问:
The equation formula for a circle laying in an arbitrary plane
I know the standard and expanded forms of the equation of the circle in the simple 2d space,

${(x-a)}^2+{(y-b)}^2=r^2$

$x^2-2ax+y^2-2by=c$

So in 3d space what are the equations for a circle laying in an arbitrary plane,
and what is the 3d version of the polar form of the circle equation

$$(a+rcos(t),b+rsin(t))$$

How the two equations (that I mentioned up) look in 3d space and are there more abstractive formulae for the circle equation in more than 3 dimensions?

答:
Choose a center, ${\bf c}$ and any two orthonormal vectors, ${\bf a}$ and ${\bf b}$ and radius $r$.  The circle so defined is the infinite set:  ${\bf c} + r \cos (\theta) {\bf a} + r \sin (\theta) {\bf b}$ for $0 \leq \theta < 2 \pi$.

If you're working in $n$-dimensional space, then just choose your orthonormal vectors in that space.


问:
How to calculate x mod (a b) if x mod a and x mod b are known
Suppose we have two relatively prime integers $a$ and $b$. Now, suppose that for some $x$ we have $x \equiv A \pmod{a}$ and $x \equiv B \pmod{b}$. Is there any way we can express $x \pmod{a \cdot b}$ in terms of $A$ and $B$? Also, can we do something similar for $x \pmod{a^n \cdot b^m}$ where $n, m\in \mathbb{Z}_{>0}$?

I suppose this would involve solving the congruence relation $(x-A)^n(x-B)^m \equiv 0 \pmod{a^n \cdot b^m}$, but I have no clue as to what to do with this...

答:
You can use the inverse isomorphism of the isomorphism of the Chinese Remainder theorem:
\begin{align}
\varphi:\mathbf Z/ab\mathbf Z &\longrightarrow \mathbf Z/a\mathbf Z\times\mathbf Z/b\mathbf Z\\
x\bmod ab &\longmapsto(x\bmod a,x\bmod b)
\end{align}
This inverse isomorphism can be expressed in terms of a Bézout's relation $ua+vb=1$. It is given, with your notations, by
$$\varphi^{-1}(A\bmod a, B\bmod b)=Bua+Avb.$$

As to the case of $a^n$ and $b^m$, you have to find a Bézout's relation for these elements (the fastest is with the extended Euclidean algorithm).

ContentLicense: CC BY-SA 4.0


问:
Unable to understand correlation coefficient / auto correlation
Suppose I have a vector say:

[5 5 5 5 4 5]

then common sense says that there is a very high auto-correlation for the vector because it is more or less the same values.
But when I try to calculate the auto-correlation coefficient, I'm getting a very low value(<0.3) for all lags. What does this mean? shouldn't it be higher because the series is very similar?

Am I missing something?

does correlatiom mean not similarity but similarity in rate(Rate of change)?

答:
A similar situation arises when you fit a regression line to perfectly horizontal data — you get a “linear” relationship that is nearly perfect but also completely uninformative.

Correlation measures how changes in X about its mean are related to changes in Y about its mean. In the case where one of the variables is (or is almost) constant, then there isn’t much variation left to attribute to the other variable:


问:
Infinite solids similar to Gabriel's Horn
Do any solids of revolution exist with properties similar to Gabriel's Horn (i.e. a geometric solid with finite volume but infinite surface area)? Please restrict your answers to functions  not in the form
$$f(x) = \frac{c}{x^p}$$
I can already think of a great number of functions in that form which satisfy properties similar to Gabriel's Horn. Also, I'd prefer examples with monotonic functions or functions that are strictly positive in the interval $[a,\infty)$ if possible. Thanks!

答:
$$f(x)=-\ln |x|$$

Take it on range [-1,1].

The volume is $4\pi$, the surface is infinite.


问:
Finding definition of an index set $A_n$ with specific conditions
I know how to find out the union and intersection of a predefined index set, but how about finding out the definition of a described set with specific conditions like its intersection?

E.g. Find out $$\{A_n\}_{n∈N}$$ where: $$ A_1 ⊃ A_2 ⊃ A_3 ⊃ ... $$ $$ \bigcap_{n∈N}A_n = R-Q $$

What would be the approach for this type of questions?

Thanks in advance.

答:
In order to make the intersection equal to $\Bbb R\setminus\Bbb Q$, you need to make sure that each $A_n$ contains every irrational number, and that each rational number is excluded from at least one $A_n$. The set of rational numbers is countably infinite, and you have countably infinitely many sets $A_n$ at your disposal, so the easiest way to do this is to use each $A_n$ to exclude a different rational number.

Let $\Bbb Q=\{q_n:n\in\Bbb Z^+\}$ be an enumeration of the rationals, and for each $n\in\Bbb N$ let $A_n=\Bbb R\setminus\{q_n\}$. That takes care of the intersection, but unfortunately, these sets aren’t nested. However, that’s easily fixed: instead of removing one rational to get $A_n$, remove the first $n$ rationals in the list: let $A_n=\Bbb R\setminus\{q_k:k=1,\ldots,n\}$. The intersection is still $\Bbb R\setminus\Bbb Q$, and now $A_{n+1}\subseteq A_n$ for each $n\in\Bbb Z^+$.


答:
For example, if $\mathbb Q = \{q_n\}_{n \in \mathbb N}$, then $$\mathbb R \setminus \{q_0\} \supseteq \mathbb R \setminus \{q_0,q_1\} \supseteq \mathbb R \setminus \{q_0,q_1,q_2\} \supseteq \cdots$$ and $$\bigcap_{n \in \mathbb N} (\mathbb R \setminus \{q_0,\dots,q_n\}) =
\mathbb R \setminus \bigg( \bigcup_{n \in \mathbb N}\{q_0,\dots,q_n\} \bigg) = \mathbb R \setminus \mathbb Q.$$
Also, you forgot the words "These conditions are not enough to identify exactly one sequence of sets" in the @Gae. S.'s comment. Another comment: there is no algorithm to approach this kind of questions.

This example came to my mind when I saw that $\bigcap_{n \in \mathbb N}A_n = \mathbb R \setminus \mathbb Q$ is equivalent to $\bigcup_{n \in \mathbb N} (\mathbb R \setminus A_n) = \mathbb Q$, thanks to De Morgan's laws.

Title: Finding definition of an index set $A_n$ with specific conditions

Body:
I know how to find out the union and intersection of a predefined index set, but how about finding out the definition of a described set with specific conditions like its intersection?

E.g. Find out $$\{A_n\}_{n∈N}$$ where: $$ A_1 ⊃ A_2 ⊃ A_3 ⊃ ... $$ $$ \bigcap_{n∈N}A_n = R-Q $$

What would be the approach for this type of questions?

Thanks in advance.


问:
Is this Laplace BVP well posed? If not, why?
Consider the boundary-value problem for the Laplacian $\nabla^2\phi(x,y)=0$ within a semi-infinite strip $0<x<a$, $0<y<\infty$, with the following boundary conditions
$$\partial_x\phi(0,y)=\partial_x\phi(a,y)=0$$
$$\phi(x,0)=f(x)$$
$$\lim_{y\rightarrow\infty}\phi(x,y)=0$$

The solution from separation of variables is
$$\phi(x,y)=\sum_{n=0}^\infty c_n \exp(-n\pi y/a)\cos(n\pi x/a)$$
With the $c_n$ determined from the Fourier cosine series of $f$
$$c_{n \neq 0}=\frac{2}{a}\int_0^a dx\ f(x)\cos(n\pi x/a)$$
$$c_0=\frac{1}{a}\int_0^a dx\ f(x)$$

For the special case $f(x)=const=\phi_0$, only the $n=0$ term is nonzero, but then $\phi(x,y)=\phi_0$ doesn't satisfy all the boundary conditions, so it seems the BVP is ill-posed.

Question: We have specified $\phi$ or $\partial_x \phi$ over the boundary of the region. In a purely Dirichlet or Neumann problem, that would be enough. In this case, the boundary conditions are mixed, and it appears that there are additional conditions that must be enforced (on the boundary data) to make the problem well posed. What are these conditions?

The same problem in a rectangle $0<x<a$, $0<y<b$, and BC $\phi(x,b)=0$ with $f(x)=\phi_0$ gives the solution that $\phi(x,y)=\phi_0 (1-y/b)$. In the sense that we are taking $\lim_{b\rightarrow\infty}$ for this problem, I see how end up with the solution to the first problem, yet this is not entirely satisfactory since that solution doesn't satisfy the BCs.

答:
I think you have shown that if there is a solution obtainable by separation of variables, with $c_0 \ne 0$, then it is not a solution. Therefore there is no solution obtainable by separation.
So, is there a solution at all? I think no, for physical reasons, but I don't have a proof. A physical argument is to take $u$ to be a steady state temperature in a long strip, with insulation on the two long edges. The heat energy flux at the long edges is zero, and the flux in at the bottom has to be equal to the flux out at the top. Plausibly if the temperature goes to zero at the far top, then the flux there is likely to be zero also, and therefore zero at the bottom. This is too much to ask. (And this is a weak argument that I hope someone can improve.)


问:
How good can a "near-miss" polyomino packing be?
Given a polyomino $P$ with $n$ cells, we can ask about its maximal packing density $\delta_P$ in the plane (perhaps the limsup if we are concerned about convergence issues, though I don't think this actually comes up).

If $\delta_P=1$, then $P$ tiles the plane: having arbitrarily good packings implies that $P$ can cover arbitrarily large $N\times N$ squares, and from there the result follows via a compactness argument.

We can then ask: among polyominoes on $n\ge 7$ cells that do not tile the plane, which one achieves the highest density? Call this maximal value $\Delta_n$. For each $n$, $\Delta_n<1$, though of course in the limit for large $n$ it approaches $1$.

As a simple lower bound, we always have $\Delta_n\ge n/(n+1)$, as exhibited by the polyomino given by taking all but the first square in a spiral of length $n+1$ around the origin; once we plug the hole, it tiles the plane without gaps, but the hole cannot be filled. For instance, with $n=7$:

In contrast, putting remotely nontrivial upper bounds on $\Delta_n$ in general should be exceedingly difficult or impossible, since any computable upper bound less than 1 would yield an algorithm to decide the tiling problem for polyominoes, which is suspected not to exist. (Non-computable upper bounds on the order of $1-1/\text{BB}(k\cdot n)$ can be done, of course, but these are rather silly.)

However, improved lower bounds and exact values for small $n$ seem pretty tractable, and I'm curious what is known in this direction. Some questions:

Is the sequence $\Delta_7,\Delta_8,\ldots$ monotonically increasing? I suspect not.

When does $\Delta_n$ first exceed $n/(n+1)$? I don't actually know that it does, but I strongly suspect so for reasons described above. When $n=7$ I have not found any packings of density greater than $7/8$, although only for one of the four non-tiling heptominoes (the one pictured above) have I proven this is optimal. (All four obtain $7/8$ by adding one square to yield a tiling octomino.)

Can $\Delta_7$ be proven to equal $7/8$, if indeed it does so? Exhausting all tilings of an $N\times N$ square by each of the four non-tiling polyominoes and counting the max density therein would at least yield an upper bound; if $\Delta_7$ exceeds $7/8$, I would guess it does so via finding a tiling $15$-omino which is the union of two copies of a non-tiling heptomino and an additional cell.

[Subjective] What are some examples of interesting "near-miss" tilings?

答:
After getting frustrated with existing tiling software, I coded up some simple algorithms myself, from which I can establish an upper bound of $64/65$ on $\Delta_7$.

Note that if copies of a polyomino $P$ cannot cover a region $R$ of size $k$ with no overlaps, then any packing of $P$ has density at most $1-1/k$.

Label the four non-tiling heptominoes $H_1,H_2,H_3,H_4$ in the order below:

Then we have (as shown via exhaustive computer search):

$H_1$ does not cover a $7\times 11$ rectangle with L-trominoes removed at the corners (image), and so $\delta(H_1)\le64/65$. Fascinatingly, $H_1$ does cover a $10\times 10$ square, as shown in the following rotationally symmetric configuration:

$\delta(H_2)=7/8$, as discussed in the OP.

$H_3$ does not cover a size-$33$ region given by deleting $2\times 2$ squares from the corners of a $7\times 7$ grid, so $\delta(H_3)\le 32/33$.

$H_4$ does not cover a $6\times 7$ rectangle with the corners removed, so $\delta(H_4)\le 37/38$.

For all of these regions, I tried a few ways to shrink them and found tilings for every smaller region investigated. While the bounds on the other three polyominoes are reasonably close to the $28/29$ obtained by Carl Schildkraut, $H_1$ seems to present a serious obstacle. Perhaps this is because it really can tile the plane in a very dense manner not yet discovered? Either way, it seems like further investigation of $H_1$ tilings is the route to good bounds on $\Delta_7$.


问:
Non-projective variety which is topologically homeomorphic to closed subset of $\mathbb{P}^n_k$
Definition by Mumford's red book:

An affine variety is defined as an irreducible algebraic subset of $\mathbb{A}^n$ with a sheaf of rings of $k$-valued functions.

A projective variety is defined as an irreducible algebraic subset of $\mathbb{P}^n$ with a sheaf of rings of $k$-valued functions (page 28).

A variety is defined as a separated prevariety (page 37) and a prevariety is a locally ringed space which is locally isomorphic to an affine variety as locally ringed space.

Question: If we have a non-projective irreducible variety, then is it possible that it is topologically homeomorphic to a closed subset of $\mathbb{P}^n_k$ in the Zariski topology?

答:
Alex's answer in the comments probably suffices, but I'll write one here. As mentioned above, $\Bbb{A}^1_k\cong \Bbb{P}^1_k$ as topological spaces with their Zariski topologies. Actually, the situation is even worse. Any pair of (quasi-compact) curves is homeomorphic because their point sets have the same cardinality and the Zariski topology coincides with the finite complement topology. In particular, given any curve in $\Bbb{P}^n_k$, it is homeomorphic to $\Bbb{A}^1_k$ and we see that your question has an affirmative answer.

As a sidenote: even if we consider curves over $\Bbb{C}$ in the classical topology, smooth curves of genus $1$ (elliptic curves) are all diffeomorphic to a torus, but this does not mean that we regard them as the same in algebraic geometry.


问:
Is $R(X'X)^- R'$ always invertible?
I'm thinking about the problem of the restricted least squares:
$$
\min_\beta (y-X\beta)'(y-X\beta) + 2\lambda'(R\beta-r),
$$
where $y \in \mathbb{R}^n$, $X \in \mathbb{R}^{n\times k}$, $\beta \in \mathbb{R}^k$, $R \in \mathbb{R}^{q\times k}$, and $r \in \mathbb{R}^q$. Of course, there are fewer than $k$ constraints on $\beta$, so $q < k$ and $\mathrm{rank}(R)=q$. Textbooks and papers comfortably write the solution
$$
\widehat{\beta}_r = \widehat{\beta} - (X'X)^- R'[R(X'X)^- R']^{-1}(R\widehat{\beta} - r),
$$
where $\widehat{\beta}$ is the OLS estimator $(X'X)^- X'y$. The question is whether $R(X^\prime X)^- R^\prime$ is always invertible. $X^\prime X$ doesn't have to be invertible hence the generalized inverse $(X^\prime X)^-$.

Edit: What I did

Two assumptions:

$R$ is full row rank.

$R\beta$ is estimable. That is, $R(X^\prime X)^-X^\prime X = R$.

I tried to approach this question through rank. For notational convenience, write the generalized inverse as $G$, i.e., $G := (X^\prime X)^-$.

Using the inequality $r(AB) \leq \min\{r(A), r(B)\}$ for conformable matrices $A$ and $B$,
$$\begin{align*}r(R) &= r(RGX^\prime X) \leq \min\{ r(RG), r(X^\prime X)\} \leq r(RG)\\ r(RG) &\leq \min(r(R), r(G)) \leq r(R). \end{align*}$$ $\therefore r(R) = r(RG)$.

I don't know where to go from here. $RGR^\prime$ should always be nonsingular if $R\beta$ is estimable, according to the problem.

答:
to accommodate an arbitrary generalized inverse, here is a somewhat unpleasant blocked argument:

via orthogonal diagonalization, we may assume WLOG that

$X^TX = \begin{bmatrix} D_m &\mathbf {0}\\ \mathbf {0} &\mathbf {0}\end{bmatrix}$

with $D_m$ being a $m\times m$ diagonal matrix with positive entries on the diagonal

and the pseudo-inverse obeys

$(X^TX)(X^TX)^{-}(X^TX)=(X^TX)=\begin{bmatrix} D_m &\mathbf {0}\\ \mathbf {0} &\mathbf {0}\end{bmatrix}$

letting $M$ be the $m\times m$ leading principal submatrix of $(X^TX)^{-}$, this implies

$D_mMD_m = D_m\implies M = D_m^{-1}$

note: $R= R(X^TX)^{-}(X^TX)$.  By taking the ranks we know $q\leq m$ i.e.

$q=\text{rank}\Big(R(X^TX)^{-}(X^TX)\Big)\leq \text{rank}\Big((X^TX)\Big) = m$

using the identity once more

$R = R(X^TX)^{-}(X^TX) = R\begin{bmatrix} I_m &\mathbf {0}\\* &\mathbf {0}\end{bmatrix}$

where $R$ has $q$ rows all of which are linearly independent-- this implies column vector $\mathbf r_{j}=\mathbf 0$ for $j\gt m$.
i.e. we have  $R=  \begin{bmatrix} R_m &\mathbf {0}\\\end{bmatrix}$, where $R_m$ is $q\times m$ and has rank $q$

finally

$R(X^TX)^{-}R^T = \begin{bmatrix} R_m &\mathbf {0}\\\end{bmatrix}\begin{bmatrix} D_m^{-1} &*\\* &*\end{bmatrix}\begin{bmatrix} R_m^T \\ \mathbf {0}\\\end{bmatrix}=\begin{bmatrix} R_m D_m^{-1}R_m^T &*\\* &*\end{bmatrix}$

the leading $m\times m$ principal submatrix is diagonalizable with the same non-zero eigenvalues as the diagonalizable matrix $D_m^\frac{-1}{2} R_m^TR_m D_m^\frac{-1}{2}$ (and hence their rank's are the same) which is congruent to $R_m^TR_m$ which has the same rank as $R_m$, i.e. rank of q.  So putting this all together we have

$q \leq \text{rank}\Big(R(X^TX)^{-}R^T\Big)\leq \text{rank}\Big(R\Big)=q$

addendum

a shorter approach, assuming generalized inverse refers to Moore-Penrose inverse:

is to note  $X^T X$ is symmetric PSD, as is its pseudo inverse, using this we can see

$R(X^T X)^-X^T X = R\implies  X^T X(X^T X)^-R^T  = R^T$

since the columns of $R^T$ are linearly independent, we know

$X^T X(X^T X)^-R^T\mathbf v  = R^T\mathbf v = \mathbf 0\implies \mathbf v =\mathbf 0$

Now for any $\mathbf v \neq \mathbf 0$

$R^T\mathbf v \not\in \ker:(X^T X)^-\implies R^T\mathbf v \not\in \ker:\Big((X^T X)^{-}\Big)^{\frac{1}{2}}\implies \Big\Vert\Big((X^T X)^{-}\Big)^{\frac{1}{2}}R^T\mathbf v\Big\Vert_2^2\gt 0$

i.e. the right hand side says

$0\lt \mathbf v^T \Big(R(X^T X)^{-}R^T\Big)\mathbf v \implies \Big(R(X^T X)^{-}R^T\Big)\succ \mathbf 0$

hence $\Big(R(X^T X)^{-}R^T\Big)$ is invertible

Title: Is $R(X'X)^- R'$ always invertible?


问:
Is every harmonic polynomial a linear combination of these?
In $N$-dimensional space, we can show by direct calculation that the polynomial
$$
 r^{2K+N-2}\nabla_a\nabla_b\nabla_c\cdots \frac{1}{r^{N-2}}
\hspace{1cm}
 \text{(with $K$ derivatives)}
$$
is harmonic (annihilated by the Laplacian $\nabla^2$), where $\nabla_n$ is the partial derivative with respect to the $n$th coordinate and $r$ is the distance from the origin.

Is every homogeneous harmonic polynomial of degree $K$ a linear combination of these? If so, how do we prove this? If not, what is a counterexample?

Motive: I'm studying physics, and this seems like a much nicer way to approach the theory of spherical harmonics (just divide this polynomial by $r^K$ to get a spherical harmonic) compared to the typical physics-textbook approach using spherical coordinates, but it's not obvious to me that all spherical harmonics are linear combinations of these for an arbitrary number of dimensions $N$.

答:
This is a theorem of Maxwell and Sylvester, see for example https://arxiv.org/abs/math-ph/0408046 , https://arxiv.org/abs/0805.1904
and https://arxiv.org/abs/astro-ph/0412231 but, it is in 3d only.

In higher dimension, this is the object of a conjecture of Shubin, see page 10 of https://arxiv.org/abs/0704.1174

Great question BTW, related to some interesting and beautiful mathematics.


问:
Complex and real matrix for eigenvalues
This is a short one but: Consider the real matrix $\alpha = \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}$ and let $\beta \in M_n (\mathbb{C})$ be the same matrix but considered as a complex matrix.

Problem

with use of a computer program calculate the eigenvalues of the two matrices. i.e $\alpha$ over the real numbers and $\beta$ over the complex numbers.

Question

What does this actually mean. They are the same but over different numbers? Is it just the same matrix and thus the same result in this problem but got new rules in different problems where the fact that it is over complex numbers makes a difference? Does it make a difference here?

What does the matrix actually look like - I am thinking it is the same but the fact it is over complex numbers do not matter here.

答:
The characteristic polynomial of your matrix is
$$-x^3+3x^2-x+3 = -(x-3)(x^2+1)$$
The real root of this polynomial is $3$ which is an eigenvalue for $\alpha$ and $\beta$:
$$\begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}
\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix} = 3\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix}$$
Non-real roots of the polynomial are $\pm i$, which are eigenvalues only for $\beta$:
$$\begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}
\begin{pmatrix}
1+i \\ -2i \\ 2
\end{pmatrix} = i\begin{pmatrix}
1+i \\ -2i \\ 2
\end{pmatrix}, \quad \begin{pmatrix}
7 &3  &-4 \\
 -2&-1  &2 \\
 6&2  &-3
\end{pmatrix}
\begin{pmatrix}
1-i \\ 2i \\ 2
\end{pmatrix} = -i\begin{pmatrix}
1-i \\ 2i \\ 2
\end{pmatrix}$$
Note that there are no real nonzero vectors $v$ such that $\alpha v = \pm i v$ so they are not eigenvalues of $\alpha$ in this sense.

Title: Complex and real matrix for eigenvalues


问:
$\frac{1}{n}\sum_{k=1}^n(X_k-\mathbb E[X_k])$ converges a.s. to $0$
Let $X_1,X_2,\dots$ be a sequence of independent random variables, such that the series $$\sum_{n=1}^\infty\frac{\operatorname{Var}(X_n)}{n^2}$$ converges. Show that as $n\to\infty$, $$\frac{1}{n}\sum_{k=1}^n(X_k-\mathbb E[X_k])$$ converges almost surely to $0$.

There is a quick solution via the martingale convergence theorem: we have that $$Y_n=\sum_{k=1}^n\frac{X_k-\mathbb E[X_k]}{k}$$ is martingale, and $\sup_n\mathbb E[|Y_n|]$ is finite, so $Y_n$ converges almost surely to some random variable $Y$, and we can finish with Kronecker's lemma.

I'm interested though in any approaches avoiding the heavy machinery of the martingale convergence theorem. I feel like defining the $Y_n$'s as I did above could be fruitful. For example, the Kolmogorov inequality gives the bound
$$\mathbb P\left(\max_{1\leq i\leq n}|Y_i|>\varepsilon\right)\leq\frac{1}{\varepsilon^2}\mathbb E\left[Y_n^2\right]=\frac{1}{\varepsilon^2}\cdot\sum_{k=1}^n\frac{\operatorname{Var}(X_k)}{k^2}.$$
But I'm unsure what to make from all this.

答: It is well known that a series of independent random variables converges almost surely iff it converges in distribution iff it converges in probability. If you are willing to assume this result the rest is very easy. To show that $Y_n$ converges in probability it is enough to prove convergence in $L^{2}$.  But convergence in $L^{2}$ follows from the following basic fact from FA: if $(x_n)$ is s sequence in a Banach space such that  $\sum \|x_n\| <\infty$ then $\sum x_n$ converges in the norm.


问:
the general linear groups - solvable
I'm doing some experimental mathematics and I'm in the situation where I need an answer to the question:

Consider the general linear groups $\operatorname{GL}(2,10)$, $\operatorname{GL}(4,10)$ (the general linear groups of degree $10$ over the fields of order $2$ and $4$) and $\operatorname{GL}(2,15)$, $\operatorname{GL}(4,15)$.

Are these groups solvable?

答:
From Wikipedia (http://en.wikipedia.org/wiki/Projective_linear_group#Finite_fields):

The projective special linear groups $PSL(n, F_q)$ for a finite field $F_q$  are finite simple groups whenever $n$ is at least $2$, with two exceptions:$PSL(2, F_2)$ and $PSL(2, F_3)$.

Since $PSL$ is a quotient group of $SL$, then $GL$ cannot be soluble.

ContentLicense: CC BY-SA 3.0

Title: the general linear groups - solvable


问:
Given an isomorphism linear transformation $T:V \to \mathbb C^3$,where $V$ is a vector space
Given an isomorphism linear transformation $T:V \to \mathbb C^3$,where $V$ is a vector space over $\mathbb C$,assume $\alpha_1,\alpha_2,\alpha_3,\alpha_4 \in V$ such that:
$$T(\alpha_1)=(1,0,i)$$
$$T(\alpha_2)=(-2,1+i,0)$$
$$T(\alpha_3)=(-1,1,1)$$
$$T(\alpha_4)=(\sqrt 2,i,3)$$

Is $\alpha_1$ in the subspace spanned by $\alpha_2,\alpha_3$?

If $W_1$ is the subspace spanned by $\alpha_1,\alpha_2$ and $W_2$ is the subspace spanned by $\alpha_3,\alpha_4$,then find the intersection of $W_1$ and $W_2$.

Find a basis for the subspace spanned by $\alpha_1,\alpha_2,\alpha_3,\alpha_4$.

$\alpha_1$ is the subspace spanned by $\alpha_2,\alpha_3$ if it can be written as a (finite) linear combination of $\alpha_2,\alpha_3$,if there exist $\lambda_2,\lambda_3 \in \mathbb C$ such that $$\alpha_1=\lambda_2\alpha_2+\lambda_3\alpha_3$$

Or equivalently (the equivalence is duo to the fact that $T$ is an isomorphism which follows that it does have an inverse) $$T(\alpha_1)=\lambda_2T(\alpha_2)+\lambda_3T(\alpha_3)$$
$$(1,0,i)=\lambda_2(-2,1+i,0)+\lambda_3(-1,1,1)$$
Which implies that $\lambda_2=i(i+1)/2,\lambda_3=i$,and so the answer to the question is yes.

So far we know that $\text{dim}(W_1)=2=\text{dim}(W_2)$,on the other hand $$ \text{dim}(W_1 \cap W_2)=\text{dim}(W_1)+\text{dim}(W_2)- \text{dim}(W_1+W_2)$$

$T$ is isomorphism ,hence a bijection,so does have an inverse ,but I don't know how to find the dimension of $W_1+W_2$.

Since the $\alpha_i$'s have not been given explicitly,hence I think we need a trick to find such a basis,but I don't know how.

答:
We have

$$\alpha_3 = \frac1{\lambda_3}(\alpha_1-\lambda \alpha_2)$$
so $\alpha_3 \in W_1$. But also $\alpha_3 \in W_2$ by definition so $\alpha_3 \in W_1\cap W_2$. This implies that $\dim (W_1 \cap W_2) \ge 1$.

It remains to see whether $\dim(W_1\cap W_1) = 1$ or $\dim(W_1\cap W_2) = 2$. In the latter case, since $\dim W_1 = \dim W_2 =2$, we would get $W_1 = W_2$. However, it is easy to see that $\alpha_4$ cannot be represented as a linear combination of $\alpha_1$ and $\alpha_2$ so clearly $W_1 \ne W_2$.

Therefore $\dim (W_1 \cap W_2) = 1$ and hence $\dim (W_1+W_2) = 3$. To find a basis for $W_1+W_2$ we have to pick three linearly independent vectors. We can pick $$\{\alpha_1, \alpha_2, \alpha_4\}.$$
Indeed, the first two are linearly independent since $\dim W_1 = 2$ and we already showed that $\alpha_4 \notin W_1.$


答:
A circle lying in the plane $Ax+By+Cz+D=0$ can be defined by getting a point P(px,py,pz) in that plane and intersecting the plane with the sphere  $(x-px)²+(y-py)²+(z-pz)²=R²$ which is centered at P

That means two equations. There's no one-expression $F(x,y,z)=k$ for a 3D circumference (as it also happens for a 3D straight).

You may think "get z=f(x,y) from plane and plug it into the sphere". What you get then is a projection on a horizontal plane, not the plane you want.

Title: The equation formula for a circle laying in an arbitrary plane

Body:
I know the standard and expanded forms of the equation of the circle in the simple 2d space,

${(x-a)}^2+{(y-b)}^2=r^2$

$x^2-2ax+y^2-2by=c$

So in 3d space what are the equations for a circle laying in an arbitrary plane,
and what is the 3d version of the polar form of the circle equation

$$(a+rcos(t),b+rsin(t))$$

How the two equations (that I mentioned up) look in 3d space and are there more abstractive formulae for the circle equation in more than 3 dimensions?


问:
Find pattern in three sequences of numbers
I am developing an algorithm to fill a matrix in a column-wise fashion. This means in a loop I iterate by column, and then by row.

For an input set:

```<code>[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
</code>```

I want to fill the lower triangular matrix, i.e.:

```<code>[0,  -,  -,  -,  -]
[1,  0,  -,  -,  -]
[2,  5,  0,  -,  -]
[3,  6,  8,  0,  -]
[4,  7,  9, 10,  0]
</code>```

But I am stuck into finding the mathematical relationship between three sequence of numbers. It's easier to explain with a simplified example with expected outputs:

My input set S has 10 elements. I will be iterating this set.

When I iterate the sequence 1 : 1, 2, 3, 4

I need two sequences: 2, 3, 4, 5 and column index 1, 1, 1, 1

For sequence 2 : 5, 6, 7

I need sequences: 3, 4, 5 and 2, 2, 2

For sequence 3 : 8, 9

I need sequences: 4, 5 and 3, 3

For sequence 4 : 10

I need sequences: 5 and 4

Since I am iterating by column, for any iteration with column_index value I know to obtain the multiset of repetitions (for example: 1,1,1,1), I need to repeat the ((size of set S) / 2) - column_index.

But I think I am overcomplicating this and I suspect there is a pattern or formulas I can't see here.

答:
For sequence 1: the difference between each element of (2,3,4,5) and that of the  required sequence (1,2,3,4) = +1

For sequence 2: the difference between each element of (3,4,5) and (5,6,7) = -2

For sequence 3: the difference between each element of (4,5) and (8,9) = -4

For sequence 4: the difference between each element of (5) and (10)=-5

Notice the pattern in the differences: +1, -2, -4, -5. They are decreasing by 3,2,1 respectively.

(ie) 1-3=-2

-2-2=-4

-4-1=-5

Now you can use an iterating variable that would decrease by 1 on each iteration.
I hope that this would be helpful for you!


问:
Help evaluating $\int_0^\infty \frac{1}{x^{1/2}(x+1)}dx$
I began solving this with U sub and partial fractions...first for $x^{1/2}$ and then for $x+1$  but neither of those methods got me the answer of $\pi$.

I know the indefinite integral should be $2\tan ^{-1} x^{1/2} + C$.

答:
The substitution $u=x^{1/2}$ works smoothly. We get $du=\frac{1}{2}x^{-1/2}\,dx$, so $dx=2x^{1/2}\,du=2u\,du$. It follows that
$$\int_0^\infty \frac{dx}{x^{1/2}(1+x)}=\int_0^\infty \frac{2\,du}{1+u^2}.$$
Now it's over, an antiderivative of $\frac{2}{1+u^2}$ is $2\arctan u$, and as $M\to\infty$, $\arctan M\to\frac{\pi}{2}$.

Remark: It we want to be formal about the calculation, consider the integral
$$\int_{\epsilon}^B \frac{dx}{\sqrt{x}(1+x)}.$$
Make the substitution as above. Then our integral is equal to
$$\int_{\sqrt{\epsilon}}^{\sqrt{B}}\frac{2\,du}{1+u^2}.$$
When we evaluate, we get $2\arctan(\sqrt{B})-2\arctan(\sqrt{\epsilon})$. This has limit $\pi$ as $b\to\infty$ and $\epsilon\to 0^+$.


问:
Mean Number of Samples from Uniform Distribution to Achieve a Desired Total

Problem Statement

I'm interested in the mean number of samples from a given uniform distribution of integers I would need to take to reach a desired total or more.

For example, a six-sided dice is a uniform distribution of integers from $1$ through to $6$. The number of times you have to roll a six-sided dice ($r$) to achieve a desired total or greater ($t$) produces a distribution. For example, here is a graph produced by sampling dice rolls, aiming for a total of $80$ per run:

Graph of six-sided dice rolls with a target of $80$

The mean of this distribution is approximately $23.332448$. This is the number I am interested in deriving. Simply 23 would be too imprecise for what I need. Right now I am determining this by simply sampling large numbers of dice rolls and then calculating the average of that data set, however, I would like to instead understand the statistics involved and compute the mean, or a reasonable approximation of the mean.

One might try to derive the mean of the number of rolls needed $P(r=t)=0.5$ by simply using the mean of the uniform distribution, and dividing the total $t$ by that mean:

$$\frac{t}{(max-min)/2+min} = \frac{80}{(6-1)/2+1} \approx 22.857$$

I will call this the 'naive' approximation. As you can see, this result can be significantly off the desired result. Sometimes, it happens to be close to the right answer, however, I am interested in many cases where it is unacceptably imprecise.

Ultimately, I want to have a method or formula that takes an arbitrary continuous uniform distribution of integers (such as a die) and the desired total, which tells me the mean number of samples from that distribution I would need to take in order to reach that total.

Prior Research

I've looked at a few posts that talk about this subject, however, none of them were able to provide the complete answer to me. Here are two of the things I've found and read:

A Similar Question on This Forum

Number of dice rolls taken to reach a certain sum

This was useful in creating an approximation of the actual distribution I'm trying to find. I used this to turn the following recursive function into the following code (in rust, dependency on the cached crate for memoizing the recursive results to improve performance to something usable):

$$P(S_t=s) = \sum_{i=1}^6P(S_{t-1} = s-i)/6$$

With the following terminators:

$P(S_0=0) = 1$

$P(S_0<0) = P(S_{t>0}=0) = 0$

```<code>#[cached]
pub fn p(t: i32, s: i32) -> f64 {
    if s <= 0 && t <= 0 {
        return 1.0;
    }

    if s <= 0 || t <= 0 {
        return 0.0;
    }

    (1..(std::cmp::min(6, s) + 1)).map(|i| p(t - 1, s - i)).sum::<f64>() / 6.0
}
</code>```

Which, when run, can be used to create a graph comparing the frequency of the sampled results to the frequency predicted by that algorithm. In the graph, the blue line is the result of the algorithm, while the orange line is sampled data from a simple PRNG using a uniform distribution (the same data as in the first graph).

Note that to get this graph, you have to divide the results of this algorithm by the inverse of the mean of the uniform distribution (so $3.5^{-1}$ in this example) The author of the original answer suggests dividing by $P(S_t=36)$ (where $36$ was their target) however fails to realise that this is just an approximation of the inverse of the mean. I'm not actually sure why the mean of the uniform pops out here, so that's an interesting tidbit.

This looks pretty good, however, it does not provide the mean number of rolls as best I can tell. The author of the answer to the linked question states that the mean in their example is "about $10.5238577$", however, they don't state how they derived that number and I haven't been able to figure it out.

It's also not generalised to any uniform, however, I don't believe it would be difficult to generalise. I don't anticipate this to be an obstacle if someone finds out how to derive the mean from that answer.

A Similar Question on Reddit

On average, how many rolls of a fair die would it take to get a cumulative score above n?

This was useful in creating a different approach to solving this problem, where I iterated $r$ (the number of rolls) upwards from 1, each time producing an approximation of the distribution of the combined rolls using a normal distribution. The first normal distribution to have a cumulative distribution function for the given target of less than $0.5$ indicated when the integer mean number of rolls was being made. Again, however, like the previous solution, this only produced the integer number of rolls. I need to know the mean number of rolls more precisely than this.

There are a number of other answers than the top up-voted answer on this Reddit post with differing amounts of clear information. The answer from the user u/possiblywrong shows an interesting cumulative distribution function which I wonder if it would solve my problem. However, I am unable to figure out how to reproduce their ideas as code or fully understand them. For completeness sake, here is an image of that response in case it is deleted in the future.

Thanks

Thank you for any time you spend on this, this is doing my head in. It feels like it should be so easy to model, and yet I've lost an entire day researching it only to find vague answers that are either too complicated for me, or are simply missing information that I need to understand them. Hopefully, you can help.

答:
I will do the reasoning with the dice, but a generalization is really straightforward. I do not have a formula for the mean. Nevertheless, I provide a piece a code in Python, which compute the precise number for given N value (in this case $80$) and number of SIDES (in this case $6$). Also, I derived an approximative formula, which might be sufficient. (Unless, I have a mistake in my reasoning.)

Mathematical view

Let us denote a Markov chain with states $S_i$ for $i$ between $0$ and $N$, where $N$ is the target value. A state says, how many dots we have thrown exactly in total. The only exception is the last state $S_N$, where we get, if we threw $\geq N$ dots.

Then we have natural transition probabilities, i.e.
$$p(S_i \to S_j) = \frac{1}{6},$$
if $i < j \leq i + 6$ and $j \not = N$.

For the last state, the probabilities differs.
$$p(S_{N-i} \to S_N) = \frac{6-i}{6},$$
for $i \in 0, \dots, 5$.

The law of total expectation

Let $X_i$ be a random variable denoting a number of throws needed to get to $S_i$. We will compute $\mathbb E[X_i]$, which will be denoted by $e_i$ for simplicity. We use the law of total expectation to compute it by induction. The partition of sample will be a configuration of the last step. The configuration $C$ consists of the last state and number of dots thrown in the last step.

That is,

$$e_i
= \mathbb E \left[\mathbb E[X_i|C] \right]
= \sum_{(j,d)} \mathbb E[X_i|C=(j,d)]P(C=(j,d)),$$

where $S_j$ was the last state and $d$ is the number of thrown dots, such that throwing $d$ dots gets us to $S_i$.

Usually, we have only $6$ configurations: the last state was $i - k$ for $k \in 1, \dots, 6$, while the last throw gave $k$ dots. This statement does not hold for only for the initial states and the last state.

The intial states have only limited number of configurations (e.g. we cannot reach state $S_3$ by throwing $6$), and the from the $S_{N-1}$ any number suffice to reach $S_N$ and so on.

Decomposing the formula

It hold the following

$$\mathbb E[X_i|C=(j,d)] = e_j + 1,$$

as the LSH is the expected numbers of throws to get to $S_i$, if the last throw will move us from $S_j$ to $S_i$.

A bit more tricky is computing the probability $P(C=(j,d))$. In one of the previous questions, you mention is computed, $P(A_j)$, where $A_j$ is the event for visiting $S_j$ during computation. We can then compute the probability that we have thrown $d$ dots while being in state $S_j$ (these two are obviously indepentent event), which is $\frac{1}{6}P(A_j)$.

The tricky part, in my opinion, is realization that we need to normalize these probalities to get $P(C=(j,d))$.

That is,

$$P(C=(j,d)) = \frac{\frac{1}{6}P(A_j)}{\sum_{(j',d')} \frac{1}{6}P(A_{j'})},$$

where $(j',d')$ are all valid configurations.

It is basically the Bayes theorem to get
$$P(\text{particular configuration $C = (j,d)$ happen}\;|\;\text{configuration $C$ is the last step})$$

$$P(\text{$C = (j,d)$}\;|\;\text{$C$ is last}) = \frac{P(C = (j,d) \land \text{$C$ is last})}{P(\text{$C$ is last})},$$

where $P(C = (j,d) \land \text{$C$ is last})$ reduces to $P(C = (j,d)$ and $P(\text{$C$ is last})$ will be computed over all particular valid configurations.

This allow us to compute $e_N$.

Programming part

There are several parts. First, we compute the probabilities $P(A_i)$ for $i \in 0, \dots, N-1$, i.e. the probability that we visit $S_i$. Then we compute all expectations except the target one, which will be computed separately.

Probabilities

It is the lengthier part, but basically it is about exploiting the reccurence relations from the mentioned answer to the previous problem.

```<code>from fractions import Fraction

N = 80
SIDES = 6

def prepare_probabilities_array():
    p = [
        [... for _ in range(N)]
        for _ in range(N)
    ]

    p[0][0] = Fraction(1)

    for i in range(1, N):
        p[0][i] = Fraction()
        p[i][0] = Fraction()

    return p

def fill_array(p):
    for throw in range(1, N):
        for num in range(1, N):
            p[throw][num] = sum(
                p[throw - 1][num - dots]
                for dots in range(1, min(num + 1,SIDES + 1))
            ) / SIDES

def get_prob_of_num(p, num):
    return sum(p[throw][num] for throw in range(0, N))

def get_probabilities_of_num(p):
    return [get_prob_of_num(p, num) for num in range(0, N)]

# [throws][num]
PROBABILITIES = prepare_probabilities_array()  # set initial conditions
fill_array(PROBABILITIES)  # do some dynamic programming
NUM_PROB = get_probabilities_of_num(PROBABILITIES)  # prob of visit, i.e. P(A_i)

print([(i,float(v)) for i,v in enumerate(NUM_PROB)])
</code>```

See that for large values of we even don't need to bother with the actual computing, as the probabilities goes quickly to $\frac{2}{7}$, i.e. $1$ over the mean of one throw.

Expectations

Then we compute the array of expectations, again for $i \in 0, \dots, N-1$, while the $e_N$ will be computed separately.

```<code>def compute_expectations(n_p):
    e = [Fraction()]
    for num in range(1, N):
        normalization_factor = Fraction()
        exp_contrib = Fraction()
        for dots in range(1, min(SIDES + 1, num + 1)):
            prob = n_p[num - dots]
            normalization_factor += prob
            exp_contrib += prob * (e[num - dots] + 1)
        e.append(exp_contrib / normalization_factor)
    return e

EXPECTATIONS = compute_expectations(NUM_PROB)
print([(i,float(v)) for i,v in enumerate(EXPECTATIONS)])
</code>```

Note that I omitted $\frac{1}{6}$ in the computation, as it always cancels out.

See that differences between expectations of $S_i$ and $S_{i+1}$ also tends to $\frac{2}{7}$ for large $i$ which is a good base for an approximating formula.

The target expectation

```<code>def compute_target(n_p, e):
    normalization_factor = Fraction()
    exp_contrib = Fraction()
    for min_dots in range(0, SIDES):
        prob = Fraction(SIDES - min_dots, SIDES) \
               * n_p[N - min_dots - 1]
        normalization_factor += prob
        exp_contrib += prob * (e[N - min_dots - 1] + 1)
    return exp_contrib / normalization_factor

print(float(compute_target(NUM_PROB, EXPECTATIONS)))  # 23.33333333333144
</code>```

Instead of computing each valid last configuration, it is more convinient to work with transition probabilities for the last state defined at the beginning.

Approximative closed formula

As hinted in previous section, we can form a closed formula which approximates the result pretty well by assuming the following:

$$e_{N-i} \approx e_c + (N-i-c)T,$$

where $T$ is inverse of the expected value from one roll, i.e. $\frac{2}{k+1}$ for $k$ being the number of sides, and $c$ is a constant large enough.

Thus,
$$ e_N \approx \frac{\sum_{i=1}^k \frac{k+1-i}{k}(e_c+(N-i-c)T + 1)}{\sum_{i=1}^k\frac{k+1-i}{k}},$$
which can be further simplified.

$$ e_N \approx \frac{\sum_{i=1}^k \frac{k+1-i}{k}(e_c+NT-cT+1-iT)}{\frac{k+1}{2}}$$

$$ e_N \approx \frac{\frac{k+1}{2}(e_c+NT-cT+1) - \sum_{i=1}^k \frac{k+1-i}{k}iT}{\frac{k+1}{2}}$$

$$ e_N \approx e_c+NT-cT+1 - \frac{\frac{T}{k}\sum_{i=1}^k (k+1-i)i}{\frac{k+1}{2}}$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2}\sum_{i=1}^k (k+1-i)i$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2} \left( \sum_{i=1}^k i(k+1)- \sum_{i=1}^k i^2 \right)$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2} \left( \frac{k(k+1)^2}{2} - \frac{k(k+1)(2k+1)}{6} \right)$$

$$ e_N \approx e_c+NT-cT+1 - \frac{2(k+2)}{3(k+1)}$$

$$ e_N \approx e_c+1 + \frac{2}{k+1} \left( N-c-\frac{k+2}{3} \right)$$

```<code>c = 40

approx = EXPECTATIONS[c] + 1 + 2/(SIDES+1)*(N - c - (SIDES+2)/3)
print(approx)  # 23.3333469998339
</code>```

Conclusion

I provided (at least I hope) a mathematical method, which lead to the code which computes precisely the desired value (at least for small $N$, such that Fraction can handle the computation).

The further observations and assumptions allow us to make a closed formula which can also do reasonably well (although I do not know the precision you can). Also, it can be scaled by selecting a convenient value of $c$.

I did not try to compute the probabilities $P(A_i)$ by solving the equations, however, I assume it will be quite a mess. Nevertheless, that might be a way how to reach the precise closed formula for your problem.

Body:
I will do the reasoning with the dice, but a generalization is really straightforward. I do not have a formula for the mean. Nevertheless, I provide a piece a code in Python, which compute the precise number for given N value (in this case $80$) and number of SIDES (in this case $6$). Also, I derived an approximative formula, which might be sufficient. (Unless, I have a mistake in my reasoning.)

Mathematical view

Let us denote a Markov chain with states $S_i$ for $i$ between $0$ and $N$, where $N$ is the target value. A state says, how many dots we have thrown exactly in total. The only exception is the last state $S_N$, where we get, if we threw $\geq N$ dots.

Then we have natural transition probabilities, i.e.
$$p(S_i \to S_j) = \frac{1}{6},$$
if $i < j \leq i + 6$ and $j \not = N$.

For the last state, the probabilities differs.
$$p(S_{N-i} \to S_N) = \frac{6-i}{6},$$
for $i \in 0, \dots, 5$.

The law of total expectation

Let $X_i$ be a random variable denoting a number of throws needed to get to $S_i$. We will compute $\mathbb E[X_i]$, which will be denoted by $e_i$ for simplicity. We use the law of total expectation to compute it by induction. The partition of sample will be a configuration of the last step. The configuration $C$ consists of the last state and number of dots thrown in the last step.

That is,

$$e_i
= \mathbb E \left[\mathbb E[X_i|C] \right]
= \sum_{(j,d)} \mathbb E[X_i|C=(j,d)]P(C=(j,d)),$$

where $S_j$ was the last state and $d$ is the number of thrown dots, such that throwing $d$ dots gets us to $S_i$.

Usually, we have only $6$ configurations: the last state was $i - k$ for $k \in 1, \dots, 6$, while the last throw gave $k$ dots. This statement does not hold for only for the initial states and the last state.

The intial states have only limited number of configurations (e.g. we cannot reach state $S_3$ by throwing $6$), and the from the $S_{N-1}$ any number suffice to reach $S_N$ and so on.

Decomposing the formula

It hold the following

$$\mathbb E[X_i|C=(j,d)] = e_j + 1,$$

as the LSH is the expected numbers of throws to get to $S_i$, if the last throw will move us from $S_j$ to $S_i$.

A bit more tricky is computing the probability $P(C=(j,d))$. In one of the previous questions, you mention is computed, $P(A_j)$, where $A_j$ is the event for visiting $S_j$ during computation. We can then compute the probability that we have thrown $d$ dots while being in state $S_j$ (these two are obviously indepentent event), which is $\frac{1}{6}P(A_j)$.

The tricky part, in my opinion, is realization that we need to normalize these probalities to get $P(C=(j,d))$.

That is,

$$P(C=(j,d)) = \frac{\frac{1}{6}P(A_j)}{\sum_{(j',d')} \frac{1}{6}P(A_{j'})},$$

where $(j',d')$ are all valid configurations.

It is basically the Bayes theorem to get
$$P(\text{particular configuration $C = (j,d)$ happen}\;|\;\text{configuration $C$ is the last step})$$

$$P(\text{$C = (j,d)$}\;|\;\text{$C$ is last}) = \frac{P(C = (j,d) \land \text{$C$ is last})}{P(\text{$C$ is last})},$$

where $P(C = (j,d) \land \text{$C$ is last})$ reduces to $P(C = (j,d)$ and $P(\text{$C$ is last})$ will be computed over all particular valid configurations.

This allow us to compute $e_N$.

Programming part

There are several parts. First, we compute the probabilities $P(A_i)$ for $i \in 0, \dots, N-1$, i.e. the probability that we visit $S_i$. Then we compute all expectations except the target one, which will be computed separately.

Probabilities

It is the lengthier part, but basically it is about exploiting the reccurence relations from the mentioned answer to the previous problem.

from fractions import Fraction

N = 80
SIDES = 6

def prepare_probabilities_array():
    p = [
        [... for _ in range(N)]
        for _ in range(N)
    ]

    p[0][0] = Fraction(1)

    for i in range(1, N):
        p[0][i] = Fraction()
        p[i][0] = Fraction()

    return p

def fill_array(p):
    for throw in range(1, N):
        for num in range(1, N):
            p[throw][num] = sum(
                p[throw - 1][num - dots]
                for dots in range(1, min(num + 1,SIDES + 1))
            ) / SIDES

def get_prob_of_num(p, num):
    return sum(p[throw][num] for throw in range(0, N))

def get_probabilities_of_num(p):
    return [get_prob_of_num(p, num) for num in range(0, N)]

# [throws][num]
PROBABILITIES = prepare_probabilities_array()  # set initial conditions
fill_array(PROBABILITIES)  # do some dynamic programming
NUM_PROB = get_probabilities_of_num(PROBABILITIES)  # prob of visit, i.e. P(A_i)

print([(i,float(v)) for i,v in enumerate(NUM_PROB)])

See that for large values of we even don't need to bother with the actual computing, as the probabilities goes quickly to $\frac{2}{7}$, i.e. $1$ over the mean of one throw.

Expectations

Then we compute the array of expectations, again for $i \in 0, \dots, N-1$, while the $e_N$ will be computed separately.

def compute_expectations(n_p):
    e = [Fraction()]
    for num in range(1, N):
        normalization_factor = Fraction()
        exp_contrib = Fraction()
        for dots in range(1, min(SIDES + 1, num + 1)):
            prob = n_p[num - dots]
            normalization_factor += prob
            exp_contrib += prob * (e[num - dots] + 1)
        e.append(exp_contrib / normalization_factor)
    return e

EXPECTATIONS = compute_expectations(NUM_PROB)
print([(i,float(v)) for i,v in enumerate(EXPECTATIONS)])

Note that I omitted $\frac{1}{6}$ in the computation, as it always cancels out.

See that differences between expectations of $S_i$ and $S_{i+1}$ also tends to $\frac{2}{7}$ for large $i$ which is a good base for an approximating formula.

The target expectation

def compute_target(n_p, e):
    normalization_factor = Fraction()
    exp_contrib = Fraction()
    for min_dots in range(0, SIDES):
        prob = Fraction(SIDES - min_dots, SIDES) \
               * n_p[N - min_dots - 1]
        normalization_factor += prob
        exp_contrib += prob * (e[N - min_dots - 1] + 1)
    return exp_contrib / normalization_factor

print(float(compute_target(NUM_PROB, EXPECTATIONS)))  # 23.33333333333144

Instead of computing each valid last configuration, it is more convinient to work with transition probabilities for the last state defined at the beginning.

Approximative closed formula

As hinted in previous section, we can form a closed formula which approximates the result pretty well by assuming the following:

$$e_{N-i} \approx e_c + (N-i-c)T,$$

where $T$ is inverse of the expected value from one roll, i.e. $\frac{2}{k+1}$ for $k$ being the number of sides, and $c$ is a constant large enough.

Thus,
$$ e_N \approx \frac{\sum_{i=1}^k \frac{k+1-i}{k}(e_c+(N-i-c)T + 1)}{\sum_{i=1}^k\frac{k+1-i}{k}},$$
which can be further simplified.

$$ e_N \approx \frac{\sum_{i=1}^k \frac{k+1-i}{k}(e_c+NT-cT+1-iT)}{\frac{k+1}{2}}$$

$$ e_N \approx \frac{\frac{k+1}{2}(e_c+NT-cT+1) - \sum_{i=1}^k \frac{k+1-i}{k}iT}{\frac{k+1}{2}}$$

$$ e_N \approx e_c+NT-cT+1 - \frac{\frac{T}{k}\sum_{i=1}^k (k+1-i)i}{\frac{k+1}{2}}$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2}\sum_{i=1}^k (k+1-i)i$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2} \left( \sum_{i=1}^k i(k+1)- \sum_{i=1}^k i^2 \right)$$

$$ e_N \approx e_c+NT-cT+1 - \frac{4}{k(k+1)^2} \left( \frac{k(k+1)^2}{2} - \frac{k(k+1)(2k+1)}{6} \right)$$

$$ e_N \approx e_c+NT-cT+1 - \frac{2(k+2)}{3(k+1)}$$

$$ e_N \approx e_c+1 + \frac{2}{k+1} \left( N-c-\frac{k+2}{3} \right)$$

c = 40

approx = EXPECTATIONS[c] + 1 + 2/(SIDES+1)*(N - c - (SIDES+2)/3)
print(approx)  # 23.3333469998339

Conclusion

I provided (at least I hope) a mathematical method, which lead to the code which computes precisely the desired value (at least for small $N$, such that Fraction can handle the computation).

The further observations and assumptions allow us to make a closed formula which can also do reasonably well (although I do not know the precision you can). Also, it can be scaled by selecting a convenient value of $c$.

I did not try to compute the probabilities $P(A_i)$ by solving the equations, however, I assume it will be quite a mess. Nevertheless, that might be a way how to reach the precise closed formula for your problem.


问:
How do I calculate the probability of drawing certain cards from a deck?
I have a deck of $40$ cards. They are equally numbered one to five. This means that there are $8$ ones, twos, threes, fours and fives. I will draw $5$ cards from the deck.

I want to find the probability that one of those cards is a three.

I want to find the probability that all five of the cards are threes.

How do I do this?

答:
Hypergeometric distribution with 8 successes and 32 failures, drawing 5 items without replacement.
Let $X$ be the number of successes (3's). [I suppose you are expected to show how to get numerical answers from combinatorial symbols, and to give a rationale for answers. I'll leave the details of that to you.]

From R: Probabilities $P(X = 1), P(X \ge 1), P(X = 5)$
from R, where dhyper is a hypergeometric PDF:

```<code>dhyper(1, 8,32, 5)
[1] 0.4371983
1 - dhyper(0, 8,32, 5)
[1] 0.6939612
dhyper(3, 8,32, 5)
[1] 0.04221225
</code>```

Combinatorial symbols:

$P(X = 1) = \frac{{8\choose 1}{32\choose 4}}{{40\choose 5}}.$

$P(X \ge 1) = 1 - P(X = 1)
=1-\frac{{8\choose 0}{32\choose 5}}{{40\choose 5}}.$

$P(X=5) = \frac{{8\choose 5}{32\choose 0}}{{40\choose 5}}.$

Simulation.

With 10 million iterations
one can expect two or three place accuracy.

```<code>set.seed(1228)
deck = rep(1:5, each = 8)
threes = replicate(10^7, sum(sample(deck, 5)==3))
mean(threes == 1)
[1] 0.4372205
mean(threes >= 1)
[1] 0.6939325
mean(threes==3)
[1] 0.0422606
</code>```

Bar chart of hypergeometric distribution.

```<code>x = 0:3;  PDF = dhyper(x, 8,32, 3)
plot(x, PDF, type="h", lwd=3)
 abline(h=0, col="green2")
 abline(v=0, col="green2")
</code>```

Body:
Hypergeometric distribution with 8 successes and 32 failures, drawing 5 items without replacement.
Let $X$ be the number of successes (3's). [I suppose you are expected to show how to get numerical answers from combinatorial symbols, and to give a rationale for answers. I'll leave the details of that to you.]

From R: Probabilities $P(X = 1), P(X \ge 1), P(X = 5)$
from R, where dhyper is a hypergeometric PDF:

dhyper(1, 8,32, 5)
[1] 0.4371983
1 - dhyper(0, 8,32, 5)
[1] 0.6939612
dhyper(3, 8,32, 5)
[1] 0.04221225

Combinatorial symbols:

$P(X = 1) = \frac{{8\choose 1}{32\choose 4}}{{40\choose 5}}.$

$P(X \ge 1) = 1 - P(X = 1)
=1-\frac{{8\choose 0}{32\choose 5}}{{40\choose 5}}.$

$P(X=5) = \frac{{8\choose 5}{32\choose 0}}{{40\choose 5}}.$

Simulation.

With 10 million iterations
one can expect two or three place accuracy.

set.seed(1228)
deck = rep(1:5, each = 8)
threes = replicate(10^7, sum(sample(deck, 5)==3))
mean(threes == 1)
[1] 0.4372205
mean(threes >= 1)
[1] 0.6939325
mean(threes==3)
[1] 0.0422606

Bar chart of hypergeometric distribution.

x = 0:3;  PDF = dhyper(x, 8,32, 3)
plot(x, PDF, type="h", lwd=3)
 abline(h=0, col="green2")
 abline(v=0, col="green2")


问:
An algebra with a strictly positive state + condition ? to become a C$^*$-algebra.
Let $(A,*)$ be an involutive $\mathbb{C}$-algebra i.e. a $\mathbb{C}$-algebra with a semi-linear anti-automorphism i.e.
identically

 $(x+y)^*=x^*+y^*$

 $(\alpha.x)^*=\overline{\alpha}.x^*$

 $(xy)^*=y^*x^*$

A strictly positive (or faithful) state $\varphi\in A^*$ is such that for all $x\in A\setminus \{0\}$, $\varphi(x^*x)> 0$.
One can prove easily that $g(x,y):=\varphi(x^*y)$ is a non-degenerate hermitian form. It can be noted that, due to the fact that $g(a.x,y)=g(x,a^*y)$ and with finite dimensionality, one gets that $A$ is semi-simple. So, one can consider $A$ as simple.

My question is the following.

Q) We suppose $A$ to be finite dimensional. Under which additional condition(s) is the Hilbert space $(A,g)$ a C$^*$-algebra ?

Late edit I insist on the fact that I am in search of conditions for
$(A,g)$ to be a C$^*$-algebra, not only $A$. This means that it should be a C$^*$-algebra for the precise norm $||x||=\sqrt{g(x,x)}$.

Late remark The fact that $A$ be a star-algebra of finite dimension, sum of matrix algebras is by no means sufficient to imply that the projectors on the blocks are *-invariant nor $A\simeq \mathbb{C}$ as shows the following counterexample. Take $B=\mathbb{C}^{n\times n}$ (algebra of complex square matrices of dimension $n>0$) and $A=B\oplus B$ with the anti-automorphism
$(X,Y)^\star=(Y^*,X^*)$. Then $(A,\star)$ is easily checked to be a star algebra. It is of finite dimension, sum of matrix algebras but $dim_\mathbb{C}=2n^2\not= 1$. Indeed, the existence of a strictly positive state is crucial as there is none over $A$.

答:
No  Hilbert space is a $C^*$-algebra in any natural way since the structure of Hilbert spaces (a vector  space with an inner
product) is quite different from that of  a C*-algebra (a Banach space equipped with an involutive algebra structure).

Nevertheless from the data provided by the OP one may give  $A$ the structure of a  $C^*$-algebra in a very natural
way.  The first step is to consider the hermitian form $g$ as an inner product on $A$, so that
$$
  H:= (A,g)
  $$
becomes a Hilbert space
(completeness follows from  finite dimensionality).
One may  then define  a map
$$
  \pi :A\to B(H),
  $$
by
$$
  \pi (a)\xi  = a\xi , \quad\forall a\in  A,\quad \forall \xi \in H,
  $$
(here $a\xi $ is nothing other than  the product of $a$ and $\xi $).

One may easily show that $\pi $ is a *-homomorphism.  It is moreover injective since, for $a\neq 0$, one has that
$\pi (a)a^*=aa^*\neq 0$, as a consequence of $\varphi (aa^*)>0$.

Finally,  identifying $A$ with its image within $B(H)$ via $\pi $, we  have  that $A$ becomes a $C^*$-algebra.

Later edit:
The late edit indeed makes the question trivial and, as noted my Martin, the only example is the complex numbers.

A perhaps more elementary proof of this fact is that every finite dimensional $C^*$-algebra is the direct sum of matrix algebras, so unless $\text{dim}(A)=1$, there exists a pair of mutually orthogonal projections $p$ and $q$.  It follows that  $\|p\pm q\|=1$, hence the parallelogram law fails.


问:
Question about diffeomorphism
Here is an assignment problem:

$f:\mathbb{S}^2 \longrightarrow \mathbb{S}^2$ is smooth and surjective. Prove $\exists$ open subset $ U $ of $\mathbb{S}^2$, such that $f|_U$ is a diffeomorphism.

I've tried to relate it to covering maps but failed.

Can someone help me with this problem? Thanks for help.

答:
Use Sard's Theorem. It gives a regular value of $f$, and since $f$ is surjective, this means that the regular value isn't vacuous (i.e. it's not just something not in the image of $f$). This gives us a point at which the derivative of $f$ is an isomorphism. Then hit it with the Inverse Function Theorem to get your diffeomorphism.

http://en.wikipedia.org/wiki/Inverse_function_theorem

Title: Question about diffeomorphism


问:
How many solutions for $37+\frac{3(x-1)x}2+3^{x+2}=y^2\,$?
How many solutions does the following exponential diophantine equation have in the positive integers?
$$37+\frac{3(x-1)x}2+3^{x+2}=y^2$$
The solutions $\,(x,y)\,$ I have found are: $\;(1,8),\;(2,11),\;(3,17),\;(4,28)$.

Are the only solutions?

Many thanks.

P.S. The equation has been built noting that:
$$27+37+0=64=8^2$$
$$81+37+3=121=11^2$$
$$243+37+9=289=17^2$$
$$729+37+18=784=28^2$$

Body:
The following can be found with the 15-digit precision of Excel.

$$
(1,8)\quad
(2,11)\quad
(3,17)\quad
(4,28)\quad
(36,1162261467)\quad
(38,3486784401)\quad
(40,10460353203)\quad
(42,31381059609)\quad
(44,94143178827)\quad
(46,282429536481)\quad
(48,847288609443)\quad
(50,2541865828329)\quad
(52,7625597484987)\quad
(54,22876792454961)\quad
(56,68630377364883)\quad
(57,118871300538602)\quad
(58,205891132094649)\quad
(59,356613901615807)\quad
(60,617673396283947)\quad
$$
But WolframAlpha disagrees with the $x\ge36$ as seen here so, if there are other pairs, it would take software with arbitrary precision to find them.


问:
Do we have non-trivial solutions for $\frac{e^{ad X_0}-1}{ad X_0}X=0$?
Let $k$ be $\mathbb{R}$ or $\mathbb{C}$ and $\mathfrak{g}$ be a finite dimensional Lie algebra over $k$. Let $X_0$ be a fixed element in $\mathfrak{g}$ and consider the equation
$$
X+\frac{1}{2!}[X_0,X]+\frac{1}{3!}[X_0,[X_0,X]]+\ldots=0.
$$
It can be rewritten as
$$
\frac{e^{ad X_0}-1}{ad X_0}X=0.
$$

If $\mathfrak{g}$ is abelian, then the only solution to the above equation is $X=0$.

My question is: in general do we have non-trivial solutions to the above equation? If so, what do they look like. It will be appreciated if you can give me relevant references.

答:
If this is for a finite-dimensional Lie algebra (or more generally a Banach Lie algebra) over the complex numbers,
$\text{ad }X_0$ is a bounded linear operator on $\mathfrak g$, and we want an element of
the null space of $g(\text{ad} X_0)$ for the entire function $g(z) = (\exp(z)-1)/z$.
By the spectral mapping theorem, $g(\text{ad} X_0)$ is invertible if and only if
$g(\lambda) \ne 0$ for all $\lambda$ in the spectrum of $\text{ad} X_0$.  On the other
hand, if $X$ is an eigenvector of $\text{ad} X_0$ for eigenvalue $\lambda$ and
$g(\lambda) = 0$, then $g(\text{ad} X_0) X = 0$.  In this case the zeros of $g$ are
$2 \pi i n$ for nonzero integers $n$.


答:
If $A$, $B$, $C$ are $(1,1)$ tensors such that $A\cdot B = C$, that is $\sum_h A^{i}_h B^h_j = C^i_j$. If we define
$A^{I}_J=A^{i_1, \ldots, i_k}_{j_1, \ldots, j_k} = A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}$
we have $\sum_H A^I_H B^H_J = C^I_J$, which we can write
$(A)(B) = (AB)$ Consider now a special $(k,k)$ tensor, the generalized $\delta$, like @Abdelmalek Abdesselam: indicated,  and the tensor $[A] = \Delta \cdot (A)=(A)\cdot \Delta$ (last equality needs a bit of thinking, not true for any $(k,k)$ tenson). The crucial euqality is
$$\Delta \Delta = k! \Delta$$
Now we want to compare $[A]\cdot [B]$ and $[AB]$. So we write
$$[A]\cdot [B]= (A)\Delta \Delta (B)= k! \Delta (A)(B)= k! \Delta (AB) = k! [AB] $$

If instead of $\Delta$ we consider $P$ ( no signs), then we have permanents, and the corresponding Binet-Cauchy for them.

Note:

We have $(\Delta \cdot (A) )^I_J= \sum \epsilon(\sigma) (A)^{\sigma(I)}_J$ while  $(  (A) \cdot \Delta)^I_J= \sum \epsilon(\sigma) (A)^{I}_{\sigma(J)}$

Now the tensor $(A)$ is such that $(A)^{\sigma(I)}_{\sigma(J)} = (A)^I_J$ ( commutativity in some product), and moreover, $\epsilon(\sigma) = \epsilon(\sigma^{-1})$.

Title: $\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}$

Body:
ACHTUNG

TO FOLLOWS I USE ENSTEIN CONVENTION TO WROTE DETERMINANTS!

Definition

If $A$ is an $n\times n$ matrix we define its $k\times k$ minotr to be the determinant of $k\times k$ submatrix of $A$, e.g.
$$
A^{i_1,...,i_k}_{j_1,...,j_k}:=\det\begin{bmatrix}A_{j_1}^{i_1}\ &\cdots&A_{j_k}^{i_1}\ \\\vdots&\ddots&\vdots\\A_{j_1}^{i_k}&\cdots&A_{j_k}^{i_k}\end{bmatrix}=\delta^{h_1,...,h_k}_{j_1,...,j_k}A^{i_1}_{h_1}...A^{i_k}_{h_k}=\delta^{i_1,...,i_k}_{h_1,...,h_k}A^{h_1}_{j_1}...A^{h_k}_{j_k}
$$
where the $\delta$-symbol is the generalised Kronecker Delta as here defined.

So I ask to prove that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
for any non-singular matrix. In particular using the above definition I calculate that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\Big(\delta^{l_1,...,l_k}_{h_1,...,h_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\Big)\Big(\delta^{h_1,...,h_k}_{m_1,...,m_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}\Big)=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{h_1,...,h_k}\delta^{h_1,...,h_k}_{m_1,...,m_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{m_1,...,m_k}A^{i_1}_{l_1}\big(A^{-1}\big)^{m_1}_{j_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_k}_{j_k}
$$
but unfortunately I do not be able to conclude anything. So could someone help me, please?


问:
Probability of success given another success (conditional probability problem)
Suppose that independent students have to solve a homework. Student A solves the homework with a probability of $70\%$, Student B solves it with a probability of $80\%$ and Student C solves it with a probability of $90\%$.

a) What is the probability that all the students solve the homework if at least two students solve the homework?

b) What is the probability that Student B solves the homework if less than two students solve the homework?

c) What is the probability that less than two students solve the homework?

I have tried this procedure:

a) There are three possible cases:

Student B and C have already solved the homework, and so student A has yet to solve the problem ($70\%$)

Student A and C have already solved the homework, and so student B has yet to solve the problem ($80\%$)

Student A and B have already solved the homework, and so student C has yet to solve the problem ($90\%$)

The probability of the first event is ($80\% * 90\% = 72\%$), for the second is ($70\% * 90\% = 63\%$), for the third is ($70\% * 80\% = 56\%$).
So the result is: ($70\% * 72\%$) + ($80\% * 63\%$) + ($90\% * 56\%$) but the result is unrealistic ($151.2\%$). What is wrong?

b) I had in mind to use a procedure very similar to point a) but I believe that I get an wrong result

c) The overall probability is given by:
($70\% * 20\% * 10\%$) + ($30\% * 80\% * 10\%$) + ($30\% * 20\% * 90\%$) = $9.2\%$
Is it correct?

答:
I'll assume that by team $A$, you meant student $A$, and similarly for $B,C$.

Define events $A,B,C,X,Y$ as

$A$ is the event that student $A$ solves the homework.$\\[4pt]$

$B$ is the event that student $B$ solves the homework.$\\[4pt]$

$C$ is the event that student $C$ solves the homework.$\\[4pt]$

$X$ is the event that all three students solve the homework.$\\[4pt]$

$Y$ is the event that at least two students solve the homework.

and for any event $E$, let $E'$ denote the complementary event "$\text{not}\,E$".

Then we have
$$
P(X)=P(A)P(B)P(C)=\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{9}{10}=\frac{63}{125}=.504
$$
and
\begin{align*}
P(Y)&=
P(A\cap B\cap C')
\,+\,
P(B\cap C\cap A')
\,+\,
P(C\cap A\cap B')
\,+\,
P(A\cap B\cap C)
\\[4pt]
&=
\left(\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{1}{10}\right)
+
\left(\frac{4}{5}\cdot \frac{9}{10}\cdot \frac{3}{10}\right)
+
\left(\frac{9}{10}\cdot \frac{7}{10}\cdot \frac{1}{5}\right)
+
\left(\frac{7}{10}\cdot \frac{4}{5}\cdot \frac{9}{10}\right)
\\[4pt]
&=
\frac{451}{500}
=
.902
\\[4pt]
\end{align*}
For parts $(a)$ and $(b)$, you need to apply the formula for conditional probability.

Part $(a)$:$\;$The probability that all three students solve the homework given that at least two students solve the homework is
$$
P(X|Y)
=
\frac{P(X\cap Y)}{P(Y)}
=
\frac{P(X)}{P(Y)}
=
\frac
{\left({\large{\frac{63}{125}}}\right)}
{\left({\large{\frac{451}{500}}}\right)}
=
\frac{252}{451}
\approx
.5587583149
$$
Part $(b)$:$\;$The probability that student $B$ solves the homework given that less than two students solve the homework is
$$
P(B|Y')
=
\frac{P(B\cap Y')}{P(Y')}
=
\frac{P(B\cap C'\cap A')}{P(Y')}
=
\frac
{
\left(
{\large{\frac{4}{5}}}
\cdot
{\large{\frac{1}{10}}}
\cdot
{\large{\frac{3}{10}}}
\right)
}
{\left({\large{\frac{49}{500}}}\right)}
=
\frac{12}{49}
\approx
.2448979592
$$
For part $(c)$, you were on the right track but you missed the case where none of the three students solve the homework.

But since we've already computed $P(Y)$, there's an easier way . . .

Part $(c)$:$\;$The probability that less than two students solve the homework is
$$
P(Y')
=
1-P(Y)
=
1-\frac{451}{500}
=
\frac{49}{500}
=
.098
$$


问:
An equality involving binomial coefitients
I am wondering why formula
$$\sum_{j=k}^n\binom{n}{j}(-1)^j = (-1)^k\binom{n-1}{k-1} $$
is correct only for $1<k<n+1$. Could it be extended to $0<k<n+1$?

I found this formula here.

答:
Yes, it can. If $k=1$ the LHS is equal to $-1$, and the RHS is also equal to $-1$.

Proof:
 $$0=\sum_{j=0}^n(-1)^j{n\choose j}={n\choose 0}(-1)^0+\sum_{j=1}^n{n\choose j}(-1)^j=1+LHS$$

In fact, it can be extended for all k, provided we agree on the common conventions that ${a\choose b}=0$ for $b<0$ or $b>a$, and that empty sums evaluate to 0.

Edit: fixed silly sign error.


问:
Prove that if $\gcd(a,b) = 1$, then $\gcd(ab,a^2+b^2)=1.$
I was reading a Proof that stated

Since $\gcd(a, b) = 1$, it follows that $\gcd(ab, a^2 + b^2) = 1$.

I can't figure out why this statement is true. I tried to factor $ab$ and $a^2+b^2$, but I don't think that you can really reduce $a^2$ and $b^2$ because it's just addition. I'm not really sure how to go about this. Any help?

答:
If $p$ is a prime and $p|ab, p|(a^2+b^2)$ then $p|(a^2+2ab+b^2)$, so $p|a+b$.

But then $p|ab+b^2$ and $p|ab+a^2$, so $p|b^2$, $p|a^2$, implying $p|MCD(a^2,b^2)=1$, contradiction.

Then $p$ doesn't exist, and the problem is solved!


问:
All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$

Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.

I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$

I think that a geometric solution might be possible, please help me to proceed.

Thanks.

答:
Remark: @VIVID gave a nice solution. I rewrote it in elementary way (without using derivative, extrema). Note that while we used calculus to motivate our solution, we do not need to include any
calculus in the solution!

Letting $x = 0$, we have $|1 - q| \le \frac{\sqrt{2}-1}{2}$ or
$$1 - \frac{\sqrt{2}-1}{2} \le q \le 1 + \frac{\sqrt{2}-1}{2}. \tag{1}$$

Letting $x = 1$, we have $|-p - q| \le \frac{\sqrt{2}-1}{2}$ or
$$-p - \frac{\sqrt{2}-1}{2} \le q \le -p + \frac{\sqrt{2}-1}{2}. \tag{2}$$

From (1) and (2), we know that $p < 0$ (reason: if $p\ge 0$ then $-p + \frac{\sqrt{2}-1}{2} < 1 - \frac{\sqrt{2}-1}{2}$).

Since $-\frac{p}{\sqrt{1+p^2}} \in [0, 1]$, by letting $x = -\frac{p}{\sqrt{1+p^2}}$,
we have $|\sqrt{p^2+1} - q| \le \frac{\sqrt{2}-1}{2}$ or
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le q \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}. \tag{3}$$

From (2) and (3), since $\sqrt{p^2 + 1} > -p$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le -p + \frac{\sqrt{2}-1}{2}$$
that is
$\sqrt{p^2 + 1} + p - \sqrt{2} + 1 \le 0$ or
$\frac{(p^2+1) - 2}{\sqrt{p^2+1} + \sqrt{2}}  + (p+1) \le 0$ or
$\frac{(p -  1 + \sqrt{p^2+1} + \sqrt{2})(p+1)}{\sqrt{p^2+1} + \sqrt{2}} \le 0$ which results in
$p \le -1$.

From (1) and (3), since $1 + \frac{\sqrt{2}-1}{2} \le \sqrt{p^2 + 1} + \frac{\sqrt{2}-1}{2}$, we have
$$\sqrt{p^2+1} - \frac{\sqrt{2}-1}{2} \le 1 + \frac{\sqrt{2}-1}{2}$$
that is $p^2 \le 1$. Thus, we have $p = -1$.

From (1) and (3), we have $q = 1 + \frac{\sqrt{2}-1}{2}$.

Finally, when $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$, it is easy to prove that $|\sqrt{1-x^2} + x - 1 - \frac{\sqrt{2}-1}{2}| \le \frac{\sqrt{2}-1}{2}, \forall x \in [0, 1]$
that is $1\le \sqrt{1-x^2} + x \le \sqrt{2}, \forall x\in [0, 1]$.

Thus, $p = -1, q = 1 + \frac{\sqrt{2}-1}{2}$ is the only one pair satisfies the statement.


问:
Find $f(U)$ and calculate $f^{-1}$.
Let $f: \mathbb R^2 \to \mathbb R^2 :  \,(x,y) \mapsto (x+y,xy)$ and $U=\{(x,y): |x| <|y| \,\}$.
Find $f(U)$ and calculate $(f|_U)^{-1}$.

My attempt : Let $(z,t)$ and find $(x,y) \in U$ s.t :
$$\begin{cases}
x+y&=z \\ xy&=t
\end{cases} \iff (x,y) \text{ solution of the equation } X^2-zX+t=0 \\\iff\Delta=z^2-4t>0 \iff X=\dfrac{z\pm \sqrt{z^2-4t}}{2}$$
so $$f^{-1}(x,y)= \left( \dfrac{x- \sqrt{x^2-4y}}{2}, \dfrac{x+ \sqrt{x^2-4y}}{2}\right) \text{ and }f(U)=\{(x,y) : x^2-4y>0\}$$
My solution is true? especially the $f(U)$ ?

答:
Your answer for $f^{-1}(x,y)$ looks right if $x>0$.  If $x<0$ then the signs of $\sqrt{x^2-4y}$ look like they need to be reversed or the absolute value condition for $U$ won’t hold, assuming you want $f^{-1}$ to map back to $U$.

If $x=0$ then it’s not possible to map back to $U$ since the two coordinates would have the same absolute value, contradicting the absolute value condition for $U$.

As for $f(U)$, it looks like a strange beast.  It looks like $R^2$ less the parabola $y=x^2/4$ and everything above it, and also less the line given by $x=0$.

Title: Find $f(U)$ and calculate $f^{-1}$.


问:
Is every harmonic polynomial a linear combination of these?
In $N$-dimensional space, we can show by direct calculation that the polynomial
$$
 r^{2K+N-2}\nabla_a\nabla_b\nabla_c\cdots \frac{1}{r^{N-2}}
\hspace{1cm}
 \text{(with $K$ derivatives)}
$$
is harmonic (annihilated by the Laplacian $\nabla^2$), where $\nabla_n$ is the partial derivative with respect to the $n$th coordinate and $r$ is the distance from the origin.

Is every homogeneous harmonic polynomial of degree $K$ a linear combination of these? If so, how do we prove this? If not, what is a counterexample?

Motive: I'm studying physics, and this seems like a much nicer way to approach the theory of spherical harmonics (just divide this polynomial by $r^K$ to get a spherical harmonic) compared to the typical physics-textbook approach using spherical coordinates, but it's not obvious to me that all spherical harmonics are linear combinations of these for an arbitrary number of dimensions $N$.

答:
I think there is an error in my calculations in my comments. The following is not a complete answer but it gives an hint on how to proceed.

Define $P(N,K)$ to be the space of homogeneous polynomials of degree $K$ in $N$ variables. It has dimension $\binom{N+K-1}{N-1}=\binom{N+K-1}{K}$ the way of distributing the degree $K$ to the $N$ variables. The space $H(N,K) \subset P(N,K)$ is the subspace of harmonic polynomials. One can show that $\dim H(N, K) = \dim P(N, K) - \dim P(N, K-2) $, because the direct sum of the small spaces is isomorphic to the big space.

Note that the expression you wrote is a map from homogeneous polynomials of degree $K$ to harmonic polynomials of degree $K$ by linear extension. For example, in the case $N=5, K=2$ we have

$$ xy +yz \mapsto r^7\nabla_x \nabla_y \frac{1}{r^3} + r^7 \nabla_y \nabla_z \frac{1}{r^3} $$

Let us call this map $\rho: P(N, K) \to H(N, K) $.
Now the relationship you wrote in the comments $\sum_a \nabla_a \nabla_a  \frac{1}{r^{\bullet}} = 0$ rewrites as $ \rho(r^2) =0$, where $ r^2 = x_1^2 +\ldots + x_N^2$ is the "norm" polynomial.

We consider the map $r^2: P(N, K-2) \to P(N, K) $ that multiplies a polynomial by $r^2$. By a similar argument, we get that the composition of the two maps

$$ P(N, K-2) \underset{r^2}{\to} P(N, K) \underset{\rho}{\to} H(N, K) $$

is zero. Also, the dimension of the first and the third space sum to the dimension of the middle space. This means that the second map is surjective if and only if the kernel is the image of the first map. Here I can't go on. In other words, we have to show that the only relations that can occur are the ones that we mentioned: contracting two indices and permuting derivatives. I think it is remarkable to mention that $P(N, K) $ writes as the direct sum of $H(N, K) $ and the image of the first map, $r^2 P(N, K-2) $; the thesis is thus equivalent to the fact that every harmonic polynomial $f$ is uniquely expressible as $\rho(f_v) $ for some harmonic polynomial $f_v$. Is this $f$ itself, or a scalar multiple? Which harmonic polynomial we could expect to be canonically associated to any $f$?


问:
Constructing non-isomorphic groups such that the counts of elements of given order match
In one of the simplest ways of proving that two groups aren't isomorphic, we can compare the orders of each element of both groups. If we find an order $\ell$ such that one of the groups has more elements of this order than the other group, we're done.

I am interested in finite groups where this fails --- ones that aren't isomorphic, but the counts for each order are the same:

$$\DeclareMathOperator\ord{ord} G \not\cong H \quad \land \quad \forall\ell.\ |\{g \in G \mid \ord g = \ell\}| = |\{h \in H \mid \ord h = \ell\}|$$

With some SageMath code, I have computed that the smallest groups where this happens are of order 16. However, I am more interested in how one could methodically find such a counterexample, apart from just trying out groups you can think of until something sticks. How would you find such a pair of groups without a computer?

答:
Here is one family of such groups.

The nonabelian group of order $p^3$ and exponent $p$ can be realized as the multiplicative group of $3\times 3$ unipotent matrices with coefficients in $\mathbb{Z}/p\mathbb{Z}$, i.e., matrices of the form
$$\left(\begin{array}{ccc}
1 & a & c\\
0 &1 & b\\
0 & 0 & 1
\end{array}\right),\qquad a,b,c\in \mathbb{Z}/p\mathbb{Z}.$$
In this group, every nontrivial element has order $p$, and so you get the same count as you do for the abelian (an dhence non-isomorphic) group $(\mathbb{Z}/p\mathbb{Z})^3$.

You can also construct further examples, known as extra-special $p$-groups (of exponent $p$), by taking central products of copies of the group above. Let us denote the group above by $\mathbf{E}_3$ ($E$ for "extra-special"; it's a technical term). Note that the center is cyclic of order $3$, given by the matrices with $a=b=0$. Now, we construct $\mathbf{E}_5$ by taking $\mathbf{E}_3\times\mathbf{E}_3/N$, where $N$ is group of elements of the form $(z,z^{-1})$ with $z\in Z(\mathbf{E}_3)$ (this "identifies" the centers of each copy of $\mathbf{E}_3$). This is again a nonabelian group of exponent $p$, with center of order $p$, of order $p^5$.

Assuming we have defined $\mathbf{E}_{2n+1}$, nonabelian group of order $p^{2n+1}$, exponent $p$, and center of order $p$, we construct $\mathbf{E}_{2n+3}$ by identifying generators $w$ and $z$ of $Z(\mathbf{E}_{2n+1})$ and $Z(\mathbb{E}_3)$, and constructing $\mathbf{E}_{2n+1}\times\mathbf{E}_3/N$, where $N$ consists of the elements of the form $(w^k,z^{-k})$. This will be a nonabelian group of order $p^{2n+3}$, exponent $p$, with center of order $p$.

Then $\mathbf{E}_{2k+1}$ and $(\mathbb{Z}/p\mathbb{Z})^{2k+1}$ will have the desired property, for all $k\geq 1$.

Now, say you have (finite) groups $G$ and $H$ as desired (nonisomorphic, but for each $k$ the number of elements of order $k$ in $G$ and $H$ are the same). Let $K$ be any finite group. I claim that $G\times K$ and $H\times K$ also satisfy the desired properties. That they are not isomorphic is not trivial, but nonetheless it is true (e.g., it follows from the Remak-Krull-Schmidt Theorem). Now I claim that for every $r$, the number of elements of order $r$ in $G\times K$ and in $H\times K$ is the same.

An element $(g,k)\in G\times K$ has order $r$ if and only if (i) the order of $g$ divides $r$; (ii) the order of $k$ divides $r$; and (iii) at least one of the orders of $g$ and $k$ is exactly equal to $r$. Thus, letting $G(d)$ be the set of elements of $G$ of order exactly $d$, and similarly for $H(d)$ and $K(d)$, we have that
$$(G\times K)(r) = \left(\sum_{d|r}|G(d)||K(r)|\right) + \left(\sum_{d|r}|G(r)||K(d)|\right) - |G(d)||K(d)|.$$
Similarly,
$$(H\times K)(r) = \left(\sum_{d|r}|H(d)||K(r)|\right) + \left(\sum_{d|r}|H(r)||K(d)|\right) - |H(d)||K(d)|.$$
Since $|G(m)|=|H(m)|$ for all $m$ by assumption, it follows that $|(G\times K)(r)|=|(H\times K)(r)|$.

An interesting question might be to determine all $n$ for which there exist two non-isomorphic groups $G$ and $H$ of order $n$ for which the count of elements of order $k$ is equal for all $k$. From the above, we see that if $n$ divisible by the cube of an odd prime, or by $16$, then such groups exist. We also know that if $n$ is cubefree and a nilpotent number, then every group of order $n$ is abelian, so no such examples exist. This reduces the problem to $n$ that is cubefree but not a nilpotent number, and to numbers of the form $2^km$ where $m$ is odd and cubefree, $k=3$. I expect counterexamples exist in that case as well.


问:
Factorial as a Integration over a Gaussian
I'm seeking a proof for a formula that I've affirmed through numerical checks. Let $n$ be a natural number. For

$$I_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, x^n D^n (e^{-x^2/2n}),$$

where $D = d/dx$, how can we show that $I_n = n!$?

The formula can be checked for first few $n$'s with Wolfram alpha. n=6 case for identity

As an alternative question, what is $K_n$ for
$$K_n = \frac{(-1)^n}{\sqrt{2\pi n}} \int\limits^{\infty}_{-\infty} dx\, \left(\frac{d}{d\ln x}\right)^n e^{-x^2/2n},$$
where $(d/d\ln x)^nf(x) = x(d/dx)x(d/dx) \cdots x(d/dx) f(x)$. For example,
\begin{align} \left(\frac{d}{d\ln x}\right)^{3}x^{3} & = x\frac{d}{dx} \left[x\frac{d}{dx} \left[ x\frac{d}{dx}\left[ x^{3} \right] \right]\right]  \\
& = 27 x^3.
\end{align}

答:
I interpret $\frac{\partial}{\partial \ln x}$ to mean $x \frac{\partial}{\partial x}$; I can't think of anything else it could mean, but with that interpretation I don't get your identity, and instead I get the following:

Consider the generating function

$$\sum_{n \ge 0} t^n \frac{(x \frac{\partial}{\partial x})^n}{n!} f(x) = e^{(x \frac{\partial}{\partial x})t} f(x) = f(e^t x)$$

(which follows formally from the observation that $xD$ acts by multiplication by $k$ on $x^k$ and hence $e^{(x \frac{\partial}{\partial x})t}$ acts by multiplication by $e^{kt}$). Setting $f(x) = \exp \left( - \frac{x^2}{2 \sigma^2} \right)$ gives

$$\sum_{n \ge 0} t^n \frac{(x \frac{\partial}{\partial x})^n}{n!} \exp \left( - \frac{x^2}{2 \sigma^2} \right) = \exp\left( - e^{2t} \frac{x^2}{2 \sigma^2} \right).$$

Integrating over $x$ gives

$$\sum_{n \ge 0} t^n \int_{-\infty}^{\infty} \frac{(x \frac{\partial}{\partial x})^n}{n!} e^{- \frac{x^2}{2 \sigma^2}} \, dx = \int_{-\infty}^{\infty} \exp\left( - e^{2t} \frac{x^2}{2 \sigma^2} \right) \, dx.$$

The $u$-substitution $u = e^t x, du = e^t \, dx$ gives that this is equal to

$$e^{-t} \int_{-\infty}^{\infty} e^{- \frac{u^2}{2 \sigma^2} } \, du = \sqrt{2\pi} \sigma e^{-t}$$

and extracting the coefficient of $\frac{t^n}{n!}$ gives

$$\int_{-\infty}^{\infty} \left( x \frac{\partial}{\partial x} \right)^n e^{- \frac{x^2}{2 \sigma^2}} \, dx = (-1)^n \sqrt{2 \pi} \sigma.$$

Setting $\sigma^2 = n$ then gives

$$\int_{-\infty}^{\infty} \left( x \frac{\partial}{\partial x} \right)^n e^{- \frac{x^2}{2n}} \, dx = (-1)^n \sqrt{2 \pi n}$$

which is off by a factor of $n!$ from your identity. I checked this explicitly for $n = 2$ though, so I don't think I just dropped the factorial somehow. Do you mean something else by $\frac{\partial}{\partial \ln x}$, and if so what?


问:
Using generating functions, Find a closed formula to next expression: $\sum_{k=0}^m{k(k+2)}$
Using generating functions, Find a closed formula to next expression:

$\sum_{k=0}^m{k(k+2)}$

If i use calculus power series rules, The question is fairly simple. But how can i find the proper relation with generating functions?

Body:
Hint: Consider the function
$$
u_m(x)=\sum_{k=0}^mx^k=\frac{1-x^{m+1}}{1-x},
$$
then
$$
u'_m(x)=\sum_{k=0}^mkx^{k-1},\quad
u''_m(x)=\sum_{k=0}^mk(k-1)x^{k-2},
$$
hence
$$
\sum_{k=0}^mk(k+2)=u''_m(1)+3u'_m(1).
$$
Can you compute $u'_m(1)$ and $u''_m(1)$?


问:
Computing the longest element of the Weyl group
I want to compute the longest element $w_0$ of the Weyl group $W$ for $A_2$, $B_2$ and $G_2$. I saw this has already been asked before here for the case of $G_2$, but the answers are still not very clear to me.  Indeed, I would like to prove this without using any notion about Coxeter groups and only using the properties of the longest element $w_0$: for instance, I know that $w_0$ is the unique element in $W$ such that $w_0$ sends all positive roots into negative ones and that $w_0^2 = id$. Finally, I know that every reduced expression of $w_0$ must contain all simple reflections (possibly appearing more than once). Any hints on how to proceed?

答: How are you describing the elements of these Weyl groups? Take $B_2$. I've drawn the standard picture below. I've picked a couple of simple reflections, $s_\alpha$ and $s_\beta$. They bound the fundamental Weyl chamber, which contains the point $v$. (Ok, I've added a square since I think of this group as symmetries of a square, but you can delete the square if you want.) I've applied the simple reflections $s_\alpha, s_\beta$ to $v$ in all possible ways, recording the shortest expression(s) for each point in the orbit. We can just read off that there is a unique longest expression, $s_\alpha s_\beta s_\alpha s_\beta(v) = s_\beta s_\alpha s_\beta s_\alpha(v)$. This is hence the longest element.


问:
Understanding Multiplication/Dot Product of Two Numbers/Two Vectors
This might be a silly question, but I just realized I don't think I understand what multiplication really is. For example I know $3\cdot4=12$ because that's what you learn in school. If we go with the idea of multiplication as repeated addition, $3\cdot4$ means add $3$ to $0$ (the zero elements) $4$ times. So here we have multiplication defined in terms of addition and the zero/identity element for natural numbers.

My confusion is around numbers that have units. Say we have oranges, how does one interpret $3$ oranges $\cdot 4$ oranges?

Does adding a unit (oranges) to the number make it a vector and therefore the multiplication operator has a different meaning because for two vectors it's dot products? If so, why do we interpret the multiplication of a basis vector with itself to be 1?

答:
It may be hard to understand from an outside point of
view but mathematics is based on precise definitions
that are decided on by mathematicians among themselves.
Thus, for example, the dot product of a unit vector
with itself is $\,1\,$ as a consequence of the precise
definition of the dot product of vectors. However, in
order to multiply oranges by oranges there is no such
precise definition of the result. Of course, you or
anyone else are free to come up with a definition if
that is desired. This new hypothetical definition may
not be useful or well defined but it could be done.
Other than that, there is no currently meaningful way
to multiply two oranges, as far as I know.

As for the precise definition of dot product of two
vectors, such as $\,\vec{i}\cdot\vec{i}\,,$ this is
closely related to the definition
of vectors themselves as geometric objects (loosely
imagined as arrows in space). The origin of vectors
has a long and interesting history. Some of this is
in Michael J. Crowe's book, A History of Vector
Analysis and several other sources. More details
are not suitable for MSE.


问:
Is $PAPA^*P \leq AA^*$?
I`m trying to figure out, if $PAPA^*P \leq AA^*$, where $A \in \mathfrak B (\mathcal H)$ and $P$ is some projector. It can be shown that $PAA^*P \nleq AA^*$ in general case. $\langle PAPA^*P x, x\rangle$ seems to be kind of a natural constriction of bilinear form $\langle AA^* x, x\rangle$ after projecting $B (\mathcal H)$ to $PB (\mathcal H)P$. That's my intuition, but I can't obtain anything more. I've tried to check explicitly, whether $\langle x , (AA^* - PAPA^*P) x\rangle $ is positive, but got nothing useful.

答:
I think
$$
P=\pmatrix{1 &0 \cr 0 &0}
$$
and
$$A=\pmatrix{ 1 & 1\cr 1 & 1}
$$
is a counter example.

Title: Is $PAPA^*P \leq AA^*$?


问:
Finite Automaton with Binary Multiple Inputs
I have created 2 finite automata that accept 00, 01,10,11 and every even place number corresponds to the 2nd number while every odd place number corrensponds to the 1st.

The first finite automaton checks if the 2nd number is strictly smaller than the 1st (also the 2nd number is not 0).

example: [1]0[0]1[1]0 ->
101/010 it accepts it since it mean 5>2 and 2 != 0.

No matter how many 00,11 we have the criterion 1st > 2nd !=0 is not met, if we read 01 before 10 then it locks to Rejection(q1) since there is no way for the 2nd number to be smaller.

Is this an acceptable finite automaton; or there are improvements to be made?

The second finite automaton accepts strings where the 2nd number is 2 times bigger than the 1st (and 2nd != 0). The q5 state is the Rejection state so there are no edges out.

example: [1]0[0]1[1]0[0]1[1]1[0]1[1]0[0]1 -> 10101010/01011101 is rejected since 170/93 != 0.5 but  [0]1[0]1[1]0[1]0[0]1[0]0[1]0[0] -> 00110010/1100100 (50/100) is accepted.

Are those 2 finite automata that solve my problems or they can be improved?
In both automatons q0 is the initial state.

答:
The first automaton is fine; the second is not. For the second problem you need only four states. The key is that when you multiply a binary number $(b_1b_2\ldots b_n)_2$ by $2$, you bet $(b_1b_2\ldots b_n0)_2$: you simply shift the number to the left one place and append a $0$. Thus, the input of such a pair must look like this:

$$[0]b_1[b_1]b_2[b_2]b_3\ldots[b_{n-1}]b_n[b_n]0\,.$$

This is easier to read if we write the inputs as vertical pairs:

$$\begin{array}{rcc}
\text{first number:}&0&b_1&b_2&\ldots&b_{n-1}&b_n\\
\text{second number:}&b_1&b_2&b_3&\ldots&b_n&0
\end{array}$$

Let $q^*$ be the initial state; the other states will be $q_0$, $q_1$, and $q_\infty$. The only acceptor state is $q_0$, and $q_\infty$ is the ‘dump’ (rejection state). The transition table is as follows:

$$\begin{array}{c|c|c}
\text{current state}&\text{input}&\text{next state}\\\hline
q^*&00&q^*\\
q^*&01&q_1\\
q^*&10,11&q_\infty\\\hline
q_0&00&q_0\\
q_0&01&q_1\\
q_0&10,11&q_\infty\\\hline
q_1&00,01&q_\infty\\
q_1&10&q_0\\
q_1&11&q_1\\\hline
q_\infty&00,01,10,11&q_\infty
\end{array}$$

The automaton remains in the initial state as long as it reads initial zeroes for both numbers. After that the automaton is in state $q_1$ when it has just read a pair $01$ or $11$, meaning that the first bit of the next pair must be $1$ for a valid input, and it is in state $q_0$ when it has just read a pair $00$ or $10$, meaning that the first bit of the next pair must be $0$ for a valid input.


问:
Is this Markov chain recurrent or transient?
A Markov chain $X_n$ $(n\ge1)$ with state space S=0,1,2,3,... has the following transition probabilities:

$$p_{i,i+1}=\frac{(i+1)^2}{2i^2+2i+1},\ p_{i,i-1}=\frac{i^2}{2i^2+2i+1}, \ p_{0,1}=1, \ and \ i\ge 1$$

I am trying to find if it is recurrent or transient. What I have is the following:

Obviously, $p_{i,i+1}$ and $p_{i,i-1}$ depends on $i$, it means that each state will have a different probability. For example, when $i=1$, then $p_{1,2}=4/5$ and $p_{1,0}=1/5$. So, it seems that $p_{i,i+1}$ is decreasing as $p_{i,i-1}$ increases on each step.

$$T=\pmatrix{0&1&0&0&...&0&0&0\\1/5&0&4/5&0&...&0&0&0\\0&4/13&0&9/13&...&0&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\0&0&0&0&...&1/2&0&1/2}_{nxn}$$

According to wiki, a state i is recurrent if and only if the expected number of visits to i is infinite:

$$ \sum_{n=0}^{\infty}p_{ii}^{(n)}=\infty\\ p_{ii}=Pr(X_n=i|X_0=0)\ for\ n\ge 1$$

for a recurrent state i, the mean hitting time is defined as:

$$M_i=E\left[ T_i \right]=\sum_{n=1}^{\infty}nf_{ii}^{(n)}$$

State i is positive recurrent (or non-null persistent) if Mi is finite; otherwise, state i is null recurrent (or null persistent).

If this Markov chain had a finite state space:

I think that I should be positive recurrent because it is possible to
get to any state from any state.

How does the infinite state space change this fact? I haven't found good information about it.

What about this?

If we start from zero, it seems that the MC may return to zero almost surely with probability 1.

Thank you!

答:
Adapting my answer from here, which asks basically the same question.

Let $h_n=\mathbb P_n(\text{hit }0)$, so $h_0=1$. Then
$$h_n=p_{n,n-1}h_{n-1}+p_{n,n+1}h_{n+1}\implies \frac{h_{n+1}-h_n}{h_n-h_{n-1}}=\frac{n^2}{(n+1)^2}.$$
Telescoping, we get $h_{n+1}-h_n=\frac{1}{(n+1)^2}(h_1-h_0)$. So summing over $0\leq n\leq m-1$, we obtain $$h_m=1+(h_1-1)\sum_{k=1}^m\frac{1}{k^2}.$$
As $\sum_{k=1}^m\frac{1}{k^2}\to\frac{\pi^2}{6}$, the fact that $(h_m)_{m=0}^\infty$ is the smallest non-negative solution to this equation implies that $h_1=1-\frac{6}{\pi^2}<1$. This is enough to show that the chain is transient.


问:
Difference between bases and topology
Be $X$ a set and $B$ a collection of subsets of $X$ satisfying the following properties:
(1) For each $U_1$, $U_2$ and each $x\in U_1\cap U_2$, there is $U\in B$ such that $x \in U\subseteq U_1\cap U_2$.
(2) For each $x\in X$, there is $U\in B$ such as $x\in U$.

So the $B^{\ast}$, formed by the $B$ subcollection unions, is a topology about $X$ and $B$ is a basis for the $(X,B^{\ast})$topological space.

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

How can I prove that given $U_1$ and $U_2$ and a point at the intersection then there is $U$ at the intersection, such as $x$ is at $U$ ?
Also, how can I show (2), which always exists $x\in X$, is there $U\in B$ such that $x\in U$?

答:
If $U_1=U_2=X$ then choose $U=X$, otherwise, assume e.g. $U_1=\{y\}$ and $x\in U_1\cap U_2$ then $x=y$ and $U=U_1$ will work.

We have $X\in B$.

Title: Difference between bases and topology


问:
A set operation involving the intersection of two complements
I'm having trouble with this question. I know that
$A\ast B=(E-A)\cap (E-B)$ is also equal to $E-(A \cup B)$ but I can't make sense of the operation described in the question and I don't know how to use it to isolate the sets required.

The question reads:

Let $E$ be a set, with subsets $A\subset E$  and $B\subset E$, and let $\ast $ be the operation $A\ast B=(E-A)\cap (E-B)$, where $E-A$ is the set of elements in $E$ that are not in $A$. Express the sets

a. $A \cup B$

b. $A \cap B$

c. $E-A$

in terms of $A,B, \ast$."

答: Once you notice that you can express complements using $*$ (i.e., part (c) of the exercise), namely as $A*A$, the rest should fall into place. You already know how to express $E-(A\cup B)$, so with the ability to express complements you can express $A\cup B$. That, with complements and De Morgan's Law, will give you $A\cap B$ as well.


问:
Irreducible Transient Markov chain with infinite state space returns to i
I have an irreducible transient Markov chain with infinite state space {0,1,2,...} and I am trying to find the probability that the chain never returns to the initial state (let's say state 0). For example, consider an asymmetric random walk, in which the states (starting from zero) have a different probability each, so the probability of going left $p_{i,i-1}$ is less than the probability of going right $p_{i,i+1}$, but when the step n goes to infinity, both probabilities are the same (1/2).

How can I prove that the chain never returns to zero?

A state is transient if there's a nonzero probability that, starting there, that state is never visited again.

In other words: $f_{ii}=P(X_n=i\ for\ some\ n \ge 1|X_0=i)>0$ In this case $i=0$

Thank you.

答: This question is false as stated. If $p_{0,1}=1$, $p_{i,i+1}=\frac{i+1}{2i+1}$ and $p_{i,i-1}=\frac{i}{2i+1}$, then here I show that the chain is in fact recurrent.


问:
Why does sin(90)=1, and not 0?

If o/h is = sin(90), wouldn't sin(90) = 0

答:
As $\theta \to 90^\circ$, $\frac{O}{H} \to 1$. Because O and H become closer and closer to the same value. You can see that on your graph (just keep track of the colors, red and green are overlapping). Another way to look at it is by generalizing the trigonometric functions beyond side lengths of triangles. Once defined on the unit circle, by their Power Series (typically Taylor), or in terms of the complex exponential, the result is fairly trivial.

On the unit circle, that point is $y=1$ $(\sin = 1)$, $x=0$ $(\cos = 0)$.

By their Power Series, $\sin(\frac{\pi}{2}) = \frac{\pi}{2} - \frac{(\frac{\pi}{2})^3}{3!} + \frac{(\frac{\pi}{2})^5}{5!} - \frac{(\frac{\pi}{2})^7}{7!}...$, which you can see after the first few terms approaches $1$.

By the complex exponential, you have $\operatorname{Im} (e^{\frac{\pi i}{2}})$, which is $1$.

Title: Why does sin(90)=1, and not 0?


答:
A slightly warped version of Ruy's example:
$$P = \frac12\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}, \quad A = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}.$$

We have $$AA^* = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \quad PAPA^*P = \frac18\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$$
and the difference has determinant $-\frac18$.

Title: Is $PAPA^*P \leq AA^*$?

Body: I`m trying to figure out, if $PAPA^*P \leq AA^*$, where $A \in \mathfrak B (\mathcal H)$ and $P$ is some projector. It can be shown that $PAA^*P \nleq AA^*$ in general case. $\langle PAPA^*P x, x\rangle$ seems to be kind of a natural constriction of bilinear form $\langle AA^* x, x\rangle$ after projecting $B (\mathcal H)$ to $PB (\mathcal H)P$. That's my intuition, but I can't obtain anything more. I've tried to check explicitly, whether $\langle x , (AA^* - PAPA^*P) x\rangle $ is positive, but got nothing useful.