问:
The maximum of (Minimum $\times$ Cardinality) Over all Subsets
We are given a set of positive reals $S$ and also a subset $S'$ of the $k$ largest numbers in $S$ such that $\min(S') \times k$ is maximum,
where $\min(S')$ is the minimum number in $S'$.

Let $sum(S)$ be the sum of numbers in $S$.
What can we say about the ratio $R=\frac{\min(S') \times k}{sum(S)}$?

Can we claim $R$ is lower bounded by a constant?

First Attempt

Choose $S'$ as the subset of numbers in $S$ at least half the average in $S$.

Here are some examples.

For $S = \{ 1,1,1,5 \}$, half of average is 1 and we get $S'=S$ and $R=4/8 = 1/2$.

For $S = \{ 1,1,5 \}$, we get $S'=\{ 5 \}$ and $R=5/7 > 1/2$.

For $S = \{ 1,1,4 \}$, we get $S'=S$ and $R=3/6 = 1/2$.

Is it always $R \geq 1/2$?

答:
No, $R$ can get smaller than any given positive value.

Consider

$$S_n=\{1, \frac1{2\times 2}, \frac1{2\times 3}, \ldots , \frac1{2\times n}\}.$$

The $k$-th maximum value equals $1$ for $k=1$, otherwise it is $\frac1{2k}$, so $k\times \min(S')$ is $1$ for $k=1$ and $\frac12$ for $k>1$. That means the extremal value is taken for $k=1$ and $S'=\{1\}.$ That means for $S_n$ the value of $R$ is

$$R=\frac{1}{1+\sum_{i=2}^n\frac1{2i}},$$

and since the denominator is basically half the partial sum of the harmonic series up to $n$, it can be made arbitrary large for large $n$, so $R$ gets arbitrary small.


问:
Contractivity of a projection on $C(\mathbb{T}^2)$
Consider the $C^\star$ algebra of all complex-valued continuous functions on the 2-torus $\mathbb{T}^2$ with the sup norm. Let us consider the set $\{z^m w^n\}_{m,n\in \mathbb{Z}}$, whose span is dense in $C(\mathbb{T}^2)$. We define the projection $\phi: C(\mathbb{T}^2) \to C(\mathbb{T}^2)$, thus:

$z^m w^n \mapsto z^m w^n$ if $mn\geq 0$

$z^m w^n \mapsto 0$ if $mn < 0$

and extend it linearly on the closure.

The question is, is this map contractive?

I really don't know what would be a way to attack this problem directly. I have tried a few indirect approaches using positivity, without much success: since the map is unital, it is contractive iff it is positive. Positive elements of $C(\mathbb{T}^2)$ are not very easy to characterize, due to the absence of a Fejer-Riesz theorem in 2 variables, but still, one may write a strictly positive trigonometric polynomial in $(z,w)$ as a finite sum of absolute squares of polynomials in $(z,w)$. Annihilating the terms with opposite signs for the powers of $z$ and $w$, it is not direct to see if it is positive...

As for my background: I am familiar with basic complex analysis, the basics of $C^\star$ algebras and the theory of positive maps.

答:
Your map is not continuous, much less contractive (so there is no way to extend it by continuity beyond the trigonometric polynomials) for the following reason.  Consider the embedding of $C(\mathbb T)$ into $C(\mathbb T^2)$ in such a way that $z^m$ goes to $z^mw$.

If your map is bounded, it's restriction to the above copy of $C(\mathbb T)$ would send $z^m$ to itself if $m\geq 0$, and to zero otherwise, and it is well known that this map doesn't exist on $C(\mathbb T)$.

ContentLicense: CC BY-SA 4.0


问:
show function satisfies Lipschitz condition on a domain
I have been given the Initial Value problem

$\frac{dy}{dx} = y^2$, $y(1)=-1$

Let $f(x,y)=y^2$.

I have defined the domain R: $|x-1|\leq a$, $|y+1|\leq b$ where $a,b>0$

I am trying to show that Lipschitz condition is satisfied on R but stuck on trying to finding a value of K. I am stuck after the following steps:

$|f(x,y_1)-f(x,y_2)|=|y_1^2-y_2^2|=|y_1+y_2||y_1-y_2|$

What value of K can I take so that $|f(x,y_1)-f(x,y_2)|\leq K|y_1-y_2|$ and hence satisfying Lipschitz condition on R?

答:
In general, the Lipschitz constant can be found by taking the maximum absolute value of the derivative (this follows from the mean value theorem) over the domain you are looking at. In your case, this means you would only need to find a bound on $|2y|$ for $y \in [-b-1, b-1]$ which should be straightforward enough.

This also coincides with a bound that you could find on $|y_1 + y_2|$, as you can easily check.


问:
continuous determination of the argument
Let $U=\{z=|z|e^{i\theta} :\theta \in (-\pi,\pi)\}$ and  $V=\{z\in \mathbb C : -\pi< \Im(z) <\pi \}$, and $$\log_{\pi}:U \to V  : z=\rho e^{i\theta} \mapsto \ln(\rho)+i\theta$$ the branch of complex logarithm,
with $z=x+iy$ and $$ \theta=
\begin{cases}
\arctan(y/x), & \text{if } x>0\\
\pi+\arctan(y/x), & \text{if } x<0,y>0\\
-\pi+\arctan(y/x), & \text{if } x<0,y<0
\end{cases}$$

I don't know where this values of  $\theta$ comes from. I inderstand the first case where $x>0$ but the I didn't understand the other two cases ($x<0,y>0 $ and $x<0,y<0$) ! So any help is highly appreciated!

答:
This may help:
$$ \theta=
\begin{cases}
\arctan(y/x), & \text{if } (x,y)\in Q_1\cup Q_4\\
\pi+\arctan(y/x), & \text{if } (x,y)\in Q_2\\
-\pi+\arctan(y/x), & \text{if } (x,y)\in Q_3
\end{cases}$$
Since $\text{arctan}$ has range $(-\pi/2,\pi/2)$, in order to recover $\theta$ in $Q_2$ or $Q_3$ you need to shift it by $\pi$.  And although it doesn't really matter which way you shift it (plus or minus), they have chosen to do it so that $\theta$ is expressed in the interval $(-\pi,\pi)$.


问:
Coherent sheaves vs sheaves locally of finite presentation
I think it is clear for most people that finitely generated and finitely presented modules are important and, in a certain sense, the "right" notions. However, the corresponding notions for sheaves are not clearly the right notions (at least to me).

For completeness, let me define some relevant notions. Let $X$ be a ringed space.

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be quasi-coherent if every point of $X$ has a neighborhood $U$ over which there is an exact sequence $$\mathscr{O}_X^{\oplus I}|_U\to \mathscr{O}_X^{\oplus J}|_U\to \mathscr{F}|_U\to 0.$$

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be locally of finite presentation if it is quasi-coherent and the indices $I$ and $J$ as above can be taken to be finite.

A sheaf $\mathscr{F}$ of $\mathscr{O}_X$-modules is said to be coherent if every point of $X$ has a neighborhood $U$ over which there is an exact sequence $ \mathscr{O}_X^{\oplus m}|_U\to \mathscr{F}|_U\to 0$ (that is, $\mathscr{F}$ is finitely generated) and for any open set $V\subset X$, any natural number $n$, and any morphism $\varphi:\mathscr{O}_X^{\oplus n}|_V\to \mathscr{F}|_V$, the kernel of $\varphi$ is finitely generated.

It is very clear to me that quasi-coherentness is a relevant notion. But I don't understand why coherent is a more usual notion than locally of finite presentation, which seems more natural to be.

In https://mathoverflow.net/questions/68125/qcoh-algebraic-stack and in Stalks of the sheaf $\mathscr{H}om$? A. Vistoli and M. Brandenburg seem to imply that the "correct" notion is indeed that of local finite presentation.

I would like to understand what are the differences between those notions. For example, over an arbitrary ringed space the category of coherent sheaves is abelian. Is the same true about locally of finite presentation? What are other advantages or problems of either notions?

答:
In order to get a sense of the differences between locally of finite presentation and coherent, it's first instructive to see when they're the same. On any locally noetherian scheme $X$, the following conditions on an $\mathcal{O}_X$-module $\mathcal{F}$ are equivalent (cf. Stacks 01XZ):

$\mathcal{F}$ is coherent,

$\mathcal{F}$ is quasi-coherent and of finite type (aka $J$ can be chosen to be finite in the exact sequence defining quasi-coherence)

$\mathcal{F}$ is locally finitely presented (aka $I$ and $J$ can be chosen to be finite in the exact sequence defining quasi-coherence).

This is just fine for lots of algebraic geometry, and it's very easy to conflate "$\mathcal{F}$ is coherent" with "$\mathcal{F}$ is quasi-coherent and $I,J$ can be taken to be finite" because they're exactly the same in this relatively nice and extremely common context. The differences start showing up when we leave the realm of locally noetherian schemes.

In this setting, there can be a strict inclusion of coherent sheaves in to locally finitely presentable sheaves: simply take any scheme on which the structure sheaf is not coherent. One example is $\operatorname{Spec} R$, where $R$ is the ring of germs of $C^\infty$ functions on $\Bbb R$ at $0$ (details here). This example also shows that locally finitely presented sheaves don't form an abelian category in general: the kernel of a morphism of locally finitely presented sheaves can fail to be locally finitely presented.

Now that we've established where these concepts start separating a bit from each other, we can start to get a sense of when and why one might prefer one over the other. The key tradeoff with coherent sheaves is that the finiteness condition of $\ker(\mathcal{O}_X^n|_U\to \mathcal{F}|_U)$ being finite type for any $U$ and any morphism $\mathcal{O}_X^n|_U\to\mathcal{F}|_U$ is very strong - it's strong enough to make for a good theory on any ringed space ($Coh(X)$ is an abelian category for any ringed space $X$), but if your space $X$ is "big" enough, coherent sheaves can fail to include the structure sheaf as seen above. Coherent sheaves also don't necessarily play well with pullback - in general, if $f:Y\to X$ is a map of ringed spaces and $\mathcal{F}$ is a coherent sheaf on $X$, there's no reason for $f^*\mathcal{F}$ to be coherent on $Y$, which is a bit of a disappointment to folks who enjoy functors.

Sheaves locally of finite presentation (and quasi-coherent sheaves more generally), on the other hand, always transform properly under pullback, since the pullback of the structure sheaf is the structure sheaf and pullback is right exact. Combining this with the fact that a fair number of nice theorems we know for coherent sheaves on schemes in the locally noetherian case admit generalizations to locally finitely presented sheaves in the non-noetherian case (for instance, the sheaf hom result you link), it is useful to relax our finiteness condition a little bit from coherence to locally finite presentation.

It may be interesting to you to know a little bit of the history around coherent sheaves to understand why they're set up the way they are. Sheaves were introduced by Jean Leray in 1946, expositing work he did while imprisoned as a POW during World War II. Cartan and Oka were major players in introducing sheaves to complex-analytic geometry in the next few years, proving Cartan's theorem A and B and Oka's coherence theorem (circa 1950-51). In 1955, Serre published FAC, introducing sheaves in to algebraic geometry, and a scant few years later (~1957), Grothendieck introduced schemes. So coherence as a condition was originally developed in the analytic framework, where the extra strength of the finiteness condition was useful.


问:
What is the number of antisymmetric relations in a set where the relations of some elements are given?

Given the set $X=\{x_1,x_2,x_3,...x_{10}\}$ find the number of antisymmetric relations in the set if $x_3\mathrel{R}x_7$, $x_7 \mathrel{R} x_9$ but $x_9 \not \mathrel{R} x_1$.

I wasn't sure if i got this right so i wanted to check. The total number of antisymmetric relation in a set with $n$ elements is calculated to be $2^n3^{\frac{n^2-n}{2}}$. If $x_3\mathrel{R}x_7$, then we are left with only one choice for $x_7$, to not be in relation with $x_1$, so we lose 3 possibilities that we had in the first place. The same thing is true for $x_7 \mathrel{R} x_9$, but for $x_9 \not \mathrel{R} x_1$, we have a possibility that $x_1$ is in relation with $x_9$ and that $x_1$ isn't in relation to $x_9$, so we have 2 possibilities instead of 3 on that field.

The total would then be: $2^{n+1}3^{\frac{n^2-n}{2}-3}$ which would make the answer for my problem $2^{11}3^{42}$.

Is this correct?

答:
Your answer is correct; good job!

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Combinations in board game, l:=landscapes, q:=folks and k:= Player
Combinations in board game, l:=landscapes, q:=folks and k:= Player
So I have struggle with this one and didn't find a similar situation. Do you have any clue on the way to solve this?

The game has l differend landscapes. In each lanscapes are q folks. At the start every player chooses a folk, where landscapes are unique.

How many different start configurations are there with k players if:

-Only the combination of folks is relevant?

Here I already have struggle because I didn't found a similar situation with a solution.

With l Landscapes and q folks I would have (l choose q) ? But it seems like each Landscape has it's own q folks.

And how do I now take the player k in account?

I guess I have to divide the space of combinations by k? Right?

My try is: $\frac{l!*q}{k}$

-If it's relevant which player plays which folk

And here multiplying by k! ?

Thanks for any help to get a better understanding

答:
Assuming that for $\ell$ landsacpes, each containing $q$ folk, each of the $k$ players choses a folk within a landscape but all players choose different landscapes...

Since the first question is looking for the options of folk choice, without distinguishing which player chose which, we can pick landscapes with $\dbinom \ell k$ and then the choice of folk in each landscape is entirely independent, so $q^k$ options once the landscapes are chosen. Together: $\dbinom \ell k q^k$

Finally if the allocation of players to folk is discriminated, we can (as you say) permute the players across each anonymous selection with an additional factor of $k!$.


问:
Is the ideal $I := (7X+14, X^3+2X^2+1) \subseteq \mathbb Z[X]$ prime? Is it maximal?
The question is fully contained in the title.

I tried to prove maximality (if that happens, $I$ is prime as well) in $\mathbb Z[X]$, but I am not able to figure a strategy out for that purpouse. Obviously, if $I$ is not maximal, I am expected to say whether $I$ is a prime ideal, which is a problem too.

How would you solve such exercise?

答:
Let's just play around and see what kinds of polynomials we can find in $I$.

First of all, we can try to cancel the cubic term from the cubic generator, and we see that
$$
7(X^3+2X^2+1)-X^2(7X+14)=7
$$
is an element of the ideal. And since $7X+14=7(X+2)$, we get
$$
I=(7,X^3+2X^2+1)
$$
which is a lot simpler.

Now what? Well, since we have a constant in our ideal, it is apt to use the third isomorphism theorem. It states that we can divide out by one generator at a time (presumably you know all the basic connections between ideals and the corresponding quotients).

So if we're interested in whether $(7,X^3+2X+1)\subseteq \Bbb Z[X]$ is maximal or prime, we might just as well look at $(X^3+2X^2+1)\subseteq \Bbb Z[X]/(7)$. And therefore we can just check whether $X^3+2X+1$ has any roots modulo 7. If it does have roots the ideal isn't prime, and if it doesn't have roots, then the ideal is prime and also maximal (because $\Bbb Z[X]/(7,X^3+2X+1)$ would then be a finite integral domain and therefore a field, or alternatively because the quotient is a degree 3 extension of the field $\Bbb Z_7$).


答:
Do you understand that, because the tangent function has period $\pi$ (rather than $2\pi$ like sine or cosine) the arctan function can only give values from $-\pi/2$ to $\pi/2$, the first and fourth quadrants.

For example, if $x= -3$ and $y= 3$, in the second quadrant, then $\arctan(y/x)= \arctan(-1)= -\pi/4$ which is in the fourth quadrant, not the second.  You have to add $\pi$ to move it to the second quadrant.  Similarly, if $(x, y)= (-3, -3)$, in the third quadrant, the $y/x= -3/-3= 1$ exactly as if it were $(3, 3)$!  The arctangent is $\pi/4$ in the first quadrant.  Adding $\pi$ moves it to the third quadrant.

Title: continuous determination of the argument

Body:
Let $U=\{z=|z|e^{i\theta} :\theta \in (-\pi,\pi)\}$ and  $V=\{z\in \mathbb C : -\pi< \Im(z) <\pi \}$, and $$\log_{\pi}:U \to V  : z=\rho e^{i\theta} \mapsto \ln(\rho)+i\theta$$ the branch of complex logarithm,
with $z=x+iy$ and $$ \theta=
\begin{cases}
\arctan(y/x), & \text{if } x>0\\
\pi+\arctan(y/x), & \text{if } x<0,y>0\\
-\pi+\arctan(y/x), & \text{if } x<0,y<0
\end{cases}$$

I don't know where this values of  $\theta$ comes from. I inderstand the first case where $x>0$ but the I didn't understand the other two cases ($x<0,y>0 $ and $x<0,y<0$) ! So any help is highly appreciated!


问:
$\cos x − 2\sin x = 1$. How to solve using $R\cos(x+a)$?
I have this question: $\cos x − 2\sin x = 1$. You have to solve for $x$, the answer being 4.069. I've tried answering but I just keep getting 2.214. This is my method:

$\cos x − 2\sin x = 1$

$R\cos(x+a)= 1$

$R = \sqrt{1^2 + (-2)^2}$

$a = \arctan(-2/1)$

therefore $\sqrt5 \cos(x - 1.107) = 1$

$\cos(x - 1.107) = \sqrt5/5$

$\arccos(\sqrt5/5) = x - 1.707$

$1.707 = x - 1.707$

therefore $x = 2.214$

Can anyone give me a hint where I'm going wrong? Thank you

答:
$$a=\arctan\left(\frac{2}{1}\right)$$
not
$$\arctan\left(\frac{-2}{1}\right)$$
This is because you are using $R\cos(x+a)$ as opposed to $R\cos(x-a)$.

If you need any more help, please don't hesitate to ask.


问:
An application of Poincaré separation theorem
Let $A\in\mathbb{R}^{N\times N}$ be a square, positive-definite, symmetric matrix. Let

$$P = I - \mathbf{v}\mathbf{v}^T$$

where $\mathbf{v}$ is a unit vector (of norm 1). Then $P$ is a projection matrix, projecting into the sub-space orthogonal to $\mathbf{v}$.

Let $\alpha_1,\alpha_2,...,\alpha_N$ be the eigenvalues of $A$, and let $\rho_1,\rho_2,\dots,\rho_N$ be the eigenvalues of $PAP$. Assume these eigenvalues are sorted in descending order,

$$\alpha_1 \ge \alpha_2 \ge \dots \ge \alpha_N$$

$$\rho_1 \ge \rho_2 \ge \dots \ge \rho_N$$

Since $PAP$ is not full rank, $\rho_N=0$.

My question is whether Poincaré's separation applies in this setting, i.e., whether:

$$\alpha_1 \ge \rho_1 \ge \alpha_2 \ge \rho_2 \ge \dots \ge \rho_{N-1} \ge \alpha_N$$

I have checked numerically that the above chain of inequalities holds, for random $A,\mathbf{v}$.

Can someone help me prove it?

答:
$P=QQ^T$ where $Q^TQ = I_{n-1}$

Then $PAP = QQ^T AQQ^T = Q\big(Q^T AQQ^T\big)$ has the same non-zero eigenvalues as  $\big(Q^T AQQ^T\big)Q=Q^T AQ\big(Q^TQ\big) =Q^T AQI_{n-1} = Q^T AQ$

$\alpha_1 \ge \rho_1 \ge \alpha_2 \ge \rho_2 \ge \dots \ge \rho_{N-1} \ge \alpha_N\gt 0$

then holds by Cauchy Eigenvalue Interlacing


答:
Hint: $I$ contains $\;7(X^3+2X^2+1)-X^2(7X+14)=7$, hence
$$\mathbf Z[X]/I\simeq \mathbf Z/7 \mathbf Z[X]/(I/7 \mathbf Z[X])= \mathbf Z/7 \mathbf Z[X]/(X^3+\bar 2X^2+\bar 1).$$
Can you show that $X^3+\bar 2X^2+\bar 1$ is irreducible in $ \mathbf Z/7 \mathbf Z[X]$?

Title: Is the ideal $I := (7X+14, X^3+2X^2+1) \subseteq \mathbb Z[X]$ prime? Is it maximal?

Body:
The question is fully contained in the title.

I tried to prove maximality (if that happens, $I$ is prime as well) in $\mathbb Z[X]$, but I am not able to figure a strategy out for that purpouse. Obviously, if $I$ is not maximal, I am expected to say whether $I$ is a prime ideal, which is a problem too.

How would you solve such exercise?


问:
How many numbers can you get by multiplying two or more distinct members of the set $\{1,2,3,5,11\}$ together?
Apparently the answer is 15,  but I got 26.

My process:

The set $\{1,2,3,5,11\}$ has five numbers.

Ways to choose two members: $_5C_2$

Ways to choose three members: $_5C_3$

Ways to choose four members: $_5C_4$

Ways to choose five members: $_5C_5$

Total: $_5C_2 + {}_5C_3 + {}_5C_4 + {}_5C_5 = 26$

Is there a detail I have misunderstood in this question?

答:
Note that, because two or more members can be multiplied, multiplying by $1$ will only make a difference if it is one of two numbers. Thus, multiplying by $1$ adds four potential numbers.

Now, we only need to consider the number of combinations that can be made from $2$, $3$, $5$, and $11$.

Choosing two from this set offers six possiblities: $2 \cdot 3$, $2 \cdot 5$, $2 \cdot 11$, $3 \cdot 5$, $3 \cdot 11$, and $5 \cdot 11$.

Choosing three offers four possibilities: $2 \cdot 3 \cdot 5$, $2 \cdot 3 \cdot 11$, $2 \cdot 5 \cdot 11$, and $3 \cdot 5 \cdot 11$.

Finally, there is one possibility with four chosen: $2 \cdot 3 \cdot 5 \cdot 11$. Thus, there are $4 + 6 + 4 + 1 = \boxed{15}$.

ContentLicense: CC BY-SA 3.0


问:
Is $\lim (\text{max}\{ f \}) = \text{max}\{(\lim f)\}$?
I'm trying to prove some statement from my differential equations class, and for some part of the proof I used that, being each $f_n(x,y)$ a continuous function defined over a compact set $\Omega$,

$$ \lim_{n \rightarrow \infty} \text{max}\{|f_n(x,y)| : (x,y)\in\Omega\} = \text{max}\{|\lim_{n\rightarrow\infty} f_n(x,y)|: (x,y)\in\Omega\ \}.$$
I thought this is true, but I don't know how to justify it, or if it's only true when $f$ is continuous or anything. Is this true in general? If not, under what conditions is it true? Any help will be appreciated, thanks in advance.

答:
Based on the useful suggestion of Zwim, I propose here a proof for your identity under the uniform convergence assumption. I.e.

Theorem. Assume that

$\Omega$ is compact

Continuous functions $f_n$ uniformly converges to continuous function $f$, i.e. $$\sup_{x\in \Omega} |f_n(x)-f(x)| \xrightarrow{n\rightarrow +\infty} 0.$$

Let $x^*_n$ and $x^*$ be the maximizer of $f_n$ and $f$, respectively
(they are well defined since $f_n$ and $f$ are continuous and $\Omega$ is
compact). Then $$\lim_n f_n(x^*_n)= f(x^*)$$

Proof.

First, notice that $x^*_n$ is a miximizer of $f_n$ and $f_n$ converges to $f$ uniformly, we obtain
\begin{align}
\lim_n f_n(x^*_n) \geq \lim_n f_n(x^*)= f(x^*).
\end{align}

Second, as $x^*$ is a minimizer of $f$ we obtain
\begin{align}
f(x^*) & \geq f(x^*_n) \Rightarrow f(x^*) \geq \lim_n f(x^*_n)= \lim_n f_n(x^*_n).
\end{align}
Here the last equality is a consequence of the uniform convergence
\begin{align}
|f(x^*_n)-f_n(x^*_n)|\leq \sup_{x\in \Omega} |f(x)-f_n(x)| \xrightarrow{n\rightarrow +\infty} 0
\end{align}

The proof is completed.


问:
Determinant of matrix with power series terms
I'm trying to find an alternative to the Taylor series which represents some function $f(x)$ as a sum of exponentials.

$$f(x) = a_0 e^{0x} + a_1 e^{1x} + a_2 e^{2x} + a_3 e^{3x} + ...$$

The coefficients $a_0, a_1, a_2...$ can be found by equating the derivatives:
$$a_0 + a_1 + a_2 + a_3 + ... = f(0)$$
$$0 + a_1 + 2a_2 + 3a_3 + ... = f'(0)$$
$$0 + a_1 + 4a_2 + 9a_3 + ... = f''(0)$$

In general you can write out $n$ equations to solve for $n$ coefficients. You can solve for the coefficients by solving the following matrix equation $[M] a = f$

$$\begin{bmatrix}
1 & 1 & 1 & 1 & .. & 1\\
0 & 1 & 2 & 3 & .. & (n-1)^{n-4}\\
0 & 1 & 4 & 9 & .. & (n-1)^{n-3}\\
0 & ..  & .. & .. & .. & (n-1)^{n-2}\\
0 & 1 & (n-4)^{n-1} & (n-3)^{n-1} & (n-2)^{n-1} & (n-1)^{n-1}
\end{bmatrix} \begin{pmatrix}
a_0\\
a_1\\
a_2\\
..\\
a_{n-1}
\end{pmatrix} = \begin{pmatrix}
f(0)\\
f'(0)\\
f''(0)\\
..\\
f^{(n-1)}(0)
\end{pmatrix}$$

Now to find all coefficients I need to solve $[M]^{-1}f$ as $n$ approaches infinity. But how can I find and simplify the inverse of this matrix? Can someone help me find the closed analytical expression for the i'th coefficient $a_i$?

答:
Note that if $u=e^x$ then $(f\circ\ln)(u)=f(x)=\sum_{n=0}^\infty a_ne^{nx}=\sum_{n=0}^\infty a_nu^n$.  Hence, provided $f\circ\ln$ has a $C^\infty$ extension $g$ to a neighborhood of zero,
$$a_n=\frac{g^{(n)}(0)}{n!}.$$
For the case where $g$ has no such extension, you will have to do more work.  I would delete this answer if I could, but you will have to un-accept it first.


问:
Find interval such that $f(x):\mathbb{R}^+\rightarrow\mathbb{R}:x\mapsto\sqrt{a+\frac{b}x}$ is a contraction.
Question: Find interval $D\subset\mathbb{R}$ such that  $f|_D(x):\mathbb{R}^+\rightarrow\mathbb{R}:x\mapsto\sqrt{a+\frac{b}x}$ is a contraction, $0<b<a$.

What I've done so far:

I've shown that:

by MVT $|f'_{max}|$ should be less than $1$,

and since $f'$ is monotonically decreasing and $\lim_{x\rightarrow\infty}=0$ I only need to find the beginning of such an interval with $|f'(x_0)|<1$.

No idea why, but the actual difficulty for me is finding such $x_0$ because I have a function that depends on both $a$ and $b$ and if I try to simplify $a$ I can't separate it from $x$.

答:
As you have probably computed already, $|f'(x_0)|<1$ needs to hold. This leads to the inequality
$$\frac{b}{2x_0\sqrt{ax_0^2+bx_0}} < 1.$$
This leads to a polynomial inquality of degree 4:
$$b^2 < 4x_0^2(ax_0^2+bx_0),$$
which is probably quite annoying to solve exactly, but should not pose any problems numerically.

However, if you are looking for a contraction, then there is another key property that you need to keep in mind, namely there must hold
$$f([x_0, \infty)) \subseteq [x_0, \infty)$$
Since $x_0 > 0$ and $f(x) \rightarrow \sqrt{a}$ for $x\rightarrow \infty$, this might fail...


问:
Let $f$ be an irreducible monic polynomial of degree 4 over $\mathbb{F}_p$. Is $f$ irreducible over $\mathbb{F}_{p^2}$?
I'm not sure whether the above is true as it came up while I was trying to answer another question. I have broken the attempted proof into two cases, but am stuck on the second.

Suppose $f$ is reducible over $\mathbb{F}_{p^2}$. This may occur in two ways: 1) $f$ has a zero $\alpha \in \mathbb{F}_{p^2}$ or 2) $f$ factors as the product of irreducible quadratic polynomials.

Case 1: Suppose $\alpha \in \mathbb{F}_{p^2}$ is a zero of $f.$ As $f$ is monic and irreducible over $\mathbb{F}_p$, it holds that $f$ is the minimum polynomial of $\alpha$ over $\mathbb{F}_p$. This implies that $\mathbb{F}_p(\alpha) \cong \mathbb{F}_p[x]/(f)$. In particular, $[\mathbb{F}_p(\alpha):\mathbb{F}_p]=4$. The field $\mathbb{F}_{p^2}$ contains both $\mathbb{F}_p$ and $\alpha$, therefore $\mathbb{F}_{p}(\alpha) \subset \mathbb{F}_{p^2}.$ Hence
\begin{equation*}
4= [\mathbb{F}_p(\alpha):\mathbb{F}_p] \leq [\mathbb{F}_{p^2}:\mathbb{F}_{p}]=2,
\end{equation*}
a contradiction.

Case 2: Suppose $f=gh \in \mathbb{F}_{p^2}[x],$ with $\deg(g)=\deg(h)=2.$ We may assume $g$ and $h$ to be irreducible, as if one is not then we are back in the first case. I don't see how to proceed from here, so I'm looking for a hint or counter-example.

答:
In fact $f$ cannot be irreducible over $\mathbb F_{p^2}$. If $\alpha$ is a root of $f$, then $\mathbb F_p(\alpha)$ is a degree 4 extension of $\mathbb F_p$, and so is a degree $2$ extension of $\mathbb F_{p^2}$, implying that the minimal polynomial of $\alpha$ over $\mathbb F_{p^2}$ has degree 2. This polynomial will divide $f$.

(In Daniel Schleper's comment above, you can see explicitly what this minimal polynomial is.)


问:
Is the converse of $ R \cong R^{op} $ true?
If R is commutative, then trivially $R \cong R^{op} $. Is the converse true ?

I understand that for every ring R we can define the opposite ring  $R^{op}$ by reversing the multiplication in R. Given the multiplication * in R, the multiplication ° in $R^{op}$ is defined as $a°b = b*a$. The "identity map" from R to $R^{op}$ , mapping each element to itself, is an isomorphism if and only if R is commutative.

答: Any group ring $RG$ -- even the group ring of a non-abelian group, which is not commutative -- is isomorphic to it's opposite via the map $g \mapsto g^{-1}$


问:
Help me gain some intuition for the simplex category
I've been reading about the simplex category on nLab and Wikipedia, and I have to say, I'm extremely confused. Maybe it's just because I'm fairly new to category theory and I don't have the requisite knowledge of homology/cohomology to know what this is actually used for, but I'm struggling to get any intuitive sense of how these objects are used.

Wikipedia defines the simplicity category $\Delta$ to have the collection of all non-empty finite ordinals as its objects, and weakly order-preserving maps as its morphisms. The augmented simplicity category $\Delta_a$ is defined to have all finite ordinal as its objects, and order preserving maps as its morphisms. nLab gives a bunch of other, fairly high level definitions that I won't reproduce here. My question is though: what does any of this have to do with simplices? I just can't see what the connection is. Beyond that, why are these categories useful to us in the way that they are in algebra and topology?

答:
My question is though: what does any of this have to do with simplices?

Write $[n]$ for the finite ordinal $\{ 0 \le 1 \le 2 \le \dots \le n \}$, considered as an object of the simplex category $\Delta$. At the most basic level the idea here is to think of this object as an $n$-(dimensional) simplex, and in particular to think of the points $0, 1, \dots n$ as its vertices, with the ordering giving the simplex an orientation. Order-preserving maps $[n] \to [m]$ then correspond to "degenerate subsimplices." Looking at how this works for small $n$ should help make this clearer:

$[0] = \{ 0 \}$ is a point / vertex / $0$-simplex.

$[1] = \{ 0 \le 1 \}$ is a directed edge, or a $1$-simplex. There are two "face" maps $[0] \to [1]$ corresponding to the two vertices of the edge, and a "degenerate" map $[1] \to [0]$ which collapses an edge down to a vertex. In fact if we take the full subcategory of the simplex category on $[0]$ and $[1]$ and consider presheaves on that we get a slight variation of the category of directed graphs in which every vertex has an "identity edge." So, as a rough first pass, the category of simplicial sets is like a category of "higher-dimensional directed graphs."

$[2] = \{ 0 \le 1 \le 2 \}$ is an oriented triangle, or a $2$-simplex. There are three "face" maps $[0] \to [2]$ corresponding to the three vertices of the triangle. There are also three injective "face" maps $[1] \to [2]$ corresponding to the three edges (with compatible orientations), and three non-injective "degenerate" maps which collapse the edge to one of the three vertices.

You may wonder why we need these "degenerate" maps; thinking just in terms of the geometry of simplices it may make sense to restrict attention to strictly order-preserving (equivalently, injective) maps only. This can be done and gives rise to the notion of a semisimplicial set (or more generally semisimplicial object). I am not the person to ask in detail about why simplicial objects turn out to be a better thing to study.

In any case here's a rough first pass describing what a simplicial set (a presheaf on the simplex category) is: in general the category of presheaves $\widehat{C} = [C^{op}, \text{Set}]$ on a (small) category is the free cocompletion of $C$, meaning that it is obtained from $C$ by "freely adjoining colimits." So the category $\widehat{\Delta}$ of presheaves on the simplex category consists of objects obtained by "freely gluing together simplices." This gives some idea of what simplicial sets are supposed to look like as spaces (namely, they look a bit like simplicial complexes), and can be formalized using the geometric realization functor $\widehat{\Delta} \to \text{Top}$, which loosely speaking turns formal colimits of simplices into actual colimits of simplices in $\text{Top}$.

Beyond that, why are these categories useful to us in the way that they are in algebra and topology?

This would require a much longer answer and I'm not really the one to write it; a full answer would really require a textbook. Loosely speaking, simplicial objects turn out to be a "nonlinear" generalization of chain complexes, and can be used to write down "resolutions" of nonlinear objects such as spaces (e.g. Cech nerves) in a way that generalizes how chain complexes can be used to write down resolutions of linear objects such as modules. But this won't be a helpful answer to you until you have some experience in homological algebra (which would be a useful prerequisite for thinking about simplicial stuff).

The simplest really generally useful application of this idea is that the free abelian group on a simplicial set can be used to write down a chain complex (via the Dold-Kan correspondence); this is how singular homology is traditionally defined, using the singular simplicial set $\text{Sing}(X)$ of a topological space. $\text{Sing}$ turns out to have a left adjoint given by geometric realization, and this was one of the motivating examples of adjoint functors when Daniel Kan introduced them in 1958.


问:
Open problems in the first order theory of the real numbers
It is a well-known result that the first-order theory of the real numbers is decidable. However the decision algorithm for this language is in double exponential time, so such an algorithm cannot be practically applied. It then seems perfectly consistent that there is some open problem expressible in the first-order theory of the reals, which the decision algorithm could solve if given enough time.

Does anyone know of any such problem?

If not, as a softer question, does anyone know of a simple statement in the first-order theory of the real numbers that is difficult to prove?

答: By some wonderful coincidence this was just covered unprompted in a lecture I attended, where it was mentioned the kissing number problem in, say, 5 dimensions is an open problem equivalent to a first order sentence in the language of the reals and hence decidable by an algorithm given enough time.</p>


问:
Simplify square of sinc functions
I need to simplify if possible the following:

$$\left(i^n\cdot \operatorname{sinc}\big(\pi(x-\tfrac{n}{2})\big)+(-i)^n\cdot \operatorname{sinc}\big(\pi(x+\tfrac{n}{2})\big)\right)^2$$

with $n \in \mathbf{N}$ and $\operatorname{sinc}(x)=\sin(x)/x$.

Thanks

答:
Note that $\;\pi\left(x-\frac n2\right)\;$ and $\;\pi\left(x+\frac n2\right)\;$ difference will be $n\pi$ so that they will be equal for $n$ even and of different sign for $n$ odd.

This allows to get :

For $n$ even :
$$\left(\frac{2\,x\sin(\pi x)}{\pi\left(x^2-(n/2)^2\right)}\right)^2$$

while for $n$ odd :
$$-\left(\frac{2\,x\cos(\pi x)}{\pi\left(x^2-(n/2)^2\right)}\right)^2$$

Title: Simplify square of sinc functions


问:
boolean logic expression: start_addr = stop_addr & ~mask
Let's say you have the following boolean expression of 16 bit words:

```<code>start_addr  = stop_addr & ~mask
</code>```

Where:  start_addr < stop_addr

example:

```<code>0xAA00 = 0xAAFF & ~0x00FF
0xAA00 = 0xAAFF & 0xFF00
0xAA00 = 0AA00
</code>```

now, suppose, I know only start_addr, and stop_addr, and mask is unknown.  How to find mask?

答:
Something like this:</p>
```<code>
start_addr = stop_addr + -mask

mask = stop_addr - start_addr

mask = 0xAAFF - 0xAA00

mask = 0xFF
</code>```

Body:
Something like this:

start_addr = stop_addr + -mask

mask = stop_addr - start_addr

mask = 0xAAFF - 0xAA00

mask = 0xFF


问:
Variance of Brownian Bridge
With $W(t)$ denoting the standard Wiener process, Brownian Bridge $B(t)$ can be defined on $t \in [0,T]$ as:

$$B(t):=W(t)-\frac{t}{T}W(T)$$

Computing the expectation gives $\mathbb{E}[B(t)]=\mathbb{E}[W(t)]-\frac{t}{T}\mathbb{E}[W(T)]=0-\frac{t}{T}0=0.$ From this we can deduce that $Var(B(t))=\mathbb{E}[B(t)^2]$. Here is my attempt:

$$\mathbb{E}[B(t)^2]=\mathbb{E}\left[\left(\sqrt{t}W(1)-\frac{t}{\sqrt{T}}W(1)\right)^2\right]=\mathbb{E}[W(1)^2]\left(\sqrt{t}\left(1-\sqrt{\frac{t}{T}}\right)\right)^2=\\=t\left(1-2\frac{\sqrt{t}}{\sqrt{T}}+\frac{t}{T}\right)=t\left(\frac{T-2\sqrt{t}\sqrt{T}+t}{T}\right)=\frac{t}{T}\left(\sqrt{T}-\sqrt{t}\right)^2$$

Now wikipedia gives the variance as $\frac{t}{T}(T-t)$, so I must be making a mistake somewhere, but I cannot figure out where? Any hints would be greatly appreciated,

答:
Notice that $\forall t \in [0,T]$ we have that: $min\left(t,\frac{t^2}{T}\right)=\frac{t^2}{T}$. Furthermore, I will make use of the following three lemmas (which I leave to prove as exercise):

Lemma 1: $cov(W_t,W_s)=min(s,t)$

Lemma 2: $Var(W_t - W_s)=Var(W_t)+Var(W_s)-2Cov(W_s,W_t)$

Lemma 3: $\frac{t}{T}W(T)=W\left(\frac{t^2}{T}\right)$ in Distribution

We then have:

$$Var(B(t))=Var\left(W_t-\frac{t}{T}W_T\right)=Var(W_t)+Var\left(W\left(\frac{t^2}{T}\right)\right)-2Cov\left(W_t,W_{\frac{t^2}{T}}\right)=\\=t+\frac{t^2}{T}-2min\left(t,\frac{t^2}{T}\right)=t+\frac{t^2}{T}-2\frac{t^2}{T}=\\=\frac{t(T-t)}{T}$$

Title: Variance of Brownian Bridge


问:
Evaluate derivative of reachability Gramian: is this correct?
So I have the reachability Gramian matrix for a linear time-invariant system:

\begin{align}
W(t_{0},t) = \int_{t_{0}}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}.
\end{align}

In this case I have $t_{0}=0$. Let us differentiate this w.r.t. to $t$:

\begin{align}
\dot{W}(0,t)
=& \frac{d}{dt}\left[\int_{0}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}ds\right] \\
=& \frac{d}{dt}\left[e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}\right] \\
=& Ae^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t} \\
&+e^{At}\left[e^{-As}BB^{\intercal}e^{-A^{\intercal}s}\bigg\vert_{0}^{t}\right]e^{A^{\intercal}t} + e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}A^{\intercal} \\[3mm]
=& AW(0,t) + BB^{\intercal} - e^{At}BB^{\intercal}e^{A^{\intercal}t}+W(0,t)A^{\intercal}.
\end{align}

So this is the result I obtain. However, in the solution to this problem that I was trying to solve the term $- e^{At}BB^{\intercal}e^{A^{\intercal}t}$ did not appear, or was zero. So I am wondering if I made some error, or if there is some control theory result that make this term vanish?

答:
When you applied the product rule, you incorrectly computed the derivative of the integral. By the fundamental theorem of calculus, we have
$$
\frac d{dt} \int_{t_0}^t e^{-As}BB^\top e^{-A^\top s}\,ds = e^{-At}BB^\top e^{-A^\top t}.
$$
There is no $BB^\top$ term that should appear here.


问:
Shifted Legendre polynomials symmetry relation
I have to prove that $p_n(1-x)=(-1)^np_n(x)$ ($x\in[0,1]$) for all $n\in\mathbb{N}$, where $(p_n)_n$ is the family of Legendre polynomials on $[0,1]$: given $(x^n)_{n=0,1,\ldots}$, $(p_n)_n$ is the family of orthogonal monic polynomials that one constructs using Gram-Schmidt process starting from $(x^n)_n$ on $L^2([0,1])$. Now, I  have to prove that $\forall n=0,1,2,\ldots$, $p_n(1-x)=(-1)^np_n(x)$ for all $x\in[0,1]$, that would follow by proving that $\tilde{p}_n\perp p_j$ for all $j\neq p$ and $(p_n,\tilde{p}_n)_2=(-1)^n$. I denoted by $\tilde{p}_n(x)=p_n(1-x)$. By induction, one has that the cases $n=0,1$ are trivial. Then, I tried to use the relation given by Gram-Schmidt, so that (using the induction)
\begin{equation}
(\tilde{p}_{n+1},p_j)_2=((1-x)^{n+1},p_j)_2-\frac{(x^{n+1},p_n)_2}{\Vert p_n\Vert_2^2}(\tilde{p}_n,p_j)_2=((1-x)^{n+1},p_j)_2+(-1)^{n+1}\frac{(x^{n+1},p_n)_2}{\Vert p_n\Vert_2^2}(p_n,p_j)_2.
\end{equation}
If $j\neq n$ the second addend is $0$ since $(p_j)_j$ is an orthogonal family. In particular, the second addend vanishes for all $j\neq n$ and if $j=n+1$, this gives $(\tilde{p}_n,p_{n+1})=((1-x)^{n+1},p_{n+1})_2=(-1)^n$, since this is obviously the coefficient of $x^{n+1}$ respect to $p_{n+1}$. If $j=n$,
\begin{equation}
(\tilde{p}_{n+1},p_n)_2=((1-x)^{n+1},p_n)_2+(-1)^{n+1}(x^{n+1},p_n)_2=((1-x)^{n+1}+(-x)^{n+1},p_n)\overset{(?)}{=}0.
\end{equation}
Since it is obvious that $(Q_j,p_k)=0$ whenever $j<k$ and $Q_j$ is a polynomial of degree $j$, it follows that $(\tilde{p}_{n+1},p_j)_2=0$ for all $j>n+1$ (we don't even need the recursive definition of $p_{n+1}$ here). If $j<n$,
\begin{equation}
(\tilde{p}_{n+1},p_j)_2=((1-x)^{n+1},p_j)_2\overset{(?)}{=}0.
\end{equation}
The question is, how do I prove the (?)? I am not supposed to know anything of the shifted Legendre polynomials except for their definition I gave at the beginning and the recursive formula $p_{n+1}(x)=(x+A_n)p_{n}(x)-B_np_{n-1}(x)$, but I cannot use the fact that (one can prove) $A_n=-1/2$ for all $n$ (actually, I have to prove this fact starting from the just given recursive formula and the symmetry of $p_n$'s - and I already did it).

答: I think I found the answer, but the induction is useless here. The proof I posted before proves that $(\tilde{p}_n,p_n)_2=(-1)^n$ for all $n$. Now, it's enough to observe that $(\tilde{p}_k,p_j)_2=(p_k,\tilde{p}_j)_2$ for all $k,j$ and use the $(Q_j,p_n)_2=0$ for all $n>j$ thing: this implies directly that $(\tilde{p}_n,p_j)_2=0$ for all $j>n$, but also $(\tilde{p}_n,p_j)_2=(p_n\tilde{p}_j)_2=0$ if $j<n$.


问:
Find exactly $3$ matching primitive Pythagorean triples for a given hypotenuse
I'm trying to find three Pythagorean triples $\quad A^2+B^2=C^2 \quad\text{where}$
$$A_1^2+B_1^2=A_2^2+B_2^2=A_3^2+B_3^2=C^2\quad \text{and} \\
 A_1\ne A_2,\ne A_3\quad\land \quad B_1\ne B_2,\ne B_3$$

It is relatively easy to find Pythagorean triples for a given hypotenuse if we solve the C-function of Euclid's formula
$ A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$
for $k$ and test a range of m-values to see which yield integers.
Here is an example using $C=65$.

\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\\
\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.

$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\\
\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$

I found $67$ C-values where
$\quad C=4n+1\space\text{ for }\space 81\le n\le 11925\quad$
with $3$ matching triples each but, in all cases, one or more of the triples had $\quad GCD(A,B,C)>1.\quad$ I ran similar tests for 4-triples, 5-tiples, 6-triples and 7-triples but, in all [my admittedly limited] cases, only an even number of them were primitive.

Does there exist $3$ and only $3$ primitive triples with the same hypotenuse?

答:
My conjecture is that there exist solutions only when $C$ is composed by $4n+1$-type primes and the number of primitive triples having $C$ as hypothenuse is $2^{p-1}$ where $p$ is the number of $4n+1$-type factors of $C$.

Exponents don't change the number of triples, i.e. $5 \cdot 13^3\cdot 29^2$ has the same number (four) of primitive triples as $5\cdot 13\cdot 17$.


问:
Number of ways to partition $2 \times N.$ Tile into $m$ parts
Given a $2 \times N$  Tile , how to find the number of ways to partition it into $M$ parts ?

Meaning of a part : Cells having same number which are adjacent to each other form a part .

Lets take $N = 5$ ,  so an example tile looks like  :

1 1 1 2 2

3 3 2 2 1

This tile has 4-parts to it :

Part-1 : (1 1 1)

Part-2 :

```<code>     (X 2 2)

     (2 2 X)
</code>```

Part-3 : (3 3)

Part-4 : (1)

Each cell can be assigned any number between $1$ and $K$ , how many ways exist to partition the tile into "M" parts ?

Number of different tile-configurations possible are : $k^x$ , where $x = 2n$

答:
Hint: You can build a recursion out of it. Notice that either the last column has the same piece or two of them. Call $A_n\in \mathbb{Z}[q]$ the polynomial such that $$A_n(q)=\sum _{i=1}^{2n}A_{n,i}q^i,$$ where $A_{n,i}$ is the number of such tilings with $i$ parts such that the last column has just one part and $B_n$ is such that
$$B_n(q)=\sum _{i=1}^{2n}B_{n,i}q^i,$$
where $B_{n,i}$ are the tilings with $i$ parts and ends with a column having two parts. You are interested in $[q^m](A_n(q)+B_n(q)).$

One can build a recurrence by understanding how to connect the last column to the right so $$A_n(q) = A_{n-1}(q)+q(k-1)A_{n-1}(q)+(k-2)qB_{n-1}(q)+2B_{n-1}(q)$$
because you can either place a column of the same color as before, or change the color(so create a new part, that is why we multiply by $q$) or you can put a new part to a column of disconnected pieces. So you get
$$A_{n}(q)=(1+q(k-1))A_{n-1}(q)+((k-2)q+2)B_{n-1}(q),$$
Analogously do it for $B_n$ getting
$$B_{n}(q)=(q^2(k-1)(k-2)+2q(k-1))A_{n-1}(q)+(1+2q(k-2)+q^2((k-1)(k-2)+1))B_{n-1}(q),$$
where $A_1(q)=kq$ and $B_1(q)=k(k-1)q^2.$

You can try generating functions or any method you know for solving these systems of recurrences.


问:
Natural Choice of Topology for Free Monoid on a Space
Suppose $X$ is a metric space, and let $X^*$ denote the free monoid on $X$, that is the monoid consisting of all finite strings of elements of $X$, with string concatenation as the monoid operation (and with the empty word as identity).

Question: Is there a natural choice of topology on $X^*$ that reflects the topology on $X$?

答:
The free monoid on a topological space $X$ is just the disjoint union

$$\bigsqcup_{n \ge 0} X^n$$

so it has a natural topology given by the disjoint union of the product topologies. Unfortunately I don't think it's useful for very much. A variation of this construction involving 1) commutativity and 2) fixing a basepoint $x \in X$ and requiring it to act as the identity produces a more interesting construction called the infinite symmetric product which, under mild hypotheses, has the property that its homotopy groups compute the reduced homology of $X$ (the Dold-Thom theorem).


答:
My answer:
D has an upper bound $x_0=\sqrt a$ since $\lim_{x\rightarrow\infty}f(x)=\sqrt a$ and $f:D\rightarrow D$.

Now I can evaluate lower bound as:
$$b < 4x_1^4(a+\frac{b}{x_1})<4x_1^4(ax_1+a)<4x_1^3(a^{3/2}+a)$$
$$x^3_1 > \frac{b}{a^{3/2}+a} $$
$$x_1>\sqrt[3]\frac{b}{a^{3/2}+a}$$
$\implies D=\Big({\sqrt[3]\frac{b}{a^{3/2}+a},\sqrt a}\Big)$

Title: Find interval such that $f(x):\mathbb{R}^+\rightarrow\mathbb{R}:x\mapsto\sqrt{a+\frac{b}x}$ is a contraction.

Body:
Question: Find interval $D\subset\mathbb{R}$ such that  $f|_D(x):\mathbb{R}^+\rightarrow\mathbb{R}:x\mapsto\sqrt{a+\frac{b}x}$ is a contraction, $0<b<a$.

What I've done so far:

I've shown that:

by MVT $|f'_{max}|$ should be less than $1$,

and since $f'$ is monotonically decreasing and $\lim_{x\rightarrow\infty}=0$ I only need to find the beginning of such an interval with $|f'(x_0)|<1$.

No idea why, but the actual difficulty for me is finding such $x_0$ because I have a function that depends on both $a$ and $b$ and if I try to simplify $a$ I can't separate it from $x$.


问:
Analytic sets and equivalence relation
Let $X$ be a polish space and $E$ an equivalence relation on $X$ such that $E\subseteq X^2$ is analytic.

Show that for every analytic set $A\subseteq X$ the set $[A]_E=\{x\in X: \exists y\in A: xE\,y\}$ is also analytic.

I struggle with this problem.
In general I have to show that $[A]_E$ is a continuous image of a Borel set.

So I have to find a continuous function $f:Y\to X$, where $Y$ is a polish space, and a Borel set $B\subseteq Y$ with $f(B)=[A]_E$.

I know some equivalent statements, but they do not seem to make it easier.

Do you have a hint on how to solve this?
Thanks in advance.

答:
Hint:

Since $E$ is analytic, there is a (continuous) map $f : \omega^\omega \to X^2$ whose image is $E$.

Moreover, notice $\pi : X^2 \to X$ which sends $(x,y) \mapsto x$ is continuous.

Then $[X]_E = \{ x \in X ~|~ \exists y . xEy \}$ is nothing more than $\pi[E]$, and so $[X]_E$ is the image of $\pi \circ f : \omega^\omega \to X$ (and thus it is analytic).

Can you modify this argument (or add a step at the end, etc. There's various approaches) to get $[A]_E$? It may be helpful to remember (or prove) that analytic sets are closed under intersections.

I hope this helps ^_^

Title: Analytic sets and equivalence relation


问:
Spectrum and compactness of $L^1$ operator
Consider a continuous complex function $g$ defined on $[0,1]$. On $L^1([0,1])$ consider operator
$$Tf(x) = g(x)f(x).$$
Calculate the norm. Determine its point spectrum and spectrum. Decide for what functions $g$ is $T$ compact.

I've managed to show that $\lVert T \rVert = \lVert g \rVert_\infty$.

To find point spectrum we want to find nontrivial solutions
$$ \forall x \in [0,1]: (\lambda - g(x)) f(x) = 0.$$

For constant function $g$ we get a nontrivial solution. From this $\sigma_p(T) = \{g = c \in \mathbb{C} \}$.

For nonconstant $g$ we have no nontrivial solutions. Which means $\sigma_p(T) = \emptyset$.

To find spectrum we choose $h \in L^1([0,1])$ and we need to find for what $\lambda \in \mathbb{C}$ the following equation has a solution
$$ \forall x \in [0,1]: \lambda f(x) - g(x) f(x) = h(x). $$

If we set $\lambda = 0$, then $f(x) = -h(x)/g(x)$ for $g \neq 0$ a. e. Since I distinct constant and noncostant functions, we get that for constant $g$ we always have a solution. Which means $0 \notin \sigma(T) $. Since a compact operator always has $0$ in its spectrum, for such funtions $g$ operator $T$ cannot be compact.

For $g$ nonconstant we need to decide if $-h(x)/g(x) \in L^1$. That is where I am unsure about how to proceed.

Is my partial attempt correct? How would I go about the compactness?

答:
Let us look at the eigenvalues first. You what $g(x)f(x)=\lambda\,f(x)$ a.e. Let $E=\{f\ne0\}$. For any $x\in E$, you have $g(x)=\lambda$; and the equality holds trivially on the complement of $E$. So $\lambda$ is an eigenvalue of $g$ if and only if $m(E)>0$. Conversely, if $m(\{g=\lambda\}>0$, we have $g\,1_E=\lambda 1_E$ and $\lambda$ is an eigenvalue. Thus
$$
\sigma_p(T)=\{\lambda:\ m(\{g=\lambda\})>0\}.
$$

For the whole spectrum, you need to ask yourself when $g-\lambda$ is invertible. It is easy to see that if $1/(g-\lambda)$ is continuous, it provides an inverse $T-\lambda I$. Because $g$ is continuous and $[0,1]$ is compact, it is enough that $g(x)\ne\lambda$ for all $x$. In other words: if $\lambda$ is not in the range of $g$, then $\lambda\not\in\sigma(g)$. That is, $\sigma(T)\subset g([0,1])$.

And conversely, suppose $g(x)=\lambda$ for some $x$. As $g$ is uniformly continuous, for each $n$ there exists $\delta_n$ such that $|x-y|<\delta_n$ implies $|g(x)-g(y)|<1/n$. Let $f_n=\frac1{\delta_n}\,1_{[x-\delta_n/2,x+\delta_n/2]}$. Then $f_n\in L^1[0,1]$, $\|f_n\|_1=1$. And

$$
\|Tf_n-\lambda f_n\|_1=\frac1{\delta_n}\,\int_{[x-\delta_n/2,x+\delta_n/2]}|g(t)-f(x)| \,dt\leq\frac1n.
$$
So $T-\lambda$ cannot be invertible, and so $\lambda=g(x)\in\sigma(T)$, showing that $g([0,1])\subset \sigma(T)$. Thus
$$\tag1
\sigma(T)=g([0,1]).
$$

Finally, compactness. As you noted, for $T$ to be compact a necessary condition is that $0\in\sigma(T)$, which means that $g(x)=0$ for some $x$. More generally a necessary condition for compactness is that the spectrum consists of $0$ and a sequence of eigenvalues that converge to zero. In light of $(1)$, this requires $g$ to take at most countably many values. By the Intermediate Value Theorem, this can only happen if $g$ is constant; but this would require $g$ to be zero. So $T$ can only be compact when $g=0$.

An even more telling way to look at compactness is to notice that, if $\lambda$ is an eigenvalue for $T$, then $\dim\ker(T-\lambda I)=\infty$. Because if $\lambda$ is an eigenvalue then $g(x)=\lambda$ on some measurable set $E$ with $m(E)>0$. If we write $E$ as a disjoint union $E=E_1\cup\cdots\cup E_m$, then $1_{E_1},\ldots,1_{E_m}$ are $m$ linearly independent eigenvectors for $\lambda$; and we can do this for any $m$. This is not a proof, though, because it may happen that $\sigma(T)=\{0\}$.


问:
Find $p$ if $(x + 3)$ is a factor of $x^3 - x^2 + px + 15$.
I'm just making sure I answered this correctly.

If $(x+3)$ is a factor, then $P(-3)$ would equal $0$, correct?

Body: Alternatively: $\,-3\,$ root of $\ x^3+27-(x^2-p\,x -4(-3))\ \Rightarrow\ p = -4 -3 $


问:
Let $G$ be a Lie group and $v\in X_L(G)$. Then $v$ is complete.
There is a proposition in my textbook that states,

" Let $G$ be a Lie group and   $v\in X_L(G)$. Then $v$ is complete. "

I believe that $X_L(G)$ is the left-invariant vector field, so then I interpret the theorem as if $G$ is a Lie group then all left-invariant vector fields are complete. I know that for this to be true then the flow must exist for all ??$\in\mathbb{R}$, would this be for all $v \in \mathbb{R}$?

How would I prove the proposition?

My idea:
Let $G$ be a Lie group with Lie algebra $\mathfrak g$ and $v \in \mathfrak g$. Write $X_L(G)$ as the left-invariant vector field on $\mathfrak g$ with $v \in X_L(G)$. Then
$$\phi_t(a) = a\gamma x(t) $$
is the flow of $X_L(G)$

Then for all $a\in G$ we have
$$\frac{d}{dt} |_{t=s} a\gamma x(t) = (dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(t))$$
$$= (dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(t+s))$$
$$=(dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s} \gamma x(s) \gamma x (t))$$
$$=(dL_a)_{\gamma x(s)}(\frac{d}{dt}_{t=s}  L_{\gamma x(s)}(\gamma x (t)))$$
$$=(dL_a)_{\gamma x(s)}1(\frac{d}{dt}_{t=s} \gamma x (t))$$
$$=(dL_a)_{\gamma x(s)}(X)$$
$$=X_L(G)(a\gamma x(s))$$
which would show that the flow of $X_L(G)$ exists for all $t$?

Any help would be appreciated. I find this topic very difficult.

答:
Yes $X_L(G)$ denotes the lie algebra of the lie group , i.e , the set of left-invariant vector fields or equivalently $T_e G$. Now for a vector field to be complete we have to show that it's flow is defined for all $t\in \mathbb{R}$, by definition. So take a left-invariant vector field $X$ and for example the identity $e$ in $G$. By results from ordinary differential equations there will exists an integral curve for $X$, $c_e(t)$ for $t\in (-\epsilon,\epsilon)$ such that $c_h(0)=e$. Now the idea is to use the left-translations of a lie group which are diffeomorphism to extend this curve to the lie group that is take $t_0 \in (-\epsilon,\epsilon)$ and we have that $c_e(t_0)c_e(t)$ will define an integral curve for $X$ at $c_e(t_0)$, where we use here the fact that $X$ is left-invariant. Now this curve is extending our previous one, and so by uniquess of the Picard-Lindelof theorem we have this curve extends, i.e, we can always extend it's interval of definition. Doing this again and again for all the points you get the desired result.

Here I will actually compute the differential so that things are clear :

Let $g=c_e(t_0)$ then we have that $\frac{d}{dt}gc_e(t)=\frac{d}{dt}L_g(c_e(t))=\frac{d}{ds}|_{s=0}L_g(c_e(t+s))=d_{c_e(t)}L_g(X_{c_e(t)})=X_{gc_e(t)}$.


问:
Question about this proof (exactness of a sequence)
In Altman-Kleinman "Introduction to Grothendieck Duality Theory", page 104 there is the following Lema:

I don't understand the underlined part, why the exactness of that sequences implies the map is the inverse of $h$?

答:
Let me rewrite the things more explicitly:

Since $\newcommand{\coker}{\operatorname{coker}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\im}{\operatorname{im}}$$0 \to \Hom(N'',\coker g) \stackrel{g^*}{\to} \Hom(N,\coker g)$ is exact, $g^*$ is injective, and then, if $\pi : N'' \to \coker g$ is the canonical projection, $g^*(\pi) = \pi \circ g = 0$ implies that $\pi = 0$. Thus $\coker g = \im \pi = 0$, meaning that $g$ is surjective.

Since $\Hom(N'',N'') \stackrel{g^*}{\to} \Hom(N,N'') \stackrel{f^*}{\to} \Hom(N',N'')$ is exact, $(g \circ f)^* = f^* \circ g^* = 0$, and in particular $0 = (g \circ f)^*(\operatorname{id}_{N''}) = \operatorname{id}_{N''} \circ\, (g \circ f) = g \circ f$, which implies that $\im f \subseteq \ker g$, and then $g$ induces a homomorphism $h : \coker f \to N''$ such that $h(n+\im f) = g(n)$ for every $n \in N$.

Now, $\Hom(N'',\coker f) \stackrel{g^*}{\to} \Hom(N,\coker f) \stackrel{f^*}{\to} \Hom(N',\coker f)$ is exact, so, if $\pi : N \to \coker f$ is the canonical projection, $f^*(\pi) = \pi \circ f = 0$ implies that $\pi \in \ker f^* = \im g^*$, meaning that $\pi = g^*(j) = j \circ g$ for some homomorphism $j : N'' \to \coker f$. In other words, $j(g(n)) = n+\im f$ for all $n \in N$, that is, $j$ is the inverse of $h$.


答:
Applying the Leibniz integral rule to your middle term yields

$$
\frac{d}{dt}\left[\int_{0}^{t} e^{-A\,s} B B^{\intercal} e^{-A^{\intercal} s}ds\right] =
\int_{0}^{t} \frac{d}{dt}\left[e^{-A\,s} B B^{\intercal} e^{-A^{\intercal} s}\right] ds + e^{-A\,t} B B^{\intercal} e^{-A^{\intercal} t} = e^{-A\,t} B B^{\intercal} e^{-A^{\intercal} t}.
$$

You can also apply the Leibniz integral rule directly to initial expression, which yields

\begin{align}
\frac{d}{dt}\left[\int_{0}^{t} e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)}ds\right]
&= \int_{0}^{t} \frac{d}{dt}\left[e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)}\right] ds + e^{A (t-t)} B B^{\intercal} e^{A^{\intercal} (t-t)}, \\
&= \int_{0}^{t} \left[A\,e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)} + e^{A (t-s)} B B^{\intercal} e^{A^{\intercal} (t-s)} A^{\intercal}\right] ds + B B^{\intercal}.
\end{align}

Title: Evaluate derivative of reachability Gramian: is this correct?

Body:
So I have the reachability Gramian matrix for a linear time-invariant system:

\begin{align}
W(t_{0},t) = \int_{t_{0}}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}.
\end{align}

In this case I have $t_{0}=0$. Let us differentiate this w.r.t. to $t$:

\begin{align}
\dot{W}(0,t)
=& \frac{d}{dt}\left[\int_{0}^{t}e^{A(t-s)}BB^{\intercal}e^{A^{\intercal}(t-s)}ds\right] \\
=& \frac{d}{dt}\left[e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}\right] \\
=& Ae^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t} \\
&+e^{At}\left[e^{-As}BB^{\intercal}e^{-A^{\intercal}s}\bigg\vert_{0}^{t}\right]e^{A^{\intercal}t} + e^{At}\int_{0}^{t}e^{-As}BB^{\intercal}e^{-A^{\intercal}s}ds\;e^{A^{\intercal}t}A^{\intercal} \\[3mm]
=& AW(0,t) + BB^{\intercal} - e^{At}BB^{\intercal}e^{A^{\intercal}t}+W(0,t)A^{\intercal}.
\end{align}

So this is the result I obtain. However, in the solution to this problem that I was trying to solve the term $- e^{At}BB^{\intercal}e^{A^{\intercal}t}$ did not appear, or was zero. So I am wondering if I made some error, or if there is some control theory result that make this term vanish?


问:
Negating there exists-for all composite
Suppose $P$ is a proposition statement about $p$ and $q$, and $Q$ is a statement about $r$. Consider:

There is a prime number $p$ such that $P$ is true for all integers $q < 10$.

$Q$ is false for all real numbers $r \leq 5$.

What are the respective negations for the above sentences?

My attempts:

For all prime numbers $p$, there is an integer $q \geq 10$ for which $P$ is false.

$Q$ is true if there exists a real number $r > 5$.

Are these negations correct? Also, is it allowed to move the "there exists" part before "$P$ is false" even though the original statement has the latter before? Any assistance is much appreciated.

答:
I find it a bit ambiguous to say that $P$ states something about $p$ and $q$, yet not denote is as $P(p,q)$. Similarly, I'd find it clearer if $Q$ was written as $Q(r)$. But ok, you use natural language to do first-order logic, so such ambiguities are only natural! :P

To the point:

Assuming that $p$ and $q$ range over integers, for the first sentence, you're almost there! You only do not take into account the case where $p$ may not be prime:

Either $p$ is not prime or (it is prime and) there is an integer $q \geq 10$ for which $P$ is false.

For the second sentence, assuming that $r$ ranges over reals, I'd say

There is a real $r > 5$ for which $Q$ is true.

The problem with your formulation is that the $r$ in your sentence

$Q$ is true if there exists a real number $r>5$

does not refer to the $r$ that $Q$ states something about, but refers to some real number for which we happen to use the same letter $r$ to name it.

Title: Negating there exists-for all composite


问:
Find the integer solutions to $4^x - 9^y = 55$
I want to find the integer solutions of:

$$ 4^x - 9^y = 55$$

For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?

I tried to solve for $y$, but to no avail. WolframAlpha tells me that it is the only solution, but it doesn't provide an explanation.

答:
This factors as $(2^x + 3^y)(2^x-3^y) = 55$, so $2^x + 3^y \in \{1, 5, 11, 55\}$. This leaves us only finitely many cases of $x$ and $y$ to check: $x \le 5$ (because $2^6$ is bigger than $55$) and $y \le 3$ (because $3^4$ is bigger than $55$).

A check of all $(x,y) \in \{0,1,2,3,4,5\} \times \{0,1,2,3\}$ gives only one solution: $(x,y) = (3,1)$.

We can be more clever and eliminate some cases, reducing the amount of casework we have to do, but that's already enough.

To minimize casework: first, because $2^x + 3^y > 2^x - 3^y$, we know that $2^x + 3^y$ is either $11$ or $55$, with $2^x - 3^y$ being either $5$ or $1$ respectively. But adding them together gives a value of either $11+5=16$ or $55+1 = 56$ for $2 \cdot 2^x$. Only the first of these is a power of $2$, so we conclude that $2^x + 3^y = 11$.

To get this, $3^y$ can be either $1$, $3$, or $9$. This leaves $10$, $8$, or $2$ for $2^x$. We reject $2^x = 10$ because $x$ must be an integer. We reject $2^x = 2$ with $3^y=9$ because the original equation $4^x-9^y=55$ is not satisfied when $x=1$ and $y=2$. Finally, the case $2^x=8$ with $3^y=3$ gives us the solution $(x,y) = (3,1)$.

It's harder, but still possible, to solve the more general equation $2^x - 3^y = 55$, to which we now know that $(x,y) = (6,2)$ is a solution.

With the aim of rejecting solutions with $y>2$, consider the equation modulo $27$: if $y$ is $3$ more, then $3^y \equiv 0 \pmod{27}$, so $2^x \equiv 55 \equiv 1 \pmod{27}$. The order of $2$ modulo $27$ is $18$, so $2^x \equiv 1 \pmod{27}$ exactly when $x$ is a multiple of $18$.

In that case, take the equation modulo $73$, which is one of the prime factors of $2^{18}-1$, so we know that $2^x \equiv 1 \pmod{73}$. This tells us that $3^y \equiv 1 - 55 \equiv 19 \pmod{73}$. This has no solutions: the powers of $3$ can take on the values $3, 9, 27, 8, 24, 72, 70, 64, 46, 65, 49, 1$, after which they repeat.

So it's impossible to have $y \ge 3$. We can try $y=0$ (in which case $2^x=56$ has no solution), $y=1$ (in which case $2^x = 58$ has no solution), or $y=2$ (in which case $2^x = 64$ gives us $(x,y) = (6,2)$).


问:
$C$ is a closed set in $X$, $X$ is $T_4$, show that $C$ is $T_4$
$X$ is a topological space, $C\subset X$. Show that if $X$ is $T_4$ and $C$ is a closed set, then $C$ is also $T_4$.

All I know is that for $X$ to be $T_4$ it has to be also $T_1$ and $\forall_{C,E\subset X;\  C\cap E=\varnothing} \exists_{U,V\subset X}:C\subset U,E\subset V: U\cap V=\varnothing$. Because $C$ is a closed set then it is $T_1$ (at least that's the way I see it, I don't know whether I'm right).
My question is how do I prove that $C$ is also $T_4$?

答:
To show that $C$ is $T_1$, let $x,y\in C$ be two distinct points. Since $X$ is $T_1$, there exists an open neighborhood $U$ of $x$ in $X$ such that $y\notin U$. Clearly, we have $x\in U\cap C$ and $y\notin U\cap C$. In particular, $U\cap C$ is open in $C$, so it is an open neighborhood of $x$ in $C$ excluding $y$. This shows that $C$ is $T_1$.

To show $C$ is normal, let $A$ and $B$ be two disjoint closed subsets of $C$. Since $C$ itself is closed in $X$, both $A$ and $B$ are also closed in $X$. Now because $X$ is normal, there are two disjoint open sets $U$ and $V$ in $X$ such that $A\subseteq U$ and $B\subseteq V$. As we can see, $U\cap C$ and $V\cap C$ are open in $C$, where $$A\subseteq U\cap C,\quad B\subseteq V\cap C,\quad(U\cap C)\cap(V\cap C)=\emptyset.$$ Therefore, $C$ is normal.


问:
Solving Coin-Weighing Problem (81 Coins, 1 Fake) Using Information Theory
So I have a coin-weighing puzzle under these situations:

There are 80 real coins and 1 fake coin (total of 81 coins).

The real coins are all the same weight, and the weight of the fake coin is different from the real coin.

The real and fake coins cannot be distinguished, other than weight.

A balance is used to identify the fake coin.

When using the balance, the same number of coins is placed on either plate. The result is either "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

According to prior research, this problem cannot be solved by using the balance 4 times or less. I tried to show this by using information theory as follows, but I feel like I am missing something here.

Label the coins 1,...,81, and let $C$ be the number of the fake coin. Also, let $W$ be a random variable defined so that $W=1$ when the fake coin is heavier than the real coin, and $W=0$ when lighter. At the initial state, $C$ and $W$ can be regarded as independent random variables following a uniform distribution.

Then, define random variable $R_k$ so that $R_k = 0$, $R_k = 1$, $R_k = 2$, when, the result after using the balance for the $k$th time is, respectively, "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

Then, $R_k$ can be regarded as uniquely determined by $R_1, ..., R_k,$ and the real values of $C$ and $W$. (Do I have to add additional proof that this is true?)

Assume that the fake coin can be surely identified by four measurements using the balance, then this means that the value of $C$ is determined when $R_1,...,R_4$ have been determined. Thus, using the entropy function and its chain rule, \begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(C) + H(R_1 | C) + H(R_2 | R_1, C) + H(R_3 | R_2, R_1, C) + H(R_4|R_3, R_2, R_1, C) + H (W| R_4, R_3, R_2, R_1, C)
\end{eqnarray*}

And, we also have\begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(W) + H(C|W) + H(R_1 | C, W) + H(R_2 | R_1, C, W) + H(R_3 | R_2, R_1, C, W) + H(R_4|R_3, R_2, R_1, C, W)
\end{eqnarray*}

Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get
$H(W) < H(W | R_1, ..., R_4, C)$

But we know that $H(W | R_1, ..., R_4, C) < H(W)$, so there is a contradiction.

OK. So I somehow arrived at a contradiction, showing that the problem cannot be solved by using the balance for times. But then I realized that the same logic applies when using the balance for any number of times, so apparently there is something wrong...

What sort of logic am I missing? I would appreciate any help.

答:
Brian has pointed out your error correctly. However, your approach is almost there (this is basically a formalised version of wendy's argument):

I'll use $R_1^4$ to denote the joint random variables $(R_1, R_2, R_3, R_4).$ Notice that if $(C,W)$ can be determined using $R_1^4$ then $H(C,W|R_1^4) = 0$ (this is the key point that you did not use in your attempt). But then $$ H(C,W, R_1^4) = H(C,W) + H(R_1^4|C,W) = H(R_1^4) + H(C,W|R_1^4) = H(R_1^4).$$

Since entropies are non-negative, this means that $$ H(C,W) \le H(R_1^4) \le 4 \log 3,$$ where I've used the fact that $R_1,R_2,R_3,R_4$ are ternary, and the upper bound on entropy provided by log of the support size. But $H(C,W) = H(C) + H(W) = 4 \log 3 + \log 2 > 4\log 3$ gives a contradiction.

Just for completeness, in this case $5$ weighings suffice - divide the $81$ coins into three groups of 27. Measure the first two groups, and then the second and the third groups. This tells you both if the anomalous coin is heavier or lighter, and which of the three groups it lies in. Divide this group into three groups of $9$, and measure the first two - since you know the direction of the anomaly, this reduces the identity of the anomalous coin to $9$ options. Doing this division into three groups two more times fixes the coin in a total of $5$ measurements.


问:
Is $\max\{-X_{(1)},X_{(n)}\}$ a one dimensional or two dimensional statistic?
Is statistic $\max\{-X_{(1)},X_{(n)}\}$ one dimension or two dimension?

I was trying to find the minimal sufficient statistic for $U(-\theta,\theta)$ from $n$ $i.i.d$ random variables $X_i$. The result is that $\theta\ge \max\{-X_{(1)},X_{(n)}\}$ thus the minimal sufficient statistic is $\max\{-X_{(1)},X_{(n)}\}$. However, the problem actually states that "Find a two dimensional minimal sufficient statistic for $U(-\theta,\theta)$". Is $\max\{-X_{(1)},X_{(n)}\}$ a two dimensional statistic?

Here $X_{(i)}$ is the $i^{th}$ smallest value of $X_1,\cdots,X_n$.

答:
I presume by $\{X_{(i)}\}$ you mean the ordered sequence constructed from $\{X_i\}$. The minimal sufficient statistics is the pair of $(min\{X_i\}, max\{X_i\})$, however, if the sequence is ordered, then you are right; the minimal sufficient statistics is $max\{-X_{(1)}, X_{(n)}\}$, but it requires an additional step of ordering.

Correction

Yes. You are right. $max\{|X_i|\}$ is the sufficient statistics.


问:
For what $x$ is there some $k$ such that $x\uparrow\uparrow k$ is an integer?
Apparently it is unknown whether $\pi \uparrow\uparrow 4$ is an integer (I learned this from this tweet). I'm curious about which real numbers have some power tower which is an integer. That is, facts about the set:

$S = \left\{x \in \mathbb{R} \vert \exists k\in\mathbb{N} \text{ s.t. }(x\uparrow\uparrow k) \in \mathbb{Z}\right\}$

What is known about this set? For instance, is it measure 0?

答:
The Gelfond-Schneider theorem implies that

if $x \upuparrows 2 = x^x \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is transcendental, and

if $x \upuparrows 3 = x^{x^x} \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is irrational.

But I don't think it's known whether or not $x \upuparrows 3 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is transcendental, or whether or not $x \upuparrows 4 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is irrational.

It is at least easy to see the following: for fixed $k$ the function $f(x) = x \upuparrows k$ satisfies $f(1) = 1$, is strictly increasing on $[1, \infty)$, continuous, and goes to infinity, so takes on every real value in $[1, \infty)$ exactly once. So for every positive integer $n \in \mathbb{N}$ there is a unique positive real $x \in [1, \infty)$ such that $f(x) = n$, which we might call the $k^{th}$ power-tower-root of $n$. But I think very little is known about these reals. I would not expect them to be expressible in terms of other familiar constants but I expect this to be quite out of reach. Here's another example of "this is obviously true and nobody has the slightest clue how to prove it": nobody knows whether or not $\pi + e$ or $\pi e$ are irrational (although at least one of them must be transcendental)!


问:
Which methods to use to integrate $\int{\frac{x^4 + 1}{x^2 +1}}\, dx$
I have this integral to evaluate:

$$\int{\frac{x^4 + 1}{x^2 +1}}\, dx$$

I have tried substitution, trig identity and integration by parts but I'm just going round in circles.

Can anyone explain the method I need to work this out?

Body:
Let's try some imaginary stuff.

$\dfrac{x^4}{x^2+1}=\dfrac{Ax^3}{x+i}+\dfrac{Bx^3}{x-i} \implies A=B= \dfrac{1}{2}$

$\dfrac{x^4+1}{x^2+1}= \dfrac{1x^3}{2(x+i)}+\dfrac{1x^3}{2(x-i)}+\dfrac{1}{x^2+1}$


问:
Introductory problem to understand Schroeder-Bernstein Theorem

Consider $A=[0,1]$ and $B=[0,1)$ and injections $f(x) = \frac13x$ from $A$ to $B$ and $g(x) =x$ from $B$ to $A$. Construct a bijection $h$ from $A$ to $B$ such that on some points of $A$, $h(x) = f(x)$ and for the other points $h(x) = g^{-1}(x).$

It seems that $f(x) = \frac13x$ is a bijection from $A \to B$ for the points $[0, \frac13]? $ If I understood the question correctly (I doubt I did...) they would want me to construct a bijection for the rest of the points? I’m reading ”topology through inquiry” and this was a problem from the book. Also I’m not sure if they mean the inverse function by $g^{-1}(x)$ or the preimage?

答:
You’re clearly going to have to let $h(x)=g^{-1}(x)=x$ for $x\in\left(\frac13,1\right)$, because those points aren’t in the range of $f$. After that you have to figure out how to piece together $f$ and $g^{-1}$ to get a bijection from $\left[0,\frac13\right]\cup\{1\}$ to itself.

HINT: Let $A=\left\{\frac1{3^n}:n\in\Bbb Z^+\right\}$. Note that $f[A]=A\setminus\left\{\frac13\right\}$. Let $h\upharpoonright A=f\upharpoonright A$ and, as already suggested above, $h\upharpoonright\left(\frac13,1\right)=g^{-1}\upharpoonright\left(\frac13,1\right)$. How can you use $f$ and/or $g^{-1}$ to complete the definition of $h$ by defining $h\upharpoonright\left(\left[0,\frac13\right]\cup\{1\}\right)$?


问:
Creating a random task with $x^2+ax+b=0$ where $x$ is an integer
I need to create random tasks with $x^2+ax+b=0$ where $x$ is an integer. What rules apply to $a$ and $b$?

答:
We know that if this equation is satisfied, $x$ must be one of two roots, $r_1$ and $r_2$ (both integers, if I understand correctly).  By Vieta's formulas, we can get that

$$ a = -r_1 - r_2$$

$$ b = r_1 r_2 $$

And, if it helps, $a$ and $b$ are both integers.


问:
Integral for cubic potential
I want to calculate $V(\vec{x})=\int_{\mathbb{R}^3}\frac{\theta(\vec{y})}{|\vec{x}-\vec{y}|}dy$, where $\theta(\vec{y})=|\vec{y}|^2-R^2$ for $|\vec{y}|\leq R$ and $\theta(\vec{y})=0$ otherwise.

I can't get any further than recognising that $V(\vec{x})=\int_{|\vec{y}|\leq R}\frac{|\vec{y}|^2-R^2}{|\vec{x}-\vec{y}|}dy$. Switching to spherical co-ordinates makes sense but then I get a messy expression for $|\vec{x}-\vec{y}|$.

答:
You have to choose the system of coordinates (directions of coordinate axes) while integrating in $\mathbb{R}^3$ - initially the Y system is not fixed in the space. You may choose the z axe in the direction of $\vec{x}$, and $|\vec{x}-\vec{y}|$ will be expressed as a function of the dot products of vectors, depending only on the polar ($\theta$) angle between $\vec{x}$ and $\vec{y}$ and not depending on radial angle $\varphi$. Integration over $\varphi$ is straightforward.

Your integral will look like $V(\vec{x})=2\pi\int_0^R \int_0^\pi\frac{y^2-R^2}{\sqrt{y^2+x^2-2xy\cos\theta}}y^2dy\sin\theta d\theta$, where $y$ and $x$ are $|\vec{y}|$ and $|\vec{x}|$ correspondingly. It is easier to integrate over $\theta$ first.

Hopefully it would help.

Title: Integral for cubic potential


问:
A sequence where the nth term is the cumulative sum over n-1
Consider the following sequence:

$u_{n}=-\frac{u_{1}}{n}-\frac{\beta\gamma}{n}(u_{1}+...+u_{n-1})$, where $u_1$, $\beta$ and $\gamma$ are some none zero constant.

In this sequence, can I obtain the expression for $u_{n}$ as a function of $u_{1}$? If I can, what would it be? At first, it seems pretty straightforward but I couldn't get a nice closed form.

答:
Clearly your recurrence relation holds for only for $n\gt1.$ Otherwise at $n=1,$ we get $u_1=-u_1$ and consequently $u_n=0$ for all $n\in\mathbb{N}.$  In particular at $n=2,$ we have $u_2=-\dfrac{u_1}{2}(1+\beta\gamma).$ Thanks for Achille hui to pointing this in comments.

There are few different ways that you can look at this question. One way is note that the quantity $$nu_n+\beta\gamma(u_1+\cdots+u_{n-1})=-u_1$$ is a constant. By rewriting the same equation for $n+1,$ we get $(n+1)u_{n+1}+\beta\gamma(u_1+\cdots+u_{n-1}+u_{n})=nu_n+\beta\gamma(u_1+\cdots+u_{n-1}),$ which simplifies to $$u_{n+1}=\left(\dfrac{n-\beta\gamma}{n+1}\right)u_n\qquad \forall n\ge2.$$ Clearly

if $u_1=0$ then the sequence is trivial.

if $\beta\gamma$ is an integer other than $1,$ then all $u_{\beta\gamma+1}=u_{\beta\gamma+2}=\cdots=0.$

Otherwise $$u_{n+1}=u_2\dfrac{u_3}{u_2}\cdots\dfrac{u_n}{u_{n-1}}\dfrac{u_{n+1}}{u_n}=u_2\prod_{k=2}^{n}\left(\dfrac{k-\beta\gamma}{k+1}\right)=-u_1(1+\beta\gamma)\dfrac{(2-\beta\gamma)\cdots(n-\beta\gamma)}{(n+1)!}$$ for all $n\ge 2.$ In fact, this last formula contain above two remarks as special cases.

Another way to solve this recurrence is compute first few terms of the sequence by hand and identify a pattern among them. Then we can use mathematical induction to justify our pattern.


问:
From DAGs to levelled DAGs
I'm looking into the literature, but I can't find anything interesting, so I thought about asking here for help or for pointers to articles/web pages.

I'm looking for the fastest algorithm that solves the following task:

Given a connected directed acyclic graph $G(V,E)$ with one root $v_0$ as an input, output a levelled DAG $G'(V',E')$, with $V \subset V'$, with the same paths of $G$. So, given $v_i,v_j \in G$, the path $v_i \rightarrow v_j$ is in $G'$, if and only if it is in $G$

In other words, given any vertex $v \in V$, there can be different paths $v_0 \rightarrow v$, and these paths can have different lengths. I need a DAG that has the same vertices (plus additionals) and the same paths as the original one, in a way that every path $v_0 \rightarrow v$ has the same length, for every $v \in V'$.

Is it actually possible? If yes, how can this be achieved?

答:
Begin with a topological sort of $G$, in which we may as well assume the root $v_0$ is the first vertex. (If there are vertices before $v_0$, there is no path from $v_0$ to those vertices anyway, and so we don't need to do anything about them.) Let $v_0, v_1, v_2, \dots, v_n$ be the sequence of vertices we get from the topological sort. With this ordering, all edges $(v_i, v_j)$ in the graph satisfy $i<j$.

Then, we're going to modify the graph to ensure that the path length property holds for $v_1, v_2, v_3, \dots$ in that order. Along the way, we're also going to save that path length: that is, for every $v_i$ we've processed, we've saved a value $\ell_i$ such that all paths $v_0 \to v_i$ have length $\ell_i$ in the modified graph. If there is no path $v_0 \to v_i$, we can set $\ell_i = \infty$. We begin by setting $\ell_0 = 0$.

Suppose that we've processed $v_1, v_2, \dots, v_{k-1}$ and we are now processing $v_k$. Begin by finding the in-neighborhood $N^-(v_k)$ in the original graph $G$: the set $\{v_i \in V(G) : (v_i, v_k) \in E(G)\}$.

Let $\ell_k = 1 + \max\{\ell_i : v_i \in N^-(v_k)\}$. This will be the target path length we shoot for.

Replace every edge $(v_i, v_k)$ by a path of length $\ell_k - \ell_i$ whose internal vertices are entirely new.

When we're done, it's true that every path $v_0 \to v_k$ in the new graph has length $\ell_k$. On any such path, take the last vertex $v_i \in \{v_0, v_1, \dots, v_{k-1}\}$ preceding $\ell_k$. We must have taken $\ell_i$ steps to get to $v_i$, and then followed the newly added path of length $\ell_k - \ell_i$, so the total length of the path is $\ell_k$.

Also, every newly added vertex $v$ has this property. If $v$ was added when we were building a path $v_i \to v_j$, then every path $v_0 \to v$ is the composition of a path $v_0 \to v_i$, plus the initial segment of the path $v_i \to v_j$ ending at $v$. All paths $v_0 \to v_i$ have the same length, so we conclude all paths $v_0 \to v$ have the same length.

Also, we haven't affected any paths from $v_0$ to previously processed vertices. We have only changed how we get from previous vertices to $v_k$, and there's no way to get back from $v_k$ to any vertices we've seen.

So when we finally process the last vertex $v_n$, the graph will have the desired property for all vertices in $V'$.

Title: From DAGs to levelled DAGs


问:
Derivation of the Fourier Sine Transform of a Rational Function
I would like to enquire about the derivation of the solution of this integral. The Fourier sine transform $F$ of function $f$ is defined as
$$F(y):=\int_0^\infty f(x)
\sin(xy)\,\mathrm dx, \ y>0.$$
In the Harry Bateman's Tables of Integral Transforms page 71, Eq.28 the Fourier sine transform of $x^{2\nu}(x^2+a^2)^{-\mu-1}$ is tabulated as:
$$\frac{1}{2}a^{2\nu-2\mu}\frac{\Gamma(1+\nu)\Gamma(\mu-\nu)}{\Gamma(\mu+1)}y \:_1\text{F}_2(\nu+1;\nu+1-\mu,3/2;a^2y^2/4)\:+\:4^{\nu-\mu-1}\sqrt{\pi}\frac{\Gamma(\nu-\mu)}{\Gamma(\mu-\nu+3/2)}y^{2\mu-2\nu+1}\:_1\text{F}_2(\mu+1;\mu-\nu+3/2,\mu-\nu+1;a^2y^2/4)
$$
How is this derived?

In particular, With $y=1;\ a=1;\ \nu=1/4;\ \mu=0$, how do we derive
$$\quad \int_{0}^{\infty} \frac{\sqrt{x}\sin(x)}{1+x^2} dx =\frac{\pi}{2\sqrt{2}\:e}\left(-e^2\text{erfc}(1)+\text{erf}(1) +1\right)$$
without using series expansion, but rather through, say, a contour integral?

For the functions erf, erfi, erfc, see : http://mathworld.wolfram.com/Erf.html , http://mathworld.wolfram.com/Erfi.html , http://mathworld.wolfram.com/Erfc.html

答:
UPDATE:

The Wikipedia article on Ramanujan's Master Theorem now mentions the method of brackets.

In order to evaluate the general case, namely  $$\int_{0}^{\infty} \frac{x^{2\ \nu}\sin(yx)}{ (x^{2}+a^{2})^{ \mu +1}} \, \mathrm dx \, , \quad (-1 < \nu < \mu+1, \ y >0, \ a >0),$$ I'm going to use a technique referred to as the method of brackets.

The method of brackets is a generalization of Ramanujan's Master Theorem.

The term bracket refers to the association of the symbol $\langle a \rangle$ with the divergent integral $\int_{0}^{\infty} x^{a-1}  \mathrm dx$.

I think some of the theory behind this technique is still a work in progress.

You can read about the method  in the following papers:

Definite integrals by the method of brackets. Part 1

The method of brackets. Part 2: Examples and applications

Evaluation of entries in Gradshteyn and Ryzhik employing the method of brackets

On the Method of Brackets: Rules, Examples, Interpretations and Modifications

The hypergeometric representation of $\sin(yx)$ is $$yx \, _{0}F_{1} \left(; \frac{3}{2}; - \frac{-(yx)^{2}}{4} \right) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)} \frac{(yx)^{2n+1}}{4^{n}} =  \sum_{n=0}^{\infty} \phi_{n} \, \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)} \frac{(yx)^{2n+1}}{4^{n}}.$$

And according to Rule 3.1 in the first paper, the function  $(x^{2}+a^{2})^{-\mu -1}$ is assigned the bracket series $$\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \phi_{k,m} \,  \frac{x^{2k} a^{2m}}{\Gamma(\mu+1)} \langle k + m+ \mu+1 \rangle.$$

Therefore, the integral $ \int_{0}^{\infty} x^{2 \nu}\sin(yx) (x^{2}+a^{2})^{- \mu -1} \, \mathrm dx$ corresponds to the bracket series $$\frac{1}{\Gamma(\mu+1)}\sum_{k=0}^{\infty} \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \phi_{k,m,n} \, \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2}+n \right)}  \frac{y^{2n+1}}{4^{n}} a^{2m} \langle k + m+ \mu+1 \rangle\langle2k+2n+2\nu+2 \rangle.   $$

And the matrix equation corresponding to the vanishing of the brackets is
$$ \begin{pmatrix}
    1 & 1 & 0 \\
    2 & 0 & 2 \\
    \end{pmatrix} \begin{pmatrix} k \\ m \\ n \end{pmatrix} = \begin{pmatrix} -\mu -1 \\ -2\nu -2 \end{pmatrix}.$$

Since the matrix equation is underdetermined, we will need to divide the evaluation of the bracket series into several cases.

Let's first let $m$ be a free parameter.

Then the brackets vanish if $k= -m-\mu-1$ and $n = m+ \mu-\nu$, with corresponding determinant $2$.

So according to Rule 3.3 in the first paper, the contribution to the integral is $$\begin{align} S_{1} &= \small \frac{1}{2\Gamma(\mu+1)} \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{\Gamma \left(\frac{3}{2} \right)}{\Gamma \left(\frac{3}{2} + \mu - \nu + m \right)} \frac{y^{2m+ 2\mu - 2\nu+1}}{4^{m + \mu - \nu}} a^{2m} \, \Gamma \left(m+ \mu + 1 \right) \Gamma \left(\nu -\mu - m\right) \\  &= \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1} } \frac{1}{\Gamma(\mu - \nu + \frac{3}{2})} \sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{\Gamma (\mu - \nu + \frac{3}{2}) }{\Gamma \left(\mu - \nu + \frac{3}{2} + m \right)} \, \frac{\Gamma( \mu + 1+ m)}{\Gamma(\mu+1)}   \Gamma \left(\nu -\mu - m\right)\left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &= \small  \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{1}{\Gamma(\mu - \nu + \frac{3}{2})}\sum_{m=0}^{\infty} \frac{(-1)^m}{m!} \frac{(\mu+1)_{m} }{(\mu-\nu + \frac{3}{2})_{m}}   \, \frac{(-1)^{m}\Gamma(\mu - \nu +1) \Gamma(\nu - \mu)}{\Gamma(\mu - \nu +1 + m)}\left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &= \small   \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{\Gamma(\nu - \mu)}{\Gamma(\mu - \nu + \frac{3}{2})}\sum_{m=0}^{\infty} \frac{1}{m!} \frac{(\mu+1)_{m} }{(\mu-\nu + \frac{3}{2})_{m}(\mu - \nu+1)_{m}}   \left(\frac{a^{2}y^{2}}{4} \right)^{m} \\ &=  \small \frac{\sqrt{\pi} \, y^{2 \mu - 2 \nu+1}}{4^{\mu - \nu+1}} \frac{\Gamma(\nu - \mu)}{\Gamma(\mu - \nu + \frac{3}{2})} \,  _{1}F_{2} \left(\mu+ 1; \mu - \nu + \frac{3}{2}, \mu - \nu+1; \frac{a^{2}y^{2}}{4} \right). \end{align}$$

Next let $n $ be a free parameter.

Then the brackets vanish if $k = - n - \nu- 1 $ and $m = n + \nu - \mu$, with corresponding determinant $-2$.

So according again to Rule 3.3, the contribution to the integral is $$\begin{align} S_{2} &= \small \frac{1}{|-2| \Gamma(\mu+1)} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{\Gamma \left(\frac{3}{2} \right) }{\Gamma \left(\frac{3}{2}+n \right)} \frac{y^{2n+1}}{4^{n}} a^{2n+2\nu - 2 \mu} \, \Gamma(n + \nu +1) \Gamma(\mu - \nu-n) \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{\Gamma \left(\frac{3}{2} \right) }{\Gamma \left(\frac{3}{2}+n \right)} \frac{\Gamma(\nu+1+n)}{\Gamma(\nu+1)} \Gamma(\mu - \nu-n) \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &= \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\frac{(\nu+1)_{n}}{\left(\frac{3}{2} \right)_{n}} \frac{(-1)^{n}\Gamma(\nu- \mu +1) \Gamma(\mu - \nu)}{\Gamma(\nu - \mu+1+n)} \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &=  \small \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \Gamma(\mu - \nu) \sum_{n=0}^{\infty} \frac{1}{n!}\frac{(\nu+1)_{n}}{\left(\frac{3}{2} \right)_{n} (\nu-\mu+1)_{n}} \left(\frac{a^{2}y^{2}}{4} \right)^{n} \\ &=  \small  \frac{a^{2 \nu - 2 \mu}y}{2\Gamma(\mu+1)} \Gamma(\nu +1) \Gamma(\mu - \nu) \, _{1}F_{2} \left(\nu+1; \nu-\mu+1, \frac{3}{2}; \frac{a^{2}y^{2}}{4} \right). \end{align}$$

Finally, if we let $k$ be a free parameter, we end up with a $_{3}F_{0}$ series. These series never converge.  So according to Rule 3.4 in the first paper, we should discard this contribution.

EDIT: A $_{3}F_{0}$ series does reduce to a finite polynomial if at least one parameter is a nonpositive integer, but that's not a possibility here.

According to Rule 3.4, the value of the integral should be $S_{1}+S_{2}$.

Indeed,  if we add $S_{1}$ and $S_{2}$ we miraculously get the result stated in Harry Bateman's Tables of Integral Transforms.

However, I'm not sure how to interpret the result when $\nu - \mu$ is zero or a negative integer.


问:
Are the numbers calculated from a log resolution birational invariants?
Let $C = V(x^2 - y^3)$ be the cuspidate cubic which sits inside $\mathbb{C}^2$. Let $\pi: \text{Bl}_0(\mathbb{C}) \to \mathbb{C}^2$ be the blow up of the origin. The reduction of the total transform of $C$ has local equation $y^2(t^2-1)=0$ and the total transform is $\widehat{C} = 2 E_1 + C_1$. Blow up again, the local equation is $t^3 u^2(t-u)=0$ with $\widehat{C} = 2E_1 + 3E_2 + C_2$. Finally, blow up again, the local equation is $t^6(v-1)^3v^2(1-2v)=0, $ with $\widehat{C} = 2E_1 + 3E_2 + 6E_3 + C_3.$

With $$K_{X/Y} := K_{\text{Bl}_0(\mathbb{C}^2)/\mathbb{C}^2} = E_1 + 2E_2 +4E_3,$$ and recalling that the log canonical threshold is given by $$\min_j \frac{k_j+1}{a_j},$$ where $$K_{X/Y} = \sum_j k_j E_j$$ and $$\widehat{C} = \sum_j a_j E_j,$$ we see that $$\text{lct} = \min \left \{  \frac{1+1}{2}, \frac{2+1}{3}, \frac{4+1}{6} \right \} = \frac{5}{6}.$$

Question: Are the numbers $\frac{k_j+1}{a_j}$ birational invariants?

答: These numbers are known as the discrepancies, and they are not even invariants of $C$, much less birational invariants ($C$ is birational to $\Bbb A^1$, for instance). To be more precise, the numbers you get can depend on the resolution of singularities you take, but they always behave somewhat nicely - if for one resolution of singularities, all $a_i$ are $>0$, $\geq 0$, $>-1$, or $\geq -1$, then the same is true for any other resolution of singularities. One says that if all $a_i$ are $>0$, the singularity is terminal; $\geq 0$, canonical;  $>-1$, log terminal; $\geq -1$, log canonical.


问:
Function is not continuous as a multivariable function but continuous as a single variable one
It's known that a multivariable function can be continuous in one variable, but doesn't necessary need to be also continuous as a multivariable one (see $\frac{xy}{x^2+y^2}$ as a counter example).
Anyway, I can't see what's wrong with the following proof, in which it seems that the claim is actually true:

Assuming $f(x,y) \stackrel{y \rightarrow 0}{\rightarrow}f(x,0)$ for every $x$, and $f(x,y) \stackrel{x \rightarrow 0}{\rightarrow}f(0,y)$ for every $y$: for every $\epsilon > 0$, there are $\delta_1 > 0, \delta_2 > 0 $ such that for every $0<|x|<\delta_1, 0<|y|<\delta_2$:
$|f(x,y)-f(0,0)| < |f(x,y)-f(0,y)| + |f(0,y)-f(0,0)| < \frac{\epsilon}{2}+ \frac{\epsilon}{2}<\epsilon$.

I used the given continuity of $f$ according to $x$ and $y$ separetly, and the triangle inequality.
I fail to see what is flawed in my line of thinking, and would like it if someone could point out what is the problem.

答: The assertion$$\lim_{y\to0}f(x,y)=f(x,0)\tag1$$lacks quantifiers. What is $x$? If what you mean is that, for each $x$, $(1)$ holds, then what this means is that, for each $x$ and for each $\varepsilon>0$, there is some $\delta>0$ (depending on $x$ and on $\varepsilon$) such that$$|y|<\delta\implies\bigl|f(x,y)-f(x,0)\bigr|<\varepsilon.$$So, the $\delta_1$ from your proof does not exist. To be more precise, there is one for each individual $x$.


问:
Evaluation of a complex integral on a closed path

Evaluate $\oint_{\gamma}\frac{f(z)}{z^{4}} dz$, where $\gamma:[0,2\pi ]\rightarrow\mathbb{C}$, $\gamma(t) :=\cos(t) +i\sin(t)$ and $f(z)=\frac{e^{z}+e^{-z}}{2}$.

For a Theorem, I know that
$f^{(n)}(z_{0})=\frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-z_{0})^{n+1}} dz$ so, because $z_{0}=0$ we have that

$\oint_{\gamma}\frac{f(z)}{z^{4}} =2\pi i  \frac{f^{'''}(z_0)}{3!}=2\pi i\frac{e^{0}-e^{0}}{3!2}=0$
Is this right?

答:
Notice that : \begin{aligned}\int_{\gamma}{\frac{f\left(z\right)}{z^{4}}\,\mathrm{d}z}=\oint_{\left|z\right|=1}{\frac{\cosh{z}}{z^{4}}\,\mathrm{d}z}=2\pi\mathrm{i}\,\mathrm{Res}\left(z\mapsto\frac{\cosh{z}}{z^{4}},0\right)\end{aligned}

Since $ \left(\forall z\in\mathbb{C}\right),\ \frac{\cosh{z}}{z^{4}}=\sum\limits_{n=0}^{+\infty}{\frac{z^{2n-4}}{\left(2n\right)!}} $, then $ \mathrm{Res}\left(z\mapsto\frac{\cosh{z}}{z^{4}},0\right) = 0$.

Thus : $$ \int_{\gamma}{\frac{f\left(z\right)}{z^{4}}\,\mathrm{d}z}=0 $$


问:
In a topology space, a set is open iff the complement is closed?
I am learning topology spaces and in the notes it states a theorem:

For a topology space $(X,\tau)$, and $A\subseteq X$, $A$ is closed iff $X\setminus A$ is open.

In the notes, the definition of topology is that it should follow the following three rules:

The empty set and the whole set should be in the topology.

Finite, countable, and uncountable union of elements in the topology is still in the topology.

Finite intersection of elements in the topology is still in the topology

And in the notes it says that the proof of this theorem is similar for that of metric space, but we don't have distance in metric space, so we cannot order all set of open sets containing $x$ to get a limit and have a contradiction. I have the proof for cases if $\tau$ is finite and countable, how to prove the uncountable case?

Also, in the notes, the definition of closed set is that any limit of any convergent sequence is still in the set, so I need to prove this definition is equivalent to what the theorem states.

答:
Another way to define closed sets in the context of metric spaces is given below.

Definitions

Let $(X,d_{X})$ be a metric space.

We say that $E\subseteq X$ is closed iff $E\supseteq\partial E$, where $\partial E$ indicates the boundary of $E$.

Moreover, we also say that $E\subseteq X$ is open iff $E\cap\partial E = \varnothing$.

Proposition

The set $E\subseteq X$ is closed iff $E^{c}$ is open.

Proof

Based on this answer, we are going to prove the proposed claim.

In order to do so, we shall consider the implication $(\Rightarrow)$ first.

If $E$ is closed (which means that $E = \overline{E} = \text{int}(E)\cup\partial E$), then we have that
\begin{align*}
E^{c} & = X - \text{int}(E)\cup\partial E = \text{ext}(E) = \text{int}(E^{c})
\end{align*}

from which we conclude that $E^{c}$ is open.

Conversely, let us prove the converse implication $(\Leftarrow)$.

If $E^{c}$ is open, then we have that

\begin{align*}
E^{c}\cap\partial E^{c} = \varnothing & \Longleftrightarrow E^{c}\cap\partial E = \varnothing\\\\
& \Longleftrightarrow \partial E\subseteq E
\end{align*}

whence we conclude that $E$ is closed.

Hopefully this helps!


问:
Solution verification for a probability question

A dental insurance policy covers three procedures: orthodontics, fillings, and extractions.
During the life of the policy, the probability that the policyholder needs

orthodontic work is $1/2$;

orthodontic work or a filling is $2/3$;

orthodontic work or an extraction is $3/4$,

a filling and an extraction is $1/8$.

The need for orthodontic work is independent of the need for a filling and is independent
of the need for an extraction. Calculate the probability that the policyholder
will need a filling or an extraction during the life of the policy.

My attempt:
Let $O, F, E$ denote the events that a policyholder would need orthodontic work, a filling, or an extraction respectively. If $A$ is an event, let $A'$ denote the complement of $A$.

Now, $P(O\cup E) = 1-P(O'\cap E') = 1-\left[P(O')\cdot P(E')\right] = 1-\left[\dfrac{ P(E')}{2}\right] \implies \dfrac{3}{4} = 1-\left[\dfrac{ P(E')}{2}\right] \implies P(E') = \dfrac{1}{2}$

Similarly, $P(O\cup F) = 1-P(O'\cap F') = 1-\left[P(O')\cdot P(F')\right] \implies \dfrac{2}{3} = 1-\left[\dfrac{ P(F')}{2}\right]\implies P(F') = \dfrac{2}{3}$

So, $P(F \cup E) = 1-P(F' \cap E') = 1-\left[\dfrac{2}{3} \cdot \dfrac{1}{2}\right] = \dfrac{2}{3}$, which is, unfortunately, not the correct response. Can someone please review my solution and let me know where I went wrong? Thanks!

答:
You correctly got $P(F')=2/3$, i.e. $P(F)=1/3$ and $P(E')=1/2$ i.e. $P(E)=1/2$.

Your mistake was then that you assumed the need for filling and the need for extraction were independent. That is not mentioned in the problem. Had those events been independent, you would've had $P(F\cap E)=P(F)P(E)=1/6$, but it is not that - it is $1/8$.

Instead, you should've just calculated $P(F\cup E)=P(F)+P(E)-P(F\cap E)=1/2+1/3-1/8=17/24$.


问:
Local martingale that is not integrable
I have just been introduced to the notion of a local martingale, and I have asked myself the question whether the fact that $(X_{t})_{t\geq 0}$ is a local martingale with associated $\{\tau_{n}\}_{n}$ stopping times that go to infinity implies that for every $X_{t}$ is indeed integrable.

I cannot prove it and thus I believe it is false but I am struggling to find counterexamples. Could anyone give me a hint as to what such a process $X$ and the stopping times would look like so that they are indeed a local martingale however not integrable.

答:
This is probably not the most exotic example, but it is indeed an example of a non-integrable local martingale:

Let $X_0$ be any non-integrable random variable and consider the proces
$$X_t = X_0 \quad \text{ for all $t\geq 0$}.$$
Now if we let
$$\tau_n = \begin{cases} 0 &, \text{ if } X_0 > n \\ \infty & , \text{ if }X_0 \leq n \end{cases},$$
then $(\tau_n)_{n\in \mathbb{N}}$ is an increasing sequence of stopping times going to infinity and the stopped proces
$$X_{\min(\tau,t)} 1_{\{\tau_n > 0 \}} = X_0 1_{\{X_0 \leq n\} },$$
is bounded, constant (with respect to $t$) and thus a martingale. We conclude, that the proces $(X_t)_{t\geq 0}$ is indeed a local martingale.

From this odd example we can actually construct a slightly more interesting example. Let $(X_t)_{t \geq 0}$ be as above and consider a true martingale $(Y_t)_{t\geq 0}$ and let
$$Z_t = X_t + Y_t.$$
Then $Z_t$ is not integrable, but since the sum of local martingales is again a local martingale, we actually get that $(Z_t)_{t\geq 0}$ is a local martingale.

Title: Local martingale that is not integrable