问:
Applying the chain rule in matrix form to prove Loss function's derivatives...
I want to prove $\nabla_A J = \nabla_Z J \cdot B^T$ where $Z=AB$. $A$ is a $m \times n$ matrix and $B$ is a $n \times k$ matrix. The function $J$ is not given to me.

I began this proof by first writing $B^T$ as

$$
    B^T = \begin{pmatrix} b_{00} & b_{10} & b_{20} \\ b_{01} & b_{11} & b_{21} \\ b_{02} & b_{12} & b_{22} \\ \end{pmatrix}
$$

Since J is a loss function, it is scalar value. Computing its derivative wrt to Z i.e. $\nabla_Z J$, I write:

$$
    \nabla_{Z} J = \begin{pmatrix} \frac{\partial L}{\partial z_{00}} & \frac{\partial L}{\partial z_{01}} & \frac{\partial L}{\partial z_{02}} \\ \frac{\partial L}{\partial z_{10}} & \frac{\partial L}{\partial z_{11}} & \frac{\partial L}{\partial z_{12}} \\ \frac{\partial L}{\partial z_{20}} & \frac{\partial L}{\partial z_{21}} & \frac{\partial L}{\partial z_{22}} \\ \end{pmatrix}
$$

Multiplying these two matrices, I get:

$$
\begin{pmatrix} \frac{\partial L}{\partial z_{00}}b_{00} + \frac{\partial L}{\partial z_{01}}b_{01} + \frac{\partial L}{\partial z_{02}}b_{02}& \frac{\partial L}{\partial z_{00}}b_{10} + \frac{\partial L}{\partial z_{01}}b_{11} + \frac{\partial L}{\partial z_{02}}b_{12} & \frac{\partial L}{\partial z_{00}}b_{20} + \frac{\partial L}{\partial z_{01}}b_{21} + \frac{\partial L}{\partial z_{02}}b_{22} \\ \frac{\partial L}{\partial z_{10}}b_{00} + \frac{\partial L}{\partial z_{11}}b_{01} + \frac{\partial L}{\partial z_{12}}b_{02}& \frac{\partial L}{\partial z_{10}}b_{10} + \frac{\partial L}{\partial z_{11}}b_{11} + \frac{\partial L}{\partial z_{12}}b_{12} & \frac{\partial L}{\partial z_{10}}b_{20} + \frac{\partial L}{\partial z_{11}}b_{21} + \frac{\partial L}{\partial z_{12}}b_{22} \\ \frac{\partial L}{\partial z_{20}}b_{00} + \frac{\partial L}{\partial z_{21}}b_{01} + \frac{\partial L}{\partial z_{22}}b_{02}& \frac{\partial L}{\partial z_{20}}b_{10} + \frac{\partial L}{\partial z_{21}}b_{11} + \frac{\partial L}{\partial z_{22}}b_{12} & \frac{\partial L}{\partial z_{20}}b_{22} + \frac{\partial L}{\partial z_{21}}b_{20} + \frac{\partial L}{\partial z_{22}}b_{22} \\ \end{pmatrix}
$$

At this point, I am stuck and I am not sure how to proceed. How do I take the next step? Note, I am not well versed in Matrix Math and it has been a while since I dealt with proves. I am eager to learn and thats why I'm asking for the next step - assuming I did it correctly so far!!

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$For typing convenience, give short names to the various matrix gradients
$$\eqalign{
\nabla_ZJ &= \p{J}{Z} &\doteq G \quad&\iff\quad &dJ = G:dZ \\
\nabla_AJ &= \p{J}{A} &\doteq H \quad&\iff\quad &dJ = H:dA  \\
}$$
The differential of $Z$ in term of $A$
$$\eqalign{
Z &= AB \qquad\implies\qquad dZ = dA\;B \\
}$$
can be used to effect a change of variable from $Z\to A$
$$\eqalign{
dJ &= G:dZ \\
 &= G:(dA\,B) \\
 &= (GB^T):dA \\
 &= H:dA &\implies\quad H = GB^T \\
}$$

In the above, a colon is used as a product notation for the trace function, i.e.
$$A:B = {\rm Tr}(A^TB)$$
The properties of the trace function allow the terms in a colon product to be rearranged in a number of equivalent ways, e.g.
$$\eqalign{
A:B &= B:A = B^T:A^T \\
CA:B &= A:C^TB = C:BA^T \\
}$$


问:
Convergence in Hilbert spaces.

Let $\mathbb{X}$ be a Hilbert space and $ \{e_ {n}: n \in \mathbb {N} \}$ an orthonomal subset. Does the series $ \sum_{n = 1}^{\infty} |\langle x, e_n\rangle |$ converge for every $x \in \mathbb{X}$?

I was thinking, can we take $x = \sum_{n \in \mathbb {N}} e_{n}$? Does that make sense? For in this case $\langle x, e_n \rangle = 1$, for every $n$ and therefore our series does not converge.

答:
Let $X=\ell^2$ and $\{e_ {n}: n \in \mathbb {N} \}$ the usual orthonormal basis of $X$.

Let $x=(\frac{1}{n})_{n \in \mathbb N}$, then $x \in X$ and $<x, e_ {n}> =|<x, e_ {n}>|=\frac{1}{n}.$

Hence the series $\sum_{n = 1}^{\infty} |<x, e_ {n}> | $ is divergent.

ContentLicense: CC BY-SA 4.0

Title: Convergence in Hilbert spaces.


问:
Solve for 2D translation, rotation and scale given two touch point movements
For a phone app I need to let users translate, rotate and scale a rectangle on the 2D plane using "pinch" gesture.

Suppose the rectangle already has TRS = $t, \theta, s $; and the user has their fingers on two touch points $p_1, p_2$.  The user then moves the touch points to  $q_1, q_2$. What is the new TRS that will map $p_1 \rightarrow q_1, p_2 \rightarrow q_2$.  In other words, the user should feel that their fingers are "stuck" to the rectangle. The scale is always the same for $x$ and $y$.

The scale is easy. Let $p=p_1 - p_2$ and  $q=q_1 - q_2$, then  $s' = s \cdot  \dfrac{||q||}{||p||}  $

I think rotation is can use the arcSine of the cross product. Let $q_\bot = (-q_y, q_x)$ and
$\Delta\theta = arcSine \dfrac{q_\bot\cdot p}{||q||\cdot||p||}$, then $\theta' = \theta + \Delta\theta$

I think the translation should preserve the midpoint. So: $t' = t +  \frac{1}{2}((q_1 - p_1)+(q_2 - p_2))$.

But when I implement this, it doesn't work very well. The rotation is not around the correct point so the rectangle seems to slide away strangely.  Any advice please?

答:
First, some background.  Let's review the underlying math and notation.

Transformation $\mathbf{T}$ has four free parameters: scale $\lambda$, rotation $\theta$, and translation by $(t_x, t_y)$.  Let's define the transformation of point $\vec{p} = (p_x , p_y)$ to point $\vec{q} = (q_x , q_y)$ as
$$\vec{q} = \mathbf{T} \vec{p} \quad \iff \quad
\left[ \begin{matrix} q_x \\ q_y \\ 1 \end{matrix} \right] = \left[ \begin{matrix}
\lambda \cos \theta & -\lambda \sin \theta & t_x \\
\lambda \sin \theta & \lambda \cos \theta & t_y \\
0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} p_x \\ p_y \\ 1 \end{matrix} \right ] \tag{1a}\label{G1a}$$
Defining the transform as a single matrix $\mathbf{T}$ as
$$\mathbf{T} = \left[ \begin{matrix} \lambda \cos \theta & -\lambda \sin \theta & t_x \\
\lambda \sin \theta & \lambda \cos \theta & t_y \\
0 & 0 & 1 \end{matrix} \right ] \tag{1b}\label{G1b}$$
is useful, because we can calculate the inverse ($\vec{p} = \mathbf{T}^{-1} \vec{q}$),
$$\mathbf{T}^{-1} = \left[ \begin{matrix}
\frac{\cos\theta}{\lambda} & \frac{\sin\theta}{\lambda} & - \frac{t_x \cos\theta + t_y \sin\theta}{\lambda} \\
- \frac{\sin\theta}{\lambda} & \frac{\cos\theta}{\lambda} & \frac{t_x \sin\theta - t_y \cos\theta}{\lambda} \\
0 & 0 & 1 \end{matrix} \right] \tag{1c}\label{G1c}$$
and combine transformations via matrix multiplication:
$$\mathbf{T}_\text{combined} = \mathbf{T}_\text{after} \mathbf{T}_\text{before} \tag{1d}\label{G1d}$$
where the rightmost transformation is applied first, and leftmost transformation last.  Note that the identity transform, "no transform" or "as-is", is
$$\mathbf{1} = \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \tag{1e}\label{G1e}$$
This way we can also "ignore" the current transform ($\mathbf{T}_\text{before}$) for the purposes here, and just calculate the transform $\mathbf{T}_\text{after}$ caused by the gesture here; and get the combined transform using $\eqref{G1d}$.

(Note that such $3 \times 3$ matrices (with only $2 \times 3$ elements actually stored) are commonly used in 2D computer graphics, including SVG transform matrix() attribute, with the third component ($1$) of vectors not stored, and just baked in to the functions doing matrix-vector multiplication.  Similarly for $4 \times 4$ matrices (with only $3 \times 4$ elements actually stored) for 3D computer graphics, including OpenGL.)

The problem is to find the transformation $\mathbf{T}$ that transforms $\vec{p}_1$ and $\vec{p}_2$ to $\vec{q}_1$ and $\vec{q}_2$,
$$\left\lbrace \, \begin{aligned}
\vec{q}_1 &= \mathbf{T} \vec{p}_1 \\
\vec{q}_2 &= \mathbf{T} \vec{p}_2 \\
\end{aligned} \right. \tag{2}\label{G2}$$
which is actually a set of four equations (two Cartesian components per vector), and four unknowns (that define $\mathbf{T}$), and therefore should have a single solution (except for degenerate cases, where $\vec{p}_1 = \vec{p}_2$ or something similar).

Note that because of finite precision of floating-point numbers, we do not want to apply the transformation incrementally.  That is, we should remember the very initial touch points $\vec{p}_1$ and $\vec{p}_2$, keeping them unchanged for the duration of the gesture, also remembering the original transformation $\mathbf{T}_\text{before}$, compute a temporary gesture transformation $\mathbf{T}_\text{gesture}$ based on the initial touch points $\vec{p}_1$ and $\vec{p}_2$ and current touch points $\vec{q}_1$ and $\vec{q}_2$, and a temporary currently applied transformation $\mathbf{T}_\text{temp} = \mathbf{T}_\text{gesture}$.  Only when the gesture is finished, will the final $\mathbf{T}_\text{temp}$ become the currently applied and stored transformation.

If you apply the transformation incrementally, i.e. $\vec{p}_1$, $\vec{p}_2$ corresponding to the previous touch locations, and $\vec{q}_1$, $\vec{q}_2$ corresponding to current touch locations, and continously apply $\mathbf{T}_\text{current} = \mathbf{T}_\text{gesture} \mathbf{T}_\text{previous}$, not only do your rounding errors compound (making the gesture control erratic), but it will also be jittery, because small movement of one finger when close to the other will cause large rotations; the faster your touch pad update rate, the more likely such small movements are.  The incremental moment-to-moment transformation approach just does not work in practice.

Let $\vec{p}_1$ and $\vec{p}_2$ be the initial touch points, and $\vec{q}_1$ and $\vec{q}_2$ the corresponding final positions:
$$\vec{p}_1 = \left[ \begin{matrix} x_1 \\ y_1 \\ 1 \end{matrix} \right], \;
\vec{p}_2 = \left[ \begin{matrix} x_2 \\ y_2 \\ 1 \end{matrix} \right], \;
\vec{q}_1 = \left[ \begin{matrix} \chi_1 \\ \gamma_1 \\ 1 \end{matrix} \right], \;
\vec{q}_2 = \left[ \begin{matrix} \chi_2 \\ \gamma_2 \\ 1 \end{matrix} \right]$$
Let $C = \lambda \cos \theta$, $S = \lambda \sin \theta$, and translation is by $(X, Y)$, so that the transformation matrix is
$$\mathbf{T} = \left[ \begin{matrix}
C & -S & X \\
S &  C & Y \\
0 &  0 & 1 \\
\end{matrix} \right] \tag{3a}\label{G3a}$$
Note that as long $C S \ne 0$, the upper left $2 \times 2$ matrix is a proper rotation and scaling matrix, for all $C \in \mathbb{R}$, $S \in \mathbb{R}$.  The vector $(C, S)$ is the new $x$ axis basis vector, and $(-S, C)$ the new $y$ axis basis vector, and the two are always perpendicular (with positive $y$ axis 90° counterclockwise from positive $x$ acis) and have the same length.

Substituting these into $\eqref{G2}$ we have four equations and four unknowns ($C$, $S$, $X$, and $Y$), which has exactly one solution:
$$\left\lbrace \, \begin{aligned}
L^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\
C &= \frac{ (\chi_2 - \chi_1)(x_2 - x_1) + (\gamma_2 - \gamma_1)(y_2 - y_1) }{L^2} \\
S &= \frac{ (\gamma_2 - \gamma_1)(x_2 - x_1) - (\chi_2 - \chi_1)(y_2 - y_1) }{L^2} \\
X &= \frac{ (x_2 - x_1)(\chi_1 x_2 - \chi_2 x_1) + (y_2 - y_1)(\chi_1 y_2 - \chi_2 y_1) + (\gamma_2 - \gamma_1)(x_2 y_1 - x_1 y_2) }{L^2} \\
Y &= \frac{ (\chi_2 - \chi_1)(x_1 y_2 - x_2 y_1) + (y_2 - y_1)(\gamma_1 y_2 - \gamma_2 y_1) + (x_2 - x_1)(\gamma_1 x_2 - \gamma_2 x_1) }{L^2} \\
\end{aligned} \right. \tag{3b}\label{G3b}$$

If we do need the angle $\theta$ and the scale factor $\lambda$, they are obviously
$$\left\lbrace ~ \begin{aligned}
\lambda &= \sqrt{C^2 + S^2} \\
\theta &= \operatorname{atan2}\left(S, C\right) \\
\cos\theta &= \displaystyle \frac{C}{\sqrt{C^2 + S^2}} \\
\sin\theta &= \displaystyle \frac{S}{\sqrt{C^2 + S^2}} \\
\end{aligned} \right. \tag{3c}\label{G3c}$$
where $\operatorname{atan2}$ is the two-parameter form of arcus tangent, provided by almost all programming languages. ($\operatorname{atan2}(y, x) = \arctan(y/x)$ for $x \gt 0$, but is valid for all $x, y \in \mathbb{R}$ except $x=0, y=0$, as it takes the signs of both $x$ and $y$ into account in other quadrants.  Typically, it returns the angle in radians between $-\pi$ (-180°) and $+\pi$ (+180°), but it may vary between programming languages.)

We can verify the above using
$$\vec{o} = \left[\begin{matrix} o_x \\ o_y \\ 1 \end{matrix}\right], ~
\vec{t} = \left[\begin{matrix} t_x \\ t_y \\ 1 \end{matrix}\right], ~
\vec{b} = \left[\begin{matrix} b_x \\ b_y \\ 1 \end{matrix}\right], ~
\vec{a} = \left[\begin{matrix} a_x \\ a_y \\ 1 \end{matrix}\right]$$
and
$$\vec{p}_1 = \vec{o} + \vec{b}, ~ ~
\vec{p}_2 = \vec{o} - \vec{b}, ~ ~
\vec{q}_1 = \vec{o} + \vec{t} + \vec{a}, ~ ~
\vec{q}_2 = \vec{o} + \vec{t} - \vec{a}$$
so that $\vec{o}$ represents the middle point between the two initial touch points, $\vec{b}$ represents the vector from the middle point to the first initial touch point and $-\vec{b}$ the vector from the middle point to the second initial touch point, $\vec{t}$ represents the translation, and $\vec{a}$ represents the vector from the middle point between the final touch points ($\vec{o}+\vec{t}$) to the first final touch point, and $-\vec{a}$ the vector from the middle point between the final touch points to the second final touch point.  The transformation matrix is then
$$\mathbf{T} = \left[ \begin{matrix}
\displaystyle \frac{a_x b_x + a_y b_y}{b_x^2 + b_y^2} & \displaystyle \frac{a_x b_y - a_y b_x}{b_x^2 + b_y^2} & \displaystyle t_x + o_x - \frac{o_x (a_x b_x + a_y b_y ) + o_y (a_x b_y - a_y b_x )}{b_x^2 + b_y^2} \\
\displaystyle \frac{a_y b_x - a_x b_y}{b_x^2 + b_y^2} & \displaystyle \frac{a_x b_x + a_y b_y}{b_x^2 + b_y^2} & \displaystyle t_y + o_y - \frac{ o_x ( a_y b_x - a_x b_y ) + o_y (a_x b_x + a_y b_y ) }{b_x^2 + b_y^2} \\
0 & 0 & 1 \end{matrix} \right]$$
and if we do the math, we indeed find that $\mathbf{T} \vec{p}_1 = \vec{q}_1$ and $\mathbf{T} \vec{p}_2 = \vec{q}_2$.

Note that we can extract rotation angle, scale factor, and translation from such a $3 \times 3$ matrix (assuming it is orthogonal, not skewed) with entries
$$\mathbf{M} = \left[ \begin{matrix} m_{11} & m_{12} & m_{13} \\ m_{21} & m_{22} & m_{23} \\ 0 & 0 & 1 \end{matrix} \right] \tag{4}\label{G4a}$$
via
$$\left\lbrace ~ \begin{aligned}
\theta &= \operatorname{atan2}\left( m_{21} - m_{22} ,\, m_{11} + m_{22} \right) \\
\lambda &= \frac{1}{2}\left(\sqrt{m_{11}^2 + m_{21}^2} + \sqrt{m_{12}^2 + m_{22}^2}\right) \\
t_x &= X = m_{13} \\
t_y &= Y = m_{23} \\
\end{aligned} \right. \tag{4b}\label{G4b}$$
In particular, if you have an existing transfrom (by $\theta_1$, $\lambda_1$, $(X_1, Y_1)$), and you wish to further transform it (by $\theta_2$, $\lambda_2$, $(X_2, Y_2)$), first construct the two matrices using
$$\mathbf{M} = \left[ \begin{matrix}
\lambda \cos\theta & -\lambda \sin\theta & X \\
\lambda \sin\theta & \lambda \cos\theta & Y \\
0 & 0 & 1 \end{matrix} \right ] \tag{4c}\label{G4c}$$
then calculate their product via matrix multiplication, original orientation rightmost, latest applied transform leftmost,
$$\mathbf{M}_\text{final} = \mathbf{M}_2 \mathbf{M}_1$$
and then extract the rotation angle $\theta$, scale factor $\lambda$, and translation $(X, Y)$ as specified in $\eqref{G4b}$.


问:
Sum of series (1 + (y/m) + (y/m)^2 + ... + (y/m)^n)
My physics textbook tells me that the sum:

(1 + (y/m) + (y/m)^2 + ... + (y/m)^n)

where m < n, equals:

(1 / (1 - y/m)) * (1 - (y/m)^n+1)

...but I can't work out how. Can anyone help?

It's part of a long argument proving the convergence of the power series exp(tX) where X is a 2x2 matrix. They're summing a series (1/m!)y^m + (1/m!m)y^m+1 ... + (1/m!)(1/m^n)y^m+n which (as I can see) equals y^m/m! times the series above.

答:
For $q \ne 1$ the following formula holds:

$$1+q+q^2+...+q^n=\frac{1-q^{n+1}}{1-q}.$$


问:
Continuity of hypergeometric function $_{1}F_1(a;2;k\pi i)$ in the argument $a$
For $0\leq a \leq 2$, are the hypergeometric functions $_{1}F_1(a;2;k\pi i)$, $_{1}F_1(a;2;-k\pi i)$ continuous in the argument $a$? where $k\in \{1,2,3 \}$. How to prove it?

Are the following equations  valid?

$$\lim_{a\to 2} {}_{1}F_1(a;2;\pm k\pi i)={}_{1}F_1(2;2;\pm k\pi i)$$

$$\lim_{a\to 0^+} {}_{1}F_1(a;2;\pm k\pi i)={}_{1}F_1(0;2;\pm k\pi i)$$

$$\lim_{a\to c} {}_{1}F_1(a;2;\pm k\pi i)={}_{1}F_1(c;2;\pm k\pi i)$$

where $0<c<1$.

答: Note that ${}_1 F_1(a;2;z)$ is an entire function of both $a$ and $z$. In general, ${}_1 F_1(a;b;z)$ is entire in $a$ and $z$, and is meromorphic in $b$. ${}_1 F_1(a;b;z)/\Gamma(b)$ is entire in all three variables.


问:
If $\left\{a_n\right\}$ is Fibonacci Number, prove that the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\}$ increase

If $a_0=1,a_1=1,a_{n+2}=a_{n+1}+a_{n}$, prove the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\},\,n\geq 1$ increase.

When $n$ is even this inequality is easy proved by Am-Gm.

If $n$ is even,we have $$\frac{\ln a_n}{\ln a_{n+1}}<\frac{\ln a_{n+1}}{\ln a_{n+2}}\Leftrightarrow \ln a_n\cdot\ln a_{n+2}<(\ln a_{n+1})^2$$

By Am-Gm, and $a_{n-1}a_{n+1}-a_n^2=(-1)^n$, we have$$LHS<\left(\frac{\ln a_n+\ln a_{n+2}}{2}\right)^2=\left(\frac{\ln(a_n\cdot a_{n+2})}{2}\right)^2=\left(\frac{\ln(a_{n+1}^2-1)}{2}\right)^2<RHS.$$

But when n is odd, the inequality is a little difficult, or I didn't use a right method.

Supplement:

Here we prove that $$a_{n-1}a_{n+1}-a_n^2=(-1)^n.$$

It is not hard to prove that if $$A=\begin{pmatrix}
0&1\\
1&1
\end{pmatrix}$$then$$A^n=\begin{pmatrix}
a_{n-1}&a_n\\
a_n&a_{n+1}
\end{pmatrix}$$by induction. So $$a_{n-1}a_{n+_1}-a_n^2=\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}^n=\left(\det\begin{vmatrix}
0&1\\
1&1
\end{vmatrix}\right)^n=(-1)^n$$

Body:
There is a primary proof by a student in Fudan University(Not me).

First, it is easy to prove that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ strictly increase.

Second, we can also prove $$0\leq\ln(1+x)\leq x,\quad(\forall\,x\geq 0).$$

And now we begin the proof.

Consider $\forall\,n\geq 3$, we have $$\frac{a_{n+1}}{a_{n-1}}=\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n+1}}{a_n}\geq\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}.$$

So
$$
\ln a_{n+1}-\ln a_{n-1}\geq\ln\frac{9}{4}=2\ln\frac{3}{2}>\frac{4}{5}\Rightarrow(\ln a_{n+1}-\ln a_{n-1})^2>\frac{16}{25}
$$

By this inequality when $n\geq 5$ we can get
$$
\begin{align*}
\ln a_{n+1}\cdot\ln a_{n-1}&=\frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{1}{4}
(\ln a_{n+1}-\ln a_{n-1})^2\\
&< \frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n+1}\cdot a_{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&\leq\frac{1}{4}(\ln(a_n^2+1))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\
&=\frac{1}{4}(\ln(a_n^2+1)-\ln a_n^2)(\ln(a_n^2+1)+\ln a_n^2)+(\ln a_n)^2-\frac{4}{25}
\\
&\leq\frac{1}{4}\left(\ln(a_n^2+1)-\ln a_n^2\right)\cdot2\ln((a_n+1)^2)+(\ln a_n)^2-\frac{4}{25}\\
&=\ln(1+\frac{1}{a_n^2})\ln(1+a_n)+(\ln a_n)^2-\frac{4}{25}\\
&\leq \frac{1}{a_n^2}\cdot a_n-\frac{4}{25}+(\ln a_n)^2\\
&\leq\frac{1}{8}-\frac{4}{25}+(\ln a_n)^2\\
&<(\ln a_n)^2
\end{align*}
$$

In a word, it is $\ln a_{n-1}\cdot\ln a_{n+1}<(\ln a_n)^2$, which is equal to$$\frac{\ln a_{n-1}}{\ln a_n}<\frac{\ln a_n}{\ln a_{n+1}},\quad n\geq 5.$$

For the rest situation ,we just need to prove it by caculating it.


问:
Smoothness of morphisms gives an exact cotangent sequence
This is Vakil 21.2 S, self-study.

We are to show that if $\pi: X \to Y$ and $\rho: Y \to Z$ are smooth morphisms of schemes, then the relative cotangent sequence

$$\pi^*\Omega_{Y/Z} \to \Omega_{X/Z} \to \Omega_{X/Y} \to 0$$

is also left-exact. By "smooth" we mean smooth of relative dimension $n$ for some $n$; see definition 12.6.2 in the linked notes; it's a local condition on the Jacobian.

We know the composition of smooth maps of relative dimension $m$ and $n$ is smooth of relative dimension $m+n$, so $\rho \circ \pi$ is smooth.  We also know that if $\pi : X \to Y$ is smooth of relative dimension $n$, then $\Omega_{X/Y}$ is locally free of rank $n$. This would give us that all our sheaves are locally free.

I do not follow the hint given in this exercise; I did not use a block upper triangular matrix in 12.6 D as the hint suggests I should have, so if there is a solution taking this hint in a different direction or opting for another route, that would be appreciated.

答:
I don't understand the proof given in stacks project either, since I am not familiar with naive cotangent complex which it uses, but I can give another easier proof.

Standard definition. If a morphism of rings is formally smooth and of finite presentation, then it is called a smooth morphism.

Proposition. Suppose $A\rightarrow B\rightarrow C$ is a sequence of rings and $C$ is formally smooth over $B$, then $0 \to C \otimes _ B \Omega _{B/A} \xrightarrow{\alpha} \Omega _{C/A} \to \Omega _{C/B} \to 0$ is split exact.

Proof. For a $B$-module $M$, we define a ring $C*M$, which is the abelian group $C\oplus M$ with multiplication given by $(c,m)\cdot(c',m'):=(cc',cm',c'm)$.

Assume $D\in Der_A(B,M)$, then we have a sequence $0\rightarrow M\rightarrow C*M\rightarrow C\rightarrow 0$ and the following commutative diagram, where $\phi: b\rightarrow (g(b),D(b))$.
\begin{array}
CC & \stackrel{}{\rightarrow} & (C*M)/M \\
\uparrow{g} & & \uparrow{} \\
B & \stackrel{\phi}{\rightarrow} & C*M
\end{array}

By the definition of formally smoothness, there exists $h: C\rightarrow C*M$ such that $h(c)=(c,D'c)$ with $D': C\rightarrow M$ making the above diagram commutate.( check that $D'$ is a derivation and $D=D'\circ g$).

Now let $M:=\Omega_{B/A}\otimes C$ and $D: B\rightarrow C\otimes\Omega_{B/A}$ given by $b\rightarrow d_{B/A}(b)\otimes1$, by what we have said above, there exists a derivation $D': C\rightarrow C\otimes\Omega_{B/A}$. By the universial property of Kähler differential, $D'$ corresponds to $\alpha':\Omega_{C/A}\rightarrow C\otimes\Omega_{B/A}$, then you can check that $\alpha'\circ\alpha=\operatorname{id}$.

Therefore the first fundamental sequence is split exact.

Remark. For the second fundamental sequence, the similar result also holds. Namely, for $B$ an $A$-algebra and $I$ is a ideal of $B$, then $I/I^2\rightarrow\Omega_{B/A}\otimes_B(B/I)$ is split injective iff $B/I$ is formally smooth over $A$.


问:
Do we calculate the power with Chinese theorem?
For a natural numbr $n$ we know that $n=3k_1+1$, $n=4k_2+1$ and $n=5k_3+4$, for natural numbers $k_1, k_, k_3$.

Calculate $5^n\pmod {13}$ .

For that, do we apply the Chinese Theorem?

答:
$n=3k_1+1 \Rightarrow 3k_1+1 \equiv 1 \pmod 4$

$3k_1+1 \equiv 1 \pmod 4 \Rightarrow k_1 \equiv0 \pmod 4$ as a result, $k_1=4m$

$n=5k_3+4 \Rightarrow 12m+1 \equiv 4 \pmod 5$

$12m+1 \equiv 4 \pmod 5 \Rightarrow m \equiv  4 \pmod 5$

$ m \equiv  4 \pmod 5 \Rightarrow m=5k+4$

$m=5k+4 \Rightarrow n=60k+49$

according to Fermat`s Theorem

$5^{12} \equiv 1 \pmod{13}$

$n \equiv 1 \pmod{12} \Rightarrow 5^n \equiv 5 \pmod{13}$

In this case,the key which was about chinese remainder theorem is the fact that we know there exists a solution for n


问:
Indefinite integral $\int \cos^k{x} dx$?
From WolframAlpha, we can see
$$\int \cos^k x \;dx = - \frac{\sqrt{\sin^2x}\csc x \cos^{k+1}(x) _2F_1\left(\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};\cos^2{x}\right)}{k+1} $$
for any $k\in\mathbb{R}$ and $_2F_1(a,b;c;z)$ is hypergeometric series.
I want to know its proof.

Body:
$$I=\int \cos^k( x) \;dx =-\int \frac{t^k}{\sqrt{1-t^2}}\,dt=-\int t^k \sum_{n=0}^\infty \frac{\left(n-\frac{1}{2}\right)! }{\sqrt{\pi }\, n!}t^{2n}\,dt$$
$$I=-\frac 1 {\sqrt{\pi }}\sum_{n=0}^\infty \int \frac{\left(n-\frac{1}{2}\right)! }{ n!}t^{2n+k}\,dt=-\frac 1 {\sqrt{\pi }}\sum_{n=0}^\infty \frac{\left(n-\frac{1}{2}\right)!}{n! (k+2 n+1)} t^{k+2 n+1}$$
$$I=-\frac{t^{k+1} }{k+1}\,
   _2F_1\left(\frac{1}{2},\frac{k+1}{2};\frac{k+3}{2};t^2\right)$$


问:
Determine if $1.001^n$ will grow faster than $n^{10}$.
We know that $n^k \leq k^n$, where $n^k$ is a polynomial and $k^n$ is an exponential function. However, what if we have a case such as $n^{10}$ and $1.001^n$. How can I determine which function will grow faster?

Background - this is a problem from my complexity class and specifically the topic about the Big O-Notation.

答: We can say that $$(1.001)^n = e^{n\log(1.001)} \geq \frac{(n\log(1.001))^{11}}{11!}  $$ using the series expansion of $e$. The result should then be clear.


问:
Derivative of function in tangent-plane
I have a sheet of answers for some of the assignments I had to complete, this is a step I don't really understand:

How does he go from $$f(x,y) = 5e^{x^2-y^2}$$
to $$fx(x,y) = 5e^{x^2-y^2}(-2y) = 10xe^{x^2-y^2}$$?

See source here:

This is the thing I don't understand: Say $$u$$ is 5 and $$v$$ is $$(e^{x^2-y^2})$$ than
the chain rule should be applied after $$u'*v*v' = (e^{x^2-y^2}) *2x(e^{x^2-y^2})$$ (I am not sure here and unfortunately
your rather formal way of doing this confuses me).

答:
Now that you've posted the source, the short answer is: he doesn't.

This is simply an application of the chain rule. Put $g(t)=5e^t$, so that $f(x,y)=g(x^2-y^2).$ Then

$$\begin{align}f_x(x,y,) &:= \frac{\partial}{\partial x}[f(x,y)]\\ &= \frac{\partial}{\partial x}[g(x^2-y^2)]\\ &= g'(x^2-y^2)\cdot\frac{\partial}{\partial x}[x^2-y^2]\\ &= g'(x^2-y^2)\cdot 2x\\ &= g(x^2-y^2)\cdot 2x\\ &=f(x,y)\cdot 2x.\end{align}$$ We can do something similar for $f_y$.

ContentLicense: CC BY-SA 3.0

Title: Derivative of function in tangent-plane


问:
Do we have $\nu_t (t!) = o(t)$?
For any positive integer $t>1$, we can define $\nu_t(n)$ as the maximal integer $k$ such that $t^k\mid n$. (Here we don't need $t$ to be a prime.)

Then I wonder if we have
$$\lim_{t\rightarrow\infty}\frac{\nu_t(t!)}{t}=0$$

For the most naive bound, I only have
$$\nu_t(t!)\leq \log_t (t!)\sim t$$
So it must use Legendre's theorem to delete some useless prime in $t!$. But I still cannot prove this limit...

答:
$$ \frac{\nu_t(t!)}{t} \leqslant \frac{\nu_p(t!)}{\nu_p(t) \cdot t} \leqslant \frac{t}{(p-1)\nu_p(t)\cdot t} = \frac{1}{(p-1)\nu_p(t)} $$

for any prime $p$ (where first inequality follows from the fact that $t^{\nu_t(t!)} \mid t!$, and the second one is consequence of Legendre's formula. Suppose $\frac{\nu_t(t!)}{t}$ will stay above some $c>0$ infinitely many times, so our sequence won't be convergent to zero. Then, for all $p \mid t$, $c \leqslant \frac{1}{p-1}$, so $p \leqslant c^{-1}+1$ and all prime divisors of $t$ are bounded. Similarly, for all $p \mid t$ we have $\nu_p(t) \leqslant c^{-1}$. But this gives bound on $t$: it can't be larger than product of all primes less than $c^{-1}+1$, raised to power $c^{-1}$. This is a contradiction with assumption about our sequence not being convergent to zero; thus

$$ \lim_{t \to \infty} \frac{\nu_t(t!)}{t} = 0 $$

as we wanted to prove.

Title: Do we have $\nu_t (t!) = o(t)$?


问:
If $f$ is a uniformly continuous function on a metric vector space, is there a continuous $g$ with $f=e^{{\rm i}g}$?
Let $E$ be a metric space and $$S^1:=\{z\in\mathbb C:|z|=1\}.$$ Say that $f\in C^0(E,S^1)$ is inessential if $$f=e^{{\rm i}g}\tag1$$ for some $g\in C^0(E,\mathbb R)$. We can show that

If $f\in C^0(E,S^1)$ with $f(E)\ne S^1$, then $f$ is inessential.

If $f_i\in C^0(E,S^1)$ with $$f_1(x)\ne-f_2(x)\;\;\;\text{for all }x\in E\tag2$$ and $f_1$ is inessential, then $f_2$ inessential.

In Appendix 2.5 of Treatise on Analysis the following corollary is shown:

If we take a look at the proof, we see that the compactness assumption is not needed as long as $f:E\times[0,1]\to S^1$ is uniformly continuous.

Now, the following is inferred from that corollary:

It seems like the only reason why the closed ball in $\mathbb R^d$ is considered
is that this is a compact metric vector space. Now I wonder, aren't we able to use precisely the same arguments to show that any uniformly continuous function $f:E\to S^1$ on a metric vector space is inessential?

EDIT

It was remarked that this can be solved by some facts about algebraic topology. However, I'm not familiar with algebraic topology. So, I'd really appreciate if someone could provide an elementary answer (maybe imposing suitable assumptions, which are satisfied in the special case of $\mathbb R^d$).

答:
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\Z}{\mathbb{Z}}$
$\renewcommand{\S}{\mathbb{S}}$

The answer to the main question is no - continuous maps from a compact metric space into $\S^1$ can be essential. See the example:

Example. Let $f \colon \S^1 \to \S^1$ be the identity: $f(z) = z$ for $x \in \S^1$. Then:

$f$ is continuous and $\S^1$ is a compact metric space, so $f$ is uniformly continuous,

there is no continuous map $g \colon \S^1 \to \R$ such that $f = e^{ig}$.

To see that the second statement is true, assume for the contrary that such a $g$ exists. Consider the continuous function $h \colon \R \to \R$ given by $h(t) = g(e^{it})$. Now for each $t \in \R$ we have
$$ e^{i h(t)} = e^{ig(e^{it})} = f(e^{it}) = e^{it}. $$
As in your Lemma 1, we see that $h(t)-t$ must be a multiple of $2\pi$. In other words, $t \mapsto h(t)-t$ is a continuous function from $\R$ to $2\pi \Z$, and as such has to be constant. Thus, $h(t) = t + 2k\pi$ for some fixed $k \in \Z$.

Coming back to our assumptions, we have $g(e^{it}) = t + 2k\pi$ for every $t \in \R$. But taking $t = 0$ and $t = 2\pi$ leads to a contradiction:
$$ 2k\pi = g(e^{0 \cdot i}) = g(1) = g(e^{2\pi \cdot i}) = (2k+1) \pi. $$

Let us dig a little deeper. You asked:

It seems like the only reason why the closed ball in $\R^d$ is considered is that this is a compact metric vector space. Now I wonder, aren't we able to use precisely the same arguments to show that any uniformly continuous function $f \colon E \to \S^1$ on a metric vector space is inessential?

Now, you already know the answer is no. But you're right that this reasoning is quite general. You should be able to check yourself that it gives us:

Theorem. Assume that the compact metric space $E$ admits a continuous function $r \colon E \times [0,1] \to E$ such that $r(\cdot,1)$ is the identity (i.e., $r(1,x) = x$), while $r(x,0)$ is constant (i.e., $r(0,x)=p$ for some fixed $p \in E$). Then every continuous map $f \colon E \to \S^1$ is inessential.

Such a map $r$ exists for the closed ball $E = \overline{B}^n$ (in fact, the proof of (Ap.2.6) relies on this) but not for the circle $E = \S^1$ (otherwise the example wouldn't work).

Finally, let me remark that there's no escaping from algebraic topology. This is algebraic topology, whether we like it or not. If you want to search for more, some vocabulary may be useful:

A space $E$ is called simply connected if all continuous maps $f \colon E \to \S^1$ are inessential.

The terms essential and inessential are usually replaced by homotopically nontrivial and homotopically trivial. These can be generalized to all possible target spaces.

In (Ap.2.5), two maps $f(x,0)$ and $f(x,1)$ as are called homotopic and the map $f$ is called a homotopy.

The map $g$ is called a lifting of $f$.

For a general target space $X$ (in place of $\S^1$), one can consider a cover $\tilde{X} \to X$ (in place of $\R \to \S^1$, $t \mapsto e^{it}$) and talk about liftings in this more general context.

Any space $E$ satisfying the assumptions of the theorem above is called contractible.

Of course, this is just vocabulary, in case you need it. Anyway, I strongly recommend A. Hatcher's brilliant and entertaining book.


问:
Solve $x^3y'= y^3 + 2xy^2$
I found the following exercise:

$$\text{Solve: } \ x^3y'= y^3 + 2xy^2$$

The idea I used:

$$\text{Let } y(x) = x\cdot v(x)$$

After some computations:

$$\int \frac{\text{dx}}{x} = \int \frac{\text{dv}}{v(v^2+2v-1)}$$

After computing the integrals, it becomes:

$$\frac{\ln(|v^2 + 2v - 1|)}{2} + \frac{\ln(|v-\sqrt{2}+1|) -\ln(|v+\sqrt{2}+1|)}{2\sqrt{2}} - \ln(|v|) = \ln(|x|) + C$$

After that we will substitute back: $v$ with $\frac{y}{x}$, in the relation above.

Question: What magic could be applied to it as to be able to find $y = \text{...}$?

Did I have a bad start? How to solve this?

Thank you very much!

答: It is not always possible to get a general solution in an explicit form. In this case, I could not find any way of doing so. You can get a feel of what's going on by drawing the direction field.

Title: Solve $x^3y'= y^3 + 2xy^2$


问:
Constructibility of regular hexagon with given area
I would like to find the algebraic translation of the following problem:

Let $r>0$ be a real number. For some $r$, the regular hexagon with area $\pi r^2$ is not constructible (with straightedge and compass).

How do I reduce this statement to a condition of some number being algebraic over the rationals?

答:
Regular hexagon with side length of $a$ has area $\frac{3\sqrt{3}}{2} a^2$. Therefore we want to find segment $a = \sqrt{\frac{2\pi r^2}{3\sqrt{3}}}$, in particular, this has to be an algebraic number.

(This is the case iff $\frac{3\sqrt{3}}{2}a^2 = \pi r^2$ is algebraic, which is easier to handle.)


问:
Is there an algorithm (on the inernet) that converts points into possible functions for the points?
Is there an algorithm where the input is a bunch of points and the output would be a possible function which includes these points? I've only found one website that gives you the correlating linear function for two points you have entered. This feature is quite limited because I have more than two points and the function shouldn't be a linear function.

In my problem, there are 5 known maxima that each have the same y value but a different x value. I know that there are infinite maxima. What I don't know is, if there are more maxima in between my known maxima or not. The x values of the points are all negative. The distances between the points shrink while you go towards positive $∞$. I didn't find any pattern (yet) regarding how they shrink. My best approach was something like

$$f(x)=a\sin \left(\frac{b}{x-c}\right)$$

but I didn't find any fitting parameters just by trial and error. That's why I asked if there is an algorithm for that.
The known maxima are:

$P_1\approx(-13 800 000 000;z)$

$P_2\approx(-4 280 000 000;z)$

$P_3\approx(-540 000 000;z)$

$P_4\approx(-2 000 000;z)$

$P_5\approx(-1 600;z)$

(These are just roundings because the numbers are very big and it doesn't matter that much/it doesn't have to be exact. It's irrelevant where the y values of the maxima are, so I'll just say the highest possible point is $z$.)

答:
To make my suggestion more real, suppose you want a function that will have relative maxima at $(-1,1)$ and $(1,2)$. As is clear, there should also be a relative minimum in the middle somewhere, let's pick it in the middle at $(a,b)$ for any $a \in \mathbb{R}$, since we seem to be able to pick anything we like for the minima.

So we need a function with $3$ extrema, hence we will have a quartic polynomial $p(x)$. Its derivative is given by
$$
p'(x) = (x-1)(x-a)(x+1) = x^3-ax^2-x+a,
$$
since we know all the roots from the extrema. Now
$$
p(x) = x^4-\frac{ax^3}{3}-\frac{x^2}{2}+ax+C
$$
and we have
$$
\begin{split}
1 &= p(-1) &= C - \frac{2a}{3} + \frac12 \\
b &= p(0)  &= C \\
2 &= p(1)  &= C + \frac{2a}{3} + \frac12 \\
\end{split}
$$
Which is a system of 3 linear equations in 3 unknowns, which you can solve and this will take care of finding $p$ exactly.

From the second equation you see $C = b$, and adding first and third together yields $3=2C+1 \iff C=1$, which implies $a=3/4$. So the result is
$$
p(x) = x^4-\frac{x^3}{4}-\frac{x^2}{2}+\frac{3x}{4}+1
$$


问:
$f(x+1)= {3x+5 \over 2x+1}, x\neq-1/2, f(x)=?$
$f(x+1)= {3x+5 \over 2x+1}, x\neq-1/2$

How can I get f(x)?

More generally, if I have $f(x+k)$ what do I have to do to get $f(x)$

答:
You can get it easily applying the equality with $x-1$.

$$
f(x)=f((x-1)+1) = \dfrac{3(x-1)+6}{2(x-1)+1}
$$

More generally,

$$
f(x) = f((x-k)+k)
$$


答:
An elementary argument.

Comparing $1.001^n$ and $n^{10}$ is the same as comparing $(\sqrt[10]{1.001})^n$ and $n$. To ease the computation, we note that $\sqrt[10]{1.001}>1.00001$, and we compare $1.00001^n$ to $n$.

Both $1.00001^n$ and $n$ grow unboundedly. (As you can check, $1.00001^{n+1}>1.00001^n+0.00001$, and this is enough to ensure growth.) Now after a while, $1.00001^n>100000$. Then $1.00001^{n+1}>100000+1$, and the exponential grows in steps larger than $1$.

Title: Determine if $1.001^n$ will grow faster than $n^{10}$.

Body:
We know that $n^k \leq k^n$, where $n^k$ is a polynomial and $k^n$ is an exponential function. However, what if we have a case such as $n^{10}$ and $1.001^n$. How can I determine which function will grow faster?

Background - this is a problem from my complexity class and specifically the topic about the Big O-Notation.


问:
Mathematical Induction Check, $n! \leq n^n/2^n $
Hello I am working on this problem and was wondering if I did the proof correct.

Use induction to prove that $n! \leq n^n/2^n $ for all $n \geq 6$.

Basic Step (n=6): 6! $\leq 6^6/2^3 = 3^6$
Thus $80 \leq 81$, so the Basic Step is true.

Assume the statement above is true for n. We need to show n+1 is also true.

$(n+1)^{n+1}/2^{n+1} = (n+1)^n (n+1)/2^{n+1} \geq n^n(n+1)/2^{n+1} \geq n^n/2^{n+1} \geq n^n/2^n \geq n!$

答: Hint: You're very close, but not quite there. Remember that $(n+1)!=(n+1)n!$.


问:
A counter example, topology question
Im looking for a counter example for the following problem:

Let $ S_{1}\neq\emptyset $ be a compact set and $S_{2}\neq\emptyset $ a closed set.
(Those are subsets of $ \mathbb{R}^n $ for some $ n $).

So there exists $ s_{1}\in S_{1},\thinspace\thinspace\thinspace s_{2}\in S_{2} $ such that for the distance between the sets:

$ d\left(S_{1},S_{2}\right):=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $

we have $ \inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right)=d\left(s_{1},s_{2}\right) $

Originally this was a prove/disprove question, but Im pretty sure that this is false, so Im looking for a counter example. Thanks in advance.

After reading the comments:

This is my attempt to prove the statement (rather then disprove it):

Let $ d\left(x_{n}^{1},x_{n}^{2}\right)$ be a sequence such that $ \lim_{n\to\infty}d\left(x_{n}^{1},x_{n}^{2}\right)=\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $.

Since $S_1 $ is compact, there exists a convergent subsequence of $ x_{n}^{1} $, say $ x_{n_{k}}^{1} $.Denote its limit by $ x_1 $. One can verify that

$ d\left(x^{1},x_{n_{k}}^{2}\right)\underset{n\to\infty}{\longrightarrow}\inf_{x_{1}\in S_{1},x_{2}\in S_{2}}d\left(x_{1},x_{2}\right) $

I tried to prove that $ x_{n_{k}}^{2} $ also have a convergent sequence and that would end the proof. But I cant see how.

答:
I think it's true now with the nonempty condition

Note that a subset of $\mathbb{R}^n$ is compact $\iff$ it is closed and bounded $\iff$ it is sequentially compact

The first important step is to realise that we can make it so that we only care about a bounded part of $S_2$

Choose any $x_1 \in S_1$ and $x_2 \in S_2$ to get an upper bound $D ≔ d(x_1, x_2)$ on $d(S_1, S_2)$

Since $S_1$ is bounded, it is contained in some large closed ball $\overline{B}(0, M)$

Hence the only points that we care about from $S_2$ are going to be within a distance $D$ of this ball

That is to say, we can w.l.o.g. consider ${S_2}' ≔ S_2 \cap \overline{B}(0, M+D)$ instead of $S_2$, which is now (sequentially) compact

From here we can take a sequence of pairs $a_n \in S_1$ and $b_n \in S_2$ with $d(a_n, b_n) \rightarrow d(S_1, S_2)$, and appeal to sequential compactness of our sets and continuity of $d$ to find the desired pair of points

Title: A counter example, topology question


答:
The fact that $n\equiv 4\pmod5$ is irrelevant.

Since $n\equiv1\pmod3$ and $n\equiv1\pmod 4$, the Chinese Remainder Theorem implies that $n\equiv1\pmod{12}$ is the unique solution mod $12$.

As $5^{12}\equiv1\pmod{13}$ by Fermat's Little Theorem, it follows that $5^n\equiv5\pmod{13}$.

Title: Do we calculate the power with Chinese theorem?

Body:
For a natural numbr $n$ we know that $n=3k_1+1$, $n=4k_2+1$ and $n=5k_3+4$, for natural numbers $k_1, k_, k_3$.

Calculate $5^n\pmod {13}$ .

For that, do we apply the Chinese Theorem?


答:
The Chinese remainder theorem will give you a single statement on the modular constraint on $n$, to the modulus $\text{lcm}(3,4,5) = 60$, but the third constraint ($\text{mod } 5$) is not really needed since by Fermat's little theorem we know that Euler's totient $\phi(13) = 13-1 = 12$ sets the periodicity of powers of $a \bmod 13$.

$n\equiv (1,1) \bmod (3,4)\implies n\equiv 1 \bmod 12$  - (constant case CRT)

$5^n \equiv 5^1 \equiv 5 \bmod 13$


问:
Questions about rings and ideals
Given a ring $R$ and an ideal $I \triangleleft R$ i'm trying to make some conclusions about the quotient
$I/R$.

Let for instance $R=\mathbb Q[x]$ and $I=(x^4-16)$. I'm looking for a zero divisor unequal to 0 to show that the quotient $R/I$ is not a field. Is it true to say $x=\pm 2$ since $(\pm 2)^4$=16?

Now let $R = (\mathbb Z / 8 \mathbb Z)[x]$ and $I=(x^3+x+1)$. Again I would like to find a zero divisor to show that $R/I$ is not a field, but I would also like to show that $x \in R/I$ is a unit. Could someone model this answer for me? That would be very kind. I have incredible difficulty working on any $\mathbb Z/n \mathbb Z$ and I am not sure how to do either of this.

Finally, I thought of letting $R = \mathbb Z[x]$ and $I=(x^2+2)$. In this case, I would like to find an element that does not have an inverse, and show that the quotient $R/I$ is a identity ring. Here, I was considering two somewhat related cases. First of all, say I had $(x^2+2)$, then that would be $\mathbb Z[i]$. If it were $(x^2-2)$ then it would just be $\{ a+b \sqrt 2 : a,b\in\mathbb Z\}.$ Can I use either of these to conclude anything about $R/I$?

答:
First example. You should find a polynomial, but not substitute any number in $x$. $x^4 - 16 = (x^2 - 4)(x^2 + 4)$ so $x^2-4$ and $x^2+4$ are zero divisors.

Second one. $x(-x^2-1) = -x^3-x = 1$ so $x$ is a unit. To find a zero divisor, try to find a factorisation of $x^3+x+1$ over $\mathbb{Z}_8$. This is a key to your problem as it follows from the previous example.

The third one. Yes, $R/I = \mathbb{Z}[\sqrt{2}i]$. So the analogy with $\mathbb{Z}[i]$ can help.

Title: Questions about rings and ideals


问:
Simplify $\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$
$$\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$$

I an trying to simplify this as much as possible for $n = 24$.

I don't know if what I'm doing is right:

First of all I replaced ${n \choose n-k}$ with ${n \choose k}$

Then I distributed the sigmas like this:

$$\sum_{k=0}^n (k*(k-1)*{n \choose k}) = \sum_{k=0}^nk *\sum_{k=0}^n(k-1)*\sum_{k=0}^n{n \choose k}$$
Am I only allowed to do this in this step ?

Body:
We have:

$$\sum_{k = 0}^{n}k(k-1)\binom{n}{n- k} = \sum_{k=0}^{n}k(k-1)\frac{n!}{k!(n -k)!} = \sum_{k=0}^{n}\frac{n!}{(k-2)!(n-k)!}$$

$$=\sum_{k=0}^{n}\frac{n(n-1)(n-2)!}{(k-2)!\big((n-2)-(k-2)\big)!} = n(n-1)\sum_{k=0}^{n}\binom{n-2}{k-2}$$

Discarding the first $2$ terms, which are $0$:

$$=n(n-1)\sum_{k-2 = 0}^{n-2}\binom{n-2}{k-2} = \boxed{n(n-1)2^{n-2}}$$


问:
Let $U\subset\mathbb{C}$ an open set and $f$ a holomorphic function in $U-\{z_0\}, z_0\in U$.
I want to prove that $\int_{\partial R}f(z)dz=0$, where $R$ is any rectangle in $U$.

I tried to use Goursat's Theorem, but I can't conclude cause $f$ is not holomorphic in $z_0$. We have as hypotheses

f is holomorphic in $U-\{z_0\}$

$\lim_{z\to z_0}(z-z_0)f(z)=0$

Any hint?

答:
Note that $f$ is bounded in $U$ by the second condition. Say $|f| < M. $Split the rectangle R into many rectangles whose $R_0, R_1, R_2 \dots$ etc. Let $R_0$ be the rectangle containing $z_0$. While splitting $R$ we can choose to make $R_0$ as small as possible.

Note that $\int_R f = \sum_i \int_{R_i} f$. However the integrals all vanish except for $i>0$ as $f$ is holomorphic in $R_1, R_2$ etc. Also, $\left |\int_{R_0}f \right|$ can be bounded by arbitrarily small $\epsilon$ through the ML inequality. So, we can see that $\int_R f = 0$


问:
In-degree and out-degree distributions of a directed graph
For an undirected random graph with n nodes and probability p that any node u connects to any node v (apart from node u, i.e. no self-loops), the degree distribution is the binomial:

$$p_k = {n-1 \choose k} p^k (1-p)^{n-k-1}$$

However, what if the graph is random but directed? Are the in-degree and out-degree distributions the same as the above? Intuitively it feels like this is the case, but how do I show it?

答:
As long as edges are independently generated, we still get a binomial distribution for the in-degree and out-degree.

Specifically, there's two ways we can try to generate a random directed graph:

For each ordered pair $(u,v)$ with $u \ne v$, add a directed edge from $u$ to $v$ with probability $p$. Then the in-degree and out-degree of a vertex have the same binomial distribution for essentially the same reason. Say we're considering the in-degree of a vertex $v$. Then there are $n-1$ potential directed edges $(u,v)$ that could contribute: one for each possible $u$. Each of them exists independently with probability $p$, and their total number is the in-degree. That's exactly what the binomial distribution is for: the probability of in-degree $k$ is $$\binom{n-1}{k} p^k (1-p)^{n-1-k}.$$

For each unordered pair $\{u,v\}$, with $u \ne v$, add an edge between $u$ and $v$ with probability $q$, and direct it in one of the two possible directions with probability $\frac12$. This avoids having edges going in both directions between $u$ and $v$, which may or may not be something you want. Anyway, in this setup, the in-degree and out-degree are binomial: the probability of in-degree $k$ is $$\binom{n-1}{k}(\tfrac q2)^k (1 - \tfrac q2)^{n-1-k}.$$

The formula $\binom{n-1}{k} p^k (1-p)^{n-1-k}$ for random graphs does not come out of nowhere: we have $\binom{n-1}{k}$ ways to pick $k$ potential edges out of a vertex, $p^k$ probability that all of them are present, and $(1-p)^{n-1-k}$ probability that all the others are absent. As long as we can do the same in the directed case (which we can), the same formula holds.


问:
If $\rho(A) = \rho(B)$ and $\rho(A^2) = \rho(B^2)$ then $A$ is similar to $B$
I need to prove that if $\rho(A) = \rho(B)$ and $\rho(A^2) = \rho(B^2)$ then $A$ is similar to $B$. Assume also that $A,B$ are of dimension 9 and $\rho$ is the spectral radius and that $A,B$ are nilpotent with the same nilpotence index.

I tried to show they are diagonalizable and have the same diagonal but having problems finding the general form without knowing all the eigenvalues. Can we prove it this way? Or by any other way?

答:
Even with the additional conditions, this doesn't seem to be true.

Consider
$$
A=\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
and
$$
B=\begin{pmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
These two matrices are already in Jordan canonical form, but since the Jordan blocks are different, they cannot be similar.
These matrices both have nilpotence index $2$.

Generically, since nilpotent matrices can only have zero eigenvalues, $\rho(A) = \rho(B)$ and $\rho(A^2)=\rho(B^2)$ is trivially true. The nilpotence index is determined by the largest block in Jordan canonical form. Thus as long as the largest Jordan block in the JCFs of $A$ and $B$ are the same, they can satisfy all the conditions you gave and still not be similar.


答:
Recognize the sum you are given is the answer to the question of "Given a committee of size $n$, how many ways can we make a subcommittee of any size with a chairperson and a vice chairperson?"

It is the answer to that question by approaching the question in the following way:  First break apart based on the size of the subcommittee.  Let the size of the subcommittee be called $k$.  Then, pick who all isn't on the subcommittee at all (thereby picking who is on the subcommittee).  From those who weren't selected (and so are on the subcommittee) pick one of them to be the chairperson and then pick someone else to be the vicechair.

An alternate approach to counting can be done by from the original $n$ person committee first picking a person to be the chairperson for the subcommittee and then picking the vicechair from those left in $n(n-1)$ ways.  Now, for the remaining $n-2$ people, pick whether each was included or not included in the subcommittee.

This second approach yields an answer of:

$$n(n-1)2^{n-2}$$

By principles of double counting, since the original expression and this second expression both count the same scenario, they must be equal.  No messy algebraic manipulation needed.

Title: Simplify $\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$

Body:
$$\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$$

I an trying to simplify this as much as possible for $n = 24$.

I don't know if what I'm doing is right:

First of all I replaced ${n \choose n-k}$ with ${n \choose k}$

Then I distributed the sigmas like this:

$$\sum_{k=0}^n (k*(k-1)*{n \choose k}) = \sum_{k=0}^nk *\sum_{k=0}^n(k-1)*\sum_{k=0}^n{n \choose k}$$
Am I only allowed to do this in this step ?


问:
Separating a set with $\alpha$-property from outside points in a Hausdorff or Urysohn space
A subspace $A$ of a space X with topology $  \tau$ has the $\alpha$-property relative to $X$ if each
$\tau$-open family which covers $A$ has a finite subfamily whose union is $\tau$-dense in $A$.

I want to show that if $X$ is a Hausdorff space and $A\subset X$ has $ \alpha$-property, then if $x_{0} \in A^{\complement}$, there exist open disjoint subsets $U, V \subseteq X$ such that $x_{0} \in U$ and $A \subseteq V$

My attempt:Since  $X$ is a Urysohn space, for any points $a \in A$ and  $x_{0} \in X$  there exist disjoint open subsets $x_{0} \in U_{a}, a \in V_{a}$ so that $\operatorname{cl}(U_{a}) \cap \operatorname{cl}(V_{a}) = \emptyset $. Consider the collection $ \{ V_{a} | a \in A \}$. This forms an open cover for
$A$ and, since $A$  has $\alpha$-property, $\{ V_{a} | a \in A \}$  has a finite subcover, such that $$A \subseteq \bigcup_{i=1}^n \operatorname{cl}(V_{a_i}) =: V$$ for some $a_1,\ldots,a_n \in A$. Recall that each $V_{a_i} ,(1\leq i \leq n)$ has a corresponding $U_{a_i}$. So consider the intersection of the collection of these $U_{a_i}$: $U:= U_{a_i} \cap \ldots \cap U_{a_n}$. We know that $U$ is open and $x_{0} \in U$, but how to say $V$ is open and $V \cap U = \emptyset$?

If this is true, can Urysohn be replaced by Hausdorff? (In topology, a Urysohn space is a topological space in which any two distinct points can be separated by closed neighbourhoods. Note that a "closed neighbourhood of $x$" is a closed set that contains an open set containing $x$.)

答:
First note that a more common name for a set $A$ with the $\alpha$-property relative $X$ is an "$H$-set in $X$", see this paper by Johannes Vermeer, who quotes an earlier paper by Velicko for this definition.

In a Hausdorff space such an $H$-set is closed, as is easily seen: Fix $p \notin A$ and for each $a \in A$, pick disjoint open neighbourhoods $U_a$ of $a$ and $V_a$ of $p$. As in your attempt, we have finitely many $F \subseteq A$ such that $$A \subseteq \bigcup_{a \in F} \overline{U_a}$$ and then $V=\bigcap_{a \in F} V_a$ is open and misses $A$: if $x \in A \cap V$ then $x \in A$ so $x \in \overline{U_a}$ for some $a \in F$, but $U_a \cap V_a = \emptyset$ implies $\overline{U_a} \cap V_a = \emptyset$ so $x \notin V_a$ and so $x \notin V$ a fortiori, contradiction.

The same argument shows that in your attempt $U \cap V = \emptyset$ too, but your $V$ is a closed set (a finite union of closed sets) so we cannot expect it to be open, and so does not work in general as the required $V$. If we leave out the closures in the union we get an open set, but it need not cover $A$...

So my hunch is that a counterexample to your main question might very well exist in some non-regular, Urysohn $X$ where the counterexample in regularity uses an $H$-set as the closed subset. I'll have a further think.


答:
I continue to do some simplifications and I got that
$$y=\frac{x^2}{(y-x(1+\sqrt{2})(y-x(1-\sqrt{2})}$$

This is not really explicit solution. I hope that might help. By the way, I did not find find the constant when I did the partial fraction.

I want the integration part

$$\int \frac{\text{dv}}{v(v^2+2v-1)}= A\text{ln}v+B~ \text{ln}(v-(-1+\sqrt{2})+C~\text{ln}(v-(-1-\sqrt{2})$$

Title: Solve $x^3y'= y^3 + 2xy^2$

Body:
I found the following exercise:

$$\text{Solve: } \ x^3y'= y^3 + 2xy^2$$

The idea I used:

$$\text{Let } y(x) = x\cdot v(x)$$

After some computations:

$$\int \frac{\text{dx}}{x} = \int \frac{\text{dv}}{v(v^2+2v-1)}$$

After computing the integrals, it becomes:

$$\frac{\ln(|v^2 + 2v - 1|)}{2} + \frac{\ln(|v-\sqrt{2}+1|) -\ln(|v+\sqrt{2}+1|)}{2\sqrt{2}} - \ln(|v|) = \ln(|x|) + C$$

After that we will substitute back: $v$ with $\frac{y}{x}$, in the relation above.

Question: What magic could be applied to it as to be able to find $y = \text{...}$?

Did I have a bad start? How to solve this?

Thank you very much!


问:
In how many ways can we distribute the black balls
We have 10 similar black balls. How many ways are there to distribute them into 6 boxes, each box is large enough for all balls.

My idea is $$\binom{10+6-1}{6-1}=\binom{15}{5}=3003$$ is that correct?

答:
Your solution is correct. I like to view it as arranging $15$ objects in a row, the objects being the $10$ balls and $5$ partitions. This is the so-called "Stars and Bars" method of counting.

One example would be $\bullet|\bullet\bullet|\bullet|\bullet\bullet\bullet||\bullet\bullet\bullet$.

This arrangement has $1$ ball in box $b_1$, $2$ balls in box $b_2$, and so on. Notice that there are $0$ balls in box $b_5$, as indicated by the two adjacent partitions.

There are $\binom{15}{5}$ ways to arrange the balls and partitions, as you said in the question.


问:
Norm of the operator and Riesz-Markov-Kakutani measure - example functional

Let $X = C[0,1]$ with supremum norm. Define linear functional:
$$
\varphi(f) = \int_0^1f(x)cos(\pi x)dx - f(\frac{1}{2})
$$
Show that $\varphi$ is bounded and compute its norm. Using the Riesz–Markov–Kakutani representation theorem give the complex measure that is associated with that functional.

My attempt was:

Bounded:
$$
\|\varphi\| = \sup_{\|f\|_{\infty}=1}|\varphi(f)| = \sup_{\|f\|_{\infty}=1}|\int_0^1f(x)cos(\pi x)dx - f(\frac{1}{2})| = \\
\leq \sup_{\|f\|_{\infty}=1}\int_0^1|f(x)cos(\pi x)|dx + |f(\frac{1}{2})| \\
\leq \sup_{\|f\|_{\infty}=1}\|f\|_{\infty}\int_0^1|cos(\pi x)|dx + \|f\|_{\infty} \leq 1 + 1 = 2
$$
Where last equation is due to cosine function being bounded by 1.

Now the norm, letting $f = 2$ (constant):
$$
\|\varphi(f)\| = |2\int_0^1cos(\pi x) - 2| = |0 - 2| = 2
$$
so the norm (taking the bound into account) is equal to 2.

Lastly, the measure seems to be:
$$
\mu(A) = \cos{(\pi x)}d\lambda(A) - \delta_{\frac{1}{2}}(A)
$$
because then
$$
\int_0^1fd\mu = \int_0^1fd\cos{(\pi x)}- \delta_{\frac{1}{2}} = \int_0^1f(x)cos(\pi x)dx - f(\frac{1}{2})
$$

It was an exam task and I only got half of the points. Could anyone show me the correct answer? I suppose my measure is wrong ($\lambda(A)$ doesn't make much sense) but I didn't have better ideas that time.

答:
Turns out:

Bound can be made tighter:

$$
|\int_0^1f(x)\cos(\pi x)dx - f(\frac{1}{2})| \leq \|f\|_{\infty}\frac{2}{\pi} + \|f\|_{\infty} \implies \sup_{\|f\|_{\infty}=1} \varphi(f) \leq \frac{2}{\pi}+1
$$

Proper measure is:

$$
\mu(A) = \int_A\cos{\pi x}dx - \delta_\frac{1}{2}(A)
$$

using Riesz-Markov-Kakutani measure we know that the norm of the functional equals to the variation of the measure

$$
|\mu|(X) = \int_0^{\frac{1}{2}}\cos{\pi x}dx + \int_{\frac{1}{2}}^1-\cos{\pi x}dx + 1 = \frac{2}{\pi} + 1 = \|\varphi\|
$$


问:
Choose a group of 5 people to create a committee
We have 10 boys, 8 girls and 4 teachers.

We want to create a group of 5 with 1 teacher, 1 girl and the remaining 3 can be anybody of them.

So is it $$\binom{4}{1}\binom{8}{1}\binom{10+(8-1)+(4-1)}{3}$$?

Tje first term for a teacher, the second for the girl and the last of the remaining three members of the group.

答:
The problem with your method is that, as pointed out in the comments, it counts groups with multiple girls or teachers multiple times. The way to go about solving this kind of problem is to first find the total number of groups of $5$, and then subtract the number of invalid groups. A group is invalid if it contains no girls or no teachers, so to find the number of invalid groups, we add the number of groups with no girls and the number of groups with no teachers, and then, since this double counts the groups with no girls and no teachers, we subtract the number of groups with only boys. This gives us:

$$\underbrace{\binom{22}{5}}_{\text{total}} - \underbrace{\binom{14}{5}}_{\text{no girls}} - \underbrace{\binom{18}{5}}_{\text{no teachers}} + \underbrace{\binom{10}{5}}_{\text{only boys}}$$


问:
Which is canonical $|D|\cong \mathbb{P}(\Gamma(X,\mathscr{O}_X(D))^{\vee}$) or $|D|\cong \mathbb{P}(\Gamma(X,\mathscr{O}_X(D))$)?
Sorry for my bad English.

Let $X$ be a smooth projective curve over an algebraically closed field $k$,
and $D$ is divisor on $X$.

By The stacks project,
$|D|\cong \mathbb{P}(\Gamma(X,\mathscr{O}_X(D))^{\vee})$ .

On the other hand, Hartshorne say
$|D|\cong \mathbb{P}(\Gamma(X,\mathscr{O}_X(D)))$.

Which is canonical?

Please tell me, thanks.

答: Your confusion here is actually just about the definition of $\mathbb{P}(V)$ for a vector space $V$.  In elementary contexts, it is common to define $\mathbb{P}(V)$ as the quotient of $V\setminus\{0\}$ by the action of nonzero scalars.  However, in algebraic geometry, it is instead defined as the quotient of $V^\vee\setminus\{0\}$ by the action of nonzero scalars.  Or more precisely, as a scheme, $\mathbb{P}(V)$ is defined as $\operatorname{Proj}(\operatorname{Sym}(V))$.  Note that this means that closed points of $\mathbb{P}(V)$ come from maximal homogeneous proper subideals of the irrelevant ideal of $\operatorname{Sym}(V)$.  Such ideals are generated by codimension $1$ linear subspaces $W$ of $\operatorname{Sym}^1(V)=V$, which (by identifying the quotient $V/W$ with $k$) correspond to nonzero homomorphisms $V\to k$ up to scalar multiples.  That is, they correspond to elements of $V^\vee\setminus\{0\}$ up to scalar multiples.  So, with this definition of $\mathbb{P}(V)$, the correct statement is $|D|\cong \mathbb{P}(\Gamma(X,\mathscr{O}_X(D))^{\vee})$


问:
Tails of family of integrable functions
It is well known that tail of an integrable function on $\mathbb{R}^d$ is small, i.e., Given $\epsilon>0$, there is  $R>0$ such that $$\int_{\{|x|>R\}}|f(x)|dx<\epsilon.$$

I was wondering whether there were any simple conditions under which a family or a sequence of integrable functions $\{f_n\}$ would have small tails. Something like: given $\epsilon>0$, there is $R>0$ such that $$\int_{\{|x|>R\}}|f_n(x)|dx<\epsilon~\mbox{for all}~n.$$

It looks similar to uniform integrability but there we have integration on a finite set.

答:
The property seems to behave bad with uniform integrability (take $f_n:=n\chi_{B(0,1)}$).

If $S_N:=\{ x,N\leqslant x<N+1\}$, then your condition means that
$$\sup_n\sum_{k\geqslant L}a(n,k)\to 0,$$
where $a(n,k):= \int_{S_k}|f_n|$. We can't say that $\{(a(n,k),k\geqslant 1),n\geqslant 1\}$ is compact in $\ell^1$ because boundedness is missing.

Title: Tails of family of integrable functions


问:
convergence of sum of "poisson many" Bernoulli RV
Let $(\xi_n)_{n \geq 0}$ be a sequence of iid. Bernoulli random variables with $\mathbb{P}(\xi_k = 1) = \frac{1}{2}$ and $\mathbb{P}(\xi_k = -1) = \frac{1}{2}$. Let $N$ be poisson with parameter $\lambda > 0$. Let $Z_{\lambda} = \xi_1 + \xi_2 + ... + \xi_N$. Compute the characteristic function of $Z_{\lambda}$ and show that $\frac{Z_\lambda}{\sqrt{\lambda}}$ converges in distribution to a standard normal random variable as $\lambda$ goes to $\infty$.

I have computed the characteristic as follows: $\mathbb{E}\left[\exp(itZ_\lambda)\right] = \mathbb{E}\left[\exp(it\xi_1)\right] \cdot...\cdot \mathbb{E}\left[\exp(it\xi_N)\right]$

and $\mathbb{E}\left[\exp(it\xi_1)\right] = \frac{1}{2}\left(\exp(it)+exp(-it)\right) = \cos(t) $

hence $\mathbb{E}\left[\exp(itZ_\lambda)\right] = \cos(t)^N$.

Concerning the convergence I thought I could use the characteristic function since I have already computed it, i.e. show that $\mathbb{E}\left[\exp(itZ_\lambda)\right] = \exp\left(-\frac{t^2}{2}\right)$. How can I get there?

We are given the HINT: $\cos(t) = 1- \frac{t^2}{2} + O(t^4)$ as $t \rightarrow 0$

答:
Your expression $\cos(t)^N$ isn't $\mathbb E[\exp(itZ_\lambda)]$ -- more precisely, we have
$$\mathbb E[\exp(itZ_\lambda)\mid N]=\cos(t)^N.$$
To actually find $\mathbb E[\exp(itZ_\lambda]$, we can use the law of total expectation to get
$$\mathbb E[\exp(itZ_\lambda)]=\mathbb E[\mathbb E[\exp(itZ_\lambda)\mid N]]=\mathbb E[\cos(t)^N]=G(\cos t)=\exp(\lambda[-1+\cos t]),$$
where $G(z)$ is the PGF of $N\sim\operatorname{Poisson}(\lambda)$. Then considering $Z_\lambda/\sqrt{\lambda}$, and taking $\lambda\to\infty$,
\begin{align*}
\mathbb E[\exp(itZ_\lambda/\sqrt\lambda)]=\exp\left(\lambda\left[-1+\cos\frac{t}{\sqrt{\lambda}}\right]\right)&=\exp\left(\lambda\left[-\frac{t^2}{2\lambda}+O\left(\frac{t^4}{\lambda^2}\right)\right]\right)\\
&\to\exp(-t^2/2),
\end{align*}
which is what we wanted.


问:
Asymptotic expression for a fraction
Assume that $x,y\in\mathbb{N}$ with $x>y+1$.

Moreoever, let $v=v(x,y)$ and $w=w(x,y)$ with $w(x,y)\to 0$ as $x\to\infty$.

Let
$$
x-y=\frac{\ln(1+v)}{\ln(1+w)}
$$
and assume that $x-y$ is constant. Does this imply that
$$
v\sim w\textrm{ as }x\to\infty?
$$

Intuitively, I think that this is correct, since if the fraction is constant, then $v$ should "behave as $w$".

In order to actually prove the statement, I have to show that $$\lim_{x\to\infty}\frac{v}{w}=1.$$

I do not get that but setting $C:=x-y$, I get
$$
x-y=\frac{\ln(1+v)}{\ln(1+w)}\Leftrightarrow (1+v)=(1+w)^C\Leftrightarrow v=(1+w)^C-1
$$
so that
$$
\frac{v}{w}=\frac{(1+w)^C}{w}-\frac{1}{w}
$$

答:
The statement $v\sim w$ as $x\to\infty$ is in general not true. More specifically,  with l'Hospital,
$$
\frac{v}{w}=\frac{(1+w)^C-1}{w}\to C\quad \textrm{ as } x\to\infty
$$
so the limit of the fraction needs not to be $1$ in general.

However,what you get is that $v\in O(w)$ as $x\to\infty$.

Title: Asymptotic expression for a fraction


问:
Polynomial differential equation
I came across this problem in an old olympiad paper (Putnam?)

Find all polynomials $p(x)$ with real coefficients satisfying the differential equation

$7\dfrac{d }{dx } [xp(x)]=3p(x)+4p(x+1)$   $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\infty<x<\infty$

I didn't find any "official" solution on internet and it would be interesting to see your approaches on this one.

My approach was the following , i don't know if i'm right.
Consider the leading coefficient $a_n$ on both sides. Then we have $7na_n=7a_n$.
Suppose $a_n\neq 0$ then we get $n=1$. I plugged in the general solution $p(x)=a_0+a_1x$ in the differential equation and solved for $a_1$ wich turned out to be $0$ , contradicition.
The other option is that $a_n=0$ but then $a_{n-1 }$ would be the leading coefficient of the polynomial wich would be again $0$ (by the same reasoning). Hence by induction the only non zero leading coefficient must be $a_0$.

Could you tell me if this is ok? Thanks in advance.

答:
By the product rule
$$ \frac d{dx}\bigl(xp(x)\bigr) = xp'(x) + p(x) $$
so we get
$$ 7xp'(x) = 4\bigl(p(x+1) - p(x)\bigr) $$
Obviously, constant $p$s are a solution, if $p$ were not constant, $n = \deg p \ge 1$, say, we have that the degree of $p(x+1) - p(x)$ is less than $n$. On the other hand, $\deg p' = n-1$, hence the degree of $xp'(x)$ is $n$. But that is not possible, so constant polynomials are the only solutions.</p>

Title: Polynomial differential equation


问:
Do you find this proof convincing?
I must show that the function $f:\mathbb R^n \to \mathbb R$ defined as $ f(\mathbf x)=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2 $ for fixed $\mathbf a_1,...,\mathbf a_k \in\mathbb R^n$ has a global minimum. Here's my reasoning:

The function $f$ is continuous.
Pick $\delta \gt 0$, by definition closed balls (neighborhoods) are compact; so define $\bar{B}_n(\mathbf 0,n\delta) =\{ \mathbf x\in\mathbb R^n :\Vert\mathbf x\Vert \le n\delta\}$ for $n \in \mathbb N$, and define $f_n:\bar{B}_n(\mathbf 0,n\delta) \to \mathbb R$ as $f_n=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2$.

By the maximum/minimum value Theorem there is a $\mathbf y \in \bar{B}_n(\mathbf 0,n\delta) $ such that $f_n(\mathbf y)\le f_n(\mathbf x)$ for all $\mathbf x \in \bar{B}_n(\mathbf 0,n\delta)$. Call $f_n(\mathbf y)= y_n$.

For all $k \in \mathbb N$ we have: $y_{n+1} \le y_n$ because $\bar{B}_k(\mathbf 0,k\delta) \subset \bar{B}_{k+1}(\mathbf 0,(k+1)\delta)$, and $y_k \ge 0$ because $f_k(\mathbf x) \ge 0$ for all $\mathbf x$s in the domain; so the sequence $\{y_n\}$ is bounded below and non increasing, hence it converges, as $n \to \infty$, to a number $y$.

$y$ is the global minimum of $f(\mathbf x)=\sum_{j=1}^{k} \Vert \mathbf x -\mathbf{a}_j\Vert^2$.

I'm not so sure if I can claim that $y$ is indeed the global minimum of the function $f$

EDIT:

I don't seem to be able to complete the proof with my method, also I'm thinking that if I prove there is a closed ball big enough (with all the $\mathbf a_i$s in it) such that every element $\mathbf q$ outside the ball has its $f( \mathbf q)$ greater than the minimum inside then all the work above is superfluous.

People in the comments and in the answers have provided a more efficient strategy:

Let $f(\mathbf a_1)=\sum_{j=2}^{k} \Vert \mathbf a_1 - \mathbf a_j\Vert^2=M$, and define the closed ball $\bar{B}(\mathbf a_1,\sqrt{M})= \{ \mathbf x \in \mathbb R^n: \Vert \mathbf x - \mathbf a_1 \Vert \le \sqrt{M} \}$.

By the minimum value theorem there is some $\mathbf y$ in $\bar{B}(\mathbf a_1,\sqrt{M})$ such that $f(\mathbf y) \le f(\mathbf x)$ for all $\mathbf x$ in $\bar{B}(\mathbf a_1,\sqrt{M})$.

Pick any element $\mathbf q \notin \bar{B}(\mathbf a_1,\sqrt{M})$, which means that $0 \le \sqrt{M} \lt \Vert \mathbf q - \mathbf a_1 \Vert$ and thus $M \lt \Vert \mathbf q - \mathbf a_1 \Vert ^2$.

By definition we have $f(\mathbf y) \le f(\mathbf a_1) = M$, but it's also true that $\Vert \mathbf q - \mathbf a_1 \Vert \le f(\mathbf q)$. Therefore $f(\mathbf y) \lt f(\mathbf q)$ which completes the proof.

答:
The sequence $\{y_n\}$ does indeed converge to some real number $y$, but that does not guarantee that $y=f(x)$ for some $x\in\Bbb{R}^n$.

Instead, continue your argument by noting that the sequence $\{y_n\}$ is eventually constant; once all the $a_i$ are contained in the closed ball, the minimum will not decrease as the ball grows further.

Here's a sketch of a slightly slicker proof using the same idea:

Take any point $p\in\Bbb{R}^n$ and let $M:=f(p)$. Let $B:=\bar{B}(a_1,\sqrt{M})$ be the closed ball of radius $\sqrt{M}$ centered at $a_1$. Then for any point $q$ not in in $B$ we have
$$f(q)\geq||q-a_1||^2>M.$$
Now apply your minimum/maximum theorem to $B$ to get a global minimum.

Title: Do you find this proof convincing?


问:
If there is a $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $\langle x,y\rangle=0$ for all $y\in V$, then $\overline V\ne\ell^q$?
This should be quite obvious, but I struggle to find a formal proof: Let $1\le p<q\le\infty$ with $\frac1p+\frac1q=1$. We know that $\ell^q$ is isometrically isomorphic to $(\ell^p)'$ via $$\iota:\ell^q\to(\ell^p)'\;,\;\;\;y\mapsto\langle\;\cdot\;,y\rangle,$$ where $$\langle x,y\rangle:=\sum_{n\in\mathbb N}x_ny_n\;\;\;\text{for }(x,y)\in\ell^p\times\ell^q.$$

Assume $x\in\ell^p\setminus\{0\}$ and $V\subseteq\ell^q$ with $$\langle x,y\rangle=0\;\;\;\text{for all }y\in V\tag1.$$ Why can we conclude that $\overline V\ne\ell^q$?

答: Let $X$ be a real or complex Banach space, $x\in X\setminus\{0\}$ and $V\subseteq X'$ with $x\in V_\perp$. By the Hahn-Banach theorem, there is a $\varphi\in X'$ with $$\langle x,\varphi\rangle=\left\|x\right\|_X\ne0.\tag1$$ If $\overline V=X'$, then there is a $(\varphi_n)_{n\in\mathbb N}\subseteq V$ with $$\left\|\varphi_n-\varphi\right\|_{X'}\xrightarrow{n\to\infty}0\tag2$$ and hence $$0=\langle x,\varphi_n\rangle\xrightarrow{n\to\infty}\langle x,\varphi\rangle\ne0\tag3;$$ which is impossible. This shows the even stronger result that even the closure of $V$ wrt the weak topology cannot be $X'$.


问:
Is the set of indistinguishable points measurable?
Let X be an uncountable set.

Let S be a countable set of uncountable subsets of X (i.e. there are countably many elements s in S, and each element s in S is an uncountable subset of X). Let R(S) be the sigma-algebra generated by S.

Let x and y be elements of X. We say that x is "indistinguishable from y" if any set s in S contains either both x and y or neither x nor y. A point x in X is "indistinguishable" if there exists some y in X from which it is indistinguishable. Let T be the subset of X which contains all indistinguishable points. (Note that some elements of T can be distinguishable from each other).

Claim 1: T is an element of R(S).

Is Claim 1 true? Thanks!

答:
No.

In this answer the $S$ in your question is denoted as $\mathcal S$.

Let $X$ be an uncountable set and let $A\subseteq X$ be countable and containing at least two elements.

Let $\mathcal S\subseteq\mathcal P(X)$ such that $B\in\mathcal S$ iff $B$ is countable with $B\cap A=\varnothing$ or $B^{\complement}$ is countable with $B^{\complement}\cap A=\varnothing$.

Then $\mathcal S$ is a $\sigma$-algebra, hence coincides with the $\sigma$-algebra generated by $\mathcal S$.

This with $A\notin\mathcal S$ while $A$ is exactly the set of indistinguishable elements.


问:
If all residue fields $A/m$ have degree $\geq2$ then $A$ is generated by elements $a-a^q$
I found the following claim which I haven't been able to prove.

Let $q$ be a power of a prime $p$. Let $A$ be a finitely generated $\mathbb{F}_q$-algebra and suppose that for every maximal ideal $m\subset A$ we have $[A/m:\mathbb{F}_q]\geq 2$. Then the set $\{a-a^q\}$ generates the unit ideal.

I was wondering if anyone could provide a hint (even for a simpler case, like $A$ being semilocal) so I can work out the details, because I haven't found a way to proceed.

Thanks in advance.

答:
Hints:

Prove that for an extension $K/\Bbb F_{q}$ with $[K:\Bbb F_q]\geq 2$ there are elements $x\in K$ such that $x-x^q\ne0$.

By localizing reduce to the case where $A$ is a local ring and use 1.

Edit: Let $I$ be the ideal in $A$ generated by the elements of the form $a-a^q$ where $a\in A$. Let $m$ be a maximal ideal of $A$. It suffices to show that the localized ideal $I_m$ is the ideal $J$ of $A_m$ generated by the elements of the form $x-x^q$ where $x\in A_m$ as we may then apply the argument in the local ring case to conclude $I_m=J=A_m$ and hence $I=A$.

Obviously we have $I_m\subseteq J$. Let $x=\frac{a}{s}\in A_m$ with $s\in A\setminus m$. We have$$x-x^q=\frac{a}{s}-\frac{a^q}{s^q}=\frac{as^q-sa^q}{s^{q+1}}=\frac{as^q-as+as-sa^q}{s^{q+1}}=\frac{1}{s^{q+1}}\left(a\underbrace{(s^q-s)}_{\in I}+s\underbrace{(a-a^q)}_{\in I}\right)$$
Hence $x-x^q\in I_m$ and we get $I_m=J$.


问:
Eigenvalues of a matrix in relation to another matrix
Let $A,B \in \mathcal{M}_n(\mathbb{K}) $  (where $\mathbb{K} \in \{\mathbb{R},\mathbb{C} \} $) invertible matrices. Prove that if $\lambda = 0 $ is an eigenvalue of matrix $C$, where $C = BA - AB ~$ then $ \lambda = 1 $ is an eigenvalue of matrix $D~$ where $D = B^{-1}A^{-1}BA$

I don't know where to start this proof, since I can't recall any eigenvalue properties that relate sum or diference of matrices.

答: Hint: Consider the determinant of $B^{-1}A^{-1}C$.


答: Hint: If $\lambda = 0$ is an eigenvalue of $C$, then there must exist an associated eigenvector. That is, there is a (non-zero) vector $v$ for which $ABv = BAv$.

Title: Eigenvalues of a matrix in relation to another matrix

Body:
Let $A,B \in \mathcal{M}_n(\mathbb{K}) $  (where $\mathbb{K} \in \{\mathbb{R},\mathbb{C} \} $) invertible matrices. Prove that if $\lambda = 0 $ is an eigenvalue of matrix $C$, where $C = BA - AB ~$ then $ \lambda = 1 $ is an eigenvalue of matrix $D~$ where $D = B^{-1}A^{-1}BA$

I don't know where to start this proof, since I can't recall any eigenvalue properties that relate sum or diference of matrices.


问:
Alternating series - absolute convergence?
The question is, whether the following series is convergent, absolutely convergent or divergent
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n-1}.$$

The term $\dfrac{(-1)^{n+1}}{2n-1}$ = $ \dfrac{(-1)\cdot(-1)^{n}}{2n-1}$ = $(-1)^{n} \dfrac{(-1)}{2n-1}$.

So according to Leibnitz-Criterion, if the series $ \dfrac{(-1)}{2n-1}$ is a Zero-sequence, then the alternating series converges, too.

The series $\sum_{n=1}^\infty \left|\dfrac{(-1)^{n+1}}{2n-1}\right|$ = $\sum_{n=1}^\infty \dfrac{1}{2n-1}$ converges.
Therefore the series is absolutely convergent.

Is this correct?

答:
$\sum_{n=1}^\infty \frac{1}{2n-1}$ is not convergent as   </p>

<p>$\sum_{n=1}^\infty \frac{1}{2n-1} \ge \sum_{n=1}^\infty \frac{1}{2n}$ and the series $\sum_{n=1}^\infty \frac{1}{n}$ is not convergent</p>

Body:
$\sum_{n=1}^\infty \frac{1}{2n-1}$ is not convergent as

$\sum_{n=1}^\infty \frac{1}{2n-1} \ge \sum_{n=1}^\infty \frac{1}{2n}$ and the series $\sum_{n=1}^\infty \frac{1}{n}$ is not convergent


问:
$\xi$ is continuous but not necessarily analytic
If $\xi:\mathbb{C}-\{0\}\to \Bbb{C}$ is a function such that $\xi(z)=\xi(\frac{z}{|z|})$ and its restriction to the unit circle is continuous, then $\xi$ is continuous but not necessarily analytic.

I didn't find anything to start this problem. Can I find some hints?

Any help will be appreciated.

答:
$\xi$ is the composition of two continous functions

An analytic function that is constant on the positive real line must be constant.


答:
$$B^{-1}A^{-1}BAx = B^{-1}A^{-1}ABx = 1x $$
for some $x\neq 0$ satisfying $ABx=BAx$.


问:
10 elements in 10 by 10 grid, no two in the same row or column
I have 10 elements that need to be arranged in a 10 by 10 grid such that no two elements occupy the same row or column.

I know the two diagonals work, and from looking at a smaller 3 by 3 grid I can find at least two other arrangements that fulfill these requirements. But I'm trying to find a general rule for this.

edit: how would this be different if they were indistinguishable vs distinct

答:
Hint:

You have $10$ elements to go into $10$ rows so one in each row

Similarly with the columns

so

Put one element in the first column in any of the $10$ rows

Put one element in the second column in any of the $9$ free rows

and so on with the rest of the elements and columns and remaining rows

until you put the last element in the last column in the last remaining row

How many ways in total can you do this?


问:
How can I "see" that calculus works for multidimensional problems?
Let's say I have some function f(x) = x^2 + b. I can see what's going on, I can count the slope geometrically even without knowing the rules of derivatives.

When I need to minimize some cost function for linear problem (linear regression with gradient descent), I just need to picture 3d, and I can "see", and be quite confident why and how it works.

How can I "see" or get intuition that calculus works for multidimensional problems? Let's say I have a problem with many variables like:

f(area_km, bedrooms, ...) = theta + theta1 * area_km + theta2 * bedrooms etc

If I want to apply gradient descent, I know I need to calculate partial derivatives and multiply it with a learning rate etc. It works. But it's kinda magical that it works.

I am sorry this is a silly question, I am just beginning.

答:
Well, I don't believe that it is easy to maintain an accurate model of a $>3$ dimensional space in one's imagination for sustained periods, but one can certainly gain an intuition for each $3$D cross-section and maneuver around that way. Pick easy shapes, like spheres or tetrahedrons, to try and visualize. The issue is mostly visualization, since our minds are in constant interaction with multidimensional spaces in day-to-day life, though those spaces are simply processed less visually.

Dreams, for example, compose many dimensions of the waking life and compress them into highly dynamic visualizations of usually $<3$ dimensions. As for generalization, typically I take a linear variable simply map it to multilinear vectors or parametric spaces. It's helpful to know that the intersection of $k$ $n$ dimensional surfaces is a $n-k+1$ dimensional surface in $n$ space, for example.

It is fundamentally very difficult to communicate understanding, so the best advice is to simply persist at trying to visualize these spaces in your own unique way. There are some who have entertained multidimensional spaces in their work for so long that they are simply, precisely familiar with $4$D (and up) landscapes. The more you persist, the more dimensions you will add. Sometimes, after a lifetime of persistence, you only end up with a half a dimension more than three. Other times, you go all the way to eight. It depends.

So, the best method, I think, is to pick simple shapes that are easy to generalize into higher dimensions (like spheres) and visualize each part until you can put it together. Then slowly expand your horizons. Manipulate the sphere here or there to change it this way or that. Soon enough what seems impossible to many will be an average consideration for you.


问:
Statistics question: Estimating mean when standard deviation is known
I am reading a textbook to learn more about statistics. This section is about estimating the mean of a population when standard deviation of the population is known.

My simple question is this: How can we know the standard deviation of a population without knowing the mean? My understanding is that to calculate the standard deviation you need to know the mean. Or is there some other way, i.e. without the formula, to calculate the standard deviation?

Any help is appreciated.

Thanks.

答: Let's say you have a normally distributed random variable that you can't measure directly, but you CAN measure the difference between two of those variables.  For example, you have two people measuring the exact same quantity (eg., two people timing the same race with stopwatches).  Their measurements have error and you don't know the true value of the quantity of interest (eg., actual race time) but you have pairs of measurements of the identical quantity.  If you assume the measurement error in a sufficiently large sample of timers is normally distributed, then you can use the measurement pairs to estimate the standard deviation of the measurement error even though you don't know the mean.  Then you can use a separate experiment (with a much smaller sample) with more expensive equipment to estimate the mean measurement error given the known deviation.</p>


问:
Push forward and pullback in products
I am reading this Questions about Serre duality, and there is one part in the answer that I'd like to know how it works. But after many tries I didn't get anywhere. So here is the problem.

Let $X$ and $B$ be algebraic varieties over an algebraically closed field, $\pi_1$ and $\pi_2$ be the projections from $X\times B$ onto $X$ and $B$, respectively.  Then it was claimed that $R^q\pi_{2,*} \pi_1^* \Omega_X^p \cong H^q(X, \Omega^p_X)\otimes \mathcal{O}_B$.

I am guessing it works for any (quasi)coherent sheaf on $X$.

Basically, I have two tools available, either Proposition III8.1 of Hartsshorne or going through the definition of the derived functors.

Thank you.

答: Using flat base-change (Prop. III.9.3 of Hartshorne), one sees that $$R^1\pi_{2 *} \pi_1^*\Omega_X^p = \pi_1^* H^q(X,\Omega^p_X) = H^q(X,\Omega^p_X)\otimes \mathcal O_B.$$</p>

Title: Push forward and pullback in products


问:
determine whether an integral is positive
Given a standardized normal variable $X\sim N\left(0,1\right)$, and constants $ \kappa \in \left[0,1\right)$  and  $\tau \in \mathbb{R}$, I want to sign the following expression:

\begin{equation}
\int_{-\infty}^\tau \left(X-\kappa \tau \right) \phi(X)\text{d}X
\end{equation}

where $\phi$ is the PDF of $X$.  Any comment will be appreciated. I would at least want to know if the sign of the expression can be determined given the information, or whether it hinges on the value of $\tau$.

答:
Note that
$$ \int_{-\infty}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX=E[X-\kappa\tau]=-\kappa\tau$$
and for $\tau>0$
$$ \int_{\tau}^\infty (X-\kappa\tau)\phi(X)\,\mathrm dX\ge\int_{\tau}^\infty (1-\kappa)\tau\phi(X)\,\mathrm dX>0.$$
So for $\tau>0$ your expression is positive.</p>


答: Put them in $(i, p(i))$ for $i = 1 \ldots 10$ where $p$ is any permutation of $1,\ldots,10$.

Title: 10 elements in 10 by 10 grid, no two in the same row or column

Body:
I have 10 elements that need to be arranged in a 10 by 10 grid such that no two elements occupy the same row or column.

I know the two diagonals work, and from looking at a smaller 3 by 3 grid I can find at least two other arrangements that fulfill these requirements. But I'm trying to find a general rule for this.

edit: how would this be different if they were indistinguishable vs distinct


问:
Why a continuous map $f : M \to \mathbb{S}^n$ can be approximated by a smooth function?
Let $M \subset \mathbb{R}^N$ be a smooth manifold of dimension $m$ where $m<n$. Suppose $f : M \to \mathbb{S}^n$ is a continuous map. How can I prove that for every $\varepsilon >0$ I can find a smooth function $f_{\varepsilon}:M \to \mathbb{S}^n$ with $\sup\limits_{x \in M} \|f_{\varepsilon}-f\| < \varepsilon$?

I guess that might help the theory of "partition of unity" to construct such a $f_{\varepsilon}$ but I do not know it and I would like to prove the assertion without it.

I started from the case when $M$ is compact and I easily concluded using Stone-Weiestrass result. But how can I conclude in general case?

I will appreciate any help or suggestions.

答: Using a smooth partition of unity, it suffices to consider the case that $M$ is $\Bbb R^n$ or even just the unit ball therein, and that the image of $M$ is contained in a hemisphere, say. Folding $f$ with a smooth kernel with tiny support (and normalizing) will then make $f$ smooth.


问:
How many eight digit numbers contains all of 2,4,6,8?
I thinking about $4^3$ $*$ $9^3$ because 4 numbers has the conditon of being only 2,4,6,8 and the other 4 has 9 digit options. How do I need to approach this kind of things?

答:
Let $X$ be the set of $8$-digit numbers.  Let $X_i$ be the set of $8$-digit numbers who include the digit $i$ as at least one of its digits.

You ask to count $X_2\cap X_4\cap X_6\cap X_8$

To do this, let us instead look at the complementary event with respect to $X$... $\overline{X_2}\cup \overline{X_4}\cup \overline{X_6}\cup \overline{X_8}$, the set of eight-digit numbers who are missing at least one of the digits $2,4,6,8$.

We can count and expand this via the Inclusion-Exclusion Principle.

 $|\overline{X_2}\cup \overline{X_4}\cup \overline{X_6}\cup \overline{X_8}| = |\overline{X_2}|+|\overline{X_4}|+\dots - |\overline{X_2}\cap \overline{X_4}|-|\overline{X_2}\cap \overline{X_6}|-\dots+|\overline{X_2}\cap \overline{X_4}\cap \overline{X_6}|+\dots-|\overline{X_2}\cap \overline{X_4}\cap \overline{X_6}\cap \overline{X_8}|$

Each term in this expansion should be trivial to calculate.

 For instance, $|\overline{X_2}\cap \overline{X_4}\cap \overline{X_6}|$ are those eight-digit numbers who have none of the digits $2,4,6$.  Pick the leading digit noting it cannot be $0$ nor any of $2,4,6$.  Then pick each remaining digit noting it may not be $2,4,6$ but may be zero.  There are $6\times 7\times 7\times 7\times \cdots \times 7 = 6\times 7^7$ such possibilities.

Subtract this amount away from $|X|$ to get the amount you were interested in, those which had all of $2,4,6,8$ as opposed to those who were missing at least one of $2,4,6,8$.
Finally, recognize the symmetry of the terms to group similar terms together to simplify the final expression and reach your final conclusion.

 $9\cdot 10^7 - 4\cdot 8\cdot 9^7 + \binom{4}{2}\cdot 7\cdot 8^7 - \binom{4}{3}\cdot 6\cdot 7^7 + 5\cdot 6^7$


问:
Binary tetrahedral group and $\rm{SL}_2(\mathbb F_3)$
The binary tetrahedral group $\mathbb T$ is an interesting 24-element group. For instance it can be expressed as the subgroup
$$ \mathbb T = \left\{ \pm 1, \pm i, \pm j, \pm k, \dfrac{\pm 1 \pm i \pm j \pm k}2 \right\} \subseteq \mathbb{H}^\times$$
of the multiplicative group of the quaternions (which also gives you a fascinating regular 24-vertex 4-polyhedron).

Alternatively, the binary tetrahedral group can be thought of as the inverse image of $\mathfrak A(4)$ through the 2:1 morphism $\rm{SU}_2 \to \rm{SO}_3$ ($\mathfrak A(4) \subseteq \rm{SO}_3$ being the group of direct isometries of the regular tetrahedron).

The property which puzzles me is the following: $\mathbb T$ is isomorphic to the matrix group $\rm{SL}_2(\mathbb F_3)$.

I'm able to give a proof of this result, but not an enlightening one. Essentially, a 24-element group with a normal subgroup isomorphic to the 8-element quaternion group $\rm Q_8$ is either isomorphic to $\rm Q_8 \times \rm C_3$ or to the unique nontrivial semidirect product $\rm Q_8 \rtimes \rm C_3$; it's not that hard to prove that both $\mathbb T$ and $\rm{SL}_2(\mathbb F_3)$ satisfy this property and belong to the second case. The only interesting thing about this proof is the description of that normal $\rm Q_8$: on the one hand $\{\pm 1, \pm i, \pm j, \pm k\} \subset \mathbb T$ is the inverse image of the quite exceptional Vierergruppe $V_4 = \{ \rm{id}, (12)(34), (13)(24), (14)(23)\} \lhd \mathfrak A(4)\subseteq \rm{SO}_3$; on the other hand, the elements of $\mathrm{SL}_2(\mathbb F_3)$ which are diagonalisable over $\mathbb F_9$ — or, equivalently, $\pm I_2$ and the matrices whose trace is zero — form a normal subgroup isomorphic to $\rm Q_8$ (and it's quite astonishing they do form a subgroup!) whose elements are
$$\begin{array}{l} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},\\ \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}, \begin{pmatrix} -1 & 1 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix}. \end{array}$$

As I said earlier, this proof isn't very satisfying. Hence my question:

Is there an enlightening proof of the isomorphism $\mathbb T \simeq \rm{SL}_2(\mathbb F_3)$?

答:
Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficients are integers or all coefficients are half-integers). The ring $H\otimes\mathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2\times 2}(\mathbb F_p)$ (with determinant as the norm).

Now the binary tetrahedral group is the group of norm $1$ elements in $H$ and it maps injectively (for $p>2$) to the group of norm $1$ elements in $H\otimes\mathbb F_p$, i.e. to  $SL_2(\mathbb F_p)$. In particular, for $p=3$ we get an isomorphism $\mathbb T\to SL_2(\mathbb F_3)$.

(Ref.: TWF 198.)


问:
If $rank(A) = rank(B)$ and $rank(A^2) = rank(B^2)$ then $A$ is similar to $B$
For which $n$ we have:

If $rank(A) = rank(B)$ and $rank(A^2) = rank(B^2)$ then $A$ is similar to $B$. Assume also that $A,B$ are of dimension $n$ and that $A,B$ are nilpotent with the same nilpotence index $m$. I've tried finding jordan form information, diagonalization, but didn't get too far.

答:
Here's a start:

The nilpotence index tells you the size of the largest Jordan block.
The rank tells you how many blocks there are, and together with the rank of the square
tells you how many blocks of size $1$.  Let's say there are $b$ blocks of which $b_1$ are of size $1$, and the largest block has size $m$.  The question then is whether there is more than one way to partition $n - b_1 - m$ into $b - b_1 - 1$ blocks, each of size from $2$ to $m$.


问:
Convert problem into SDP
I have a optimization problem like:
$$min_{R,z} -\log det R $$
$$s.t. R > 0, \quad (c_i - z)^TR(c_i - z) \leq 1 \quad (i=1,\dots,k)$$.

In above problem, $c_i$ is constant $R^n$.

How can I convert the problem in the form of SDP (semidefinite programming)?

答:
I am not going to show how to convert it to a pure SDP problem, as the conversion of the geometric mean operator (to replace logdet unless you use an SDP solver which handles logdet term natively, currently only SDPT3 as far as I know) is absolutely horrendous. Instead, I simply give the initial start required to linearize the model.

The constraints  $(c_i - z)^TR(c_i - z) \leq 1 $ can be written as  $(R(c_i - z))^TR^{-1}R(c_i - z) \leq 1 $ which means a Schur complement leads you to $\begin{bmatrix}1 & (R(c_i-z))^T\\ R(c_i-z) & R\end{bmatrix}\succeq 0$. Define new variable $q = Rz$ and the constraint is a linear semidefinite constraint, and you have a so called MAXDET problem in $(R,q)$.

If you use a modelling layer, all the boring stuff around the logdet/geomean operator will be taken care of. If you use YALMIP in MATLAB (disclaimer: developed by me) and SDPT3 as a solver, it will be solved natively using the logdet objective, and for any other SDP solver the objective is converted to a geometric mean.
https://yalmip.github.io/tutorial/maxdetprogramming/

Title: Convert problem into SDP


问:
Question on continuous injection between $L^p$ spaces
Let ($\Omega, M, \mu$) be a complete measure space. Is it true that $$L^1(\Omega, M, \mu) \cap L^\infty(\Omega, M, \mu) ↪ L^p(\Omega, M, \mu)$$
for all $p \in [1,\infty]$?

May you help me answering this question please? I have no clue

答:
In fact something more general is true. If $p < r < q$ then one has the inequality
$$||f||_r \leq ||f||_p^{1/r - 1/q) /(1/p - 1/q)} ||f||_q^{(1/r-1/q)/(1/p-1/q)}.$$
See for instance here. This implies that $L^p \cap L^q \subseteq L^r$ always. See also a related discussion here.


问:
Can you draw a curve that captures light?
Suppose we had a differentiable curve $C$ in $\mathbb{R}^2$ that serves as our "light container". Light is shining in from all directions, so the space of incoming light-beams is $\mathbb{R} \times S^1$ where the $S^1$ parametrises the angle that the light is coming in from, and the $\mathbb{R}$ parametrises the offset from the origin. When the light hits $C$ it bounces as you would expect it to off a mirror. Some of the incoming light beams will be trapped inside $C$, and keep bouncing forever without escaping, however most will just bounce off.

Is there such a finite curve $C$ such that it actually captures a non-zero amount of light? (take the natural measure on $\mathbb{R} \times S^1$, is there a curve $C$ that captures a non-null subset of the light?) My intuition seems to be telling me that there isn't (because I can't seem to write down any convincing examples), but I can't really think of a reason why not.

If so, what's the curve $C$ that captures the most light? (for a given length $L$. Or, I guess any collection of curves $C_i$ whose total length is $L$).

Sorry about the confusion, I mean that the light is shining in from infinity. Here for example, the black curve C is our jar, and the blue lines represent incoming light. Most just reflect off, some may get trapped forever. We can draw such C that captures a discrete amount of light, but can we draw C that captures a non-trivial amount of light?

答:
A possible answer is given by S. Tabachnikov in his book "Geometry and billiards" and originally proposed by R. Peirone (Reflections can be trapped. Amer. Math. Monthly 101 (1994), 259–260).

First of all, one can construct a trap for light rays parallel to a given direction, see figure below (taken from Tabachnikov's book).

Curve $\gamma$ is an ellipse with foci $F_1$ and $F_2$, curve $\Gamma$ is a parabola with focus $F_2$. These curves are joined in a smooth way to produce a trap: light entering with a direction parallel to the axis of the parabola is reflected to focus $F_2$, and one can prove (exercise 4.3 in that book) that the path of reflected light through the foci of an ellipse converges to its major axis, so that a captured ray never escapes (provided the hole in the ellipse is not too large).

Unfortunately, such a trap only works if incident rays exactly have the given direction: it is shown in the same book (p. 115) that it is not possible to trap a set of divergent rays of light.

Edit by OP: the proof in the book that you can't capture a non-null amount of light is by Poincare Recurrence theorem; since the evolution of light-beams inside the container is given by a measure-preserving map. So Poincare Recurrence  tells you that almost all ingoing light beams will eventually escape.

Edit. Just to clarify the above observation: some light CAN be trapped, but only if coming from parallel rays. Such light has vanishing volume IN PHASE SPACE (because all rays have the same momentum), hence that doesn't contradict Poincaré's theorem.


问:
Third statement in example 2.3.3 in Velleman's How to Prove It.
I am self-studying this text to try to finally learn good proof writing and to learn some more abstract math. It's been going fairly well, but I'm stuck on the statement mentioned in the title. Velleman asks the reader to analyze the logical form of the statement $$B \in \left\{\mathscr{P}(A)\mid A \in \mathcal{F}\right\}.$$ This means that $B \in \mathscr{P}(A)$ where $A \in \mathcal{F}$. If $B \in \mathscr{P}(A)$, then $B$ is a subset of $A$. Then, this means the same thing as $\exists A \in \mathcal{F} (B = \mathscr{P} (A))$. This comes from the opening of the section. He says that $$x \in \left\{n^2 \mid n \in \mathbb{N}\right\}$$ is equivalent to $$\exists n \in \mathbb{N} (x = n^2).$$

This sort of makes sense, but when I think about it harder it becomes confusing. Since $n$ is a natural number, the square of any $n$ will be a perfect square, and thus I don't understand why this statement isn't instead $\forall n \in \mathbb{N} (x = n^2).$ Does this have something to do with the fact that there are an infinitely many natural numbers?

Further, regarding 2.3.3, if $B \subseteq A$, then $B$ is made up of elements of $A$. But the power set of $A$ is made up of sets of elements in $A$. To me this makes this seem like $B$ is two different types of objects and the main source of my confusion. I couldn't find anything on this example beyond the textbook. I see that Velleman is applying the equivalence given in the opening of the section, but I haven't convinced myself that it's true yet. I'm thankful for a nudge in the right direction. I've been going in circles with this one.

答:
$B \in \{\mathscr{P}(A)\mid A \in \mathcal{F}\}$ means that $\exists A \in \mathcal{F}: B = \mathscr{P}(A)$, so $B$ is a powerset of some element of $\mathcal{F}$.

The set on the right consists of a collection of powersets after all, and $B$ is one one them. Sets (also powersets of sets) can be members of other sets, as in this case.

It's different if we would have said $B \in \bigcup \{\mathscr{P}(A)\mid A \in \mathcal{F}\}$, because then $B \in \mathscr{P}(A)$ for some $A \in \mathcal{F}$ (by the definition of a union!) and then we'd have the equivalent formulation
$\exists A \in \mathcal{F}: B \subseteq A$ instead. But from your remarks Velleman favours my interpretation from the beginning. I cannot check, as I don't own the book myself.

Just as $x \in \{n^2\mid n \in \Bbb N\}$ means that $x$ is the  square of some natural number, so $\exists n \in \Bbb N: (x = n^2)$.
We just say that $x$ must be of the form $n^2$, as the set consists of all squares of naturals, i.e. $\{1,4,9,16,25,\ldots \}$ (maybe $0$ too, if $0 \in \Bbb N$, I don't know what Velleman uses)..

Using "$\forall n\in \Bbb N$" instead wouldn't make sense, as a single number cannot be  the square of more than natural number, all such squares are different.


问:
Finding equation of a circle given three non - collinear points
A circle is given which passes through three non collinear points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ then prove that equation of this circle is given by
$$\begin{vmatrix}
x^2+y^2&x&y&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\\
\end{vmatrix} = 0$$
It is just like one of answers to this question Finding an equation of circle which passes through three points
I would like to know how we get to this determinant and I have no idea from where I can start?

答:
The quadratic form $Ax^2+By^2+Cxy+Dx+Ey+F=0$ represents always a conic (eventualy degenerate in two right lines if it is a product of two linear equations) and it represents a circle if and only if $A=B\ne0$ and $C=0$. Then for a circle  we have the equation
$$A(x^2+y^2)+Dx+Ey+F=0$$  When we want a circle passing through three non collinear points $(x_i,y_i)$ we have a linear system of three equations  whose solution gives the three values $ a=\dfrac{D}{A}, b=\dfrac{E}{A}, c=\dfrac{F}{A}$ of the required circle.

But when we want to have the equation of the circle without getting explicitely $a,b,c$ we consider the linear system of four equations with three unknowns $a,b,c$
$$(x^2+y^2)+ax+by+c=0\\(x_1^2+y_1^2)+ax_1+by_1+c=0\\(x_2^2+y_2^2)+ax_2+by_2+c=0\\(x_3^2+y_3^2)+ax_3+by_3+c=0$$
and it is well known that the condition of compatibility of this four equations is that its determinant be equal to zero. In other words, making more explicitely, we have  the matricial equation

$$\begin{pmatrix}
x^2+y^2&x&y&1\\
x_1^2+y_1^2&x_1&y_1&1\\
x_2^2+y_2^2&x_2&y_2&1\\
x_3^2+y_3^2&x_3&y_3&1\\
\end{pmatrix}\cdot\begin{pmatrix}1\\a\\b\\c\end{pmatrix}=\begin{pmatrix}0\\0\\0\\0\end{pmatrix}$$

There are two evident solutions, $(0,0,0,0)$ and $(1,a,b,c)$ so the determinant must be equal to $0$ because if not then the solution should be unique.