问:
Write all sub matrices of $\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\ \end{bmatrix}$

Write all submatrices of  $A=\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\
 \end{bmatrix}$

Here is my answre:

To solve this problem I wrote all $1\times1$ ,$1\times2$ , $1\times3$, $2\times1$ , $2\times2$ , $2\times3$ , $ 3\times1$ ,$3\times2$ , $3\times3$ matrices from $A$:

$1\times1$ matrices:

$$\begin{bmatrix}1
\end{bmatrix},\begin{bmatrix}2
\end{bmatrix},\begin{bmatrix}3
\end{bmatrix},\begin{bmatrix}4
\end{bmatrix},\begin{bmatrix}5
\end{bmatrix},\begin{bmatrix}6
\end{bmatrix},\begin{bmatrix}7
\end{bmatrix},\begin{bmatrix}8
\end{bmatrix},\begin{bmatrix}9
\end{bmatrix}$$

$1\times2$matrices:

$$\begin{bmatrix}1&2
\end{bmatrix},\begin{bmatrix}2&3
\end{bmatrix},\begin{bmatrix}4&5
\end{bmatrix},\begin{bmatrix}5&6
\end{bmatrix},\begin{bmatrix}7&8
\end{bmatrix},\begin{bmatrix}8&9
\end{bmatrix}$$

$1\times3$ matrices:

$$\begin{bmatrix}1&2&3
\end{bmatrix},\begin{bmatrix}4&5&6
\end{bmatrix},\begin{bmatrix}7&8&9
\end{bmatrix}$$

$2\times1$ matrices:

$$\begin{bmatrix}1\\4
\end{bmatrix},\begin{bmatrix}4\\7
\end{bmatrix},\begin{bmatrix}
2\\5\end{bmatrix},\begin{bmatrix}
5\\8\end{bmatrix},\begin{bmatrix}
3\\6\end{bmatrix},\begin{bmatrix}
6\\9\end{bmatrix}$$

$2\times2$ matrices:

$$\begin{bmatrix}1&2\\4&5
\end{bmatrix},\begin{bmatrix}2&3\\5&6
\end{bmatrix},\begin{bmatrix}4&5\\7&8
\end{bmatrix},\begin{bmatrix}5&6\\8&9
\end{bmatrix}$$

$2\times3$ matrices:

$$\begin{bmatrix}1&2&3\\4&5&6
\end{bmatrix},\begin{bmatrix}4&5&6\\7&8&9
\end{bmatrix}$$

$3\times1$ matrices:

$$\begin{bmatrix}1\\4\\7
\end{bmatrix},\begin{bmatrix}2\\5\\8
\end{bmatrix},\begin{bmatrix}3\\6\\9
\end{bmatrix}$$

$3\times2$ matrices:

$$\begin{bmatrix}1&2\\4&5\\7&8
\end{bmatrix},\begin{bmatrix}2&3\\5&6\\8&9
\end{bmatrix}$$

$3\times3$ matrices:

$$\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\
 \end{bmatrix}$$

Is my answer right? (did I write all sub matrices of $A$ correctly?)

答:
Nope. Submatrices don't have to be "connected", so you're missing, among others,
\begin{bmatrix}1&3\\7&9
\end{bmatrix}


问:
Grouping natural numbers into arithmetic progression
I need to find the number of ways of dividing the first 12 natural numbers into 3 equal groups (4 numbers each), so that the numbers in any particular group can be arranged in AP (Arithmetic progression).

This is what I did:

Consider the group which contains the number $1$

For this group the possibilities are:

$1- 2 -3- 4$ or,
$1- 3 -5- 7$ or,
$1- 4 -7- 10$

Clearly 12 is not included in any one of these groups, so $12$ must be part of another group.
For this group too we get three possibilities (counting backwards from $12$).
After that, I just checked every possible combination of the two groups to see if the third was in AP. (it was kind of easy, just 9 cases)

I got my final answer as 4. I'd like to know if there is more simple and elegant method to solving this, and if possible generalizing for the first n natural numbers.

ContentLicense: CC BY-SA 3.0

Body:
This is an interesting question. I don't have a real answer but I wrote a program to calculate values for different $n$ and $m$, counting the number of ways to split the first $n$ numbers into groups of arithmetic sequences of size $m$ . Here are some results with the pairs are written as $(n,m)$:

$(9,3)=5$,

$(12,3)=15$, $(12,4)=4$,

$(15,3)=55$, $(15,5)=4$,

$(16,4)=11$,

$(18,3)=232$, $(18,6)=4$, $(18,6)=4$,

$(20,4)=23$, $(20,5)=10$,

$(24,3)=6643$, $(24,4)=68$, $(24,6)=10$, $(24,8)=4$.

I skipped writing pairs that satisfy these equations:

If $m$ is not a divisor of $n$ then $(n,m)=0$. If $n$ is even then

$$(n,\frac{n}{2})=2,$$

$$(n,2)=\frac{n!}{2^{n/2}\frac{n}{2}!}.$$

EDIT

This wouldn't fit in a comment, but here are the ways of grouping (12,3):

$$
\begin{array}{ccc}
\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 9 & 11 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 6 & 8 \\
5 & 7 & 9 \\
10 & 11 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 7 & 10 \\
5 & 8 & 11 \\
6 & 9 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 2 & 3 \\
4 & 8 & 12 \\
5 & 6 & 7 \\
9 & 10 & 11
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
7 & 8 & 9 \\
10 & 11 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
7 & 9 & 11 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 6 & 10 \\
4 & 8 & 12 \\
7 & 9 & 11
\end{array}\right] & \left[\begin{array}{ccc}
1 & 3 & 5 \\
2 & 7 & 12 \\
4 & 6 & 8 \\
9 & 10 & 11
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 4 & 7 \\
2 & 5 & 8 \\
3 & 6 & 9 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 3 & 4 \\
6 & 7 & 8 \\
10 & 11 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 4 & 6 \\
3 & 7 & 11 \\
8 & 10 & 12
\end{array}\right]\\
\left[\begin{array}{ccc}
1 & 5 & 9 \\
2 & 6 & 10 \\
3 & 7 & 11 \\
4 & 8 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 6 & 11 \\
2 & 3 & 4 \\
5 & 7 & 9 \\
8 & 10 & 12
\end{array}\right] & \left[\begin{array}{ccc}
1 & 6 & 11 \\
2 & 7 & 12 \\
3 & 4 & 5 \\
8 & 9 & 10
\end{array}\right]
\end{array}
$$


答:
You found them all according to wolfram. However, when we speak about submatrices we have to state which kind of submatrix we are considering, for example, look at this definition. This last definition is the most frequently used. (The first kind is called block submatrix).

So, at the end of the day, you found all block submatrices (which I frankly think was what you were asked to do). Anyway, you should check the definition given in your material to be sure.

Here there is a similar question if you want to read about it.

ContentLicense: CC BY-SA 4.0

Title: Write all sub matrices of $\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\ \end{bmatrix}$

Body:
Write all submatrices of  $A=\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\
 \end{bmatrix}$

Here is my answre:

To solve this problem I wrote all $1\times1$ ,$1\times2$ , $1\times3$, $2\times1$ , $2\times2$ , $2\times3$ , $ 3\times1$ ,$3\times2$ , $3\times3$ matrices from $A$:

$1\times1$ matrices:

$$\begin{bmatrix}1
\end{bmatrix},\begin{bmatrix}2
\end{bmatrix},\begin{bmatrix}3
\end{bmatrix},\begin{bmatrix}4
\end{bmatrix},\begin{bmatrix}5
\end{bmatrix},\begin{bmatrix}6
\end{bmatrix},\begin{bmatrix}7
\end{bmatrix},\begin{bmatrix}8
\end{bmatrix},\begin{bmatrix}9
\end{bmatrix}$$

$1\times2$matrices:

$$\begin{bmatrix}1&2
\end{bmatrix},\begin{bmatrix}2&3
\end{bmatrix},\begin{bmatrix}4&5
\end{bmatrix},\begin{bmatrix}5&6
\end{bmatrix},\begin{bmatrix}7&8
\end{bmatrix},\begin{bmatrix}8&9
\end{bmatrix}$$

$1\times3$ matrices:

$$\begin{bmatrix}1&2&3
\end{bmatrix},\begin{bmatrix}4&5&6
\end{bmatrix},\begin{bmatrix}7&8&9
\end{bmatrix}$$

$2\times1$ matrices:

$$\begin{bmatrix}1\\4
\end{bmatrix},\begin{bmatrix}4\\7
\end{bmatrix},\begin{bmatrix}
2\\5\end{bmatrix},\begin{bmatrix}
5\\8\end{bmatrix},\begin{bmatrix}
3\\6\end{bmatrix},\begin{bmatrix}
6\\9\end{bmatrix}$$

$2\times2$ matrices:

$$\begin{bmatrix}1&2\\4&5
\end{bmatrix},\begin{bmatrix}2&3\\5&6
\end{bmatrix},\begin{bmatrix}4&5\\7&8
\end{bmatrix},\begin{bmatrix}5&6\\8&9
\end{bmatrix}$$

$2\times3$ matrices:

$$\begin{bmatrix}1&2&3\\4&5&6
\end{bmatrix},\begin{bmatrix}4&5&6\\7&8&9
\end{bmatrix}$$

$3\times1$ matrices:

$$\begin{bmatrix}1\\4\\7
\end{bmatrix},\begin{bmatrix}2\\5\\8
\end{bmatrix},\begin{bmatrix}3\\6\\9
\end{bmatrix}$$

$3\times2$ matrices:

$$\begin{bmatrix}1&2\\4&5\\7&8
\end{bmatrix},\begin{bmatrix}2&3\\5&6\\8&9
\end{bmatrix}$$

$3\times3$ matrices:

$$\begin{bmatrix} 1&2&3\\ 4&5&6\\ 7&8&9\\
 \end{bmatrix}$$

Is my answer right? (did I write all sub matrices of $A$ correctly?)


问:
Preimage under a continuous positive definite function
Consider a continuous positive definite function:
$$
f:D \rightarrow \mathbb{R}_+\\
f(0) = 0\\
\forall x \in D \setminus \{0\}: f(x) > 0
$$
where $0 \in D \subseteq \mathbb{R}^n$ and $D$ is unbounded.

I want to show that
$$
\exists \epsilon > 0: \exists r> 0: \forall x \in D:(f(x)=\epsilon \Rightarrow \Vert x \Vert \leq r)
$$
which says that there exists $\epsilon > 0$ such that the preimage $f^{-1}[\epsilon]$ is bounded. This is a seemingly true statement but I don't know how to show that. The continuity gives that $f^{-1}[\epsilon]$ is closed in $D$ for all $\epsilon > 0$ but can we say that it is also bounded under the above conditions?

答: You can do this in 1D. Start with $y=x^2$, then cut it off by multiplying by $e^{-x^2}$. Then add $x^2\sin^2 x$. Voila. The function $f(x)=x^2e^{-x^2}+x^2\sin^2(x)$  is clearly non-negative, and is only zero at $x=0$. But for any $\varepsilon>0$ there there will be $N$ large enough that for any $x>N$ we have $x^2e^{-x^2}<\varepsilon$, while $x^2>\varepsilon$, so that the function will cross below and above $\varepsilon$ with every oscillation (i.e., at $x= \pi k$ the value is $x^2e^{-x^2}< \varepsilon$, at $x=\pi/2+\pi k$ the value is at least $x^2>\varepsilon$).


问:
Find a polynomial $P$ of the smallest degree such that $\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$
Find a polynomial P of the smallest degree such that $$\int_{-1}^{1}\left|x^{2 / 3}-P(x)\right|^{2} d x<0.01$$

I am ususally solve this kind of "best approximation" problems by taking using $GS$ but I think this time it should be solved by Chebyshev polynomials since that was the topic of the lecture, I am not sure this is the idea here but I would like for some help here (Im not even sure if I can tag it under "chebyshev polynomials"

答:
It is standard $L^2$ on $[-1,1]$, and the orthogonal polynomials are the Legendre polynomials.  Or you can calculate the distance from your function to the space $\langle 1, x, \ldots, x^n\rangle$ using Gram determinants.

$\bf{Added:}$

Let us do some calculations.  The Legendre $P_n(x)$ polynomials $P_n(x)$ on $[-1,1]$ are an orthogonal system. $P_n(x)$ is of degree $n$, and orthogonal to $1$, $x$, $\ldots$, $x^{n-1}$. This leaves it uniquely determined up to a normalizing constant, which it is such that $P_x(1)=1$.  Here is a list of the first $6$ polynomials.

$$ 1, x, \frac{1}{2}(-1 + 3 x^2), \frac{1}{2}(-3 x + 5 x^3), \frac{1}{8}(3 - 30 x^2 + 35 x^4), \frac{1}{8}(15 x - 70 x^3 + 63 x^5)$$

The projection of a vector on a subspace spanned by several orthogonal vectors $w_k$ is
$$\pi(v) = \sum_k \frac{\langle v, v_k\rangle v_k}{\langle v_k, v_k\rangle}$$
and the distance squared from $v$ to this subspace is
$$d^2 =\|v\|^2 - \sum_k \frac{|\langle v, w_k\rangle|^2}{\|w_k\|^2}$$

Now, for the Legendre polynomials we have $\|P_n(x)\|^2 = \frac{2}{2n+1}$.

Another approach is with Gram matrices and determinants. The distance squared from $x^{\alpha}$  to a space of  functions $\langle  x^{\alpha_i} \rangle$ equals  $G(x^{\alpha}, x^{\alpha_i})/ G( x^{\alpha_i})$.

Notice that $x^{2/3}$ is even, so it will project into even functions. Recall that for $\beta$ even we have $\int_{-1}^1 x^{\beta} = 2/(\beta+1)$. Therefore, for $\alpha_i$ even the Gram determinant $G(x^{\alpha_i})$ is a Cauchy dererminant $$\det(\frac{2}{\alpha_i + \alpha_j+1}) = 2^{|I|} V(\alpha_i)^2 \cdot \prod_{i,j} \frac{1}{\alpha_i + \alpha_j + 1}$$
where $V$ is a Vandermonde determinant.

We conclude that the distance squared from even $x^{\alpha}$ to $\langle 1, x^2, \ldots, x^{2n}\rangle$  equals
$$d^2 =  \frac{2}{2\alpha+ 1} \prod_{k=0}^n \frac{(2k -\alpha)^2}{(2k +\alpha + 1)^2}$$


问:
About the induced vector measure of a Pettis integrable function(part 2)
Notations: In what follows, $X$ stands for a Hausdorff LCTVS and $X'$ its topological dual. Let $(T,\mathcal{M},\mu)$ be a finite measure space, i.e., $T$ is a nonempty set, $\mathcal{M}$ a $\sigma$-algebra of subsets of $T$ and $\mu$ is a nonnegative finite measure on $\mathcal{M}$.

Definition. A function $f:T\to X$ is said to be Pettis-integrable if

for each $x'\in X'$, the composition map
$$x'\circ f:T\to \mathbb{R}$$
is Lebesgue integrable and

for each $E\in \mathcal{M}$, there exists $x_E\in X$ such that
$$x'(x_E)=\int_E(x'\circ f)d\mu$$
for all $x'\in X'$. In this case, $x_E$ is called the Pettis integral of $f$ over $E$ and is denoted by
$$x_E=\int_E fd\mu.$$

Remark. Let $f:T\to X$ be Pettis-integrable. Define
$$m_f:\mathcal{M}\to X$$
by
$$m_f(E)=\int_E fd\mu$$
for any $E\in \mathcal{M}.$ Hahn-Banach Theorem ensures that $x_E$ in the above definition is necessarily unique and so $m_f$ is a well-defined mapping. Moreover, Orlicz-Pettis Theorem imply that the induced vector measure $m_f$ is countably additive, see for instance this.

Question. With the above discussions, how do we show that $m_f$ is $\mu$-continuous?

I would be thankful to someone who can help me...

答:
Let $\mu(E)=0$. Take arbitrary $x'\in X'$, then
$$
x'(m_f(E))=\int_E (x'\circ f)d\mu=0
$$
then. Since $x'\in X'$ is arbitrary by corollary of Hahn-Banach theorem $m_f(E)=0$. Thus, $m_f\ll\mu$</p>


问:
$\tan(z)$ with residue theorem
Calculate
$$\oint_{|z|=2}\tan(z)\,dz$$
because $\tan(z)=\dfrac{\sin(z)}{\cos(z)}$ the poles are when $\cos(z)=0$

at $z=\pm\pi/2\pm n\pi, \;n\in\mathbb{Z}$

Poles inside $|z|=2$ are $\pm\pi/2$ and are of first order. Can I know use the theorem for simple poles and just calculate the residuals
$$\operatorname{Res} \limits_{z=\pi/2} = \lim_{z\to \pi/2}(z-\pi/2)\frac{\sin(z)}{\cos(z)}$$
But I get stuck here. Hints?

OR
$$\cos (z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}\cdots $$
How do I know which coefficient to choose for the residual. Is it $1$ because the coefficient when $n=0$ is $1$ in the series? So the residuals for $\pm \pi/2$ are $1$.

So the integral is $2\pi i (1+1)=4\pi i$ ?

答:
You can split up the limit into:

$$2\pi i\lim_{z\rightarrow \frac{\pi}{2}} \frac{\sin z \cdot (z-\pi/2)}{\cos z} \: = 2\pi i(\lim_{z\rightarrow \frac{\pi}{2}} \sin z) (\lim_{z\rightarrow \frac{\pi}{2}} \frac{z - \frac{\pi}{2}}{\cos z}) =
2\pi i(\lim_{z\rightarrow \frac{\pi}{2}} \sin \frac{\pi}{2})(\lim_{z\rightarrow \frac{\pi}{2}} \frac{\frac{d}{dx}(z - \frac{\pi}{2})}{\frac{d}{dx} \cos z}) =
2\pi i (1)(\lim_{z\rightarrow \frac{\pi}{2}}(\frac{1}{\text{-} \sin z})) =
2\pi i (\frac{1}{\text{-}\sin\frac{\pi}{2}}) =
\text{-}2\pi i$$

The other residue is solved the same way and also ends up being -1 so it sums up to -4$\pi i$.

Title: $\tan(z)$ with residue theorem


问:
Hadamard theorem and length of geodesics
I read  sevral times  in articles that:

"By Hadamard theorem, there is no closed  geodesics of length $2\pi$ in a space of sectional curvature $<0$?

Can someone give me a reference of this theorem, because i don't find it !

Thank you in advance

答:
You can find a proof of the Cartan-Hadamard theorem for instance in do Carmo's book "Riemannian Geometry" or in Peterson's book with the same title. (Actually, most advanced RG textbooks will contain a proof.) Here is the full statement:

Theorem. Suppose that $M$ is a complete simply-connected Riemannian manifold of sectional curvature $\le 0$ everywhere. Then for each point $p\in M$ the exponential map $\exp_p: T_pM\to M$ is a diffeomorphism.

To conclude that such $M$ cannot contain a closed geodesic of positive length ($2\pi$ is irrelevant), note that each nonconstant geodesic in $M$ (regarded as a set!) has the form $\exp_p({\mathbb R} u)$ for some $p\in M$ and a unit tangent vector $u\in T_pM$. By the CH theorem,  $\exp_p({\mathbb R} u)$ is diffeomorphic to the real line, hence, cannot be a loop.


问:
Proving using mean value theorem
Let f be a function continuous on [0, 1] and twice differentiable on (0, 1).

a) Suppose that f(0) = f(1)  =0 and f(c) > 0 for some c ∈ (0,1).
Prove that there exists $x_0$ ∈ (0,1) such that f′′($x_0$) < 0.)

b) Suppose that $$\int_{0}^{1}f(x)\,\mathrm dx=f(0) = f(1) = 0.$$

Prove that there exists a number x$_0$ ∈ (0,1) such that f′′(x$_0$) = 0.

How do I solve the above two questions? I have tried using Mean Value Theorem but it gives me zero when I differentiate for the first time. Not sure how I can get the second derivative to be less than zero.

Any help is much appreciated! Thanks!

答:
We can choose either $[0,c]$ or $[c,1]$ to be the domain of $f(x)$ for our purposes. Let us write mean value theorems for each of the domains.

$$f:[0,c]\mapsto\Bbb R, \frac{f(c)-f(0)}{c-0}\space=\space\frac{f(c)}{c}\space=\space f'(a)$$

$$f:[c,1]\mapsto\Bbb R, \frac{f(1)-f(c)}{1-c}\space=\space f'(b)$$ Now we can switch to the original domain and write $$\frac{f'(b)-f'(a)}{b-a}\space=\space -\frac{f(c)(c+1)}{c(1-c)(b-a)}\space=\space f''(x_0)\space\lt\space 0$$ We are done with the first part, we can cover the second part. The integral vanishes, this means $f(x)$ should take on values below and above the $x$ axis. We can easily deduce that $f(x)$ has at least three roots. $$f(0)\space=\space f(1)\space=\space f(d)\space=\space 0, d\in(0,1)$$By Rolle's Theorem we can deduce that for some particular point $f''(x_1)=0$

Title: Proving using mean value theorem


答: No. You will get all submatrices by deleting some (0 to 3) rows and columns. Property of elements "to be in the same column with" will be preserved (the same for rows).</p>


问:
An equality involving a moving surface $\Sigma (t)$
Given a surface $\Sigma(t)$ that moves in space over time, a vector field $ \boldsymbol{B}=\boldsymbol{B}\left( x,y,z,t \right) $, and $\mathrm{div} \boldsymbol{B}=0$. $ \boldsymbol{V}$ is the velocity field on the surface.

Prove that
$$
\frac{\text{d}}{\text{d}t}\iint_{\Sigma \left( t \right)}{\boldsymbol{B}\cdot \text{d}\boldsymbol{S}}=\iint_{\Sigma \left( t \right)}{\left[ \frac{\partial \boldsymbol{B}}{\partial t}+\left( \nabla \cdot \boldsymbol{V} \right) \boldsymbol{B}+\left( \boldsymbol{V}\cdot \nabla \right) \boldsymbol{B}-\left( \boldsymbol{B}\cdot \nabla \right) \boldsymbol{V} \right] \cdot \text{d}\boldsymbol{S}}
$$

It seems difficult for me to deal with problems involving moving surfaces. Also the formula above is seemed to relate to Stokes' theorem. How to prove the formula? And can we generalize it to manifolds?

答:
THe formula is in accordance with the material derivative idea:
$$\frac{\partial}{\partial t} \int_{\Sigma(t)} \vec{B} d\vec{S} = \int_{\Sigma (t)} \left(\frac{\partial \vec{B}}{\partial t} + \nabla \vec{B} \cdot \vec{v} \right) + \int_{\partial \Sigma (t)} \left[\vec{B} \times \vec{v}\right]d\vec{s} $$
Now use the formula for transforming line integral into Surface one. How? We know that
$$\int_\gamma \vec{F} \cdot d\vec{s} = \int_{\Sigma} \left[\nabla \times \vec{F}\right] \cdot d\vec{\Sigma}$$
Then you transform the last integral and use that
$$\vec{\nabla} \times \left[\vec{B} \times \vec{v}\right] = \vec{B} \cdot \left(\nabla \cdot \vec{v}\right) - \vec{v} \cdot \left(\nabla \cdot \vec{B}\right)$$
Plugging this into the equality above yields desired formula.
NB! This is not material derivative, given by $\frac{D}{Dt} = \frac{\partial}{\partial t} + \vec{u}\nabla$ where $u$ is vector velocity of fluid. In your example, the velocity of the domain that changes over time.


问:
A question on countability of isolated points of a subset of R
The question is to prove that with respect to the euclidean metric on the Real numbers prove that if A is any subset of R, isoA is countable and hence deduce  that if A is uncountable the A' is uncountable.

I am still not that comfortable with proving things like countability so any answers or help is appreciated

thanks

答:
Let A $ \subset \mathbb{R} $ . Now
iso(A) $\subset A$. Let a $\in$ iso(A). We can choose a nbd N(a) of a such that N(a) $\cap$ A= { a } and N(a) $\cap$ N(b) = $\emptyset$ if b $\in$ iso(A) and b$\neq$a. Since $\mathbb{Q }$ is dense in R, N(a) contains a rational number, say q. $\forall a\in$ iso(A), choosing a rational number in such way we can construct a mapping f:iso(A)$\mapsto \mathbb{Q}$.
Since f is injective, cardinality of iso(A) is $\leq$ cardinality of $\mathbb{Q}$. As $\mathbb{Q}$ is countable, iso(A) is also countable.
Let A be uncountable.  Then A-iso(A) is also uncountable. Now A-iso(A) $\subseteq A' $. Therefore $A'$ is also uncountable. Hence every uncountable subset of $\mathbb{R}$ has uncountable number of limit points in $\mathbb{R}$.


问:
The confidence interval is $[2.663; 2.937]$?
The concentration of calcium in the blood for a given population follows a normal law of mean μ = 2.8 mmol / L and standard deviation σ = 0.7 mmol / L. If we take a sample of 100 people, what is the confidence interval of the calcium concentration at the 95% confidence level.

I got the confidence interval [2.663; 2.937] but I think this is wrong.

答:
We know the population standard deviation $\sigma = 0.7$ so we can find $z^*$ by:
$$
z^* =  \Phi^{-1}(1-\frac{\alpha}{2})
$$
As we want a 95% confidence level we have $\alpha = 0.05$, this means $z^* = 1.96$. We also know the population mean $\mu = 2.8$ so we can get the confidence interval by plugging in the values (including $n=100$):
$$
\left(\mu - z^* \frac{\sigma}{\sqrt{n}}, \mu + z^* \frac{\sigma}{\sqrt{n}} \right) = (2.6628, 2.9372)
$$

I hope it helps :)


问:
Derivation of refined Ratio test of convergence with Bernoulli inequallity
I am asked to derive the "General Ratio Test" (by using the Bernoulli inequality) with is stated as follows:

$(a_j)_{j\in\mathbb{N}}$ is a sequence in $\mathbb{R}_{>0}$. Show that the series $\sum_j{a_j}$ converges in the case there is a $\theta \in \mathbb{R}$ with $\theta > 1 $ and an $N\in\mathbb{N}$, so that:
\begin{equation}
\frac{a_{j+1}}{a_j}\leq1-\frac{\theta}{j+1},\text{ for all }j\geq N
\end{equation}

However, my problem is that I must use the Bernoulli Inequality, which is as follows:

a) For $q=n/m$, with $n,m \in \mathbb{N}_{\geq 1}$ with $n\leq m$
\begin{equation}
(1+x)^q\leq 1+qx
\end{equation}
b) For $q\in\mathbb{Q}_{\geq 1}$
\begin{equation}
(1+x)^q\geq 1+qx
\end{equation}

My approach was the following:
\begin{equation}
\frac{a_j}{a_{j-1}}\leq 1-\frac{\theta}{j} \\
a_j \leq a_{j-1}\left(1-\frac{\theta}{j}\right)\leq a_{j-2}\left(1-\frac{\theta}{j}\right)^2\leq ... \leq a_N\left(1-\frac{\theta}{j}\right)^{j-N}
\end{equation}

With a variable $j'=j-N$ change, I get:
\begin{equation}
\sum_{j=N}^n a_j \leq \sum_{j=N}^n a_N\left(1-\frac{\theta}{j}\right)^{j-N}=\sum_{j'=0}^{n-N} a_N \left(1-\frac{\theta}{j'+N}\right)^{j'}
\end{equation}

I thout I could use b) but apparently it wouldn't give me any insights into the convergence. I suppose my approach is wrong, or there is something I am missing.

I appreciate your answers.

答:
At this point your approach is incorrect since

$$\left(1-\frac{\theta}{j}\right)^{j-N} = \left(1-\frac{\theta}{j}\right)^{-N}\left(1-\frac{\theta}{j}\right)^{j} \underset{j \to \infty}\longrightarrow 1 \cdot e^{-\theta} \neq 0,$$

and the upper bound $\displaystyle\sum_{j=N}^n\left(1-\frac{\theta}{j}\right)^{j-N}$ diverges as $n \to \infty$.

Straightforward Proof

Suppose $a_j > 0$ for all $j$ and we have $\displaystyle\frac{a_{j+1}} {a_j} \leqslant 1 - \frac{\theta}{j+1}$ for all $j \geqslant N$, where $\theta> 1$.  It follows that

$$\frac{(j+1)a_{j+1}}{a_j} \leqslant j+1 - \theta = j - (\theta-1),$$

which implies that $(\theta-1) a_j \leqslant ja_j - (j+1)a_{j+1}.$

Summing over $j = N, \ldots, M$, we get

$$S_M= \sum_{j = N}^Ma_j = (\theta-1)^{-1}[Na_N - (M+1) a_{M+1}] < (\theta-1)^{-1}Na_N,$$

and the partial sums $S_M$ are increasing and bounded for all $M > N$ and, therefore, convergent as $M \to \infty$.

Therefore,

$$\sum_{j=1}^\infty a_j = \sum_{j=1}^{N-1} a_j + \sum_{j=N}^\infty a_j < +\infty$$


问:
Infinity as supremum for a sequence of numbers
Let $X_1, X_2, · · ·$ be sequence of independent nonnegative random variables having a common distribution function $F$ such that $P(X_1=0)<1, E[X_1]=a<\infty$ and $\mathrm{Var}[X_1]=\sigma^2<\infty$. Let's denote $$S_0 = 0, S_n = X_1 + \cdots + X_n$$ for $n = 1, 2, \ldots $ and $$N_t =\sup\left\{n:S_n ≤t\right\}$$ for any $t≥0$.

I want to show that $N_t\rightarrow\infty$ as $t\rightarrow\infty$.

By common sense, this is obvious but I do not know how to write it down in a mathematical way. Any help would be appreciated!

答:
We want to prove that $P(lim_{t\rightarrow\infty}N_t=\infty)=1$. $lim_{t\rightarrow\infty}N_t<\infty$ can happen only when $X_i=\infty$ for some $i$.
$$\{lim_{t\rightarrow\infty}N_t<\infty\}\subseteq\cup_{i=1}^\infty\{X_i=\infty\}$$
On the other hand we have $P(X_i=\infty)=0$, therefore:
$$P(lim_{t\rightarrow\infty}N_t<\infty)\leq P(\cup_{i=1}^\infty\{X_i=\infty\})\leq \Sigma_{i=1}^\infty P(X_i=\infty)=0$$


问:
What does this question mean by Diagonal Matrix

Consider the vector space $M_{2\times2}$. Its vectors are $2 \times 2$ matrices with real
entries. The addition operation is addition of matrices, and scalar multiplication is
scalar multiplication of matrices.

(a) What is the zero vector of this vector space?

(b) Is the set of all $2 \times 2$ diagonal matrices a subspace of $M_{2\times2}$?

(c) Is the set of all $2 \times 2$ invertible matrices a subspace of $M_{2\times2}$?

What does this question mean by diagonal matrices?

答: When you are dealing with square matrices of order n (i.e. with n rows and n columns), a matrix $A=(a_{ij})$ is said to be diagonal if the entry $a_{ij}$ is zero for all indices $i\neq j$. Thus only the diagonal entries (that's how we call the entries $a_{ij}$ with i=j) can be nonzero.


问:
Monte Hall Problem - Question on Decision Tree Construction (Conditional Probability)

(The Monte Hall Problem, also known as the 3 door problem):  On the game show Let's Make a Deal, there are 3 doors hiding 3 prizes.  Two of the prizes are goats, 1 prize is an expensive new automobile.  Naturally you would like the car, but only Monte knows which door it is behind.  You select one of the doors.  Rather than opening the door you chose, Monte opens one of the two other doors to reveal a goat.  You are now given the option to stick with your original selection or to switch to the other opened door.

(a)  What is the conditional probability that you win the car if you stick with the original choice of doors?

(b)  What is the (conditional) probability that you win the car if you change yoour mind and select the other opened door?

I read about the Monte Hall problem and understand the principle behind it:  The door you choose is random, but the door Monte chooses is NOT.  This is why switching doors gives you a higher probability.  What I really had a question on is the construction of the decision tree for part (b).

(a):  I understand this solution is 1/3.

(b)  This question is asking you $Pr(win_{auto}|\Delta doors)=\cfrac{Pr(win_{auto}) \cap Pr(\Delta doors)}{Pr(\Delta doors)}$.  I was just confused on the decision tree provided as a solution.  Originally, I thought a decision tree gives you ALL possible outcomes in the sample space.  But in this problem, a decision tree is drawn ONLY IF you change doors.  The solutions tree is below:

Above you can correctly calculate  the Probability by summing the branches:  (1/3)*1*1 *2 = 2/3.

Now, lets say I said draw the tree assuming that you always keep your initial door choice.  What is the probability that you win an auto given that you always stay with your initial door (I know its 1-2/3 just go with me for a second)?  I thought I would draw something like this:

Above you see the probability of winning is = 2*(1/3* 0) + 1/3 * 1 = 1/3 (which is the correct answer).

Is it correct to state that decision trees do not necessarily show the entire sample space of a problem?  Instead, sometimes they only show a specific scenario or subset of the entire sample space when calculating probabilities?

Thank you!

答:
You can draw a decision tree that gives you all possible outcomes.

There are only two moves in the game so the tree has depth $2$. Your first move is to guess where the prize is. The two alternatives are that you guessed correctly or you did not. Thus, one branch corresponds to a correct guess with a probability of $\frac{1}{3}$ and the other corresponds to an incorrect guess with a probability of $\frac{2}{3}$.

The second move is to decide whether to stay or switch. On the branch corresponding to an initially correct guess, the "stay" branch is labeled with a $1$ meaning if you were originally correct and you stay you win all the time. The "switch" branch is labeled with a $0$. The opposite values are given for "stay" and "switch" on the branch corresponding to an initially incorrect guess.


问:
I have been trying this inequality for weeks and I just can't seem to solve it
If $a,b,c>0$ prove that $\frac{{(1+a)}^{2\ }{(1+b)}^2}{1+c^2}\ +\ \frac{{(1+b)}^2{(1+c)}^2}{1+a^2}\ +\frac{{(1+c)}^2{(1+a)}^2}{1+b^2}\ \geq\ 8(a+b+c)$
I have tried factorising, using the mean inequality, Schur's inequality, Cauchy's and even tried a trigonometric approach but nothing seems to work.

Body:
Using the AM-GM inequality, we have
$$(1+a)^2(1+b)^2 = \left[(1+ab)+(a+b)\right]^2 \geqslant 4(a+b)(1+ab) = 4\left[b(1+a^2)+a(1+b^2)\right].$$
Therefore
$$\sum \frac{{(1+a)}^{2\ }{(1+b)}^2}{1+c^2} \geqslant 4 \sum \frac{b(1+a^2)+a(1+b^2)}{1+c^2}$$
$$ = 4 \sum a\left(\frac{1+b^2}{1+c^2}+\frac{1+c^2}{1+b^2}\right) \geqslant 8(a+b+c).$$


问:
Tilings of the plane and quotient spaces
A square lattice has two discrete translation symmetries, isormorphic to $\mathbb{Z}^2$. If we take the quotient of the plane by this group, $\mathbb{R}^2/\mathbb{Z}^2$, we get the square with opposite sides identified, which is the torus.

Can a similar argument be developed for a regular hexagonal lattice? I mean, the square has two independent translational symmetries; so does the hexagon; what is the difference, topologically? To put it differently, the hexagon with opposite sides identifies can be seen as the plane "quotiented" by some equivalence relation. But can it be seen as the quotient of two groups? Or at least two spaces?

Same as above but for equilateral triangle lattice.

Also, the symmetry group of the square lattice is bigger than just translations, it has rotations and reflections. Can we consider a different quotient in which in the numerator $\mathbb{R}^2$ is replaced by the whole 2D Euclidian group, and in the denominator $\mathbb{Z}^2$ is replaced by the whole symmmetry group of the square lattice? What do we get?

Where can I read an accessible introduction to these themes?

答:
There's a difference depending on what structure about the lattice you want to consider.
As groups, every lattice you mentioned is the same. They're all free Abelian groups generated by two elements, so they're isomorphic to $\mathbb Z^2$. Thus,  all the corresponding quotient spaces are the same $\mathbb R^2 /\mathbb Z^2$.

If you are interested in the geometrical structures of the lattices, then you should look at their symmetry groups. In this sense, there are five different kinds of lattices in $\mathbb R^2$:

Their symmetry groups are some of the Wallpaper Groups. These groups have the form $G = \mathbb Z^2 \times D$, where $D$ is a discrete group generated by rotations and reflections.

Then you consider the symmetry group $G$ acting on $\mathbb R^2$.
When $D$ is the trivial group, you'll get that $$\mathbb R^2 / G = \mathbb R^2/ \mathbb Z^2 \times D \cong \mathbb R^2/\mathbb Z^2.$$ So you are in the case you mentioned at the very beginning of your question. In general the quotient space $$\mathbb R^2 / G \cong \mathbb T^2/D$$  will be an Orbifold, $i.e.$ a geometric space which is a manifold except for a finite set of points (take a look at the section "orbifold notation" in the previous link). You can perform the intuitive identifications on the fundamental region of your pattern. Moreover, these orbifolds characterize the available symmetries for wallpapers.

All this is very nicely explained in the book "The symmetries of things" by J. Conway, C. Goodman-Strauss and H. Burgiel.


问:
Separating $\frac{1}{1-x^2}$ into multiple terms
I'm working through an example that contains the following steps:

$$\int\frac{1}{1-x^2}dx$$
$$=\frac{1}{2}\int\frac{1}{1+x} - \frac{1}{1-x}dx$$
$$\ldots$$
$$=\frac{1}{2}\ln{\frac{1+x}{1-x}}$$

I don't understand why the separation works. If I attempt to re-combine the terms, I get this:

$$\frac{1}{1+x} \frac{1}{1-x}$$
$$=\frac{1-x}{1-x}\frac{1}{1+x} - \frac{1+x}{1+x}\frac{1}{1-x}$$
$$=\frac{1-x - (1+x)}{1-x^2}$$
$$=\frac{-2x}{1-x^2} \ne \frac{2}{1-x^2}$$

Or just try an example, and plug in $x = 2$:
$$2\frac{1}{1-2^2} = \frac{-2}{3}$$
$$\frac{1}{1+2} -\frac{1}{1-2} = \frac{1}{3} + 1 = \frac{4}{3} \ne \frac{-2}{3}$$

Why can $\frac{1}{1-x^2}$ be split up in this integral, when the new terms do not equal the old term?

Body:
you wrote wrong fraction :
$$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1-x}-\dfrac{1}{1+x})$$

instead of :$$\dfrac{1}{1-x^2}=\dfrac{1}{2}(\dfrac{1}{1+x}-\dfrac{1}{1-x})$$
Now check it by plug in $x=2$

LHS=$\dfrac {1}{-3}\,$ ; RHS=$\dfrac{1}{2}(\dfrac{1}{3}-{1})\implies \dfrac {1}{-3}$


问:
p-value for test regarding sample belonging to gaussian process
Suppose we know that $X$ is a stationary mean zero gaussian process with known parameters. Suppose an experiment provides me with a collection of samples $(t_i, x_i)$ for $i = 1, 2, . . . , N$. How could I compute a p-value for the hypothesis that the $x_i$ are the values of $X$ at the times $t_i$?.

I was thinking of using the cholesky decomposition of the covariance matrix $\Sigma=LL^T$ to get iid random variables, i.e., $Y=L^TX \sim N(0,I)$ but not too confident on how to proceed. How can I compute the p-value for the specified test?

答: If knowing the parameters of $X$ means knowing the mean function $\mu(t)$ and the covariance kernel $k(t,t')$ then we can easily test for the values $(t_i,x_i)$. In fact, by consistency, we know that $X(t_i) = N(\mu(t_i), k(t_i,t_i))$ so we only have to test if $x_i$ comes form $N(\mu(t_i), k(t_i,t_i))$ for each $i = 1,...,N$.


问:
Proving if this function is bijective
Suppose we have a function: $f:A\rightarrow A$ such that $f(f(a)) = a$ for all $a$. Prove that this is bijective.

For me to prove this is one to one: $f(f(b))=f(f(a)) \rightarrow f^-1f(f(b))=f^-1f(f(a)) \rightarrow f(b) = f(a)$, thus one to one has been proven.

Confused how to do onto...

答: $f$ is onto because $a = f(f(a)) \in f(A)$

Title: Proving if this function is bijective


问:
What would we get if we differentiate $\exp(x)$ infinitely many times?
If we considered the exponential functions $\exp(x)$ which is differentiable infinitely many times, what can we say on the limit
$$ \lim_{n\to\infty}\frac{d^n}{dx^n} (\exp(x))\ ? $$

If we write $\exp(x)$ as the Taylor extension can we say that the limite is approaching $0$? Since:

答:
No, you can't say that. If $f$ is the exponential function and $x \in \mathbb R$, then $f^{(n)}(x) = e^x$ for every positive integer $n$. So, $\lim_{n \to \infty} f^{(n)}(x) = \lim_{n \to \infty} e^x = e^x$.

Comment: If
$$
a_n = a_{n,1} + a_{n,2} + a_{n,3} + \cdots
$$
for every positive integer $n$, then it might be tempting to think that
$$
\lim_{n \to \infty} a_n = \lim_{n\to\infty} a_{n,1} + \lim_{n\to\infty} a_{n,2} + \cdots.
$$
However, as this example shows, that is simply false.


答:
For your first question all the derivatives are the same thing $e^x$ so the limit would be that too.

For your second question all the series will be the same as $e^x$ since inserting any finite number of zeroes before an infinite sum has no effect on the sum value.

Title: What would we get if we differentiate $\exp(x)$ infinitely many times?

Body:
If we considered the exponential functions $\exp(x)$ which is differentiable infinitely many times, what can we say on the limit
$$ \lim_{n\to\infty}\frac{d^n}{dx^n} (\exp(x))\ ? $$

If we write $\exp(x)$ as the Taylor extension can we say that the limite is approaching $0$? Since:


问:
A question about vector representation in polar coordinates.
Suppose we want to write the vector:
$$\vec{A}=4\hat{x} +3\hat{y}$$
in polar coordinates $(\rho, \phi)$, by making use of the identities:
$$
\hat{x}=\cos\phi\,\hat{\rho} -\sin{\phi}\,\hat{\phi}, \quad \hat{y}=\sin\phi\,\hat{\rho} +\cos{\phi}\,\hat{\phi}
$$
which simply yield:
$$
\vec{A}= \left(4\cos{\phi}+3\sin{\phi}\right)\hat{\rho} + \left(-4\sin{\phi}+3\cos{\phi}\right)\hat{\phi}
$$
Now expressing $\vec{A}$ as above does not make much sense to me because $\vec{A}$'s orientation is fixed so there should appear no free parameter such as the angle $\phi$. If $\phi$ indeed represents the orientation of the given vector then:
$$
\cos\phi=\frac{A_x}{||A||}=\frac{4}{5}, \quad \sin\phi=\frac{A_y}{||A||}=\frac{3}{5}
$$
If I plug these in $\vec{A}$, I end up with:
$$A=5\hat{\rho}$$
By using the same reasoning above, a generic vector's, $\vec{A}=A_x \hat{x} + A_y\hat{y}$, representation in polar coordinates appears to be
$$\vec{A}=||A|| \hat{\rho}, \quad ||A||=\sqrt{A_x^2+A_y^2}$$

But on the other hand, a generic vector in polar coordinates is given in the form:
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
If my steps of converting $\vec{A}$ from cartesian to polar coordinates is correct, then why does the $\hat{\phi}$ component seems to disappear ?

EDIT

Thank you all for your answers. I need to clear up few more things. In one of the answers below it is stated the representation (although I have seen such notation on the web):
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
with constant coefficients is wrong. Is this also true for the representation of
say:
$$
\vec{A}=4\hat{x} +3\hat{y} + 2\hat{z}
$$
in cylindrical and spherical coordinates ? If so, then the proper conversion of $\vec{A}$ must read
$$
\vec{A}=5\hat{\rho} + 2\hat{z}, \quad \rho=\sqrt{x^2 + y^2}
$$
in cylindrical coordinates and
$$
\vec{A}=\sqrt{29}\hat{r}, \quad r=\sqrt{x^2+y^2+z^2}
$$
in spherical coordinates. Therefore the representation of a generic vector with arbitrary but constant coefficients:
$$
\vec{A}= A_{\rho} \hat{\rho} + A_{\phi}\hat{\phi} + A_{z}\hat{z}\\
\vec{A}= A_{r} \hat{r} + A_{\phi}\hat{\phi} + A_{\theta}\hat{\theta}
$$
is wrong. Is this conclusion correct ?

答:
Probably I'm not understanding your notation or your question properly, but if what you want is to express the point $(4,3)$ in the Cartesian plane in polar coordinates, I'll use the identities
$r=\sqrt{x^2+y^2}$ and $\tan(\theta)=\frac{y}{x}$
Substituting your values, the answer will be $\left(5, \arctan\left(\frac{3}{4}\right)\right)$


问:
Clarification on proof that for a linear map T, being continuous and being Lipschitz are equivalent statements
I know that in general for any function $f:V \longrightarrow W $ from a normed vector space $V$ to a second normed vector space $W$, being a Lipschitz function implies being continuous, is the second implication on my question the one I'm having troubles to understand.

答: Hint: Consider an open ball around $0$ within $f^{-1}(B_1(0))$ to obtain that $f$ is bounded and use linearity.


问:
Hatcher Algebraic Topology: I have all the prereqs, so why is this book unreadable for me?
Before, I scoured stack exchange for the prerequisites to read this book:

(See: Topology Prerequisites for Algebraic Topology, Module theory for chapters 1-3 of Hatcher Algebraic Topology, Learning Roadmap for Algebraic Topology, Algebra prerequisites for Hatcher's Algebraic Topology)

I got 100 on my point set topology course and am very comfortable with groups, rings, and modules so I thought I should be able to comfortably start learning out of Hatcher. I've heard great things about this book - how it's by far the most readable book in introductory algebraic topology, beautiful typesetting, builds up from basics, assumes little of the reader, etc.

But a couple pages in and I'm completely lost!

Hatcher keeps talking about orientable surfaces and genus - both concepts which are not once mentioned anywhere in Munkres, or indeed, in the majority of introductory topology courses. Neither of these concepts are defined in Hatcher either. So I thought - maybe I'll learn these concepts first and come back to the book! But no - every reference on genus I've found is another book on algebraic topology, one that needs prereqs beyond the ones demanded by Hatcher. All mentions of orientable surfaces lead me to differential geometry references - but I know 0 things about differential geometry.

The kicker is the prof I talked to who is running the course based on this book said I'll be fine in the course given my background knowledge. But from what I've skimmed from the book, I'm sure to fail the course at this rate.

I'll give a different example. The first time Hatcher defines the real projective space is here:

But is it just me or is this 'proof' extremely unrigorous and handwavy? I feel like half this proof relies on the reader's intuition about $\mathbb{R}^n$ for $n<=3$ and somehow this translates to general $\mathbb{R}^n$. The other half is massive leaps in logic that took me a long time just to think of a proof sketch. And this was just one example from the book, mind you.

Honestly, I'm thinking I've just hit a wall in my limits in pure math. What am I doing wrong? What makes this text readable to others but not me?

Or rather, my question should be - how do new people read this book in such a way that they don't get hung up in the way I am doing right now?

答: I highly recommend that you do not start with chapter 0, and if you really want to read Hatcher, just start with chapter 1. Chapter 0 is supposed to be extremely informal in spirit and can be skipped (he says this in the first para), and so it isn't meant to be scrutinized in that way. You are absolutely NOT hitting your limits in pure math; please don't be discouraged. I think a more gentle introduction to algebraic topology is Massey's "Algebraic Topology, an Introduction." It doesn't cover homology or cohomology, but it does the fundamental group very well. There are nice pictures in the book and it is a good continuation from point-set. Then you can pick up Hatcher at chapter 2 and start with homology.</p>


问:
Generalizing the Borsuk problem: How much can we shrink a planar set of diameter 1 by cutting it into $k$ pieces?
Borsuk's problem asks whether a bounded set in $\mathbb{R}^n$ can be split into $n+1$ sets of strictly smaller diameter. While true when $n=1,2,3$, it fails in dimension $64$ and higher; I believe all other $n$ are open as of this writing.

However, it turns out that at least in the $n=2$ case we can be more precise than "strictly smaller diameter"; if the original set has diameter 1, we can ensure that each piece has diameter at most $\frac{\sqrt{3}}{2}\approx 0.866$, a bound attained by the circle of diameter $1$. To see that this holds, we note that the regular hexagon of width $1$ is a solution to Lebesgue's universal covering problem, and can be split into three sets of diameter $\frac{\sqrt{3}}2$ as well:

I am interested in putting bounds on such dissections with more than $3$ pieces: what is the minimum diameter one can ensure when cutting a planar set of unit diameter into $k$ pieces?

Using the same approach as above (finding specific sets with a lower bound, and dissecting a universal cover for sets of diameter 1), I have some bounds for higher $k$ as well, though only for $k=3,4,7$ are they exact:

(Extending this table beyond $k=7$ would be difficult, as working out optimal dissections for the circle would become much more complicated.)

Edit: By taking spokes at $72^\circ$ angles on a regular hexagon (with one spoke meeting the hexagon at the midpoint of a side), I think I can get a slightly better upper bound of around $0.6434$ for the case $k=5$. Optimizing spoke placement further (so that the distances between spoke endpoints are equal) gets me around $0.6223$.

In the limit, I think the diameter of each piece is asymptotic to $\sqrt{\frac{2\pi}{3\sqrt{3}k}}\approx \frac{1.1}{\sqrt{k}}$ by tiling with regular hexagons. Certainly one can do no better than $1/\sqrt{k}$ when dividing the circle, using the isodiametric inequality (if the pieces were any smaller, they would have too little area). Using a trivial dissection of the square, one also has an upper bound of $\frac{\sqrt{2}}{\lceil\sqrt{k}\rceil}$.

Some questions I have about this problem:

Has this question been investigated before in the literature? If so, what is known?

Are there any $k$ for which the circle does not present the worst-case scenario for dissection?

Can the $k=5,6$ upper bounds be substantially improved? I think using Pal's slightly smaller solution to the universal covering problem would allow for a few adjustments when $k=6$, but haven't worked out the details.

答:
what is the minimum diameter one can ensure when cutting a planar set of unit diameter into $k$ pieces?

This problem is considered in 1974 in Problem 102 from [SCY], where the minimum diameter is denoted $\delta_2(k)$. Unfortunately, there are given not much more bounds than in your question. A main tool for the evaluation of $\delta_2(k)$ there is $\delta(k, A)$, the minimum diameter one can ensure when cutting a planar set $A$ of unit diameter into $k$ pieces. Special for $S$ are cases are a disk $D$, a square $S$, and an equilateral triangle $T$. In Problems 103 and table at p. 97 (referenced to paper [Gra] from 1967) bounds $\delta(k, A)$ are shown for $D$ for $k\le 5$, for $T$ and $k\le 10$, and for $S$ and $k\le 4$. Also in [Gra] are evaluated $\delta(k, T)$ for $k\le 15$. When I was a schoolboy, in 1991 I read the article [KK] where were calculated $\delta(2,S)=\tfrac {\sqrt{10}}4$, $\delta(3,S)=\tfrac {\sqrt{130}}{16}=0.712\dots$, and $\delta(5,S)=\tfrac {5\sqrt{34}}{64}=0.455\dots$, found an upper bound $0.4200\dots$ on $\delta(6, S)$, and noted that $\delta(k, D)$ for $k\ge 8$ and $\delta(k,T)$ for $k\ge 16$ are unknown. At pages 96 and 98 are written rather pessimistic thoughts about this approach and in Problem 104 are shown values $\delta_2(2)$, $\delta_2(3)$, $\delta_2(4)$, and $\delta_2(7)$, which you already know. There is noted that no other exact values for  $\delta_2(k)$ when $k\ge 2$ are known. Value of $\delta_2(3)$, was, in fact, found by Borsuk [Bor1, Bor2] in 1932–1933 (see also [Gal]). In 1956 a German geometer Lenz [Len1, Len2] thoroughly studied values of $\delta_2(k)$ for small $k$ and calculated $\delta_2(4)$, $\delta_2(5)$ and $\delta_2(7)$. Value of $\delta_2(4)$ was found also by Selfridge [Sel]. In [Gru] is observed that if $G_{11}$ is a regular $11$-gon of diameter $1$ then $\delta_2(6)\ge \delta(6, G_{11})=\frac 1{2\cos (\pi/22)}=0.505141\dots$.

Unfortunately, I don’t speak German, but I guess that in [Len1] at p. 34 are provided bounds $\delta_2(k)\le\tfrac {\sqrt{2}}{\lfloor \sqrt{k}\rfloor}$ for $k\ge 2$ and $\delta_2(k)<\tfrac 1{k-8\pi/\sqrt{27}}\left\lfloor\tfrac {4\pi}{\sqrt{27}}+\sqrt{\tfrac{2\pi k}{\sqrt{27}} }\right\rfloor$ for $k\ge 5$, and at p. 36 a bound $\delta_2(k)\le\tfrac 1{k-1}\left(\tfrac {2}{\sqrt{3}}+\sqrt{\tfrac 43+ \frac{2\pi}{\sqrt{27}}(k-1) }\right)$. Both latter bounds are about $\sqrt{\frac{2\pi}{\sqrt{27}}}k^{-1/2}\approx 1.1 k^{-1/2}$.

But these references are old and some progress could be made from that time.

We should have $\delta_2(k)\approx \sqrt{\frac{2\pi}{\sqrt{27}}}k^{-1/2}$ asymptotically, see below.

A lower bound. Given $k$, Pigeonhole principle implies $\delta_2(k)\ge d(k+1)/2$, where $d(k+1)$ be a maximal possible minimum distance between $k+1$ points of the unit disk, see this thread. This approach should provide an asymptotic bound $\delta_2(k)\ge\approx \sqrt{\tfrac {2\pi}{3\sqrt{3}k}}\approx 1.1 k^{-1/2}$.

An upper bound. Let $C$ a be a (not necessarily convex) subset of the plane that contains a congruent copy of every planar set of unit diameter and $a$ be an area of $S$. The best known bounds for $a$ are about $0.8441$, see a thread about a hard and ungrateful quest for them. If $C$ can be covered by $k$ cells of a hexagonal grid with side $d$ then $\delta_2(k)\le 2d$. This approach should provide an asymptotic bound $\delta_2(k)\le\approx 2\sqrt{\tfrac {2a}{3\sqrt{3}k}}\approx 1.14 k^{-1/2}$.

But Lenz’ bound suggest that we don’t need to use an universal covering set, because at p.11 of [Lit] it is shown that “an area of (greatest) diameter not greater than $1$ is at most $\tfrac{\pi}4$”.

This observation should point to an asymptotically tight upper bound.

References

[Bor1] K. Borsuk, Über die Zerlegung einer euklidischen $n$-dimensionalen Vollkugel in $n$ Mengen, Verhandlungen Intern. Math. Kongr., Zürich 2 (1932) 192.

[Bor2] K. Borsuk, Drei Sätze über die $n$-dimensional  Späre, Fundamenta Math. 20 (1933), 177–190.

[Gal] D. Gale, On inscribing $n$-dimensional sets is a regular $n$-simplex, Proc. Amer. Math. Soc. 4  (1953) 222–225.

[Gra] R.L. Graham, On partitions of a equilateral triangle, Canadian Journ. Math. 19 (1967)  394–409.

[Gru] B. Grünbaum, Etudes in combinatorial geometry and the theory of convex bodies, Moskow, Nauka, 1971, in Russian.

[KK] I. Kokorev, L. Kurlyandchik, A big cake on small plates,  Kvant 7 (1991) 13–17.

[Len1] H. Lenz, Über die Bedeckung ebener Punktmengen durch solche kleineren Durchmessers, Archiv Math. 7 (1956) 34–40, doi:10.1007/bf01900521.

[Len2] H. Lenz, Zerlegung ebener Bereiche in konvexe Zellen von möglichst kleinem Durchmessers, Jahresber. Deutsch. Math. Vereinigung 58 (1956) 87–97.

[Lit] J.E. Littelwood, A Mathematician’s Miscellany, Methued & Co, London, first published in 1953.

[SCY] D.O. Shklyarskiy, N.N. Chentsov, I.M. Yaglom, Geometrical estimations and combinatorial geometry problems, Moskow, Nauka, 1974, in Russian.

[Sel] J.L. Selfridge, An informal seminar on the coverings of convex sets (Report of the Inst. in the Theory  of Numbers), Colorado, 1959. 334.


问:
Element of minimum norm in a closed subset of $L^2[-\pi,\pi]$
Let

$$Y=\Bigg\{ f\in L^2[-\pi,\pi]\, :\, \int_{-\pi}^\pi xf(x)\, dx=1,\,\, \int_{-\pi}^\pi (\sin x)f(x)\, dx=2 \Bigg\}.$$

My goal is to calculate the minimum norm element in $Y$.

The indication that appears to me is: to calculate $f\in Y$ of minimum norm, find a function of the type $g(x)=sx+t\sin x$ such that $f\in Y$ is fulfilled if and only if $f-g\in Z^\perp$ where $Z=\textrm{Span}[x,\sin x]$, that is, that $Y=g+Z^\perp$ is fulfilled. Then $f=g$.

But I don't understand why with this procedure I will get an $f$ element of minimum norm...

答:
I will just solve the problem, I think that at the end it will be clear what form the minimum must have. Define $f$ to be the minimum of $I(f):=\|f \|_{L^2[-\pi, \pi]}^2$ over $Y$, this is well defined since the ambient space $L^2[-\pi, \pi]$ is reflexive and $Y$ is closed in the strong topology of $L^2[-\pi, \pi]$ and convex. Indeed, $Y$ weak sequentially closed and $I$ weak sequentially lower semi-continuous and coercive would have been sufficient, but nevermind. Let $Z= \text{span}_{\mathbb{R}}\{x, \sin(x) \}$ and $g \in Z^{\perp}$, then in particular
$$\int_{-\pi}^{\pi} xg(x) \,dx = 0 \qquad\text{and} \qquad\int_{-\pi}^{\pi} \sin(x)g(x) \,dx = 0 \,.$$
Thus, for every $\varepsilon \in \mathbb{R}$ and $g \in Z$, we have $f+\varepsilon g \in Y$ since
$$ \int_{-\pi}^{\pi} x(f(x)+\varepsilon g(x)) \, dx = \int_{-\pi}^{\pi} xf(x) \,dx +\varepsilon \int_{-\pi}^{\pi} xg(x) \,dx =1 ,$$
and in the same way
$$ \int_{-\pi}^{\pi} \sin(x)(f(x)+\varepsilon g(x)) \, dx = \int_{-\pi}^{\pi} \sin(x)f(x) \,dx +\varepsilon \int_{-\pi}^{\pi} \sin(x)g(x) \,dx =2 .$$
Then the function
$$ J(\varepsilon) := \| f+\varepsilon g\|_{L^2[-\pi, \pi]}^2 = \int_{-\pi}^{\pi} (f(x)+\varepsilon g(x))^2 \, dx $$
has a minimum at $\varepsilon =0$, hence by the derivative test
$$ 0= \frac{\partial}{\partial \varepsilon} \Bigg|_{\varepsilon =0} J(\varepsilon) = 2\ \int_{-\pi}^{\pi} f(x)g(x) \, dx ,$$
and this holds for all $g \in Z^{\perp}$. Thus $f \in Z^{\perp \perp} = \overline{Z}=Z$ as $Z$ is closed. Hence $f=\alpha x + \beta \sin(x)$ for some $\alpha, \beta \in \mathbb{R}$, and imposing that $f \in Y$ you find $\alpha$ and $\beta$ from a system of two equations in two unknowns. In particular $f$ is necessarily a linear combination of $x$ and $\sin(x)$, which I think is what you were asking for.


答:
If $\dim V$ and $\dim W$ are finite, here it is a proof for the sake of curiosity.

Based on such assumption, let us consider a basis $\mathcal{B} = \{b_{1},b_{2},\ldots,b_{n}\}$ for $V$ and a basis $\mathcal{B}' = \{b'_{1},b'_{2},\ldots,b'_{m}\}$ for $W$, where $n = \dim V$ and $m = \dim W$.

Then one has the following matricial representation for $T$:
\begin{align*}
[T]_{\mathcal{B}}^{\mathcal{B}'} = [[T(b_{1})]_{\mathcal{B}'}\, [T(b_{2})]_{\mathcal{B}'} \ldots\,[T(b_{n})]_{\mathcal{B}'}]
\end{align*}

Consequently, if $v = a_{1}b_{1} + a_{2}b_{2} + \ldots + a_{n}b_{n}$, one has that
\begin{align*}
T(v) = [T]_{\mathcal{B}}^{\mathcal{B}'}[v]_{\mathcal{B}} = a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}}
\end{align*}

Finally, according to the triangle inequality as well as the Cauchy-Schwarz inequality, we have that

\begin{align*}
\|T(v)\| & = \|a_{1}[T(b_{1})]_{\mathcal{B'}} + a_{2}[T(b_{2})]_{\mathcal{B'}} + \ldots + a_{n}[T(b_{n})]_{\mathcal{B'}} \|\\\\
& \leq \|a_{1}[T(b_{1})]_{\mathcal{B'}}\| + \|a_{2}[T(b_{2})]_{\mathcal{B'}}\| + \ldots + \|a_{n}[T(b_{n})]_{\mathcal{B'}}\|\\\\
& = |a_{1}|\|[T(b_{1})]_{\mathcal{B'}}\| + |a_{2}|\|[T(b_{2})]_{\mathcal{B'}}\| + \ldots + |a_{n}|\|[T(b_{n})]_{\mathcal{B'}}\|\\\\
& = \langle(|a_{1}|,|a_{2}|,\ldots,|a_{n}|),(\|[T(b_{1})]_{\mathcal{B'}}\|,\|[T(b_{2})]_{\mathcal{B'}}\|,\ldots,\|[T(b_{n})]_{\mathcal{B'}}\|)\rangle \leq K\|v\|
\end{align*}

Finally, due to the linearity of $T$, the desired result holds:
\begin{align*}
\|T(v) - T(w)\| = \|T(v - w)\| \leq K\|v - w\|
\end{align*}

Hopefully this helps!

Title: Clarification on proof that for a linear map T, being continuous and being Lipschitz are equivalent statements

Body: I know that in general for any function $f:V \longrightarrow W $ from a normed vector space $V$ to a second normed vector space $W$, being a Lipschitz function implies being continuous, is the second implication on my question the one I'm having troubles to understand.


问:
Showing that $x (1-x) y (1-y) (1 - xy)^{-2}$ is less than or equal to 1.
Let $x,y \in [0,1]$ and consider the inequality $ f(x,y) \equiv x (1-x) y (1-y) (1 - xy)^{-2} \leq 1$. One can show this easily by finding $argmax_x f(x,y)$ and then plugging into $f(x,y)$ to show that $\max_y \max_x f(x,y) \leq 1$. But I wonder if this can be shown using a more powerful approach? Is this an implication of some well known inequality or general mathematical result? My ultimate goal is to extend this to an analogous inequality for matrices.

Body:
This is
$x (1-x) y (1-y) \le (1 - xy)^{2}
$
or,
with help from Wolfy,
$(1 - xy)^{2}-x (1-x) y (1-y)
=1/4 y (2 x + y - 3)^2 + 1/4 (-y^3 + 6 y^2 - 9 y + 4)
\ge 0
$.

$-y^3 + 6 y^2 - 9 y + 4
=4-y(y^2-6y+9)
=4-y(y-3)^2
$
and for
$0 \le y \le 1$,
if $f(y) = y(y-3)^2$,
$f'(y)
=3 (y^2 - 4 y + 3)
=3(y-3)(y-1)
\ge 0$
for $0 \le y \le 1$.

Since
$f(0) = 0$
and $f(1) = 4$,
$-y^3 + 6 y^2 - 9 y + 4
\ge 0$
so
$(1 - xy)^{2}-x (1-x) y (1-y)
\ge 0
$.


答:
Just expand and compare terms.

 $0\leq x(1-y) = x - xy \leq 1 - xy$

 $0\leq  y (1-x) = y - xy \leq 1 - xy$.

Multiply these and we are done.

Title: Showing that $x (1-x) y (1-y) (1 - xy)^{-2}$ is less than or equal to 1.

Body: Let $x,y \in [0,1]$ and consider the inequality $ f(x,y) \equiv x (1-x) y (1-y) (1 - xy)^{-2} \leq 1$. One can show this easily by finding $argmax_x f(x,y)$ and then plugging into $f(x,y)$ to show that $\max_y \max_x f(x,y) \leq 1$. But I wonder if this can be shown using a more powerful approach? Is this an implication of some well known inequality or general mathematical result? My ultimate goal is to extend this to an analogous inequality for matrices.


问:
What exactly is $\mathcal{O}_X$ as featured in the exact sequence of a hypersurface?
For instance say that $X$ is a degree-$d$ hypersurface in $\mathbb{P}^n$.  To be precise, put $R = k[x_0, \ldots, x_n]$, $d > 0$, $f \in R_d$ squarefree, and $X = V(f) \subset \mathbb{P}^n$.  Then we have an exact sequence of graded $R$-modules

$0 \to R(-d) \xrightarrow{-\cdot f} R \to R/(f) \to 0$;

when we promote this to sheaves, apparently we have

$0 \to \mathcal{O}_{\mathbb{P}^n}(-d) \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_X \to 0$.

Now normally I'd think of "$\mathcal{O}_X$" as representing a sheaf on $X$, but this construction seems to have produced a sheaf on $\mathbb{P}^n$ instead.

答:
Just for the sake of resolving this question, the sequence should read

$0 \to \mathcal{O}_{\mathbb{P}^n}(-d) \to \mathcal{O}_{\mathbb{P}^n} \to i_\ast \mathcal{O}_X \to 0$,

where $i : X \to \mathbb{P}^n$ is the inclusion.  Using $\mathcal{O}_X$ in place of $i_\ast \mathcal{O}_X$ is apparently an accepted abuse of notation in this context.

Thanks to Zhen Lin for clearing this up in the comments.


问:
Why does the maximum of the normal distribution ($\mu$) trace out $\frac{1}{\sin(x)}$ with varying $p$
As I was recently playing around with the graph of the normal distribution, I noticed that as I varied $p$, the maximum of the graph traced out $\frac{1}{\sin(x)}$.

In the following graphic I plotted the Gaussian bell function:
$$
{\displaystyle g(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}\exp {\left(-{\frac {1}{2}}{\frac {(x-\mu )^{2}}{\sigma ^{2}}}\right)}.}
$$
with $\mu=n\cdot p$ and $\sigma=\sqrt{\mu(1-p)}$. As I varied $p$ from $[0, 1]$ the maximum traced out $\frac{1}{\sin(x)}$:

Image of graph

I was wondering where the trigonometric function originates from and why it shows up here.

答:
For a given $\mu$, the maximum value of the normal density is located at $x = \mu$:  $$g(\mu) = \frac{1}{\sqrt{2\pi} \sigma}.$$  Therefore, for $\sigma = \sqrt{np(1-p)}$, the locus of the maximum as $p$ varies on $(0,1)$ is $$f(p) = \frac{1}{\sqrt{2\pi n p(1-p)}}.$$  It is not $\csc p$ nor is it proportional to such a curve, as can be seen in a comparison here:

The blue curve is $2 \csc (\pi p)$, and the orange curve is $(p(1-p))^{-1/2}$.  There is no linear transformation between the two that will map one curve to the other:  they have different shapes.


问:
Why is a miracle happening on this double integral?
I am trying to solve a double integral of the form

$$\int_0^\infty du\int_0^\infty dx \frac{1}{x}\left(\frac{\partial}{\partial x}\right)^{2n+1}f\left(\sqrt{x^2+u}\right),$$

where $f$ is a well-behaved even function that decays rapidly at infinity. The content of this question (and miracle) does not depend on the specific $f$, as long as it is even and well behaved, but $f(t)=e^{-t^2}$ is a fine example. Note that since $1/x$ multiplies an odd function, the integrand is nonsingular as $x\to0$.

I would like to have a solution for a generic integer $n$, in terms of the known quantities $f^{(m)}(0)$ (i.e., arbitrary $m$th derivatives of $f$ evaluated at zero).

Amazingly, such a solution seems to always exist (I have tested up to $n = 5$), but the method I am using is extremely tedious, relies on Mathematica, and results in infinite terms that miraculously cancel. So the question is, what's a more efficient and clear method to get the same result?

Step 1: Evaluate all $2n+1$ derivatives.

Step 2: Transform the first integral with the change of variables $u=y^2,du=2ydy$.

Step 3: Convert to polar coordinates.

Step 4: Evaluate the $\theta$-integral.

Step 5: What remains is an integral of the form $\int_0^\infty dr\left(c_1\frac{1}{r^{p_1}}f^{(q_1)}(r) + \cdots + c_r r^{p_N}f^{(q_N)}(r)\right)$. Some of the terms with powers of $x$ in the denominator are infinite, but integrating by parts on some terms always makes these cancel. The terms with positive powers of $x$ can be integrated by parts repeatedly until they become $\int_0^\infty dr f^{(m)}(r) = - f^{(m-1)}(0)$.

Steps 1–4 can be performed by Mathematica using the single line (here $n=2$):

```<code>Integrate[
 Assuming[{r, \[Theta]} \[Element] PositiveReals,
  2r Tan[\[Theta]] D[f[Sqrt[x^2 + u]], {x, 2*2 + 1}] /. {x ->
      r Cos[\[Theta]], u -> r^2 Sin[\[Theta]]^2} //
   Simplify], {\[Theta], 0, \[Pi]/2}]
</code>```

I invite you to copy/paste this line in, examine the output and verify what I have claimed in step 5. It may seem pointless for me to ask this question, given that I already have a method for getting to a solution; however, Mathematica starts taking a very long time to compute this even for $n=5$, and even still the above line does not consider the final integrations by parts that make several more terms vanish and combine, so I think a better method is needed for both efficiency and understanding's sake.

答:
Let $I$ denote the integral. Then by the Fubini's theorem and Leibniz's integral rule,

\begin{align*}
I
&= \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n+1}}{\mathrm{d}x^{2n+1}} \int_{0}^{\infty} \mathrm{d}u \, f(\sqrt{x^2+u}) \\
&= \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n+1}}{\mathrm{d}x^{2n+1}} \int_{x}^{\infty} \mathrm{d}s \, 2s f(s) \tag{$s=\sqrt{x^2+u}$} \\
&= -2 \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \frac{\mathrm{d}^{2n}}{\mathrm{d}x^{2n}} x f(x) \\
&= -2 \int_{0}^{\infty} \mathrm{d}x \, \frac{1}{x} \bigl( x f^{(2n)}(x) + 2n f^{(2n-1)}(x) \bigr) \\
&= - 4n \int_{0}^{\infty} \mathrm{d}x \, \frac{f^{(2n-1)}(x)}{x}
\end{align*}

In the fourth step, we utilized the general Leibniz's rule, and in the last step $f^{(2n-1)}(0) = 0$ is used. Given this formula, I see no reason for this to reduce to a linear combination of $f^{(k)}(0)$'s. For instance,

If $f(x) = e^{-x^2}$ and $n = 2$, then $I = -32\sqrt{\pi}$.

If $f(x) = \frac{1}{1+x^2}$ and $n = 2$, then $I = -24\pi$.

If $f(x) = \operatorname{sech} x$ and $n = 2$, then $I = 8\pi - \frac{1}{\pi^3}\psi^{(3)}\bigl(\tfrac{1}{4}\bigr) $, where $\psi^{(m)}$ is the polygamma function of order $m$.


问:
Check if a student will pass an MCQ exam with negative marking
Here is a scenario:
We have an MCQ exam with 100 questions.

Every 3 wrong answers will remove 1 correct answer.

The Minimum needed score to pass this exam is 50

I'm using the following formula :
score = ( (correct * 3) - wrong) / (3 * total)) * 100;

now, I will calculate the student's score on each question, at what point should I decided that this student will not pass the exam no matter what he would do on the remaining questions?

I have not slept for almost 96 hours and I can't think clearly at this point

答:
If the student answers all the questions, and the number of questions answered correctly is $r$ and the number answered incorrectly is $w$ we have $$r+w=100$$ and in order to pass, the student needs $$r-\frac w3\geq50$$  Eliminating $w$ gives $r\geq 62.5$ and since $r$ must be an integer, we have $r\geq63$, which implies $w\leq37$.  When the student has answered $38$ questions incorrectly, there is no possibility that he will pass.

The foregoing only applies if the student answers all the questions, but a student may skip some questions.  I think the easiest thing to do is to keep a running score.  If the student has answered $r$ questions correctly, $w$ incorrectly, and skipped $s$ questions, his score up to this point is $$r-\frac w3$$ and there are $100-r-w-s$ questions remaining, so the highest score he can possibly achieve is $$r-\frac w3+100-r-w-s=100-s-\frac{4w}{3}$$ If this is less than $50$, the student has no chance.  Furthermore, the minimum number of questions he must answer correctly in order to pass is $50$ minus his current score, round up to the next integer, if necessary.

For example, suppose the student has seen $60$ questions, answered $40$ correctly, skipped $15$, and answered $5$ wrong.  The student's score to this point is $$40-\frac53=38.333$$ so the student needs to answer at least $12$ more questions correctly.

If it were me, I would be reluctant to tell a student, "You must answer at least $12$ more questions correctly in order to pass," for fear he would interpret it, or at least say he interpreted it, as "If you answer at least $12$ more questions correctly, you will pass," which isn't true, since he might answer $12$ right and $28$ wrong.  I would rather show the current score and the number of questions remaining.  Then if a student sees that his current score is $40$ and there are only $9$ questions remaining, he will easily understand that he has no possibility of reaching $50$ and passing.


答:
$Z$ is closed, being finite dimensional so in any case have $f\in Z\cup Z^{\perp}$ so $f(x)=ax+b\sin x +g(x)$ for some real numbers $a,b$ and some $g\in Z^{\perp}.$ Now

$\tag1 \|f\|^2=\frac{2a^2\pi ^3}{3}+4\pi ab+\pi b^2+\|g\|^2.$

Let $\delta$ be the desired minimum. There is a sequence $(f_n)\subseteq Y$ such that $\|f_n\|\to \delta.$ $Y$ is closed convex so the parallelogram law shows that $(f_n)$ is Cauchy and so  converges to some $f\in Y\ $(because $Y$ is closed and $L^2$ is complete). $(1)$ implies that $g=0$ a.e.

By continuity of the inner product, $\langle x,f_n\rangle\to \langle x,f\rangle$ and $\langle \sin x,f_n\rangle\to \langle \sin x,f\rangle.$  It follows that $\langle x, ax+b\sin x \rangle=2$ and $\langle \sin x,ax+b\sin x\rangle=1.$ This gives the equations $\frac{a\pi^3}{3}+b\pi =1$ and $2a\pi+b\pi=1.$ It follows that $a=0$ and $b=\frac{1}{\pi}$ and so

$\tag 2 f(x)=\frac{1}{\pi}\sin x.$

Title: Element of minimum norm in a closed subset of $L^2[-\pi,\pi]$

Body:
Let

$$Y=\Bigg\{ f\in L^2[-\pi,\pi]\, :\, \int_{-\pi}^\pi xf(x)\, dx=1,\,\, \int_{-\pi}^\pi (\sin x)f(x)\, dx=2 \Bigg\}.$$

My goal is to calculate the minimum norm element in $Y$.

The indication that appears to me is: to calculate $f\in Y$ of minimum norm, find a function of the type $g(x)=sx+t\sin x$ such that $f\in Y$ is fulfilled if and only if $f-g\in Z^\perp$ where $Z=\textrm{Span}[x,\sin x]$, that is, that $Y=g+Z^\perp$ is fulfilled. Then $f=g$.

But I don't understand why with this procedure I will get an $f$ element of minimum norm...


答:
You need to be very careful with notation here. What exactly do you mean by ${\frac{d^{\infty}}{dx^{\infty}}}$?

Without defining what you mean here the question doesn't even make sense. It's certainly true (and you can prove by induction) that for any ${n \in \mathbb{N}}$ we have ${\frac{d^n}{dx^n}e^x=e^x}$, but that's it.

Another point to make is that you cannot sum an "infinite" number of zeroes and conclude the answer is $0$.

EDIT: I guess just to really explicitly answer the question "What would we get if we differentiate ${e^x}$ infinitely many times?" - the answer is it depends how you define taking the derivative infinitely many times. We cannot physically do things infinitely many times in Mathematics, and so have to create definitions to work around it. The same sort of reasoning applies here. You must define what you mean first.


问:
Vector spaces with no Complete norms
Can anyone give me an example of a vector space $V$ such that there is no norm which is complete on $V$? </p>

答:
I assume you want $V$ to be a vector space over $\mathbb{R}$ or $\mathbb{C}$.  Since the latter can be viewed as a special case of the former, let's say $V$ is an $\mathbb{R}$-vector space to fix ideas.

Hints:

1) A finite-dimensional subspace is always closed.

2) A proper closed subspace is nowhere dense.

3) There is a famous theorem about countable unions of nowhere dense subsets of a complete space.

Addendum: I decided that my comment calling for a lack of other answers after having already given an answer myself was something of a conflict of interest, so I have made this answer CW.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5

Title: Vector spaces with no Complete norms


问:
Is there a name for the intersection of the maximal ideals containing an ideal?
Clearly it's a prime ideal containing the radical.  Do you just call it the "Jacobson radical of I?"

答: For the sake of resolving this question, it apparently is sometimes known as the "Jacobson radical of $I$."  Thanks to JSchlather for supplying a reference in the comments.</p>


问:
Does the norm of a (1,1) tensor depend on the conformal factor?
If we have a $(1,1)$ tensor $T$ on a (pseudo-)Riemannian manifold $(M,g)$, its norm is
$$
g(T, T) = g_{AB}g^{CD}T^A_C T^B_D.
$$
If we let $h = \phi^2 g$ be a conformally equivalent metric, then its inverse is $h^{-1} = \phi^{-2}g^{-1}$ and thus the norm of $T$ as measured by $h$ is identical, as the conformal factors cancel out. This only happens when $T$ is a mixed tensor which is covariant in the same number of indices as it is contravariant.

Questions: Is this proof correct? Is there any geometric reason why these types of tensors have their norm preserved by conformal transformations?

答: The proof is correct. Giuseppe Negro's comment about $(1,1)$-tensors being regarded as linear maps is also right on point. The same works for $(r,r)$-tensors, as $$h_{i_1j_1}\cdots h_{i_rj_r}h^{k_1\ell_1}\cdots h^{k_r\ell_r} T^{i_1\cdots i_r}_{\qquad k_1\cdots k_r}T^{j_1\cdots j_r}_{\qquad \ell_1\cdots \ell_r} = g_{i_1j_1}\cdots g_{i_rj_r}g^{k_1\ell_1}\cdots g^{k_r\ell_r} T^{i_1\cdots i_r}_{\qquad k_1\cdots k_r}T^{j_1\cdots j_r}_{\qquad \ell_1\cdots \ell_r}, $$since cancellations happen in pairs. Linear algebra: a multilinear map $(V^*)^{\times r} \times V^{\times r} \to \Bbb R$ can be seen as a linear map $V^{\otimes r} \to V^{\otimes r}$, so there you have it. If the quantity of covariant indices is not equal to the quantity of contravariant indices, you'll have extra $V$ or $V^*$ factors which will break the conformal invariance.


答:
But on the other hand, a generic vector in polar coordinates is given in the form:
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$

The formula
$$\underline{v}=v_r \hat{\underline{r}}+v_\theta \hat{\underline{\theta}}$$
Is wrong. The radius unit vector is defined such that the position vector $\underline{\mathrm{r}}$ can be written as
$$\underline{\mathrm{r}}=r~\hat{\underline{r}}$$
That's what makes polar coordinates so useful. Sometimes we only care about things that point in the direction of the position vector, making the theta component ignorable.

Title: A question about vector representation in polar coordinates.

Body:
Suppose we want to write the vector:
$$\vec{A}=4\hat{x} +3\hat{y}$$
in polar coordinates $(\rho, \phi)$, by making use of the identities:
$$
\hat{x}=\cos\phi\,\hat{\rho} -\sin{\phi}\,\hat{\phi}, \quad \hat{y}=\sin\phi\,\hat{\rho} +\cos{\phi}\,\hat{\phi}
$$
which simply yield:
$$
\vec{A}= \left(4\cos{\phi}+3\sin{\phi}\right)\hat{\rho} + \left(-4\sin{\phi}+3\cos{\phi}\right)\hat{\phi}
$$
Now expressing $\vec{A}$ as above does not make much sense to me because $\vec{A}$'s orientation is fixed so there should appear no free parameter such as the angle $\phi$. If $\phi$ indeed represents the orientation of the given vector then:
$$
\cos\phi=\frac{A_x}{||A||}=\frac{4}{5}, \quad \sin\phi=\frac{A_y}{||A||}=\frac{3}{5}
$$
If I plug these in $\vec{A}$, I end up with:
$$A=5\hat{\rho}$$
By using the same reasoning above, a generic vector's, $\vec{A}=A_x \hat{x} + A_y\hat{y}$, representation in polar coordinates appears to be
$$\vec{A}=||A|| \hat{\rho}, \quad ||A||=\sqrt{A_x^2+A_y^2}$$

But on the other hand, a generic vector in polar coordinates is given in the form:
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
If my steps of converting $\vec{A}$ from cartesian to polar coordinates is correct, then why does the $\hat{\phi}$ component seems to disappear ?

EDIT

Thank you all for your answers. I need to clear up few more things. In one of the answers below it is stated the representation (although I have seen such notation on the web):
$$
\vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}
$$
with constant coefficients is wrong. Is this also true for the representation of
say:
$$
\vec{A}=4\hat{x} +3\hat{y} + 2\hat{z}
$$
in cylindrical and spherical coordinates ? If so, then the proper conversion of $\vec{A}$ must read
$$
\vec{A}=5\hat{\rho} + 2\hat{z}, \quad \rho=\sqrt{x^2 + y^2}
$$
in cylindrical coordinates and
$$
\vec{A}=\sqrt{29}\hat{r}, \quad r=\sqrt{x^2+y^2+z^2}
$$
in spherical coordinates. Therefore the representation of a generic vector with arbitrary but constant coefficients:
$$
\vec{A}= A_{\rho} \hat{\rho} + A_{\phi}\hat{\phi} + A_{z}\hat{z}\\
\vec{A}= A_{r} \hat{r} + A_{\phi}\hat{\phi} + A_{\theta}\hat{\theta}
$$
is wrong. Is this conclusion correct ?


答:
Cartesian coordinates have some very convenient properties.
The vector that expresses the distance and direction from point
$P = (p_x,p_y)$ to point $Q = (q_x,q_y)$
is simply $(q_x - p_x)\hat x + (q_y - p_y)\hat y$;
that is, the vector's $x$ coordinate is simply the difference of the two points'
$x$ coordinates, and similarly with $y$ coordinates.

This works exactly the same no matter where you are in the plane.

We don't have this convenience with polar coordinates.
You cannot just take the polar coordinates of $P$ and $Q$,
find the difference in $\rho$ coordinates, and use this as the $\rho$ coordinate of the vector between the points.
It doesn't work for the $\phi$ coordinate either.

Usually when someone uses $\hat\rho$ and $\hat\phi$ as unit vectors like $\hat x$ and $\hat y$, what they are actually doing is (in at least some sense) looking at the grid of the polar coordinate system at some point in the plane and building a Cartesian coordinate system with an origin at that point and axes aligned with the radial and tangential directions.
That's why you can construct a vector with components like
$A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}$;
it is actually a vector in a locally defined Cartesian coordinate system where $\hat\rho$ is the unit vector pointing along one axis and $\hat\rho$ is the unit vector pointing along the other axis.

It turns out we can define the same locally defined coordinates for any point along the same radial line out from the origin--that is, for all points with a given $\phi.$ But if you take a point whose $\phi$-coordinate is different you have a different local coordinate system.
That's where the "free" $\phi$ in your vector expression comes from.

That "free" $\phi$ is not really free; it is a function of the point on which you decided to construct the locally-defined coordinate system.
It expresses the angle between the $\hat x$-$\hat y$ coordinate axes
and the $\hat\rho$-$\hat\phi$ coordinate axes for those two local Cartesian coordinate systems at that point, which you must know in order to correctly convert the coordinates of a given vector between those two coordinate systems.


问:
Clopen Sets and Sets being Neither Open Nor Closed
I am very new to Topology. I do not have a college math degree, only highschool math. I have a real hard time understanding how a set can be both open and closed. And set being neither open nor closed.

Is it correct to think of a set being both open and closed as below?

Let's consider a topology over two sets (A and B) that are disjoint. For example, A = (1,2) and B = (4,5). Both the sets are open as they don't contain their limit points. However, by definition, a set is closed if that set's compliment is open. Therefore, A is closed because A's compliment (which is B) is open and B is closed because its compliment is open.

I am yet to come up with an example for sets that are neither open nor closed. It would be really great if someone can give a simple example for that.

答: In the real numbers $\mathbb{R}$, any half-open interval of the form $[a,b)$ or $(a,b]$ where $a<b$ is neither open nor closed.


问:
Transcendental number definition query
The wiki page says that a transcendental number is defined as "a number that is not the root of a non-zero polynomial with rational coefficients".  When they provide this definition, shouldn't they add that the polynomial order has to be finite also?</p>

答: Any polynomial is by definition of finite order, so it’s not necessary to include it in the definition. If it has infinite order, then it is a Taylor series or something similar, and not a polynomial.</p>

Title: Transcendental number definition query


问:
Carleman's condition for Catalan numbers
I've been going through some elementary random matrix theory and in these lecture notes we say a probability measure is completely determined by its moments so long as it obey's Carleman's condition i.e. the sequence of moments $m_k$ are such that
$$
\sum_{k \geq 1} m_k^{-1/(2k)}= \infty.
$$ Later on when the semicircle distribution is brought up, whose moments are 0 for odd moments and the Catalan numbers for even numbers, they use a complex analytic method to prove that the moments are completely determined by the probability measure which is natural following Hadamard's formula. However I was wondering if a pure real analysis proof could still be constructed as I haven't found anything online, i.e. is the criterion
$$
\sum_{k \geq 1} C_k^{-1/(2k)}= \infty
$$
obeyed without analytic continuation? I've tried using Sterling's Formula which I hoped would cancel the exponent as well but no success as this series tends to zero.

答:
Asymptotically, $C_k \sim \frac{4^k}{k^{3/2}\sqrt{\pi}}$.  Therefore,
$$ \lim_{k \to \infty} C_k^{-\frac{1}{2k}} = \frac{1}{2} \lim_{k \to \infty} \pi^{\frac{1}{4 k}} k^{\frac{3}{4k}} = \frac{1}{2} \lim_{k \to \infty}  \pi^{\frac{1}{4 k}} \exp\left(\frac{3 \log k}{4 k}\right) = \frac{1}{2} \left( 1 \cdot \exp(0) \right) = \frac{1}{2} $$
Since the terms in the series do not in fact tend toward zero, the series diverges.


问:
Intuition on the quality of the approximation of the sample average in the central limit theorem
Let's say that a sequence of random variables $(X_n)_{n \in \mathbb{N}}$ is $\mathcal {o}(Y_n)$ for some other sequence of random variables $(Y_n)_{n \in \mathbb{N}}$, if $\lim_{n \to \infty}\frac{X_n}{Y_n}=0$ in distribution.

We can use this to interpret the central limit theorem for such a sequence of random variables of mean $\mu$ and variance $\sigma^2$ as an asymptotic expansion (of the distribution)
$$ \frac{1}{n}\sum_{j=1}^{n} X_j = \mu + \frac{\xi}{\sqrt{n}} + \mathcal{o}\left(\frac{1}{\sqrt{n}}\right), $$
where $\xi$ is a random variable distributed as $\mathcal{N}(0,\sigma^2)$.

Intuitively the first term comes from the law of large numbers, and the second term is a correction of order $\frac{1}{\sqrt{n}}$ coming from the central limit theorem.
Is there some intuition in why this term should be of order $\frac{1}{\sqrt{n}}$? Why isn't the expansion in some other power of $n$, e.g. $\frac{1}{n}$ as in the definition of differentiability.

Assuming that it is intuitively clear that for large $n$, the sample average should be approximately distributed as a normal distribution, why should the quality of this approximation be exactly $\frac{1}{\sqrt{n}}$ and not some other numerical value?

答:
For simplicity set $\sigma^2=1,\mu=0$.If you take $X_1,...,X_n$ as iid $N(0,1)$ random variables then $\xi=(X_1+...+X_n)/\sqrt n$ has the $N(0,1)$ distribution and the expansion
$$
\frac{1}{n}\sum_{i=1}^n X_i = \frac{\xi}{\sqrt n}
$$
is exact, so indeed there is not much possible choice for the scaling in front of $\xi$.


答: The definition of an open set in the real line is a set that is equal to the set of its inner points, that is: a set $A$ is open if for every $a\in A$ there exist a small number $\varepsilon$ such that $(a-\varepsilon,a+\varepsilon)$ is contained in $A$, a set is closed if its complement is open. Take $A=[0,1)$ in $\mathbb R$, this set is not open since $(0-\varepsilon,0+\varepsilon)$ will never be contained in $A$ but $0\in A$. And it's complement is not open since $1\in A^c$ but $(1-\varepsilon,1+\varepsilon)$ cant be contained in $A^c$, so $[0,1)$ is neither open or closed.

Title: Clopen Sets and Sets being Neither Open Nor Closed

Body:
I am very new to Topology. I do not have a college math degree, only highschool math. I have a real hard time understanding how a set can be both open and closed. And set being neither open nor closed.

Is it correct to think of a set being both open and closed as below?

Let's consider a topology over two sets (A and B) that are disjoint. For example, A = (1,2) and B = (4,5). Both the sets are open as they don't contain their limit points. However, by definition, a set is closed if that set's compliment is open. Therefore, A is closed because A's compliment (which is B) is open and B is closed because its compliment is open.

I am yet to come up with an example for sets that are neither open nor closed. It would be really great if someone can give a simple example for that.


问:
Find $E(X)$ for random variable $X$: the number of seconds it took a frog to reach its target
A jumping frog is currently at $x=0$ and $t=0$(time). With each second, the frog jumps to the right to its target, by randomly choosing the leap length from $\{0, 1, \dots, n\}$.

We want to find the time it takes the frog to get out of the region $[0, n-1]$. Let $X$ be a random variable denoting the number of seconds it took the frog to reach at least $n$.

Find $E[X]$ (the expectation of $X$).

So, first of all I know that $$E[X] = \sum_{i=0}^\infty P(X\ge i)$$
$$P(X\ge i) = 1 - P(X<i)$$
Is this true?
We can loot at the total number of steps as the sum of steps at each second $i$.
So we use elementary combinatorics and it leads us to this:
$$X_1 + X_2 + ... + X_{i-1} \le n-1$$
$$P(X\ge i) = \frac{n+i-2 \choose n-1}{(i-1)^{n+1}} $$
I also tried calculating the sum but it didn't go anywhere.

Thanks in advance :)

答: For $j\in\{0,1,\ldots,n-1\}$ let $E_j$ denote the expected time it takes for the spider to reach at least $n$ given the spider resides at position $j$. You're trying to find $E_0$. Using the total law of expectation with recursion we get $$E_j=1+\frac{1}{n+1}\sum_{k=j}^{n-1}E_k$$ This system of $n$ equations boils down to $$E_j=\Bigg(\frac{n}{n+1}\Bigg)^jE_0$$ $$E_{n-1}=\frac{n+1}{n}$$ Combining the two gives us an expression for $E_0$: $$E_0=\Bigg(\frac{n+1}{n}\Bigg)^n \longrightarrow e$$ as $n\longrightarrow \infty$.


问:
Divergence of the secant method
Those ten iterations are obtained using the secant method for the function $\tan(x\pi)-6$ starting with an interval of $[0\,\,\,0.48].$

The actual root is about $0.447431543$.

Can you explain why the method exhibited poor performance?

答: This Desmos demo might help you see what's going on.  The image of the plot is also below, the green and red dots are the initial interval.  It has the first four secant lines plotted, but you can probably follow the pattern/add more if you want.  Essentially the iterates are jumping around between different pieces of your function.  Because your function has so many vertical asymptotes you're going to need your initial interval to be closer to the root if you want any hope of the method converging.

Title: Divergence of the secant method


问:
Is there a closed form for an integral $\int \{\frac{1}{x}\} dx$?
Is there a closed form for an integral: $$\int \left\{\frac{1}{x}\right\} dx$$

I know $\int_0^1 \left\{\frac{1}{x}\right\} dx=1-\gamma$

答:
Use the same idea:

Split up $ \{ 1 / x \} = 1/x - \lfloor 1/x \rfloor $, and then integrate on regions $ [ 1/(n+1), 1/n]$.


问:
Calculation of integral around circle
I had the equation$$\int(xdy-ydx)$$around the circle
$$x^2+y^2=1$$
I tried this:
$$\begin{align}x&=(1-y^2)^{1/2}\\y&=(1-x^2)^{1/2}\end{align}$$
$$\int(1-y^2)^{1/2}dy-\int(1-x^2)^{1/2}dx=0$$
The limits of both $x$ and $y$ are same so both should cancel out and give ans$= 0$
but the answer is $2\pi$.

答:
Alternatively we can use Green's theorem:  $$\int_C Pdx + Qdy=\int_D(Q_x-P_y)dA$$
where C is a simple closed curve, D is the area bounded by C.  In your integral, we have $$P=-y$$, $$Q=x$$, C is the unit circle, D is the unit disk. So $$P_y=-1$$$$Q_x=1$$
and our integral switches to just
$$\int_D(1-(-1))dA$$
and since the area of the unit circle is $\pi$,  there you go.

Title: Calculation of integral around circle


问:
Analysis: limit notation when the function value approaches some value
Is an expression such as the following mathematically admissible?

$$\lim_{h(x) \to h_1} f(x)$$

where $f, h: X \subset \mathbb{R}^n \to \mathbb{R}$ are some continuous functions and $h_1 \in \mathbb{R}$.

Motivating examples are

$$\lim_{x + y \to 0} (x+y) = 0$$

and

$$\lim_{x^2 + y^2 \searrow 0} (x^2 + 2 y^2 + y ) = 0$$

Thank you.

答:
I learned topology and logic recently. I tried to apply my new knowledge to this question.

Let $f: X \rightarrow \mathbb{R}$ and $h: X \rightarrow \mathbb{R}^n$. For some $\alpha \in \mathbb{R}^n$ and $l \in \mathbb{R}$,
$$
\lim_{h(x) \rightarrow \alpha} f(x) = l
$$
is equivalent to
$$
\begin{align}
&\forall \epsilon > 0 :\exists\delta > 0 : \forall x \in X:\Vert h(x) -\alpha\Vert < \delta \Rightarrow |f(x)-l| < \epsilon\\
&\forall \epsilon > 0 :\exists\delta > 0 : \forall x \in X: h(x) \in B_\delta(\alpha) \Rightarrow f(x) \in B_\epsilon(l)\\
&\forall \epsilon > 0 :\exists\delta > 0 : \forall x \in X: x \in h^{-1}[B_\delta(\alpha)] \Rightarrow x \in f^{-1}[B_\epsilon(l)]\\
&\forall \epsilon > 0 :\exists\delta > 0 : h^{-1}[B_\delta(\alpha)] \subseteq f^{-1}[B_\epsilon(l)]
\end{align}
$$
If we can find such $\delta$ for each $\epsilon$, whatever $f$ and $h$ are, we can conclude that $f(x)$ converges to $l$.

This question was originally posted after seeing the notation in Wikipedia


问:
Trying to help my dad stop gambling using combinatorics
My dad plays keno but always (mostly) loses. I want to explain to him in a way the odds are clear.

20 numbers are picked per game from the draw and the player picks 10. There are 80 numbers total to choose from. I hope I explained that clearly.

So, I thought it would be 80 choose 10 but I’ve seen another answer say it’s (equivalently) 20 choose 10/ 80 choose 10. Much appreciation to get an answer for this.

Here is the payout table:

Keno payout table

答:
For your dad to win the biggest prize, he would have to pick ten numbers of the winning twenty, so there are $20\choose 10$ ways he can win. The total number of ways he can make his pick is $80\choose 10$, so his chance of winning a million dollars is ${20\choose 10}/{80\choose 10}\approx 1.12\cdot 10^{-7}$.

However, that's not the only way he can win. He might also have anywhere from four to nine of the winning numbers. For him to have $k$ of the winning numbers, he would have $20\choose k$ ways to pick those numbers, and then $60\choose 10-k$ ways to pick the rest. Note that when $k=10$, that ${60\choose 10-k}={60\choose 0}=1$, so this generalizes the previous formula. There are still $80\choose 10$ ways he can make his choice.

With this, we can compute the expected value of this game, which means for each winning outcome, computing the probability that he wins, multiplying the probability by the prize money, and adding these values up. Let $p_k$ denote the prize money for having $k$ winning numbers (assuming $p_{10}=10^6$). Then the expected value is

$$\frac{\sum_{k=4}^{10} {20\choose k}{60\choose 10-k}p_k}{80\choose 10}$$

By running a short python script, we can see that the expected value is around $\$0.65$. If your dad is paying more than $65$ cents when he buys ten numbers (which he probably is), then on average, he'll lose money per ticket.


问:
Curvature form computations
Let:

$M$ be a differential manifold,

$\pi:E\longrightarrow M$ a vector bundle on $M$ with $E_x\cong \mathbb{R}^k$,

$D:\Gamma(E)\longrightarrow \Gamma(T^*M\otimes E)$ a connection on $E$,

$d$ is the exterior derivative $d:\wedge^k T^*M\longrightarrow \wedge^{k+1}T^*M$.

Now the operator $D$ can be extended to higher cotangent sections of $\wedge^k T^*M\otimes E$ by setting $D(\omega\otimes\sigma):=d(\omega)\otimes\sigma+\omega\wedge D(\sigma)$

(the $\wedge$ symbol in the second summand is an abuse but it can be defined in a reasonable way).

Now i am trying to calculate the local expression of the curvature, given by a local expression for $D(D(s))$.

One need a local expression for $E$:

Let $\{s_1,\dots,s_k\}$ be a base for $E|_{U}$ with $U$ open in $M$.

Now $s=\sigma^k s_k$ (Einstein convention adopted and $\sigma^k$ are smooth functions).

From the property of $D$ one has $D(s)=d(\sigma^k)\otimes s_k+ \sigma^k\omega^i_k\otimes s_i$ (where $D(s_k)=\omega^i_k\otimes s_i$).

For the final step a can calculate separately the two summand becouse $D$ is linear:

$D(d(\sigma^k)\otimes s_k)=d^2(\sigma^k)\otimes s_k+d(\sigma^k)\wedge D(s_k)=d(\sigma^k)\wedge\omega^j_k\otimes s_j$.

$D(\sigma^k\omega_k^i\otimes s_i)=d(\sigma^k\omega^i_k)\otimes s_i+\sigma^k\omega^i_k\wedge D(s_i)=[d(\sigma^k)\wedge\omega^i_k+\sigma^kd(\omega^i_k)]\otimes s_i+\sigma^k\omega^i_k\wedge(\omega^j_i\otimes s_j)$.

This is not what I want, I need to find $D(D(s))= \sigma^kd(\omega^i_k)\otimes s_i+\sigma^k\omega^i_k\wedge(\omega^j_i\otimes s_j)$.

Questions:

Set $\theta^i_k:=\omega^i_k$ and $\omega:=\theta$.

I don't think this is a good way to do computations because I can't see what's going on.

I tried to make sense, this is not formalized so I am gonna give the idea.

Fix the base $\{s_i\}$ so that $s$ is a vector of smooth functions on $U$, $(s_1,\dots,s_k)$ and if I fix the elements $\{1\otimes s_i\}$ i can say that $D(s)=ds+\theta s=(d+\theta)s$ indeed $ds:=d(\sigma^k)*(1\otimes s_k)$ and $\theta s=s^k\theta^i_k*(1\otimes s_i)$

(star stands for an action or something like that).

Now I look at $D$ when it goes from $T^*M\otimes E$ in $T^*M\wedge T^*M\otimes E$ and if $\theta=\theta^k\otimes s_k$ i have that $D(\omega)$ is $d+\theta$ indeed $d(\theta)=d(\theta^k*(1\otimes s_k)$ and $\theta\omega=\theta^k\wedge\omega^i_k*(1\otimes s_i)$ (again the star is something not defined).

Now $D(D(s))=D(d+\theta(s))=D(ds)+D(\theta s)=(d+\theta)(ds)+(d+\theta)(\theta s)=d^2s+\theta ds+d\theta s+\theta^2s$ in the meddle I can see due to the wedge product involved that there is a cancellation and so I get the correct form I expect.

Where is the mistake in the first calculation?

Does my observation stands for a general way to proceed and how can I formalize it?

Any hints or suggest would be appreciated, thanks.

答:
First, note that you have a small mistake. The formula for the extension of $D$ should read $$D(\omega \otimes s) = {\rm d}\omega \otimes s + (-1)^{{\rm deg}(\omega)} \omega \wedge Ds$$instead.

Let $s \in \Gamma(E) = \Omega^0(M;E)$. Then $Ds \in \Omega^1(M;E)$ and $D^2s \in \Omega^2(M; E)$. Here, I use the "shorthand" $\Omega^k(M;E) = \Gamma( (T^*M)^{\wedge k}\otimes E)$ as usual.

Claim: $(D^2s)(X,Y) = R^D(X,Y)s$.

Proof: Choose a local trivialization $(e_i)$ for the bundle and write $De_j = \omega^i_{~j}\otimes e_i$, as well as $s= \sigma^ie_i$. Then $$Ds = D(\sigma^je_j) = {\rm d}\sigma^j \otimes e_j + \sigma^j \omega^i_{~j}\otimes e_i = ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\otimes e_i.$$Then $$\begin{align} D^2s &= D(({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\otimes e_i) \\ &= {\rm d}({\rm d}\sigma^i + \omega^i_{~j}\sigma^j) \otimes e_i - ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\wedge De_i \\ &=  (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j})\otimes e_i - ({\rm d}\sigma^i + \omega^i_{~j}\sigma^j)\wedge (\omega^k_{~i}\otimes e_k) \\ &\stackrel{(\ast)}{=}  (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j})\otimes e_i - ({\rm d}\sigma^k + \omega^k_{~j}\sigma^j)\wedge (\omega^i_{~k}\otimes e_i) \\ &= (\sigma^j {\rm d}\omega^i_{~j} + {\rm d}\sigma^j \wedge \omega^i_{~j}
 - {\rm d}\sigma^k \wedge \omega^i_{~k} - \sigma^j\omega^k_{~j}\wedge \omega^i_{~k})\otimes e_i \\ &= (\sigma^j {\rm d}\omega^i_{~j} + \sigma^j \omega^i_{~k}\wedge \omega^k_{~j})\otimes e_i \\ &= \sigma^j({\rm d}\omega^i_{~j} + \omega^i_{~k}\wedge \omega^k_{~j})\otimes e_i \\ &= \sigma^j \Omega^i_{~j}\otimes e_i,\end{align}$$as required. In $(\ast)$ we have renamed $i\leftrightarrow k$ in the second term (to make $\_\otimes e_i$ appear on both terms, so it could be factored) and in the last step, the second structure equation $\Omega^i_{~j} = {\rm d}\omega^i_{~j} + \omega^i_{~k}\wedge \omega^k_{~j}$. Here, the curvature $2$-forms $\Omega^i_{~j}$ are defined by the relation $R^D(\_,\_)e_j = \Omega^i_{~j} \otimes e_i$. The sign convention is $R^D(X,Y) = [D_X,D_Y] - D_{[X,Y]}$. $\quad\qquad\square$

Personally, I'm not a fan of this type of computation, and I think this covariant exterior derivative is better understood if you compare it with the usual exterior derivative in a coordinate free way. For example: $$\begin{align}{\rm d}\alpha(X,Y) &= X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X,Y]), \qquad \alpha \in \Omega^1(M) \\ {\rm d}\alpha(X,Y) &= D_X(\alpha(Y)) - D_Y(\alpha(X)) - \alpha([X,Y]), \qquad \alpha \in \Omega^1(M; E), \end{align}$$and similarly for higher degree forms. In particular, when $E = TM$, $D({\rm Id}) = \tau$ is the torsion of $D$, given by $\tau(X,Y) = D_XY-D_YX-[X,Y]$, and $$D^2\tau(X,Y,Z) = R^D(X,Y)Z + R^D(Y,Z)X + R^D(Z,Y)X$$says that the content of the first Bianchi identity for torsion-free connections is "the derivative of zero is zero". For arbitrary $E$ again, we have the second Bianchi identity $DR^D = 0$, where $R^D \in \Omega^2(M; {\rm End}\,E)$ is regarded as a ${\rm End}\,E$-valued $2$-form. See my short notes for proofs.

Title: Curvature form computations


问:
Proving that a particular CFG grammar generates language L
$$L=\left\{w\in\{x,y\}^*\mid N(y)=2N(x)\right\}$$

and $G$ is

$$S\to yySx\mid ySxSy\mid xSyy\mid SS\mid\epsilon$$

I need to proof that the grammar generates this language.
I know that I need to use induction to prove it, but I can't follow through

so, could you show how to prove it?

答:
There are two parts to this: you have to show that every word generated by $G$ is in $L$, and you have to show that every word in $L$ is generated by $G$. The first of these is straightforward. Suppose that $w$ is generated by $G$. Then it has a derivation

$$S=w_0\Rightarrow w_1\Rightarrow w_2\Rightarrow \ldots\Rightarrow w_{n-1}\Rightarrow w_n=w\,.$$

Clearly $|S|_x=|S|_y=0$, so $|S|_y=2|S|_x$. Now suppose that $0\le k<n$ and that $|w_k|_y=2|w_k|_x$, and verify that $|w_{k+1}|_y=2|w_{k+1}|_x$ no matter which production of $G$ is used to derive $w_{k+1}$ from $w_k$. Once you’ve done that, you can conclude by induction that $|w|_y=|w_n|_y=2|w_n|_x=2|w|_x$ and hence that $w\in L$.

The second part is a bit harder, but it can be done by induction on $|w|$. If $w\in L$ has length $0$, then $w=\epsilon$, which is generated by $G$. Now for the induction step let $n\in\Bbb Z^+$, suppose that every word of $L$ of length less than $n$ is generated by $G$, and let $w\in L$ be of length $n$; you need to show that $w$ is generated by $G$.

We’ll need to break the argument into cases, and we’ll use a technique that is useful but not obvious in order to do so. Say $w=u_1u_2\ldots u_n$, where each $u_k$ is either $x$ or $y$. Let $c_0=0$, and for $k=0,\ldots,n-1$ let

$$c_{k+1}=\begin{cases}
c_k+1,&\text{if }u_k=y\\
c_k-2,&\text{if }u_k=x\,.
\end{cases}$$

This is less complicated than it looks. Imagine reading $w$ from left to right, starting a counter at $0$, increasing it by $1$ each time you read a $y$, and decreasing it by $2$ each time you read an $x$; $c_k$ is the value of the counter after you’ve read $k$ letters of $w$. Because $w\in L$, it has twice as many $y$s as $x$s, so the counter must reach $0$ at the end of the word: $c_n=0$.

Suppose first that $c_k=0$ for some $k$ with $0<k<n$. Let $w_1=u_1\ldots u_k$ and $w_2=u_{k+1}\ldots u_n$. Show that $w_1,w_2\in L$. Clearly $|w_1|,|w_2|<n$, so the induction hypothesis then ensures that $w_1$ and $w_2$ have derivations in $G$. Show that $w$ has a derivation in $G$ that starts $S\Rightarrow SS$.

Now suppose that $c_k\ne 0$ for $1\le k\le n-1$.

If $u_1=x$, then $c_1=-2$. Suppose that $c_k>0$ for some $k>1$. Then there is a smallest $k$ such that $c_k>0$. When the counter increases, it always increases by $1$, so it must be that $c_k=1$ and $c_{k-1}=0$. But this is impossible (why?), so it must be that $c_k<0$ for $1\le k\le n-1$. Since $c_n=0$, this implies that $c_{n-1}=-1$ and $c_{n-2}=-2$ (why?) and hence that $u_{n-1}=u_n=y$. Thus, $w=xw_1yy$, where $w_1=u_2\ldots u_{n-2}$. Clearly $|w_1|<n$, so $w_1\in L$, and $w_1$ has a derivation in $G$. Show how to modify this derivation to get a derivation of $w$ in $G$.

Now suppose that $c_k\ne 0$ for $1\le k\le n-1$, and $u_1=y$, so that $c_1=1$. There are now two possibilities.

Suppose that $u_n=x$, so that $c_{n-2}=2$. Argue as in the previous bullet point to show that if some $c_k<0$, then there must be an $\ell$ such that $k<\ell<n-1$ and $c_\ell=0$ and that this is impossible. Conclude that $c_k>0$ for $1\le k\le n-1$. In particular, $c_2>0$, so $u_2=y$ (why?). Thus, $w=yyw_1x$, where $w_1=u_3\ldots u_{n-1}$. Show that $w_1$ has a derivation in $G$, and show how to modify this derivation to get a derivation of $w$ in $G$.

Finally, suppose that $u_n=y$. Then $c_1=1$, and $c_{n-1}=-1$ (why?), but there is no $k$ such that $1<k<n-1$ and $c_k=0$. This means that there must be a $k$ such that $c_k=1$ and $c_{k+1}=-1$; why? Clearly this means that $u_{k+1}=x$. Let $w_1=u_2\ldots u_k$ and $w_2=u_{k+2}\ldots u_{n-1}$. Show that $w_1,w_2\in L$; clearly $|w_1|,|w_2|<n$, so $w_1$ and $w_2$ have derivations in $G$. Show how to modify those derivations to get a derivation of $w$ in $G$.

Note that each of the four cases is based on one of the non-terminal productions in $G$.