightarrow{t} \right) = \int_{\mathbb{R}^{3}} e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v},$$

where $f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)$ is the probability density funciton for $\overrightarrow{V}$. We will differentiate this function with respect to the first and second entries of $\overrightarrow{t}$, denoted $t_1$ and $t_2$ respectively

\begin{align*}
  \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1} &= \int_{\mathbb{R}^{3}} u\,e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v},\\
   \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_2} &= \int_{\mathbb{R}^{3}} v\,e^{\overrightarrow{t}^{T}\,\overrightarrow{v}}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v}.
\end{align*}

Evaluating these partial derivatives at $\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}$, we obtain

\begin{align*}
  \left.\frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. &= \int_{\mathbb{R}^{3}} u\,e^{b\,x}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v} = \left\langle U\,e^{b\,X} \right\rangle,\\
   \left. \frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_2}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}} \right. &= \int_{\mathbb{R}^{3}} v\,e^{b\,x}\,f_{\overrightarrow{V}}\left( \overrightarrow{v} \right)\,d\overrightarrow{v} = \left\langle V\,e^{b\,X} \right\rangle.
\end{align*}

Now, since $\overrightarrow{V}$ is a normally distributed, its moment-generating function is

$$M_{\overrightarrow{V}} \left( \overrightarrow{t} \right) = e^{\overrightarrow{t}^{T}\,\left(\overrightarrow{\mu}_{\overrightarrow{V}} + \frac{1}{2}\,\Sigma\,\overrightarrow{t}\right)}$$

where $\Sigma$ is the covariance matrix of $U$, $V$, and $X$. Hence,

\begin{align*}
   \left\langle U\,e^{b\,X} \right\rangle &= \left.\frac{\partial M_{\overrightarrow{V}} \left( \overrightarrow{t} \right)}{\partial t_1}\middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. \\
   &= \left. \left( \left\langle U \right\rangle + t_1\,\text{Cov}\left( U,\ U \right) + t_2\,\text{Cov}\left( U,\ V \right) + t_3\,\text{Cov}\left( U,\ X \right) \right)\,e^{\overrightarrow{t}^{T}\,\left(\overrightarrow{\mu}_{\overrightarrow{V}} + \frac{1}{2}\,\Sigma\,\overrightarrow{t}\right)} \middle|_{\overrightarrow{t} = \left[ 0\ 0\ b \right]^{T}}\right. \\
   &= \left( \left\langle U \right\rangle + b\,\text{Cov}\left( U,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)},
\end{align*}

and similarly

\begin{align*}
   \left\langle V\,e^{b\,X} \right\rangle &= \left( \left\langle V \right\rangle + b\,\text{Cov}\left( V,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)},
\end{align*}

Therefore,

\begin{align*}
   \left\langle Z\,e^{b\,X} \right\rangle &= \left\langle U\,e^{b\,X} \right\rangle + i\,\left\langle V\,e^{b\,X} \right\rangle \\
   &= \left( \left( \left\langle U \right\rangle + i\,\left\langle V \right\rangle \right) + b\,\left( \text{Cov}\left( U,\ X \right) + i\,\text{Cov}\left( V,\ X \right) \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)} \\
   &= \left( \left\langle Z \right\rangle + b\,\text{Cov}\left( Z,\ X \right) \right)\,e^{b\,\left\langle X \right\rangle + \frac{1}{2}\,b^2\,\text{Var}\left( X \right)}
\end{align*}

as desired.


问:
Let $(\mathbb{R\times R,+,\otimes)}$ a commutative and unitary ring where $(a,b)\otimes(c,d)=(ac,ad+bc)$ Find all the proper ideals
Let $(\mathbb{R\times R,+,\otimes)}$ a commutative and unitary ring where

$(a,b)\otimes(c,d)=(ac,ad+bc)$

Find all the proper ideals

I found $I=(0,b),b\in\mathbb{R}$ because

$(0,b)-(0,c)=(0,b-c)\in I$

$(e,f)\otimes (0,b)=(0,eb)\in I$

Is there another proper ideal?

答:
One important skill is being able to recognize when something seemingly unfamiliar is actually just a different guise of something familiar. In this case, I personally think the easiest way to think of this ring is as the quotient $R:=\mathbb{R}[x]\big/(x^2)$. I'll leave it to you to verify that the map from $(\mathbb{R}\times\mathbb{R},+,\otimes)$ to $R$ given by $(a,b)\mapsto \overline{a+bx}$ is an isomorphism. Now, the ideals of $R$ are in bijective correspondence with the ideals of $\mathbb{R}[x]$ containing $(x^2)$, so let $I\leqslant\mathbb{R}[x]$ be such. Since $\mathbb{R}[x]$ is a principal ideal domain, we have $I=(f)$ for some polynomial $f\in\mathbb{R}[x]$. Since $x^2\in I$, $f$ must divide $x^2$, so it must be an associate of $1$, of $x$, or of $x^2$. The first and last of these correspond to the ideals $R$ and $\left\{\overline{0}\right\}$, respectively, so the only non-trivial proper ideal of $R$ is $\left(\overline{x}\right)$. Pulling this back along our isomorphism $\mathbb{R}\times\mathbb{R}\rightarrow R$, you can see that this corresponds exactly to the ideal that you've found.

This might seem like a roundabout solution to the problem, but I think there's a lesson to it... the ring $(\mathbb{R}\times\mathbb{R},+,\otimes)$ at first glance seems unfamiliar (and, in fact, a priori it's not even completely clear that it forms a ring, at least to me), but by recognizing it as a different guise of the ring $R$, we're suddenly able to apply all our intuition and knowledge about polynomials rings over a field and their quotients. You could prove the result directly if you wanted, but recognizing this isomorphism allows you to reduce it to a situation you're already familiar with. It's not a very significant simplification in this particular case, but you can imagine how much uglier the problem would initially seem if you were instead given the ring $(\mathbb{R}\times\mathbb{R}\times\mathbb{R}\times\mathbb{R},+,\otimes)$ with structure corresponding to the ring $\mathbb{R}[x]\big/(x^4)$, for instance. When you're solving a problem, it's worth taking a step back at the start and thinking about whether it's a special case of something you already know, even if it doesn't look immediately like it is.


问:
Percentage of all natural numbers divisible by a prime $p_n$ and no smaller prime
The ratio of the count of all natural numbers divisible by $2$ to the count of all natural numbers is of course $\frac{1}{2}$ as there is no smaller prime than $2$.

The ratio of a count of all natural numbers divisible by $3$, but not by $2$ to the count of all natural numbers is $$\frac{1}{3}-\left(\frac{1}{2}\times\frac{1}{3}\right)=\frac{1}{6}$$

The ratio of a count of all natural numbers divisible by $5$, but not by $3$ or $2$ to the count of all natural numbers is $$\frac{1}{5}-\left(\frac{1}{2}\times\frac{1}{5}\right)-\left(\frac{1}{3}\times\frac{1}{5}\right)+\left(\frac{1}{2}\times\frac{1}{3}\times\frac{1}{5}\right)=\frac{1}{15}$$

The ratio of a count of all natural numbers divisible by $7$, but not by $5$, $3$ or $2$ to the count of all natural numbers is

$$\frac{1}{7}-\left(\frac{1}{2\ 7}+\frac{1}{3\ 7}+\frac{1}{5\ 7}-\frac{1}{2\ 3\ 7}-\frac{1}{2\ 5\ 7}-\frac{1}{3\ 5\ 7}+\frac{1}{2\ 3\ 5\ 7}\right)=\frac{4}{105}$$

This gets harder and harder to calculate with each larger prime. (I'm not completely sure this last one is correct)

Is there perhaps an easier way to think about this calculation to help find a general fraction formula for any prime, with the total sum of all fractions generated adding up to $1$?

答:
There's a much easier way to think about it. The natural density of natural numbers divisible by $p$ is $\frac{1}{p}$, so the density of numbers not divisible by $p$ is $1 - \frac{1}{p}$. Moreover the events of being divisible by different primes are "independent" in a suitable sense. So the density of natural numbers divisible by $p_n$ but not by $p_1, \dots p_{n-1}$ is

$$\left( \prod_{i=1}^{n-1} \left( 1 - \frac{1}{p_i} \right) \right) \frac{1}{p_n}.$$

Expanding this out produces the sum you get via inclusion-exclusion.

We need to be a little careful with this kind of "independence" reasoning because density is not actually a probability measure and so e.g. reasoning about infinitely many primes simultaneously this way is dodgy (although can, when done carefully, produce useful heuristics and conjectures), but you can prove rigorously that for finitely many primes everything works out. This is because the desired divisibility conditions are equivalent, via the Chinese remainder theorem, to a condition on the remainder $\bmod p_1 \dots p_n$, so the problem reduces to a counting problem in $\mathbb{Z}/(p_1 \dots p_n) \simeq \prod \mathbb{Z}/p_i$, which is finite.


问:
Unconstrained matrix factorization vs SVD
SVD decomposes a matrix $M$ as the product of two orthogonal matrices $U$ and $V$ (plus a diagonal matrix of singular values $\Sigma$).

$ M = U \Sigma V^*$

In this case, $U$ and $V$ are truncated to the first $n$ singular vectors for compressing $M$.

I'm exploring an alternative unconstrained factorization of $M$ using gradient descent to find $U$ and $V$ with the same dimensions as before:

$ \underset{UV}{\mathrm{argmin}} \: \mathbf{MSE}(M-UV) $

My results indicate that both methods for decomposing $M$ achieve the same mean square error ($\mathbf{MSE}$) score up to a surprising level of accuracy. Obviously, the resulting factorisations are different in each case, where the $U$ and $V$ found with the gradient descent method are not orthogonal.

My questions are:

Given that the second method has no constraints, I'd have supposed it should be able to find better factorisations than SVD -- in a $\mathbf{MSE}$ sense. Is this assumption right?

Is there any property of $M$, such as its rank or dimensions, that can be extracted from this result? The orthogonality constraint is apparently not limiting the SVD factorisation but I'm not sure if this happens in general or just for certain types of $M$.

Any thoughts about this or links to relevant literature would be greatly appreciated.

答:
Your first question is not technically a question, but your supposition is incorrect. No, it is not true that you can find a factorization that comes closer in the mean square error sense (I assume that's what MSE stands for). The EYM theorem guarantees that the approximation attained by truncating the SVD is optimal.

Your second question is indeed a question, but I don't know what you mean by "extracting property of $M$ from this result". In particular, I don't know what you mean by "extracting", "a property of $M$", or "this result".


问:
$\sum \frac{\ln(n)}{\sqrt{n^5}}$ test for convergence
Let $\sum a_{n}=\sum \frac{\ln(n)}{\sqrt{n^5}}$. To find if the serie is convergence or not, I had some difficult on finding the proper serie to test the given one.

After some work around, I found this sequence $b_{n}=\frac{1}{\sqrt{n^3}}$, whose serie converges!

Then I applyed the limit test comparasion.

$$\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n^5}}\cdot \sqrt{n^3}=\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n^{2}}}=\lim_{n \to \infty} \frac{\ln(n)}{n}=0$$

If the limit is equal $0$ and the $\sum b_{n}$ converges, then $\sum a_{n}$ converges to.

My main difficult was to fint out the proper $b_{n}$. Is there an easy way to find it? Thanks

答:
Once you know that
$\ln(n)/n \to 0$ as $n \to \infty$,
by putting $n = m^a$
for any real $a > 0$,
you get
$\ln(m)/m^a \to 0$
as $m \to \infty$.</p>


问:
Derive Rodrigues’ formula for Laguerre polynomials

Derive Rodrigues’ formula for Laguerre polynomials
$$
L_n(x)=\frac{e^x}{n!}.\frac{d^n}{dx^n}(x^ne^{-x})
$$

The Rodrigues’ formula for Hermite polynomials  can be obtained by taking $n^{th}$ order partial derivatives of its generatig function
$$
g(x,t)=\sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}=1+tH_1(x)+\frac{t^2}{2!}H_2(x)+\cdots\cdots\cdots+\frac{t^n}{n!}H_n(x)+\cdots\cdots\cdots\\
\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)=H_n(x)+\frac{(n+1)n(n-1)\cdots2}{(n+1)!}tH_{n+1}(x)+\cdots\\
H_n(x)=\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{2xt-t^2}\Big)\Bigg]_{t=0}=e^{x^2}\Bigg[\frac{\partial^n}{\partial t^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}\\
=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-(x-t)^2}\Big)\Bigg]_{t=0}=(-1)^ne^{x^2}\Bigg[\frac{\partial^n}{\partial x^n}\Big(e^{-x^2}\Big)\Bigg]
$$

Generating function for laguerre polynomials is $g(x,t)=\dfrac{e^{-\frac{xt}{1-t}}}{1-t}=\sum_{n=0}^\infty L_n(x) t^n$

I do not think the same technique applies in the case of Laguerre polynomials. So how do I derive that for Laguerre polynomials ?

答:
Note that the Taylor-Maclaurin formula cannot be used directly (as in the case of Hermite polynomials) as there is some $n$ dependency inside the $n^{th}$ derivative. How can we get around this ? ... Let use a coefficient extrator
\begin{eqnarray*}
x^n= [u^0]: \frac{u^{-n}}{1-xu}.
\end{eqnarray*}
We have
\begin{eqnarray*}
\sum_{n=0}^{\infty} L_n(x) t^n &=&[u^0]: e^x \sum_{n=0}^{\infty} \left(\frac{t}{u} \right)^n \frac{1}{n!}  \frac{d^n}{dx^n} \frac{e^{-x}}{1-xu} \\
&=&[u^0]: e^x  \frac{e^{-(x+t/u)}}{1-u(x+t/u)} \\
&=& \frac{1}{1-t} [u^0]:  \frac{e^{-t/u}}{1-\frac{ux}{1-t}}. \\
\end{eqnarray*}
Expand these two functions and observe that the central term is ... what we want
\begin{eqnarray*}
\left( \sum_{i=0}^{\infty} \frac{(-t/u)^i}{i!} \right) \left( \sum_{j=0}^{\infty} \left( \frac{ux}{1-t} \right)^j \right) =
\cdots+ \sum_{k=0}^{\infty} \frac{1}{k!} \left( \frac{-xt}{1-t} \right)^k + \cdots
\end{eqnarray*}
Now reverse engineer all this ... and your result follows.


问:
Proving sequence is bounded and increasing for Monotone Convergence
I have to prove that the sequence $a_{k+1}=\frac{5a_k}{2+a_k}$ with $a_0=1$ is convergent.

I think I can use the Monotone Convergence Theorem for this, but I can't show that it is bounded and convergent inductively.

答:
We have $a_n > 0$ (by induction).

Moreover, let us prove that $a_n \in (0,3)$ by induction.

$$a_{n+1}  \vee  3$$
$$\frac{5a_n}{2+a_n} \vee 3$$
$$5a_n \vee 6 + 3a_n$$
$$a_n \vee 3$$
$$a_n < 3$$
Hence $a_{n+1} < 3$.

Let us prove that $a_{n+1} > a_n$ by induction.

$$a_{n+1} \vee a_n$$
$$\frac{5a_n}{2+a_n} \vee a_n$$
$$5 \vee 2 + a_n$$
$$5 > 2 + a_n$$
Hence $a_{n+1} > a_n$.

Thus $0 < a_n < a_{n+1} < 3$ and by monotone convergence theorem we get that $a_n$ converges.

It's not hard to show that $a_n \to 3$.

Any questions?


问:
The coefficient and asymptotic in generating function

Let $L_n$ be the set of all the paths from $(0,0)$ to $(n,0)$ such that every step is $u=(1,1)$ , $d=(1,-1)$ and $r=(2,0)$.
Notice that the path could go under the $x$ axis.

a. Write a generating function of the number of paths in $L_n$.

b. Analyze the asymptotic of $|L_n|$ for a large n.

My trial:
in part a, I wrote:

Every path = (empty path) or (r every path) or (u Schröder path d every path) or (d Schröder path u every path).

Now, denote our g.f. by $A(x)$ and Schröder g.f. by $S(x)$, and denote $x$ to count half of the length, get: $A(x)=1+xA(x)+xS(x)A(x)+xS(x)A(x)$. Thus,

$$A(x)=\frac{1}{1-x-2xS(x)}$$

Then if we substitute the formula of $S(x)=\frac{1-x-\sqrt {(1-x)^2-4x}}{2x}$ :

$$A(x)=\frac{1}{\sqrt {(1-x)^2-4x}}$$

In part b, to analyze the asymptotic of $|L_n|$, I guess I should find the coeifficient of $[x^n]$ in $A(x)$ where $A(x)=\sum_{n\geq 0} |L_n| x^{2n}$.
However if we solve denominator of $A(x) = 0$ we get

\begin{align*}
(1-x)^2-4x=0
\quad\Rightarrow\quad
x^2-6x+1=0
\end{align*}

so, $x=3 \pm 2\sqrt2$. So the g.f. will be analytic if we delete the intervals $(3+\sqrt2, \infty)$ and $(-\infty,3-\sqrt2)$.
Write:
$A(x)=\frac{1} {\sqrt{(x- r_1)(x-r_2)}}=\frac{1}{\sqrt{x-r_1}} \frac{1}{\sqrt{x-r_2}}= \frac{1}{\sqrt{1-x/r_1}} \frac{1}{\sqrt{1-x/r_2}}$ where $r_1=3+\sqrt{8}$ and $r_2=3-\sqrt{8}$.
Then, $A(x)=(\sum_{n\geq 0}  (1/r_1)^n x^n \sum_{n\geq 0} (1/r_2)^n x^n)^{1/2}= \sum_{n\geq 0} \sum_{j=0}^{n} (1/r_1)^j (1/r_3)^{n-j} x^n$ ,
here I used Caushy formula.

So, $[x^n] A(x)= \sum_{j=0}^{n} (1/r_1)^j (1/r_2)^{n-j}$

$L_n = \sum_{j=0}^{n} (r_2/r_1)^j (1/r_2)^n$.
But this was not much useful!

Edited:
Another approach, which does not use the Legendre polynomial:

*I found in oeis the sequence of this problem is A001850 and it shows that the coefficient of the series is $\sum_{k=0}^{n} {n\choose k}{n+k \choose k}$ .In addition,  wikipedia page : https://www.google.com/url?sa=t&source=web&rct=j&url=https://en.m.wikipedia.org/wiki/Delannoy_number, mentioned the coefficient and the asymptotic of the above  g.f : $\frac{1}{\sqrt{1-6x+x^2}}$  which is the g.f of the  Delannoy number, but without explanation/proof.
How they find the coefficient of this generating function, and how they analayzed its asymptotic then?

Many thanks!

答:
We show the following asymptotic expansion of $A(x)=\frac{1}{\sqrt{1-6x+x^2}}$ is valid:
\begin{align*}
\color{blue}{[x^n]A(x)\sim\frac{\left(3+2\sqrt{n}\right)^n}{2\sqrt{\left(3\sqrt{2}-4\right)\pi n}}
\left(1-\frac{8-3\sqrt{2}}{32n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)}\tag{1}
\end{align*}
$A(x)$ has the isolated singularities
\begin{align*}x_0=3-2\sqrt{2}\qquad x_1=3+2\sqrt{2}, \qquad x_0x_1=1
\end{align*}
with $x_0$ being the dominant singularity determining the asymptotic expansion (1).

Expansion at $x_0=3-2\sqrt{2}$:

At first we consider
\begin{align*}
A(x)&=\frac{1}{\sqrt{1-6x+x^2}}=\frac{1}{\sqrt{\left(x-x_0\right)\left(x-x_1\right)}}\\
&=\frac{1}{\sqrt{x_0x_1}}\,\frac{1}{\sqrt{\left(1-\frac{x}{x_0}\right)}}\,\,\frac{1}{\sqrt{\left(1-\frac{x}{x_1}\right)}}\\
&=\frac{1}{\sqrt{1-\left(3-2\sqrt{2}\right)x}}\,\frac{1}{\sqrt{1-\left(3+2\sqrt{2}\right)x}}\tag{2}
\end{align*}

We expand (with some help of Wolfram Alpha) the first term in (2) at $x_0$ and obtain
\begin{align*}
\frac{1}{\sqrt{1-\left(3-2\sqrt{2}\right)x}}&=\frac{1}{2\sqrt{3\sqrt{2}-4}}+\frac{1}{16\sqrt{2}\sqrt{3\sqrt{2}-4}}\left(x-\left(3-2\sqrt{2}\right)\right)\\
&\qquad+\mathcal{O}\left(\left(x-\left(3-2\sqrt{2}\right)\right)^2\right)\\
&=\frac{1}{2\sqrt{3\sqrt{2}-4}}-\frac{3-2\sqrt{2}}{16\sqrt{2}\sqrt{3\sqrt{2}-4}}\left(1-\left(3+2\sqrt{2}\right)x\right)\\
&\qquad+\mathcal{O}\left(\left(1-\left(3+2\sqrt{2}\right)x\right)^2\right)\tag{3}
\end{align*}

The other term $\frac{1}{\sqrt{1-\left(3+2\sqrt{2}\right)x}}$ in (2) is already in the expanded form we need.

Asymptotic expansion:

We follow chapter VI in Analytic Combinatorics by P. Flajolet and R. Sedgewick and get from Theorem VI.1 an asymptotic expansion of
\begin{align*}
[x^n]f(x)=(1-x)^{-\alpha}\qquad\qquad\alpha\in\mathbb{C}\setminus\mathbb{Z}_{\leq 0}\tag{4}
\end{align*}
We take terms up to order $2$ and get
\begin{align*}
[x^n]f(x)\sim\frac{n^{\alpha-1}}{\Gamma(\alpha)}\left(1+\frac{\alpha(\alpha-1)}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\tag{5}
\end{align*}
We need from (5) the following expansions
\begin{align*}
[x^n](1-x)^{-\frac{1}{2}}&\sim\frac{n^{-\frac{1}{2}}}{\Gamma\left(\frac{1}{2}\right)}
\left(1+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&=\frac{1}{\sqrt{n\pi}}\left(1-\frac{1}{8n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\tag{6}\\
[x^n](1-x)^{\frac{1}{2}}&\sim\frac{n^{-\frac{3}{2}}}{\Gamma\left(-\frac{1}{2}\right)}
\left(1+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&=-\frac{1}{2\sqrt{n^3\pi}}\left(1+\frac{3}{8n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&=-\frac{1}{2\sqrt{n\pi}}\left(\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\tag{7}
\end{align*}

Note, that (4) is given in standard form, but we have something like $(1-x_1x)^{-\alpha}$ instead. We overcome this by recalling a Taylor series expansion and noting that
\begin{align*}
[x^n]f(x)&=x_{1}^{-n}[x^n]f\left(x_1x\right)\\
[x^n]f(x_1x)&=x_1^n[x^n]f(x)\\
&=\left(3+2\sqrt{2}\right)^n[x^n]f(x)\tag{8}
\end{align*}

Putting all together:

We take from (3) the expansion at $x_0$ and get
\begin{align*}
A(x)&=\frac{1}{\sqrt{1-\left(3-2\sqrt{2}\right)x}}\,\frac{1}{\sqrt{1-\left(3+2\sqrt{2}\right)x}}\\
&=\Bigg\{\frac{1}{2\sqrt{3\sqrt{2}-4}}-\frac{3-2\sqrt{2}}{16\sqrt{2}\sqrt{3\sqrt{2}-4}}\left(1-\left(3+2\sqrt{2}\right)x\right)\\
&\qquad\qquad+\mathcal{O}\left(\left(1-\left(3+2\sqrt{2}\right)x\right)^2\right)\Bigg\}\,\left(1-\left(3+2\sqrt{2}\right)x\right)^{-\frac{1}{2}}\\
&=\frac{1}{2\sqrt{3\sqrt{2}-4}}\left(1-\left(3+2\sqrt{2}\right)x\right)^{-\frac{1}{2}}\\
&\qquad-\frac{3-2\sqrt{2}}{16\sqrt{2}\sqrt{3\sqrt{2}-4}}\left(1-\left(3+2\sqrt{2}\right)x\right)^{\frac{1}{2}}
+\mathcal{O}\left(\left(1-\left(3+2\sqrt{2}\right)x\right)^\frac{3}{2}\right)
\end{align*}
Using (6) and (7) together with (8) we obtain
\begin{align*}
\color{blue}{[x^n]A(x)}&\color{blue}{\sim}\frac{\left(3+2\sqrt{2}\right)^n}{2\sqrt{3\sqrt{2}-4}}
\frac{1}{\sqrt{n\pi}}\left(1-\frac{1}{8n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&\qquad+\frac{\left(3+2\sqrt{2}\right)^n\left(3-2\sqrt{2}\right)}{32\sqrt{2}\sqrt{3\sqrt{2}-4}}\frac{1}{\sqrt{n\pi}}\left(\frac{1}{n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&=\frac{\left(3+2\sqrt{2}\right)^n}{2\sqrt{\left(3\sqrt{2}-4\right)\pi n}}\left(1-\frac{1}{8n}
+\frac{3-2\sqrt{2}}{16\sqrt{2}n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)\\
&\,\,=\color{blue}{\frac{\left(3+2\sqrt{2}\right)^n}{2\sqrt{\left(3\sqrt{2}-4\right)\pi n}}\left(1
-\frac{8-3\sqrt{2}}{32n}+\mathcal{O}\left(\frac{1}{n^2 }\right)\right)}
\end{align*}
and the claim (1) follows.

Note: The expansion (1) can be found e.g. in Asymptotics of the weighted Delannoy numbers by R. Noble.


答:
Note that $\frac{a_{k+1}}{a_k}=\frac5{2+a_k}$ is greater than $1$ when $-2<a_k<3$. Since $a_0=1$, it’s clear that $a_k>0$ for all $k$, so the sequence is increasing as long as $a_k<3$. Suppose that $a_k<3$. Then

$$\frac3{a_{k+1}}=\frac{6+3a_k}{5a_k}=\frac35+\frac6{5a_k}>\frac35+\frac6{15}=1\,,$$

so $a_{k+1}<3$. Thus, the sequence is increasing and bounded above by $3$, so it converges to some $L\le 3$. Now use the recurrence to show that $L=3$.

Title: Proving sequence is bounded and increasing for Monotone Convergence

Body:
I have to prove that the sequence $a_{k+1}=\frac{5a_k}{2+a_k}$ with $a_0=1$ is convergent.

I think I can use the Monotone Convergence Theorem for this, but I can't show that it is bounded and convergent inductively.


问:
Question Regarding $\int_0^1\int_0^∞ e^{-xy}-xye^{-xy}dxdy$
Specifically for $\int_0^{\infty}e^{-xy}-xye^{-xy}dx$,
I see that this inner integral evaluates to $xe^{-xy}$ from x = zero to x = infinity, which after plugging in, comes out to be zero (according to my textbook). However, I stuck on how to evaluate $xe^{-xy}$  for x = infinity. I tried to solve it by rephrasing it to $$\lim \limits_{x \to {\infty}}\frac{x}{e^{xy}}$$
But I realized there is no l'hopital's rule for multiple variables.

Any hints or answers would be greatly appreciated. Thanks.

答: When you are integrating w.r.t $x$ you are treating $y$ as a  constant. So L'Hopital's Rule is applicable and $\lim \frac x {e^{xy}}=\lim \frac  1{ ye^{xy}} =0$ for any $y>0$.


答: You can write $xe^{-xy}=x(e^y)^{-x}$.  For any $y \in (0,1), e^y$ is a constant $k$ greater than $1$, and we know $\frac x{k^x} \to 0$.  As changing the value of the integrand at one point does not change the integral, we can ignore what happens at $y=0$.

Title: Question Regarding $\int_0^1\int_0^∞ e^{-xy}-xye^{-xy}dxdy$

Body:
Specifically for $\int_0^{\infty}e^{-xy}-xye^{-xy}dx$,
I see that this inner integral evaluates to $xe^{-xy}$ from x = zero to x = infinity, which after plugging in, comes out to be zero (according to my textbook). However, I stuck on how to evaluate $xe^{-xy}$  for x = infinity. I tried to solve it by rephrasing it to $$\lim \limits_{x \to {\infty}}\frac{x}{e^{xy}}$$
But I realized there is no l'hopital's rule for multiple variables.

Any hints or answers would be greatly appreciated. Thanks.


问:
Are the indices in $\Sigma_j A_{ji}e_j$ correct?
A lot of times when I see a matrix-vector multiplication written in index form, it seems like the dimensions of the matrix are inverted or that the index they are using for the vector is the wrong one. I have seen this in many courses, but its never clear.

In my lecture notes for instance as part of a larger proof, they start like this:

Let $\{e_1,e_2,...e_n \}$ be an orthonormal basis of $V$ and $f:V\mapsto V $
an endormorphism. Let $A$ be the matrix representing the
endomorphism in this basis.

The columns of the matrix are: $f(e_i)=\Sigma_j A_{ji}e_j$

They didn't say it explicitly but I am pretty sure they are using column vectors as it is standard in linear algebra. So if the bases have $n$ elements I expect the matrix to be a $n\times n$ matrix, and therefore if the matrix $A$ is indexed as $A_{ji}$, $e$ should be indexed with $i=i\times1$ and not with $j$ as they are doing, that is I expect

$$f(e_i)=\sum_i A_{ji}e_i$$ or $$f(e_j)=\sum_j A_{ij}e_j$$

I don't think is a typo since they use it everywhere.

Could it be that I am unaware of some common usage? Could it be that they just don't care about the vector being a row or a column all the time and thus when they  write $\Sigma_j A_{ji}e_k$ with either $k=i$ or $k=j$   the vector $e_k$ is supposed  to be whatever it needs to be to make the expresion meaningful ( that is a column vector if they are using $j$ or  a row vector if they are using $i$) ?

Note: I found this hand-out: https://sites.math.washington.edu/~lee/Courses/443-2013/bilinear.pdfwhich is actually about the bilinear forms which is the context of my question and they do the same again, right from the first equation

$$A\mathbf{x}_j=\sum_{i=1}^nA_{ij}\mathbf{x}_j$$

答:
The notation is correct. In the equation
$$
Ax_j=\sum_{i=1}^n A_{ij}x_i\tag1
$$
the $i$ is an index of summation, whereas $j$ is "unbound". When the summation on the RHS is performed, the index $i$ disappears and the result (the RHS) involves $j$ only, same as the LHS.  If the index of summation in (1) were $j$ instead of $i$, the RHS would depend on $i$ after the dust settles, whereas the LHS would depend on $j$, which wouldn't make sense. (In fact if the index of summation in (1) were $j$ instead of $i$, we'd be able to factor out $x_i$ from the sum.)

In other words, the specific $x_j$ on the LHS is not the same as the generic $x_i$ on the RHS. In the RHS as $i$ ranges from $1$ to $n$ we visit all vectors $x_1,\ldots,x_n$, whereas in the LHS we stay with the specific vector $x_j$.

You should interpret (1) as a statement that holds for each $j$, i.e., it's an identity in $j$.


答:
$$\dfrac{1}{a_{n+1}}=\dfrac{2+a_n} {5a_n}=\dfrac{2}{5a_n}+\dfrac{1}{5}$$
Let $b_n = \dfrac1{a_n}$.  Then $$b_n = \frac25 b_{n-1} + \frac15$$ with $b_0 = 1$, so
\begin{aligned}
b_n &= \left[\frac15\sum_{i=0}^{n-1} \left(\dfrac25\right)^i \right] + \left(\frac25\right)^n b_0 \\
&= \frac15 \frac{1-(2/5)^n}{1-2/5} + \left(\frac25\right)^n = \frac{1-(2/5)^n}{3} + \left(\frac25\right)^n \\
&= \frac13+\frac23 \left(\frac25\right)^n,
\end{aligned}
thus $\lim_\limits{n \to +\infty} b_n = \dfrac13$, and $\lim_\limits{n \to +\infty} a_n = \dfrac{1}{\lim_\limits{n \to +\infty} b_n} = 3$.

Note: In this approach, it's clear that the initial value $b_0$ has no effect on the final limit, as its coefficient is $(2/5)^n$, which quickly diminishes as $n$ grows large.


答:
This is not a "literal" answer, but a sort of "meta-answer".

These indexing issues with matrices are an ever-recurring source of trouble.

Part of the problem is that these indexing things are not truly intrinsic, but only some sort of fallout from notational choices and conventions about the language involved.

Some decades ago, I did think there'd be some underlying ground about order of indices... but, after working hard to get down to it, found that the true answer is/was that the problem is not about mathematics, not even quite about "normal language", but about very volatile conventions of notation.

So, my serious advice to you is to re-appraise each situation...

Title: Are the indices in $\Sigma_j A_{ji}e_j$ correct?

Body:
A lot of times when I see a matrix-vector multiplication written in index form, it seems like the dimensions of the matrix are inverted or that the index they are using for the vector is the wrong one. I have seen this in many courses, but its never clear.

In my lecture notes for instance as part of a larger proof, they start like this:

Let $\{e_1,e_2,...e_n \}$ be an orthonormal basis of $V$ and $f:V\mapsto V $
an endormorphism. Let $A$ be the matrix representing the
endomorphism in this basis.

The columns of the matrix are: $f(e_i)=\Sigma_j A_{ji}e_j$

They didn't say it explicitly but I am pretty sure they are using column vectors as it is standard in linear algebra. So if the bases have $n$ elements I expect the matrix to be a $n\times n$ matrix, and therefore if the matrix $A$ is indexed as $A_{ji}$, $e$ should be indexed with $i=i\times1$ and not with $j$ as they are doing, that is I expect

$$f(e_i)=\sum_i A_{ji}e_i$$ or $$f(e_j)=\sum_j A_{ij}e_j$$

I don't think is a typo since they use it everywhere.

Could it be that I am unaware of some common usage? Could it be that they just don't care about the vector being a row or a column all the time and thus when they  write $\Sigma_j A_{ji}e_k$ with either $k=i$ or $k=j$   the vector $e_k$ is supposed  to be whatever it needs to be to make the expresion meaningful ( that is a column vector if they are using $j$ or  a row vector if they are using $i$) ?

Note: I found this hand-out: https://sites.math.washington.edu/~lee/Courses/443-2013/bilinear.pdfwhich is actually about the bilinear forms which is the context of my question and they do the same again, right from the first equation

$$A\mathbf{x}_j=\sum_{i=1}^nA_{ij}\mathbf{x}_j$$


问:
Trigonometric problem
I'm trying to get the roots for a complex number $x^2+1$

$x^2+1=0\rightarrow x^2=-1 \rightarrow x = \sqrt{-1} \rightarrow i$

So, $w^2 = 0 + 1i$

$p = \sqrt{0^2+1^2} = 1$

$\theta = \tan^{-1} \left( \frac{1}{0} \right )$

But I don't know what I can do to get the $\tan^{-1}(\frac10)$

答:
If you want to solve $x^2 = -1$,
note that both
$x = i$ and $x = -i$
satisfies this equation.

Remember that, aside from $0$,
all complex numbers have
2 square roots, 3 cube roots,
and $n$ $n$-th roots for any positive integer $n$.

Title: Trigonometric problem


问:
Reduce power modulo a semiprime quickly by hand
I worked through problem 7 in this GRE prep material (https://math.uchicago.edu/~min/GRE/files/week1.pdf), and I got the right answer, but I'm wondering if there is a more efficient way to do this, particularly considering that the GRE math subject test doesn't permit using a calculator. The question is to find the remainder upon dividing $ 13 ^{2019} $ by $95$. So we must solve for $x$ in $ 13^{2019} \equiv_{95} x.$ Since 95 factors into 5*19, I first found each of those residues.

$ 13^{2019} \equiv_{5} {{(13 \pmod 5})}^{2019 \pmod{\phi(5)}}\equiv_{5} 3^{2019 \pmod{4}}\equiv_{5} 3^3=27\equiv_{5}2.$
$ 13^{2019} \equiv_{19} {13}^{2019 \pmod{\phi(19)}}\equiv_{19} {13}^{2019 \pmod{18}}= 13^3\equiv_{19}(-6)^3=36*(-6)\equiv_{19}(-2)*(-6)\equiv_{19}12.$

Since $2 \equiv_5 12$, that means $13^{2019}$ is equivalent 12 with respect to each modulus, which means that $13^{2019} \equiv_{95} 12.$

Are there any other techniques I could have used to get the answer more quickly?

Body:
$\phi(95)=\phi(5) \cdot \phi(19) = 4 \cdot 18=72$.

$2019 \equiv 3 \pmod{72}, 13^{72} \equiv 1 \pmod{95}$, therefore

$$13^{2019} \equiv 13^3 \equiv 169 \cdot 13 \equiv  (-21) \cdot 13 \equiv 12 \pmod{95}$$


问:
What is the difference between "infinite" and "transfinite"?
I have never properly got my head round exactly what the difference is between "transfinite" and "infinite".

Wikipedia says: transfinite numbers are numbers that are "infinite" in the sense that they are larger than all finite numbers, yet not necessarily absolutely infinite.

Huh?

For example, the set of all natural numbers $\mathbb N$ is "infinite" in cardinality, in fact "countably infinite" -- but its cardinal $\aleph_0$ and the ordinal $\omega$ which is the "order type" of $\mathbb N$ are defined as being "transfinite".

The rest of that wikipedia article on "transfinite number" does not help too much, except to explain that Cantor coined the term "transfinite" as a sweetener, so to speak, so as to make the medicine of his work go down easier.

But apart from historical reasons to do with battling with hidebound modes of thought (and I am familiar with Cantor's difficulties with Kronecker and those of his way of thinking), does there exist a concrete definition that one can go to that says: "this is what transfinite means: ... and this is what infinite means: ... and the difference between the two is ..."?

答: When Cantor first outlined his theory of transfinite numbers, he wanted to stress that there are indeed distinct numbers beyond the finite numbers. He was clear that there are numbers that measure infinite size (infinite cardinal numbers) as well as numbers that measure infinite (well) orderings (infinite ordinal numbers). Cantor did not define these numbers out of intellectual curiosity, but because they provided new proof techniques, especially in the subject that we now call set-theoretic topology. For example, if a set is thought of as comprising branches (sequences) of a tree with a root, and if a branch is called "isolated" if there is a node of the branches beyond which there are no other branches, then by iteratively removing isolated branches from a tree any finite number of times, we see that a set comprises a countable set of branches and a remainder set (which could be empty). In the case of a set of real numbers with all of its limit points (a closed set), Cantor showed that the remainder set is a set of limit points of the same size as the set of real numbers (called a "perfect" set). The technique can be generalised to sets where branches transfinite sequences and (dropping the use of trees) to metric spaces and certain topological spaces. For further reading on Cantor's mathematics I would recommend the classic books by J. Dauben and M. Hallett, and for a readable take on what would now be called descriptive set theory, F. Hausdorff's Set Theory (from the 1930s).</p>


问:
Limit of Itô-Integral
We consider the following Itô-Integral $$Z_t := \int_0^t \exp(-\mu s) W_s ds$$ for $\mu\geq0$. I wonder if I could calculate the limit as $t\to\infty$ in some convergence, but how should I start? Some ideas?

Edit: Is it possible to get convergence almost surely to some random variable? Or in $L^2$?

答:
As shown by @fesman, $Z_t$ differs from the stochastic integral $\int_0^t e^{-\mu s}\,dW_s$ by an amount that converges a.s (and also in $L^2$) to $0$. It's not hard to show that $\int_0^t  e^{-\mu s}\,dW_s$ converges in $L^2$ to $\int_0^\infty  e^{-\mu s}\,dW_s$ (the $L^2$ norm of their difference is ${e^{-2\mu t}\over t}$). In fact, by the martingale convergence theorem, the convergence is also almost sure, because
$$
\int_0^t  e^{-\mu s}\,dW_s = E\left[\int_0^\infty  e^{-\mu s}\,dW_s\mid\mathcal F_t\right].
$$
It follows that $Z_t$ converges to $\int_0^\infty  e^{-\mu s}\,dW_s$ a.s. and in $L^2$.

Title: Limit of Itô-Integral


问:
Nature of cycle space
In graph theory, the cycle space of a connected, undirected finite graph consists of all Eulerian (all vertices have only even degrees) subgraphs. It's easy to see why cycles generate (using symmetric difference as 'addition') only such subgraphs, but I cannot find a proof of the opposite, namely that each Eulerian subgraph is a member of the cycle space - how easy is it to show that?
Since I have not received any answer yet, let me try my own proof: any Eulerian subgraph must contain at least one cycle (start at any positive-degree vertex of the subgraph and continue 'walking' through consecutively adjacent vertices till re-vising a vertex, and thus completing a cycle; this is always possible since every new vertex you enter has at least one extra incident edge to leave, until you re-visit a vertex).
Then, remove all the edges of this cycle from the original subgraph; what remains is another Eulerian subgraph (since the degree of every vertex we have visited has been reduced by two). This can be repeated till no edges remain.
This shows that a Eulerian subgraph consist of a union of cycles (potentially sharing vertices but not edges); such a union is clearly a member of the cycle space.</p>

答:
A silly answer is that if we define the cycle space to have all Eulerian subgraphs as its elements, then each Eulerian subgraph is in the cycle space by definition. That's obviously not what you mean. You want to define the cycle space to be the vector space generated by symmetric-difference addition of all cycles. Then, the question you want can be stated purely graph-theoretically as:

Can (the edge set of) every Eulerian graph be represented as the symmetric difference of one or more cycles?

Your proof is exactly the proof I would give that the answer is yes. It actually shows a stronger claim: that every Eulerian graph has a decomposition into edge-disjoint cycles, for which symmetric difference can be replaced with plain union.

Another way to prove this claim is to prove that a fundamental cycle basis spans the cycle space (as in your previous question). The elements of a fundamental cycle basis are all cycles, and what we'd prove there is exactly that every Eulerian graph is the symmetric difference of one or more cycles in the fundamental cycle basis.

This is a bit overkill for this question, though it is a valid argument. We limit our choice of cycles to only the cycles in the cycle basis. This means we have to do more work to finish the proof, and that we are forced to use symmetric difference and not just edge-disjoint union.

Title: Nature of cycle space


问:
Is the category of sheaves of objects from an abelian category abelian?
Suppose we have an abelian category $\mathcal{A}$ and a topological space $X$. Let $\mathscr{F}^{\mathcal{A}}$ be the category of all sheaves of $\mathcal{A}$ objects over $X$. Is $\mathscr{F}^{\mathcal{A}}$ guaranteed to be an abelian category? I don't really have any evidence that this should be true other than that it happens to be true when $\mathcal{A}$ is Ab or $R$-Mod, and those are the principle abelian categories.

Of course, by the Freyd–Mitchell embedding theorem $\mathcal{A}$ is always equivalent to a full subcategory of $R$-Mod, but I'm not familiar enough with sheaves or the embedding theorem itself to know if the "subcategory" factor could play a problem here.

In the event that the answer to the posed question is "no", is there some extended criterion for $\mathscr{F}^{\mathcal{A}}$ to be abelian?

答:
In the first place, it is unclear to me what a sheaf of $\mathcal{A}$-objects on $X$ should be. The obvious candidate, if you are used to the functor definition of sheaf, is that a sheaf of $\mathcal{A}$-objects on $X$ is a functor $\textrm{Open} (X)^\textrm{op} \to \mathcal{A}$ that satisfies the usual sheaf condition. I think this is the wrong definition and will fail to give an abelian category in general – at least, it is not obvious to me where cokernels are going to come from. On the other hand, there is another candidate, more in line with the original definition of sheaf in terms of stalks, and this one yields an abelian category. In more detail:

Let $\mathcal{A}$ be an abelian category and suppose $\mathcal{A}$ is an exact full subcategory of an abelian category $\mathcal{B}$ that satisfies Grothendieck's axioms AB3, AB3*, and AB5. (For example, if $\mathcal{A}$ is a Grothendieck abelian category we may take $\mathcal{B} = \mathcal{A}$; or if $\mathcal{A}$ is an essentially small abelian category we may take $\mathcal{B} = \textbf{Ind} (\mathcal{A})$.)

Definition. A sheaf of $\mathcal{A}$-objects on $X$ is a functor $F : \textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ with the following properties:

$F : \textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ satisfies the sheaf condition.

For all $x \in X$, the stalk $F_x$ (computed in $\mathcal{B}$) is (isomorphic to) an object in $\mathcal{A}$.

Remark. It does not follow that $F (U)$ is (isomorphic to) an object in $\mathcal{A}$! For example, if $X$ is an infinite discrete set and $\mathcal{A}$ is the category of finite abelian groups (and $\mathcal{B}$ is the category of all abelian groups), then $F (X)$ can be infinite even if every $F_x$ is finite.

Proposition. The category of sheaves of $\mathcal{A}$-objects on $X$ is an abelian category.

Proof. The category of sheaves of $\mathcal{B}$-objects on $X$ is a full subcategory of the category of all functors $\textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ and has an exact reflector. The category of all functors $\textrm{Open} (X)^\textrm{op} \to \mathcal{B}$ is an abelian category, so it follows that the category of sheaves of $\mathcal{B}$-objects on $X$ is also an abelian category. Taking stalks is also exact, so the category of sheaves of $\mathcal{A}$-objects is an exact full subcategory of the category of sheaves of $\mathcal{B}$-objects, hence is also an abelian category. ■

What is not clear to me is whether the category of sheaves of $\mathcal{A}$-objects on $X$ defined in this way is independent of the choice of embedding $\mathcal{A} \hookrightarrow \mathcal{B}$. If it is then I could be convinced this is the correct definition.


问:
Path Independence of Line Integrals
The Cauchy-Goursat theorem states that if a function $f$ is analytic on, and in the interior of, a contour $C$ then: $$\int_{C} f(z) \hspace{1mm} dz = 0.$$

From here, a lot of lecture notes deduce the following.

Cor. Suppose $f$ is analytic on some simply connected region $A$. Then for any two points $a, b \in A$ we have that $$\int_{C} f(z) \hspace{1mm} dz$$ is independent of the contour $C$ in $A$ joining $a$ and $b$.

However, the 'proof' often only shows $$\int_{C_1} f(z) \hspace{1mm} dz = \int_{C_2} f(z) \hspace{1mm} dz.$$ when $C_1$ and $C_2$ are non-intersecting (because then $C := C_1\cup C_2$ is a closed contour). What is the best way to prove independence for all $C_1$ and $C_2$? I can 'see' that one can just pick a third contour $C_3$ which doesn't intersect either $C_1$ or $C_2$, and the proof follows from that. But how can I make the existence of $C_3$ rigorous without 'advanced machinery'?

答: If $\gamma_1$ and $\gamma_2$ are two paths going from $a$ to $b$ and if $\gamma$ is the path $\gamma_1$ followed by the inverse path of $\gamma_2$ (usually, $\gamma$ is denoted by $\gamma_2^-\vee\gamma_1$, then $\gamma$ is a closed path and therefore $\oint_\gamma f(z)\,\mathrm dz=0$. But this is the same thing as asserting that$$\int_{\gamma_1}f(z)\,\mathrm dz=\int_{\gamma_2}f(z)\,\mathrm dz.$$Whether the paths $\gamma_1$ or $\gamma_2$ intersect or not doesn't matter.

Title: Path Independence of Line Integrals


问:
topology over $X$
Definition. A topology $\tau$ is a collection of open X such that:

(i) $\emptyset$ and $X \in \tau$;

(ii) arbitrary union of $A$ open is opened;

(iii) finite intersection of open is opened.

To prove that the union of topologies is a topology this stems from the topology axiomas, because:

(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the intersection.

(ii) arbitrary intersection of finite intersection is opened. How to prove this fact?

(iii) arbitrary intersection of arbitrary union is opened.How to prove this fact?

Already for the union arbitraria, have:  What's the flaw?

(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the union.

(ii) arbitrary union of finite intersection is opened. How to prove this fact?

(iii) arbitrary union of arbitrary union is opened. How to prove this fact?

Can someone help me please?

答:
The part about $\bigcap C$ being a topology is not too difficult if you write out the definitions and follow your way through.

For an example where $\bigcup C$ is not a topology, take $C = \{\tau_1, \tau_2\}$ where $\tau_i$ comprises all the subsets of $\{1, 2, 3\}$ that are either empty or contain $i$. Then $\{3\} = \{1, 3\} \cap \{2, 3\} \not\in \bigcup C$, but it would have to be if $\bigcup C$ were a topology.

Title: topology over $X$


问:
Prove that if f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n) + f2(n) = O(g1(n) + g2(n)).
I've been working on this problem for a quite a period of time and I don't even know where to start,I tried to follow the basic upper bound rule but it did not help me.
Any suggestions please?</p>

答:
You can't do it because it's false. For example if $f_1(n)=f_2(n)=g_1(n)=1$ and $g_2=-1$ for all $n.$

The definition of the Landau Big-O is that $f(n)=O(g(n))$ as $n\to \infty$ means $\exists K\in \Bbb R^+\,\exists n_0\in\Bbb R\,(\,n\ge n_0\implies |f(n)|\le K|g(n)|\,).$


答:
Suppose that $\mathscr{C}$ is a family of topologies on $X$. Let $\tau_0=\bigcap\mathscr{C}$. You showed that $\varnothing,X\in\tau_0$. To show that $\tau$ is closed under taking arbitrary unions, let $\mathscr{U}\subseteq\tau$; we want to show that $U\in\tau_0$. Let $\tau\in\mathscr{C}$; then $\mathscr{U}\subseteq\tau_0\subseteq\tau$, so $\mathscr{U}\subseteq\tau$. And $\tau$ is a topology, so $\bigcup\mathscr{U}\in\tau$. Thus, $\bigcup\mathscr{U}\in\tau$ for each $\tau\in\mathscr{C}$, and therefore

$$\bigcup\mathscr{U}\in\bigcap\mathscr{C}=\tau_0\,,$$

as desired.

The proof that $\tau_0$ is closed under taking finite intersections is very similar; see if you can carry it out using this as a model.

To show that $\bigcup\mathscr{C}$ is not necessarily a topology on $X$, you need to produce a counterexample: you want a set $X$ and a family of topologies on $X$ whose union is not a topology on $X$. You can actually do this with just a pair of topologies, say $\tau_1$ and $\tau_2$ on a small set.

Try to find topologies $\tau_1$ and $\tau_2$ on $X=\{0,1,2\}$ such that there are sets $U_1\in\tau_1$ and $U_2\in\tau_2$ whose intersection is not in $\tau_1\cup\tau_2$; this will show that $\tau_1\cup\tau_2$ is not a topology on $X$.

Title: topology over $X$

Body:
Definition. A topology $\tau$ is a collection of open X such that:

(i) $\emptyset$ and $X \in \tau$;

(ii) arbitrary union of $A$ open is opened;

(iii) finite intersection of open is opened.

To prove that the union of topologies is a topology this stems from the topology axiomas, because:

(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the intersection.

(ii) arbitrary intersection of finite intersection is opened. How to prove this fact?

(iii) arbitrary intersection of arbitrary union is opened.How to prove this fact?

Already for the union arbitraria, have:  What's the flaw?

(i) $\emptyset$ and $X\in \cap C$, because $\emptyset$ and $X$ belongs to each topology, so it belongs to the union.

(ii) arbitrary union of finite intersection is opened. How to prove this fact?

(iii) arbitrary union of arbitrary union is opened. How to prove this fact?

Can someone help me please?


问:
Calculating the Fundamental Group of a CW Complex with Attaching Maps of Varying Degrees
I'm reviewing old homework questions for an upcoming topology exam, and I came across a question that I did not fully understand at the time. In the solution to this homework question, I had to compute the fundamental group of the following $CW$ complex.
$X$ is the CW complex obtained from $S^1$ with its usual cell structure by attaching two 2-cells by maps of degrees 2 and 3 respectively.

I approached the problem as follows.
Let $U$ be an small open neighborhood of $S^1\subset X$ unioned with the first 2 cell attached, and let $V$ be the same small open neighborhood of $S^1\subset X$ unioned with the second 2-cell
attached. Then $U\cap V$ is that small open neighborhood of $S^1$, so we get that $U,V,U\cap V$ and are open and path connected, and $X=U\cup V$. I wish to apply SVK's theorem to $U$ and $V$ to calculate the fundamental group of
$X$. However, to find $\pi_1(U)$ and $\pi_1(V)$, I think I have to apply SVK's theorem to each first. In order to avoid duplicate work, I will prove that
the fundamental group of CW complex $X'$ obtained from $S^1$ by attaching a two
cell with maps of degree $n$  isomorphic
to $\mathbb{Z}/ n\mathbb{Z}$ and apply it separately to $U$ and $V$ which are
special cases. If one lets $U'$ be some small open neighborhood of $S^1$ (and hence it contains $\partial D^2$) and let $V'$ be the interior of the attached 2-cell, $U'\cap V'$ is connected and
retracts to $S^1$ so it has fundamental group isomorphic to $\mathbb{Z}$. Let $i_{U'}:U'\cap V' \hookrightarrow U'$ be the inclusion map and $\alpha$ be a
generator for $\pi_1(U'\cap V')$. Then $(i_{U'})_*(\alpha)=n\alpha$ because the attaching map maps $\partial D^2$ $n
$ times around $S^1$. The interior of $D^2$ is contractible, so we get
$$\pi_1(X')=\pi_1(U')*_{\pi_1(U'\cap V')}\pi_1(V')=\mathbb{Z}*\{0\}/ n\mathbb{Z}\cong \mathbb{Z}/n\mathbb{Z}$$

The way I understand this result geometrically is that if I have a loop that wraps around $S^1\subset X'$ once, I cannot contract it to a constant loop because it wraps around $S^1$ and $S^1$ is not contractible. However, if I wrap around $S^1$ $n$ times, then the loop
traverses the boundary of the attached 2 cell $D^2$ and can``escape"  (via homotopy) to the 2-cell $D^2$. $D^2$ is contractible of course, so this loop contracts to the trivial loop.

Back to the original question. My geometric arguments seems to give the fundemantal group of $X$ almost immediately. Indeed, if I have a loop that wraps around $S^1\subset X$ once, it is not trivial (like before). However, if I wrap around $S^1$ twice, then my loop can "escape"to the 2-cell
attached with the map of degree 2. Thus if a loop wraps around $S^1$ 3 times, then the loop first wrapping arounds twice, which is homotopic to the identity, so it is homotopic to a loop that wraps around once. Thus attaching a
2-cell with a map of higher degree adds NOTHING to the fundamental group. More generally, if I attach 2-cells to $S^1$ with maps of varying degree, the
fundamental group is simply isomorphic
to $\mathbb{Z}/ m\mathbb{Z}$ where $m$ is the lowest degree of the attaching
maps. Is this intuition correct?
This
reasoning leads me to believe that the
fundamental group of $X$ is simply
$\mathbb{Z}/2\mathbb{Z}$. This seems
like a fairly rigorous argument to me,
but I would like to calculate directly
using SVK's theorem.

When I intersect
$U$ and $V$, I will get a space that contracts to the $S^1$ (the 1-skeleton). Let $1$ be the generator for $\pi_1(U\cap V)\cong\mathbb{Z}$.  Let $i_U:U\cap V\hookrightarrow U$ and $i_V:U\cap V\hookrightarrow V$ be the corresponding inclusion maps. Then $(i_U)_*(1)= \overline{1}\in \mathbb{Z}/2\mathbb{Z}$ and  $(i_V)_*(1)= \tilde{1}\in \mathbb{Z}/3\mathbb{Z}$.
SVK's theorem gives
$$ \pi_1(X)=\pi_1(U)*_{\pi_1(U\cap
 V)}*\pi_1(V)=\langle \overline 1,
 \tilde 1| \overline 1 ^2=0, \tilde 1
 ^3=0 , \overline 1=\tilde
 1\rangle\cong 0 $$
So, there seems to be a problem with my computation. How might I fix my computation? Is there a different (perhaps shorter) way of calculating $\pi_1(X)$? Originally I was thinking about removing a point in the interior of each attached 2-cell and proceeding similarly to the way one would show the fundamental group of the sphere is trivial using SVK's theorem.

答:
Look at my answer to Can this counter example disprove the statement? The following is shown:

Let $X$ be any space. Take an element $g \in \pi_1(X,x_0)$, represent it by a map $\gamma : S^1 \to X$ and attach a $2$-cell to $X$ via $\gamma$. If $i : X \to X' = X \cup_\gamma D^2$ denotes inclusion, then
$$i_* : \pi_1(X,x_0) \to \pi_1(X', x_0) $$
is an epimorphism whose kernel is the normal subgroup $N(g)$ of $\pi_1(X,x_0)$ generated by $g$.

Now let $f_k : S^1 \to S^1$ be a map of degree $k$ (we may take $f_k(z) = z^k$). Consider $X_2 = S^1 \cup_{f_2} D^2$ with inclusion $i : S^1 \to X_2$ and $X = X_2 \cup_{if_3} D^2$ with inclusion $j : X_2 \to X$. The space $X$ is the CW complex of your question. Let $g = [f_1]$ be the canonical generator of $\pi_1(S^1) \approx \mathbb Z$. By the above result $i_* : \pi_1(S^1) \to \pi_1(X_2)$ is an epimorphism whose kernel is generated by $[f_2] = 2g$. Thus $i_*(g)$ is the generator of $\pi_1(X_2) \approx \mathbb Z_2$. The map $if_3 : S^1 \to X_2$ represents the homotopy class $i_*([f_3]) = i_*(3g) = 3i_*(g) = i_*(g)$. But $j_* : \pi_1(X_2) \to \pi_1(X)$ is an epimorphism whose kernel is generated by $i_*(g)$. Therefore $\pi_1(X) = 0$.


答: If $C_1$ and $C_2$ only intersect finitely many times then you can break them both up at each intersection point and use the fact that path integrals are additive with respect to concatenating paths. Then the statement reduces to the non-self-intersecting case. A limiting argument can be used to handle the possibility that they intersect countably many times, I think.

Title: Path Independence of Line Integrals

Body:
The Cauchy-Goursat theorem states that if a function $f$ is analytic on, and in the interior of, a contour $C$ then: $$\int_{C} f(z) \hspace{1mm} dz = 0.$$

From here, a lot of lecture notes deduce the following.

Cor. Suppose $f$ is analytic on some simply connected region $A$. Then for any two points $a, b \in A$ we have that $$\int_{C} f(z) \hspace{1mm} dz$$ is independent of the contour $C$ in $A$ joining $a$ and $b$.

However, the 'proof' often only shows $$\int_{C_1} f(z) \hspace{1mm} dz = \int_{C_2} f(z) \hspace{1mm} dz.$$ when $C_1$ and $C_2$ are non-intersecting (because then $C := C_1\cup C_2$ is a closed contour). What is the best way to prove independence for all $C_1$ and $C_2$? I can 'see' that one can just pick a third contour $C_3$ which doesn't intersect either $C_1$ or $C_2$, and the proof follows from that. But how can I make the existence of $C_3$ rigorous without 'advanced machinery'?


问:
Polynomial $f(X) g(Y)- g(X) f(Y)$ cannot be divided by a non-constant $h(X) \in F[X]$

Let $F$ be a field and $f(X), g(X) \in F[X]$ two coprime
non constant polynomials.
Now the question is how to show that
the symmetric polynomial $P(X,Y):= f(X)  g(Y)- g(X)  f(Y)$
can not be divided by a non constant polynomial
$h(X) \in F[X] \setminus F$ in one variable $X$?

By  this question $P(X,Y)$ need not to be irreducible, so we can probably find divisors in $F[X,Y]$. Now I think that it's not possible to find a divisor in $F[X]$ but I'm not convinced I can show it.

Is there standard approach to show it? I used following
method about which I quite non convinced if this could lead me to
a correct way to prove it: We know that for an arbitrary ring $A$ and ideal $I$ of $A$
there exist a reduction modulo $I$ isomorphism
$A[Y]/I \cdot A[Y] \cong (A/I)[Y]$. For any $f = \sum^n a_i Y^i$
and $a \in A$ this implies that following statements are
equivalent:

$I$ "divides" $f$ in the sense $f \in I \cdot A[Y]$ iff
$a_i \in I$ for every $i $.

Now assume that $A$ is a $F$-algebra (in uor case $A=F[X]$) and non constant
$f(Y),g(Y) \in F[Y]$ are $F$-linear independent.
Let $a, a_f, a_g \in A$.

Is then $a | (a_f f(Y) + a_g g(Y))$ equivalent to
$a | a_f$ & $ a| a_g$? I try think about it as an application
of reduction modulo isomorphism to different basis instead
the standard basis $1, Y, Y^2,...$. Does it make sense?

Otherwise, is there a more conceptional way to prove that $f(X)  g(Y)- g(X)  f(Y)$
can not be divided by a non constant polynomial
$h(X) \in F[X] \backslash F$.

答:
Let $f,g\in F[t]$ be non-constant and suppose $p(x,y):=f(x)g(y)-g(x)f(y)\in F[x,y]$ is divisible by $h(x)$ for some $h\in F[t]$. We will show that $f$ and $g$ share a common irreducible factor, which will show the desired result. Now, $p(x,y)$ is divisible by every factor of $h(x)$, so we may assume that $h$ is irreducible and monic. Embed $F[x,y]$ in $\overline{F}[x,y]$, where $\overline{F}$ is the algebraic closure of $F$, and let $\alpha\in\overline{F}$ be a root of $h$. (Note that, by construction, $h$ is the minimal polynomial of $\alpha$ over $F$.) Then we have $p(\alpha,y)=0$, so $f(\alpha)g(y)=g(\alpha)f(y)$. In particular, since $g(y)$ and $f(y)$ are both non-zero, $f(\alpha)=0$ if and only if $g(\alpha)=0$.

We thus have two cases; first suppose that $f(\alpha)$ and $g(\alpha)$ are both non-zero. Then we have $f(y)=\lambda g(y)$, where $\lambda=f(\alpha)\big/g(\alpha)\in\overline{F}$ is non-zero. In particular, since both $f(y)$ and $g(y)$ lie in $F[y]$, we must have $\lambda\in F^\times$, so $f(t)$ and $g(t)$ are associates and hence share all irreducible factors.

On the other hand, suppose that $f(\alpha)$ and $g(\alpha)$ are both zero. We claim that $h(t)$ is a common factor of both $f(t)$ and $g(t)$. To see this, note that, by the division algorithm, there are $q,r\in F[t]$ such that $f=qh+r$, where $\deg r<\deg h$. Then, embedding $F[t]$ in $\overline{F}[t]$, we have $$r(\alpha)=f(\alpha)-q(\alpha)h(\alpha)=0.$$ Since $h$ is the minimal polynomial of $\alpha$ over $F$, and $\deg r<\deg h$, this means $r$ must be the zero polynomial, whence $f=qh$ in $F[t]$. Exactly the same argument shows that $g$ is divisible by $h$ in $F[t]$, so $f$ and $g$ share a common irreducible factor, as desired.


问:
Unconstrained matrix factorization vs SVD
SVD decomposes a matrix $M$ as the product of two orthogonal matrices $U$ and $V$ (plus a diagonal matrix of singular values $\Sigma$).

$ M = U \Sigma V^*$

In this case, $U$ and $V$ are truncated to the first $n$ singular vectors for compressing $M$.

I'm exploring an alternative unconstrained factorization of $M$ using gradient descent to find $U$ and $V$ with the same dimensions as before:

$ \underset{UV}{\mathrm{argmin}} \: \mathbf{MSE}(M-UV) $

My results indicate that both methods for decomposing $M$ achieve the same mean square error ($\mathbf{MSE}$) score up to a surprising level of accuracy. Obviously, the resulting factorisations are different in each case, where the $U$ and $V$ found with the gradient descent method are not orthogonal.

My questions are:

Given that the second method has no constraints, I'd have supposed it should be able to find better factorisations than SVD -- in a $\mathbf{MSE}$ sense. Is this assumption right?

Is there any property of $M$, such as its rank or dimensions, that can be extracted from this result? The orthogonality constraint is apparently not limiting the SVD factorisation but I'm not sure if this happens in general or just for certain types of $M$.

Any thoughts about this or links to relevant literature would be greatly appreciated.

答:
Given that the second method has no constraints, I'd have supposed it should be able to find better factorisations than SVD -- in a MSE sense. Is this assumption right?

Ben is right about EYT but I'd just note that the EYT says

$$ \| A - A_k \|_2 =  \bigg\|  \sum_{i=k+1}^n  \sigma_i u_i v_i^*\bigg\|_2 = \sigma_{k+1} $$

Uses the 2-norm which is actually the RMSE. Mean square error is $\| \cdot \|_2^2$,no? But that's neither here nor there.

I'm exploring an alternative unconstrained factorization of M using
gradient descent to find U and V with the same dimensions as before

The general way the SVD is done actually is iterative. It uses Golub-Kahan-Bidiagonalization but there is another algorithm by Hestene and Stiefels that works through the gradient.


问:
$B = \{\{x\}: x \in X -\{x_0\}\} \cup \{X\}$ is a base for a topology over $X$.
Let $X$ be a non-empty, fixed set $x_{0} \in X$ arbitrary. Show that $B = \{\{x\}: x \in X -\{x_0\}\} \cup \{X\}$ is a base for a topology over $X$.

Definition. We say that a collection $B$ of openings of $X$ is a base of openings for the topological space $X$ if all openings can be written as union and a subcollection of $B$.

Does anyone know how to prove this fact? doesn't make sense to me, because this base will have only the set $X$. Can someone help me? appreciate.

答:
You can use this elementary lemma:

A collection $B$ of subsets of a set $X$ is a base for a topology on $X$ iff

(i). $\;\cup B=X\;$ (i.e. each $x\in X$ belongs to at least one $b\in B$ ), and

(ii). If $b,b'\in B$ and $x\in b\cap b'$ then there exists $b''\in B$ with $x\in b''\subseteq b\cap b'.$


问:
How to calculate the range [0,1] of a line section in ratio to the total line length?
(Disclaimer: I am a programmer, not great in maths, so sorry if I don't explain my question in proper mathematical terms. Moreover English is not my first language, which makes it even harder for me to explain. But I will try.)

I have a line with a start point $S=0$ and an end point $E$. That end point can have any positive number, e.g. 100 or 256. Since the length of this line (segment betweeen $S=0$ and $E$) can change, we always represent that segment with $0.0$ at point $S$ and $1.0$ at point $E$. So in other words: regardless of the real length of the segment between $S$ and $E$, we can always represent that line in the range $[0,1]$.

As an example: If $E=256$ (equals to a segment length of 256 as well), half of that length is 128, so this would be at 0.5 in our "uniformed" representation with range $[0,1]$. A point at 64 in this example would be at 0.25, and so on. These are the blue numbers in the picture above.

Then there is a a sub section of that line segment, and I know the start ($A$) and end ($B$) points of that sub section as well. They are always within the line segment, i.e. both $A$ and $B$ are between $S$ and $E$. In our example, let's say $A=45$ and $B=128$.

Now I also want to have its own range $[0,1]$ for this sub section, as shown in orange above. So regardless of the line segment length between $S$ and $E$ and also regardless of the sub section length between $A$ and $B$, I want to "map" the "outher" range $[0,1]$ (blue) to the "inner" range $[0,1]$ (orange). So in our example $B$ would be at $0.5$ of the outher range (blue), which corresponds to $1.0$ (orange) for the inner range.

I hope that was understandable. I do have $S$, $E$, $A$, $B$ and a number in the range $[0,1]$ from the outher range. How can I calculate the corresponding value of the inner range for that given number from the outher range?

答:
You're looking for a linear function $x_{outer} = u x_{inner} + v$, parameterized by $u$ and $v$, that transforms the inner points A and B to the outer points S and E, respectively.

We have:

$S (= 0) = uA + v$ and $E = uB + v$

By solving the system for $u$ and $v$, you get $x_{outer} = \frac{x_{inner} - A}{B - A} E$.


问:
Why the terms "unit" and "irreducible"?
I'm trying to understand why in a ring we choose the names unit to an invertible element and irreducible element in this definition

Maybe historical reasons?

For example, I suppose the second definition are named as prime elements because of the analogy to prime numbers.

Thanks in advance.

答:
For example, in $\Bbb Z$, the units are $\pm 1$, and the primes and irreducibles coincide. Moreover, it has unique factorization, and unique here means 'up to a unit', e.g. $-10=(-1)\cdot 2\cdot 5$, but $-1$ doesn't count here, as it is only a unit. Among the Gaussian integers, $\Bbb Z[i]$ also $\pm i$ are units.

It is easy to see that if $u$ is a unit (invertible), then for any element $a$, the generated ideals $(a)$ and $(ua)$ coincide.

Title: Why the terms "unit" and "irreducible"?


问:
Is there a formal name for this recursive sequence?
Suppose I have two numbers, A,B. Suppose I am allowed to add them together, but can only add one at a time. So, for example:

```<code>Seed Case: (A,    B)
Step 1:    (A+B,  B)  or  (A,   A+B)
Step 2:    (A+2B, B)  or  (A+B, A+2B)  or  (2A+B,  B+A)  or  (A, B+2A)
Step 3:    (A+3B, B)  or  (A+B, 2A+3B) or  (2A+3B, A+2B) or ...
... etc
</code>```

I am wondering if anyone knows of the formal name of this sequence, as I am trying to find information on its characteristics. Specifically, I am interested in the minimum distance between the base case and a given term (in terms of intermediate additions which must occur). But I don't know where to start researching it.

答:
I'll write $a$ and $b$ in lowercase. Let $X = \left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$ and $Y = \left[ \begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array} \right]$. Each expression you describe here can be obtained by starting from the vector $v = \left[ \begin{array}{cc} a \\ b \end{array} \right]$ and repeatedly applying $X$ or $Y$, since

$$Xv = \left[ \begin{array}{cc} a + b \\ b \end{array} \right], Yv = \left[ \begin{array}{cc} a \\ a + b \end{array} \right].$$

So the terms in Step 2 correspond to $X^2 v, XY v, YX v, Y^2 v$, for example. In general you are looking at the monoid generated by these two matrices, and you want to compute the length of an element of this monoid (the shortest word in $X$ and $Y$ that produces it).

This is actually quite easy, for the following reason: if you start from an expression $(n_1 a + m_1 b, n_2 a + m_2 b)$ (equivalently, a matrix $\left[ \begin{array}{cc} n_1 & m_1 \\ n_2 & m_2 \end{array} \right]$) and want to work backwards to the base $(a, b)$, given that this is possible there is a unique way to do it at each step, because exactly one of the two expressions can be subtracted from the other and not vice versa. For example, given $(a + b, 2a + 3b)$ in Step 3,

the first term can be subtracted from the second but not vice versa, so the previous step must be $(a + b, a + 2b)$. Then,

again the first term can be subtracted from the second but not vice versa, so the previous step must be $(a + b, b)$. Then,

the second term can be subtracted from the first but not vice versa, so the previous step must be $(a, b)$ and we're done.

In other words, the sequence of $X$s and $Y$s generating a given expression is unique; this says that $X$ and $Y$ generate the free monoid on two generators.

When performing these reductions the coefficients $n_i, m_i$ are being subjected to a slight variation of the Euclidean algorithm so that's a place you can look for bounds on the total number of steps; generally speaking they will be logarithmic although there'll be edge cases, e.g. $(a + nb, b)$ takes $n$ steps. You get basically the Euclidean algorithm if you decide that adding $n$ copies of one of the numbers to the other counts as one step and this simplifies the counting of the total number of steps.


问:
Is local circularity an aspect of differentiable functions?
I understand that differentiable functions approach linearity at very small scales. But since they never really lose their curvature (excluding linear functions), I was wondering whether their curvature approaches circular arcs at very small scales. It seems that it should not be true, but is there a simple way to prove or disprove this conjecture? I got to thinking about this while considering the mean value theorem and whether the point c approaches the midpoint of the interval as the interval size goes to zero, which I suppose is true for a circular arc.</p>

答: If you are looking at the graph of 2nd order differentiable functions, you can always find an osculating circle. (The flat case is just one with infinite radius).</p>


答:
A straight line doesn't have a corresponding circle... If you are going around a figure 8 track, most of the time you are circulating around one of two centers, but there is a transition point where you are not circulating anything.

Frequently, we can talk about a "kissing circle" to a curve and the radius of curvature.  But, that radius may not be always be defined as in the examples mentioned.

Title: Is local circularity an aspect of differentiable functions?

Body: I understand that differentiable functions approach linearity at very small scales. But since they never really lose their curvature (excluding linear functions), I was wondering whether their curvature approaches circular arcs at very small scales. It seems that it should not be true, but is there a simple way to prove or disprove this conjecture? I got to thinking about this while considering the mean value theorem and whether the point c approaches the midpoint of the interval as the interval size goes to zero, which I suppose is true for a circular arc.</p>


问:
Prove that integral converges
How can I prove that this integral converges?

$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$

I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.

Are there better methods?

答:
First of all $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\frac{\ln{x}}{\left(x-1\right)\sqrt{x+1}}\cdot\frac{1}{\sqrt{x-1}}\underset{x\to 1}{\sim}\frac{1}{\sqrt{2}\sqrt{x-1}} $, and we know that $ \int_{1}^{a}{\frac{\mathrm{d}x}{\sqrt{x-1}}} $ converges for any $ a>1 $.

We also have that $ \frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{\sqrt{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\right)=\underset{\overset{x\to +\infty}{}}{\scriptsize\mathcal{O}\normalsize}\left(\frac{1}{x\sqrt{x}}\right) $, and $ \int_{b}^{+\infty}{\frac{\mathrm{d}x}{x\sqrt{x}}} $ converges for any $ b>1 $.

Thus our integral $ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x} $ converges.

Let's use a substitution $ \left\lbrace\begin{matrix}x=\sec{y}\\ \sec{y}\,\mathrm{d}y=\frac{\mathrm{d}x}{\sqrt{x^{2}-1}}\ \ \ \ \ \ \ \end{matrix}\right. $, we get : \begin{aligned}\int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\int_{0}^{\frac{\pi}{2}}{\frac{\sec{y}\ln{\left(\sec{y}\right)}}{\sec{y}-1}\,\mathrm{d}y}\\ &=-\int_{0}^{\frac{\pi}{2}}{\frac{\ln{\left(\cos{y}\right)}}{1-\cos{y}}\,\mathrm{d}y}\\ &=\left[\left(\cot{\left(\frac{y}{2}\right)}-1\right)\ln{\left(\cos{y}\right)}\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=\int_{0}^{\frac{\pi}{2}}{\left(\cot{\left(\frac{x}{2}\right)}-1\right)\tan{x}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{x}}{\sin{x}+\cos{x}}\,\mathrm{d}x}\\ &=4\int_{0}^{\frac{\pi}{4}}{\frac{\cos{\left(\frac{\pi}{4}-x\right)}}{\cos{\left(\frac{\pi}{4}-x\right)}+\sin{\left(\frac{\pi}{4}-x\right)}}\,\mathrm{d}y}\\ &=2\int_{0}^{\frac{\pi}{4}}{\left(1+\tan{x}\right)\mathrm{d}x}\\ &=2\left[x-\ln{\left(\cos{x}\right)}\right]_{0}^{\frac{\pi}{4}}\\ \int_{1}^{+\infty}{\frac{\ln{x}}{\left(x-1\right)\sqrt{x^{2}-1}}\,\mathrm{d}x}&=\frac{\pi}{2}+\ln{2}\end{aligned}

Title: Prove that integral converges


问:
Prove that $5^n \equiv 8n^2-4n+1 \mod 64$ using induction
I am trying to prove that $5^n \equiv 8n^2-4n+1 \mod 64 $ using induction. I have seen the post which solves this problem using other methods, but I really need to be able to do it using induction.

Here is what I have so far:

Base step:
$5 \equiv 5 \mod 64$

Then, I assumed that $5^k \equiv 8k^2 -4k+1 \mod 64.$

I want to prove that $5^{k+1} \equiv 8(k+1)^2-4(k+1)+1 \mod 64$.

Note that $8(k+1)^2-4(k+1)+1=8k^2+12k+5$.

Hence,
$5^{k+1} \equiv 5^k *5 \mod 64
\equiv (8k^2-4k+1)*5 \mod 64 \equiv 40k^2 -20k+5 \mod 64.$

I am stuck here.
Any insight would be very useful!!
Thanks

Body: Hint: $5(8n^2-4n+1)-(8(n+1)^2-4(n+1)+1)= 32 n (n - 1) \equiv 0 \bmod 64$


问:
finding all possible. numbers
When a number is divided by $11$,the obtained number equals the sum of the squares of the digits of the starting number. If the starting number has three digits, find all such numbers.

Detailed solution please.

Title: finding all possible. numbers

Body:
Hint: You can utilize the fact that $11 | \overline{abc} \implies 11 | (a-b+c)$

Since $-8 = 1-9+0\le a-b+c\le 9-0+9 = 18$, either $a-b+c=0$ or $a-b+c=11$.

Case 1: $a+b-c=0 \implies a^2+(a+c)^2+c^2 = (100a+10b+c)/11\\=(100a+10(a+c)+c)/11=10a+c \\\implies 2a^2+(2c-10)a+2c^2-c=0\\
\implies \Delta = (2c-10)^2 -  8(2c^2-c) \ge 0\\
\implies 3c^2+8c-25 \le 0 \implies c \le (\sqrt{91}-4)/3 < 2$

Because $c$ is even, then $c=0, a=5, b=5, \overline{abc}=550.$

Case 2: $a+b-c=11 \implies a^2+(a+c-11)^2+c^2 = (100a+10b+c)/11\\=(100a+10(a+c-11)+c)/11=10a+c-10 \\\implies 2a^2+(2c-32)a+2c^2-23c+131=0\\
\implies \Delta = (2c-32)^2 - 8(2c^2-23c+131) \ge 0\\
\implies 3c^2-14c+6 \le 0 \implies c \le (\sqrt{31}+7)/3 < 5$

Because $c$ is odd, $c=1$ (no integer solution) or $c=3, a=8, b=0, \overline{abc}=803.$