问:
Do there exist elements in cyclotomic fields with certain absolute value?
Consider the $m$-th cyclotomic field $K$ over $\mathbb Q$. For an integer $n\ge2$, we know it is possible that there does not exist integer $\alpha\in\mathcal O_K$ such that $\alpha\bar\alpha=n$. What if we let $\alpha\in K$? Does it hold for arbitrary $n$ and $m$?

I suppose this is not a hard problem, but I can't find references. Thank you in advance.

答:
Of course you cannot factor an arbitrary integer $n$ this way.

Write $(\alpha) = {\mathfrak a}/{\mathfrak b}$ for coprime integral ideals ${\mathfrak a}$ and ${\mathfrak b}$. Then $(\overline{\alpha}) = \overline{\mathfrak a}/\overline{\mathfrak b}$. Since $\alpha \overline{\alpha} = n $ is an integer,   ${\mathfrak b}$ must divide $\overline{\mathfrak a}$, say $\overline{\mathfrak a} = {\mathfrak b} {\mathfrak c}$, and then $(\alpha)(\overline{\alpha}) = {\mathfrak c} \overline{\mathfrak c}$. If $K$ has class number $1$, then ${\mathfrak c}$ is principal, hence there do not exist any integers $n$ that can be written as $\alpha \overline{\alpha}$ for proper fractional elements $\alpha$. In any case $n$ must split into two ideal factors in the maximal real subfield.

A more challenging question is whether the equation  $\alpha\overline{\alpha} = n$ is possible for $\alpha \in K$ when it is impossible in the ring of integers. We know that ${\mathfrak a} \sim {\mathfrak b}$, hence ${\mathfrak c} = \overline{\mathfrak a}/{\mathfrak b} \sim \overline{\mathfrak a}/{\mathfrak a} = {\mathfrak a}^{\sigma-1}$, where $\sigma$ denotes complex conjugation. This implies that ${\mathfrak c}^{1+\sigma} \sim 1$. Thus the ideal class of ${\mathfrak c}$ must be in the kernel of the relative norm to the maximal real subfield. This is a necessary condition; I do not know whether it is sufficient.


问:
Height of a tetrahedron
How do I calculate the height of a regular tetrahedron having side length $1$ ?

Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table?

答: Here is an answer without words following the lines of J.M.'s answer:

ContentLicense: CC BY-SA 3.0

Title: Height of a tetrahedron

ContentLicense: CC BY-SA 4.0


问:
Generalization of Bézout's theorem
Let $$p_1(x_1,x_2,\dots,x_n),p_2(x_1,x_2,\dots,x_n),\dots,p_n(x_1,x_2,\dots,x_n)$$
be polynomials with $n$ variables. I would like to to find out if
$$p_1(x_1,x_2,\dots,x_n)=0\\
p_2(x_1,x_2,\dots,x_n)=0\\
\vdots\\
p_n(x_1,x_2,\dots,x_n)=0
$$

have finite number of solutions?

Is there something like  Bézout's theorem for this problem? I know that if $n=2$, it means that I have only $2$ polynomials of $2$ variables $p_1(x_1,x_2)$ and $p_2(x_1,x_2)$ which are coprime and homogeneous, where $deg(p_1)=m$ and $deg(p_2)=n$ then number of solutions of
$$p_1(x_1,x_2)=0\\
p_2(x_1,x_2)=0\\
$$
is at least $m\cdot n$.

Are there something general which can be use for my problem? It is enough to say that if polynomials $$p_1(x_1,x_2,\dots,x_n),p_2(x_1,x_2,\dots,x_n),\dots,p_n(x_1,x_2,\dots,x_n)$$
are coprime and homogeneous, that number of solutions of equations
$$p_1(x_1,x_2,\dots,x_n)=0\\
p_2(x_1,x_2,\dots,x_n)=0\\
\vdots\\
p_n(x_1,x_2,\dots,x_n)=0
$$
is finite.

Any help will be appreciated. Thank you very much.

答:
Question: "Are there something general which can be use for my problem?"

Answer: In general if $p_1,\dots,p_l \in k[x_1,\dots,x_n]$ are polynomials over a field $k$, you may consider the quotient ring $A:=k[x_1,\dots,x_n]/(p_1,\dots,p_l)$. Let $I$ denote the system of polynomial equations $p_1=0,\dots,p_l=0$. The following holds:

Lemma: The system $I$ has a finite set of solutions (in a finite extension of $k$) if and only if $\dim_k(A)<\infty$.

You'll find a "proof" of this Lemma at the following link:

When does a system of polynomial equations have infinitely many solutions?

Title: Generalization of Bézout's theorem


问:
Is it possible to find the Padé approximation for $\ln{(1+x)}$?
On the Padé approximant wiki, people present various form of padre approximant for $\sin{(x)}, \exp{(x)}$, erf$(x)$ but does not provide instruction on how to create them.

What would be the correct way to find the Padé approximant for $\ln{(1+x)}$ ?

Thank you very much !

答:
I shall make the solution as simple as possible. It is simplistic but it works.

The $[n,m]$ Padé approximant write
$$f(x)=\frac{\sum_{k=0}^n a_k\,x^k}{1+\sum_{k=1}^m b_k\,x^k}$$ But, you also have
$$f(x)=\sum_{k=0}^\infty c_k\,x^k$$ Cross multiply
$$\left(1+\sum_{k=1}^m b_k\,x^k\right)\left(\sum_{k=0}^\infty c_k\,x^k \right)=\sum_{k=0}^n a_k\,x^k$$

Expand and identify the coefficients $a_k$ and $b_k$. That is all !


问:
Infimum and supremum of the set $Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land 4^n n^m < m^m < 27^n n^m \right\}$
The problem is to find supremum and infimum of set $$Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land \space4^n n^m < m^m < 27^n n^m \right\}$$

I have an answer to this question, infimum is $2$ and supremum is $3$ but I don't see how to actually do the problem, so I would like to ask for some help with that. Thank you.

答:
We have

\begin{align}4^nn^m<m^m&<27^nn^m \implies 4^{n/m}n<m<27^{n/m}n\\[1ex]
&\implies 4^{n/m}<m/n<27^{n/m}\\[1ex]
&\implies 4<(m/n)^{m/n}<27\\[1ex]
&\implies 2^2<q^q<3^3
\end{align}

where $\;q=m/n$.

Now, it is easy to see that the function $\;q\longmapsto q^q\;$ is strictly increasing for $q\geq 1/e$, and it is $\;\geq 1\;$ iff $q\geq1$. Hence, from the previous calculation we deduce that $\;2<q<3$. The bounds $2$ and $3$ are respectively $\;\it inf\;$ and $\;\it sup\;$ because

$$ \lim_{q\to2^+}q^q=2^2\qquad \text{and}\qquad \lim_{q\to3^-}q^q = 3^3.$$


问:
Conditions for which $P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z')$
Given two independent draws $(X,Z,Y)$ and $(X',Z',Y')$ from some distribution $P$, I would like to understand under which conditions does the following hold:

$$
P(Y\le Y'|X,Z,X',Z') = P(Y\le Y'|Z,Z').
$$

This clearly holds when $X \perp Y | Z$ for example, but I want to be able to say that the above statement holds if and only if some condition is met.

答:
Partial answer / a dependent example:

First of all, lemme simplify by removing $Z, Z'$ from the problem -- i.e. consider examples where $Z$ is independent of everything else.

So, can we have $P(Y \le Y' \mid X, X') = P(Y \le Y')$ even though $X,Y$ are dependent?  Here is a classic example, one of the textbook examples illustrating two dependent variables with zero correlation:

$(X,Y)$ is a random point on the unit circle.

Clearly, $X,Y$ are dependent, and slightly less clearly, they have zero correlation.  It is also easy to verify that, for any $X = x, X' = x'$, both probabilities equal $1/2$.

If you want to add back a non-trivial dependence on $Z$, simply pick:

$(X,Y,Z)$ is a random point on the unit sphere (surface, not interior).

Again, both probabilities equal $1/2$.

Come to think of it, picking from the unit disk (including interior), or from the unit ball (including interior) also satisfies your equation.

Back to your general question, "Conditioned on $Z$, then $X,Y$ have zero correlation" is not sufficient to guarantee your equation, as many examples of zero correlation show.  So I still don't know what the general criterion is.


问:
Show that $\sum_{k=0}^N\binom{N}{k}x^k(1-x)^{N-k}(1-\alpha k/N)^{-1}$ is decreasing in $N$
I've come across the following sum of binomial coefficients:
$$S = \sum_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \frac{1}{1 - \alpha \frac{k}{N}},$$
where $x \in [0, 1]$ and $\alpha \in [0, 1)$.

I need to show $S$ is decreasing in $N$ (as the last term becomes smaller when $N$ increases), regardless of values of $x$ and $\alpha$ within the range specified above (that's what I obtained from simulation). I understand that $S$ ranges between $1$ and $\frac{1}{1 - \alpha}$ but don't know how to proceed further.

Much appreciated.

答:
(For comparison purposes, this was the direction I was considering...  It does not seem to provide a satisfactory conclusion.)

If $\alpha=0$ the conjecture is false.  If $\alpha=1$ the last term in the sum is indeterminate.  So we assume that $0\lt\alpha \lt 1$ and choose $N\gt \frac 1{1-\alpha}$ and note that this series is finite and geometric series with $|r|\lt 1$ are absolutely convergent, so we can write

$$\frac 1{1-\alpha\frac kN}=\sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\\
\frac 1{1-\alpha\frac k{N-1}}-\frac 1{1-\alpha\frac {k+1}N}=\frac{\alpha\frac k{N-1}-\alpha\frac {k+1}N}{(1-\alpha\frac k{N-1})(1-\alpha\frac {k+1}N)}\\
=\alpha\frac{\frac {Nk-Nk-N+k+1}{N(N-1)}}{(1-\alpha\frac k{N-1})(1-\alpha\frac kN)}=\frac{\alpha}{N}\frac{\frac k{N-1}-1}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}$$
which can then be used as
$$S_N = \sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k}\frac 1{1-\alpha\frac kN}=\sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\\
=1+\sum\limits_{k = 0}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=1}^\infty\left(\alpha\frac kN\right)^i\qquad \text{inner sum has first term $1$}\\
=1+\sum\limits_{k = 1}^N \binom{N}{k}x^k (1-x)^{N-k} \sum_{i=1}^\infty\left(\alpha\frac kN\right)^i\qquad k=0\to \text{outer first term zero}\\
=1+\alpha\sum\limits_{k = 1}^N \binom{N}{k}\frac kNx^k (1-x)^{N-k} \sum_{i=0}^\infty\left(\alpha\frac kN\right)^i\qquad \text{divide all inner-sum terms by $\alpha\frac kN$}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \sum_{i=0}^\infty\left(\alpha\frac {k+1}N\right)^i\qquad \text{re-index $k$}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \frac 1{1-\alpha\frac{k+1}N}\qquad\text{revert inner sum}\\
=1+\alpha x\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \left(\frac 1{1-\alpha\frac{k}{N-1}}+\frac \alpha N\frac{1-\frac k{N-1}}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}\right)\\
=1+\alpha xS_{N-1}+\frac {\alpha^2x} N\sum\limits_{k = 0}^{N-1} \binom{N-1}{k}x^{k} (1-x)^{N-k-1} \frac{1-\frac k{N-1}}{\left(1-\alpha \frac k{N-1}\right)\left(1-\alpha \frac kN\right)}
$$
This is almost a good way to compare $S_N$ with $S_{N-1}$.


问:
What is 8 times 9 in 12-hour clock arithmetic?
What is $8\times 9$ in $12$ hour arithmetic?

Will it be $0$, or will it be $6$?

$8*9 = 72$ and $72/12 = 6$

So is the answer $0$ or $6$?

答:
Modular arithmetic deals with remainders upon division.  The remainder of $72$ upon dividing by $12$ is $0$.  So we have

$$8\cdot9\equiv72\equiv0\operatorname{mod}12$$


问:
Solution using $f\left( x \right) = x\cos \frac{1}{x},x \ge 1$
If $f\left( x \right) = x\cos \frac{1}{x},x \ge 1$.
Then which of the following is the correct option for the domain $x \ge 1$

(A) $f(x+2)-f(x)<2$

(B) $f(x+2)-f(x)>2$

My approach is as follow

$T\left( x \right) = f\left( {x + 2} \right) - f\left( x \right) = \left( {x + 2} \right)\cos \frac{1}{{x + 2}} - x\cos \frac{1}{x}$

$T\left( x \right) = x\left( {\cos \frac{1}{{x + 2}} - \cos \frac{1}{x}} \right) + 2\cos \frac{1}{{x + 2}} \Rightarrow T\left( x \right) = x\left( {2\sin \left( {\frac{{\frac{1}{{x + 2}} + \frac{1}{x}}}{2}} \right)\sin \left( {\frac{{\frac{1}{x} - \frac{1}{{x + 2}}}}{2}} \right)} \right) + 2\cos \frac{1}{{x + 2}}$

$T\left( x \right) = x\left( {2\sin \left( {\frac{{2x + 2}}{{2x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{2}{{2x\left( {x + 2} \right)}}} \right)} \right) + 2\cos \frac{1}{{x + 2}} \Rightarrow T\left( x \right) = x\left( {2\sin \left( {\frac{{x + 1}}{{x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{1}{{x\left( {x + 2} \right)}}} \right)} \right) + 2\cos \frac{1}{{x + 2}}$

$T\left( x \right) = 2\left( {\left( x{\sin \left( {\frac{{x + 1}}{{x\left( {x + 2} \right)}}} \right)\sin \left( {\frac{1}{{x\left( {x + 2} \right)}}} \right)} \right) + \cos \frac{1}{{x + 2}}} \right)$

From here onward how do I proceed

答:
From your expression $T(x)$, now we find roots for

$$
T(x) - 2 = 0 \\
x \sin\left( \frac{x+1}{x(x+2)} \right) \sin\left( \frac{1}{x(x+2)} \right) + \cos \left(\frac{1}{x+2}\right) - 1 = 0
$$

Comparing denominators and numerators inside every sin and cos expresion:

$$
x+1 < x(x+2)\\
1 < x(x+2)\\
1 < x+2
$$

So sin and cos expresions are monotonic. Check that:

$$
x \sin\left( \frac{x+1}{x(x+2)} \right) \sin\left( \frac{1}{x(x+2)} \right) \to 0  \text{
 (monotonically increassing)}\\
\cos \left(\frac{1}{x+2}\right) \to 1  \text{
 (monotonically decreasing)}
$$

Then note that

$$
1 \sin\left( \frac{1+1}{1(1+2)} \right) \sin\left( \frac{1}{1(1+2)} \right) = 0.20232\dots \\
\cos \left(\frac{1}{1+2}\right) = 0.94495\dots
$$

Also
$$
T(1) - 2 = 0.20232 + 0.94495 - 1 = 0.14727 > 0\\
T(x) - 2 \to 0 \text{(monotonically)}
$$

Thus

$T(x) > 2$

Not that rigorous but hope it helps,
Bests!


问:
Is there any property of the set of nilpotent matrices?
Let k be a field, $M_nk$ its matrix ring. We denote the set of nilpotent matrices by $N$. What properties can we know about $N$? Is there any analogy of "nilradical" in the matrix ring? Could you share any ideas on this problem?

答:
There are numerous nilradicals of rings, all of which are zero anyways in $M_n(k)$ since it is a simple ring.

The set of nilpotent elements in a noncommutative ring does not have many useful properties.  They are not even closed under addition or multiplication, as evidenced by:

$\begin{bmatrix}0&1\\0&0\end{bmatrix}+\begin{bmatrix}0&0\\1&0\end{bmatrix}=\begin{bmatrix}0&1\\1&0\end{bmatrix}$

and $\begin{bmatrix}0& 1\\0&0\end{bmatrix}\begin{bmatrix}0&0\\1&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$

The only properties I can come up with at all right now is that they are closed under scalar multiplication and exponentiation, both of which are pretty trivial.


问:
Polynomials that generate given ideal
We define an ideal $I$ to be the set of all polynomials with integer coefficients, with constant and linear terms divisible by $5$. (If $f(x) = a_0+a_1 x + \cdots \in I$, both $a_0$ and $a_1$ are divisible by $5$). I have already shown that $I$ is an ideal in $\mathbb Z[x]$. How do I go about finding polynomials that generate $I$?

Title: Polynomials that generate given ideal

Body:
Hint. $$5b_0+5b_1x+a_2x^2+\cdots=5(b_0+b_1x)+x^2(a_2+\cdots).$$
Any guess now?


问:
Left invariant Affine connections
I just started to study Affine connections and I'm trying to prove the following;

Let $\nabla$ be an Affine connection on a Lie group $G$, and $X_1, X_2, \dots, X_n$ be a basis of the Lie algebra $\mathfrak{g}$ of $G$. (Note that $X_i$ are left-invariant vector fields on $G$ by definition.) If $\nabla_{X_i} X_j$ is a left-invariant vector field for any $i, j = 1, 2, \dots, n$, then $\nabla$ is left-invariant.

To show that $\nabla$ is left-invariant, I need to show that for every vector fields $Z$ and $Z'$, it satisfies that $\nabla_{dL_{\sigma}Z} (dL_{\sigma}Z') = dL_{\sigma} \nabla_Z Z'$ for each $\sigma \in G$. Since $X_1, X_2, \dots, X_n$ is a basis, we may write $Z = \sum_{i=1}^n f_i X_i$ and $Z' = \sum_{j=1}^n g_j X_j$ for some smooth functions $f_i, g_j \in C^{\infty}(G)$. And then by linearity I can write

$\nabla_{Z^{L_{\sigma}}}(Z')^{L_{\sigma}} = \nabla_{\sum_{i=1}^n f_i^{L_{\sigma}} X_i^{L_{\sigma}}} \sum_{j=1}^n g_j^{L_{\sigma}} X_j^{L_{\sigma}} = \sum_{i=1}^n f_i^{L_{\sigma}} \nabla_{X_i^{L_{\sigma}}} \sum_{j=1}^n g_j^{L_{\sigma}} X_j^{L_{\sigma}}$

where $Z^{L_{\sigma}}$ is the push-forward $dL_{\sigma}Z$, and same for $f^{L_{\sigma}}$.

But then I stuck here, since the axioms on Affine connections does not tell me that $\nabla$ is linear in the second variable. I want to apply the Leibniz rule on $g_j^{L_{\sigma}} X_j^{L_{\sigma}}$, but I should first separate those summands somehow.

How can I resolve this situation?

答:
Let $X_1,\dots, X_n$ be a basis of left invariant vector fields on $G$ and let $\nabla$ be a connection on $G$ such that
$$
\nabla_{X_i} X_j
$$
is left invariant.  We want to show that for all $g\in G$ and all vector fields $A$ and $B$ on $G$ we have
$$
\nabla_{dL_gA}(dL_gB)=dL_g \nabla_AB.
$$
Write (using Einstein summation notation)
$$
A=f^iX_i,~B=h^iX_i
$$
Let $k\in G$.  Then
$$
\begin{align}
(dL_gA)_{k}&=dL_g~A_{g^{-1}k}\\
&=f^i(g^{-1}k)dL_g(X_i)_{g^{-1}k}\\
&=f^i(g^{-1}k)(X_i)_k
\end{align}
$$
Hence,
$$
dL_g A= (f^i\circ L_{g^{-1}})X_i
$$
Likewise,
$$
dL_g B= (h^i\circ L_{g^{-1}})X_i
$$
Write $\tilde{f}^i:=f^i\circ L_{g^{-1}}$ and $\tilde{h}^i:=h^i\circ L_{g^{-1}}$.
Then
$$
\begin{align}
\nabla_{dL_gA}(dL_gB)&=\tilde{f}^i\nabla_{X_i}(\tilde{h}^jX_j)\\
&=\tilde{f}^i(X_i\tilde{h}^j) X_j+\tilde{f}^i\tilde{h}^j\nabla_{X_i} X_j
\end{align}
$$
On the other hand,
$$
\nabla_{A}B=f^i(X_ih^j) X_j+f^ih^j\nabla_{X_i} X_j
$$
Using the fact that $\nabla_{X_i} X_j$ is left invariant, we apply $dL_g$ to $\nabla_{A}B$
$$
\begin{align}
dL_g \nabla_{A}B&=(f^i\circ L_{g^{-1}})[(X_ih^j)\circ L_{g^{-1}}] X_j+(f^i\circ L_{g^{-1}})(h^j\circ L_{g^{-1}})\nabla_{X_i} X_j\\
&=\tilde{f}^i[(X_ih^j)\circ L_{g^{-1}}] X_j+\tilde{f}^i \tilde{h}^j\nabla_{X_i} X_j
\end{align}
$$
The left invariance of $\nabla$ is proved if we can show that the function
$$
(X_ih^j)\circ L_{g^{-1}}=X_i\tilde{h}^j
$$
Apply $k\in G$ to the left side and using the left invariance of $X_i$ gives:
$$
\begin{align}
(X_ih^j)\circ L_{g^{-1}}(k)&=(X_ih^j)(g^{-1}k)\\
&=(X_i)_{g^{-1}k}h^j\\
&=[dL_{g^{-1}}(X_i)_k]h^j\\
&=(X_i)_k (h^j\circ L_{g^{-1}})\\
&=(X_i)_k \tilde{h}^j\\
&=(X_i \tilde{h}^j)(k)
\end{align}
$$
Hence, $(X_ih^j)\circ L_{g^{-1}}= (X_i \tilde{h}^j)$ which proves the left invariance of $\nabla$.

Title: Left invariant Affine connections


问:
Composition Method for negative term
I'm a undergraduate student now learning a graduate simulation course, and I met some problems.

The Professor ask us to sample this function by composition method. Its CDF and PDF are like:
$$
F(x)=1-e^{-x}-e^{-2x}+e^{-3x} \\
f(x)=-3e^{-3x}+2e^{-2x} +e^{-x}
$$
I found out the PDF has a negative term $-3x$, and I don't know how to deal with it.

P.S. The Professor taught us the examples but there're all positive terms.

thx btw.

答:
Rewrite your density as $(e^{-x} - e^{-3x}) + 2(e^{-2x} - e^{-3x})$. Neither $e^{-x} - e^{-3x}$ nor $e^{-2x} - e^{-3x}$ are proper densities, but you can check that
\begin{align}
f_1(x) &= \tfrac{3}{2}(e^{-x} - e^{-3x}),\\
f_2(x) &= 6(e^{-2x} - e^{-3x})
\end{align}
are. That means you can write your $f$ as
\begin{align*}
f(x) &= \tfrac{2}{3}f_1(x) + (2)(\tfrac{1}{6})f_2(x) \\
&= \tfrac{2}{3}f_1(x) + \tfrac{1}{3}f_2(x).
\end{align*}
So one possible composition method is:

Generate $U$ uniformly from $[0, 1]$.

If $U \leq \tfrac{2}{3}$ generate a variate from $f_1$; otherwise from $f_2$.

The latter step is easy since you can use sums of two exponentially distributed variates. Generate $E$ and $E'$ as independent exponentially distributed, each with mean 1. If you need an $f_1$ return $E + \tfrac{1}{3}E'$. Otherwise $\tfrac{1}{2}E + \tfrac{1}{3}E'$ gives you an $f_2$.

Title: Composition Method for negative term


答:
Remember,
$$a\equiv b\bmod n\;\iff \;n\text{ goes into }a-b,$$
or in other words,
$$a\equiv b\bmod n\;\iff \;\text{dividing }a-b\text{ by }n\text{ leaves no remainder}.$$
Using this definition, which statement is true:
$$72\equiv 0\bmod 12$$
or
$$\;\;\;72\equiv 6\bmod 12 \;?$$</p>

Title: What is 8 times 9 in 12-hour clock arithmetic?

Body:
What is $8\times 9$ in $12$ hour arithmetic?

Will it be $0$, or will it be $6$?

$8*9 = 72$ and $72/12 = 6$

So is the answer $0$ or $6$?


问:
Geodesic flow and Killing vector field
Let $\gamma$ be a geodesic, $\gamma'$ its tangent vector, $X$ a Killing vector field and $X_\gamma$ the restriction of $X$ to the curve $\gamma$. Let $g$ be the metric on the manifold considered.

Prove that $g(\gamma',X_\gamma)$ is constant along $\gamma$.

The definition I have for a Killing field $X$ is that it satisfied $\mathcal{L}_Xg=0$ where $\mathcal{L}$ denotes the Lie derivative.

I thought about using the induced covariant derivative on the curve $\gamma$, but all I have is a definition and no real explanation on how to put this into pratice.

答: Here's an overkill solution: the flow of $X$ consists of isometries, and so it preserves the Lagrangian $L\colon TM \to \Bbb R$ given by $L(x,v) = g_x(v,v)/2$, whose fiber derivative is $\mathbb{F}L(x,v) = g_x(v,\cdot)$. Critical points of the action functional of $L$ are geodesics, so Noether's Theorem says that along geodesics, the Noether charge $$\mathbb{F}L(x,v)X_x = g_x(v,X_x)$$is constant. So $t \mapsto g_{\gamma(t)}(\gamma'(t), X_{\gamma(t)})$ is constant if $X$ is Killing and $\gamma$ is a geodesic.

Title: Geodesic flow and Killing vector field


答:
If $\overline{\alpha}$ means the complex conjugate then $\alpha\overline{\alpha}=n$ gives that
$$N_{\Bbb{Q}(\zeta_m)/\Bbb{Q}}(\alpha) =N_{\Bbb{Q}(\zeta_m+\zeta_m^{-1})/\Bbb{Q}}(n)=n^{\phi(m)/2}$$

If $n=p$ is an inert prime number (ie. it stays prime in $\Bbb{Q}(\zeta_m)$ which happens when $p$ has order $\phi(m)$ modulo $m$) then this is impossible,

Because any $\beta\in \Bbb{Q}(\zeta_m)^*$ is of the form $p^r u$ with $r\in \Bbb{Z}$ and $u$ a unit of the DVR $\Bbb{Z}[\zeta_n]_{(p)}$, the conjugates of $u$ are units too, so $$N_{\Bbb{Q}(\zeta_m)/\Bbb{Q}}(\beta)=p^{\phi(m) r} N_{\Bbb{Q}(\zeta_m)/Q}(u)\in p^{\phi(m) r} \Bbb{Z}_{(p)}$$

Title: Do there exist elements in cyclotomic fields with certain absolute value?

Body:
Consider the $m$-th cyclotomic field $K$ over $\mathbb Q$. For an integer $n\ge2$, we know it is possible that there does not exist integer $\alpha\in\mathcal O_K$ such that $\alpha\bar\alpha=n$. What if we let $\alpha\in K$? Does it hold for arbitrary $n$ and $m$?

I suppose this is not a hard problem, but I can't find references. Thank you in advance.


问:
How to check if a power series formula is correct?
Given the next power series convergent within the interval $(-3,-1)$:

$$\sum_{n=0}^\infty (-1)^{n+1}\frac{2^{2n+1}(x+2)^{2n}}{7^{2n}}$$

I need to express it as a formula, so using the Geometric series:

First I reestructure the series to make it similar to the geometric series

$\sum_{n=0}^\infty \frac{(-1)^{n+1}2^{2n+1}}{7^{2n}}(x+2)^{2n}$

$(-2)\sum_{n=0}^\infty \frac{(-1)^{n}2^{2n}}{7^{2n}}(x+2)^{2n}$

$(-2)\sum_{n=0}^\infty (-1)^{n}\biggl(\frac{2^2}{7^2}(x+2)^2\biggr)^n$

$(-2)\sum_{n=0}^\infty \biggl((-1)\frac{2^2}{7^2}(x+2)^2\biggr)^n$

According to the geometric series $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ so $z=-\frac{4}{49}(x+2)^2$

That leaves us with:

$$S(x)=\frac{-2}{1+\frac{4}{49}(x+2)^2} $$

Now my question is: How do I check this formula is correct?

答:
As @RobertTheTutor pointed in the answer above, a rigorous way to prove it will be to compute the derivatives of both functions and compare them (that is, by checking if both have the same Taylor polynomial). But if you are looking for a non-rigorous way of doing it, I would recommend you to try these options:

Check the series representation section in the WolframAlpha site.

Check by hand the first values of their derivatives at a certain $x\in(-3,-1)$.

Code a simple script where you calculate an approximated value of the original series $$\sum\limits_{n=0}^{100}\cdots$$ and compare it with the expected value of your answer.


问:
Null Hypothesis formulation from a "Fewer than" statement
The aim of this exercise is to translate statements into Null and Alternative Hypotheses.

The example statement is: "Fewer than 5% of adults ride the bus to work in Los Angeles."

My Answer was : 0: ≤0.05;: >0.05.

Given that 0 always has a symbol of =, ≤, or ≥.

Given that  always has a symbol of >, <, ≠.

However, the correct answer is : 0: =0.05;: <0.05

Why is the 0 statement 0: =0.05 and not : >0.05 ?

Why is the  statement : <0.05 and not : >0.05 ?

答:
"Fewer than 5% of adults ride the bus to work in Los Angeles."

The null hypothesis to be verified is the complement: $H_0\geq 5\%$

To set up a test, we expect that the data contraddict this Hypothesis, i.e. sample mean is $4\%$. The test is performed to tell us if actually $4<5$ or the difference has to be attributed to the phenomena's variability.

The system is the following

$$\begin{cases}
H_0: p\geq 5\%\\
H_1: p<5\%
\end{cases}$$

This is a good answer and correct too.

For some reasons that should be clear to you, the previous Hypothesis System is equivalent to the following one

$$\begin{cases}
H_0: p= 5\%\\
H_1: p<5\%
\end{cases}$$


问:
Geometric Mirror Problem with light beam
Two mirrors (represented by the blue lines) share an acute angle in the Point $A$. A beam of light $Y$ that is parallel to the horizontal mirror is reflected in the Point $C$. Then it is again reflected in $E$ and in $F$ and touches the horizontal mirror in a right angle in $G$.

We know that $AG$ has the length $5$, so  which of the options is the lenght of $m=GH$?

$4 \sqrt{2}$

$5\sqrt{2}$

$10$

My Progress:

We label the angles and because of the law of regular reflection and the parallelity of $Y$ to the horizontal mirror we get:

Now my thinking was, that $m$ could be any value that is bigger $FG$, because we can make $\alpha$ as big as we want, as long as it is acute. Where am I wrong and how do I find the right solution?

答:
If $\small \angle GAF = \alpha$, $\small \angle HCE = 90^0 - 2 \alpha \implies \angle GEF = \angle HEC = 2\alpha$

and $\small \angle GFE = 90^0 - 2 \alpha \implies \angle AFG = \angle CFE = 45^0 + \alpha$

So we have, $\small \alpha = \displaystyle \frac{\pi}{8}$

Now $\small GF = AG \tan \alpha, GE = GF \cot 2\alpha, FE = GF \csc 2\alpha$

As $\small \angle CEF = 90^0, CE = FE \tan (45^0+\alpha), EH = CE \cos 2\alpha$

Simplifying,  $\small GH = GE + EH = AG (1 + \tan \alpha) = 5 \sqrt2$


问:
Connection between a.s. equality of conditional expectations and a.s. equality of conditional probabilities
Let $P$ be some distribution and let $(X,Y,Z), (X',Y',Z')$ be two random draws from $P$. I would like to understand the connection between the following two conditions

(1) $\mathbb{E}[Y|X,Z] = \mathbb{E}[Y|Z]$ almost surely

(2) $\mathbb{P}_P(Y<Y'|X,Z,X',Z') = \mathbb{P}_P(Y<Y'|Z,Z')$ almost surely

Does one imply the other, or is there some case where one holds and the other doesn't? I know that both must hold if $Y \perp X|Z$, but I would like to be able to characterise the connection between the two. This is related to my unanswered question here

答:
Equality of expectations seems like a very weak constraint.  Certainly (1) does not imply (2), per this example:

$Z$ is independent of everything else, so I will drop it from consideration

First do this coin-flip: $P(X = 1) = P(X = 2) = 1/2$

Then $Y$ depends on $X$ this way:

If $X=1$ then $P(Y = 1) = 2/3, P(Y = -2) = 1/3, E[Y|X=1] = 0$.

If $X=2$ then $P(Y = -1) = 2/3, P(Y = 2) = 1/3, E[Y|X=2] = 0$.

Clearly $E[Y|X] = E[Y] = 0$ surely, so (1) is true. However, (2) is false:

$P(Y<Y' \mid X=1, X'=2) = P(Y=-2 \cup (Y = 1, Y' = 2) \mid X=1,X'=2) = 1/3 + 2/9 = 5/9$

$P(Y<Y' \mid X=1, X'=1) = P(Y=-2, Y'=1 \mid X=1, X'=1) = 2/9$

UPDATE: (2) also does not imply (1), per this other example:

Again, $Z$ is independent of everything else.

First draw $X \sim Uniform(0,1)$

Then $Y$ depends on $X$ this way: $P(Y = 2X) = P(Y = -X) = 1/2$, so $E[Y\mid X] = X/2$, and (1) is false.

However, (2) is true (almost surely) because

$$\forall x \neq x': P(Y < Y' \mid X=x, X'=x') = 1/2 = P(Y < Y')$$


问:
Structures that are reciprocally homomorphic images but are not isomorphic
Consider two algebras $\mathscr{A}=\langle A,O_i\rangle$ and $\mathscr{B}=\langle B,P_i\rangle$ with the same type. Suppose that there is a homomorphism from $A$ onto $B$, and a homomorphism from $B$ onto $A$. Clearly (edit: not really), $|A|=|B|$. It is easy to prove that if the algebras are finite, they must be isomorphic, but can we conclude the same thing for infinite algebras? I can find a counterexample for the same question about relational structures (sets equipped with relations) but my counterexample uses a one-many relation (the structures are $\big\langle\mathbb{N},\{\langle1,1\rangle,\langle1,2\rangle\}\big\rangle$ and $\big\langle\mathbb{N},\{\langle1,1\rangle\}\big\rangle$).

答:
Consider the abelian groups  $G = \mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies) and $H = \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}\oplus \dots$ (countably many copies).

There is a surjective homomorphism $G\to H$ which is the quotient map $\mathbb{Z}/4\mathbb{Z}\to \mathbb{Z}/2\mathbb{Z}$ on the first component and the identity on all other components: $(a_1,a_2,a_3,\dots)\mapsto (\overline{a}_1,a_2,a_3,\dots)$ where $\overline{a}_1$ is the residue of $a_1$ mod $2$.

There is also a surjective homomorphism $H\to G$ which projects out the first component:  $(a_1,a_2,a_3,\dots)\mapsto (a_2,a_3,a_4,\dots)$.

But these groups are not isomorphic. Note that the element $(1,0,0,\dots)$ in $H$ has order $2$ but is not divisible by $2$. On the other hand, every element of order $2$ in $G$ is divisible by $2$, since it has a $0$ or a $2$ in each component.


答: Let $G$ be an infinite group and let $H$ be a nontrivial proper subgroup. Let $A$ be the $G$-set that has countably infinitely many singleton orbits and countably infinitely many regular orbits (i.e., orbits isomorphic to ${}_GG$).  Now let $B$ be constructed exactly like $A$, but include one more orbit isomorphic to ${}_G(G/H)$. $A$ is not isomorphic to $B$, since $B$ contains a point whose stabilizer is $H$ and $A$ does not. But there is a homomorphism of $A$ onto $B$ that collapses one regular orbit to ${}_G(G/H)$, and there is a homomorphism of $B$ onto $A$ that collapses the orbit ${}_G(G/H)$ onto a singleton.

Title: Structures that are reciprocally homomorphic images but are not isomorphic

Body: Consider two algebras $\mathscr{A}=\langle A,O_i\rangle$ and $\mathscr{B}=\langle B,P_i\rangle$ with the same type. Suppose that there is a homomorphism from $A$ onto $B$, and a homomorphism from $B$ onto $A$. Clearly (edit: not really), $|A|=|B|$. It is easy to prove that if the algebras are finite, they must be isomorphic, but can we conclude the same thing for infinite algebras? I can find a counterexample for the same question about relational structures (sets equipped with relations) but my counterexample uses a one-many relation (the structures are $\big\langle\mathbb{N},\{\langle1,1\rangle,\langle1,2\rangle\}\big\rangle$ and $\big\langle\mathbb{N},\{\langle1,1\rangle\}\big\rangle$).


问:
How know which direction a particle is moving on a polar curve
I have being doing problems from the released AP BC Calcululs Free-Response questions, and I have come to realize that I don't have a very good idea of explain or a deep understanding of how to tell if a particle is moving to the orgion (pole?).

Example 2005 AP BC Free-Response Question #2.

The curve above is drawn in the $xy$-plane and is described by the equation in polar coodinates $r =\theta +\sin\left(2\theta\right)$ for $0\leq \theta \leq \pi$, where $r$ is measured in meters and $\theta$ is measured in radians. The derivative of $r$ with respect to $\theta$ is given by $\frac{\mathrm{d}r}{\mathrm{d}\theta} = 1 + 2\cos\left(2\theta\right)$.

(c) for $\frac{\pi}{3} < \theta < \frac{2\pi}{3}$, $\frac{\mathrm{d}r}{\mathrm{d}\theta}$ is negative. What does this fact say about $r$? What does this fact say about the curve?

I have learned from doing problems and reading the answers that if the derivative is negative its getting closer to the origin.  However, I have no clear understanding why?

I don't trust my justification:

Just as with the Cartesian coordinate system, when the 1st derivative is negative the particle is decreasing, moving downward, thus in the polar coordinate system that is moving to the origin.

答: There is a subtlety missed here. If the sign of the radius' rate of change is the same as the sign of the radius at a given point, the particle is moving away from the pole. If, however, the sign of the radius' rate of change and the sign of the radius are different, the particle is moving  towards the pole.</p>


答:
Applying exterior angle theorem gives $\delta=\alpha+\alpha=2\alpha$ and similarly proceeding will give $\beta=3\alpha$. $\alpha+\beta=90^{\circ}$ and thereafter $\alpha=\left(\frac{45}{2}\right)^{\circ}$.  $FG$ can be found from $\triangle AGF$.

$\triangle FGE$ and $\triangle CHE$ are both right isosceles triangles. Hence,  $AH=AG+GE+EH=AG+FG+GH$.

$\triangle AGF\sim \triangle AHC$ and thus $\frac{AG}{AH}=\frac{FG}{CH}$. Put the values of $AG$ and $FG$ alongside $m=CH$ and solve for it .

Title: Geometric Mirror Problem with light beam

Body:
Two mirrors (represented by the blue lines) share an acute angle in the Point $A$. A beam of light $Y$ that is parallel to the horizontal mirror is reflected in the Point $C$. Then it is again reflected in $E$ and in $F$ and touches the horizontal mirror in a right angle in $G$.

We know that $AG$ has the length $5$, so  which of the options is the lenght of $m=GH$?

$4 \sqrt{2}$

$5\sqrt{2}$

$10$

My Progress:

We label the angles and because of the law of regular reflection and the parallelity of $Y$ to the horizontal mirror we get:

Now my thinking was, that $m$ could be any value that is bigger $FG$, because we can make $\alpha$ as big as we want, as long as it is acute. Where am I wrong and how do I find the right solution?


问:
Derivative of exponential function $f(x)=e^{-1/x^2}$
I'm supposed to determine $f'(0)$ given $f(0)=0$

$$f(x)=e^{-1/x^2}.$$

My method was taking log and differentiating which on rearranging gives

$$f'(x)=\frac{2e^{-1/x^2}}{x^3}.$$  How should I proceed further?

答:
Hint:

I would simply use the definition of  a derivative:
$$f'(0)=\lim_{x\to 0}\frac{\mathrm e^{-\tfrac 1{x^2}}}{x},$$
and do some asymptotic analysis – setting $u=\frac1{x^2}$, it becomes
$$f'(0)=\lim_{u\to \infty} \pm\frac{\sqrt{u}}{\mathrm e^u},$$
which is a standard limit from high school.


问:
Intuition of law of iterated logarithm
Let $X_i$ be iid random variables with $EX_i = 0$ and $Var X_i=1$ and $S_n=X_1+\cdots+X_n$. Then the law of the iterated logarithm says almost everywhere we have

$$\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} = \sqrt{2}$$

On the other hand the central limit theorem says

$$\frac{S_n}{\sqrt{n}} \to N(0,1)$$

Can anyone explain why dividing by an extra $\sqrt{\log{\log{n}}}$ should go from giving $N(0,1)$ to something bounded by the constant $\sqrt{2}$?

To try to understand I considered the simple case when each $X_n$ is $N(0,1)$ so that $S_n/\sqrt{n}$ is also normally distributed as $N(0,1)$. Then $S_n/\sqrt{n\log{\log{n}}}$ is distributed as $N(0,1/\log{\log{n}})$. Then it would seem to me that to even have just $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ requires either

$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) < \infty$$

or if

$$\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right) = \infty$$

then to achieve $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}$ the sets {$ \omega : \frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}$} cannot for example cover the probability space over and over infinitely forever. I don't know the value of $\sum_{n=3}^\infty P\left(\frac{S_n}{\sqrt{n\log{\log{n}}}} > \sqrt{2}\right)$ but since it is the sum of the probability of the tail ends of a bunch of normal distributions you would expect there to be no closed form even for partial sums.

In the other direction for $\limsup_{n\to\infty}\frac{S_n}{\sqrt{n\log{\log{n}}}}$ to not have a value lower than $\sqrt{2}$ isn't it necessary that something like the following holds

$$\sum_{n=3}^\infty P\left(\sqrt{2}-\epsilon < \frac{S_n}{\sqrt{n\log{\log{n}}}} \le \sqrt{2}\right) = \infty$$

Can anyone explain why this number $\sqrt{2}$ should pop up?

答:
The main difference is that a convergence is almost sure and the other one is in distribution. To grasp how both can occur simultaneously, consider an independent sequence of random variables $(\xi_n)$ such that $P(\xi_n=\sqrt{2\log\log n})=1/n$ and $P(\xi_n=0)=1-1/n$.

Then $\xi_n\to0$ in distribution and, by Borel-Cantelli lemma,
$$
\limsup_{n\to\infty}\frac{\xi_n}{\sqrt{\log\log n}}=\sqrt2\quad \text{almost surely}.
$$

Title: Intuition of law of iterated logarithm


问:
Why do we have $E_0(\Bbb N^\Bbb N)\sim_c E_0$?
Let $E_0$ and $E_0(\Bbb N^\Bbb N)$ be the relations of eventual agreement on $2^\Bbb N$ and $\Bbb N^\Bbb N$, so $$xE_0y\iff\exists m\forall n\geq m x(n)=y(n)$$ and similarly for $E_0(\Bbb N^\Bbb N)$. Clearly $2^\Bbb N\hookrightarrow\Bbb N^\Bbb N$ is a continuous reduction of $E_0$ to $E_0(\Bbb N^\Bbb N)$. Why do we have a continuous reduction in the opposite direction as well?

答:
Here's one way to do it:

Let $(A_i)_{i\in\mathbb{N}}$ be a partition of $\mathbb{N}$ into infinitely many infinite pieces, and let $a_{i,j}$ be the $j$th smallest element of $A_i$. Given $f\in\mathbb{N}^\mathbb{N}$, we'll build a set $S_f\in 2^\mathbb{N}$ by coding $f(i)$ into $S_f\upharpoonright A_i$: $$S_f(\langle i,j\rangle)=\begin{cases}
1 & \mbox{ if $f(i)<j$,}\\
0 & \mbox{ otherwise.}\\
\end{cases}$$

Here $\langle\cdot,\cdot\rangle$ is your favorite pairing function. The infinite binary sequence $S_f$ is built by interweaving infinitely many infinite binary sequences $S(\langle i, -\rangle)$, each of which is eventually zero; this latter point guarantees $(S_f,S_g)\in E_0$ whenever $(f,g)\in E_0(\mathbb{N}^\mathbb{N})$, while the other direction follows since for all $m$ there is some $n$ such that for all $k>n$ we have $A_k\cap\{0,1,...,m\}=\emptyset$.


问:
Question on integration/differentiation of function series
Let be $S_n:[a,b]\to\mathbb{R}$ a differentiable function for all $n\in\mathbb{N}$ and $(S_n)_{n\in\mathbb{N}}$ a function series that converges for at least one point $c\in[a,b]$. Further, we assume that that the function series of $(S_n)_{n\in\mathbb{N}}$ derivatives, $(S'_n)_{n\in\mathbb{N}}$ converges uniformly.

Then we know that $(S_n)_{n\in\mathbb{N}}$ also converges uniformly, $\lim\limits_{n\to\infty}S'_n$ is differentiable and $\lim\limits_{n\to\infty}S'_n= (\lim\limits_{n\to\infty}S_n)'$.

(see https://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability)

Now let's assume we want to find a closed form of $\lim\limits_{n\to\infty}S_n$ but the partial sums $S_n$ are somewhat complicated and it is hard to figure out how $\lim\limits_{n\to\infty}S_n$ looks like. However, we know that $\lim\limits_{n\to\infty}S'_n$ attains a simple form which allows us to easily find the antiderivative of $\lim\limits_{n\to\infty}S'_n$.

My question is, though we are able to find the antiderivative of $\lim\limits_{n\to\infty}S'_n$ it's not unique, so how do I know which of those antiderivatives is my desired expression for $\lim\limits_{n\to\infty}S_n$? Or am I mistaken and this strategy is not applicable in general?

答:
An antiderivative of $\lim S_n'$ isn't unique but the limit of $(S_n)$ is because for some $c \in [a,b]$ the series of function converges at point $c$. Here, $\lim S_n$ is the antiderivative of $\lim S_n'$ taking the value $\lim S_n(c)$ at $c$ :
$$ (\lim S_n) (x) = \lim (S_n(c)) + \int_c^x (\lim S_n')(t)dt. $$
Also, denoting
$$ S_n = \sum_{k=0}^n f_k $$
this last formula becomes clearer :
$$ \sum_{n=0}^{+ \infty} f_n(x) = \sum_{n=0}^{+ \infty} f_n(c) + \int_c^x \sum_{n=0}^{+ \infty} f_n'(t)dt$$
where one can swap the series and the integral. This strategy is very useful when dealing with a series of the kind you described : the series of the derivatives is nice but the series itself isn't.


问:
John M. Lee's Introduction to smooth manifolds: push forward in local coordinates
On page 50 of John M Lee's Introduction to smooth manifolds about the local coordinates of the pushfoward $F_*$ it says

$$\left( \left. F_*\frac{\partial}{\partial x^i}\right|_{p} \right)f = \frac{\partial(f\circ F)}{\partial x^i}(p) = \frac{\partial f}{\partial y^j}(F(p))\frac{\partial F^j}{\partial x^i}(p) = \left( \left.\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial y^j}\right|_{F(p)}\right)f.$$

I don't fully understand the very last equation. Could someone elaborate how the last equation comes up?

Prior to posting this question i searched for an answer but couldn't find any, i hope this is not a duplicate.

Thanks for any help.

答:
Well, what you gotta understand is that $$\left( \left.\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial y^j}\right|_{F(p)}\right)$$ is a tangent vector at $T_{F(p)}N$, and as such it "eats" germs of functions at $F(p)$ and spits out a real number. Tangent vectors are linear operators, so if $v = v^{j} \left.\frac{\partial}{\partial y^{j}}\right|_{F(p)}$, then $v(f) = v^{j} \left.\frac{\partial f}{\partial y^{j}}\right|_{F(p)}$. In your case, the $v^j$ are given by $\frac{\partial F^j}{\partial x^i}(p)$, therefore:

$$\left( \left.\frac{\partial F^j}{\partial x^i}(p)\frac{\partial}{\partial y^j}\right|_{F(p)}\right)f = \frac{\partial f}{\partial y^j}(F(p))\frac{\partial F^j}{\partial x^i}(p)$$

Keep in mind we're using Einstein notation here.


问:
Baire space is zero dimensional
Here is the basis of Baire's topology with $s\in A^{<\mathbb{N}}$ and $\mathcal{B}=\{N_{s}:s\in A^{<\mathbb{N}}\}$  I was trying to show that elements of $\mathcal{B}$ is both open and closed. I assumed that complement of Ns is not open but I am confused about choosing initial segments etc. Can you please help me

答:
Hint. Let $s\in A^{<\mathbb{N}}$ be a finite sequence over $A$, of length $n$. Can you prove that
$$A^\mathbb{N} \setminus N_s=\bigcup_{}\{N_t \mid t\neq s, \text{ length}(t)=n\}?$$

Title: Baire space is zero dimensional


问:
How can it be true that $5\geq -2$?
I know that it is true that $5\geq  -2$, but I really do not understand why. Let me explain how I think.

$5\geq -2$ means that $5$ is greater than or equal to $-2$. I am perfectly fine with the fact that $5$ is greater than $-2$. I am not so fine with the fact that $5=-2$, as $5\neq -2$. I think that I have misunderstood something in the concept of inequalities.

With consideration to what I have written in the quote, why is it true that $5\geq -2$?

答: $\geq$ is greater than OR equal to.  If it were greater than AND equal to, you would be right in thinking there were a problem.

Title: How can it be true that $5\geq -2$?


问:
Infimum and supremum of the set $Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land 4^n n^m < m^m < 27^n n^m \right\}$
The problem is to find supremum and infimum of set $$Z = \left\{ \frac{m}{n}: m, n \in \mathbb{N} \space \land \space4^n n^m < m^m < 27^n n^m \right\}$$

I have an answer to this question, infimum is $2$ and supremum is $3$ but I don't see how to actually do the problem, so I would like to ask for some help with that. Thank you.

答:
Suppose $n,m \ne 0$, let's manipulate the second condition of the elements of that set:
$$\begin{align}
4^n n^m < m^m < 27^n n^m & \iff 4^n < \left(\frac{m}{n}\right)^m < 27^n \\[8pt]
& \iff 4^{\frac{n}{m}} < \frac{m}{n} < 27^{\frac{n}{m}}
\end{align}$$

Now, if you call for simplicity $a := \frac{m}{n}$ the last equation gives $$4^{\frac{1}{a}} < a < 27^{\frac{1}{a}}$$
that holds if and only if
$$2 < a < 3$$

With this conclusion we arrived to the following set:
$$Z := \left\{a \in \mathbb{Q} \ : \  2 < a < 3 \right\}$$


问:
prerequisites for understanding game theory
I am from programming background but with very limited knowledge of maths.

I am very much eager to learn and apply game theory to understand dynamics of International Politics and economics. But I am facing difficulties to understand the maths involved in it.

So,

What are prerequisites for understanding the math in the game theory?

What are the authentic sources to understand game theory mathematically and
and its application in the international politics and economics?

答:
Secondary-school math. And since game theory is essentially math, along the way you'll learn more math.

Ken Binmore's Playing for Real: A Text on Game Theory is suitable for undergraduates and doesn't get heavily mathematical at all.

What is game theory about?

How do I apply game theory?

Why is game theory right?

There's even a marker in the margin in places where you're allowed to skip some of the math and move on to the next bit.

It isn't easy to write in a light-hearted style when presenting mathematical material, but I did my best.

I think he did an excellent job.


问:
Integral of the natural logarithm of a function
How do I solve this?
$$
\int \ln(f(x)) dx = ?
$$
I know that
$$
\int \ln(x) dx = x\ln(x) - x
$$

In particular, I am interested in

$$
\int \ln\left(\frac{1}{a^2 x^2}\right) dx
$$

答:
Hint: [Converted from comment] Note that in your case, $$\ln \frac{1}{a^2x^2} = -2\ln a -2\ln x$$
assuming $a,x>0$.


问:
How do I use the chain rule here?
I have this problem, where the function is to derived with respect to $\theta$

$$f(\theta) = \frac{1}{2} (y-N(x, \theta))^2
$$

I know that I am supposed to use the chain rule here, but I am still a little confused. How can $h(x)' = g(x)*g(x)'$ be of any use here. Any tips are appreciated.

答:
The chain rule is used if there is a concatenation of functions that you want to find the derivative of. In this case, you have the function

$\theta \mapsto y-N(x,\theta)$

first, and that you concatenate with the function

$z \mapsto \frac{1}{2} z^2$.

Now if $f: X\rightarrow Y$ and $g: Y\rightarrow Z$ are functions, then the chain rule says

$(g\circ f)'(x) = (g'\circ f)(x) \cdot f'(x)$,

which in this case means

$f'(\theta) = (y-N(x,\theta)) \cdot \left(-\frac{\partial N(x,\theta)}{\partial \theta}\right)$.

Edit: OP said that $x$ and $y$ also should be considered as functions of $\theta$. In this case, we get

$f'(\theta) = (y(\theta) - N(x(\theta),\theta)) \cdot (y'(\theta) - N'(x(\theta),\theta)) = (y(\theta) - N(x(\theta),\theta)) \cdot (y'(\theta) - \left(\frac{\partial N(x(\theta),\theta)}{\partial x} \cdot \frac{dx}{d\theta} + \frac{\partial N(x(\theta),\theta)}{\partial \theta}\right)$.

Here I used the notation with the prime to denote total derivatives.

Title: How do I use the chain rule here?


问:
$f:\mathbb{R} \to \mathbb{R}$ a function with continious derivatives $f',f^{(2)},f^{(3)},f^{(4)}$.
$f:\mathbb{R} \to \mathbb{R}$ a function with continious derivatives $f',f^{(2)},f^{(3)},f^{(4)}$.

Take $a \in \mathbb{R}$ so that $f'(a)=f^{(2)}(a)=f^{(3)}(a)=0$.

Now I want to prove that $f$ has a local minimum in $a$ if $f^{(4)}(a)>0$.

答:
Use Taylor or mean-value theorem, e.g.,
$$
f(x) = f(a) + \frac1{4!} f^{(4)}(\xi)\cdot (x-a)^4
$$
where $\xi$ is between $x$ and $a$. By continuity, there is a neighborhood of $a$ such that $f^{(4)}(\xi) \ge \frac12f^{(4)}(a)>0$ for all $\xi$ in this neighorhood.
Now choose $x$ in that neighborhood to conclude $f(x)>f(a)$.


问:
Graph puzzles: Constructing graphs from tiles
There is a vast literature on the reconstruction conjecture which says that two graphs with the same deck $D$ are isomorphic. The deck is the multi-set of vertex-deleted subgraphs of a graph (which are called cards).

I'd like to take the other perspective and start with the multi-set $T$ of so-called tiles of a graph which are the vertex-induced subgraphs of the graph containing a distinguished vertex, all of its neighbours, and all the edges between them. As for $D$, $T$ contains as many elements as there are nodes in the graph: one tile for each.

Different from $D$, the tile set $T$ of a graph can be the tile set of many other non-isomorphic graphs. Needless to say that for a given multi-set $T$ of possible tiles there may be no graph at all which it is the tile set of.

In case there are many graphs with the same tile set $T$, these may share or may share not a lot of graph properties like being connected or having a small diameter. The only property theses graphs do share for sure is the degree distribution which is immediately given by the tile set.

Some tile sets may define a unique graph. I assume this is the case for this one, but I may be wrong:

Puzzle 1

Can you find a solution to this puzzle? Solving the puzzle means to put together tiles by identifying  nodes or edges while the red dots must not match black ones. Note, that the Euclidean lengths of edges in the final graph doesn't have to match the lengths of edges in the tiles when embedding them in some Euclidean space: you can stretch them however you want. (By the way, this is not necessary in these examples.)

Try out this one:

Puzzle 2

My questions are:

If $T$ is guaranteed to be the tile set of a graph which you don't know (like in the examples above): by which algorithm would you construct any graph which $T$ is the tile set of?
While this algorithm is probably some kind of a search algorithm, I added the [searching] tag.

How to prove that for any given set $T$ of tiles

there is no graph with tile set $T$

there are exactly $n$ graphs with tile set $T$ (for small $n$)

there are really many graphs with tile set $T$ (e.g. $\mathcal{O}(|T|^\alpha)$)

there is a graph with tile set $T$, but only with a small fraction $\epsilon$ of tiles left.

In case there are really many graphs with the same tile set $T$, these may be considered as a family of random graphs (= all graphs with tile set $T$). What may be said about this family of random graphs? Note, that in case you have an algorithm that generates a solution (question 1) you will have to randomize it to generate other random graphs with tile set $T$.

Since these questions seem so natural, I would imagine that they have been asked before. In this (probable) case, I would be happy to get a reference where I will find them answered. And a search term I can scholar-google for.

For those who like to puzzle and find out from which (non-random) graphs some given tile sets originated, here are two more puzzles. (Spoiler alert: Solutions are given in an answer below.)

Puzzle 3

Puzzle 4

答:
Solution to Puzzle 3: These are the vertex-induced subgraphs of a 5x5x5 cubic grid.

Solution to Puzzle 4: These are the vertex-induced subgraphs of a torified 8x8 square grid.


问:
Fréchet derivative of scalarfield operator $ f \mapsto \nabla f \cdot \sigma $?
Let $\sigma \in \mathfrak{X}(\Omega)$ be a fixed smooth vector field on an open domain $\Omega \subset \mathbb{R}^d$. Consider the operator $T\colon C^\infty(\Omega) \to C^\infty(\Omega)$ on the space of smooth scalar fields $f\colon \Omega \to \mathbb R$ given by $Tf := \nabla f \cdot \sigma$. Is there a simple expression for the Fréchet derivative of $T$? (or perhaps some other sensible interpretation of the derivative of $T$).

Since $T$ is linear the basepoint of the derivative should not matter. I also have a suspicion that it might be given by pointwise mutliplication with $-\operatorname{div} \sigma$, but I am not sure. I am only concerned with the cases $\Omega \in \{\mathbb R^d, (0,1)^d\}$ for $d=2,3$, and I also do not care about regularity conditions, but I do not think it will be relevant.

答:
As a remark note that $Tf = \langle \nabla f , \sigma\rangle = df(\sigma) = \sigma(f)$ by unpacking definitions.

Concerning the Frechet derivative of $T$, note that $T$ is a continuous linear map and as such equal to its own derivative (remember that the derivative is the linear part of the best affine approximation of a map, hence the derivative of any continuous linear map must be this linear map itself).


问:
Proving that every term of the sequence is an integer
Let $m,n$ be nonnegative integers.

The sequence $\{a_{m,n}\}$ satisfies the following three conditions.

For any $m$, $a_{m,0}=a_{m,1}=1$

For any $n$, $a_{0,n}=1$

For any $m\ge0, n\ge1$, $a_{m+1,n+1}a_{m,n-1}=a_{m+1,n-1}a_{m,n+1}+a_{m+1,n}a_{m,n}$

Prove that $a_{m,n}$ is an integer for any $m\ge0, n\ge0$.

Could you tell me how to prove it?

答:
I'm posting an answer just to inform that the question has received an answer by Timothy Chow on MO.

https://mathoverflow.net/questions/132191/proving-that-every-term-of-the-sequence-is-an-integer

He gives a link to the paper http://arxiv.org/abs/math/0104241 where it is shown that if one replaces the initial conditions $a_{m,0} = 1$, $a_{m,1} = 1$ and $a_{0,n} = 1$ by indeterminates, $a_{m,0} = x_{m,0}$, $a_{m,1} = x_{m,1}$ and $a_{0,n} = x_{0,n}$, then each $a_{m,n}$ is a Laurent polynomial in these indeterminates with coefficients in $\mathbb{Z}$. In particular if one then substitutes $1$ for each of these indeterminates, the result is an integer.


问:
Integral involving product of exponentials of powers
This integral comes from a toy model I was interested in, from statistical mechanics. Essentially, it involves solving an integral:

$$
\int_{-\infty}^\infty \varsigma^{-1} \exp(-\beta x)\exp\left(-\left|\frac{x}{\varsigma}\right|^m\right)\,dx
$$

Where $\beta>0$, $\varsigma >0$ and $m\geq 1$. Note that $m=1$ is a special case. For specific integers values of $m$, say $m=1, 2, ...$, I am able to find a solution. However, I'd like to understand general features of this system with $m$ and this would require a better solution than simply enumerating values of $m$. Is there a known general form for this integral ? I tried looking in Gradshteyn & Ryzhik and in the Erdelyi, but didn't find it.

答:
remark

Write
$$
F(n,\beta,\zeta) = \int_{-\infty}^\infty \zeta^{-1} \exp(-\beta x)\exp\left(-\left|\frac{x}{\zeta}\right|^m\right)\,dx
$$
I used zeta $\zeta$ instead of varsigma $\varsigma$.  Then
$$
F(m,\beta,\zeta) = \int_0^\infty \frac{e^{\beta x}+e^{-\beta x}}{\zeta}
\exp\left(-\left(\frac{x}{\zeta}\right)^m\right)
$$
change variables $y=x/\zeta$,
$$
F(m,\beta,\zeta) = \int_0^\infty (e^{\beta\zeta y} + e^{-\beta\zeta y}) e^{-y^m}\;dy = \beta\zeta F(m,\beta\zeta,1)
$$
So it suffices to do the case $\zeta=1$.

As noted it can be done in individual cases $m=1,2,3,\dots$.  It seems that $F(m,b,1)$ is a linear combination of terms like
$$
b^k\;{}_0F_{m-2}\left(;\cdots;\frac{b^m}{m^m}\right);
$$
where $\cdots$ represents $m-2$ rationals with denominator $m$.


问:
How would you go about learning the combination of coins the man has?
I am trying to identify the branch of math that would help to solve the following problem:

A man has picked $10$ coins out of a bag and has laid them in a row. You cannot see them for yourself, and the heads/tails ratio is unknown. Your goal is to identify the correct heads/tails sequence of the coins by repeatedly picking $10$ coins of your own from the bag, with the man telling you how many match the position of his each time. You cannot decide whether the coins are heads or tails (for the sake of the problem assume they are the same on both sides) and your sequence depends on the order you remove them from the bag. Note that the man only selects his coins once and his never change.

For example, if the man has $HTTHHHHTTT$ and you propose $TTTHTHTHHT$ he will let you know that $5$ are a match, however he will not tell you which ones. If on your second try you pull out $THHTTTHTHT$ he will inform you that $3$ match. Without trying every possible combination $(2^{10})$, how would you go about learning the combination of coins the man has?

Comparing the two sample results I see that they share $3$ in common, but I'm not sure if or how that is useful information since I don't know which ones actually match.

I suspect it might require statistics/combinatorics, and I was told it could be solved with "either high school or college math". This isn't a homework problem so it's possible I'm just being trolled by the person. Any information you can provide for approaching this type of problem is much appreciated. Thank you.

答:
The problem described in the Question, to determine exactly the opponent's fixed sequence of heads and tails, is not fully formulated because it needs some probability distribution for the sample sequences drawn by the player.

Given that probability distribution, one can ask the expected number of trials until the hidden sequence is determined.  For some (trivial) distributions it will be impossible to make that determination, e.g. if all the drawn sequences are tails only, then all that can be known is what the split of the ten coins is between heads and tails.

Of a more combinatorial nature is how to extract information from the samples that are drawn.  One way to think about it is in terms of the set of possibilities for the hidden sequence.  In this connection having zero matches (or ten matches) would be ideal, because the exact sequence is uniquely determined.

In general getting $k$ matches will imply that there are $\binom{10}{k}$ possible solutions.  Once the intersection of those subsets of possible solutions has a single common sequence, the problem is solved.

Let's consider how much information can be extracted from the example given in the Question.  The first sample:

```<code>TTTHTHTHHT
</code>```

gives five matches, and the second sample:

```<code>THHTTTHTHT
</code>```

has only three matches.

Therefore in looking at the positions where there was a change:

```<code>T***T***HT
</code>```

we know that two matches were lost (two that matched in the first sample but did not match in the second sample).  So at least two of the five matches in the first sample are in those six "*" (changed) positions.  Similarly at most three of the five matches in the first sample are in the four unchanged positions.

But more can be deduced from the net change of $-2$ matches from the first sample to the second sample.  If the first sample has $k$ matches in the six changed positions, then the second sample would have $6-k$ matches there (since all have changed, and there are only heads or tails).  So that would give us a net change of $-k + (6-k)$, and in order for this to be $-2$, we require $k=4$.

This reduces the number of possible solutions accordingly.  The first sample alone gives $\binom{10}{5} = 252$ solutions.  The second sample alone gives $\binom{10}{3} = 120$ solutions.  Combining them gives us just $\binom{6}{4} \cdot \binom{4}{1} = 60$ possible solutions (pick four of the six changed positions for matches in the first sample, and one of the four unchanged positions for the final fifth match in the first sample).

Body:
The problem described in the Question, to determine exactly the opponent's fixed sequence of heads and tails, is not fully formulated because it needs some probability distribution for the sample sequences drawn by the player.

Given that probability distribution, one can ask the expected number of trials until the hidden sequence is determined.  For some (trivial) distributions it will be impossible to make that determination, e.g. if all the drawn sequences are tails only, then all that can be known is what the split of the ten coins is between heads and tails.

Of a more combinatorial nature is how to extract information from the samples that are drawn.  One way to think about it is in terms of the set of possibilities for the hidden sequence.  In this connection having zero matches (or ten matches) would be ideal, because the exact sequence is uniquely determined.

In general getting $k$ matches will imply that there are $\binom{10}{k}$ possible solutions.  Once the intersection of those subsets of possible solutions has a single common sequence, the problem is solved.

Let's consider how much information can be extracted from the example given in the Question.  The first sample:

TTTHTHTHHT

gives five matches, and the second sample:

THHTTTHTHT

has only three matches.

Therefore in looking at the positions where there was a change:

T***T***HT

we know that two matches were lost (two that matched in the first sample but did not match in the second sample).  So at least two of the five matches in the first sample are in those six "*" (changed) positions.  Similarly at most three of the five matches in the first sample are in the four unchanged positions.

But more can be deduced from the net change of $-2$ matches from the first sample to the second sample.  If the first sample has $k$ matches in the six changed positions, then the second sample would have $6-k$ matches there (since all have changed, and there are only heads or tails).  So that would give us a net change of $-k + (6-k)$, and in order for this to be $-2$, we require $k=4$.

This reduces the number of possible solutions accordingly.  The first sample alone gives $\binom{10}{5} = 252$ solutions.  The second sample alone gives $\binom{10}{3} = 120$ solutions.  Combining them gives us just $\binom{6}{4} \cdot \binom{4}{1} = 60$ possible solutions (pick four of the six changed positions for matches in the first sample, and one of the four unchanged positions for the final fifth match in the first sample).


问:
Is $3 \ge 1$ or is it just $3 > 1$?
Well, probably this might seem a really simple question (and it might be so too!), but off late me and my friends have been debating quite hard over this question.

Is $3 \ge 1$ or is it just $3 > 1$?

Actually it all started here: if functions $f$ and $g$ satisfy $f>g$ then do they also satisfy $f \ge g$? All of us agreed. (are we wrong here itself?). After all, the $\ge$ symbol is basically 'greater than' OR 'equal too'. So if a function $f$ takes values greater than $g$, then definitely it satisfies the $\ge$ relation.

But then somehow the above mentioned question came up. My friends say that $3 \ge 1$ is absolutely wrong. Their reasoning: "under no circumstances can 3 be equal to 1", so $3 \ge 1$ is wrong.

My argument: $3 \ge 1$ is a mathematical statement which can either be true or false. It states that 3 is either greater than OR equal to 1. One of the conditions is satisfied and the two conditions are linked by an 'OR'. So $3\ge 1$ is true.

Who's right?

(please edit or add tags as i don't know what to tag the question as!)

答: You're correct. It's true that $3\geq 1$, however, if one wish to obtain the stronger statement, one can always specify that $3>1$. Both are correct, but the latter is more preferable.

Title: Is $3 \ge 1$ or is it just $3 > 1$?


问:
We have $n$ numbers less than a given number $k$ such that none of the numbers divides the product of other $n-1$ numbers.
How do you prove that $n$ can't be more than the number of primes below $k$?

My try

Suppose not. Then $n$ is more than the number of primes below $k$, so one of the given numbers has to be composite. But then how would I get a contradiction?

答:
This is actually quite simple. Suppose there exists some $m$ in the group such that it has prime decomposition $p_1^{m_1}p_2^{m_2}\ldots p_d^{m_d}$, and for each $p_i$, there exists another number in the group with prime factorization including $p_i$ to a power at least as high as $m_i$. Then, $m$ clearly divides the product of the other numbers. So this is a contradiction.

This shows that each number $m$ in the group is associated with a prime $p$ such that there exists some $k\in\mathbb N$ such that $p^k\mid m$, and for any $j\neq m$ in the group, $p^k\not\mid j$. Thus, we have an injection from the group of numbers into the prime numbers.


问:
Addition of 2 subspaces.
The questions goes like this:

Write down the subspace $U + V$ explicitly if

$U = {(t, 2t, 3t)}$ and $V = {(t, -2t, 0)}$ where $t$ is a member of the real number

My answer for the question is:

$U + V$ $=$ {$(x + y, 2x - 2y, 3x)$ | $x$, $y$ are member of the real number}

However, I am confused as other have mentioned that the answer should be instead:

$U + V\{(2t, 0, 3t)\mid t \text{ is a member of the real number}\}$

In my opinion, the I feel that the second answer is wrong as the two sub-spaces are not a subset of each other. Is my thinking wrong?

答:
The first answer is the correct one. The reason being that $t$ is a dummy variable used to represent both linear subspaces $U, V$, which are lines by the way.

When you make the sum, there is no link between those dummy variables that have to be separated. So using two different variables $x,y$ as in the first answer is the correct approach.

Title: Addition of 2 subspaces.


问:
Isomorphic graphs definition, understanding of the application of permutation matrix
I was thinking about the definition for two graphs being isomorphic:

Basically considering two graphs $\mathcal{G}, \, \mathcal{S}$ they are known to be isomorphic if they are essentially identical in terms of their structure. In other words, denoting $A_{\mathcal{G}}, \, A_{\mathcal{S}}$ we have that

$$A_{\mathcal{S}} = PA_{\mathcal{G}}P^T$$

for some permutation matrix $P$.

This is essentially clear but I don't understand well why we would express the above thing that way, what is the meaning of applying the operation $P (\cdot)P^T$. What is the sense of applying $P$ to $A_{\mathcal{G}}$ and then applying again $P^T$?

Thank you!

答: Note that $P$ represents a permutation $\pi\colon \{1,\ldots,n\}\to \{1,\ldots,n\} $ and also that $P^T=P^{-1}$. If $\pi$ is the permutation that corresponds to how the vertices of $\mathcal G$ are identified with those of $\mathcal S$, then $PA_{\mathcal G}P^{T}$ maps some base vector $e_{\pi(k)}$ to $e_k$, then to the sum of all neighbours $e_{j_1},\ldots,e_{j_d}$ of $e_k$, and finally to the sum of all neigbours $e_{\pi(j_1)},\ldots,e_{\pi(j_d)}$ of $e_{\pi(k)}$, as desired.


问:
Euler's number inherently in $\mathbf{x^a=a^x}$ and $\mathbf{x.a=a^x}$?
I was solving some equations graphically, when I came across an identity of Euler's number in two cases:

Case 1: $\mathbf{a.x = a^x}$

Ever wondered how for two numbers $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}}
 :\space a.x = a^x= y}$?

For example natural numbers: $\mathbf{2.2=2^2=4}$

What is the general formula for $\mathbf{x}$, given that $\mathbf{x.a=a^x}$?

It has multiple solutions. One is always $\mathbf{x=1}$ And interestingly, if
$\mathbf{x_1=x_2: \space a=e}$. In this case, actually: $\mathbf{x_1=x_2=1}$. So as $\mathbf{a\to e\implies x_1=x_2\to1}$.

$\mathbf{e}$ "came out of nowhere". I am sure the analytical solution would introduce something like Lambert W function which would explain the presence of $\mathbf{e}$.

Case 2: $\mathbf{x^a = a^x}$

Or second case when: $\mathbf{\left\{{ a,x,y} \right \}\in \mathbb{R_{> 0}}
 :\space x^a = a^x= y}$?

For example $\mathbf{2^2=2^2=4}$ or how $\mathbf{3^{2.478...}\approx2.478...^3\approx15.216}...$

What is the general formula for $\mathbf{x}$, given that $\mathbf{x^a=a^x}$?

It has multiple solutions. And interestingly, as $\mathbf{x\to e\implies x_1\to x_2}$.

=======================================

What are the two formulas for $x$ and how to obtain them?

Solution attempt

How to solve for $x$?

I tried this:

$\mathbf{log_ax^a = log_aa^x\\a.log_ax=x\\a.x^{-1}log_ax=1\\a.log_ax^{x^{-1}}=1}\\$

Lambert W function comes into mind. But how?

答:
It is the answer that provides the solution to your second equation:

$$ax=a^x$$

$$\frac{a^x}{x}=a$$

Let, $a^x=t$ and $x\ln a=\ln t$, then we get

$$\frac{t}{\frac {\ln t}{\ln a}}=a$$

$$\frac {t}{\ln t}=\frac {a}{\ln a}$$

$$\frac {\ln t}{t}=\frac {\ln a}{a}$$

Based on the first answer, we have

$$t=e^{-W\left(-\frac{\ln a}{a}\right)}$$

$$x=\frac {\ln t}{\ln a}=\frac {-W\left(-\frac{\ln a}{a}\right)}{\ln a}.$$


问:
Integral test for convergence
Show that
$$\sum_{n=3}^\infty \frac1{n \log n(\log \log n)^p}$$
converges if and only if $ p > 1$.

I know the hint for this: we have to use integral test so we know that the sum of $f(n)$ converges $\iff$  integral of $f(x)$ converges.

Anyone know how exactly to do this?

答:
Using Cauchy condensation test,

For $p=1$,

Here $$a_n=\frac{1}{n\log n \log (\log n)}$$

By Cauchy condensation test $\sum a_n $& $\sum 2^n a_{2^n}$ converges or diverges together,

\begin{align*}
\sum 2^n a_{2^n}&=\sum 2^n \frac{1}{2^n \log {2^n} \log (\log 2^n)}\\
                &=\sum \frac{1}{n\log 2 \log (n \log 2)}
\end{align*}

Since,

$\log 2 < 1$

$n \log 2 < n$

$\displaystyle \frac {1}{n \log 2} > \frac {1}{n}$

$\displaystyle \frac{1}{\underbrace{n \log 2}_{m} \log\underbrace{(n \log 2)}_{m}} > \frac{1}{n \log n}$

Check the behavior  $\displaystyle \sum \frac{1}{m \log m} $ it is divergent for p=1,

There is one example which you have to solve,

$$\sum_{n=2}^\infty \frac{1}{n(\log n)^p} \ \  \text{if} \ \  p>0 $$

For $P>1$ converges and $P\leq 1$ diverges.

$\displaystyle \therefore \sum 2^n a_{2^n}$ is divergent

$\displaystyle \therefore \sum_{n=3}^\infty \frac{1}{n \log n \log (\log n)} $ is divergent.

I hope you can conclude behavior for $P>1$.

Title: Integral test for convergence


问:
Number of types in simple theories

Let $T$ be the theory of random graphs. I want to count the number of 1-types in this theory, i.e $|S_1(T)|.$

My claim is this: $|S_1(T)|=1$, because every two points have the same types by extension axiom. (I cannot prove my claim precisely!)

We know the theory of random graph is simple. So there is a following natural question.

If $T$ is an arbitrary simple theory, what can we say about the number of $n$-types. In particular what can we say about the size of $S_1(T)$?

I have no idea about question 2.

答:
Looking at $S_1(T)$ usually tells you very little about the theory $T$. This is because $S_1(T)$ only captures the behavior of formulas with one variable and no parameters, so it says nothing about how elements of a model relate to other elements. It's much more interesting to look at types in multiple variables ($S_n(T)$ for all $n$) or types with parameters ($S_1(A)$ where $A\subseteq M\models T$).

Here are some properties that are captured by the spaces $S_n(T)$. Assume the language is countable.

The Ryll-Nardzewski theorem says that $T$ is $\aleph_0$-categorical if and only if $S_n(T)$ is finite for all $n$.

A theory is called small if $S_n(T)$ is countable for all $n$. A theory is small if and only if it has a countable saturated model, and every small theory also has a countable prime model.

Smallness and $\aleph_0$-categoricity are essentially orthogonal to Shelahian dividing lines like stability and simplicity, e.g. there are $\aleph_0$-categorical theories on both sides of all the major dividing lines. (The one exception to this orthogonality that I can think of is that $\omega$-stability implies smallness.)

Instead, the Shelahian dividing lines tend to have interactions with the spaces $S_1(A)$ of types with parameters (and often the local type spaces $S_\varphi(A)$ where we only consider $\varphi$-types). For example, the following are equivalent (again, assume a countable language): (1) $T$ is stable, (2) there is some cardinal $\kappa$ such that $|S_1(A)|\leq \kappa$ for all $A$ with $|A|\leq \kappa$, (3) for all formulas $\varphi(x;y)$, $|S_\varphi(A)|\leq |A|$.

This characterization of stability generalizes somewhat nicely to NIP theories: If $T$ is NIP, then for all formulas $\varphi(x;y)$, $|S_\varphi(A)|\leq \mathrm{ded}(|A|)$, where $\mathrm{ded}(\kappa)$ is the supremum of the cardinalities $|L|$ over all linear orders $L$ which contain a dense set of cardinality $\kappa$. Now it is consistent that $\mathrm{ded}(\kappa) = 2^\kappa$ for all $\kappa$, in which case this theorem tells us nothing. But it is also consistent that $\mathrm{ded}(\kappa) < 2^\kappa$ for some cardinals $\kappa$, and in such a model of set theory, the condition $|S_\varphi(A)|\leq \mathrm{ded}(|A|)$ characterizes NIP.

Simplicity also has a counting types characterization, but it's more complicated to state, and it involves counting  partial types (closed sets in the space $S_1(A)$).  Given cardinals $\kappa$ and $\lambda$, let $\mathrm{NT}(\kappa,\lambda)$ be the supremum of the cardinalities $|P|$ over all families $P$ of pairwise-contraditory partial types in one variable of cardinality $\leq\kappa$ over a set of parameters of cardinality $\leq \lambda$. Then $T$ is simple if and only if $\mathrm{NT}(\kappa,\lambda) \leq \lambda^{|T|}+2^\kappa$ for all cardinals $\lambda$ and $\kappa$.

See the paper The number of types in simple theories by Casanovas for much more on this theme.

Title: Number of types in simple theories


问:
Polar set of orthogonal matrices set is nuclear norm ball
Reltated problems:

Show that the dual norm of the spectral norm is the nuclear norm

Prove that the nuclear norm is convex

The set of orthogonal matrices is defined as:

$$\mathcal{O}(n) = \{X\in \mathbb{R}^{n\times n}:X^TX=I\}$$

The polar of $\mathcal{O}(n)$:

$$\mathcal{O}(n)^o = \{Y\in \mathbb{R}^{n\times n}:\langle Y,X\rangle\leq 1, \forall X\in \mathcal{O}(n)\}$$

i.e., the set of linear functionals that take value at most one on $\mathcal{O}(n)$.  The definition of polar cone is the general definition.

How to prove the polar of $\mathcal{O}(n)$ is the nuclear norm ball?

  i.e. $$\mathcal{O}(n)^o = \{Y\in \mathbb{R}^{n\times n}:\sum_{i=1}^n\sigma_i(Y)\leq 1\}$$

I try to  consider how to go to $\sum\sigma_i\leq1$ from $\langle Y,X\rangle=\text{tr}(YX)\leq 1$; however, I cannot find a way to break through.

By user1551's suggestion:

Let $YX=U\Sigma V^T$, where $Y\in \mathcal{O}(n)^o, X\in \mathcal{O}(n)$.

By Convex hull of orthogonal matrices, can I say:

$\|YX\|_2^2=\langle Y,X \rangle \leq 1$ so I get $\Sigma$'s diagonal elements $\in [0,1]^n$.   If this is true, how to say the SVD of $Y$, particularly $\Sigma(Y)$?

Note: $\|\cdot\|_2$ is the spectral norm (the largest singular value).

答:
I will call $\mathcal{A} = \{Y\in \mathbb{R}^{n\times n}:\langle Y,X\rangle\leq 1, \forall X\in \mathcal{O}(n)\}$ and $\mathcal{B} = \{Y\in \mathbb{R}^{n\times n}:\sum_{i=1}^n\sigma_i(Y)\leq 1\}$, denote the nuclear norm by $\|\cdot\|_1$ and the operator norm by $\|X\|_{\infty}$ (the indexing comes from viewing them as special cases of the Schatten norms).

Part 1: $\mathcal{B} \subseteq \mathcal{A}$

This direction follows from the Hölder's inequality. Let $Y \in \mathcal{B}$ and $X \in \mathcal{O}(n)$ then by Hölder's inequality we have
$$
\begin{aligned}
\langle Y, X \rangle &\leq \|Y\|_1 \|X\|_{\infty}\\
&= \|Y\|_1\\
&\leq 1.
\end{aligned}
$$
Thus $Y \in \mathcal{A}$ and $\mathcal{B} \subseteq \mathcal{A}$.

Part 2: $\mathcal{A} \subseteq \mathcal{B}$

This direction is a little more tricky. To solve it we need the following lemma which gives a variational characterization of the nuclear norm.

Lemma: Let $Y \in \mathbb{R}^{n\times n}$ then
$$
\|Y\|_1 = \sum_{i=1}^n\sigma_i(Y) = \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle.U
$$
Proof$\quad$ Consider the singular value decomposition $Y = UDV$ where $U,V \in \mathcal{O}(n)$ and $D$ is the diagonal matrix containing the singular values of $Y$. Then let $X = UV$ and we see
$$
\begin{aligned}
\langle Y, X\rangle &= \mathrm{Tr}[Y^T X] \\
&= \mathrm{Tr}[V^T D U^T UV] \\
&= \mathrm{Tr}[VV^T D U^T U] \\
&= \mathrm{Tr}[D] \\
&= \sum_{i=1}^n \sigma_i(Y).
\end{aligned}
$$
Thus $\sum_i \sigma_i(Y) \leq \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle$. The other inequality follows directly from the Hölder's inequality used in Part 1, i.e. $\langle Y, X\rangle \leq \|Y\|_1$ for any $X \in \mathcal{O}(n)$. $$\tag*{$\blacksquare$}$$

Now the result follows fairly quickly. Given $Y \in \mathcal{A}$ we know $\langle Y , X \rangle \leq 1$ for all $X \in \mathcal{O}(n)$. However, from the above lemma we also know that
$$
\begin{aligned}
\sum_{i=1}^n \sigma_i(Y) &= \max_{X \in \mathcal{O}(n)} \langle Y,X\rangle \\
&\leq 1
\end{aligned}
$$
Thus $Y \in \mathcal{B}$ and $\mathcal{A} \subseteq \mathcal{B}$ and we are done!


问:
Integral basis of $\mathbb{Q}(\sqrt{2+\sqrt{2}})$?
So far, I know the Galois group is $C_4$, the conjugates are $\sqrt{2+\sqrt{2}}$, $-\sqrt{2+\sqrt{2}}$, $\sqrt{2-\sqrt{2}}$, $-\sqrt{2-\sqrt{2}}$. At first I thought these would form the integral basis, but then I realised it's not that simple, and I'm not sure how to proceed.

答:
Let $a = \sqrt{2+\sqrt2}$, so $(a^2 - 2)^2 - 2 = 0$, i.e. $a^4 - 4a^2 + 2 = 0$. Also let $K = \Bbb Q(a)$.

This polynomial is Eisenstein at $2$, so the ring of integers of $K_2$ is $\Bbb Z_2[a]$, i.e. $\{1, a, a^2, a^3\}$ is a basis. I'm not sure if this helps.

The discriminant of $\{1,a,a^2,a^3\}$ is $N(f'(a))$ $=$ $N(4a^3-8a)$ $=$ $2^8 N(a) N(a^2 - 2) = 2^8 \times 2 \times N(\sqrt2)$ $=$ $2^8 \times 2 \times 2^2 = 2^{11}$. This means $[\mathcal O_K : \Bbb Z[a]]$ is a power of $2$, so for every $x \in \mathcal O_K$ there is $n$ such that $2^n x \in \Bbb Z[a]$.

However, we know that $\mathcal O_{K_2} = \Bbb Z_2[a]$, so if $x \in \mathcal O_K$, then $x \in \Bbb Z_2[a]$, so $2^n x \in 2^n \Bbb Z_2[a] \cap \Bbb Z[a] = 2^n \Bbb Z[a]$, so $x \in \Bbb Z[a]$. Therefore, $\mathcal O_K = \Bbb Z[a]$.

PS: The equality $2^n \Bbb Z_2[a] \cap \Bbb Z[a] = 2^n \Bbb Z[a]$ is because:
$$\begin{array}{cl}
& 2^n \Bbb Z_2[a] \cap \Bbb Z[a] \\
=& 2^n \Bbb Z[a]_{(a)} \cap \bigcap_{\mathfrak p \ne (a)} \Bbb Z[a]_\mathfrak p \\
=& 2^n \Bbb Z[a]_{(a)} \cap \bigcap_{\mathfrak p \ne (a)} 2^n \Bbb Z[a]_\mathfrak p \\
=& 2^n \bigcap_\mathfrak p \Bbb Z[a]_\mathfrak p \\
=& 2^n \Bbb Z[a]
\end{array}$$


问:
2 questions about convergence
I had a couple of questions about convergence and divergence:

Does $\sum_{n=1}^{\infty}ne^{-an^2}, a > 0$ converge? The main problem I have with this function is that it has an unknown variable a. Thankfully a>0, so I graphed it to see that the function always starts off increasing, then decreases. This seems to satisfy all the conditions for an integral test. Should I do that?

Does $\sum_{n=1}^{\infty}(-1)^n \frac{(2n)!}{(5n)^n}$ converge? This one has $(-1)^n$ in it, so it is a changing series. I remember there was a way to express a fraction of factorials and exponents in a different way, like expanding the factorials and simplifying it, but I don't know what it is.

答:
The first converges by the integral test:  $\int_1^\infty ne^{-an^2}=-1/(2a)e^{-a}\lt\infty $.

The second is alternating and decreasing, so converges by the Leibniz test.

Title: 2 questions about convergence


问:
Murphy Chapter 1 Exercise 5

I was able to do (a) and (b). Just stuck on (c). I noticed that this is quiet similar to Theorem 1.3.7. But I'm not sure here if $\sigma(a_1, \cdots, a_n)$ is compact here. Also, let $\hat{a}: \Omega(A) \rightarrow \sigma(a_1, \cdots, a_n)$, where $\tau \rightarrow (\tau(a_1), \cdots, \tau(a_n))$. Can we still say that $\ Ran (\hat{a}) = \sigma(a_1, \cdots, a_n)$?

Thank you for your help!

答:
The equality $\sigma(a_1,\ldots,a_n)=\operatorname{Ran}(\hat a)$ is the definition of $\sigma(a_1,\ldots,a_n)$.

As for compactness, $\sigma(a_1,\ldots,a_n)$  is the image of the compact set $\Omega(A)$. So all you need to show that the map $\tau\longmapsto (\tau(a_1),\ldots,\tau(a_n))$ is continuous.

Title: Murphy Chapter 1 Exercise 5


问:
Let $A,B$ be $C^*$-algebras. Does every homomorphism $A\to Q(B)$ determine an extension of $A$ by $B\otimes\mathcal K$?
$\mathcal K$ is the compact operator algebra on a $\omega$-dimensional Hilbert space.

$M$ denotes the multiplier algebra. $Q(B)=M(B\otimes\mathcal K)/B\otimes\mathcal K$.

An extension of $A$ by $B\otimes\mathcal K$ is a short exact sequence:

$0\to B\otimes\mathcal K\to E\to A\to 0$.

Since there is a canonical homomorphism $A=E/B\otimes \mathcal K\to M(B\otimes\mathcal K)/B\otimes\mathcal K=Q(B)$, every extension determines a homomorphism $A\to Q(B)$.

Does every homomorphism $A\to Q(B)$ determine a extension? Or at least, corresponds to an extension?

If $B\otimes \mathcal K$ is essential so that $\varphi:A\to Q(B)$ is injective, $E$ can be chosen to be $\pi^{-1}(\varphi(A))$ in $M(B\otimes\mathcal K)$, and homomorphism $E\to A$ is $\varphi^{-1}\circ \pi$.

Also, if $\varphi:A\to Q(B)$ is $0$, every element of $E$ acts on $B\otimes\mathcal K$ like an element in $B\otimes\mathcal K$. This shows $E=B\otimes\mathcal K\oplus A$, by mapping $x$ to $(\lim xe_\lambda,x+B\otimes\mathcal K)$, where $e_\lambda$ is the approximate unit of $B\otimes\mathcal K$.

What if $\varphi$ is not injective? How do I construct $E$?

答:
(I'm am going to assume $B$ is stable, solely to avoid writing $B\otimes\mathcal K$ so many times.  The results hold in the non-stable case, just by tensoring with $\mathcal K$ in the appropriate places.)

Each $*$-homomorphism $\varphi:A\to Q(B)$ determines (an equivalence class of) extensions of $A$ by $B$.  Indeed, let $\pi:M(B)\to Q(B)$ denote the quotient map, and define the $C^*$-algebra
$$E(\varphi)=\{(a,x)\in A\oplus M(B):\varphi(a)=\pi(x)\}.$$
and define a $*$-homomorphism $q:E(\varphi)\to A$ by  $q(a,x)=a$.  Then $q$ is surjective, and the kernel of $q$ is $0\oplus B\cong B$.  Thus we obtain a short exact sequence
$$0\to B\to E(\varphi)\overset{q}{\to} A\to 0.$$

Moreover, if $\eta:0\to B\to E\to A\to 0$ is an extension of $A$ by $B$ and $\varphi:A\to Q(B)$ is its Busby invariant, then the extension $E(\varphi)$ of $A$ by $B$ is isomorphic to $\eta$, in the sense that there is a $*$-isomorphism $E\to E(\varphi)$ making the following diagram commute:


答:
The condition can be rewritten as $a_i^2\not|\displaystyle\prod_{j=1}^n{a_j}$ where $a_i$'s are our $n$ chosen numbers.

Then for each $a_i$ there exists a prime number $p_i$ such that $$V_{p_i}(a_i^2)>V_{p_i}(\displaystyle\prod_{j=1}^n{a_j})$$ Otherwise $a_i^2|\displaystyle\prod_{j=1}^n{a_j}$

Name $p_i$ as the troubleshooting prime of $a_i$. Then if some prime number $p$ is the troubleshooting prime of both $a_s$ and $a_t$, then we'll have :
$$V_{p}(a_s^2)=2V_{p}(a_s)>V_{p}(\displaystyle\prod_{j=1}^n{a_j})=\displaystyle\sum_{j=1}^n{V_p(a_j)}$$
$$V_{p}(a_t^2)=2V_{p}(a_t)>V_{p}(\displaystyle\prod_{j=1}^n{a_j})=\displaystyle\sum_{j=1}^n{V_p(a_j)}$$
Now WLOG assume that $V_p(a_s)\geqslant V_p(a_t)$. Then :

$$2V_p(a_t)\leqslant V_p(a_s)+V_p(a_t)\leqslant\displaystyle\sum_{j=1}^n{V_p(a_j)}$$
And this contradicts the second result. So we conclude that each prime number can be the troubleshooting prime of at most one of the $a_i$ and therefore there can at most be $\pi(k)$ different numbers.

Title: We have $n$ numbers less than a given number $k$ such that none of the numbers divides the product of other $n-1$ numbers.

Body:
How do you prove that $n$ can't be more than the number of primes below $k$?

My try

Suppose not. Then $n$ is more than the number of primes below $k$, so one of the given numbers has to be composite. But then how would I get a contradiction?


问:
Modes of free vibration in a fixed-fixed beam
I am trying to find the mode shapes of vibration on a fixed-fixed beam.
this can be characterised as a PDE:
$$ EI \frac{\partial^4 v(x,t)}{\partial x^4} + \rho A \frac{\partial^2 v(x,t)}{\partial t^2} = 0 $$
with the boundary conditions
$$v(0,t) = 0  \hspace{2mm} \hspace{2mm} v_x(0,t) = 0 $$
$$v(l,t) = 0 \hspace{2mm}  \hspace{2mm} v_{x}(l,t) = 0 $$
We know that $E$, $I$, $\rho$ and $A$ are constant along the length

I have used the separation of variables method and then substituted in my boundary conditions and reached two different solutions:
$$\text{Re}\left[e^{ItB^2} (\cosh[Bx] - \cos[Bx]) + \frac{\sinh[Bl] + \sin[Bl]}{\cosh[Bl] -\cos[Bl]} (\sin[Bx] - \sinh[Bx])\right]$$
Plot of above showing fixed at l
$$\text{Re}\left[e^{ItB^2} \left((\sinh[Bx] - \sin[Bx]) - \frac{\sinh[Bl] - \sin[Bl]}{\cosh[Bl]
+ \cos[Bl]} (\cosh[Bx] - \cos[Bx])\right) \right]$$
Plot of above showing no slope at l

where $B$ is a constant that combines the natural frequency and the other constants

Each of these solutions satisfies a different three of the boundary conditions but I cant work out how to satisfy them all at the same time

答:
Too long for a late comment.

Starting from @DinosaurEgg's answer, if we look for the positive zero's of
$$f(x)=\cos (x) \cosh (x)+\sin (x) \sinh (x)-1$$
$$f'(x)=2 \cos (x) \sinh (x)$$ we have minima and maxima for $x=(2n+1)\frac \pi 2$.

Building the $[2,2]$ Padé approximant round these points, we just need to solve a quadratic equation in $\big[x-(2n+1)\frac \pi 2\big]$ to have a rather good estimate. For the first solutions
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{solution} \\
 0 & 2.19426 & 2.18354 \\
 1 & 5.52114 & 5.50353 \\
 2 & 8.65614 & 8.63913 \\
 3 & 11.7980 & 11.7810
\end{array}
\right)$$ and
$$x_{n+1} \sim x_n+\pi$$ One iteration of Newton method is more than sufficient.


问:
How to write subtraction of 2 matrix as multiplication
In a paper I saw the following equation.

If I express the equation in a simple way, It comes down to the following equation.
$(I-A)^{-1} - (I-B)^{-1} = (I-B)^{-1} [A-B] (I-A)^{-1} $

How can we prove these are equal to each other?

Thank you

Body:
Thanks for the hint @JMoravitz

With help of the hint ,the solution is

$$(I-B)^{-1} (I-B) (I-A)^{-1} - (I-B)^{-1} (I-A) (I-A)^{-1}$$
$$(I-B)^{-1} [(I-B)-(I-A)](I-A)^{-1}$$
$$(I-B)^{-1} [A-B] (I-A)^{-1}$$


问:
Show, that if the removal of a vertex $v$ reduces the depth of a minimal acyclic orientation, every longest path contains $v$
Let $G = (V, E)$ be a graph, that contains directed and undirected edges.

$G$ has the following properties:

let $(v_1, \dots, v_k)$ be a directed Path form $v_1$ to $v_k$,
then ther exists a directed edge $(v_1, v_k) \in E$

it follows, that the nodes of ever directed path form a clique

Only edges that are directed in $G$ have the transitive property

$G$ is a perfect graph

$G$ is acyclic

Now, let $H_G$ be a minimal acyclic orientation of $G$ so that.

the length of the longest path is minimal

every path of length $j$ cannot be shortened without creating a new path of at least length $j$

Shortening of a path: Let $p$ be a path in an acyclic Orientation $H_G$ of $G$.
If follows, that p is directed in $H_G$. If $p$ contains an edge that is undirected in $G$, we could flip the orientation of said edge in $H_G$, and cut/shorten the path at this edge. However, this does not always work, as it is possible to create a new and even longer path.

I want to prove the following statement:

If the longste path in $H_G$ contains at least $k$ vertices and
there exists a vertex $v \in E(G)$, so that the longest path in a minimal Orientation of the graph
$G' = G[V \setminus \{v\}]$ contains less than $k$ vertices, then every path of length $k$ in $H_G$ contains v.

答:
EDIT : The setup changed, this is no longer correct

Mixing directed and undirected edges is generally strange. And here, it turns out it doesn't really make sense.

First, your conditions imply that the graph is acyclic and transitive ($u \to v \wedge v \to w \Rightarrow u \to w$) so the undirected edges are all independent (no vertex is contained in two undirected edges).

Then, if two vertices are connected by an undirected edge ($u \leftrightarrow v$), then for every other vertex $w$, we have $u \to w \Leftrightarrow v \to w$ and $w \to u \Leftrightarrow w \to v$. The two vertices are indistinguishable, which means that if we swap $u$ and $v$, we get the same graph up to isomorphism.

Thus, all the orientations of the unoriented edges are isomorphic to each other. So, for every point of view, it doesn't matter which orientation you choose -- apart from one point of view, which is the labeling of the vertices that are part of an undirected edge.

For clarity, if $u \leftrightarrow v$ in $G$, I will write $s(u) = v$ and $s(v) = u$ : $s$ is the swapping function.

Now, the initial question becomes rather easy, because we don't have to worry about which orientation is minimal -- they're all the same, apart maybe from the position of the vertex $v$ that you're trying to remove.
But since this vertex $u$ is indistinguishable from its counterpart $s(u)$ in $G$, the only way to tell them apart is the orientation of the edge $u \leftrightarrow s(u)$. When you remove either $u$ or $s(u)$, this way to tell them apart disappear. So $H_G \setminus \{u\}$ is isomorphic to $H_G \setminus \{s(u)\}$, whatever orientation we choose for $H_G$. So all orientations are really the same for your problems.

Now, forget the ambiguity of the choice of orientation, and fix any orientation. You're asking to prove that, if removing a vertex forbids the existence of a path of length $k$, then this vertex is part of every path of length $k$. This seems rather straightforward, no ?


问:
$E[X^2] = \sum_{x=0}^{\infty}(2x+1)P[X>x]$
How to prove $E[X^2] = \Sigma_{x=0}^{\infty}(2x+1)P[X>x]$? It is easy to use generating function to prove: $E[X]=\sum_{x=0}^{\infty}P[X>x]$. Given that $X$ is discrete random variable and with countable elements.

答:
Also, $X$ need to be positive.

Hint

Briefly, \begin{align*}
\mathbb E[X^2]&=\int_\Omega X^2\,\mathrm d \mathbb P\\
&=\int_\Omega \int_0^X2x\,\mathrm d x\,\mathrm d \mathbb P\\&=\int_\Omega \int_0^\infty 2x\boldsymbol 1_{\{X>x\}}\,\mathrm d x\,\mathrm d \mathbb P\\
&\underset{\text{Fubini}}{=}\int_0^\infty 2x\mathbb P\{X>x\}\,\mathrm d x.
\end{align*}
Using this previous formula yields :
$$\mathbb E [X^2]=\sum_{k=0}^\infty \mathbb P\{X>k\}\int_k^{k+1}2x\,\mathrm d x=...$$


问:
Construct example of non 2-connected graph with a given property
I've been asked to give an example of a graph in which each vertex has at least degree 2017 and which is not 2-connected.

At first, I was thinking about a fully-connected graph on 2017 vertices which is 2016-connected. But this answer is wrong since any n-connected graph is also n-1, n-2, ..., 1-connected.

Could somebody give me a hint? Thanks.

答: Take two copies of $K_{2018}$ and join them by one edge.


问:
Why is the difference between two odd squares multiples of $3$ is divisible by $72$?
Why is the difference between two odd squares multiples of $3$ is divisible by $72$?

Here is my solution and I am not sure what should be next.

Note that any odd number is in the form of  $2m+1$

Here we take  $m$,  $n$  so that  $m>n$

$(2m+1)^2−(2n+1)^2$

$=4(m^2−n^2+m−n)$

$=4(m-n)(m+n+1)$

Since $m$ and $n$ are odd, the difference will be even.
Hence, $m-n$ is divisible by $2$, which makes the difference between two odd squares divisible by $8$.
Now, since the given difference between two odd squares must be multiples of $3$, $(m+n+1)$ must be divisible by $3$. This makes the given difference between two odd squares divisible by $24$.

From here, I am not sure how to show that it is divisible by $72$.
Please feel free to share your ideas on how to solve this problem.
Thank you in advance.

答:
Hint:

A square multiple of $3$ is necessarily a multiple of $9$ by the uniqueness of the decomposition into prime factors. Hence you only have to prove the difference of odd squares is divisible by $8$.

It's simpler to do it with congruences. Note that an odd integer is congruent to $\pm 1$ or $\pm 3\bmod 8$. What can you deduce for its square?


问:
Solve the inequality $a^3/(2b^3+ab^2)+b^3/(2c^3+bc^2)+c^3/(2d^3+cd^2)+d^3/(2a^3+da^2)\geq 4/3$ for $a,b,c>0$.

Solve the inequality
$\frac{a^3}{2b^3+ab^2}+\frac{b^3}{2c^3+bc^2}+\frac{c^3}{2d^3+cd^2}+\frac{d^3}{2a^3+da^2}\geq \frac{4}{3}$ for $a,b,c>0$.

I tried to use the inequality $a^3+2b^3\geq 3ab^2$, etc., adding $1$ to each fraction. I tried with Hölder too, but nothing works. Please help me.

答:
Study the function $f(x) = \frac{x^3}{x+2}  $ with  $x>0$. This function is convex because  $f''(x) =\frac{2x(x^2+6x+12)}{(x+2)^3}>0$. Then appling the Jensen's inequality:

$$\frac{1}{4}f\left(\frac{a}{b}\right)+\frac{1}{4}f\left(\frac{b}{c}\right)+\frac{1}{4}f\left(\frac{c}{d}\right)+\frac{1}{4}f\left(\frac{d}{a}\right) \ge f \left(\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+\frac{d}{a}\right)\right) \tag{1}$$

It's easy to notice that

$$\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+ \frac{d}{a}\right) \ge 1 \tag{2}$$

and as $f(x)$ is increasing  because $f'(x) = \frac{2x^2(x+3)}{(x+2)^2}>0$ then from $(2)$, we have

$$f \left(\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+\frac{d}{a}\right)\right) \ge f(1) =\frac{1}{3} \tag{3}$$

From $(1)$ and $(3)$ we conclude that

$$\frac{a^3}{2b^3+ab^2}+\frac{b^3}{2c^3+bc^2}+\frac{c^3}{2d^3+cd^2}+\frac{d^3}{2a^3+da^2}\geq \frac{4}{3}$$

The equality occurs when $a = b= c = d$.


答: I think you're asking why two quantities of the form $(3(2a+1))^2$ differ by a multiple of $72$. This is equivalent to$$8|(2a+1)^2-(2b+1)^2=4(a-b)(a+b+1).$$Compare the parities of $a-b,\,a+b+1$.

Body:
Why is the difference between two odd squares multiples of $3$ is divisible by $72$?

Here is my solution and I am not sure what should be next.

Note that any odd number is in the form of  $2m+1$

Here we take  $m$,  $n$  so that  $m>n$

$(2m+1)^2−(2n+1)^2$

$=4(m^2−n^2+m−n)$

$=4(m-n)(m+n+1)$

Since $m$ and $n$ are odd, the difference will be even.
Hence, $m-n$ is divisible by $2$, which makes the difference between two odd squares divisible by $8$.
Now, since the given difference between two odd squares must be multiples of $3$, $(m+n+1)$ must be divisible by $3$. This makes the given difference between two odd squares divisible by $24$.

From here, I am not sure how to show that it is divisible by $72$.
Please feel free to share your ideas on how to solve this problem.
Thank you in advance.


问:
Universal Property of the functor $q:\mathbf{TOP}\rightarrow \mathbf{HO(TOP)}$
Let $\mathbf{TOP}$ be the category of topological spaces and $\mathbf{HO(TOP)}$ be the category whose objects are topological spaces and morphisms are equivalence classes of continuous maps. We have a canonical functor $q: \mathbf{TOP}\rightarrow \mathbf{HO(TOP)}$ that sends an object to itself and a map to it's equivalence class.

Now I am trying to see that this map has the universal property that if $F:\mathbf{TOP}\rightarrow C$ is a functor that sends homotopy equivalences to isomorphisms, then it extends uniquely to a functor $F':\mathbf{HO(TOP)}\rightarrow C$ such that $F=F'\circ q$. Now to do this I need to show that if $f\cong f'$ then $F(f)=F(f')$.

Now I am bit lost on how to achieve this even for a simple case where I assume that $f\cong id$ then I would like to see tht $F(f)=id$. I know that since $f\cong id$ then $f$ is an homotopy equivalence and so $F(f)$ is an isomorphism  and after playing around with it and $F(f^k)$ I was not able to see why I would have that $F(f)=id$.

Now I am not sure if I am confusing something or just forgetting to use some propery but any hint or help is appreciated. Thanks in advance.

答: Hint : consider the projection map $X\times [0,1]\to X$


问:
Proof that an involutory matrix has eigenvalues 1,-1
I'm trying to prove that an involutory matrix (a matrix where $A=A^{-1}$) has only eigenvalues $\pm 1$.

I've been able to prove that $det(A) = \pm 1$, but that only shows that the product of the eigenvalues is equal to $\pm 1$, not the eigenvalues themselves.

Does anybody have an idea for how the proof might go?

Thanks.

答:
You can easily prove the following statement:

Let $f: V\to V$ be an endomorphism. If $\lambda$ is an eigenvalue of $f$, then $\lambda^k$ is an engeinvalue of $\underbrace {f\ \circ\ ...\ \circ f}_{k \text{ times}}$

In this case, let $A$ be a matrix of an endomorphism $f$ such that $f\circ f = I$. This means that $A$ is an involutory matrix (Because $AA=I$). So if $\lambda$ is an eingenvalue of $f$, then $\lambda ^2$ is an eigenvalue for $f \circ f = I$. The only eigenbalue of the identity funcion is $1$, so $\lambda^2 = 1$ meaning that $\lambda = \pm1$.


问:
Greedy Optimized Subset-Sum Problem
Given positive integers $a_1,...,a_n,b$, find $x_1,...,x_n \in \{0,1\}$ such that $a_1x_1 + ... + a_nx_n \lt b$ but is as large as possible.

How do I show that there is a greedy algorithm to this problem and how do I estimate its performance versus the optimized problem.

Any help would be appreciated, not really sure where to start with the problem.

Thanks.

答:
Each variable  $x_j$ is boolean in spirit: it indicates if you include $a_j$ in the sum or not. The greedy algorithm would do the following:

Order $a_j$ by size: $a_1\ge a_2\ge \dots$

Introduce $s=0$, current sum and $j=1$, current index

If $s+a_j<b$, include $a_j$: that is, $s=s+a_j$

Increment $j$

If $j<n$, then return to step 3, otherwise stop.

An example where the greedy algorithm performs poorly: $a_1=51$, $a_2=50$, $a_3=50$, $b=101$. The optimal solution is $a_2+a_3=100$, but the greedy algorithm chooses $a_1=51$.

You can try to prove that the greedy algorithm will always get at least the half of the maximal possible amount.

Title: Greedy Optimized Subset-Sum Problem


问:
Let P denote the product of first n prime numbers (with n > 2). Find all values of $n$ .

Let P denote the product of first n prime numbers (with $n > 2$). For what values of n we
have.

$1.$ $P - 1$ is a perfect square.

$2.$ $P + 1$ is a perfect square.

What I Tried: I have almost solved Part $2$, and I have some ideas of Part $1$.

For Part $2$, let $P + 1 = k^2$ .

$\rightarrow P = (k + 1)(k - 1) . $

So, after experimenting with some values of $n$, I observed that $P + 1$ is a perfect square when $n = 2,4$ .

When it is $3*5$ it works, and in the next $n$ it works as we get $3 * 5 * 7 * 11 = 33 * 35$ .

I was not able to get any other $n$, and if there is a proof, I need help to get it.

For Part $1$, I didn't know how to do it algebraically, so I experimented with some $n$. Obviously $P - 1$ won't be prime for $n = 2,4$. However, I noticed that Part $1$ does not work for any $n$, but I was not able to get a proof for it.

Can someone help me with these two parts? Thank You.

Edit:- I just now noticed that I thought myself that $n$ meant the starting prime value, which is wrong. It meant that $n$ is the total no. of starting primes, so my approach for Part $2$ was wrong.

Body:
Hints :

For $n>2$,

$(P-1)$ is $2$ modulo $3$.

$(P+1)$ is $3$ modulo $4$.