问:
Minimizing number of multiplication signs in Computational linear algebra for kind of such these examples

"The main purpose of this note is pedagogical."

i.e.
$$a^{2}+ b^{2}+ c^{2}- ab- bc- ca=\left ( c- a \right )^{2}- \left ( a- b \right )\left ( b- c \right )$$
$$b^{2}- 4ac=\left ( c- a \right )^{2}- \left ( a- b+ c \right )\left ( a+ b+ c \right )$$
The right sides has fewer multiplication signs than the left ones. A famous application, what I know most that related to these results is shortening the running time of programming. I wondered about the method that Strassen did use to minimize the number of multiplication signs like that. I need to the help, thanks a real lot !

Her view (@VeronicaPhan, June 9 '21). However, should not be up to which way of decomposition. The only solution is to find the perfect-fitting decomposition in each scenario.

For the above example of mine, that is $a^{2}+ b^{2}+ c^{2}- ab- bc- ca= M^{2}+ N\left ( M+ N \right )\quad{\rm with}\;M:=c- a\!,{\rm and}\;N:=a- b.$
Interpolative decomposition kills it in an unnatural way
$$\begin{bmatrix} 1 & -1 & 0\\ 0 & 1 & -1\\ -1 & 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & -1\\ 0 & 1\\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & -1 \end{bmatrix}= \begin{bmatrix} -1 & 1\\ 0 & -1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 0 & 1\\ 0 & -1 & 1 \end{bmatrix}\!.$$
Hence $a^{2}+ b^{2}+ c^{2}- ab- bc- ca= \left ( c- a \right )\!^{\!2}- \left ( a- b \right )\!\left ( b- c \right )\!.$

答:
$$b^2 - 4ac = \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \begin{bmatrix} 0 & t_1 & t_2 - 4 \\ -t_1 & 1 & t_3 \\ -t_2 & - t_3 & 0\end{bmatrix} \begin{bmatrix} a\\ b \\ c\end{bmatrix} $$

We would like to find $t_1, t_2, t_3 \in \Bbb R$ such that the rank of the matrix is minimized. The easy first step would be to force the matrix to be rank-$2$ by making its determinant vanish. Note that

$$\det \begin{bmatrix} 0 & t_1 & t_2 - 4 \\ -t_1 & 1 & t_3 \\ -t_2 & - t_3 & 0\end{bmatrix} = t_2 (t_2 - 4) - 4 t_1 t_3$$

Via visual inspection, after some work, we conclude that $(t_1, t_2, t_3) = \color{blue}{(-1, 2, 1)}$ does make the determinant vanish. Via the eigendecomposition,

$$\begin{bmatrix} 0 & -1 & -2\\ 1 & 1 & 1\\ -2 & -1 & 0\end{bmatrix} = \begin{bmatrix} \color{red}{-1}\\ 0 \\ \color{blue}{1}\end{bmatrix} \begin{bmatrix} \color{red}{-1}\\ 0 \\ \color{blue}{1}\end{bmatrix}^\top - \begin{bmatrix} \color{red}{1}\\ \color{magenta}{-1} \\ \color{blue}{1}\end{bmatrix} \begin{bmatrix} \color{red}{1}\\ \color{magenta}{1} \\ \color{blue}{1}\end{bmatrix}^\top$$

and, thus,

$$b^2 - 4 a c = \left( \color{blue}{c} - \color{red}{a} \right)^2 - \left( \color{red}{a} \color{magenta}{- b} + \color{blue}{c} \right) \left( \color{red}{a} + \color{magenta}{b} + \color{blue}{c} \right)$$

SymPy code

```<code>>>> from sympy import *
>>> Q = Matrix([[ 0,-1,-2],
                [ 1, 1, 1],
                [-2,-1, 0]])
>>> Q.rank()
2
>>> (V,D) = Q.diagonalize()
>>>
>>> V
Matrix([
[ 1,  1, -1],
[-1, -2,  0],
[ 1,  1,  1]])
>>>
>>> D
Matrix([
[-1, 0, 0],
[ 0, 0, 0],
[ 0, 0, 2]])
>>>
>>> V**-1
Matrix([
[   1,  1,    1],
[-1/2, -1, -1/2],
[-1/2,  0,  1/2]])
</code>```

Body:
$$b^2 - 4ac = \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \begin{bmatrix} 0 & t_1 & t_2 - 4 \\ -t_1 & 1 & t_3 \\ -t_2 & - t_3 & 0\end{bmatrix} \begin{bmatrix} a\\ b \\ c\end{bmatrix} $$

We would like to find $t_1, t_2, t_3 \in \Bbb R$ such that the rank of the matrix is minimized. The easy first step would be to force the matrix to be rank-$2$ by making its determinant vanish. Note that

$$\det \begin{bmatrix} 0 & t_1 & t_2 - 4 \\ -t_1 & 1 & t_3 \\ -t_2 & - t_3 & 0\end{bmatrix} = t_2 (t_2 - 4) - 4 t_1 t_3$$

Via visual inspection, after some work, we conclude that $(t_1, t_2, t_3) = \color{blue}{(-1, 2, 1)}$ does make the determinant vanish. Via the eigendecomposition,

$$\begin{bmatrix} 0 & -1 & -2\\ 1 & 1 & 1\\ -2 & -1 & 0\end{bmatrix} = \begin{bmatrix} \color{red}{-1}\\ 0 \\ \color{blue}{1}\end{bmatrix} \begin{bmatrix} \color{red}{-1}\\ 0 \\ \color{blue}{1}\end{bmatrix}^\top - \begin{bmatrix} \color{red}{1}\\ \color{magenta}{-1} \\ \color{blue}{1}\end{bmatrix} \begin{bmatrix} \color{red}{1}\\ \color{magenta}{1} \\ \color{blue}{1}\end{bmatrix}^\top$$

and, thus,

$$b^2 - 4 a c = \left( \color{blue}{c} - \color{red}{a} \right)^2 - \left( \color{red}{a} \color{magenta}{- b} + \color{blue}{c} \right) \left( \color{red}{a} + \color{magenta}{b} + \color{blue}{c} \right)$$

SymPy code

>>> from sympy import *
>>> Q = Matrix([[ 0,-1,-2],
                [ 1, 1, 1],
                [-2,-1, 0]])
>>> Q.rank()
2
>>> (V,D) = Q.diagonalize()
>>>
>>> V
Matrix([
[ 1,  1, -1],
[-1, -2,  0],
[ 1,  1,  1]])
>>>
>>> D
Matrix([
[-1, 0, 0],
[ 0, 0, 0],
[ 0, 0, 2]])
>>>
>>> V**-1
Matrix([
[   1,  1,    1],
[-1/2, -1, -1/2],
[-1/2,  0,  1/2]])


问:
Why doesn't the multiplication rule work here?
I have the following question:

How many ways are there to select $3$ candidates from $8$ equally
qualified recent graduates for openings in an accounting firm?

I was given the following solution to it: $\binom{8}{3}$, however, I got the answer: $8 \times 7 \times 6$ by doing the multiplication rule, I was wondering why my answer doesn't work.
My initial reason is that in the multiplication rule there is the underlying assumption that order matters and here obviously order doesn't matter. However, I am posting this here as that is the first time I come to such a realization and I am not sure that my reasoning is correct, especially since that I never learned multiplication rule with the idea of order in my mind. Any clarification would be appreciated.

答:
For example, let's say that the candiates' names are A, B, C, and so on. Choosing A,B,C is the same as choosing B,A,C or C,A,B and so on. There are $3! = 6$ ways to pick the same candiates, so the answer will be $\frac{8\times7\times6}{3\times2\times1}$ = $56$.
This is due to picking candidates to do for the same thing. You need to divide by the factorial of the number of elected only when you're picking the same things or people. If not, the multiplication rule works without needing to divide by something.

ContentLicense: CC BY-SA 4.0


答:
$\binom{8}{3}$ is the same as $\frac{8!}{3!5!}$ which is $\frac{8.7.6}{3!}$, so you just need to divide by the extra $3!$

That's because you have overlapping choices by e.g choosing ABC, is the same as ACB,BAC,BCA,CAB and CBA, 6 ways to get the same ABC.

Title: Why doesn't the multiplication rule work here?

Body:
I have the following question:

How many ways are there to select $3$ candidates from $8$ equally
qualified recent graduates for openings in an accounting firm?

I was given the following solution to it: $\binom{8}{3}$, however, I got the answer: $8 \times 7 \times 6$ by doing the multiplication rule, I was wondering why my answer doesn't work.
My initial reason is that in the multiplication rule there is the underlying assumption that order matters and here obviously order doesn't matter. However, I am posting this here as that is the first time I come to such a realization and I am not sure that my reasoning is correct, especially since that I never learned multiplication rule with the idea of order in my mind. Any clarification would be appreciated.


答:
Yes you are right.
The formula for the 'choose' function is $$\frac{n!}{k!(n-k)!}$$ where k is the number of things you choose and n is the number of things you choose from.
If you do $8 \times 7 \times 6$ you are working out the number of permutations (where order matters). But you're trying to work out the number of combination and therefore you need to exclude all combinations where the order is the same.
E.g. abc is the same as acb = bac = bca = cab = cba

(n-k)! takes care of how many people to choose each time (aka 3). But k! allows you to neglect the order and make sure each combination is distinct from another.

If you did $\frac{8!}{8-3!}$ you'd get $8 \times 7 \times 6$ but that would only give you the number of permutations

You want combinations so you should do $\frac{8!}{3!8-3!}$

Hope this helps!


问:
Prove sum of arithmetic series is $\frac{1}{4}n(x + y)$ given sum of first two and last two members.
I am working through a maths text book as a hobby. I have come across this problem:

Given that the sum of the first and second terms of an arithmetic progression is x and the sum of the $(n -1)$th  and nth terms is y, prove that the sum of the first n terms is $\frac{1}{4}n(x + y)$

If $u_{n}$ refers to the nth term of the series, I know $u_{1} +  u_{n} = u_{2} + u_{n - 1}$ and also that $S_{n} = \frac{1}{2}n\{2a + (n -1)d\}$, where a refers to the first member and d is the common difference. But I cannot put this knowledge together to work out the solution.

答:
$$S_n=\dfrac n2(a+a_n)$$
$$S_n=\dfrac n2(a_2+a_{n-1})$$
Add these two equations together and use definition of $x$ and $y$.


问:
Meaning of tangent space to $\mathbb{R}$
If the tangent space  $T_{x_0}\mathbb{R}^n$  is the tangent space to $\mathbb{R}^n$ at the point $x_0$, as the set of all vectors applied to $x_0$ and so it is isomorphic to $\mathbb{R}^n$, that is the set of all vectors applied in $0$: $$\textbf{what is the meaning of $T_{x_0}\mathbb{R}$, and why it is isomorphic to $\mathbb{R}$?}$$

I mean in $\mathbb{R}$ there is no a notion of vector so I can't imagine what is the tangent space in this case. Can you help me?

答:
If you define tangent spaces for any differentiable manifold, you'll notice that when your manifold is a (finite-dimensional) real vector space $V$ you get to identify $T_p V$ with $V$ at any point $p$ (to see why, see for example this question).

The case of $\Bbb R$ is simply the one-dimensional case of this phenomenon, since $\Bbb R$ is (the) one-dimensional real vector space.

(I don't really get why you would say "there's no notion of vector in $\Bbb R$", but you can easily "visualize" what a vector in $\Bbb R$ is as you can do in any other $\Bbb R^n$: as an arrow going from the origin, i.e. $0$, to any element of the space. That is, you're looking at every real number $t$ as a vector of length $|t|$, going left or right depending on the sign of $t$, respectively negative or positive sign.)


问:
There don't exist positive integers m and n such that $\varphi(n) = 2 · 7^{5m+4}$
Suppose there are $m,n\in \mathbb{N}$ such that $\varphi(n)=2\cdot 7^{5m+4}$.

Then $\varphi(n)=\varphi(p_1^{k_1}p_2^{k_2}...p_r^{k_r})=\varphi(p_1^{k_1})\varphi(p_2^{k_2})...\varphi(p_r^{k_r})=p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=2\cdot 7^{5m+4}$

In another words, $2=\frac{p_1^{k_1-1}(p_1-1)p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)}{7^{5m+4}}$. This means that one of $p_i$'s is 2 with $k_i=2$. WLOG, assume that $p_1=2$ and $k_1=2$. Then  $p_2^{k_2-1}(p_2-1)...p_r^{k_r-1}(p_r-1)=7^{5m+4}$ but this is not possible since we must have that $p_i=7$ and $p_i-1=6$ also appears.

Does this make sense?

答:
I would suggest the following approach :

If $p$ is a prime factor of $n$, then $p-1$ divides $\varphi(n)=2\cdot 7^{5m+4}$
The only divisors $d$ of $2\cdot 7^{5m+4}$ for which $d+1$ is prime, are $1$ and $2$, since $2\cdot 7^k+1$ is divisible by $3$ and greater than $3$ for every positive integer $k$, whereas $7^k+1$ is obviously not prime for positive integer $k$. Hence $n$ cannot have a prime factor exceeding $3$ and hence its totient value cannot be divisible by $7$.


问:
Approximation of subtraction of modified Bessel functions of first kind
I need to find asymptotic expansion of $e^{-x}[I_{0}-I_{1}](x)$ when $x\rightarrow \infty$. Where $I_{0}(x)$ and $I_{1}(x)$ are modified Bessel functions of first kind of order zero and one respectively. I know that $I_{0}$ and $I_{1}$ both are asymptotically equal to $\dfrac{e^{x}}{\sqrt (2\pi x)}$ when $x\rightarrow \infty $. Also that if $f_{1}\sim g_{1} $ and $f_{2}\sim g_{2}$ then it won't imply $f_{1}-f_{2}\sim g_{1}-g_{2}$. I am not sure how to solve this.

Any help is appreciated.

Body:
It is known that
$$
I_\nu  (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 - \frac{{4\nu ^2  - 1}}{{8x}} + \frac{{(4\nu ^2  - 1)(4\nu ^2  - 9)}}{{128x^2 }} -  \cdots } \right)
$$
as $x\to +\infty$ with fixed $\nu$ (cf. http://dlmf.nist.gov/10.40.E1). Thus
$$
I_0 (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 + \frac{1}{{8x}} + \frac{9}{{128x^2 }} +  \cdots } \right),\;\, I_1 (x) \sim \frac{{e^x }}{{\sqrt {2\pi x} }}\left( {1 - \frac{3}{{8x}} - \frac{{15}}{{128x^2 }} -  \cdots } \right),
$$
and therefore,
$$
e^{ - x} (I_0 (x) - I_1 (x)) \sim \frac{1}{{\sqrt {2\pi x} }}\left( {\frac{1}{{2x}} + \frac{3}{{16x^2 }} +  \cdots } \right) = \frac{1}{{2\sqrt {2\pi } x^{3/2} }}\left( {1 + \frac{3}{{8x}} +  \cdots } \right)
$$
as $x\to +\infty$.


问:
Application of the principle of conservation of angular momentum and the principle of conservation of energy
A satellite is launched in a direction parallel to the surface of the earth with a velocity of $36900km/hr$ from an altitude of $500km$. Determine

(a) the maximum altitude reached by the satellite.

(b) the maximum allowable error in the direction of launching if the satellite is to go into orbit and come no closer than $200km$ to the surface of the earth.

I know that I have to apply the principle of conservation of angular momentum

$r_1 m v_1 \sin \phi_1 = r_2 m v_2 \sin \phi_2$

where
$r =$ distance of the satellite from the center of the earth
$m =$ mass of the satellite
$v =$ velocity of the satellite
$\phi =$ the angle the velocity makes with the radius vector

and the principle of conservation of energy

$T_1 + V_1 = T_2 + V_2$

Where $T =$ kinetic energy and $V =$ potential energy.

I know I have to use the two principles but I don't know how to use them. May someone please help me with how to use them or offer an alternative solution to the question.

答:
The Newton laws of gravity, on the way to establishing the Kepler laws, give a parametrization
$$
r=\frac{R}{1+e\cos(φ-ψ)} ~~ \text{ and } ~~ \dot r=\sqrt{\frac{GM_E}{R}}e\sin(φ-ψ),~~r^2\dot\phi = -\sqrt{RGM_E} .
$$
for the polar coordinates $(r(t),\phi(t))$, $x+iy=re^{i\phi}$, of the satellite (moving clock-wise if moving horizontally to the right). $ψ$ is the angle for the smallest radius, at periapsis, $\frac{R}{1+e}$.

The initial data in polar coordinates can be obtained from the given initial data that can easily be applied to Cartesian coordinates. As $x_0+iy_0=r_0e^{iφ_0}$, and the interpretation of the launch conditions as $x_0=0$, $y_0=R_E+H_{sat}$ gives $r_0=y_0$ and $φ_0=0$. The constants in the reduced equations follow from
$$
\frac{\dot x_0+i\dot y_0}{x_0+iy_0}=\frac{\dot r_0}{r_0}+i\dotφ_0
$$
with $\dot x_0=V_{sat}\cos\theta$, $\dot y_0=V_{sat}\sin\theta$, so that
$$
\dot r_0=\dot y_0=V_{sat}\sin\theta, ~~~ \dot φ_0=-\frac{\dot x_0}{y_0}=-\frac{V_{sat}}{r_0}\cos\theta
$$
Then
$$
r_0V_{sat}\cosθ=\sqrt{RGM_E}\implies R=\frac{r_0^2V_{sat}^2\cos^2θ}{GM_E}\\
e\cosψ=\frac{R}{r_0}-1  ~\text{ and }~
e\sinψ=-\dot r_0\sqrt{\frac{R}{GM_E}}
$$
This is sufficient to calculate $e$ and $ψ$ as polar coordinates and thus the minimal and maximal elevations, periapsis and apoapsis (or perigee and apogee)
$$
\frac{R}{1+e}-R_E  ~\text{ and }~  \frac{R}{1-e}-R_E
$$

```<code>G       = 6.67408e-11*1e-9 # km^3 s^-2 kg^-1
M_E  = 5.9721986e24 # kg
R_E  = 6378.137 # km
H_sat    = 500 # km above Earth
V_sat    = 36900/3600 # km/s
theta = 0

r0 = R_E + H_sat
dotr0 = V_sat*m.sin(theta)
dotphi0 = -V_sat/r0*m.cos(theta)

R = (r0**2*dotphi0)**2/(G*M_E)
wx = R/r0-1; wy = -dotr0*(R/(G*M_E))**0.5
E = (wx*wx+wy*wy)**0.5; psi = m.atan2(wy,wx)
print(f"R={R:.8f} km, e={E:.8f}, psi={psi:.8f} rad")
print(f"H_peri={R/(1+E)-R_E:.8f} km, H_apo={R/(1-E)-R_E:.8f} km")
</code>```

with the result

```<code>R=12469.92166223 km, e=0.81297954, psi=-0.00000000 rad
H_peri=500.00000000 km, H_apo=60298.64124618 km
</code>```

Setting theta=0.2 gives

```<code>R=11977.74005179 km, e=0.82116894, psi=-0.44435938 rad
H_peri=198.81463324 km, H_apo=60599.82661294 km
</code>```

which is slightly outside the admissible range.

Body:
The Newton laws of gravity, on the way to establishing the Kepler laws, give a parametrization
$$
r=\frac{R}{1+e\cos(φ-ψ)} ~~ \text{ and } ~~ \dot r=\sqrt{\frac{GM_E}{R}}e\sin(φ-ψ),~~r^2\dot\phi = -\sqrt{RGM_E} .
$$
for the polar coordinates $(r(t),\phi(t))$, $x+iy=re^{i\phi}$, of the satellite (moving clock-wise if moving horizontally to the right). $ψ$ is the angle for the smallest radius, at periapsis, $\frac{R}{1+e}$.

The initial data in polar coordinates can be obtained from the given initial data that can easily be applied to Cartesian coordinates. As $x_0+iy_0=r_0e^{iφ_0}$, and the interpretation of the launch conditions as $x_0=0$, $y_0=R_E+H_{sat}$ gives $r_0=y_0$ and $φ_0=0$. The constants in the reduced equations follow from
$$
\frac{\dot x_0+i\dot y_0}{x_0+iy_0}=\frac{\dot r_0}{r_0}+i\dotφ_0
$$
with $\dot x_0=V_{sat}\cos\theta$, $\dot y_0=V_{sat}\sin\theta$, so that
$$
\dot r_0=\dot y_0=V_{sat}\sin\theta, ~~~ \dot φ_0=-\frac{\dot x_0}{y_0}=-\frac{V_{sat}}{r_0}\cos\theta
$$
Then
$$
r_0V_{sat}\cosθ=\sqrt{RGM_E}\implies R=\frac{r_0^2V_{sat}^2\cos^2θ}{GM_E}\\
e\cosψ=\frac{R}{r_0}-1  ~\text{ and }~
e\sinψ=-\dot r_0\sqrt{\frac{R}{GM_E}}
$$
This is sufficient to calculate $e$ and $ψ$ as polar coordinates and thus the minimal and maximal elevations, periapsis and apoapsis (or perigee and apogee)
$$
\frac{R}{1+e}-R_E  ~\text{ and }~  \frac{R}{1-e}-R_E
$$

G       = 6.67408e-11*1e-9 # km^3 s^-2 kg^-1
M_E  = 5.9721986e24 # kg
R_E  = 6378.137 # km
H_sat    = 500 # km above Earth
V_sat    = 36900/3600 # km/s
theta = 0

r0 = R_E + H_sat
dotr0 = V_sat*m.sin(theta)
dotphi0 = -V_sat/r0*m.cos(theta)

R = (r0**2*dotphi0)**2/(G*M_E)
wx = R/r0-1; wy = -dotr0*(R/(G*M_E))**0.5
E = (wx*wx+wy*wy)**0.5; psi = m.atan2(wy,wx)
print(f"R={R:.8f} km, e={E:.8f}, psi={psi:.8f} rad")
print(f"H_peri={R/(1+E)-R_E:.8f} km, H_apo={R/(1-E)-R_E:.8f} km")

with the result

R=12469.92166223 km, e=0.81297954, psi=-0.00000000 rad
H_peri=500.00000000 km, H_apo=60298.64124618 km

Setting theta=0.2 gives

R=11977.74005179 km, e=0.82116894, psi=-0.44435938 rad
H_peri=198.81463324 km, H_apo=60599.82661294 km

which is slightly outside the admissible range.


答:
I think your question seems to be when you can assume order doesn't matter, and when it actually does.

I'll give you two examples which might make it easier.

The first is like your own question. How many ways of choosing 3 people out of 8 people? You're just choosing people, and not assigning any special role or task to them to distinguish between individual members of your chosen triple. Here, use $^8C_3$.

The second is a variation of your question. How many ways of choosing 3 club office holders (namely President, Secretary and Treasurer) out of 8 people? From the wording, you're not just choosing three people, you're also assigning each to a particular role. In this case, order definitely matters since $(a, b, c)$ in the respective roles of President, Secretary and Treasurer is a very different selection from $(c, b, a)$ in those respective roles. So here you use $^8P_3$ (which is equal to $8\cdot 7 \cdot 6$.


问:
Finding complex number for intersection of a line and a circle?
How to find complex number for intersection of a line and a circle?

In particular, if $M=$ midpoint $BC$, then I want $AM ∩ (ABC)$ in complex numbers.

Let me describe why I needed this,

I was doing a problem which I reduced to proving $4$ points $(B,T,M',J)$ are cyclic where $ABC$ is an acute scalene triangle, $T$ is the intersection of tangents to $(ABC)$ from $B$ and $C$, $M' = AM ∩ (ABC)$ ;($M =$ midpoint $BC$) , and $J = AB ∩ CM'$.

I tried to prove this synthetically but I could not, so I decided to bash. First I tried coordinate but that was too long, so I decided to go with complex because that felt feasible. But then I remembered that I have not yet studied complex numbers (except when I would read EGMO complex numbers for fun, like a storybook). So I opened EGMO and could get complex numbers for everything like how to prove cyclic, intersection, $T$ etc. but Point $M'$ was a problem. I couldn't find how to get it anywhere so I come to MSE.

Please help me, Thanks!

答:
Let $A,B,C$ lie on the unit circle, with complex coordinates $a,b,c$. Since you know EGMO, you should be able to find a proof of the fact that the line through the points $a,b$ on the unit circle has equation
$$z + ab\overline{z} = a+b$$

Now denoting $D = AM \cap (ABC)$ with coordinate $d$, the line $AD$ is given by $z + ad \overline{z} = a+d$.

However you know that the midpoint of $BC$, namely $ \frac{b + c}{2}$ lies on line $AD$, so

$$\frac{b + c}{2} + ad \frac{\overline{b} + \overline{c}}{2} = a+d$$

You can now just solve for $d$ to find the required intersection.


Title: Prove sum of arithmetic series is $\frac{1}{4}n(x + y)$ given sum of first two and last two members.

Body:
I am working through a maths text book as a hobby. I have come across this problem:

Given that the sum of the first and second terms of an arithmetic progression is x and the sum of the $(n -1)$th  and nth terms is y, prove that the sum of the first n terms is $\frac{1}{4}n(x + y)$

If $u_{n}$ refers to the nth term of the series, I know $u_{1} +  u_{n} = u_{2} + u_{n - 1}$ and also that $S_{n} = \frac{1}{2}n\{2a + (n -1)d\}$, where a refers to the first member and d is the common difference. But I cannot put this knowledge together to work out the solution.

Body:
In the spirit of Gauss:

$$\begin{array}{rc}
S_n &=& u_1 &+& u_2 &+& u_3 &+& \cdots &+& u_{n-2} &+& u_{n-1} &+& u_n \\
S_n &=& u_n &+& u_{n-1} &+& u_{n-2} &+& \cdots &+& u_3 &+& u_2 &+& u_1 \\
2S_n &=& C &+& C &+& C &+& \cdots &+& C &+& C &+& C
\end{array}$$
where:
$$2C = u_1+u_n+u_2+u_{n-1} = x+y$$
so:
$$C = \dfrac{x+y}2$$
so:
$$2S_n = nC = \dfrac12 n(x+y)$$
so:
$$S_n = \frac14n(x+y)$$


问:
Transition from Upper Bound definition for Sets to Sequences using Logical transitions & Set-Builder notation
Upper bound definition for sets: $ M \in \mathbb{R} $ is an upper bound of set $ A $  if $ \forall \alpha\in A. \alpha \leq M$

Upper bound definition for sequences: $ M \in \mathbb{R} $ is an upper bound of sequence $ (a_n)$  if $ \forall n \in \mathbb{N}. a_n \leq M$

Suppose we look at the set $ A = \{ a_n | n \in \mathbb{N} \} $ .

I've been pondering for a while about the following 2 questions related to mathematical-writing , logic and set builder notation:

Questions:

1. How do we get with logic from the defintion of upper bound for sets to the definition of the upper bound for sequences using the set $ A = \{ a_n | n \in \mathbb{N} \} $? My reasoning:

$ \forall \alpha \in A . \alpha \leq M  \iff \forall \alpha.( \alpha\in A \rightarrow \alpha \leq M )  \iff  $ $\forall \alpha.( \alpha \in A \rightarrow \exists n \in \mathbb{N}. \alpha = a_n \rightarrow \alpha \leq M ) \iff $   $ \forall \alpha.( \alpha \in A \rightarrow \exists n \in \mathbb{N}. \alpha = a_n \rightarrow \alpha \leq M \rightarrow a_n \leq M) $ Hence $ \forall \alpha \in A \exists n \in \mathbb{N}.( \alpha=a_n \land a_n\leq M $ ), Now this is not equivalent to the defintion of upper bound of sequences above ( $ \forall n \in \mathbb{N}. a_n \leq M$  ), why? can you please give correct transitions? I've made mistakes but It's confusing me to see how to write them correctly.

2. Someone told me that since every element of $ A = \{ a_n | n \in \mathbb{N} \}  $ is generated by every element of $ n \in \mathbb{N} $ so therefore $ \forall \alpha\in A. \alpha \leq M$ is equivalent to $ \forall n \in \mathbb{N}. a_n \leq M$ .

How is that possible that the two statements are equivalent? Since for arbitrary $ \alpha \in C $ there exists a specific $ n \in N $ ( not arbitrary ) therefore it appears false to write $ \forall n \in \mathbb{N}. a_n \leq M$ but seems more reasonable to write $ \exists n \in \mathbb{N}. a_n \leq M$ .

答:
In part the question asks why $\forall \alpha \in A : P(\alpha)$ is logically equivalent to $\forall n \in \mathbb{N} : Q(n)$ where:

$A = \{a_n \in \mathbb{R} : n \in \mathbb{N}\}$,

$P(x)$ is some property of real numbers, and

$Q(n)$ is the property of natural numbers defined by $Q(n)$ holds iff $P(a_n)$ holds.

Start by supposing $\forall \alpha \in A : P(\alpha)$. We need to justify $\forall n \in \mathbb{N} : Q(n)$. To do that, pick an arbitrary $n \in \mathbb{N}$, and consider the element
$a_n \in A$. We have that $P(a_n)$ holds, so by definition of $Q$, also
$Q(n)$ holds. Our choice of $n$ was arbitrary, hence $\forall n \in \mathbb{N} : Q(n)$ as required.

This shows
$$\forall \alpha \in A : P(\alpha) \Rightarrow \forall n \in \mathbb{N} : Q(n).$$

We still have to justify the converse implication. Suppose that
$\forall n \in \mathbb{N} : Q(n)$. Pick an arbitrary $\alpha \in A$.
For this $\alpha$ there must exist some $n \in \mathbb{N}$ such that
$\alpha = a_n$. Of course there may be many such $n$, but all that matters is that we can find at least one. We know that $Q(n)$ holds for this particular $n$ because it holds for all of them. By the definition of $Q$, this means that
$P(a_n)$ holds. Hence $P(\alpha)$ holds as $\alpha = a_n$. Since $\alpha$ was arbitrary, we conclude $\forall \alpha \in A : P(\alpha)$.

This shows
$$\forall \alpha \in A : P(\alpha) \Leftarrow \forall n \in \mathbb{N} : Q(n).$$

In predicate logic the rule that allows you to deduce $\forall x: P(x)$ when you have established $P(x)$ for an arbitrary $x$ is called generalization. Expressing the whole argument formally and symbolically is entirely possible though might make it less clear.


问:
$Ra=\langle a\rangle$ for ring $R$ and $a \in R$
I am reading some proofs of "$R$ is a field iff only ideals of $R$ is $\{0\}$ and $R$" and in the process of reading, I have a question.

In the converse parts, the textbooks say

Let $a \in R$, then $Ra$ is the ideals of $R$. If $a=0$, then $Ra=\{0\}$. If $a\neq 0$, then $Ra=\langle a\rangle \neq \{0\}$. From the assumption $Ra=\langle a\rangle=R$ and since $1\in R=Ra$, $1=ra$, $r \in R$, so $r=a^{-1}$, .i.e., $a$ have an inverse.

What I can not understand is $Ra=\langle a\rangle$. I understand $\langle a\rangle \subset Ra$ But how one can understand $Ra \subset \langle a\rangle$?

答:
The statement that $R$ is a field is trivially false if $R$ is not commutative (since it can never be a field); and a counterexample is given by both the quaternions (a division ring), and the ring of $2\times 2$ matrices over a field (which is not a field and not a division ring, but is simple).

The statement is false if we do not assume $R$ to have a $1$: the cyclic group of order $2$ with zero multiplication has no ideals other than the trivial ones, but is not a field.

So you need to state that $R$ is commutative with unity, otherwise the whole thing collapses.

By definition, $\langle a\rangle$ is the smallest ideal of $R$ that contains $a$. For an arbitrary ring (not necessarily commutative, not necessarily with $1$), you should prove that
$$\langle a\rangle = \left.\left\{na + ra + as + \sum_{i=1}^kr_ias_i\,\right|\, n\in\mathbb{Z},k\in\mathbb{N},r,s,r_i,s_i\in R\right\}.$$

By definition,
$$\begin{align*}
Ra &= \{ra\mid r\in R\}\\
aR &= \{ar\mid r\in R\}\\
RaR &= \left\{\sum_{i=1}^k r_ias_i\,\Bigm|\, k\in\mathbb{N}, r_i,s_i\in R\right\}.
\end{align*}$$
You should verify that $Ra$ is a left ideal, $aR$ is a right ideal, and $RaR$ is a two-sided ideal.

When $R$ has a unity, you should prove that $RaR=\langle a\rangle$.

When $R$ is commutative, you should prove that $Ra=aR=RaR$.

When $R$ is commutative and has a unity, you should prove that $Ra=aR=RaR=\langle a\rangle$.


问:
Convergence in probability of random variables and continuous function
I want to prove that if $\xi_n \xrightarrow {\mathbb{P}} \xi$ and for some continuous function $\mathit{f}:\mathbb{R}\to\mathbb{R}$ then $$\mathit{f}(\xi_n)\xrightarrow {\mathbb{P}} \mathit{f}(\xi)$$
May somebody can help me? I would be grateful.

答: Using the definition of continuity, let  $\varepsilon > 0$, then ther exists $\delta >0$ s.t. $\lvert X_n - X \rvert \leq \delta \implies \lvert f(X_n) - f(X) \rvert \leq \varepsilon$ or equivalently (taking the negation of previous statement), $\lvert f(X_n) - f(X) \rvert > \varepsilon \implies \lvert X_n - X \rvert > \delta$. Then by inclusion of events, $\mathbb{P}(\lvert f(X_n) - f(X) \rvert > \varepsilon) \leq \mathbb{P}(\lvert X_n - X \rvert > \delta)$ and the last quantity goes to $0$ as $n\to \infty$ by definition of convergence in probability.


问:
Show that the second derivative of $y=(x^2−3x)^3$ is $6(x^2-3x)(5x^2-15x+9)$
How do you find the second derivative of $$y=(^2−3)^3$$  and show that it is $6(x^2-3x)(5x^2-15x+9)$

Every time I worked it out, I used the chain rule first to get the first derivative which is
$ {3(x^2-3x)^3}\cdot{2x-3}$. Then I used the product rule + chain rule but I only managed to get the expanded form of what I was supposed to "show" which is $30x^4-180x^3+324x^2-162x$

Any help would be appreciated.

Body:
You did not get the first derivative correctly.

The first derivative is $3(x^{2}-3x)^{2}(2x-3)$ and the second derivaive is $6(x^{2}-3x) (2x-3)^{2}+ 6(x^{2}-3x)^{2}$. Factor out $(x^{2}-3x)$ to get $6(x^{2}-3x) (4x^{2}-12x+9+x^{2}-3x)=6(x^{2}-3x)  (5x^{2}-15x+9)$


问:
Let $(a_{n})$ be a sequence that does not converge in the wide-sense. Then it has at-least two different partial limits in the wide-sense
Problem: Let $ ( a_n )  $ be a sequence that does not converge in the wide-sense. Then it has at-least two different partial limits in the wide-sense.

Proof:
Let $ a $ be a partial limit of $ ( a_n ) $. Suppose $ a $ is real. From our assumption, $ a_n \nrightarrow a $ therefore there exists $ \epsilon>0$ such that for every $ N \in \mathbb{N} $ there exists $ n>N  $ s.t. $ | a_n - a| \geq \epsilon $. We can build a subsequence $ (a_{n_{k}}) $ of $ (a_n) $ s.t. for all $ k \in \mathbb{N} $ , $ | a_{n_k} - a| \geq \epsilon $. Let $ b $ be a partial limit in the wide sense of the sequence $ ( a_{n_k} ) $, therefore it is also a partial limit of $ (a_n) $. Then $ b \neq a $ since for all $ k \in \mathbb{N} $ , $ | a_{n_k} - a| \geq \epsilon $.

[ Another 2 proofs here for the cases where $ a = \pm\infty $ ]

Note:  I disregarded the other 2 proofs where $ a = \pm\infty $ since I had difficulty with the case where $ a $ is real. And " In the wide sense " means the limit is either a real number or $ \pm \infty $

Questions ( About the case where $ a $ is real ) :

To be sure, we essentially have in total 3 subsequences in the above proof? one whose limit is $ a $, second is $ ( a_{n_k} ) $, third is a subsequence of $ (a_{n_k} ) $ whose limit is $ b $ .

However , I don't fully understand why $ b \neq a $? My attempt was as follows:

Denote $ (a_{n_{k_j}}) $ a subsequence of $ (a_{n_k} ) $ s.t. $ a_{n_{k_j}} \rightarrow b $. Then, since we already have introduced an $ \epsilon > 0 $ in the proof above, then there exists $ J\in \mathbb{N} $ for every $ j>J $ s.t. $ | a_{n_{k_j}} - b | < \epsilon $. Since also for all $ k \in \mathbb{N} $ , $ | a_{n_k} - a| \geq \epsilon $ then [ here I need to proceed but I got stuck ]

答:
1.

We technically have three sub-sequences and a sub-sub-sequence. The sequence $\left(a_n\right)_{n\in\mathbb{N}}$ is a sub-sequence of it self, the implicit sub-sequence $\left(a_{n_i}\right)_{i\in\mathbb{N}}$ which converges to the partial limit $a$, the sub-sequence $\left(a_{n_k}\right)_{k\in\mathbb{N}}$ which was created in the process of the proof, and $\left(a_{n_{k_j}}\right)_{j\in\mathbb{N}}$ which converges to $b$.

2.

You almost there in understanding why the sub-sequence $\left(a_{n_{k_j}}\right)_{j\in\mathbb{N}}$ does not converge to $a$.

A sequence, of real numbers, is really just a function $f:\mathbb{N}\to\mathbb{R}$ and a sub-sequence requires some strictly increasing function $g:\mathbb{N}\to\mathbb{N}$ where the sub-sequence is defined as the composition $f\circ g:\mathbb{N}\to\mathbb{R}$. So if we define $a_n=f(n)$, then we should define a sub-sequence as $a_{n_k}=f(g(k))$. If we add in another strictly increasing function, $h:\mathbb{N}\to\mathbb{N}$, then we should define a sub-sub-sequence as $a_{n_{k_j}}=f(g(h(j)))$.

We have that there exists an $\varepsilon>0$ such that for each $k\in\mathbb{N}$, $|a_{n_k}-a|\geq\varepsilon$. We also have that $\{n_{k_j}:j\in\mathbb{N}\}\subseteq\{n_k:k\in\mathbb{N}\}$. This is seen from the fact that, using our example functions above, we would have $n_{k}=g(k)$ and $h(j)=k_j$ as well as $n_{k_j}=g(h(j))$. Since $h$ is strictly increasing, then it is injective, but it may not be surjective, so the image of $h$, denoted $h(\mathbb{N})$ is a sub-set of $\mathbb{N}$. Then $g(h(\mathbb{N}))\subseteq g(\mathbb{N})$. Using this fact, $|a_{n_{k_j}}-a|\geq\varepsilon$ for each $j\in\mathbb{N}$, since each $n_{k_j}$ is one of the $n_k$'s for which $|a_{n_k}-a|\geq\varepsilon$.

Alternatively, if you want to prove this result indirectly, you can define the following object:

Let $\left(a_n\right)_{n\in\mathbb{N}}$ be a sequence of real numbers. Define the set
$$E=\{l\in\mathbb{R}:\exists\text{ a sub-sequence }\left(a_{n_k}\right)_{k\in\mathbb{N}}:\lim\limits_{k\to\infty}a_{n_k}=l\}$$

$E$ is called the set of sub-sequential, or partial limits of $\left(a_n\right)_{n\in\mathbb{N}}$.

One can then show that $\left(a_{n}\right)_{n\in\mathbb{N}}$ converges to a real number if and only if $E$ is a singleton set containing a real number. In the case where $\left(a_n\right)_{n\in\mathbb{N}}$ "converges" to $\pm\infty$, then this happens if and only if $E=\{\pm\infty\}$, respectively.


问:
Does $4b^3pq=2a^2+b$ has a solution for non-zero integers $a,b$ and odd primes $p<q$?
Just recently, I had an equation on my hands that I thought had no solutions:
Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?

I was on a wrong path and now have an equation $4b^3pq=2a^2+b$ that really do not brings me any non-zero integer solution $a,b$ where $2<p<q$ are odd primes. What I see is that $b$ must be even and when I substitute $b=2t$ I have to solve $0=16t^3pq-t-a^2$ for $t$ (quadratic equation) and analyze the discriminant?

答:
More generally, the equation
$$
2b^3c=2a^2+b
\qquad\;\;\;\;
$$
has no integer solutions with $c > 0$ and $a,b\ne 0$.

Suppose otherwise.
\begin{align*}
\text{Then}\;\;&
2b^3c=2a^2+b
\qquad\qquad\qquad\;
\\[4pt]
\implies\;&
2b^3c-b=2a^2
\\[4pt]
\implies\;&
b(2b^2c-1)=2a^2
\\[4pt]
\implies\;&
b(2b^2c-1) > 0
\\[4pt]
\implies\;&
b > 0\;\;\text{(since $2b^2c-1 > 0)$}
\end{align*}
Next let $d=\gcd(a,b)$.

Then
$$
\left\lbrace
\begin{align*}
a&=da_1\\[4pt]
b&=db_1\\[4pt]
\end{align*}
\right.
\qquad\qquad\qquad\;\;\;
$$
for some integers $a_1,b_1$ with $b_1 > 0$ and $\gcd(a_1,b_1)=1$.
\begin{align*}
\text{Then}\;\;&
2b^3c=2a^2+b
\qquad\qquad\qquad\;\;\;\;
\\[4pt]
\implies\;&
2(db_1)^3c=2(da_1)^2+db_1
\\[4pt]
\implies\;&
2d^3b_1^3c=2d^2a_1^2+db_1
\\[4pt]
\implies\;&
2d^2b_1^3c=2da_1^2+b_1
\\[4pt]
\implies\;&
2d{\,\mid\,}b_1
\\[4pt]
\end{align*}
hence $b_1=2db_2$ for some positive integer $b_2$.

Then since $\gcd(a_1,b_1)=1$ and $b_2{\,\mid\,}b_1$, it follows that $\gcd(a_1,b_2)=1$.
\begin{align*}
\text{Then}\;\;&
2d^2b_1^3c=2da_1^2+b_1
\qquad\qquad\qquad
\\[4pt]
\implies\;&
2d^2(2db_2)^3c=2da_1^2+2db_2
\\[4pt]
\implies\;&
16d^5b_2^3c=2da_1^2+2db_2
\\[4pt]
\implies\;&
8d^4b_2^3c=a_1^2+b_2
\\[4pt]
\implies\;&
b_2{\,\mid\,}a_1^2\
\\[4pt]
\implies\;&
b_2=1\;\;\text{(since $\gcd(a_1,b_2)=1$)}
\\[4pt]
\implies\;&
8d^4c=a_1^2+1
\\[4pt]
\implies\;&
a_1^2\equiv -1\;(\text{mod}\;4)
\\[4pt]
\end{align*}
which is not possible.


问:
Support of a measure on Borel sets not well defined.
Let $(X, \tau$) be a topological space and let $B$ be the Borel $\sigma$-algebra generated by $\tau$. Let  $(X,B,\mu)$ be a measure space. Then the support of the distribution $\mu$ is defined as :
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all U open neighborhood of }x\}.$$ Now in the wikipedia article about  Radon measures, it is written

A common problem is to find a good notion of a measure on a
topological space that is compatible with the topology in some sense.
One way to do this is to define a measure on the Borel sets of the
topological space. In general there are several problems with this:
for example, such a measure may not have a well defined support.
Another approach to measure theory is to restrict to locally compact
Hausdorff spaces[...]

But I don't see why the support is not well defined when we use the above expression.

答: Support  is defined in that article as the smallest closed set $S$ such that $\mu (S^{c})=0$. This is the usual definition of supprt and it is the most convenient. With this definition the complement of the support is the largest open set of measure $0$. If you consider the collection  of all open sets of measure $0$ their union is cetainly open but it may not have measure $0$ since an uncountable union of sets of measure $0$ need not have measure $0$. That is the reason why not every Radon measure has a support.  On any second countable topological space every Radon measure has  a support.


问:
Limits of an Improper Integral: $\lim\limits_{n\to\infty} \int_{-\infty}^{+\infty}f(x)\sin(nx)dx$
Is

$$
\lim\limits_{n\to\infty} \int_{-\infty}^{+\infty}f(x)\sin(nx)dx=0
$$

when $\int_{-\infty}^{+\infty}f(x)dx$ is convergent but not absolutely convergent?

答:
Try the inverse Fourier transform of $$\hat{f}(\omega)= \sum_{k\ge 1} e^{-(k^2 (\omega-k))^2}e^{i \omega e^k}\in L^1\cap L^2$$

We have $\lim_{n\to \infty}\hat{f}(n)e^{-i n e^k}=1,\lim_{n\to -\infty}\hat{f}(n)=0$.

The $e^{i \omega e^k}$ term is there to ensure that $\lim_{A\to \infty}\int_{-A}^A e^{-inx} f(x)dx$ converges to $\hat{f}(n)$.


问:
$X$th Permutations of $1,2,3,4,5$ is $25314$. Find X
The permutations of $1,2,3,4,5$ are lexicographically ordered.
$X$th permutation is $25314$. Find $X$.

I am getting $1*4! + 3*3! + 1*2! + 1=45$. Is it correct?

Reasoning:
There are $1*4!$ numbers of the form of $1X_1X_2X_3X_4$.
Similarly, there are $3*3!$ numbers of the form of $2YX_1X_2X_3$ where $Y \in \{1,3,4\}$. We continue the pattern this way to get the answer.

答:
Your answer is correct.

There are $4!$ permutations which have a $1$ in the first position, all of which precede the first permutation which begins with a $2$.

Each permutation which begins with $21$, $23$, or $24$ precedes the first permutation which begins with $25$.  Since there are $3!$ permutations which begin with $21$, $3!$ permutations which begin with $23$, and $3!$ permutations which begin with $24$, there are $3 \cdot 3!$ permutations which begin with $2$ that precede the first permutation which begins with $25$.

Hence, there are $4! + 3 \cdot 3!$ permutations which precede the first permutation which begins with $25$.

Of the permutations which begin with $25$, those that begin with $251$ precede the first permutation which begins with $253$.  Since there are $2!$ permutations which begin with $251$, there are $4! + 3 \cdot 3! + 1 \cdot 2!$ permutations which precede the first permutation which begins with $253$.

Of the permutations that begin with $253$, $25314$ is the first.  Hence, the position of the permutation $25314$ if the permutations of $1, 2, 3, 4, 5$ are listed in increasing order is indeed
$$4! + 3 \cdot 3! + 1 \cdot 2! + 1 = 45$$


问:
Calculate $\int_{\rm C}(e^{x^2}- y{x}^{2}){\rm d}x+ \cos({y}^{2}){\rm d}y$
I am trying to calculate the curve integral
$$\int_{\rm C}(e^{x^2}- y{x}^{2}){\rm d}x+ \cos({y}^{2}){\rm d}y$$
where ${\rm C}$ is the unit circle passed through one lap counter clockwise. I am using Green's theorem which renders me the new integral
$$\iint x^2{\rm d}x{\rm d}y$$
and using polar coordinates I get the integral
$$\iint r^3\cos^2\theta{\rm d}\theta{\rm d}r$$
where I integrate $\theta$ from $0$ to $2\pi$ and $r$ from $0$ to $1$. This gives me $0$ as the result but that is not the correct answer. What have I done wrong in my calculations ?

答:
Everything else is fine but you have a mistake in your integral.

$\displaystyle \iint_R r^3 cos^2\theta \ d\theta \ dr = \int_0^1 \int_0^{2\pi} r^3 \frac{1+cos2\theta}{2} \ d\theta \ dr$

$ = \displaystyle \int r^3 \bigg[\frac{2\theta+ \sin2\theta}{4}\bigg]_0^{2\pi} \ dr  = \int_0^1 \pi r^3 dr = \frac{\pi}{4}$


问:
Uniform Asymptotic Expansion of a ratio modified Bessel functions of the first kind
Let $I_{\nu}(\cdot)$ be the modified Bessel function of the first kind of order $\nu$. Can someone please confirm if, when $\nu\rightarrow\infty$, the following approximation

$$\frac{I_{\frac{\nu}{2}}(\frac{\nu}{2}x)}{I_{\frac{\nu}{2}-1}(\frac{\nu}{2}x)}=\frac{x}{1+(1+x^{2})^{1/2}}+O\!\left(\frac{1}{\nu}\right)$$

holds uniformly with respect to $x$? How can I show this?

Thanks for any help.

答:
Suppose that $x>0$ and $\nu$ is large and positive. First note that, by $(10.29.2)$,
$$
\frac{{I_\nu  (\nu x)}}{{I_{\nu  - 1} (\nu x)}} = \frac{{I_\nu  (\nu x)}}{{I'_\nu  (\nu x) + (1/x)I_\nu  (\nu x)}} = \cfrac{1}{{\displaystyle\frac{{I'_\nu  (\nu x)}}{{ I_\nu  (\nu x)}} + \frac{1}{x}}}.
$$
If we denote
$$
\eta  = \sqrt {1 + x^2 }  + \log \left( {\frac{x}{{1 + \sqrt {1 + x^2 } }}} \right),
$$
then
$$
I_\nu  (\nu x) = \frac{{\mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } (1 + x^2 )^{1/4} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right)
$$
and
$$
I'_\nu  (\nu x) = \frac{{(1 + x^2 )^{1/4} \mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } x}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right),
$$
as $\nu\to +\infty$, uniformly with respect to $x>0$. These follow from $\S10.41.\mathrm{(ii)}$ and the double asymptotic properties of these expansions (cf. $\S10.41.\mathrm{(iv)}$). Accordingly,
$$
\frac{{I'_\nu  (\nu x)}}{{I_\nu  (\nu x)}} = \cfrac{{\displaystyle\frac{{(1 + x^2 )^{1/4} \mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } x}}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right)}}{{\displaystyle\frac{{\mathrm{e}^{\nu \eta } }}{{\sqrt {2\pi \nu } (1 + x^2 )^{1/4} }}\left( {1 + \mathcal{O}\!\left( {\frac{1}{{\nu \sqrt {1 + x^2 } }}} \right)} \right)}} = \frac{{\sqrt {1 + x^2 } }}{x} + \mathcal{O}\!\left( {\frac{1}{{\nu x}}} \right),
$$
and therefore,
\begin{align*}
\frac{{I_\nu  (\nu x)}}{{I_{\nu  - 1} (\nu x)}} & = \cfrac{1}{{\displaystyle\frac{{\sqrt {1 + x^2 } }}{x} + \mathcal{O}\!\left( {\frac{1}{{\nu x}}} \right) + \frac{1}{x}}} = \cfrac{x}{{\displaystyle\sqrt {1 + x^2 }  + 1 + \mathcal{O}\!\left( {\frac{1}{\nu }} \right)}}
\\ & = \frac{x}{{\sqrt {1 + x^2 }  + 1}} + \mathcal{O}\!\left( {\frac{x}{{\nu (\sqrt {1 + x^2 }  + 1)^2 }}} \right) = \frac{x}{{\sqrt {1 + x^2 }  + 1}} + \mathcal{O}\!\left( {\frac{1}{\nu }} \right).
\end{align*}
Throughout the whole derivation, the implied constants in the big-$\mathcal{O}$'s are independent of $x$ and $\nu$.


问:
Projective finitely generated module over noetherian ring
Let $A$ be a noetherian ring and $M$ be a finitely generated $A$-module, I want to prove the following

M is projective $\iff$ for all $P\subset A$ prime ideal the localisation $M_P$ is a free $A_P$-module

As far as I know since $A$ is noetherian and $M$ is finitely generated then it is also finitely presented. Moreover the localisation of a noetherian ring is still noetherian and so $A_P$ are all noetherian rings. It seems to me that I have to use Nakayama's lemma but I don't see how.

Is there a direct proof of this statement by using directly properties of projective modules? (Such as they are direct summand of a free module) Thanks.

答:
If $M$ is projective, then there is module $K$ s.t. $M \oplus K \cong R^{(\Lambda)}$. For any multiplicative set $S \subset R$, one gets $S^{-1}M \oplus S^{-1} K \cong (S^{-1}R)^{(\Lambda)}$. Hence, any localization of a projective module is projective.

If $(R,m)$ is local and $M$ is finitely generated projective then $M$ is free. Here is a proof. The short exact sequence
$$
0 \to K \to R^n \to M \to 0,
$$
splits, hence $K$ is finitely generated and sequence
$$
0 \to K \otimes R/m \to R^n \otimes R/m \to M \otimes R/m \to 0,
$$
also splits. By Nakayama's lemma we can choose generators of $M$ in such way that the map $R^n \otimes R/m \to M \otimes R/m$ is an isomorphism, then $K \otimes R/m \cong K/mK=0$. By Nakayma's lemma $K=0$, since it is finitely generated.

Fact: if $M$ is finitely presented, then $S^{-1}Hom_R(M,N) \cong Hom_{S^{-1}R} (S^{-1}M, S^{-1}N)$.

To check that $M$ is projective is enough to check that for any surjection $N \to N'$ map $Hom(M,N) \to Hom(M, N')$ is surjective, but this can be checked stalkwise if $M$ is finitely presented (by the fact).

There are plenty of other proofs, but often they involve flat modules, in this proof we only use two ways to characterize projective modules and the fact. Notice that this proof doesn't use that $R$ is Noetherian.


问:
How can you solve this probability problem using combinatorics instead of diagrams?
A game involves rolling $2$ $6$-sided dice, Another identical die is then rolled, and the game is won if the number rolled on the third die is in between the previous two rolls. What is the probability that a player has no chance of winning before the third die roll? And what is the probability that the player wins the game

I tried a method of eliminating the sample space ie identical numbers are banned, consecutive numbers are banned. I'm still having trouble keeping track of all the cases though - is there a general method for these types of problems? I also want to know of a way that uses combinatorics $6C1$ but I'm having trouble with thinking of one up myself. I also know I could draw a sample space diagram but I even have trouble with that! I drew a $6$ by $6$ grid but I don't know how to consider the fact that it has to be "between" - it seems I need a $3$D table!

Edit: I have solved part one of the problem using a table. $P = \frac{4}{9}$. I still welcome other solutions using other methods

答:
There are six outcomes in which the first two rolls are identical.  There are five pairs of consecutive numbers.  Each of these five pairs can occur in two ways, depending on which of the dice has the higher outcome.  Hence, there are
$$6 + 2 \cdot 5$$
outcomes which make it impossible to win on the third roll.  Thus, the probability that it is impossible for the player to win on the third roll is indeed
$$\frac{6 + 2 \cdot 5}{36} = \frac{16}{36} = \frac{4}{9}$$
as you found.

For the player to win, there must be three distinct outcomes on the three dice, which can occur in $\binom{6}{3}$ ways, and the third roll must be the middle value, which can occur in two ways, depending on whether the first or second die shows the highest value.  Hence, there are
$$2\binom{6}{3}$$
favorable outcomes among the $6^3$ outcomes of the three rolls.  Hence, the probability that the player wins the game is
$$\frac{2\dbinom{6}{3}}{6^3} = \frac{40}{216} = \frac{5}{27}$$

Addendum:  An alternate way of phrasing the above argument is that there are $6 \cdot 5 \cdot 4$ ways of obtaining three distinct outcomes in a sequence of three rolls, one third of which have the third roll as the middle value, giving
$$\frac{6 \cdot 5 \cdot 4}{3} = 40$$
favorable outcomes.


问:
Probability that sampled data will fall within range?
What mathematical concept do I need to read about to answer the following question?

Given:

A population with an unknown mean, variance.

A sample with a known size, mean, variance.

What is the range of values for 95% of the population data?

I initially thought that confidence intervals answered this question, but per https://math.stackexchange.com/a/1967239/25814 they provide a statement about the population mean, not its data.

答:
Perhaps Chebyshev's  inequality is what you're looking for.
https://en.m.wikipedia.org/wiki/Chebyshev%27s_inequality</p>


问:
Probability question involving colored dice
This question is really getting to me, a friend of mine has suggested a solution but I believe it is incorrect.
Anyway here's the question:

You throw 9 coloured dice. Every dice has 2 white sides, 2 red sides and 2 blue sides. What is the chance to get every color three times?

I'd be over the moon if you can help me find the solution as it's been getting to me. Thanks.

答:
Since each colour is on two sides, the probability for each colour on each die is $1/3$. (Unfortunately we can't say that the dice are equivalent to $3$-sided dice because there's no such thing.)

The probability to get every colour thrice is the number of ways of colouring nine dice with three dice coloured in each colour, over the number of ways of colouring nine dice with three colours. The latter is simply $3^9$, and the former is the multinomial coefficient

$$
\left({9\atop3,3,3}\right)=\frac{9!}{3!3!3!}=1680\;,
$$

so the probability is

$$
\frac{1680}{3^9}=\frac{560}{3^8}=\frac{560}{6561}\approx0.085\;.
$$

ContentLicense: CC BY-SA 3.0


问:
Integral inequality with a function twice differentiable
Let $f:[0,1]\longrightarrow\mathbb{R}$ be a function twice differentiable with continous second derivative and $f(1)=f(0)$. The inequality:
$$\int_{0}^{1}(f''(x))^2dx\geq 120\left(\int_{0}^{1}xf'(x)dx\right)^2$$ holds?</p>

答:
Since $f(1)=f(0)\Rightarrow \int_0^1f'(x)\,\mathrm{d}x=0$, By integration by parts,
\begin{equation}\label{MSE-730580}
\int_0^1g(x)f''(x)\mathrm{d}x=g(x)f'(x)\bigg|_0^1-\int_0^1g'(x)f'(x)\mathrm{d}x \tag{1}
\end{equation}
Let $g(x)$ be a polynomial defined as,
$$g(x)=ax^2+bx+c\Rightarrow g'(x)=2ax+b$$
Let
$$g(x)f'(x)\bigg|_0^1=0\Rightarrow g(0)=g(1)=0\Rightarrow b=-a$$
we can get,
$$g(x)=ax(x-1)\Rightarrow g'(x)=2ax-a \tag{2}$$
by $(1),(2)$, we have
$$\int_0^1x(x-1)f''(x)\mathrm{d}x=\int_0^1(1-2x)f'(x)\mathrm{d}x=-2\int_0^1xf'(x)\mathrm{d}x$$
by the Cauchy-Schwarz inequality,
\begin{align*}
\left(\int_0^1xf'(x)\mathrm{d}x\right)^2
&=\left(\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)f''(x)\mathrm{d}x\right)^2\\
&\leqslant\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x
\end{align*}
Therefore,
$$120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x$$
because,
$$\int_0^1\bigg(-\frac{1}{2}x(x-1)\bigg)^2\mathrm{d}x=\frac{1}{120}$$


问:
Dimension of the complex projective space $\mathbb{C}\mathbb{P}^n$ as almost complex manifold
I have already shown that the complex projective space $\mathbb{C}\mathbb{P}^n$ is a complex manifold by checking the required properties of the transition maps. Since every complex manifold is an almost complex manifold, $\mathbb{C}\mathbb{P}^n$ is also an almost complex manifold. I wonder what is its dimension. I have proved that almost complex manifolds always have even dimension. But how can I determine the dimension of the complex projective space $\mathbb{C}\mathbb{P}^n$ as an almost complex manifold? And how can I prove this answer? It would be great if someone could help me with this proof.  Thanks in advance for your help!

答:
An almost complex manifold is a real smooth manifold $M$ with a bundle map
$$
J: TM\rightarrow TM
$$
such that $J^2=-id$.  When you ask what is the dimension of $\mathbb{CP}^n$ as an almost complex manifold, what you're asking is what is its dimension as a real manifold.  If $M$ is a complex manifold of complex dimension $n$, then the dimension of $M$ as a real manifold is $2n$.  In other words,
$$
\dim_{\mathbb{R}}M=2\dim_{\mathbb{C}}M
$$
Since $\dim_{\mathbb{C}}\mathbb{CP}^n=n$, then $\dim_{\mathbb{R}}\mathbb{CP}^n=2n$.


问:
Finding complex number for intersection of a line and a circle?
How to find complex number for intersection of a line and a circle?

In particular, if $M=$ midpoint $BC$, then I want $AM ∩ (ABC)$ in complex numbers.

Let me describe why I needed this,

I was doing a problem which I reduced to proving $4$ points $(B,T,M',J)$ are cyclic where $ABC$ is an acute scalene triangle, $T$ is the intersection of tangents to $(ABC)$ from $B$ and $C$, $M' = AM ∩ (ABC)$ ;($M =$ midpoint $BC$) , and $J = AB ∩ CM'$.

I tried to prove this synthetically but I could not, so I decided to bash. First I tried coordinate but that was too long, so I decided to go with complex because that felt feasible. But then I remembered that I have not yet studied complex numbers (except when I would read EGMO complex numbers for fun, like a storybook). So I opened EGMO and could get complex numbers for everything like how to prove cyclic, intersection, $T$ etc. but Point $M'$ was a problem. I couldn't find how to get it anywhere so I come to MSE.

Please help me, Thanks!

答:
$AT$ is the symmedian of $\triangle ABC$ through $A$. Hence, $\angle BAT=\angle MAC=\angle M'AC$ and also $\angle CAT=\angle M'AB$.

Now, notice that,  $\angle JAT=\angle BAT=\angle M'AC=\angle M'CT=\angle JCT$ and thus quadrilateral $AJTC$ is cyclic.

Now,  $\angle M'JT=\angle CJT=\angle CAT=\angle BAM'=\angle M'BT$ and therefore quadrilateral $BJTM'$ is cyclic.

P.S. The symmedian through a vertex is the reflection of the median over the internal angle bisector through that vertex. It is provable by euclidean geometry that the symmedian through a vertex goes through the intersection point of the tangents to the circumcircle at the other two vertices.
Check out this link for more:  https://brilliant.org/wiki/symmedian/#:~:text=A%20symmedian%20of%20a%20triangle,D%3D%E2%88%A0CAM.


问:
How many different menus can the customer order?
There is a restaurant that has 5 different options in the menu. A customer can only choose 3 of them for a meal. So how many different orders can the customer give for 3 meals?</p>

答:
For a single meal he can choose one option in $5$ ways, two options in $\binom{5}{2}$ ways, and three options in $\binom{5}{3}$ ways. Add these to get number of ways to choose a single meal.

If he is to order three meals, he can do it in the cube of the previous total.

Here the "binomial coefficient" $\binom{n}{k}$ is the number of $k$ element subsets of an $n$ element set. It can be calculated as $n!/(k!(n-k)!)$ where $b!$ is "b factorial" and is the product of the integers from $1$ to $b.$


答:
Hint:
Set $u=x^2-3x$ and use that $\:\bigl(u^3\bigr)'=3u^2u',\:$ hence
$$\bigl(u^3\bigr)''= 6u\mkern2mu u'^2+3u^2u''=\cdots$$

Title: Show that the second derivative of $y=(x^2−3x)^3$ is $6(x^2-3x)(5x^2-15x+9)$

Body:
How do you find the second derivative of $$y=(^2−3)^3$$  and show that it is $6(x^2-3x)(5x^2-15x+9)$

Every time I worked it out, I used the chain rule first to get the first derivative which is
$ {3(x^2-3x)^3}\cdot{2x-3}$. Then I used the product rule + chain rule but I only managed to get the expanded form of what I was supposed to "show" which is $30x^4-180x^3+324x^2-162x$

Any help would be appreciated.


问:
Lagrange polynomial interpolation maximum degree
I want to prove that no polynomial of degree $1$ that passes through $(0, \cos(0))$, $(0.6, \cos(0.6))$ and $(0.9, \cos(0.9))$. By the following theorem:

Theorem 1. If $x_{0}, x_{1}, \ldots, x_{n}$ are $n+1$ distinct numbers and $f$ is a function whose values are given at these points, then a unique polynomial $P(x)$ of degree at most $n$ exists with
$$
f\left(x_{k}\right)=P\left(x_{k}\right), \quad \text { for each } k=0,1, \ldots, n
$$
This polynomial is given as
$$
P(x)=f\left(x_{0}\right) L_{n, 0}(x)+\cdots+f\left(x_{n}\right) L_{n, n}(x)=\sum_{k=0}^{n} f\left(x_{k}\right) L_{n, k}(x), \quad \text { where } L_{n, k}=\prod_{i=0 \atop i \neq k}^{n} \frac{x-x_{i}}{x_{k}-x_{i}}
$$

Suppose that I picked $x_0$ and $x_1$ for the degree $1$ polynomial, that by the above theorem, $f(x_3)$ does not necessarily equal $P(x_3)$, so it does not guarantee to pass through all 3 points. However, this only shows that it is guaranteed to pass through 2 points but it does not state that it must not pass through all 3 points, how can I prove that in this case? Thank you

答:
You get using the equivalent Newton interpolation formula
$$
p_2(x)=f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)=p_1(x)+f[x_0,x_1,x_2](x-x_0)(x-x_1)
$$
So at $x=x_2$ you get
$$
f(x_2)=p_2(x_2)=p_1(x_2)+f[x_0,x_1,x_2](x_2-x_0)(x_2-x_1)
$$
Now $f[x_0,x_1,x_2]=\frac12f''(\xi)$ for some point in the middle, which means that the error
$$
p_1(x_2)-f(x_2)=-f[x_0,x_1,x_2](x_2-x_0)(x_2-x_1)
$$
is decidedly positive for the given situation, as $\cos x>\frac12$ for $x\in[0,0.9]$.


问:
Absolute error formula
Trying to figure out a problem in my textbook.
In my exercise in a textbook, a problem says that Determine the error in the approximation given that the actual length is $3.7437137$.

And in the solution in the back of the book, it says that The approximation is $3.743713701$ with an absolute error of $0.2226462$

How is $|3.743713701-3.7437137|= 0.2226462$ ? I am confused. Any help is appreciated

答: There's a typo. The actual length is $3.96635989732264...$

Title: Absolute error formula


问:
What is meant by $\subset$ minimal?
Consider a $\sigma-$algebra $\mathcal{A}$ and the $\bf{unique}$ set  of $\subset$-minimal elements of  $\mathcal{A} \setminus\emptyset.$

What does this mean?

Is there an inutitive explanation why this set is unique?

答:
The word "unique" here is very weird. It would be like saying "the unique set of divisors of 12". Sure, it's unique, because we defined exactly what its elements are, but it's a weird thing to say.

Here you have a set $\mathcal{A}$ of subsets of some $X$, and you are considering the subset $\mathcal{B}\subset \mathcal{A}$ of subsets $Y\subset X$ such that $Y\in \mathcal{A}$, and you can't have $Z\subset Y$ if $Z\in \mathcal{A}$ unless $Z=\emptyset$.

By the way it should be "$\mathcal{A}\setminus \{\emptyset\}$" and not "$\mathcal{A}\setminus \emptyset$", which is just $\mathcal{A}$.

Title: What is meant by $\subset$ minimal?


问:
Find $\arg(z),$ if $|z|=|z-|z||$ and $z$ - non-zero number
$z$ - non-zero number. I need to find the argument $z,$ when $|z|=|z-|z||.$ I'm literally stopped. How can I solve this problem?

答:
Let $z = re^{i\theta}, r>0 \Rightarrow |z| =r$

So,

$\begin{align}&r = |re^{i\theta}- r| = r|e^{i\theta} - 1| \\\Rightarrow &1 = |e^{i\theta} - 1| = \sqrt{(\cos\theta -1)^2 + \sin^2\theta} = 2\left|\sin\frac\theta2\right|\\\Rightarrow& 1 = 2\left|\sin\frac\theta2\right|\end{align}$

Can you finish now?


问:
Simple Probabilities Problem - Ιndependence
I have this problem:

Possibilities:

$A,P(A)=0.3$

$B,P(B)=0.6$

$C,P(C)=0.1,$ $C$ is independent of $A$ and independent of $B$

$P(A$ and $ B)=0.2$

F: A or B or C

Find $P(F)$

I know that it easy, but I cannot find the right solution.
Here is what I made so far:
$P(F)= P(A) + P(B) + P(C) - 2*P(A$ and $B)=0.6$ but this answer isn't the right one.
Anny suggestions? Thanks.

答:
Hint: Recall your Inclusion-Exclusion Principle.

For three sets that would be:

$$\Pr(A\cup B\cup C) =\Pr(A)+\Pr(B)+\Pr(C)-\Pr(A\cap B)-\Pr(A\cap C)-\Pr(B\cap C)+\Pr(A\cap B\cap C)$$


问:
Why is the endomorphism ring of $\mathbb{Z}\times\mathbb{Z}$ noncommutative?
So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?</p>

Body:
HINT $\rm\ \ \ a\ x + y\:\ \ne\:\ a\ (x+y)\ \ $ if $\rm\ \ a\ne 1$

$\quad\ $ i.e. $\rm\ \ \ \ \ (x,\:y)\ \mapsto\ (a\:x,\:y) $

$\quad$ and $\rm\ \ \ \ \:(x,\:y)\ \mapsto\ (x+y,\:y)\ \ $ do not commute.


问:
$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \bar A$, $f(a)\in \overline{ f(A)}$.
Definiton.

$f$ is continuous at $a$ provided that for each open set $V$ in $Y$ containing $f(a)$ there is an open set $U$ in $X$ containing $a$ such that $f(U) \subset V$.

Problem.

$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \overline{A}$, $f(a)\in \overline{ f(A)}$.

答:
The condition $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ can be taken as the definition of continuity (provided one defines a closure operator before one defines open sets, of course). Let's assume the

DEF $f:(X,\mathscr T)\to(Y,\mathscr U)$ is continuous if for each open $A\in Y$, $f^{-1}(A)$ is open in $X$.

We know that this is equivalent to

DEF* $f:(X,\mathscr T)\to(Y,\mathscr U)$ is continuous if for each closed $F\in Y$, $f^{-1}(F)$ is closed in $X$.

Then we have the

THM $f:(X,\mathscr T)\to (Y,\mathscr U)$ is continuous if and only if for each $A\subseteq X$, $f(\bar A)\subseteq \overline{f(A)}$

PROOF Suppose that $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq  f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.

Conversely, suppose that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$. Let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.

See this for more.


答:
Hint:

Consider the Argand-Cauchy plane $\:Oxy\:$ and let $M$ be the affix of $z$ and $A$ be the point on the real axis with abscissa $|z|$. Observe that $\bigl|z-|z|\bigr|$ is the length of $\:\overrightarrow{\!AM}$. What can you deduce for triangle $OAM$?

Title: Find $\arg(z),$ if $|z|=|z-|z||$ and $z$ - non-zero number

Body: $z$ - non-zero number. I need to find the argument $z,$ when $|z|=|z-|z||.$ I'm literally stopped. How can I solve this problem?


问:
What are examples of almost complex structures on Kahler Manifolds
I am studying Kahler Manifolds for applications in non-Hermitian Quantum Mechanics. I am struggling to get an intuitive understanding of the almost complex structure $J$.

I would like to formulate $J$ for specific Kahler Manifolds in particular coordinate systems. For example:

$\mathbb{C}P^1$ is a Kahler manifold and parametrised by $z$. The Kahler form for this manifold is $K=\tfrac{i}{2}\frac{dz\wedge{d}\bar{z}}{(1+z\bar{z})^2}$ and metric is defined via $ds^2=\frac{Re(dz\otimes{d}\bar{z})}{(1+z\bar{z})^2}$. In this example, what is $J$ such that $\omega(X,Y)=g(JX,Y)$?

As $\mathbb{C}P^1\cong{S^2}$, we can parametrise $S^2$ with coordinates $x,y,z$. Can we express a symplectic form, metric tensor and almost complex structure in terms of $x,y,z$ on $S^2$? Would this now be a Poisson Manifold and would the techniques have to be adjusted?

Thank you in advance for your help,

Wasim

答:
The simplest Kahler manifold is $\mathbb{C}^n=\mathbb{R}^{2n}$.  Write the $i$th complex coordinate on $\mathbb{C}^n$ as $z_i=x_i+\sqrt{-1}y_i$. The natural complex structure on $\mathbb{C}^n$ is
$$
J\frac{\partial}{\partial x_i}=\frac{\partial}{\partial y_i},~~J\frac{\partial}{\partial y_i}=-\frac{\partial}{\partial x_i}
$$
The standard symplectic form $\omega_0$ on $\mathbb{R}^{2n}$ is then a Kahler form for $(\mathbb{C}^n,J)$.  Explicitly, $\omega_0$ is
$$
\omega_0=\sum_i dx_i\wedge dy_i
$$
One can then check that $\omega_0(J\cdot,J\cdot)=\omega_0(\cdot,\cdot)$ and
$$
\omega_0(u,Ju)>0
$$
for all nonzero $u\in T\mathbb{R}^{2n}$.


问:
How do I show that if the ratio of the periods of an elliptic function is irrational then the function reduces to a constant?
I know that taking $\omega_1$ and $\omega_2$ periods of an elliptic function, the ratio $\frac{ω1}{ω2}$
has to be not real, in fact when $\frac{ω1}{ω2}$ is real the parallelogram collapse, when $\frac{ω1}{ω2}$
is rational the function reduces to a singly periodic function but how do I show that if $\frac{ω1}{ω2}$ is irrational then the function reduces to a constant?

答:
For $f$ meromorphic and $w_1,w_2$ periodic and $w_1/w_2$ real irrational.

If $f$ has a pole on the real line at $a$ then it also has a pole a $a+ \Bbb{Z}w_1+\Bbb{Z} w_2$, which is dense in $\Bbb{R}$, whence $f$ isn't meromorphic, a contradiction.

Therefore $f$ has no pole on the real line, it is continuous, so that $f(a)=f(a+ \Bbb{Z}w_1+\Bbb{Z} w_2)$ and the density of $a+ \Bbb{Z}w_1+\Bbb{Z} w_2$ in $\Bbb{R}$ implies that $f(\Bbb{R})=f(a)$, ie. $f$ is constant on the real line whence on the whole of $\Bbb{C}$ by analytic (meromorphic) continuation.


答:
$p(A \vee B) = p(A) + p(B) - p(A \wedge B) = 0.3 + 0.6 - 0.2 = 0.7$

Consider the event $E$ to represent $A \vee B$.  Then

$p(E) = 0.7.$

Since event $C$ is independent of both $A$ and $B$, $C$ is independent of $E$.

It is desired to compute $p(E \vee C).$

Since $C$ is independent of $E$, $p(E \wedge C) = 0.7 \times 0.1 = 0.07.$

Thus, $p(E \vee C) = p(E) + p(C) - p(E \wedge C) = 0.7 + 0.1 - 0.07 = 0.73.$

Title: Simple Probabilities Problem - Ιndependence

Body:
I have this problem:

Possibilities:

$A,P(A)=0.3$

$B,P(B)=0.6$

$C,P(C)=0.1,$ $C$ is independent of $A$ and independent of $B$

$P(A$ and $ B)=0.2$

F: A or B or C

Find $P(F)$

I know that it easy, but I cannot find the right solution.
Here is what I made so far:
$P(F)= P(A) + P(B) + P(C) - 2*P(A$ and $B)=0.6$ but this answer isn't the right one.
Anny suggestions? Thanks.


问:
For each $n \in \mathbb{N}$ consider $A_n = \{1/n\}\times [0,1]$ and let $X=\bigcup_{n \in \mathbb{N}}A_n \cup \{(0,0)(0,1)\}$

For each $n \in \mathbb{N}$ consider $A_n = \{1/n\}\times [0,1]$ and let $X=\bigcup_{n \in \mathbb{N}}A_n \cup \{(0,0)(0,1)\}$.

Show that $\{(0,0)\}$ and $\{(0,1)\}$ are connected components.

I have an Idea of how to proceed with the proof but It's kind of messy. Let $C$ be a connected component containing $(0,1)$, I want to see that $C=\{(0,1)\}$. Suppose that there exist $y \in X$, $y \neq (0,1)$, $y\neq (0,0)$, such that $y \in C$. Then there exist $n_0 \in \mathbb{N}$ such that $y \in A_{n_0}$. Then since $A_{n_0}$ is a (path)-connected set containing $y$ then $A_{n_0} \subseteq C$. Which is a contradiction since I could discconect $C$ in $C=\{(0,1)\}\cup A_{n_0}$. The problem I'm having is that when writting the disconnection, Im assuming no other element is in $C$ (other than the ones explicitly written in the disconnection). How can I fix this?

答:
The connected components of $X$ are the sets $A_n:=\{\frac1n\}\times [0,1]$ and also both singletons $\{(0,0)\}$ and $\{0,1)\}$:

Start by observing that $A_n$ is closed-and-open (clopen) in $X$: if $n > 1$, pick $a_n, b_n \in \Bbb R$ such that $\frac{1}{n+1} < a_n < \frac1n < b_n < \frac{1}{n-1}$ and note that $A_n = ((a_n, b_n) \times \Bbb R) \cap X$, so $A_n$ is open in $X$, and $A_n$ is already closed in $\Bbb R^2$ as the product of two closed sets $\{0\}$ and $[0,1]$, and so also closed in $X$. That $A_1$ is clopen is similar, $A_1 = (\frac34, \rightarrow) \times \Bbb R)) \cap X$.

$A_n$ is also connected (homeomorphic to $[0,1]$) and the combination of these two facts implies that each $A_n$ is maximally connected: if $A_n$ is a proper subset of $B$ then $A_n$ is itself a non-trivial clopen set of $B$ so that $B$ cannot be connected. This shows maximality of $A_n$ as a connected subspace.

Next, the components of a space $X$ always form a closed partition of $X$ of connected subspaces. $X$ consists of all the $A_n$ and the only points left are $\{(0,0),(0,1)\}$. This is a disconnected subspace (as is any two point set in a $T_1$ space: both singletons are closed in it, and disconnect it) so the remaining components must be $\{(0,0)\}$ and $\{(0,1)\}$, which are indeed closed and connected and complete the partition of $X$ into components.


问:
Another proof of $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2\cdot|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$.
I know how to prove the formula $|\vec{a}\times\vec{b}|^2 = |\vec{a}|^2\cdot|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2$, but my teacher asked me to prove it by using the formula $(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$, how to prove it?

答: Assume without loss of generality $\vec{a}\ne\vec{0}$. Set $\vec{c}=\vec{a}$ so$$(\vec{a}\times\vec{b})\times\vec{a}=a^2\vec{b}-\left(\vec{a}\cdot\vec{b}\right)\vec{a}.$$This is a cross product of perpendicular vectors, so its square modulus is$$|\vec{a}\times\vec{b}|^2a^2=a^4b^2+\left(\vec{a}\cdot\vec{b}\right)^2a^2-2a^2\left(\vec{a}\cdot\vec{b}\right)^2=a^2\left(a^2b^2-\left(\vec{a}\cdot\vec{b}\right)^2\right).$$Now cancel the $a^2$ factors.


问:
Simplifying $f(\sqrt{7})$, where $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$

If $f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}}$ ; then $f(\sqrt {7})=\; ?$

I tried solving this equation through many methods, I tried rationalizing, squaring, etc. But after each of them, the method became really lengthy and ugly.

I also noted that once we substitute $\sqrt {7}$ the inner part of the radical becomes imaginary. How to proceed with this piece of information?

Please help me with this problem. Any more innovative methods would be appreciated.

Answer: $4$.

Edit $[7^{th}$ March, $2021]$:

I was told yesterday that this question was wrong and that we were supposed to find $f(2\sqrt5)$. Although I can solve with $2\sqrt5$ the same way as given below in the answers.... What I don't understand is that why $\sqrt7$ doesn't work. Can someone please help?

答:
Using the denesting formula

$$\sqrt{a\pm\sqrt{b}} = \sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
We have
$$f(x) = \sqrt{x-4\sqrt{x-4}}+\sqrt{x+4\sqrt{x-4}} \\= 2\sqrt{\frac{x+\sqrt{x^2-16(x-4)}}{2}}=2\sqrt{\frac{x+\sqrt{(8-x)^2}}{2}}$$

Now $\sqrt7<8$, therefore
$$f(\sqrt 7)=2\sqrt{\frac{\sqrt 7+8- \sqrt 7}{2}}=2\cdot 2=4.$$

However as one of the members pointed out, the square root of a nonreal complex number is not well defined. We may get around this by requiring that the "square root" have nonnegative real part, but this is not standard practice.

Update: You can just square both sides directly:

$$(f(x))^2=2x+2\sqrt{x^2-(4-x)^2} = 2x+2\sqrt{(8-x)^2}$$
So if $x\leq 8$ then $f(x)=\sqrt{2x+2(8-x)}=4.$


问:
Determine the matrix of a quadratic function
I'm given a quadratic form $\Phi:\mathbb{R}^3\longrightarrow\mathbb{R}$, for which we know that:

$(0,1,0)$ and $(0,1,-1)$ are conjugated by the function

$(1,0,-1)$ belongs to the kernel

$\Phi(0,0,1)=1$

The trace is $0$

From here, I know the matrix must be symmetric, so it will have up to six unique numbers, I have four conditions that I can apply, and I will get four equations that relate the numbers of the matrix, but I still need two to full determine it. Applying the above I get that the matrix must be of the form:

$$A=\pmatrix{2c-1 & b & c\\b&-2c&-2c\\c & -2c & 1}$$

How do I determine $b$ and $c$?

答: Ok, solved it, the last two equations came from knowing the vector that was in the kernel, so it should be that $f_p[(1,-0,-1),(x,y,z)]=0$, being $f_p$ the polar form of $\Phi$: $f_p=\frac{1}{2}[\Phi(x+y)-\Phi(x)-\Phi(y)]$</p>


问:
Prove that $\int_0^1 \big(1-x^2\big) \big(f'(x)\big)^2\,dx \ge 24 \left(\int_0^1 xf(x)\,dx\right)^{\!2}$

Prove that if $f:[0,1] \to \mathbb{R}$ is a continuously differentiable function with $\int_0^1 f(x)\,dx=0$, then $$\int_0^1 \big(1-x^2\big) \big(f'(x)\big)^2\,dx \ge 24 \left(\int_0^1 xf(x)\,dx\right)^{\!2}.$$

I think that I should somehow use the Cauchy-Schwarz inequality, but I wasn't succesful in doing this. I know that
$$\left(\int_0^1 xf(x)\,dx\right)^2\le \int_0^1 x^2\,dx \cdot \int _0^1 f^2(x)\,dx=\frac{1}{3}\int _0^1 f^2(x)\,dx,$$ but this is clearly not enough.

答:
by integration by parts,
$$\int_0^1g(x)f'(x)\,\mathrm{d}x=g(x)f(x)\bigg|_0^1-\int_0^1g'(x)f(x)\,\mathrm{d}x \tag{1}$$
Let $g(x)$ be a polynomial defined as,
$$g(x)=ax^2+bx+c\Rightarrow g'(x)=2ax+b$$
Let
$$g(x)f(x)\bigg|_0^1=0\Rightarrow g(0)=g(1)=0\Rightarrow b=-a$$
we can get,
$$g(x)=ax(x-1)\Rightarrow g'(x)=2ax-a \tag{2}$$
by $(1),(2)$, we have
$$\int_0^1x(x-1)f'(x)\,\mathrm{d}x=-\int_0^1(2x-1)f(x)\,\mathrm{d}x=-2\int_0^1xf(x)\,\mathrm{d}x$$
by the Cauchy-Schwarz inequality,
\begin{align*}
\left(\int_0^1xf(x)\mathrm{d}x\right)^2
&=\left(\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)f'(x)\mathrm{d}x\right)^2\\
&=\left(\int_0^1\frac{-x(x-1)}{2\sqrt{1-x^2}}\cdot\sqrt{1-x^2}f'(x)\mathrm{d}x\right)^2\\
&\leqslant\int_0^1\bigg(-\frac{x(x-1)}{2\sqrt{1-x^2}}\bigg)^2\mathrm{d}x\int_0^1(1-x^2)(f'(x))^2\mathrm{d}x
\end{align*}
Therefore,
$$\int_0^1 \big(1-x^2\big)\big(f'(x)\big)^2\,\mathrm{d}x \geqslant\frac{6}{3\ln2-2}\left(\int_0^1 xf(x)\,\mathrm{d}x\right)^{\!2}.$$
Because,
$$\int_0^1\bigg(-\frac{x(x-1)}{2\sqrt{1-x^2}}\bigg)^2\mathrm{d}x=\frac{1}{2}\ln2-\frac{1}{3}$$


问:
The formula to calculate Euler Phi function
I know that $$\phi(m)=m\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right)\text{  Where }m=\prod_{i=1}^n p_i^{a_i}$$

But when i tried to find a formula of $\phi(n)$ i got this:
$$\phi(m)=\phi \left(\prod_{i=1}^n p_i^{a_i}\right) = \prod_{i=1}^n \phi(p_i^{a_i}) $$ Now since $\phi(p^m)=p^m -p^{m-1}$, Thus:
$$\phi(m)=\prod_{i=1}^n \left(p_i^{a_i}-p_i^{a_i-1}\right )$$

Is this’s a valid proof? and if it is why most people are using this formula:
$$\phi(m)=m\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right) $$
Side note:

I’ve used the same method to find the formula of $\sigma(n)$ and $\sigma_m(n)$

答: It would be worth writing this as $\phi(n)=n\prod_{p\in\Bbb P,\,p|n}(1-1/p)$. Your proof is the standard one, viz.$$\begin{align}\phi(n)&=\prod_{p\in\Bbb P,\,p|n}\phi(p^{\operatorname{ord}_p(n)})\\&=\prod_{p\in\Bbb P,\,p|n}p^{\operatorname{ord}_p(n)}(1-1/p)\\&=\prod_{p\in\Bbb P,\,p|n}p^{\operatorname{ord}_p(n)}\prod_{p\in\Bbb P,\,p|n}(1-1/p)\\&=n\prod_{p\in\Bbb P,\,p|n}(1-1/p).\end{align}$$


问:
How to prove a PDE preserves mass?
My general question is: suppose you are given a PDE (possibly with boundary conditions). What does it mean to say the PDE "conserves mass"?

Specifically, if you are given the PDE
$$- \nabla \cdot (a(x) \nabla u(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u = 0 \quad \text{ on } \partial \Omega,$$
how would you determine whether "mass is conserved"?

My guess would be that the mass is the integral $\int_\Omega u(x) \, dx$. But if this is the case, I don't know how to prove whether it is constant or not.

Edit: In this specific problem, there is no time involved, so you can't actually use a time derivative. However, it is the starting problem of a homogenization procedure, so actually it reads
$$- \nabla \cdot (a_\epsilon(x) \nabla u_\epsilon(x)) = f(x) \quad \text{in a domain $\Omega$,}$$
$$u_\epsilon = 0 \quad \text{ on } \partial \Omega.$$

So the mass conservation might be related to the $\epsilon$, but I don't think so (because from the context I gather that determining whether the problem conserves mass is supposed to be done for a fixed $\epsilon$).

Any help is appreciated!

答:
The time independent PDE you gave:
\begin{align}
- \nabla \cdot (a(x) \nabla u(x)) &= f(x) \quad \text{in  }\Omega,
\\
u &= 0 \quad \text{ on } \partial \Omega,
\end{align}
is normally the limit equilibrium of the time dependent problem:
$$
\frac{\partial u}{\partial t}- \nabla \cdot (a(x) \nabla u(x,t)) = f(x),
$$
where we let $t\to \infty$, and
$$
\lim_{t\to \infty}\frac{\partial u}{\partial t} = 0,
$$
this roughly means $u$ won't be changed along with time anymore in the equilibrium state. If $u$ is the mass density in $\Omega$, then it won't change anymore with respect to time, so I am guessing the "conservation of mass" in OP means this.

We can let
$$
u(x)= \lim_{t\to \infty}u(x,t)
$$
which solves the time independent problem. We assume everything is nice and smooth we can interchange the limit and the integral, then
$$
\lim_{t\to \infty}\frac{d}{dt}\int_{\Omega} u = \lim_{t\to \infty}\int_{\Omega}\frac{\partial u}{\partial t} = \int_{\Omega} \nabla \cdot (a \nabla u) + \int_{\Omega} f = 0.
$$

In the equation,
\begin{align}
- \nabla \cdot (a_{\epsilon}(x) \nabla u_{\epsilon}(x)) &= f(x) \quad \text{in  }\Omega,
\\
u_{\epsilon} &= 0 \quad \text{ on } \partial \Omega,
\end{align}
the mean of "conservation" is rather straightforward: For any open subset $M\subset \Omega$
$$
\text{Constant} = \int_{M}f = -\int_{M}\nabla \cdot (a_{\epsilon} \nabla u_{\epsilon}) = \int_{\partial M} a_{\epsilon} \nabla u_{\epsilon}\cdot \nu \,dS,
$$
where $-a_{\epsilon} \nabla u_{\epsilon}$ is the flux of this mass density $u_{\epsilon}$. In an arbitrary region $M$ within this domain, The amount of total mass flux on the boundary of $M$ is constant independent of $\epsilon$ for any $\epsilon$.

Title: How to prove a PDE preserves mass?


问:
How is the pullback of a Riemannian metric defined?
Let $M, N$ are two manifolds. Now I am interested in defining the pullback of an arbitrary tensor field of type $(r,s)$ under the diffeomorphism $\phi : M \rightarrow N$ as follows:

$\phi^* T(\eta_1,\dots, \eta_r, X_1, \dots, X_s) = T( (\phi^{-1})^*(\eta_1), \dots, (\phi^{-1})^*(\eta_r), \phi_* X_1, \dots, \phi_* X_s)$.

where $\eta_i \in T_p^*(M)$ is a covector and $X_j \in T_p(M)$ is a vector.

Actually I am interested in pullback of a metric tensor.

We know that it is a $(0, 2)$ tensor. Therefore, the above formula becomes $$\phi^*g(X,Y) = g(\phi_*X, \phi_*Y)$$.

Now consider $g_{\alpha \beta}$ is a $(0, 2)$ tensor on $N$. Now my question is can one write the formula as follows

$$(\phi^*g)_{\mu \nu} = \frac{\partial y^{\alpha} }{\partial x^{\mu}} \frac{\partial y^{\beta} }{\partial x^{\nu}} g_{\alpha \beta}$$.

Please help me. Thanking in advanced.

答:
Let $x=(x^1,\dots, x^m)$ and $y=(y^1,\dots,y^n)$ be local coordinates on $M$ and $N$ respectively.

Locally, $\phi: M\rightarrow N$ is
$$
\phi(x)=(\phi^1(x),\dots,\phi^n(x))
$$
Then
$$
\begin{align*}
\phi_\ast \frac{\partial}{\partial x^j}= \frac{\partial \phi^i}{\partial x^j}\frac{\partial}{\partial y^i}
\end{align*}
$$
Locally, $g=g_{kl}dy^k\otimes dy^l$.  So
$$
\begin{align*}
(\phi^\ast g)(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})&=
\frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g(\frac{\partial}{\partial y^k}\frac{\partial}{\partial y^l})\\
&=\frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g_{kl}
\end{align*}
$$
So
$$
(\phi^\ast g)_{ij}=\frac{\partial \phi^k}{\partial x^i}\frac{\partial \phi^l}{\partial x^j}g_{kl}
$$


问:
Prove $-12^{482}$ has a multiplicative inverse modulo 175
As in the title, I'm trying to prove $-12^{482}$ has a multiplicative inverse modulo 175. Generally, I believe I could do this by showing that $gcd(-12^{482}, 175)=1$, (i.e that they're relatively prime) but I'm not sure about how to deal with the exponent without fully computing $-12^{482}$.

答:
Hint: $x^k$ is invertible modulo $n$ iff $x$ is. This is because $x$ and $x^k$ have the same prime factors, just with different multiplicities.

In your case, $n = 175 = 5^2 \cdot 7$ while $x = 12 = 2^2 \cdot 3$ are coprime.


问:
How do I know what the gain is of a second order system which is expressed in this format?
Can someone please explain how to know the gain of a second order system which is expressed as such
$$\frac{number_{1}}{(s+ number_{2})(s+ number_{3})}$$
For example
$$\frac{40}{(s+0.2)(s+20)}$$
Thank you !

答:
The standard form of second order transfer function is ,

$$k\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$

where,

$k$ is the dc gain

$\zeta$ is the damping factor and

$\omega_n $ is the undamped natural frequency.

So first expand the denominator and try to make it of the standard form.

$$\frac{40}{(s+0.2)(s+20)} = \frac{40}{s^2+20.2s+4}$$

Could you find the gain now?


问:
If $a$ ($a\ne 1$) has order $t\mod p$, show that $a^{t-1}+a^{t-2}+\cdots+1\equiv 0\pmod{p}$.
If $a$ ($a\ne 1$) has order $t\mod p$, show that $a^{t-1}+a^{t-2}+\cdots+1\equiv 0\pmod{p}$.

Since $a$ has order $t$ we have that $a^t\equiv1\pmod{p}$ or $a^t-1 \equiv 0\pmod{p}$. We also know that $a^t-1=(a-1)(a^{t-1}+a^{t-2}+\cdots+1)$.

So $0=a^t-1=(a-1)(a^{t-1}+a^{t-2}+\cdots+1)\pmod{p}$. How do I get rid of the $(a-1)$ term on the left ? Can I simply say that since $(a-1)\ne0$ and divide both sides by it and get the desired result ? Any solutions/hints are greatly appreciated.

答:
Since $p$ is a prime number, we know that $p\mid AB$ implies either $p\mid A$ or $p\mid B$.

Apply this with $A = a - 1$ and $B = a^{t - 1} + \dots + 1$ to conclude.


问:
Calculating the intersection points of two circles, of which the centers are located on the x axis.
I'm trying to solve the problem mentioned in the title, and the answer detailed here provides exactly what I was looking for: https://stackoverflow.com/a/3349134/4777480...except for the fact that the author skips several steps, leaving me confused as to how he ended up with the results he obtained!

For starters, there's this line:

a = (r02 - r12 + d2 ) / (2 d)

Thankfully, a reply to the answer does clarify this one and I did verify it using the steps they detailed.

But leaves me with the last two steps...

How did we end up with this result?

P2 = P0 + a ( P1 - P0 ) / d

And how did that give us the intersection coordinates? (I think in the first line it should be 'a' instead of 'h' but I'm not sure, can someone clarify?)

x3 = x2 +- h ( y1 - y0 ) / d

y3 = y2 -+ h ( x1 - x0 ) / d

A step-by-step explanation would be greatly appreciated, I want to understand the full process so I can implement it correctly in my code.

答:
So I tried verifying the formula Yves posted and I'm not sure if I made a mistake or if it's incorrect...

But, I managed to find a decent & simple answer from this page: https://planetcalc.com/8098/

Essentially, we have:

$$$$
$$ a=(r_1^2-r_2^2+d^2)/2d$$

and:

$$h=√(r_1^2-a^2)$$

Therefore:

$$x_4=a+x_1$$

$$y_4=h$$

The coordinates for the second intersection point would simply be:

$$(a+x_1,-h)$$


问:
Expected number of elements before 1 out of 2 hash tables are full?
From a previous asked question it became clear to me that the expected number of elements to fill up one hash table (assuming ideal randomness) is just the Harmonic number: $mH_m$ where $m$ is the size of the table (collisions could just be solved with chaining or whatever. Don't care.)

However, a more tricky question is:

What is the expected number of elements we need before one of the hash tables are full? I.e. I want to know how fast it takes before one of the tables are full. We can assume that choosing either table $a$ or $b$ has probability $P[a|b]= \frac{1}{2}$ (see figure below).

答:
There might be an elegant explanation of the result, but here is the bruteforce calculation.

We fix $m$ and define $E(a, b)$ as the expected number of elements to fill one table, if there are still $a$ and $b$ empty entries in the two tables.

Thus $E(0, *) = E(*, 0) = 0$ and we are looking for $E(m, m)$.

For $a, b > 0$, by considering the possibilities of the next element, we have: $$E(a, b) = 1 + \frac a {2m}E(a - 1, b) + \frac b {2m} E(a, b - 1) + (1 - \frac{a + b}{2m})E(a, b)$$ which simplifies to: $$(a + b) E(a, b) = 2m + aE(a - 1, b) + bE(a, b - 1).$$

If we define $F(a, b) = \frac 1 {2m} E(a, b)$, then we have the recurrence relation $$(a + b) F(a, b) = 1 + aF(a - 1, b) + bF(a, b - 1).$$

It is not totally obvious, but one can show by induction that $F(a, b) = H_a + H_b - H_{a + b}$ for all $a, b \geq 0$, where $H_n = \frac 1 1 + \dots + \frac 1 n$ denotes the harmonic number.

Proof: We simply need to verify that $$(a + b)(H_a + H_b - H_{a + b}) = 1 + a(H_{a - 1} + H_b - H_{a + b - 1}) + b(H_a + H_{b - 1} - H_{a + b - 1})$$ which is a routine calculation.

Therefore we have $F(m, m) = 2H_m - H_{2m}$ and hence $$E(m, m) = 2m F(m, m) = 4mH_m - 2mH_{2m}.$$


答: By Fermat-Euler $12^{120}=1$, so the inverse is $-12^{118}$.

Title: Prove $-12^{482}$ has a multiplicative inverse modulo 175

Body: As in the title, I'm trying to prove $-12^{482}$ has a multiplicative inverse modulo 175. Generally, I believe I could do this by showing that $gcd(-12^{482}, 175)=1$, (i.e that they're relatively prime) but I'm not sure about how to deal with the exponent without fully computing $-12^{482}$.


问:
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$
Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?

答:
Recall the well-known fact: If $g(u)$ is a convex function on $\mathbb{R}^n$, then
$$g(u) \ge g(v) + \nabla g(v)^\mathsf{T}(u - v), \ \forall u, v \in \mathbb{R}^n.$$

Clearly, $f(x, y)$ is a convex function on $\mathbb{R}^2$. Thus, we have
$$f(x, y) \ge f(\tfrac{5}{12}, \tfrac{4}{3}) + \tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (x - \tfrac{5}{12})
+ \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) \cdot (y - \tfrac{4}{3}) = \sqrt{10}.$$
Also, $f(\tfrac{5}{12}, \tfrac{4}{3}) = \sqrt{10}$.

The minimum of $f(x, y)$ is $\sqrt{10}$ achieved at $x = 5/12$ and $y = 4/3$.

Another way: Since $f(x, y)$ is a convex function on $\mathbb{R}^n$ and $\tfrac{\partial f}{\partial x}(\tfrac{5}{12}, \tfrac{4}{3})
= \tfrac{\partial f}{\partial y}(\tfrac{5}{12}, \tfrac{4}{3}) = 0$, $f(x,y)$ achieves its minimum at $x = 5/12$ and $y = 4/3$.


问:
Does group of deck tranformations acts transitively on each fibre if it acts traansitively on one fiber?
i am reading bredon "Topology and Geometry "

It states that if we have a covering map p : X ->Y s.t. p(x) = y.X,Y are Hausdorff, path connected and locally path connected etc.

I have 2 questions:

It states that There exist $\alpha \in N(p^*(\pi _1(X,x))$ s.t lift of loop $\alpha$ at x takes it to $x_0$ iff There is a deck tranformation taking x to $x_0$ whose proof i understood.

He says using above theorem one can say that (a)={$p^*(\pi _1(X,x))$ is normal in $\pi _1(Y,y)$}  iff (b)={deck transformations act transitively on fiber of x}. a  => b is ok, how to prove b  => a?

2.Does deck transformations act transitively on each fiber ,if they act transitively on one of the fibers?

```<code>     can anybody hint a proof or a  related link?
</code>```

答: Let $G$ be the group of deck transformations of the cover $p$. Let $y, y' \in Y$, and let $F, F'$ be the fibres above $y$ and $y'$ respectively. The sets $F$ and $F'$ are $G$-sets, i.e. sets equipped with an action of $G$. I claim that they are isomorphic $G$-sets. Indeed, let $\sigma$ be a path in $Y$ taking $y$ to $y'$. Let $x \in F$. By the path lifting property, there is a unique path $\sigma_x$ in  $X$, starting at $x$ and lifting $\sigma$. Define $f_\sigma(x) = \sigma_x(1)$ as the end-point of this path. Then $f_\sigma(x) \in F'$, so $f_\sigma$ is a map $F \to F'$. By the uniqueness of path lifting, it is easy to see that $f_\sigma(g \cdot x) = g \cdot f_\sigma(x)$ for every $g \in G$, so $f_\sigma$ is a morphism of $G$-sets. Moreover, the inverse path $\sigma^{-1}$ provides a double-sided inverse $f_{\sigma^{-1}}$ to $f_\sigma$. Thus, $F \simeq F'$ as $G$-sets; in particular, the action is transitive on $F$ if and only if it is transitive on $F'$.</p>


答:
Your proof is valid (assuming you already proved the facts you used about $\phi$). Both formulas are practically equivalent

$$\prod_{i=1}^n (p_i^{a_i}-p_i^{a_i-1} ) =\prod_{i=1}^n p_i^{a_i}\left(1-\frac{1}{p_i}\right) = m \prod_{i=1}^n \left(1-\frac{1}{p_i}\right) $$

and both are widely used.

Title: The formula to calculate Euler Phi function

Body:
I know that $$\phi(m)=m\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right)\text{  Where }m=\prod_{i=1}^n p_i^{a_i}$$

But when i tried to find a formula of $\phi(n)$ i got this:
$$\phi(m)=\phi \left(\prod_{i=1}^n p_i^{a_i}\right) = \prod_{i=1}^n \phi(p_i^{a_i}) $$ Now since $\phi(p^m)=p^m -p^{m-1}$, Thus:
$$\phi(m)=\prod_{i=1}^n \left(p_i^{a_i}-p_i^{a_i-1}\right )$$

Is this’s a valid proof? and if it is why most people are using this formula:
$$\phi(m)=m\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right) $$
Side note:

I’ve used the same method to find the formula of $\sigma(n)$ and $\sigma_m(n)$


问:
Solving a linear system over $\mathbb{Z}_{5}$
I would like to ask how to solve this matrix, if I substitute for z (x, y, z) z = 4, I can't calculate the matrix. I tried to adjust the matrix to this shape, but I don't know how to proceed.

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
1 & 0 & 0 & 4\\
1 & 3 & 4 & 0
\end{array} } \right)$ in $\mathbb{Z}_{5}$

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 3 & 2\\
0 & 0 & 0 & 0
\end{array} } \right)$

答:
start:

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
1 & 0 & 0 & 4\\
1 & 3 & 4 & 0
\end{array} } \right)$

first steps: subtract 1st row from 2nd and 3rd, to eliminate the other 1's in the first column

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 3 & 4 & 1\\
0 & 1 & 3 & 2
\end{array} } \right)$

where I use that $-1 \equiv 4, -2 \equiv 3, -3 \equiv 2$ in $\Bbb Z_5$.
The pivot 3 on place $(2,2)$ has an inverse 2 over $\Bbb Z_5$ so I multiply the 2nd row by $2$ (modulo $5$) to make for easier computations:

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 3 & 2\\
0 & 1 & 3 & 2
\end{array} } \right)$

We subtract the two identical rows at the end to get:

$\left( {\begin{array}{ccc|c}
1 & 2 & 1 & 3\\
0 & 1 & 3 & 2\\
0 & 0 & 0 & 0
\end{array} } \right)$

So we have a consistent system (so there are solutions) that is  underdetermined. So there will be $5$ solutions, depending on the value we pick for the free variable $z$, we can solve it in terms of this free variable:

$y + 3z = 2$ so $y = 2-3z = 2 + 2z$

$x+2y+z = 3$ so $x+ 4z + 4+ z = 3$ so $x+4=3$ so $x=4$, as $5z\equiv 0$.

So we get the solutions $\{(4,2+2z, z) \mid z \in \Bbb Z_5\}$, a set of $5$ vectors, as expected.


问:
How to find the class of a function
Let $$f(x) = \begin{cases}
 x^3\sin(\frac{1}{x}); & \text{ if, } x \neq 0 \\
 0; & \text{ $x=0$ }
\end{cases}$$

How do I find the greatest value of n such that $f \in C^n ([-1,1])$?

I know to find n, I must find the greatest n such that $f^{(n)}(x)$ is differentiable in the interval $[-1,1]$. But how do I prove its differentiability?

答:
You have to check if $f^{(n)}$ is continuous $n$ by $n$.

Start with $n=0$.

The function is continuous because
$$\lim_{x \to 0^+} x^3 \sin{\left(\tfrac{1}{x}\right)} = \lim_{x \to 0^-} x^3 \sin{\left(\tfrac{1}{x}\right)}=0.$$

For $n=1$,
$f'(x) = \begin{cases} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} , & \text{ if } x \neq 0\\
0, & \text{ if } x = 0
\end{cases}$
is continuous because
$$\lim_{x \to 0^+} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} = \lim_{ x \to 0^-} -x \cos{\left(\tfrac{1}{x}\right)} + 3x^2 \sin{\left(\tfrac{1}{x}\right)} = 0.$$
For $n = 2$,
$f''(x) = \begin{cases} -4 \cos{\left(\tfrac{1}{x}\right)} - \tfrac{1}{x}\sin{\left(\tfrac{1}{x}\right)}+6x\sin{\left(\tfrac{1}{x}\right)} , & \text{ if } x \neq 0\\
0, & \text{ if } x = 0
\end{cases}$ is not continuous because
$$\lim_{ x \to 0}-4 \cos{\left(\tfrac{1}{x}\right)} - \tfrac{1}{x}\sin{\left(\tfrac{1}{x}\right)}+6x\sin{\left(\tfrac{1}{x}\right)}$$ does not exist.

Thus the greatest $n$ for which $f \in C^n([-1,1])$ is $n=1$.

Title: How to find the class of a function


问:
Math behind RSA
Now I very recently wrote an investigation paper on RSA Encryption and how it works. After completing it, I thought of looking up real life key generators. To my surprise, they all have characters within them, and not just numbers... Everything I've read till now has been based on the fact that RSA works solely on prime numbers + Euler's Totient function + Modular Arithmetic. I know how to encrypt when the public and private keys are solely numbers, but how is it done with characters?

Just to make it clear, I do know how to use RSA encryption and know the math behind it.

PS: Here are a few links :

https://travistidwell.com/jsencrypt/demo/

https://www.devglan.com/online-tools/rsa-encryption-decryption

答:
This is simply a way of encoding numbers.

Consider writing down a number of size $1024$ bits. If we use decimal system, then the number is around $308$ digits long, and hence needs $308$ characters to print out.

Now if we use a higher base such as base 64, then the number of "digits" will be reduced to around $171$, which is much more compact.

In theory, one can use any sort of "number-to-string" encoding, but there are some standard ones that are commonly used.

Title: Math behind RSA


问:
Write down the probability mass function of $N$ and apply it to get $P[N = 2]$
A box has one white ball, $2$ black balls and $3$ grey balls. Balls are drawn one by one
and with replacement from the box. Let $N$ be the number of draws until all colors have
been observed (after any $2$ colors have been seen, we keep trying until the first time the
third color is seen).

Write down the probability mass function of $N$ and apply it to get $P[N = 2]$.

My attempt:

The last ball must be of a third color. So three cases arise:

$1-$ $(W$ or $B)$ in first $(N-1)$ draws and $G$ in $N$th draw.

$2-$ $(B$ or $G)$ in first $(N-1)$ draws and $W$ in $N$th draw.

$3-$ $(G$ or $W)$ in first $(N-1)$ draws and $B$ in $N$th draw.

I'll call these $P(1), P(2)$ and $P(3)$ respectively.

Then $P(1)=(1/6+1/3)^{(n-1)} *1/2$.

Similarly, $P(2)=(1/3+1/2)^{(n-1)} *1/6$ and

$P(3)=(1/2+1/6)^{(n-1)}*1/3$.

Summing them up gives us the required value for $P(N=n)$

However, when I try it for $P(N=2)$, I get $0.61$, which is wrong, because $P(N=2)$ should be $0$ as we cannot get $3$ different colors in $2$ draws.

Can someone please point out where I went wrong? Thank you.

答:
Take the first case where you want only $W$ and $B$ in the first $(n-1)$ draws and then $G$ in the last draw.

$\small \displaystyle P(W) = \frac{1}{6}, P(B) = \frac{1}{3}, P(G) = \frac{1}{2}$.

$\small (1- P(G))^{n-1}$ will give you probability of only black or white in first $(n-1)$ draws but you need to subtracts two cases where all $(n-1)$ draws are either black or white.

So the probability should be $P_G(n) = \displaystyle \big(\frac{1}{2^{n-1}} - \frac{1}{3^{n-1}} - \frac{1}{6^{n-1}}\big) \cdot \frac{1}{2}$

Now similarly find probability where the last drawn ball is $W$ and $B$.

Can you take it from here?


问:
Show that $Z/\sqrt{n^{-1}\sum_{i=1}^n Y_i^2}$ converges in distribution
Let $Z, Y_1, Y_2, \dots, Y_n$ be iid Gaussian random variables. I have to show that $$\frac Z {\sqrt{n^{-1}\sum_{i=1}^n Y_i^2}}$$ converges in distribution.

Since $n$ only appears in the denumerator only, the convergence of the quantity there has to be verified. Indeed, $\frac{1}{\sqrt n}\sum_{i=1}^n(Y_i^2 - 1)$ converges in distribution to $\mathcal N(0, 2)$. However, that is not quite what is given here. I have therefore tried to rewrite the expression as follows: $$n^{-1}\sum_{i=1}^nY_i^2 = \frac{1}{\sqrt n} T_n + 1,$$ where $T_n = \frac{1}{\sqrt n}\sum_{i=1}^n(Y_i^2 - 1)$. So $T_n$ converges to $\mathcal N(0, 2)$ and the $1$ remains constant. But $\frac 1 {\sqrt n}T_n$ seems to convergence to a degenerate distribution as $\frac 1 {\sqrt n}$ converges to zero (Slutzky's theorem). Is my reasoning here correct? And what does this result imply for the original question?

答:
A short answer is this expression converges to $Z$. Effectively, $\frac{1}{n}\sum_{i=1}^n Y_i^2 \xrightarrow{n \to +\infty} 1$ almost surely (by law of large numbers) then you can deduce that $\frac{1}{\sqrt{n^{-1}\sum_{i=1}^n Y_i^2}}$ converge to 1 almost surely. Consequently, $\frac{Z}{\sqrt{n^{-1}\sum_{i=1}^n Y_i^2}}$ converges to Z almost surely (and so converges to $Z$ in distribution).

If you want to find the asymptotic law of this expression, you can do as follows. From CLT, we have
$$\sqrt{n} \left( \frac{1}{n}\sum_{i=1}^n Y_i^2 -E(Y^2)  \right) \xrightarrow{n \to +\infty} \mathcal{N}(0,V(Y^2))$$
or for $n \to +\infty$
$$\frac{1}{n}\sum_{i=1}^n Y_i^2 \sim 1+\frac{1}{\sqrt{n}}\mathcal{N}(0,2)$$
($\sim$ is "the equality in distribution")

By Taylor approximation, we have for $n \to +\infty$
\begin{align}
\frac{Z}{\sqrt{n^{-1}\sum_{i=1}^n Y_i^2}} &\sim \frac{Z}{(1+\frac{1}{\sqrt{n}}\mathcal{N}(0,2))^{\frac{1}{2}}} \\
&\sim Z(1-\frac{1}{2}\frac{1}{\sqrt{n}}\mathcal{N}(0,2) + \mathcal{O}(\frac{1}{n})) \\
&\sim Z(1-\frac{1}{2}\frac{1}{\sqrt{n}}\mathcal{N}(0,2) + \mathcal{O}(\frac{1}{n})) \\
&\sim Z-\frac{1}{\sqrt{2n}}ZT + \mathcal{O}(\frac{1}{n})) \\
\end{align}
where $Z,T$ are two iid gaussian random variables.


问:
For every pair of points $x,y\in S\subset\mathbb{R}^2,\exists z\in S$ s.t. $z$ lies on the straight line formed by $x,y.$ Can $S$ be finite?
Actual question: If $S\subset\mathbb{R}^2$ contains at least $3$ noncollinear points and for every pair of points $x_1,x_2$ of $S$, there exists a third point $x_3$ which lies on the (extended) straight line formed by $x_1$ and $x_2$, then can $S$ be finite?

I have more questions on this topic (in particular, what can we say about $S$ if $S$ is bounded - e.g. does $S$ have to be uncountable, or connected, or open...?), but it's one question per question, so leave it as this for now.

My attempt so far (I suspect the answer is no): Suppose, by way of contradiction that the answer is yes. Then $S$ is finite and is therefore bounded. Consider the convex hull, $H$, of $S.$ The boundary of $H$ has $n\geq3$ vertices. On each straight line joining adjacent vertices of the boundary of $H$, there is at least one other point of $S$. Now what?

答:
No such finite set exists. The Sylvester-Gallai theorem asserts that for any finite set of points $S$ in the plane, there is either a line passing through exactly two of them, or all points are collinear. Since we want $S$ to have at least 3 non-collinear points, if $S$ is finite then there is a line passing through exactly two points, i.e. there are two points $P,Q \in S$ for which there isn't another point in the set lying on the line $PQ$.

Now it's not necessary for a bounded set $S$ with the presented property to be uncountable. Let $a,b,c$ be the complex numbers corresponding to the vertices of a triangle in the plane, and let
$$S = \{\alpha a + \beta b + \gamma c \mid \alpha + \beta + \gamma = 1, \alpha , \beta, \gamma \in \mathbb{Q}_{\geq 0} \}$$

This set is clearly bounded and countable, since the number of triples of $(\alpha, \beta,\gamma)$ is smaller than $\mathbb{Q}^3,$ which is countable. To see that $S$ has the required property, let $x, y \in S$ be two points in $S$, so that $x = \alpha_1 a + \beta_1 b + \gamma_1 c$, and $y = \alpha_2 a + \beta_2 b + \gamma_2 c$. Then the midpoint of the segment connecting these two points, aka the point
$$\frac{\alpha_1 + \alpha_2}{2} a + \frac{\beta_1 + \beta_2}{2} b +  \frac{\gamma_1 + \gamma_2}{2} c $$
is also clearly in $S$, so $S$ has the required property.

Note that this set is dense in the region bounded by the triangle; After some scribbling it seems to me rather clear (although I haven't showed it rigorously) that any bounded such $S$ has at least one limit point (as opposed to, say, an infinite square lattice, which also satisfies the conditions but does not have any limit points).