问:
Property of abelianization
This is related to this old MO question which wasn't answered properly, though I don't feel I phrased the question in the best way (or posted it on the right site)

Define the abelianization of a group $G$ to be the quotient group $G_{ab} := G/[G,G]$, where $[G,G]$ is the commutator subgroup. I want to know how this definition implies the following property of abelianization.

Let $\phi: G \to G_{ab}$ be the canonical surjection. For any abelian group $H$ and homomorphism $f:G\to H$, there exists a unique homomorphism $F: G_{ab} \to H$ such that $f = F\circ \phi$.

This is the ability to 'descend to a homomorphism' that I explained so badly in my initial post, though at the time it was the only terminology I heard being used.

I am aware that these two definitions are equivalent but I have not yet seen a proof, nor managed to prove it myself. Please do point me to an online proof if you know of one.

答:
This generalizes to any number of other situations: e.g., if $G$ is a group and $n\gt 0$ a positive integer, let $N$ be the subgroup of $G$ generated by all elements of the form $g^n$, with $g\in G$. Then $G/N$ has exponent $n$, and if $\pi\colon G\to G/N$ is the canonical surjection, then for any group $H$ of exponent $n$ and any group homomorphism $f\colon G\to H$, there exists a unique homomoprhism $F\colon G/N \to H$ such that $f=F\circ \pi$.

For the general context:

Definition. Let $F_{\infty}$ be the free group of countably infinite rank, and let $w\in F_{\infty}$. We say that $w$ is an identity of $G$ if and only if for every group homomorphism $f\colon F_{\infty}\to G$, we have $f(w)=1$. Equivalently, every evaluation of $w$ in $G$ is the identity.

Definition. Let $S\subseteq F_{\infty}$. The variety of groups determined by $S$ is the collection of all groups $G$ for which $w$ is an identity of $G$ for every $w\in S$.

Proposition. Let $S\subseteq F_{\infty}$. The variety of groups $\mathfrak{V}$ determined by $S$ satisfies:

$\mathfrak{V}$ is closed under subgroups: if $G\in\mathfrak{V}$ and $H\lt G$, then $H\in\mathfrak{V}$.

$\mathfrak{V}$ is closed under homomorphic images: if $G\in\mathfrak{V}$ and $\pi\colon G\to K$ is a surjective group homomorphism, then $K\in\mathfrak{V}$.

$\mathfrak{V}$ is closed under arbitrary direct products: if $\{G_i\}_{i\in I}$ is a family of groups (of arbitrary size), and $G_i\in\mathfrak{V}$ for every $i\in I$, then $\prod\limits_{i\in I}G_i\in\mathfrak{V}$.

Birkhoff's HSP Theorem. Let $\mathcal{C}$ be a nonempty class of groups. Then $\mathcal{C}$ is a variety of groups if and only if $\mathcal{C}$ is closed under (H) homomorphic images, (S) subgroups, and (P) arbitrary direct products.

Proposition. Let $G$ be a group, and let $\mathfrak{V}$ be a variety of groups. Then there is a smallest normal subgroup of $G$, $\mathfrak{V}(G)$, such that $G/\mathfrak{V}(G)\in \mathfrak{V}$. If $\mathfrak{V}$ is determined as a variety by the set of identities $S\subseteq F_{\infty}$, then $\mathfrak{V}(G)$ is the subgroup of $G$ generated by the images of $S$ under all homomorphisms $F_{\infty}\to G$.

Proof. Clearly, there is at least one normal subgroup $N$ such that $G/N\in\mathfrak{V}$, namely $N=G$. If $\{N_i\}$ is a family of normal subgroups of $G$ such that $G/N_i\in\mathfrak{V}$ for every $i$, then $\prod (G/N_i)\in\mathfrak{V}$, since it's a product of groups in $\mathfrak{V}$. And the canonical map of $G$ into $\prod(G/N_i)$ has kernel $\cap N_i$, hence $G/N_i$ is isomorphic to a subgroup of a group in $\mathfrak{V}$, hence lies in $\mathfrak{V}$. Thus, we can take $\mathfrak{V}(G)$ to be the intersection of all normal subgroups $N\triangleleft G$ such that $G/N\in\mathfrak{V}$.

For the second description, let $f\colon F_{\infty}\to G$ be a homomorphism. Then $\pi\circ f\colon F_{\infty}\to G\to G/\mathfrak{V}(G)$ is a map from $F_{\infty}$ into a group in $\mathfrak{V}$, and since $S$ determines $\mathfrak{V}$, then $S$ must lie in th ekernel of this map. Hence $f(S)\subseteq \mathfrak{V}(G)$. That is, $\mathfrak{V}(G)$ contains all images of $S$ under homomorphisms $F_{\infty}\to G$. Now let $N$ be the subgroup generated by all such images. This group is normal, because if $f\colon F_{\infty}\to G$ is any homomorphism, then $f(S)^g$ is the image of $S$ under the homomorphism $\varphi_{g}\circ f$ (where $\varphi_g$ is the inner automorphism of $G$ determined by $g$), hence $f(S)^g\subseteq N$ for all $g\in G$. This holds for all $f$, so the generating set of $N$ is mapped to itself by conjugation. This shows that $N$ is normal; and $G/N\in\mathfrak{V}(G)$, because every homomoprhism $F_{\infty}\to G/N$ lifts to a homomorphism $F_{\infty}\to G$, and then the image of $S$ in $G/N$ is trivial. Therefore, $\mathfrak{V}(G)\subseteq N$, proving equality. QED

The normal subgroup $\mathfrak{V}(G)$ is called the verbal subgroup of $G$ corresponding to $\mathfrak{V}$.

Theorem. Let $G$ be a group, and let $\mathfrak{V}$ be a variety of groups. Then $\mathfrak{V}(G)$ is characterised by the following universal property: $\mathfrak{V}(G)\triangleleft G$, $G/\mathfrak{V}(G)\in\mathfrak{V}$,  and for every $H\in\mathfrak{V}$ and every group homomorphism $f\colon G\to H$, there is a unique group homomorphism $F\colon G/\mathfrak{V}(G)\to H$ such that $f=F\circ \pi$, with $\pi\colon G\to G/\mathfrak{V}(G)$ the canonical projection.

Proof. First, we show $\mathfrak{V}(G)$ has this property. Let $H\in\mathfrak{V}(G)$, and let $f\colon G\to H$ be any group homomorphism. By the Isomorphism Theorem, $G/\mathrm{ker}(f)\cong f(G)\lt H$; since $f(G)$ is a subgroup of a group in $\mathfrak{V}$, then $f(G)\in\mathfrak{V}$, hence $G/\mathrm{ker}(f)\in\mathfrak{V}$ (it is isomorphic to a group in $\mathfrak{V}$). Since $\mathfrak{V}(G)$ is the smallest normal subgroup of $G$ with a quotient in $\mathfrak{V}$, then $\mathfrak{V}(G)\subseteq\mathrm{ker}(f)$, by the Universal property of the quotient, $f$ factors uniquely through $G/\mathfrak{V}(G)$, yielding $F$. That $\mathfrak{V}(G)\triangleleft G$ and $G/\mathfrak{V}(G)\in\mathfrak{V}$ has already been established.

Finally, we show that a normal subgroup $N$ satisfying this universal property is in fact $\mathfrak{V}(G)$. We know that $\mathfrak{V}(G)\subseteq N$ by construction. And by the universal property, the canonical projection $G\to G/\mathfrak{V}(G)$ factors through $G/N$, so $N\subseteq \mathfrak{V}(G)$, proving equality. QED

Now, the class of all abelian group is a variety (closed under homomorphism images, subgroups, and arbitrary direct products). In fact, the class is determined by the single identity $x^{-1}y^{-1}xy$: a group $G$ is abelian if and only if for every $g,h\in G$, $g^{-1}h^{-1}gh = 1$; if we let $\mathfrak{A}$ denote the variety of all abelian groups, then $\mathfrak{A}(G)$ is precisely the subgroup generated by all values of $x^{-1}y^{-1}xy$, i.e., the commutator subgroup. So the commutator subgroup has the desired universal property, hence $G^{\rm ab} = G/[G,G]$ has the desired universal property.

If you want to replace "abelian group" with "abelian group of exponent $n$", then you would use the set $S=\{x^{-1}y^{-1}xy, z^n\}$, and use the subgroup of $G$ generated by the commutators and all $n$th powers. If you replace "abelian group" with "nilpotent of class $c$", then you replace the commutator subgroup with the $(c+1)$st term of the lower central series, determined by the word
$$[[\cdots [x_1,x_2],x_3]\cdots x_{n+1}].$$
And so on.

See also this discussion on commutator-center duality for more on varieties and verbal subgroups.

Title: Property of abelianization


问:
Find the asymptotic distribution of δ
Let $X_1, \ldots, X_n$ i.i.d. from the $\mathrm{Bernoulli}(p)$ distribution. Then $X = X_1 +\ldots+X_n$ follows the $\mathrm{Binomial}(n,p)$ distribution.
We wish to estimate $g(p)=p(1-p)$ whose UMVUE is
$$\delta=\frac{X(n−X)}{n(n−1)}
$$
Find the asymptotic distribution of $\delta$.

I tried:

let
$$
\delta^*=\frac{X(n-X))}{n^2}=\frac{X}{n}-\left(\frac{X}{n}\right)^2=\frac{\sum_{i=0}^n X_i}{n}-\left(\frac{\sum_{i=0}^n X_i}{n}\right)^2=\bar X-(\bar X)^2
$$
so,
$$
\delta=\frac{X(n-X)}{n(n-1)}=\frac{n}{n-1}\big(\bar X - (\bar X)^2\big)
$$
Let, $$h(\mu) = \left(\frac{n}{n-1}\right)\big(\mu-(\mu)^2\big) $$
So,
$$h^\prime(\mu)=\frac{n}{n-1}-2\mu\left(\frac{n}{n-1}\right)$$
hence, the asymptotic distribution of $\delta$ is
$$
\sqrt n {\big(h(\bar x) - h(\mu)\big)}\xrightarrow{L} N\big(0, (h^\prime(\mu))^2\sigma^2\big)=N\big(0, \big(\frac{n}{n-1}-2\mu(\frac{n}{n-1})\big)^2\big(np(1-p)\big)^2\big)
$$
By the Delta method.

But, to use the Delta method, I must prove $\delta$ and $g(p)$ are consistent. How do I prove it? Thank you very much!

答: To show that $\delta$ and $g(p)$ are consistent, use the partial fraction decomposition $\delta=\frac{X(n-X)}{n(n-1)}=\frac{X}{n-1}-\frac{X^2}{(n-1)n}$. Taking the limit, we get $\lim_{n\rightarrow\infty}\delta=p-p^2=p(1-p)=g(p)$.

ContentLicense: CC BY-SA 4.0

Title: Find the asymptotic distribution of δ


问:
How to evaluate $ \int_0^{2 \pi} \sin((2n-1)x) \ln(\sin(x)+5/4) dx $?
Is there a closed form for

$$f(2n-1) = \int_0^{2 \pi} \sin((2n-1)x) \ln(\sin(x)+5/4) dx $$

Of course this comes from searching the fourier series of $ \ln(\sin(x)+5/4) dx $.

In fact, it is equivalent to searching for that Fourier series since

$\int_0^{2 \pi} \sin((2n)x) \ln(\sin(x)+5/4) dx = \int_0^{2 \pi} \cos(nx) \ln(\sin(x)+5/4) dx = 0$

I know $f(4n + 1)$ has the opposite sign as $f(4n + 3)$.

Maybe using the fourier series of $\sin(x)^n$ and the taylor series of $\ln(\sin(x)+5/4) $ helps.

Or maybe we can use formulas such as $\cos(nx) = 2 \cos((n-1)x) \cos(x) - \cos((n-2) x) $ and the sine analogue.

Many thanks.

答:
Without Fourier series.

Just a bit of patience with integrals to see obvious patterns.
$$I_n=\int_0^{2\pi}\sin [(2 n-1) x]\, \log \left(\sin (x)+\frac{5}{4}\right)\,dx=\color{blue}{(-1)^{n+1}\frac{  4^{1-n}}{2 n-1}\pi}$$


问:
Determinant of unspecified matrices
Suppose A and B are $5\times 5$ matrices with $\det(A) = -1/3$ and $\det(B) = 6$, find the determinant of $ 2AB$.

Solution:

$$= \det(2AB) $$
$$= 2^5 \det(A)\det(B) $$
$$= (32)(-1/3)(6)$$
$$= -64$$

I understand how all of this works, except for where $2^5$ comes from, can anyone explain how this happens?

答: When you multiply a matrix by $2$, that multiplies all the rows of the matrix by $2$. The determinant is linear in each row, and the matrix $AB$ has five rows, so it multiplies the resulting determinant by $2^5$.</p>

ContentLicense: CC BY-SA 3.0

Title: Determinant of unspecified matrices


问:
hyperbola curve formula in 3 dimensions
Cartesian formula for 2d hyperbola curve is $x^2/a^2-y^2/b^2 = 1$.

What is the formula for a 3d hyperbola curve?

答:
What is a 3D hyperbola curve?  How about $\frac {x^2}{a^2}-\frac {y^2}{b^2}=1, z=0$.  If you want a curve in 3D, you need two equations, just like a line.  I suspect this is not what you want, so please explain.

Alternately, you might want $\frac {x^2}{a^2}-\frac {y^2+z^2}{b^2}=1$, which is a surface that rotates the 2D curve around the $x$ axis.

Added in response to comment:  Given your three points, you can solve for the $a,b,c$ that are the equation of the plane.  Be careful-we have been using $a,b$ as parameters of the hyperbola and now you are reusing them.  The $m$ in my previous comment corresponds to $\frac ba$ here, but my plane went through the $z$ axis while if $c \ne 0$ it will miss the $z$ axis by that much.

A hyperbola with its axis parallel to $z$ that is in the plane $ax+by=0$ is $\frac {z^2}{d^2}-\frac {x^2+y^2}{e^2}=1,y=-\frac ba x$  To get it into the plane $ax+by=c$ we add $\frac c{\sqrt {1+(\frac ab)^2}}$ to $x$ and $\frac ab\frac c{\sqrt {1+(\frac ab)^2}}$ to $y$ giving $$\left\{(x',y',z)\left|\frac {z^2}{d^2}-\frac {x^2+y^2}{e^2}=1,y=-\frac ba x,x'=x+,y'=y+\frac ab\frac c{\sqrt {1+(\frac ab)^2}}\right.\right\}$$

Title: hyperbola curve formula in 3 dimensions


问:
Build a confidence interval
I am studying descriptive statistic and I am trying to do this exercise but I am not getting.

A certain manufacturer guarantees that his machines produce only $5\%$ of defective parts. In order to verify the veracity of this statement, $1000$ pieces produced by a machine were analyzed, of which $40$ pieces had some type of defect. Build a $95\%$ confidence interval for the proportion of defective parts and discuss the manufacturer's claim.

答: $\hat p=\frac{40}{1000}$, so a 95% confidence interval for the true population proportion is $\hat p\pm z*s.e.=\frac{40}{1000}\pm 1.96 \sqrt{\frac{\frac{40}{1000}\left(1-\frac{40}{1000}\right)}{1000}}=(0.02785432, 0.05214568)$. 5% is contained in this interval, so we believe the manufacturer's claim that 5% of his items are defective is reasonable.

Title: Build a confidence interval


问:
Two mysterious missing angles in the sine values of acute angle list?
There is a well known list of 5 for  trig values of special angles between 0$^o$ to $90^o$:

$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$

If wee add 15$^o$ and 75$^o$ , we can  make the angles evenly spaced by 15$^o$ and expand the list of 5 to a list of 7:

$\sin 0^o =\frac{\sqrt {\color{blue}{0}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{red}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{blue}{1}}}{2}$
, $\sin 45^o =\frac{\sqrt {\color{blue}{2}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{blue}{3}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{red}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{blue}{4}}}{2}$

Reformat the list :

$\sin 0^o =\frac{\sqrt {\color{green}{2-\sqrt {4}}}}{2}$,$\sin 15^o =\frac{\sqrt {\color{green}{2-\sqrt3}}}{2}$, $\sin 30^o =\frac{\sqrt {\color{green}{2-\sqrt{1}}}}{2}$

, $\sin 45^o =\frac{\sqrt {\color{green}{2-\sqrt{0}}}}{2}$
, $\sin 60^o =\frac{\sqrt {\color{green}{2+\sqrt{1}}}}{2}$,$\sin 75^o =\frac{\sqrt {\color{green}{2+\sqrt 3}}}{2}$, $\sin 90^o =\frac{\sqrt {\color{green}{2+\sqrt{4}}}}{2}$

Now, we see a nice pattern of $\frac{\sqrt {{2\color{red}{\pm}\sqrt i}}}{2}$ with i=0,1,3,4.

But wait, why is 2 missing from this i list?  Seems like we have two special angles lost. With them, we can expand this to list of 9. Where are these two mysterious angles?  Can you find them and prove they  fit into the pattern? Or, do you find ways to expand this list even further to a list of 11,13,...and find something amazing?

答:
The missing angles are:

$$\sin 22.5º = \frac{\sqrt{2 - \sqrt2}}{2}$$
$$\sin 67.5º = \frac{\sqrt{2 + \sqrt2}}{2}$$

There are other values listed here on Wikipedia, but the pattern is not obvious. For instance, $\sin 18º = \frac{\sqrt5 - 1}{4}$, but $\arcsin \frac{\sqrt5 + 1}{4}$ is not $90º - 72º = 36º$ but $54º$. In fact, $\sin 36º = \frac{\sqrt{10 - \sqrt{20}}}{4}$ and $\sin 72º = \frac{\sqrt{10 + \sqrt{20}}}{4}$.

The nice numbers for the sines of multiples of $18º$ come from the fact that $\sin (3\pi/10) = \cos (2 \pi/10)$, and solving $\sin(3x) = \sin(2x)$  using the triple-angle and double-angle formulas gives a nice quadratic in disguise after substituting $u = x^2$. The formulas involving $\sin(4x)$ and so on do not reduce into quadratics, and so if they can be reduced into radicals at all, the forms will be more complicated.


问:
Imposing conditions on orders of cyclic group to create a group homomorphism
I am trying to solve:

Let $G,H$ be cyclic groups, generated by elements $x,y$. Determine the condition on the orders $m,n$ of $x$ and $y$ so that the map sending $x^i \rightsquigarrow y^i$ is a group homomorphism.

Here is my attempt.

We require that the map $\varphi: \underset{x^i}{G} \underset{\longmapsto}{\to} \underset{y^j}{H}$ is well-defined. That is, if $x^i = x^j$ for $i,j \in \mathbb{Z}$, then $y^i = y^j$. That is, if $i \equiv j \text{ (mod $m$)}$, then $i \equiv j \text{ (mod $n$)}$, i.e., if $i - j = \alpha m$ for some $\alpha \in \mathbb{Z}$, then $i - j = \beta n$ for some $\beta \in \mathbb{Z}$. As $i,j$ are arbitrary and any integer can be written in the form $i - j$, this means that for all $z \in \mathbb{Z}$, $m \mid z$ implies that $n \mid z$, which is true provided that $n \mid m$. To show that this condition is sufficient, let $m = nk$ for some $k \in \mathbb{Z}$ and suppose that $x^i = x^j$ for some $i,j \in \mathbb{Z}$. Then we have
\begin{align*}
i - j = \gamma m = \gamma (nk) = n(\gamma k)
\end{align*}
for some $m \in \mathbb{Z}$, so $i \equiv j \text{ (mod $n$)}$, so $y^i = y^j$, hence $f(x^i) = f(x^j)$, so $\varphi$ is well-defined. I claim that $\varphi$ is then a homomorphism. Indeed, for any $i,j \in \mathbb{Z}$, we have
\begin{align*}
\varphi\left(x^i x^j\right) = \varphi\left(x^{i+j}\right) = y^{i+j} = y^i y^j = \varphi\left(x^i\right) \varphi\left(x^j\right),
\end{align*}
as desired.

Is the logic correct? The only thing I am still not completely certain about is the pairing of $i-j$ with some arbitrary integer $z$, but there is surely at least a surjective mapping from differences of integers with the integers themselves, so I believe the logic is valid.

答:
I believe you have the right condition there:  $n\mid m $.

That's because the order of $\varphi  (x) =y $ is $n $; but it must also divide $m $, since $(\varphi  (x))^m=\varphi  (x^m)=\varphi  (e)=e $.

On the other hand,  well-definedness doesn't seem to be much of an issue (anymore)  because the homomorphism property will ensure it.

So, the condition is necessary and sufficient.


问:
How to prove that $\Gamma(x) \leq x^x$ for $x \geq 1$
Where can I find a (simple?) proof that $\Gamma(x) \leq x^x \,\,\,\, \forall x \geq 1$?

$\Gamma(x)$ is the Gamma function defined as

$$ \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, \mathrm{d}t
$$

答:
With the substitution $t = xu$, we have
\begin{align*}
\Gamma(x) &= x^x\int_0^\infty u^{x - 1} \mathrm{e}^{- xu}\mathrm{d}u \\
&= x^x\int_0^\infty \mathrm{e}^{(x - 1)\ln u - xu}\mathrm{d}u \\
&\le x^x\int_0^\infty \mathrm{e}^{(x - 1)u - xu}\mathrm{d}u \tag{1}\\
&= x^x
\end{align*}
where in (1) we have used $\ln u \le u$ for all $u > 0$.

We are done.


问:
A particular relation between two matrices
I'd like to consider the consequences of the following relation between two symmetric square matrices $A$ and $B$
$$SA^{-1}S^T=B$$
$$S^TB^{-1}S=A$$
There are no inversion conditions on the matrix $S$.

Now, say we have the eigenvalues/vectors of $A$, can we say anything about the eigenvalues/vectors of $B$?

Edit: I might also add the conditions on $S$
$$S^TS=aA^2$$
$$SS^T=bB^2$$
where $a$ and $b$ are real numbers.

答:
When $A$ and $B$ are real symmetric matrices of the same sizes, the conditions $SA^{-1}S^T=B$ and $S^TB^{-1}S=A$ imply that $S$ is invertible. Hence the two conditions are equivalent and they simply mean that $A$ is congruent to $B$. It follows from Sylvester's law of inertia that $A$ and $B$ have the same inertia, but the exact values of their eigenvalues are completely unrelated. E.g. any two positive definite matrices are congruent to each other, but they have different eigenvalues in general.

With also the conditions $S^TS=aA^2$ and $SS^T=bB^2$, a more precise relation between the eigenvalues of $A$ and $B$ can be obtained. Since $S$ is invertible, $aA^2$ and $bB^2$ are positive definite. Hence $a$ and $b$ are positive real numbers. By taking determinants on both sides of $SA^{-1}S^T=B,\,S^TS=aA^2$ and $SS^T=bB^2$, we further see that $ab=1$.

Now, from $S^TS=aA^2$, we obtain
$$
S=\sqrt{a}Q|A| \text{ and } S=\sqrt{b}|B|U^T\tag{1}
$$
for some real orthogonal matrices $Q$ and $U$. Thus
$$
B=SA^{-1}S^T
=(\sqrt{a}Q|A|)A^{-1}(\sqrt{a}Q|A|)^T
=aQAQ^T.\tag{2}
$$
That is, $B$ is orthogonally similar to $aA$.

In addition, $(2)$ implies that $\sqrt{b}|B|=\sqrt{a}Q|A|Q^T$, while $(1)$ implies that $\sqrt{b}|B|=SU=\sqrt{a}Q|A|U$. It follows that $U=Q^T$. Thus, assuming that $A$ and $B$ are nonsingular symmetric matrices, the general solution to the set of equations $S^TB^{-1}S=A,\ S^TS=aA^2$ and $SS^T=bB^2$ is given by $b=\frac1a>0,\,S=\sqrt{a}Q|A|$ and $B=aQAQ^T$, where $Q$ is real orthogonal. Unless at least two of the variables among $A,B$ and $S$ are known, $Q$ remains arbitrary and we cannot determine the relationship between the eigenvectors of $A$ and $B$.


问:
Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$
I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.

So that means that $\frac{10^a-1}{9}$ must be divisible by 3.

$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.

When is that last statement true?

答:
You can use induction to prove

$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$

is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,

$$\begin{equation}\begin{aligned}
\frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\
& = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\
& = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$

The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for $n = 3^{m}$. For the second factor, note

$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$

This means

$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$

i.e., the second factor is also an integer. With the product of $2$ integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for $n = 3^{m+1}$. This proves by induction that \eqref{eq1A} is an integer for $n = 3^k$ for all integers $k \ge 0$.


问:
Is the set of points ${i/n},\,n=1,2,3\ldots$ in the complex plane a closed set?
A set is said to be closed if it contains all of its boundary points, where a boundary point of a set in the complex plane is any point such that every neighborhood of the boundary point contains at least one point in the set and at least one point not in the set. At first glance, I assumed that the set $S$ defined by $S=\{ i/n\,\vert \,n\in\mathbb{Z}^+\}$ is closed, since each point $i/n$ is a boundary point of the set and is an element of the set. However, the more I think about it, the more I feel that the complex number $z=0$ is also a boundary point of the set, since any neighborhood of $0$ must contain a complex number that is arbitrarily close to $0$ on the positive imaginary axis (and therefore in $S$) and $0$ itself (which is not in $S$). Since this boundary point $z=0$ is not contained in the set, $S$ does not contain all of its boundary points, and is therefore not closed. I know that this is not a rigorous proof of my claim, but is this reasoning correct?

答: You do not have to deal with complex numbers as you have a mapping $\mathbb{C} \mapsto \mathbb{R}$ given by $x: x^2$ or $S_{\mathbb{N}}=\{n \in \mathbb{N}:-\frac{1}{n^2}\}$, but $S_{\mathbb{N}}$ is not a closed set so it cannot map to a closed set.


答: Hint: $\, 2AB = (2I)AB,\,$ so $\det(2AB) = \det(2I) \det(A) \det(B).\,$ $\,2I\,$ is diagonal with fives $\,2$'s.</p>

Title: Determinant of unspecified matrices

Body:
Suppose A and B are $5\times 5$ matrices with $\det(A) = -1/3$ and $\det(B) = 6$, find the determinant of $ 2AB$.

Solution:

$$= \det(2AB) $$
$$= 2^5 \det(A)\det(B) $$
$$= (32)(-1/3)(6)$$
$$= -64$$

I understand how all of this works, except for where $2^5$ comes from, can anyone explain how this happens?


问:
Does the multiplication by diagonal matrix preserve the signs of eigenvalues?
My question is as follows:

Let $A$ be a real symmetric $n\times n$ matrix and $D=\mathrm{diag}(d_1,\ldots,d_n)$ be a diagonal matrix whose diagonal entries are all positive real numbers(denoted as $d_1,\ldots,d_n$). If $A$ has $k$ positive eigenvalues, $m$ zero eigenvalues(i.e. $0$ is an eigenvalue with multiplicity $m$) and $n-m-k$ negative eigenvalues. Then does $DA$ have the same number of positive, zero, negative eigenvalues?

Intuitively, I think this statement is true because, heuristically, if we assume that $A$ is also diagonal, its diagonal entries are exactly the eigenvalues of $A$, and the identity $\mathrm{diag}(d_1,\ldots,d_n)\cdot \mathrm{diag}(a_1,\ldots,a_n)=\mathrm{diag}(a_1d_1,\ldots,a_nd_n)$ directly gives the result. Also it is well-known that the statement is true if $A$ is positive semi-definite. However, I'm not sure that this is also true for the general cases where $A$ is not necessarily positive semi-definite. I tried to find a counterexample, but it also didn't work.

Does anyone have ideas?

Thanks in advance!

答: Yes. Since $B=D^{1/2}AD^{1/2}$ is congruent to $A$, they have the same inertia, by Sylvester's law of inertia. However, $B$ is similar to $D^{1/2}BD^{-1/2}=DA$. Therefore $DA$ also has the same number of positive/zero/negative eigenvalues as $A$.


问:
Ordinary Differential Equation system with constant decay rates
I am trying to understand how two differential equations interact, in the context of flows between discrete compartments in a population model.

Suppose I have three compartments, $A$, $B$, $C$, which have values of $10, 10, 0$ at $t=0$.

The rate of change in both A and B, when taken in isolation, are such that 50% of the initial value decays in one unit of time, i.e.,

$\dfrac{dA}{dt} = \mathrm{ln}(0.5)A(t)$

$\dfrac{dB}{dt} = \mathrm{ln}(0.5)B(t)$

Now I connect up the compartments, so that the outflow from A goes to B and the outflow from B goes to C,

A -> B -> C

implying

$$\dfrac{dA}{dt} = \mathrm{ln}(0.5)A(t)$$
$$\dfrac{dB}{dt} = -\mathrm{ln}(0.5)A(t) + \mathrm{ln}(0.5)B(t)$$
$$\dfrac{dC}{dt} = -\mathrm{ln}(0.5)B(t)$$

In my naive thinking, I would have thought that at time $t=1$, this should lead to $A=5, B=10, C=5$, given that $A$ and $B$ have the same initial values and rates of change.

However, for example, using numerical solvers such as R's deSolve and Julia DifferentialEquations.jl shows this not to be the case.

Could someone give me an intuitive explanation why? Thank you.

答:
First of all, notice that
$$\dfrac{dA}{dt} +\dfrac{dB}{dt} +\dfrac{dC}{dt} =0\implies A+B+C=k$$

Now, the first equation is simple
$$\dfrac{dA}{dt}=-\log(2) A\implies A=A_0\, 2^{-t}$$ Plug in the second equation and solve
$$B=B_0\, 2^{-t}+A_0 \log(2)\,\color{red}{t}\, 2^{-t}$$
$$C=(A_0+B_0+C_0)-B_0\, 2^{-t}-A_0\,( \log (2)t+1)\, 2^{-t}$$

With $A_0=B_0=10$ and $C_0=0$, after $1$ minute $A=5$, $B=\sim 8.47$ and $C\sim 6.53$.

The why is in $\color{red}{red}$.


问:
Different approach to a seating permutation problem.
How many ways can $24$ students be sat such that $10$ particular students never sit next to each other?

The key to solving this is to see first that the $14$ other students can be rearranged within themselves in $14!$ ways. Then, there are $15$ other "slots" (or spaces) where the $10$ in particular can sit, allowing $^{15}P_{10}$ rearrangements in total. Henceforth, the total number of permutations are $14!\ ^{15}P_{10} \approx 9.5 \times 10^{20}$.

But I was wondering if I could approach it the other way. $10$ students can be rearranged without restrictions in $10!$ ways. The $9$ slots in-between could be filled with $14$ students in $^{14}P_{9}$ ways. Now that I've placed $19$ students while the criterion of placing those $10$ in particular separately is fulfilled, I can place the remaining $5$ students wherever I please. Seeing as that there are $20$ "slots" where they could be placed, and that the number of slots increase by $1$ every time I place a student, I should be able to rearrange them in $20\times21\times22\times23\times24=24!/19!= {}^{24}P_{5}$ ways. Henceforth, the total number of allowable permutations should be $10! \times {}^{14}P_{9} \times {}^{24}P_{5} \approx 1.345 \times 10^{22}$ too, but it isn't.

In fact, it is larger than the actual number of permutations. Why? Where did I go wrong?

答:
The problem with your second working is that it leads to duplicates. To understand it, when you have chosen $9$ students of $14$ and arranged them in between $10$ students who cannot sit together, say there is a student A who is in the group of $9$. Now you are left with group of $5$ students. Say there is a student B in this group of $5$. Now you seat B in one of the twenty available slots.

But you also have arrangements where B is chosen to be in the first group of $9$ and A is in the second group of $5$. Now when you try and seat A as part of second group, it leads to duplicate arrangements.

If you want to seat $10$ students first and then seat $14$ students, here is one way to count arrangements -

First use stars and bars to count number of possible arrangements of seats. There are $11$ slots available after seating $10$ students. $9$ slots in between cannot be empty and that leaves only $5$ seats that must go to one of the $11$ slots. That can be done in $\displaystyle {15 \choose 5}$ ways.

Now that we have $14$ seats, all is left is to seat the students,  which can be done in $14!$ ways.

So total number of permissible arrangements =  $\displaystyle 10! \cdot {15 \choose 5} \cdot 14!$


问:
How to evaluate these types of definite integral?
$$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{1}}_{\cssId{lower-bound-mathjax}{0}} \sin^{2k}\left(\dfrac{{\pi}}{2}x\right)\,\cssId{int-var-mathjax}{\mathrm{d}x}$$.
I tried integration by parts but couldn't go far. The answer given in textbook is $$(2k)!/((2^
(2k))(k!)^2)$$

答:
This is Wallis’integrals, as you can see by substituting $\frac{\pi}{2}x$ with $u$.

The given formula is obtained by induction.

See Wikipedia.


答:
The determinant is a multilinear form so
$$\det(\lambda_1 C_1,\ldots,\lambda_n C_n)=\lambda_1\times\cdots\times\lambda_n \det(C_1,\ldots,C_n)$$

in your case take $\lambda_i=2,\quad i=1,\ldots,5$.

Title: Determinant of unspecified matrices


问:
Limsup and in probability convergence
Suppose I have two sequences $X_{n}$ and $Y_{n}$. We know that $$\limsup_{n\rightarrow\infty}X_{n}=\infty\quad a.s.\text{ and }|X_{n}-Y_{n}|\overset{p}{\rightarrow}0\text{ as }n\rightarrow\infty.$$ Then can we conclude that
$$
\limsup_{n\rightarrow\infty}Y_{n}=\infty\text{ a.s.}
$$

答: This is not true. Let $(X_n)$ be independent with $P(X_n=n)=\frac 1 n$ and $P(X_n=-n)=1-\frac 1 n$. Let $Y_n=-n$ when $X_n=-n$ and $0$ when $X_n=n$. Since $\sum P(X_n=n)=\sum \frac 1n =\infty$ it follows by Borel Cantelli Lemma that $P (\lim \sup X_n=\infty)=1$. Since $Y_n\leq 0$ it follows that  $P (\lim \sup Y_n=\infty)=0$. Note that $P(|X_n-Y_n| >\epsilon) \leq P(X_n \neq Y_n)=\frac 1 n \to 0$.

Title: Limsup and in probability convergence


问:
Desmos plot only for integral $x$
I'm trying to plot $2\operatorname{floor}\left(\log_{2}x\right)+1$ but only if $x$ is a natural number.

Desmos plots correctly

$2\operatorname{floor}\left(\log_{2}x\right)+1$

I would like this however:

I achieved this using a list of points $[(1,1),(2,3),(3,3),(4,5)\ldots]$ but I would like to plot up to infinity and I don't think desmos supports list comprehensions like that.

So instead I tried using desmos-style piecewise functions like $y=\left\{\operatorname{mod}\left(x,1\right)=0:2\operatorname{floor}\left(\log_{2}x\right)+1\right\}$ but this has no output at all.

So, how can I create a graphic like the second for all natural $x$? (At least up to $x=2^{2700}$)

答:
This will get you what you want.

The idea is to define a function, then plot points using a list.

In this case, we define a function $$f(x)=2\text{floor}(\log_2x)+1$$ and make the calculator plot points $(a,f(a))$ where $a\in(1,2,3,...)$.

Title: Desmos plot only for integral $x$


Title: Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$

Body:
I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.

So that means that $\frac{10^a-1}{9}$ must be divisible by 3.

$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.

When is that last statement true?

Body:
Using induction,

If $10^r=1+3^sk$ where $3\nmid k$

$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$

for $s+1\le2s+1\iff  s\ge0,3s\ge s+1\iff 2s\ge1$

Now for the base case $r=1=3^0, s=2$

So, by using weak induction, $$10^{3^a}-1$$ is divisible by $3^{a+2}$ for $a\ge0$


问:
Question regarding Zeta function representation
We know the following from Euler product:

$$\sum_{n,p}\frac{1}{n(p)^{nz}}= \log(\zeta(z))$$

$$\sum_{n,p} \frac{\log p}{p^{nz}} = -\frac{\zeta'(z)}{\zeta(z)}
$$

I want to find the alternate representation for the following sum:

$$\sum_{n,p}\frac{\log(p)}{n(ap)^{nz}}$$

Here '$a$' is a constant greater than 1

What is the  representation of above sum in terms of Zeta and other allied function?

答:
The first formula is incorrect: it's halfway in between
$$
\sum_{p^n} \frac1{np^{nz}} = \log\zeta(z) \quad\text{and}\quad
\sum_{p^n} \frac{\log p}{p^{nz}} = -\frac{\zeta'(z)}{\zeta(z)}
$$
for $\Re z>1$. Analogous to the first formula, we would have
\begin{align*}
\sum_{p^n} \frac1{n(ap)^{nz}} &= \sum_p \log\big( 1-(ap)^{-z} \big)^{-1} \\
&= \log \prod_p \big( 1-(ap)^{-z} \big)^{-1} = \log \sum_{n=1}^\infty \big( a^{-\Omega(n)}n \big)^{-z}
\end{align*}
in the same half-plane, where $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity.


问:
Stable point of group action
Group $G$ of order $55$ acts on a set $M$ of cardinality $39$. Prove, that there is an element $x \in M$, such that $\forall g \in G$ $gx=x$.

I tried to apply some combinations of Burnside's lemma or Orbit-stabilizer theorem, to compute sizes of stabilizer subgroups but didn't achieve anything.

答: Lengths of orbits under $G$ can be $1,5,11,55$. As $M$ cannot have an orbit of length $55$, assume that there are only $a$ orbits of lengths $5$ and $b$ orbits of length $11$. Then $5a+11b=39$. This has no solution in $a,b\ge0$.

Title: Stable point of group action


问:
How do I compute the distribution of $\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$?
Let $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$. Let $$M = \min\{X,Y,1-Y,1-X,|X-Y|,1-|X-Y|\}.$$ Supposedly $\operatorname{image}(M) \subseteq (0,\frac13)$ and distribution of $M$ is $$F_M(m)= (3m(2-3m))1_{(0,\frac13)}(m) + 1_{(\frac13,\infty)}(m).$$ See here.

Question: How do I compute the distribution of $M$? I wonder if there's a clever way of doing this like with $W$ here. Below is what I've tried so far.

Well $|X-Y| \ge m$ and $1-|X-Y| \ge m$ tell me $m \le |X-Y| \le 1-m$. So it looks like I have 4 cases

$x+(m-1) \le y \le x-m$

$x+m \le y$

$y \le x-m$

$x-(m-1) \le y \le x+m$ (but I think I rule out this last case because here I have $m \ge \frac12$.

Then $X,Y,1-Y,1-X \ge m$ tell me $m \le X \le 1-m$ and $m \le Y \le 1-m$. So I guess the 3 cases become

$x+m-1 = \max\{m,x+m-1\} \le y \le \min\{x-m,1-m\} = x-m$

$x+m = \max\{m,x+m\} \le y \le 1-m$

$m \le y \le \min\{x-m,1-m\} = x-m$

But it seems if I do $\int_0^1 \int_{\cdot}^{\cdot} 1\, dy \,dx$ for each case and then add them up I get only $2-6m$.

答:
As a function of $X$ and $Y$, $M$ has the following symmetries:

$$
\begin{align}
M(X,Y)&=M(Y,X), \\
M(X,Y)&=M(1-X,1-Y), \\
M(X,Y)&=M(|X-Y|,\max(X,Y)).
\end{align}
$$

Being the minimum of piecewise-linear functions, its graph is piecewise-linear (the symmetries can help understand this graph):

Alternatively, here is the corresponding contour plot:

Having found the maximum of $M$ to be $1/3$, we can see that the survival function, $P(M(X,Y)>m)$ is obtained by integrating over 2 congruent right triangles in the uniform $(X,Y)$ space.

Note that the length of a leg of one of these triangles, $l(m)$, satisfies
$$l(0)=1, \\l(1/3)=0. $$  Linear interpolation then gives

$$l(m)=1-3m, $$
and thus
$$P(M(X,Y)>m) = (1-3m)^2. $$

Finally, the CDF is

$$1-(1-3m)^2 , 0 \leq m \leq 1/3.$$


问:
How to understand cluster points?
We talked about cluster points yesterday in class and I am now very confused. I know the formal definition of the limit: $a$ is a cluster point, if for every $\epsilon > 0$, there exist infinitely many natural numbers $n ∈ ℕ$ such that $|a_n -a|<\epsilon$

Okay. I think I sort of understand this intuitively. If we have some sort of open interval around $a$, then a is a cluster point because we can fit infinitely many elements from a set or sequence into that interval no matter how small it is ($\epsilon$ is). The integer set does not contain any cluster points because it every element is isolated. Okay cool.

But I still cannot do this:

$1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, ... .$

Shouldn't there be no cluster points? If I draw an open interval around 1, there are no other elements from the sequence except 1 that I can fit inside it since every elements of the sequence are integers.

Consider:

In this case, I think $1$ and $-1$ are the cluster points. I don't think $0$ can be a cluster point for that I can't construct a small interval to include elements from the sequence.

and what happens if we have a something like this

my teacher gave an explanation of the definiton of cluster point and that's it :(

答:
You're confusing the cluster points of a sequence with the accumulation points of a set. For the former, the $n$'s play a role! Say $(a_n)_{n\in \mathbb{N}}$ is the constant sequence $a_n=a_0$, then $a_0$ is a cluster point of the sequence.

Why? Well, let's check the definition. In order to prove that $a_0$ is a cluster point of the sequence, we let $\varepsilon>0$ be given and consider the set

$$
A_{\varepsilon}=\{n|\; |a_n-a_0|\leq \varepsilon\}
$$
and we're supposed to prove that this set is infinite.

However, clearly, $A_{\varepsilon}=\mathbb{N}$, which is obviously infinite, so $a_0$ would be a cluster point of the sequence. Similar things are true for the examples you listed. We do count repeat occurences of the same element in a sequence with multiplicity.

Title: How to understand cluster points?


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
A way of beginning understanding it is for example:

Take any number in the series, multiply by $30$, add the next one. You get the one that is five to the right of where you started.

Because multiplying by $30$ then adding the next one is the same as just multiplying by $32$. Which is the same as multiplying by $2$ five times.


问:
There are two German couples, two Japanese couples and one unmarried person

There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?

I tried using the string method, but that will be only counting the cases where the wife speaks just after her husband.

There are too many elements to take care of, unlike a similar question which involved only two people, so by symmetry the answer is half of the total number of ways of arranging those $n$ people.

Then how is this one solved

答:
This problem is unusual : you solve a Combinatorics problem by pretending that it is a probability problem.  Normally, the reverse is true, that you solve a probability problem by pretending that it is a Combinatorics problem.

Let $A$ be the event that the German wife precedes the German husband.

Let $B$ be the event that the Japanese wife precedes the Japanese husband.

$A$ and $B$ are independent events, each of which has probability $(1/2).$

Therefore, the answer is $(1/4) \times 9!$.

Edit

Thanks to Aditya Gupta for pointing out my mistake.  For some reason, I thought that there was only 1 Japanese and 1 German couple, rather than 2 of each.

So the correct enumeration is $(1/16) \times 9!$.


问:
Lower bounds of the entropy function
Let $H(x,p)$ be the entropy function. That is
$$H(x,p) :=x\log(\frac{x}{p})+(1−x)\log(\frac{1-x}{1-p})\quad(0<x<1,\,0<p<1)$$

A) Show if $x\neq p$ then $H(x,p) \geq 2(x-p)^2$

B) Show if $\frac{1}{2}\leq p < x$ (or if $x<p\leq\frac{1}{2}$) then $H(x,p) \geq \frac{(x-p)^2}{2p(1-p)}$

I can't seem to figure either of these out. In a preceding problem we found the first and second derivative of $H,$ which may factor into the solution, but I cannot see where. Does anyone have any idea ? I think there is something simple I am missing for both here.

Edit: Proof for (A)

First see that for $0<x<1$, $\frac{d^2H}{dx^2} = \frac{1}{x(1-x)}\geq 4$ Now, integrating both sides we see $\frac{dH}{dx} = 4x$, and integrating again we have $H \geq 2x^2 \geq 2(x-p)^2$ since $0<p<1$.

I am fairly sure the same argument applies to part (B). I thought I had it figured out, but while writing I realized my attempt was incorrect. I attempted to do roughly the same exact thing but since $2p(1-p)$ varies from 0 to 1/2, it is larger than $2(x-p)^2$.

答:
There are some problems with your proof of part (A). First, you probably mean $\frac{dH}{dx} \ge 4x$, not $\frac{dH}{dx} = 4x$. But that conclusion would be valid only if $\frac{dH}{dx}$ is zero for $x=0$, which it isn't: For fixed $p$, the function decreases first and then increases with increasing $x$. The same problem is with the conclusion that $H \geq 2x^2$, and $2x^2 \geq 2(x-p)^2$ does not hold for $x < p/2$.

These problems can be solved. With $h(x) = H(x, p)$ for fixed $p$ you can show that $h(p) = h'(p) = 0$. Then $h''(x) \ge 4$ implies
$$
h(x) = h(p) + h'(p) (x-p) + \frac{h''(c)}{2}(x-p)^2\ge 2 (x-p)^2 \, ,
$$
using Taylor's theorem. You can proceed similarly for part (B): If $\frac{1}{2}\leq p < x$ then
$$
 h''(t) = \frac{1}{t(1-t)} \ge \frac{1}{p(1-p)}
$$
for $p \le t \le x$.

Here is a different approach, using an integral representation, similarly as in Prove Positivity of a Function Involving $\log$..

We have
$$
 H(x, p) = x\int_p^x \frac{dt}{t} + (1-x)\int_{1-p}^{1-x}\frac{dt}{t} \\
= \int_p^x  \left( \frac xt - \frac{1-x}{1-t}\right) \, dt
= \int_p^x   \frac{x-t}{t(1-t)} \, dt \, .
$$

For part (A) we use $t(1-t) \le 1/4$, that gives
$$
 H(x, p) \ge 4\int_p^x (x-t) \, dt = 2(x-p)^2 \, .
$$

(Remark: Yes, that works if $p \le x$ and if $p \ge x$.)

For part (B), if $\frac{1}{2}\leq p < x$ or $x<p\leq\frac{1}{2}$ then $t(1-t) \le p(1-p)$ for all $t$ in the integration interval, and therefore
$$
 H(x, p) \ge \frac{1}{p(1-p)}\int_p^x (x-t) \, dt = \frac{(x-p)^2}{2p(1-p)} \, .
$$

Title: Lower bounds of the entropy function


问:
Multiplicity of integer partitions in iterative process
Let $(M_k)_{k\geq0}$ be a sequence of multisets. The multiset $M_0=\{[\:]\}$ has only one element, which is an empty sequence. For positive $k$, $M_k$ is a multiset of sequences of integers sorted in non-increasing order. For each sequence $s=\{s_1,s_2,\dots\,s_j\}\in M_{k-1}$, the multiset $M_k$ includes the sequences $\{s_1+1, s_2, \dots, s_j\}$, $\{s_1, s_2+1, \dots, s_j\}$, $\dots$, $\{s_1, s_2, \dots, s_j+1\}$, and $\{s_1, s_2, \dots, s_j, 1\}$ that one can obtain by adding 1 to one of the elements of $s$ or appending a new 1 at its end. If this process generates integer sequences that are not in non-increasing order, reorder the sequence.

Every element of $M_k$ is an integer partition of $k$. I am interested in the multiplicity $m$ of each integer partition in $M_k$, preferably without drawing out the entire tree for the generating process. Is there a way to express $m$ as a function of $s_1, s_2, \dots$?

It seems to me that this problem should be related to multinomial coefficients and/or some multinomial version of Pascal's triangle. But so far, I haven't been able to make that connection.

The example below shows the tree for the generating process of integer partitions and their resulting multiplicities $m$ up to $k=5$.

答:
This is not an answer, and too long for a comment, but I have been struck by the similarity of your drawing with this one:

that was published in this paper:
The Lattice of integer partitions and its infinite extension

In this work, we were interested in Brylawski's lattice of integer partitions already mentioned by Brian Hopkins in his answer. We encoded all partitions of any integer into an infinite tree, in which the $i$-th level are the partitions of $i$, and, more interesting, a sub-tree with same root also gives this set. In the picture above, for instance, the two dotted sets correspond to the partitions of $7$.

Our goal was not to count multiplicity, but I think some results in this paper may help.


问:
Number of index $p^k$ subgroups
I'm trying to find the number of subgroups of $\mathbb{Z}^n$ such that the quotient is a $\bf{cyclic}$ group of order $p^k$ ($p$ a prime).

I guess I don't know enough about the finitely generated abelian groups, though I consulted some results about that. So I hope for a hint, thank you very much.

答:
Subgroups $K \le {\mathbb Z}^d$ with ${\mathbb Z}^d/K \cong C_{p^r}$ are kernels of surjective homomorphisms from ${\mathbb Z}^d$ to $C_{p^r}$, so let's find the number of surjective  homomorphisms.

The total number of homomorphisms is $(p^r)^d = p^{rd}$. A homomorphism is surjective if the image of at least one generator generates $C_{p^r}$. So the number that are not surjective is $p^{(r-1)d}$, and the total number of surjective homomorphisms is
$$p^{rd} - p^{(r-1)d}.$$
But some of these will have the same kernel. In fact homomorphisms $\phi$ and $\psi$ have the same kernel if and only if there exists $\alpha \in {\rm Aut}(C_{p^r})$ with $\alpha\circ \phi = \psi$. Since $|{\rm Aut}(C_{p^r})| = p^{r-1}(p-1)$, the total number of kernels, which is the required number of subgroups,  is
$$\frac{p^{rd} - p^{(r-1)d}}{p^{r-1}(p-1)}.$$
I agree that this part of the questions seems to be significantly more challenging than the other parts!

Title: Number of index $p^k$ subgroups


问:
hyperbola curve formula in 3 dimensions
Cartesian formula for 2d hyperbola curve is $x^2/a^2-y^2/b^2 = 1$.

What is the formula for a 3d hyperbola curve?

答:
What do you mean by a "3d hyperbola curve"?

If you really mean a curve, something like
$$\begin{cases}
\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1 \\ z= 0 \end{cases}$$
is one example. (There are others.)

If you mean the "3d version" you get hyperboloids of one and two sheets respectively, but these are surfaces, not curves:

$x^2+y^2-z^2 = 1$

$x^2-y^2-z^2 = 1$

Notice that these surfaces are obtained by rotating a planar hyperbola around one of its two lines of symmetry.

Title: hyperbola curve formula in 3 dimensions


问:
Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n} n$ converges, then does the following also converge?
I was wondering if the following is true.

Suppose a series $a_n$ is greater than 0 for all positive integer n, and that $\sum \frac {a_n}n$ converges, then is $\displaystyle \lim_{m\to \infty}\sum_{n= 1}^m {a_n \over m+n} = 0$?

It seems to be true because if $\sum {a_n\over n}$ converges, then that means that ${a_n\over n }\to 0$ for $n\to \infty$. This means, neglecting $n$, ${a_n \over m+n}$ will also tend to 0, and thus the summation would be equal to 0, but I don't know if this is true.

答:
Let $\varepsilon$ be a strictly positive real.

$\sum\dfrac{a_n}{n}$ converges, so: $\displaystyle \ \ \exists N \in \mathbb N^{\star} \ , \ \sum_{n=N}^{+\infty} \dfrac{a_n}{n} \leqslant \dfrac{\varepsilon}{2}  $

Then:  $\ \ \displaystyle \forall m \geqslant N \ , \ \ \sum_{n=1}^{m} \dfrac{a_n}{m+n} \leqslant  \sum_{n=1}^{N-1} \dfrac{a_n}{m} + \sum_{n=N}^m \dfrac{a_n}{n} \leqslant \dfrac{\varepsilon}{2}+\dfrac{1}{m} \sum_{n=1}^{N-1} a_n $

Now: $\\ \displaystyle \exists M\geqslant N \ , \ \forall m\geqslant M , \dfrac{1}{m} \sum_{n=1}^{N-1} a_n \leqslant \dfrac{\varepsilon}{2}$

We can conclude that:
$$ \forall \varepsilon > 0 \ , \ \exists M \in \mathbb N \ , \ \forall m\geqslant M \ ,\ 0\leqslant \sum_{n=1}^m\dfrac{a_n}{n+m} \leqslant \varepsilon    $$

And  $\ \ \displaystyle \lim_{m\rightarrow +\infty} \sum_{n=1}^m\dfrac{a_n}{n+m} = 0$


问:
Global, local and linearized observability
I am working on a nonlinear system and i want to know how well its states can be estimated by a time-discrete estimator. Which kind of observability, i.e. global or local (for every possible state and input) would the system need to fullfill such that an estimator would converge to the true states for every possible trajectory?

I learned that local observability at a point can be sufficiently proved by the rank test using Lie derivatives. Is it possible to make a statement about the ability of the estimator to estimate the correct states if i use this rank test just at every timestep for the current states and inputs?

Alternatively, do i gain any information about this ability by linearizing the system at every timestep and using the linear rank test? I already read that a nonlinear system can be observable at a point while its linearization is not.

For an Extended Kalman Filter which uses the linearization of the system in its algorithm, is it necessary that the linearized system is observable at every time?

答:
To evaluate the observability of the nonlinear system, you must use Lie derivatives. However, you might be interested to look at the following paper:

Chen, Zhe, Ke Jiang, and James C. Hung. "Local observability matrix
and its application to observability analyses." Conference of IEEE
Industrial Electronics Society. 1990.

The linearized system is time-varying due to the computation of the Jacobians, so the local observability might be useful depending on your application. In some cases, the local observability of the linearized system is equivalent to the observability of the nonlinear system if the system is linearized using the true states. Unfortunately, the true state are not available in practice, so the local observability likely does not match the observability of the nonlinear system.

Nevertheless, the local observability is still a useful, but these things should be kept in mind.


答:
There are $2$ possible orders of every couple - H W or W H but in this case it is fixed as H W. As they cannot be permuted within, the answer is

$\displaystyle \frac{9!}{ (2!)^4}$ arrangements.

Take an analogy with a string of $9$ characters. Say the string is 1 2 3 4 5 6 7 8 9. There are $9!$ ways to arrange the string.

Now if we have a string 1 2 3 4 5 6 7 8 1, there are $\displaystyle \frac{9!}{2!}$ ways to arrange them as $1$ and $1$ can be arranged in only one way instead of usual $2!$ ways for a pair of distinct characters.

Body:
There are two German couples, two Japanese couples and one unmarried person. If all 9 persons are two be interviewed one by one then the total number of ways of arranging their interviews such that no wife gives an interview before her husband is?

I tried using the string method, but that will be only counting the cases where the wife speaks just after her husband.

There are too many elements to take care of, unlike a similar question which involved only two people, so by symmetry the answer is half of the total number of ways of arranging those $n$ people.

Then how is this one solved


答:
I'm not sure how defining $H$ is relevant, so I'm going to ignore that part of your question.

We start by observing that if $$M(X,Y) = \min\{X, Y, 1-Y, 1-X, |X-Y|, 1 - |X-Y|\},$$ then $$M(X,Y) = M(Y,X),$$ so we can restrict our attention to the case $0 \le Y \le X \le 1.$, which describes a triangular domain on the unit square.  Then on this domain,  $$M(X,Y) = \min\{Y, 1-X, X-Y\}.$$  This is because:

We can eliminate $X$ since $X \ge Y$.

We can eliminate $1-Y$ because $1-Y \ge 1-X$.

$|X-Y| = X-Y$ since $X \ge Y$.

$1-|X-Y| = 1-X+Y \ge 1-X$, so we can eliminate this also.

Consequently, the probability $$\Pr[M > m \mid 0 \le Y \le X \le 1] \\ = \Pr[(Y > m) \cap (1-X > m) \cap (X-Y > m) \mid 0 \le Y \le X \le 1], \tag{1}$$ since in order for $M > m$, each of $Y, 1-X, X-Y$ must exceed $m$; conversely, if each of $Y, 1-X, X-Y$ exceed $m$, then $M > m$.  But the simultaneous inequalities on the RHS of $(1)$ can be interpreted as the region of the $(X,Y)$-plane comprising the interior of an isosceles triangle with vertices at $$(2m,m), (1-m, m), (1-m, 1-2m).$$  The first vertex comes from combining the first and last inequality:  $Y > m$ and $X - Y > m$ implies $X > m + Y > 2m$.  The second vertex is direct from the first and second inequalities.  And the third vertex comes from the second and third inequalities:  $X < 1-m$ and $Y < X-m$ implies $Y < 1-2m$.  So, remembering the symmetry about $X = Y$, the desired probability is equal to twice the area of this triangle, namely $$\Pr[M > m] = \begin{cases} 1, & m < 0 \\ (1 - 3m)^2, & 0 \le m \le 1/3 \\ 0, & m > 1/3. \end{cases} $$ hence $$f_M(m) = 6(1-3m) \mathbb 1(0 \le m \le 1/3).$$

Title: How do I compute the distribution of $\min\{X,Y,1-Y,1-X$ $,|X-Y|,1-|X-Y|\}$, where $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$?

Body:
Let $X,Y \overset{\text{iid}}{\sim} \operatorname{Unif}(0,1)$. Let $$M = \min\{X,Y,1-Y,1-X,|X-Y|,1-|X-Y|\}.$$ Supposedly $\operatorname{image}(M) \subseteq (0,\frac13)$ and distribution of $M$ is $$F_M(m)= (3m(2-3m))1_{(0,\frac13)}(m) + 1_{(\frac13,\infty)}(m).$$ See here.

Question: How do I compute the distribution of $M$? I wonder if there's a clever way of doing this like with $W$ here. Below is what I've tried so far.

Well $|X-Y| \ge m$ and $1-|X-Y| \ge m$ tell me $m \le |X-Y| \le 1-m$. So it looks like I have 4 cases

$x+(m-1) \le y \le x-m$

$x+m \le y$

$y \le x-m$

$x-(m-1) \le y \le x+m$ (but I think I rule out this last case because here I have $m \ge \frac12$.

Then $X,Y,1-Y,1-X \ge m$ tell me $m \le X \le 1-m$ and $m \le Y \le 1-m$. So I guess the 3 cases become

$x+m-1 = \max\{m,x+m-1\} \le y \le \min\{x-m,1-m\} = x-m$

$x+m = \max\{m,x+m\} \le y \le 1-m$

$m \le y \le \min\{x-m,1-m\} = x-m$

But it seems if I do $\int_0^1 \int_{\cdot}^{\cdot} 1\, dy \,dx$ for each case and then add them up I get only $2-6m$.


问:
estimate population percentage within an interval, given a small sample
Given a small sample from a normally-distributed population, how  do I calculate the confidence that a specified percentage of the population is within some bounds [A,B]?

To make it concrete, if I get a sample of [50,51,52] from a normally-distributed set, I should be able to calculate a fairly high confidence that 50% of the population lies within the range of 0-100, even with such a small sample.

This is certainly related to the "tolerance interval", but differs in an important way. In all of the examples I can find for tolerance intervals, the required percentile and confidence is given, and the interval is found. In my problem, the interval and percentile are given, and I need to find the confidence.

The relevant equation is this one: (Guttman 1970)

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$

With definitions:

$1 - \gamma$ is the confidence

$n$ is the number of samples

$100p$ is the percentage of the population required to be within the interval, as estimated from the sample mean and sample variance.

$t_0 = x - K_{S, z_p}$ is the $(1 - p) 100$th percentile of the standard normal distribution

$T_v^*(\delta)$ is the noncentral Student’s t distribution with $v$ degrees of freedom and noncentrality parameter $\delta$.

This solves the one-sided problem, but I'm having trouble extending this to the two-sided problem. In confidence-interval land, I'd use the fact that $P(t_1 \leqq X \leqq t_2) = 1 - P(t_1 \gt X) - P(X \gt t_2)$, to break this into two one-sided problems, but in tolerance-interval land I need to relate these back to the confidence ($1-\gamma$), and I don't see how.

$$1 - \gamma = P\left[P(t_1 \geqq X \geqq t_2) \geqq 1 - p\right] = ??? $$

If I attempt to turn this into two one-sided problems:

$$1 - \gamma = P\left[1 - P(t_1 \lt X) - P(X \lt t_2) \geqq 1 - p\right] = ??? $$

And I'm utterly stuck there. I don't see how to relate this back to the one-sided tolerance interval solution.

I'm not certain this is useful for people to understand the question, but it might, so I'm putting it in this addenda.

In scipy, I'm able to pretty easily calculate $K$ given $p$ $\gamma$ and $n$ as:

```<code>def K(p, gamma, n):
    from scipy import stats
    return stats.nct.ppf(1-gamma, n-1, sqrt(n) * stats.norm.ppf(1-p)) / sqrt(n)
</code>```

I'm also able to find $\gamma$ given $K$ $p$ and $n$ as:

```<code>def gamma(p, n, K):
    from scipy import stats
    z_p = stats.norm.ppf(1-p)
    return 1 - stats.nct.cdf(sqrt(n) * K, n-1, sqrt(n) * z_p)
</code>```

Much less important, but is this a valid simplification of the Guttman's formula?

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$
$$\gamma = P\left[P(X \geqq t_0) \lt 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$
$$\gamma = P\left[P(X \lt t_0) \lt p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$

If so, this form seems way easier to understand, to me.

答: I would suggest using bootstraping. It is especially very useful for small sample size case. You can get more information about it here: http://en.wikipedia.org/wiki/Bootstrapping_(statistics)</p>


问:
Clearly, $3 \in \mathbb{Z}$ is not a unit, because $1/3 \notin \mathbb{Z}$. What theorem does this kind of reasoning appeal to?
Intuitively, we may conclude that $3$ is not a unit in $\mathbb{Z}$ simply by observing that $1/3 \notin \mathbb{Z}$. However, what does this reasoning actually appeal to??? Is it true that:

Conjecture. For all rings $R$ and all $r \in R$, if there exists a ring $S$ and a homomorphism $\varphi : R \rightarrow S$ such that $\varphi(r)$ has a multiplicative inverse $s' \in S$ such that $s' \notin \mathrm{im}\, \varphi$, then $r$  is not a unit in $R$.

If not, what is the correct statement of the theorem?

答: Your conjecture is true! Suppose to the contrary that $r$ is a unit in $R$, say $rs=1$, then $\varphi(rs)=\varphi(r)\varphi(s)=1$, contradicting $\varphi(r)s'=1$.</p>


问:
Relation between Picard group of affine curve and the projective closure
Suppose that we have an smooth affine curve $C$ over an algebraically closed field of finite characteristic $p$, defined by $f(x,y) = 0$, and let $D$ be the smooth projective curve defined by the homogenisation of $f$; $F(X,Y,Z) = 0$.

I know that there is an exact sequence
$$
\mathbb{Z}^{|S|} \rightarrow \renewcommand{\Pic}{\operatorname{Pic}} \Pic(D)  \rightarrow \Pic(C) \rightarrow 0,
$$
where $S$ is the finite set of points $D \setminus C$.

I am wondering if I know that the $p$-torsion of $\Pic(D)$ vanishes, can I conclude immediately that $\Pic(C)[p] = 0$ too?

If not, are there standard techniques for establishing this?

Thanks.

答:
I want to talk about two special case, and prove that you can't necessarily say $\renewcommand{\Pic}{\operatorname{Pic}} \Pic C$ does not have any $p$-torison but I think the question about what you can say about $p$-torison is interesting.

First if $S$ consists of a single point $P$ then $\Pic C$ doesn't have any $p$-torison: if we have $p\bar x=0$ in $\Pic C$,then we should have $px=nP$ in $\Pic D$ for some $n$, by looking at the degree you get that $p|n$ and because you don't have $p$-torison we have $x=n'P$ hence $\bar x=0$.

Second if $S$ consists of points $P,Q$ such that $P-Q$ is not torison, then take a $p^{th}$ root of $P-Q$, which exists because $p$ is an isogeny, and call it $T$. this point is clearly a $p$ torsion in $C$, but if $T=aP+bQ$ in $\Pic D$ you get a relation $mP=nQ$ and by looking at the degree you get that $m(P-Q)=0$ a contradiction. hence $T$ is a nontrivial $p$-torison in this case.


问:
Expectation of a betting game involving generalized Catalan numbers
I've recently become interested in a type of biased random walk. In the original context, the simplest problem I cannot answer is as follows:

Suppose you enter a simple but lousy betting game. The price to enter
is \$3, and a fair coin is flipped. If it lands heads you win \$4,
otherwise you win nothing. So, each game either increases your wealth by
\$1 or decreases it by \$3. If you begin with just \$3 and repeatedly
enter the game until you can no longer pay the fee, what is the
expected number of games you will play?

Actually, I'd like to be able to answer this question for any game where the cost is $m$, the prize is $m+1$, and we have any given amount of starting wealth $w$. This similar question+answer shows that the expectation diverges when $m=1$ for any starting wealth.

Another way to phrase the question would be as a random walk. If you consider our position on the number line to be our total wealth after paying the fee, the case $m=w=1$ is a random walk from the origin that ends whenever the position becomes negative. The general case $m$ is a sort of random walk, but really it's more of a "stumble", since we either advance one step or reverse $m$ steps. I am interested in the expected length of these $m$-stumbles. Here is what I know so far:

The expected number of games, $g_{m,w}$, satisfies an easy recurrence relating your starting wealth:

$$ g_{m,w}  = 1 + \frac{1}{2}g_{m,w+1} + \frac{1}{2}g_{m,w-m}$$

Which I find has generating function:

$$\sum_{w \ge 0} g_{m,w}z^n = \left(\alpha_m - \frac{2z}{1-z}\right)\left(\frac{1}{1 - 2z + z^{m+1}}\right)z^{m}$$

where $\alpha_m = g_{m,m}$ represents the average length of an $m$-stumble from the origin. The $g$ terms are related to the Fibonacci n-step numbers by $A_{w}\alpha_m - 2B_{w}$ where A is a partial sum of the first $w$ n-step numbers $\sum_{k \le w}F_{k}^{(n)}$, and B is a double convolution, i.e. the sum of the first $w-1$ partial sums $\sum\sum_{k < w}F^{(n)}_{k}$. The easiest way to find $g_{m,w}$ given some $g_{m,s}$ that I know of is to use the recurrence directly.

I have found the following double sum formula for general $\alpha_m$:

$$\alpha_m = \sum_{n = 0}^{\infty} \sum_{k=1}^{m} k\binom{mn + n + k}{n} / 2^{mn + n + k}$$

which more-or-less follows from the definition of the Fuss-Catalan numbers. Each term represents the contribution to the expectation of games ending at flip number $(m+1)n+k$. The inner sum does not include the term $k=0$ because it is not possible to lose after flip numbers divisible by $m+1$. (After flip $n$ your total wealth must be congruent to $n+m \mod{(m+1)}$, since it increases by exactly $1$ or $-m \equiv +1 \mod{(m+1)}$ each flip, and it begins at $w = m \equiv -1 \mod{(m+1)}$, so if you haven't already lost your wealth must be at least $m$ meaning you are guaranteed to have enough to pay the fee for these games.)

I don't know how to evaluate this sum, but I do know at least one value and have good guesses for others. We know from the linked question $\alpha_1 = \infty$, but for $m > 1$, $\alpha_m$ converges. The one value I know is:

$$\alpha_2 = \sum_{n=0}^{\infty} \binom{3n+1}{n}/2^{3n+1} + 2\sum_{n=0}^{\infty}\binom{3n+2}{n}/2^{3n+2} = 1 + \sqrt{5}$$

Which is the value Mathematica gives me. I don't how how it arrives at this value, but it appears correct and matches my simulations. Unfortunately, Mathematica chokes on $\alpha_3$ and higher. I guessed that $\alpha_m$ is the root of an $m$'th order polynomial, so $\alpha_2 = 1 + \sqrt{5} = (x^2 - 2x - 4)_{+}$, meaning $\alpha_2$ is the unique positive real root of $x^2 - 2x - 4 = 0$.

A value for $\alpha_3$ would answer the initial problem statement in this question. I find the following guesses match the value of my sum with very high accuracy, to at least 100 digits:

$$
\begin{align}
\alpha_3 &= (x^3 - 4x - 4)_{+} \\
\alpha_4 &= (3x^4 + 4x^3 - 8x^2 - 24x - 16)_{+} \\
\alpha_5 &= (2x^5 + 5x^4 - 20x^2 - 32x - 16)_{+}
\end{align}
$$

but even knowing these values I have no idea how to prove them, even for the apparently simple cases of $\alpha_2$ and $\alpha_3$. Does anyone know how to find $\alpha_3$ or general $\alpha_m$?

答:
I'm gonna show that $\alpha_m$ is the (unique) solution of $\alpha^{m+1}=(\alpha-2)(\alpha+2)^m$ with $\alpha>2$.

My point of departure is your double sum, rewritten as $$\alpha_{m-1}=\sum_{k=1}^{m-1}k\cdot 2^{-k}G_{m,k}(2^{-m}),\qquad G_{m,k}(z)=\sum_{n=0}^\infty\binom{mn+k}{n}z^n.$$

Since you're familiar with the Fuss–Catalan numbers, let's recall that if $$F_{m,k}(z)=\sum_{n=0}^\infty\frac{k}{mn+k}\binom{mn+k}{n}z^n$$ and $F_m(z)=F_{m,1}(z)$, then $F_{m,k}(z)=\big(F_m(z)\big)^k$ and $F_m(z)=1+z\big(F_m(z)\big)^m$.

Now $G_{m,k}(z)$ can be expressed in terms of $F_m(z)$. We have $$G_{m,k}(z)=F_{m,k}(z)+\frac{mz}{k}F_{m,k}'(z)=\big(F_{m}(z)\big)^k R_m(z),$$ where $R_m(z)=1+mzF_m'(z)/F_m(z)$. On the other hand, $F_m'(z)=\big(F_m(z)\big)^m R_m(z)$ by taking the derivative of $F_m(z)=1+z\big(F_m(z)\big)^m$. Hence $$R_m(z)=1+\frac{mz}{F_m(z)}\frac{F_m(z)-1}{z}R_m(z);$$ solving this, we get $R_m(z)=F_m(z)/\big(m-(m-1)F_m(z)\big)$ and $$G_{m,k}(z)=\frac{\big(F_m(z)\big)^{k+1}}{m-(m-1)F_m(z)}.$$

Thus, $2^{-k}G_{m,k}(2^{-m})=2x_m^{k+1}/\big(m-2(m-1)x_m\big)$, where $x_m=F_m(2^{-m})/2\color{blue}{\in(1/2,1)}$ implied by $m-2(m-1)x_m>0$. And $x_m$ is the unique solution $x\in(1/2,1)$ of $$0=x^m-2x+1=(x-1)(x^{m-1}+\ldots+x-1).$$

Then $\beta_m=x_{m+1}$ is the unique solution $\beta>1/2$ of $\beta^m+\ldots+\beta=1$, and $$\alpha_m=\frac{2\beta_m\gamma_m}{1-m(2\beta_m-1)},\qquad\gamma_m=\sum_{k=1}^m k\beta_m^k.$$ Now $\beta_m\gamma_m=m\beta_m^{m+1}+\sum_{k=1}^m(k-1)\beta_m^k=m(2\beta_m-1)+\gamma_m-1$ gives $$\gamma_m=\frac{1-m(2\beta_m-1)}{1-\beta_m}\implies\alpha_m=\frac{2\beta_m}{1-\beta_m}\implies\beta_m=\frac{\alpha_m}{\alpha_m+2},$$ and if we put $\beta=\alpha/(\alpha+2)$ into $\beta^{m+1}=2\beta-1$, we get the claimed equation for $\alpha=\alpha_m$.


问:
Pictorial puzzle
Can someone suggest a pictorial representation of a mathematical puzzle or problem that could be represented in coloured tiles on an orangery floor $4489\times7525$mm with an insert of $907 \times1996$ for steps  This has to be assembled so simple but complex is preferred</p>

答:
An apollonian gasket could be fun, but it might be difficult to get your hands on suitable tiles. A variant of the Sierpinski carpet would be easier, though less aesthetically pleasing.

If you're using tiles small enough to draw smooth curves, I think a table of knots might be cool. Or you could split the floor up into 17 sections and fill each one with a representation of the wallpaper groups.

Title: Pictorial puzzle


问:
Expressing $\frac{x-y}{w-z}$ as a function of $\frac{x}{w}$ and $\frac{y}{z}$
I have the expression $\frac{x-y}{w-z}$ and I would like to rewrite it as a function of two fractions, $\frac{x}{w}$ and $\frac{y}{z}$. Ideally a linear combination of $\frac{x}{w}$ and $\frac{y}{z}$ and their powers. If that is not possible, a non-linear expression in $\frac{x}{w}$ and $\frac{y}{z}$ will also do.

(Apologies in advance for possible tagging this improperly ..)

答:
Thanks to all for sharing. Given that the expansion I seek is not possible in closed form, let me ask further whether it is possible as an approximation.

[Moderators please advise whether I need to post under a new thread]


问:
$n$-th homology group of a particular product space
1- Let $I$ be a finite set and for every $i \in I$, let $S_i$ be an arbitrary subspace of $[0,1]$ (endowed with its usual topology).
Moreover, let $S := \prod_{i \in I} S_i$ (endowed with product topology).

Does for every $n > 0$, $H_n(S) \simeq \{0\}$ (where $H_n(S)$ is the $n$-th singular homology group of $S$) ? If not, do we have sufficient conditions on $(S_i)_{i \in I}$ to obtain that $H_n(S) \simeq \{0\}$ ?

2- More generally, let $I$ be a set and for every $i \in I$, let $X_i$ be an arbitrary topological space.
Moreover, let $X := \prod_{i \in I} X_i$ (endowed with product topology).

How do we determine, for every $n > 0$, $H_n(X)$ ?

Thank you !

答:
The path components of any subset $S_i \subset I$ are intervals (which may be degenerate to points), in particular they are contractible.

The path components $P_\alpha$ of $S$ are the products of the path components of the $S_i$, hence contractible. Since
$$H_n(S) = \sum_\alpha H_n(P_\alpha) ,$$
we get $H_n(S) = 0$ for $n > 0$ and $H_0(S) =  \sum_\alpha \mathbb Z$ which is a free abelian group whose rank equals the cardinality of the sets of path components of $S$.


问:
Can we solve $\frac{\cos x \sin x+\cos y \sin y}{\cos x+\sin y}$ further?
I was solving a mechanics question in physics, where I came across the following expression which I am not able to evaluate.

How to evaluate this:

$$\frac{\cos x \sin x+\cos y \sin y}{\cos x+\sin y}$$

Help would be genuinely appreciated. Thanks.

I know it is not a very well-framed question, but I have no approach. Sorry.

答:
Your expression is equivalent to

$$\frac{\sin 2x+\sin 2y}{2(\cos x+\sin y)}.$$

There's little more that you can say.


问:
Find $f(n)$ where $f(2n)= 2f(n)+ 1$
A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the following for all $n$ natural.

$f(1)= 1$

$f(f(n))= n$

$f(2n)= 2f(n)+ 1$

Find $f(n)$

I'm really stuck, new to functions

All I've gotten is :

$f(2(f(f(n)))= 2(f(f(f(n)))+ f(1),$

which isn't very helpful..

Title: Find $f(n)$ where $f(2n)= 2f(n)+ 1$

Body:
WLOG Let $f(m)=g(m)+b$

$$1=f(2n)-2f(n)=g(2n)+b-2(g(n)+b)=g(2n)-2g(n)-b$$

Let $b=-1$  to find $$g(2n)=2g(n)$$

$1=f(1)=g(1)+(-1)\implies g(1)=2$

$$g(2^r)=2g(2^{r-1})=\cdots=2^rg(1)=2^r(2)$$


答:
Hint: Think about the following true statement and how it relates to your conjecture.

Suppose $r\in R$ is a unit.  Then, for all ring homomorphisms $\varphi:R\to S$, $\varphi(r)$ is a unit in $S$, with inverse $\varphi(r^{-1})$.

Title: Clearly, $3 \in \mathbb{Z}$ is not a unit, because $1/3 \notin \mathbb{Z}$. What theorem does this kind of reasoning appeal to?

Body:
Intuitively, we may conclude that $3$ is not a unit in $\mathbb{Z}$ simply by observing that $1/3 \notin \mathbb{Z}$. However, what does this reasoning actually appeal to??? Is it true that:

Conjecture. For all rings $R$ and all $r \in R$, if there exists a ring $S$ and a homomorphism $\varphi : R \rightarrow S$ such that $\varphi(r)$ has a multiplicative inverse $s' \in S$ such that $s' \notin \mathrm{im}\, \varphi$, then $r$  is not a unit in $R$.

If not, what is the correct statement of the theorem?


问:
Probability of sum of two discrete uniform random variables modulo k
Let U and V be two dependent discrete random variables, each
being uniformly distributed on {1, 2, . . . , k}. Let W be another
random variable having the same uniform distribution but independent of U and V . Define a random variable X = (V + W)
mod k. Show that
(a) X is uniformly distributed on {0, 1, 2, . . . , k − 1},
(b) U and X are independent.

P(X=0)=P(V+W=k)+P(V+W=2k)
P(X=i)=P(V+W=i)+P(V+W=k+i)
Nit being able to bring it in the form of discrete uniform (0,1,2,..,k-1)
Please help

And also how to approach the second part

答:
a) For $x\in \{0,1,\ldots, k-1\}$,
$$
\mathsf{P}(X=x)=\sum_{\substack{1\le, v,w\le k \\ (v+w \mod k)=x}}\mathsf{P}(V=v)\mathsf{P}(W=w)=\sum_{\substack{1\le, v,w\le k \\ (v+w \mod k)=x}}\frac{1}{k^2}.
$$
Since there are $k$ combinations of $v$ and $w$ s.t. $(v + w \mod k) = x$,
$$
\mathsf{P}(X=x)=\frac{1}{k}.
$$

b) Since $W$ is independent of $U$ and $V$,
\begin{align}
\mathsf{P}(X=x\mid U=u)&=\sum_{\substack{1\le, v,w\le k \\ (v+w \mod k)=x}}\mathsf{P}(V=v\mid U=u)\mathsf{P}(W=w) \\
&=\sum_{\substack{1\le, v,w\le k \\ (v+w \mod k)=x}}\mathsf{P}(V=v\mid U=u) \times \frac{1}{k} \\
&=\sum_{1\le v\le k}\mathsf{P}(V=v\mid U=u)\times \frac{1}{k} \\
&= 1\times \frac{1}{k}=\mathsf{P}(X=x)
\end{align}
because $\sum_{1\le v\le k}\mathsf{P}(V=v\mid U=u)=1$.


答:
$$\cfrac{\cos x \sin x+\cos y \sin y}{\cos x+\sin y}=\cfrac{\frac 12[\sin (2x)+ \sin (2y)]}{\cos x+\sin y}=\dfrac{1}{2(\cos x+\sin y)}\left[2\sin\frac{2x+2y}{2}\cos\frac{2x-2y}{2}\right]=\dfrac{\sin (x+y)\cos (x-y)}{\cos x+\sin y}$$
Formulae: $$\sin\: x\cdot \cos\:y=\frac{1}{2}[\sin(x+y)+\sin(x-y)]\\\sin\: x+\sin\: y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$$

Title: Can we solve $\frac{\cos x \sin x+\cos y \sin y}{\cos x+\sin y}$ further?

Body:
I was solving a mechanics question in physics, where I came across the following expression which I am not able to evaluate.

How to evaluate this:

$$\frac{\cos x \sin x+\cos y \sin y}{\cos x+\sin y}$$

Help would be genuinely appreciated. Thanks.

I know it is not a very well-framed question, but I have no approach. Sorry.


问:
Making H the subject $Vp=L\left[R^2\cos^{-1}\frac{R-H}{R}-(R-H)\sqrt{2RH-H^2}\right]$
I am looking at the paper "How to Calculate the Volumes of Partially Full Tanks" (via researchgate.net). On page 2 and equation 5 you are given the formula:

$$Vp=L\left[R^2\cos^{-1}\frac{R-H}{R}-(R-H)\sqrt{2RH-H^2}\right]$$

What I am aiming to do is re-arrange the formula in order to get $H$, height, the subject of the equation. i.e., when a user enters the desired volume, they can find the target height.

I've tried myself and have had a lot of difficulty. If this isn't possible I would like to know if there are any numerical solutions that you could advise.

I greatly appreciate any help with this matter.

答:
In this case you cannot get a formula, where $H$ would be explicitly given. In order to get $H$ you should use Newton's method on function:
\begin{align}
f(H)&=L\left[R^2\text{arccos}\left(\frac{R-H}{R}\right)-(R-H)\sqrt{2RH-H^2}\right]-V_p=0
\end{align}
By using
\begin{align}
H_k&=H_{k-1}-\frac{f(H_{k-1})}{f'(H_{k-1})},~~~   k=1,2,\dots\\
f'(H)&=2L\sqrt{2RH-H^2}\\
\end{align}
for the reasonably chosen starting value $H_0$ you get a sequence converging to $H$.


答:
You know that $f\left(1\right)=1$. Then
$$
f\left(2\right)=2f\left(1\right)+1=3
$$
$$
f\left(4\right)=2f\left(2\right)+1=7
$$
$$
f\left(8\right)=2f\left(4\right)+1=15
$$
You can then show that
$$
f\left(2^n\right)=2^{n+1}-1
$$
Then what about odd numbers ?
$$
f\left(3\right)=f\left(f\left(2\right)\right)=2
$$
Then $$f\left(6\right)=2f\left(3\right)+1=5$$
$$
f\left(5\right)=f\left(f\left(6\right)\right)=6
$$
I let you continue to find a pattern, knowing how to compute $f\left(n\right)$ for all $n$.

Title: Find $f(n)$ where $f(2n)= 2f(n)+ 1$

Body:
A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the following for all $n$ natural.

$f(1)= 1$

$f(f(n))= n$

$f(2n)= 2f(n)+ 1$

Find $f(n)$

I'm really stuck, new to functions

All I've gotten is :

$f(2(f(f(n)))= 2(f(f(f(n)))+ f(1),$

which isn't very helpful..


问:
Is the product of metric spaces normal?
The product of a countable collection of metric spaces is always metric. Is it true in general that the product space of any collection of metric spaces is always normal?</p>

答:
It is a classic fact, due to A.H. Stone, that $\Bbb N^{\omega_1}$ is not a normal space in the product topology, for a proof online see here.

So taking a non-compact metric space $X$ (products of compact metric spaces are always normal, as a compact Hausdorff space is, and we can apply Tychonoff's theorem) and looking at $X^I$ where $I$ is uncountable, we note that $X$ has a closed subspace $A$ that is homeomorphic to $\Bbb N$ (this follows from being a non-compact metric space) and so $X^I$ has a closed subset that is homeomorphic to $\Bbb N^{\omega_1}$ (as $\omega_1$ injects into $I$), and so cannot be normal as well (as normality is closed-hereditary).

So $\Bbb N^{\omega_1}$ is the simplest counterexample and it extends to a lot of other cases where we have enough copies of $\Bbb N$ into factors. I like $\Bbb R^{\omega_1}$ because it is even a separable topological vector space that is not normal. Of course all products of metric spaces will be Tychonoff (completely regular, or $T_{3\frac12}$) but as we saw, that's the best we can expect.

Title: Is the product of metric spaces normal?


问:
Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$
I am trying to evaluate

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$

My solution is as follow

$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n} \Rightarrow T = {e^{\mathop {\lim }\limits_{n \to \infty } n\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}} - 1} \right)}} = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{n}} \right)}}$$

$$T = {e^{\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdots + \frac{1}{{{n^2}}}} \right)}} = {e^{\left( {0 + 0 + \cdots + 0} \right)}} = {e^0} = 1$$

The solution is correct  but I presume my approach $\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdot + \frac{1}{n}}}{n}} \right) \Rightarrow \mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{n} + \frac{1}{{2n}} + \frac{1}{{3n}} + \cdot + \frac{1}{{{n^2}}}} \right) = 0$ is wrong.

Is there any generalized method

答:
Problem with the Approach in the Question

One must take extra care when taking the limit of the sum of an increasing number of terms, each of which is tending to $0$. For example,
$$
\overbrace{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\cdots+\frac1{n+n}}^{\text{$n$ terms, each of which tends to $0$}}
$$
If we sum the limits of each term, we get $0$. However, each of the $n$ terms is $\ge\frac1{2n}$, therefore, the $\liminf\limits_{n\to\infty}$ of the sum is $\ge\frac12$.

Inequalities and the Limit

First, we have the bound
$$
\begin{align}
\sum_{k=1}^n\frac1k
&\le1+\sum_{k=2}^n\int_{k-1}^k\frac{\mathrm{d}x}x\tag{1a}\\
&=1+\int_1^n\frac{\mathrm{d}x}x\tag{1b}\\[6pt]
&=1+\log(n)\tag{1c}
\end{align}
$$
Explanation:

$\text{(1a)}$: $\frac1k\le\int_{k-1}^k\frac{\mathrm{d}x}x$

$\text{(1b)}$: the sum of an integral over disjoint intervals

$\phantom{\text{(1b):}}$ is the integral over the union of those intervals

$\text{(1c)}$: integrate

Furthermore, for $x\gt-1$ and $n\ge1$ and $nx\lt1$,
$$
\begin{align}
(1+x)^n
&\le\left(\frac1{1-x}\right)^n\tag{2a}\\
&\le\frac1{1-nx}\tag{2b}
\end{align}
$$
Explanation:

$\text{(2a)}$: $(1+x)(1-x)=1-x^2\le1$

$\text{(2b)}$: Bernoulli's Inequality applied to the reciprocal

Thus,
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n
&\le\lim_{n\to\infty}\left(1+\frac{1+\log(n)}{n^2}\right)^n\tag{3a}\\
&\le\lim_{n\to\infty}\frac1{1-\frac{1+\log(n)}n}\tag{3b}\\[3pt]
&=1\tag{3c}
\end{align}
$$
Explanation:

$\text{(3a)}$: apply $(1)$

$\text{(3b)}$: apply $(2)$

$\text{(3c)}$: $\lim\limits_{n\to\infty}\frac{1+\log(n)}n=0$

Therefore, since $\lim\limits_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n\ge1$,
$$
\lim_{n\to\infty}\left(1+\frac{\sum_{k=1}^n\frac1k}{n^2}\right)^n=1\tag4
$$


问:
Gauss sums ray class group
In Neukirch's book, page 503, in  remark 1 he says that

He gives a reference for that.  I'm not able to get it.

Could someone at least draw the shape of the indicated Gauss sum, beause I have several unsuccessful attempts.

答:
The citation for the article given by Neukirch is

Allgemeine Theorie der Gaußschen Summen in algebraischen Zahlkörpern. Abh. Deutsch. Akad. Wiss. Berlin. Math.-Nat. Klasse 1 (1951) 4–23

I couldn't find the article itself (not that it would've helped me much since I don't know German), but I do have access to the MathSciNet review, which I will attempt to paraphrase because I am not confident that a direct quote would fall under fair use:

We choose $\chi$, a congruence character of a number field $K$, whose conductor is $\mathfrak{f}\mathfrak{u}$ where $\mathfrak{f}$ is the finite part and $\mathfrak{u}$ is the infinite part, as well as a multiple of the conductor, $\mathfrak{m}\mathfrak{w}$, where similarly, $\mathfrak{m}$ is the finite part and $\mathfrak{w}$ is the infinite part. We choose a divisor $\mathfrak{a}$ of $K$, and define its complement $\widetilde{\mathfrak{a}}$ by the relation $\mathfrak{a}\widetilde{\mathfrak{a}}=\mathfrak{d}^{-1}$ where $\mathfrak{d}$ is the different of $K$. Lastly, we choose $n$, which is a non-zero element of $\widetilde{\mathfrak{a}}\mathfrak{m}^{-1}$.

The Gauss sums considered by Hasse in this article are, apparently, those of the form
$$\tau_n(\chi,\mathfrak{m}\mathfrak{w}|\mathfrak{a})=\textstyle\sum\chi(\mathfrak{x})e(xn).$$
The sum is over the residue class representatives $x$ which satisfy (and here I am quoting, because I don't understand what it means):

$$x\bmod \mathfrak{m}\mathfrak{a},\;x\in\mathfrak{a}; \;x=\mathfrak{x}\mathfrak{a}, \;(\mathfrak{x},\mathfrak{m})=1\,\text{ and }\,x\equiv 1\,\mathord{(}\!\bmod\mathfrak{w}).$$
Hopefully this is helpful to you. If not, all I can suggest is to find a library close to you with access to MathSciNet, so that you can look at the full review yourself.

Title: Gauss sums ray class group


问:
$d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t $
I keep reading about one proof on internet and I found there a specific implication which I don't understand. Let $W_t$ be a Wiener process. Could you please explain to me why from

$$d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t$$

We get

$$\int_0^t f(s)dW_s = f(t)W_t - \int_0^tf'(s)W_s ds$$

答:
$$d(f(t)W_t)=f'(W_t)dt+f(t)dW_t,$$
is a short notation for
$$f(t)W_t-f(0)W_0=\int_0^t f'(W_s)ds+\int_0^t f(s)dW_s.$$

Title: $d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t $


问:
The order of $C=\sum_{k=0}^{90}(90-k)\cos(k ^{\circ}), S=\sum_{k=0}^{90}(90-k)\sin(k ^{\circ})$, and $T=\sum_{k=0}^{90}(90-k)\tan(k ^{\circ})$.
This is multiple choice questions, where using calculators is not allowed. Candidates have, on average, $2$ minutes $30$ seconds to solve it.

MY ATTEMPT:

$k ^{\circ} = (\text{positive constant} \times k)$ radians. That positive constant is $\pi/180 ^{\circ}$.

So, for simplicity, we can replace $k ^{\circ}$ by $k$ since the sign of inequality will not be changed when we multiply by a positive constant.

$$C \approx 90 \int_0^{\pi/2}\cos(x)dx-\int_0^{\pi/2}x \cos(x)dx=90(1)-(\frac{\pi}{2}-1) \approx 89.4$$

$$S \approx 90 \int_0^{\pi/2}\sin(x)dx-\int_0^{\pi/2}x \sin(x)dx=90(1)-(1) = 89$$

From here, $S<C$. Option D is the only valid one. That is; $S<C<T$.

Just to check, I used Excel, I found that D is the correct one. However, I am not sure if my approach was okay or not.

Your help to solve this problem in an easier and faster way would be appreciated. THANKS!

答:
As $k$ increases, $\cos k^\circ$ and $90-k$ both decrease and $\sin k^\circ$ increases.

Using the fact that $\cos k^\circ = \sin (90^\circ - k^\circ)$ and the Rearrangement inequality, $C>S$ follows immediately.

For $T$, $\tan 90^\circ$ is not defined. Excluding the case $k=90$, $S < T$ follows from $\cos x \le 1$. The inequality $T>C$ will require a bit more work.


答:
Answer assuming that Ito's formula holds.

You can consider the process $X$ as $d X_t = A_t dt+B_td W_t$. In this setting, your process $X$ is the brownian motion $W$, i.e. $A_t=0$ and $B_t=1$ for all $t$. In order to apply Ito's formula we should compute the derivatives of the function $g(t,x):= f(t)x$. In this case we have $g_t(t,x)=f'(t)x$, $g_x(t,x)=f(t)$ and $g_{xx}(t,x)=0$. Applying Ito's formula to the function $g$ we get
$$g(t,W_t)=g(0,W_0)+\int_0^t(g_t(s,W_s)+A_sg_x(s,W_s)+\tfrac{1}{2}B_s^2g_{xx}(s,W_s))ds+\int_0^t B_sg_x(s,W_s)dW_s.$$
Using the previous information and thanks to the fact that $W_0=0$ you have
$$f(t)W_t=\int_0^tf'(s)W_s ds+\int_0^t f(s)dW_s.$$

Title: $d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t $

Body:
I keep reading about one proof on internet and I found there a specific implication which I don't understand. Let $W_t$ be a Wiener process. Could you please explain to me why from

$$d(f(t)W_t) = f'(t)W_t dt + f(t)dW_t$$

We get

$$\int_0^t f(s)dW_s = f(t)W_t - \int_0^tf'(s)W_s ds$$


问:
Equilibrium distribution of a Markov Chain
Can anyone please tell me or help me with this question shown below?

A drunken chess grandmaster dials a long string of digits on a standard telephone keypad (laid out as shown below). It takes more than alcohol to make a grandmaster forget the rules of chess, so each digit he dials is a knight's move away from the previous one (e.g. 4 can be followed by 3, 9, or 0). The choices are made at random, independently of previous ones, with all available knight's moves being equally likely to be chosen for each digit.

Find the equilibrium distribution of this Markov chain. (Hint: it helps to use the symmetry, e.g. 1 and 3 must have the same equilibrium probability.)

```<code>1 2 3
4 5 6
7 8 9
  0
</code>```

I seem not to understand the question at all. Unlike finding other equilibrium distribution of a Markov Chain question where it is shown as a transition matrix with state space given, could anyone please help on trying to "read" and "identify" this question?

Your helps would be much appreciated.

Body:
Hint: the transition matrix looks like this (1 is the first entry and 0 is the last in each row/column). I have assumed that if the grand master presses $5$ first, we will randomly pick the next digit uniformly.

$$\left[\begin{array}{cccccccccc}
0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0\\
0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 & 0\\
0 & 0 & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & \frac{1}{3}\\
\frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10} & \frac{1}{10}\\
\frac{1}{3} & 0 & 0 & 0 & 0 & 0 & \frac{1}{3} & 0 & 0 & \frac{1}{3}\\
0 & \frac{1}{2} & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0\\
\frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & 0 & 0
\end{array}\right]$$


问:
Condensation points in metric space $\mathbb Q \times \mathbb Q$
A condensation point is defined in Exercise (2.)27 on Page 45 of Walter Rudin's Principles of Mathematical Analysis, 3rd Edition as follows:

Define a point $p$ in a metric space $\mathbb X$ to be a condensation point of a set $\mathbb E \subset \mathbb X$ if every neighborhood of $p$ contains uncountably many points of $\mathbb E$.

In light of this definition, the following question came to my mind:

Let $\mathbb X$ be a metric space consisting of all ordered pairs $(\alpha_1, \alpha_2)$ where $\alpha_1, \alpha_2 \in \mathbb Q$ (rational numbers). Let $\mathbb E$ be any subset of $\mathbb X$. Does $\mathbb E$ have a condensation point?

I tend to think that the answer is "no". That's because, $\mathbb X = \mathbb Q \times \mathbb Q$ is countable $\implies$ each neighborhood of each point $p$ in $\mathbb E$ does not include uncountably many points of $\mathbb E$ $\implies$ $p$ fails to satisfy the requirement of being a condensation point of $\mathbb E$. In other words, no subset of $\mathbb Q \times \mathbb Q$ has a condensation point. I'm not sure somehow if this argument is correct, and would appreciate help. Thanks.

答:
You're right: in a countable space $\Bbb X$ there are no condensation points, for any subset, for the simple cardinality reason that all subsets of $\Bbb X$ are countable.

There is a more "cardinal-neutral" definition: if $A$ is an infinite subset of $\Bbb X$ and $x \in X$, then $x$ is called a point of total accumulation of $A$ iff every neighbourhood $U$ of $x$ satisfies $|A \cap U| = |A|$ (so is as infinite as it can be, given the size of $A$). Then in $\Bbb X = \Bbb Q \times \Bbb Q$ all points are points of total accumulation of $\Bbb X$. Fun fact: $\Bbb X$ is compact iff every infinite set $A \subseteq \Bbb X$ has a point of total accumulation in $\Bbb X$. Replcaing "infinite" by merely "countably infinite" we get a characterisation of $\Bbb X$ being countably compact. So it's quite a natural notion IMHO, and better behaved than "condensation point"


问:
Combinatorial identity on decreasing dice throws
Suppose I repeatedly throw fair $n$-sided dice until I throw a $1$, at which point I stop.

I want to know the probability $p(n)$ that my sequence of throws will be decreasing, such as $5-4-2-1$ or just $1$ immediately, but not $2-3-1$ which would count as a failure.

The answer clearly depends on whether I am considering strictly decreasing sequences in which case $4-2-2-1$ would count as a failure, or weakly decreasing sequences in which case it would be a success. Allowing weakly decreasing sequences would increase the probability and would mean there is no bound on the potential length of successful sequences.

For strictly decreasing sequences I have $$p_s(n)=\sum\limits_{k=1}^n {n-1 \choose k-1}\frac{1}{n^k} = \frac{(n+1)^{n-1}}{n^n}$$

For weakly decreasing sequences and $n>1$ I have $$p_w(n)=\sum\limits_{k=1}^\infty {n+k-3 \choose k-1}\frac{1}{n^k} = \frac{n^{n-2}}{(n-1)^{n-1}}$$

For example with $n=6$, I get $p_s(6)=\frac{7^5}{6^6}\approx 0.36023$ and $p_w(6)=\frac{6^4}{5^5}\approx 0.41472$. In general $$p_s(n)=p_w(n+1).$$  Is there a combinatorial argument that leads to this equality between the finite sum and the infinite sum?

答:
By making the substitution $m=n-1$, the RHS of $p_w$ becomes the RHS of $p_s$.

Use the binomial identity to change the binomial coefficient of $p_w$ to:

$$\binom{n+k-3}{n-2}$$

The binomial coefficient of $p_w$ after the substitution represents one more die:

$$\binom{n+k-2}{n-1}$$

There is a $1-1$ mapping from the collection of throws of specific numbers in  $p_w$ and the single same specific throw in $p_s$.

$\frac{1}{m^k}=\frac{1}{(n+1)^k}$ represents the adjusted probability, and $p_w$ uses the probability of an infinite run length, given by:

$$\frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots$$
$$=\frac{1}{1-\frac{1}{(n+1)}}-1$$
$$=\frac{n+1}{n}-1$$
$$=\frac{1}{n}$$

Equality of $p_s$ and $p_w$ then comes from equating the two LHS's by $\frac{1}{n^k}$.


答:
(a) Fix any $0\leq \ell\leq k-1$.
$$\begin{align}
\Pr[X=\ell] &= \Pr[V+W=\ell \bmod k] \\
&= \sum_{i=1}^k \Pr[V=i, (W+i=\ell\bmod k)] \\
&= \sum_{i=1}^k \Pr[V=i]\Pr[W+i=\ell\bmod k] \tag{$V,W$ independent}\\
&= \sum_{i=1}^k \frac{1}{k}\cdot\frac{1}{k} \tag{$V,W$ uniform}\\&=\frac{1}{k}
\end{align}$$
where we used the fact that, for any $a\in \mathbb{Z}$, there exists a unique $1\leq b\leq k$ such that $a = b \bmod k$.

This shows that $X$ is indeed uniform on $\{0,1,2,\dots, k-1\}$.

(b) To prove that $U$ and $X$ are independent, let us compute, for $0\leq i \leq k-1$, $1\leq j\leq k$,
\begin{align*}
\Pr[X=i, U=j]
&= \sum_{\ell=1}^k \Pr[X=i, U=j, V=\ell]\\
&= \sum_{\ell=1}^k \Pr[V+W=i \bmod k, U=j, V=\ell] \\
&= \sum_{\ell=1}^k \Pr[W=i-\ell \bmod k, U=j, V=\ell] \\
&= \sum_{\ell=1}^k \Pr[W=i-\ell \bmod k]\Pr[U=j, V=\ell]  \tag{$W\perp \!\!\! \perp (U,V)$}\\
&= \frac{1}{k}\sum_{\ell=1}^k \Pr[U=j, V=\ell]  \tag{$W$ uniform}\\
&= \frac{1}{k}\Pr[U=j] \\
&= \Pr[X=i]\Pr[U=j]  \tag{$X$ uniform}
\end{align*}
showing the independence.

Title: Probability of sum of two discrete uniform random variables modulo k

Body:
Let U and V be two dependent discrete random variables, each
being uniformly distributed on {1, 2, . . . , k}. Let W be another
random variable having the same uniform distribution but independent of U and V . Define a random variable X = (V + W)
mod k. Show that
(a) X is uniformly distributed on {0, 1, 2, . . . , k − 1},
(b) U and X are independent.

P(X=0)=P(V+W=k)+P(V+W=2k)
P(X=i)=P(V+W=i)+P(V+W=k+i)
Nit being able to bring it in the form of discrete uniform (0,1,2,..,k-1)
Please help

And also how to approach the second part


问:
How to find the maximum value of this expression?
expression:$$\sum_{i=1}^{n-1} \frac{\sqrt{1+\cos(\theta_i)}}{\sqrt{2}} +\frac{\sqrt{1-\cos(\sum_{i=1}^{n-1}\theta_i)}}{\sqrt{2}}$$ I guess its maximum value is obtained when all $\theta_i=\pi/n$ are the same and the the max value is $n\cos(\frac{\pi}{2n})$.The partial derivatives is so complex that I can't analytically  solve it.

答:
Here is an attempt at solving the problem using Lagrange multipliers. There is some angle $\theta_n$ such that $\sum_{i = 1}^n \theta_i = \pi$. Using this $\theta_n$, the problem becomes equivalent to maximizing:
\begin{equation}
f(\theta_1,\dots,\theta_n) := \sum_{i = 1}^n \frac{\sqrt{1 + \cos(\theta_i)}}{\sqrt{2}},
\end{equation}
subject to the condition $g(\theta_1,\dots,\theta_n) := \sum_{i = 1}^n \theta_i = \pi$. For this one can use the method of Lagrange multipliers. So we calculate the partial derivatives:
\begin{align}
\frac{\partial f}{\partial \theta_j} & = \frac{-\sin(\theta_j)}{2\sqrt{2}\sqrt{1 + \cos(\theta_j)}} \\
\frac{\partial g}{\partial \theta_j} & = 1.
\end{align}

Now we need to solve the following system of $n+1$ equations in $n+1$ unknowns ($\theta_1,\dots,\theta_n,\lambda$):

\begin{align}
\lambda & = \frac{-\sin(\theta_j)}{2\sqrt{2}\sqrt{1 + \cos(\theta_j)}}, \quad j = 1,\dots, n \\
\pi & = \sum_{i = 1}^n \theta_i.
\end{align}

Observe that $\theta \mapsto \frac{\sin(\theta)}{\sqrt{1 + \cos(\theta)}}$ is a bijection between $(-\pi,\pi)$ and $(-\sqrt{2},\sqrt{2})$. Hence WLOG, we may assume that $\theta_j = \theta^* = \theta + 2\pi m_j$ for integers $m_j$ and some fixed angle $\theta$. The last equation then becomes:
\begin{equation}
n\theta + 2\pi \sum_{i = 1}^n m_j = \pi \implies \theta = \pi/n + 2\pi N/n,
\end{equation}
for some integer $N$. Evaluate $f$ at these angles to get:
\begin{equation}
f(\theta^*,\dots,\theta^*) = \frac{n}{\sqrt{2}}\sqrt{1 + \cos(\pi/n + 2\pi N/n)}.
\end{equation}
But this is maximized when $N = 0$ (the argument is a close as possible to $2\pi m$, there are other possible choices of $N$ that maximize $f$) and in this case we have:
\begin{equation}
f(\pi/n,\dots, \pi/n) = \frac{n}{\sqrt{2}}\sqrt{1 + \cos(\pi/n)} = n\cos\left(\frac{\pi}{2n}\right),
\end{equation}
where we have used the double angle formula in the end, just like OP suspected. A remark is in order, the map $\theta \mapsto \sin(\theta)/\sqrt{1 + \cos(\theta)}$ is discontinuous at the points $\pi + 2\pi m$ for $m \in \mathbb{Z}$, so there might be some issues in justifying the use of Lagrange multipliers.


答:
Assume* $f$ is uniquely defined and consider the function $g : \mathbb{N} \to \mathbb{N}$ defined by $$g(1) = 1 \\ g(2n) = 2g(n)+1 \\ n\geq 1 \implies g(2n+1) = 2g(n)$$
then it's not hard to see that $g$ satisfies $g(g(n)) = n$ so that $f=g$.

The reason this characterization is nicer is that it can be described simply in terms of the binary expansion of $n$.  That is, to find $g(n)$, write $n$ in binary and flip every bit other than the leading bit.  That is, we keep the most significant $1$, and every less significant bit is flipped, $1$ becomes $0$ and $0$ becomes $1$.

This may be written succinctly as $$f(n) = g(n) = 3\cdot 2^{\lfloor\log_2 n\rfloor}- n - 1$$

*To show $f$ is uniquely determined, it's convenient to define $N(n) = \lfloor\log_2n\rfloor$ and then induct on $N(n)$ to show $f(n)$ is uniquely determined and $N(f(n))=N(n)$.

The base case of $N(n)=0 \iff n=1$ is immediate from $f(1)=1$.

For the inductive step, first assume $n>1$ is even, so that $n=2m$ where $N(m) = N(n)-1$. Then $f(n) = 2f(m)+1$ is uniquely determined and $$N(f(n)) = N(f(2m)) = N(2f(m)+1) = 1+N(f(m)) = 1+N(m) = N(n).$$  Here, we also note that $f(n)$ is odd, where $f(f(n)) = n$ and $N(f(f(n))) = N(n) = N(f(n))$.

This effectively finishes the inductive step, because $N^{-1}[N(n)] = \{k \in \mathbb{N} : N(k) = N(n)\}$ has an equal number of odd numbers and even numbers.  Since $f$ is injective and the above argument can show $k\in N^{-1}[N(n)]$ is even implies $f(k) \in N^{-1}[N(n)]$ is odd, it follows that every odd number in $N^{-1}[N(n)]$ is the image of an even number in $N^{-1}[N(n)]$.

Title: Find $f(n)$ where $f(2n)= 2f(n)+ 1$


问:
Null space of a rotation matrix
If we have a rotation matrix of the kind:

link to the rotation matrix

how do i compute the null space of this matrix?

I know that to obtain the null space we need to write the matrix in echelon form, but in this case we have an orthonormal matrix, so my question is: how do we ge the null space of a rotation matrix knowing that it is an orthonormal matrix? Is there a different, and perhaps faster and easier way to do it?

Thank's in advance.

答:
For the $3 \times 3$ rotation matrix
$$A =
\begin{pmatrix}
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0 \\
0 & 0 & 1
\end{pmatrix},
$$
we compute its determinant. As $det(A) = 1 \neq 0,$ the matrix $A$ is invertible.
This implies $rank(A)=3,$ so the null space only contains the zero vector.

Alternatively, we proceed via trigonometry. Namely, the the null space condition
$$N(A) =
\bigg\{\begin{pmatrix} x \\ y \\ z \end{pmatrix} : \;
\begin{pmatrix}
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\bigg\}.$$
implies
$$\begin{pmatrix} cos(\theta) x - sin(\theta) y \\ sin(\theta) x + cos(\theta) y \\ z \end{pmatrix} =
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.$$
Hence
$$cos(\theta) x = sin(\theta) y,$$
$$sin(\theta) x = -cos(\theta) y,$$
$$z = 0,$$
and the latter equation, $z = 0,$ suggests we focus on the preceding equations, which arise in the study of $2 \times 2$ rotation matrices.

Now, multiply the first equation by $cos(\theta)$ and the second equation by $sin(\theta),$ then add the modified equations to obtain
$$x = 1x = (cos^{2}(\theta) + sin^{2}(\theta)) x = (sin(\theta)cos(\theta) - sin(\theta)cos(\theta))y = 0y = 0.$$ Similarly, $y=0.$ Therefore the null space only contains the zero vector, i.e., $$N(A) = \bigg\{\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\bigg\}.$$

Title: Null space of a rotation matrix