问:
Show that $(n+1)^{n-1}<n^n$ using the binomial theorem
I want to show that $(n+1)^{n-1}<n^n$ using the binomial theorem.

I have done the following:\begin{align*}\left (n+1\right )^{n-1}&=\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1-k}1^k=\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1-k}\\ & <\sum_{k=0}^{n-1}\binom{n-1}{k}n^{n-1}=n^{n-1}\sum_{k=0}^{n-1}\binom{n-1}{k} \\ & =n^{n-1}\cdot 2^{n-1}<n^{n-1}\cdot n^{n-1}=n^{2n-2}\end{align*} This is not correct.

What could I change to get the correct result?

答:
This is equivalent to showing that
$\left(1+\frac1n\right)^{n-1}<n,$ that is, that

$$1+n\left(\frac1n\right)+\frac{n(n-1)}{2}\left(\frac1n\right)^2+\cdots+\left(\frac1n\right)^{n-1}<n.$$

But this is true since the expansion can be rearranged thus

$$ 1+1+\frac12\left(1-\frac1n\right)+\frac{1}{3!}\left(1-\frac1n\right)\left(1-\frac2n\right)+\cdots+\frac{1}{n^{n-1}},$$ from which it is obvious that the inequality follows, since the expansion is less than $$\underbrace{1+1+1+\cdots+1+1}.$$


问:
Find the limit of the product series ...
$$
\lim_{n\to \infty}\frac{(2^n + 1)(2^n+2)\cdots(2^n+n)}{\sqrt 2^{3n}}
$$

I tried using Cesaro-Stolz on it and I still could not get in a human form.

ContentLicense: CC BY-SA 4.0

Body:
$$
\frac{(2^n + 1)(2^n+2)\cdots(2^n+n)}{\sqrt 2^{\ 3n}}
> \frac{(2^n)(2^n)\cdots(2^n)}{\sqrt 2^{\ n} \times \sqrt2^{\ n} \times \sqrt 2^{\ n}} = \left(\frac{ 2}{\sqrt2}\right)^{3n} \times \left({2^n}\right)^{n-3}
> \sqrt 2^{\ 3n}
$$

for $n > 3$.


问:
A circumference as intersection of a quadric with a plane
I have to prove that the following intersection, is a circumference:
$$ \mathcal{C} = \begin{cases} x^2+y^2+xy-3x-4y+z+\frac{5}{2}=0 \\ x+y+z=2 \end{cases} $$
The method I would like to use is to find the improper points, and observe that they fulfill the relationship $x^2+y^2+z^2=0$. Is this a right procedure?
For improper points:
$$ \begin{cases} x^2+y^2+xy=0 \\ z=2-x-y \\ t=0 \end{cases} \Rightarrow \begin{cases} x=(\frac{-1 \pm i\sqrt{3}}{2})y \\ z=2-(\frac{1 \pm i\sqrt{3}}{2})y \\ t=0 \end{cases}  $$
$P_{\infty} = (\frac{-1 \pm i\sqrt{3}}{2},1,\frac{3 \mp i\sqrt{3}}{2},0)$
and $(\frac{-1 \pm i\sqrt{3}}{2})^2+1+(\frac{3 \mp i\sqrt{3}}{2})^2$ is not zero.

答:
Your idea is fine, but you didn't get the points at infinity right. The correct equations are

$$ \begin{cases} x^2+y^2+xy=0 \\ z=-x-y \\ t=0 \end{cases}$$

which give

$$
P_{\infty} =
\left(\frac{-1 \pm i\sqrt{3}}{2},1,\frac{1 \mp i\sqrt{3}}{2},0\right).
$$


问:
Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?
WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:

$$
 \lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) =  \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
=  \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1  \\
= 1?
$$

I wonder where I'm wrong...

答:
$$\lim_{h\to0}\frac{\sin h}h=1$$ but $$\lim_{h\to0}\frac{\sin \frac1h}{\frac1h}=\lim_{h\to0}h\cdot  \sin \frac1h$$

Now, $-1\le \sin \frac1h\le 1$

$$\implies \left|h\cdot \sin \frac1h\right|\le \left|h\right| $$

Now, $\lim_{h\to0}h=0$

$$\implies  \lim_{h\to0}h\cdot \sin \frac1h =0 $$

ContentLicense: CC BY-SA 3.0


问:
Find the norm of a linear functional on $C[0,1] $
Find the norm of a linear functional on $C[0,1]$
$l(x) = x(0) + \int \limits_0^1 (2t-1)x(t)dt$ To find the functional norm I need to evaluate and give an example.
I tried to apply the second mean theorem for the integral, but I didn't get any good results

答:
Just go brute force on it:

If $\Vert x\Vert\leq 1$, then
$$|l(x)|\leq |x(0)|+\int_0^1|2t-1|dt=\frac{3}{2}$$
On the other hand, given $\epsilon>0$, take a $x\in C[0,1]$ satisfying:

$-1\leq x\leq 1$ on all of $[0,1]$

$x(0)=1$;

$x=-1$ on $(\epsilon,\frac{1}{2}-\epsilon)$;

$x=1$ on $(\frac{1}{2}+\epsilon,1]$

and evaluate $l(x)=\frac{3}{2}\pm 3\epsilon$.


问:
Numerically Solving 2nd Order System of Differential Equations
This is equations of motion in $3D:$ How can we numerically solve this system of the equations? (preferably Euler). can you give a sample of $2$ step sizes how is it possible?

$\frac{d^2x}{dt^2} = -vk_d\frac{dx}{dt} + vk_l\frac{dy}{dt}$

$\frac{d^2y}{dt^2} = -vk_d\frac{dy}{dt} - vk_l\frac{dx}{dt}$

$\frac{d^2z}{dt^2} = -vk_d\frac{dz}{dt} - g$

$v = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 +(\frac{dz}{dt})^2}$

$g, k_d, k_l$ are constants.

step size = $0.01$ sec = $t$

Intial Conditions:

x,y,z = 0

$\frac{dx}{dt}=10.45$

$\frac{dy}{dt}=5.45$

$\frac{dz}{dt}=33.45$

$v_0=25$

答:
In python (and similar in other languages) you would encode the ODE function of the right side, for the first order system with $v_x=\dot x$ etc., as

```<code>def motioneqns(t,u):
    x,y,z,vx,vy,vz = u
    v = (vx**2+vy**2+vz**2)**0.5
    ax = -kd*v*vx + kl*v*vy
    ay = -kd*v*vy - kl*v*vx
    az = -kd*v*vz - g
    return vx,vy,vz,ax,ay,az
</code>```

and pass it along to any general-purpose solver, which might be one implementing the Euler method, or RK4, or some higher order method with adaptive step size.

If you use implementations similar to those in Does fourth-order Runge-Kutta have an higher accuracy than the second-order one? you need to wrap the result list in a vector type, for instance numpy.array, either in the return line or in the function call. For further inspiration on getting the trajectories as list see https://stackoverflow.com/questions/61333687/how-do-i-solve-a-2nd-order-differential-equation-for-projectile-motion, https://stackoverflow.com/questions/53645649/cannot-get-rk4-to-solve-for-position-of-orbiting-body.

Body:
In python (and similar in other languages) you would encode the ODE function of the right side, for the first order system with $v_x=\dot x$ etc., as

def motioneqns(t,u):
    x,y,z,vx,vy,vz = u
    v = (vx**2+vy**2+vz**2)**0.5
    ax = -kd*v*vx + kl*v*vy
    ay = -kd*v*vy - kl*v*vx
    az = -kd*v*vz - g
    return vx,vy,vz,ax,ay,az

and pass it along to any general-purpose solver, which might be one implementing the Euler method, or RK4, or some higher order method with adaptive step size.

If you use implementations similar to those in Does fourth-order Runge-Kutta have an higher accuracy than the second-order one? you need to wrap the result list in a vector type, for instance numpy.array, either in the return line or in the function call. For further inspiration on getting the trajectories as list see https://stackoverflow.com/questions/61333687/how-do-i-solve-a-2nd-order-differential-equation-for-projectile-motion, https://stackoverflow.com/questions/53645649/cannot-get-rk4-to-solve-for-position-of-orbiting-body.


问:
permutation containing substring
So, i just want to check if my answer to this question is correct or not.

Question : Determine the number of non-repeating strings with length 9 consisting only of numbers (0-9) or vowels (a, i, u, e, o, A, I, U, E, O) that contain the substrings "B16" and "uA".

To answer this question, we treat "B16" and "uA" as a single element, so that there will be 6 elements to arrange. 2 position is already filled by "B16" and "uA" and the remaining 4 position can be filled by 15 x 14 x 13 x 12. And then the arrangements of 6 elements is 6!. So the final answer is 15 x 14 x 13 x 12 x 6!. Is my answer correct?

答:
If none of the characters can repeat (i.e they can appear only once in your substrings of $9$ characters), you are correct till the point that there are $2$ positions already filled by (B$16$) and (uA).

But if you are calculating remaining $4$ position to be filled as $15 \times 14 \times 13 \times 12$, you are permuting them as well. So to again arrange them as part of $6!$ will bring in duplicates.

You should rather select $4$ characters out of remaining $15$ as $15 \choose 4$ and then multiply by $6!$.

So your answer should be ${15 \choose 4} \times 6!$

Title: permutation containing substring


答: The final sequence diverges. Note that $a_{n+1}=\left(\frac{1+i}{\sqrt2}\right)a_n$ and that therefore, if it converges to some $l\in\Bbb C$, then\begin{align}l&=\lim_{n\to\infty}a_{n+1}\\&=\lim_{n\to\infty}\left(\frac{1+i}{\sqrt2}\right)a_n\\&=\left(\frac{1+i}{\sqrt2}\right)l\end{align}and that therefore $l=0$. But you cannot possibly have $l=0$, since $(\forall n\in\Bbb N):|a_n|=1$.

Title: Convergence of sequences: What dowe do in case of complex numbers?

Body:
I want to check if the following sequences converge or not.

\begin{equation*}a_n=\frac{n^4-n^3-\frac{1}{n}}{\left (2n^2+2\right )^2}=\frac{n^4-n^3-\frac{1}{n}}{4n^4+8n^2+4}=\frac{n^4\cdot \left (1-\frac{1}{n}-\frac{1}{n^5}\right )}{n^4\cdot \left (4+\frac{8}{n^2}+\frac{4}{n^4}\right )}=\frac{1-\frac{1}{n}-\frac{1}{n^5}}{4+\frac{8}{n^2}+\frac{4}{n^4}}\rightarrow \frac{1-0-0}{4+0+0}=\frac{1}{4}\end{equation*} So it converges to $\frac{1}{4}$.

$ \ $ \begin{align*}a_n&=\frac{n(2n-1)(n+2)+\sqrt{n}-\left [\sqrt{2n}(n+1)\right ]^2}{4n^2+1}=\frac{n(2n^2+4n-n-2)+\sqrt{n}-[2n(n+1)^2]}{4n^2+1}\\ & =\frac{2n^3+3n^2-2n+\sqrt{n}-2n(n^2+2n+1)}{4n^2+1}=\frac{2n^3+3n^2-2n+\sqrt{n}-2n^3-4n^2-2n}{4n^2+1}\\ & =\frac{-n^2-4n+\sqrt{n}}{4n^2+1} =\frac{n^2\cdot \left (-1-\frac{4}{n}+\frac{1}{\sqrt{n}^3}\right )}{n^2\cdot \left (4+\frac{1}{n^2}\right )}=\frac{-1-\frac{4}{n}+\frac{1}{\sqrt{n}^3}}{4+\frac{1}{n^2}} \rightarrow\frac{-1-0+0}{4+0}=-\frac{1}{4} \end{align*} So it converges to $-\frac{1}{4}$.

\begin{equation*}a_n=\frac{\displaystyle{\sum_{j=0}^n n^j}}{n^{n+1}}=\frac{\frac{1-n^{n+1}}{1-n}}{n^{n+1}}=\frac{1-n^{n+1}}{(1-n)n^{n+1}}=\frac{1-n^{n+1}}{n^{n+1}-n^{n+2}}=\frac{n^{n+2}\cdot \left (\frac{1}{n^{n+2}}-\frac{1}{n}\right )}{n^{n+2}\cdot \left (\frac{1}{n}-1\right )}=\frac{\frac{1}{n^{n+2}}-\frac{1}{n}}{\frac{1}{n}-1} \rightarrow \frac{0-0}{0-1}=0\end{equation*} So it converges to $0$.

\begin{equation*}a_n=(-1)^n\cdot \frac{2n^2+1}{(n+1)(\sqrt{2n}-1)(\sqrt{2n}+1)}\end{equation*}

Let $b_n:=\frac{2n^2+1}{(n+1)(\sqrt{2n}-1)(\sqrt{2n}+1)}$.

We have that
\begin{align*}b_n&=\frac{2n^2+1}{(n+1)(\sqrt{2n}-1)(\sqrt{2n}+1)}=\frac{2n^2+1}{(n+1)(2n-1)}=\frac{2n^2+1}{2n^2-n+2n-1}=\frac{2n^2+1}{2n^2+n-1}=\frac{n^2\cdot \left (2+\frac{1}{n^2}\right )}{n^2\cdot \left (2+\frac{1}{n}-\frac{1}{n^2}\right )}\\ & =\frac{2+\frac{1}{n^2}}{2+\frac{1}{n}-\frac{1}{n^2}}\rightarrow \frac{2+0}{2+0-0}=1\neq 0 \end{align*} So the sequence $a_n$ doesn't converge.

\begin{equation*}a_n=\left (\frac{1+i}{\sqrt{2}}\right )^n\end{equation*}
I tried to check the absolute value: \begin{equation*}|a_n|=\left |\left (\frac{1+i}{\sqrt{2}}\right )^n\right |=\left |\frac{1+i}{\sqrt{2}}\right |^n=\left (\frac{|1+i|}{\sqrt{2}}\right )^n=\left (\frac{\sqrt{1^2+1^2}}{\sqrt{2}}\right )^n=\left (\frac{\sqrt{2}}{\sqrt{2}}\right )^n=\left (1\right )^n=1\end{equation*} but this doesn't help? What can we do in this case?


答: Since $-1\leq \sin(1/x)\leq 1$, we have $-x\leq x \sin(1/x)\leq x $. Hence, by squeze theorem we have $\lim_{x\rightarrow 0} x \sin(1/x)=0$.</p>

Title: Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?

Body:
WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:

$$
 \lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) =  \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
=  \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1  \\
= 1?
$$

I wonder where I'm wrong...


问:
What does it mean for a discontinuous piecewise function to have the same slope from the left and right at a point?
I came across a piecewise function which isn't continuous but the slope at a certain point from the left and right are the same at $x=0$. What does this mean for the function, why wouldn't we consider this to be differentiable?

$$f(x) = \begin{cases}
7x-6x^{3}-7 & \text{if } x \geq 0,\\
2x - \frac{20}{x+2}  & \text{if } x < 0.
\end{cases}$$

I'm familiar with the theorem which states if a function is differentiable then it's continuous but this has puzzled me a bit.

答:
Tge derivative is
$$\lim_{h\to 0}\frac{f(h)-f(0)}{h-0}$$
and there is no value of $f(0)$ where that limit exists


问:
Find the acceleration of a particle and its maximum speed
I study maths for a hobby and am stuck with this problem:

A particle moves in a straight line such that after t seconds its displacement from a fixed point is s metres, where $s = 4\cos t - \cos2 t$. If the particle first comes to rest after T seconds, where $T \gt 0$, find (a) its acceleration at time T, (b) the maximum speed it attains for $0 \lt t \lt T$

My working is as follows:

$s = 4\cos t - \cos2t$

let V = velocity and A = acceleration

$V = \frac{ds}{dt} = 2\sin 2t - 4\sin t$

$A = \frac{dV}{dt} = 4\cos 2t - 4\cos t$

V = 0 when $2 \sin 2t = 4\sin t$

Now, $\sin2 t = 2\sin t\cos t$

So V = 0 when $4\sin t \cos t = 4 \sin t$

$\cos t = \frac{4\sin t}{4\sin t} = 1$

When I substitute this into my equation for acceleration I get 0.

The book says the answers are (a) 8 m/sec (b)$3\sqrt3 $ m/sec

答:
As you found out, when $v = 0$, $\sin t \cos t =  \sin t$.

This is true when $\sin t = 0 \,$.

So the first time when it happens is when $t = 0$ as $\sin t = 0$. But we need to find the first time $T$ it comes to rest for $t \gt 0$.

$\sin t = 0 \,$ when $t = \pi$.

Please note when $\cos t = 1, $ $\sin t = 0$ but if you go by $\cos t = 1$ that only happens at $t = 2\pi \,$ for $t \gt 0$.

Now if you plug in $t = \pi$, you get the acceleration.

Also for max speed $|v_{max}|$, set $\frac{dV}{dt} = 0$. That gives you,

$\cos 2t = \cos t \implies 2cos^2t - \cos t - 1 = 0$

i.e. $(2\cos t + 1) (\cos t - 1) = 0$

That leads to only one possible point $t = \frac{2 \pi}{3}$ for $0 \lt t \lt T$.


答:
Hint:

$ 4\sin t\cos t = 4\sin t$  implies either $\cos t=1 $ or $\sin t =0$

The smallest $t>0$ satisfying this is $t=\pi$ in which case $\cos 2t=1$ and $\cos t = -1$, leading to $A=8$

For (b), maximum velocity occurs when acceleration is $0$ and is changing from positive to negative

Title: Find the acceleration of a particle and its maximum speed

Body:
I study maths for a hobby and am stuck with this problem:

A particle moves in a straight line such that after t seconds its displacement from a fixed point is s metres, where $s = 4\cos t - \cos2 t$. If the particle first comes to rest after T seconds, where $T \gt 0$, find (a) its acceleration at time T, (b) the maximum speed it attains for $0 \lt t \lt T$

My working is as follows:

$s = 4\cos t - \cos2t$

let V = velocity and A = acceleration

$V = \frac{ds}{dt} = 2\sin 2t - 4\sin t$

$A = \frac{dV}{dt} = 4\cos 2t - 4\cos t$

V = 0 when $2 \sin 2t = 4\sin t$

Now, $\sin2 t = 2\sin t\cos t$

So V = 0 when $4\sin t \cos t = 4 \sin t$

$\cos t = \frac{4\sin t}{4\sin t} = 1$

When I substitute this into my equation for acceleration I get 0.

The book says the answers are (a) 8 m/sec (b)$3\sqrt3 $ m/sec


问:
Convert infinite series with infinity limit to integral
I have to prove this equality :

$\sum_{k=0}^{+\infty} \frac{1}{16^k(p+8k+1)} = \int_0^1 \frac{16x^p}{16-x^8}dx$

I tried to use Riemann sums but I don't really know how to convert a sum without any "$n$" inside, I also tried to convert the integral into a Riemann sum without any success either.

Body:
$$S=\sum_{k=0}^{\infty} \frac{16^{-k}}{8k+p+1}=\sum_{k=0}^{\infty} 16^{-k}\int_{0}^{1} x^{p+8k} dx=\int_{0}^{1}\sum_{k=0}^{\infty}x^p(x^8/16)^k dx$$ $$=\int_{0}^{1}\frac{x^p}{1-(x^8/16)} dx, ~if~ x^8<16 \implies x <\sqrt{2}.$$
Here we have used sum of an infinite GP.


问:
How do I prove that $A^{T} A$ is not invertible when n<m?
$A$ is an $ \times $ matrix.
Prove that if $n < m$ then $A^{T}A$ is not invertible.

I know that when $Ax=0$ does not have trivial solution the matrix is not invertible.

So by my understanding I need to show that $x$ has a non trivial solution.

I have trouble proving this with out using dimensions and ranks since I have not yet learned those  in detail.

答: A $n \times m$ matrix is a map from $\mathbb{R}^m \rightarrow \mathbb{R}^n$. If $n<m$, then this linear map cannot be injective. Similarly, $A^T$ would be a map from $\mathbb{R}^n \rightarrow \mathbb{R}^m$. This map cannot be surjective. So $A^{T}A$ would be a map from $\mathbb{R}^m \rightarrow \mathbb{R}^m$.  This map is not bijective and hence an inverse cannot exist.


答: In the second fraction, you're confusing $\lim_{x\to 0}$ and $\lim_{x \to\infty}$.</p>


问:
Is $\left (\sum_{i=1}^{n}a_{i}b_{i} \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$
Is it true that
$$\left (\sum_{i=1}^{n}a_{i}b_{i}  \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$$

That is, can we always combine sums in that way?

答:
This is true. You can, for example by induction on $m$, prove that
$$\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^m b_j\right)=\sum_{i=1}^n\sum_{j=1}^m a_ib_j,$$
which implies your claim.

If $m=0$, both sides are trivially $0$, so we're good.

Assume that the formula holds for $m$. Let's prove it for $m+1$.
We have
$$\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^{m+1} b_j\right)=\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^{m} b_j+b_{m+1}\right)=\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^m b_j\right)+\sum_{i=1}^na_ib_{m+1}.$$

Now, we apply the induction hypothesis, to obtain
$$\sum_{i=1}^n\sum_{j=1}^m a_ib_j+\sum_{i=1}^na_ib_{m+1}=\sum_{i=1}^n\left(\sum_{j=1}^m a_ib_j+a_ib_{m+1}\right)=\sum_{i=1}^n\sum_{j=1}^{m+1}a_ib_j.$$


问:
Poisson process - expected reward until time t
The calls to the fire department occur according to a Poisson process with a rate of three per day. The fire department must respond to each call. Of the calls that come into the fire department, on the average one third turn out to be false alarms.
Firemen are presently paid 140 Dollars a day. A new plan is proposed where they will be paid a random amount of money per fire that they actually fight. In the new scheme the expected pay per fire fought is 60 Dollars. Determine the expected reward until time t.

[Resnick, Adventures in Stochastic Processes]

My approach is the following: Let $T _n=\sum\nolimits_{i=1}^{N_ t}X_ i $ where $T _n$ is the time of the nth call and $X _i $ the time between the calls. The number of calls in $[0,t]$ is given as $N _t = \sum\nolimits_{n=1}^{\infty}1 _{(T_ n\leq t)}$. Let $Y _n$ the reward associated with $X_ n$. Then we have for the total reward at time t:
$R _t = \sum\nolimits_{i=1}^{N _t}Y_ i $.

So what I have to do now is to compute the expected value of $R_t$ . How can I do that? Can I use Wald's identity:
$E(R _t)=E(N _t)E(Y _i)$?

答:
In the usual case, the rewards  are independant from the fire arrival times. That's what implied in A new plan is proposed where they will be paid a random amount of money per fire that they actually fight.

Therefore you can assume that the rewards process $(Y_i)_{i}$ is independant from the time of the calls $(T_i)_i$ and $(N_t)_{t>0}$ as well, because it is a measurable function of $(T_i)_i$. Then you can apply your result (Wald's formula) which is easily derived using iterated expectation  as follows :

$\mathop{\mathbb{E}}[R_t] = \mathop{\mathbb{E}}[\mathop{\mathbb{E}}[R_t|N_t]] = \mathop{\mathbb{E}}[\mathop{\mathbb{E}}[Y_1]N_t] = \mathop{\mathbb{E}}[Y_1]\mathop{\mathbb{E}}[N_t]$


答:
I think you're confusing $\,x\to 0\,$ with $\,x\to\infty\,$ in the case $\,\frac{\sin x}x\,$ . The general case is

Proposition: Assming $\,f,g\,$ are functions defined in a punctured neighbourhood I$_0\;$ of some point $\,x_0\,$ and s.t.

$$\lim_{x\to x_o}f(x)=0\;\;,\;\;|g(x)|\le M\;,\;\forall\,x\in\text{I}_0\;,\;\;\text{for some constant}\;M\in\Bbb R\;,\;\text{then}$$

$$\lim_{x\to x_0}f(x)g(x)=0$$


答:
Direct proof: $$\begin{align}\sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j} &\overset 1=\sum_{j=1}^{n}a_{1}b_{1}c_{j}d_{j} +  \sum_{j=1}^{n}a_{2}b_{2}c_{j}d_{j}+\dots +\sum_{j=1}^{n}a_{n}b_{n}c_{j}d_{j} \\&\overset2=a_{1}b_{1}\sum_{j=1}^{n}c_{j}d_{j}+a_{2}b_{2}\sum_{j=1}^{n}c_{j}d_{j} +\dots + a_{n}b_{n}\sum_{j=1}^{n}c_{j}d_{j} \\ &\overset3= (a_1b_1 + a_2b_2 + \dots +a_nb_n)\sum_{j=1}^{n}c_{j}d_{j} \\ &\overset4= \left
(\sum_{i=1}^{n}a_{i}b_{i}  \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right )\end{align}$$
Explanation:

Expand the first sum (by writing out all terms individually)

Exclude the factors that are not dependent on index $j$

Factor out the common factor (the $j$-sum)

Rewrite the first factor with the sigma notation

Title: Is $\left (\sum_{i=1}^{n}a_{i}b_{i} \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$

Body:
Is it true that
$$\left (\sum_{i=1}^{n}a_{i}b_{i}  \right )\left ( \sum_{j=1}^{n}c_{j}d_{j} \right ) = \sum_{i=1}^{n}\sum_{j=1}^{n}a_{i}b_{i}c_{j}d_{j}$$

That is, can we always combine sums in that way?


答: More generally, $\sum_ix_i\sum_jy_j=\sum_{ij}x_iy_j$, by distributivity in both directions. (However, as @Zuy noted, distributing over a sum still needs induction on the number of terms.) That $x_i=a_ib_i, \, y_j=c_jd_j$ is a distraction.


答:
In 5. we have $a^2=i$ so $a_{8n}=(i^4)^n=1^n=1$ and $a_{8n+1}=a_1a_{8n}=a_1$ so the sequence does not converge.

$\{a_n:1\le n\le 8\}$ is the set of the 8 complex solutions of $z^8-1=0.$

Your work on the other 4 problems is all correct and clearly presented.


问:
Show that these inclusions are correct
Let say that S,T and U are subspaces of V. How can I show these inclusions?

a) $S\cap U\subseteq(S+T)\cap U$ and $T\cap U\subseteq(S+T)\cap U \\$

b) $(S\cap U)+( T\cap U) \subseteq(S+T)\cap U$

I have no idea how to start. Any help?

答: a). $S\subseteq S+T$ for each $S,T$. Intersection is monotonous w.r.t. inclusion. Thus $S\cap U\subseteq (S+T)+U$. The other inclusion is similar.

Title: Show that these inclusions are correct


答:
For anyone interested, here is a link to what the graph of OP's piecewise function looks like.

To determine whether or not $f'(c)$ exists, one can see if the following is true:

$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}.$$

If either the LHS or RHS to the above equation does not exist, or the LHS and RHS exists but have different values, then $f'(c)$ does not exist.

For your piecewise function, let's determine whether or not the derivative of $f$ at $c = 0$ exists.

$\lim_{x\to 0^+}\left(\frac{f(x)-f(0)}{x-0}\right) = \lim_{x\to 0^+}\left(\frac{7x-6x^3 -7 - (-7)}{x}\right) = \lim_{x\to 0^+}\left(7 - 18x^2\right) = 7.$

However,

$\lim_{x\to 0^-}\left(\frac{f(x)-f(0)}{x-0}\right) =  \lim_{x\to 0^-}\left(\frac{2x-\frac{20}{(x+2)} - (-7)}{x}\right) = \lim_{x\to 0^-}\left(2 + \frac{7-\frac{6}{x}}{x+2}\right) \to +\infty,$

or rather it's more proper to say that the latter limit does not exist.

I also think it would be improper to say, " $-7 \neq +\infty$ hence $f'(0)$ does not exist."

Rather, it is proper to say that since the second limit does not exist, $f'(0)$ does not exist.

Title: What does it mean for a discontinuous piecewise function to have the same slope from the left and right at a point?

Body:
I came across a piecewise function which isn't continuous but the slope at a certain point from the left and right are the same at $x=0$. What does this mean for the function, why wouldn't we consider this to be differentiable?

$$f(x) = \begin{cases}
7x-6x^{3}-7 & \text{if } x \geq 0,\\
2x - \frac{20}{x+2}  & \text{if } x < 0.
\end{cases}$$

I'm familiar with the theorem which states if a function is differentiable then it's continuous but this has puzzled me a bit.


问:
Prove that $\int\limits_0^\pi {\frac{{\sin \left( {xt} \right)}} {t} \mathrm dt} $ is continuous
How can I prove that this function is continuous? $$
f\left( x \right) = \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
$$
Some hint?
Don´t consider the zero in the endpoint of the integration zone, just take it as a limit $$
f\left( x \right) = \mathop {\lim }\limits_{\varepsilon ^ +   \to 0} \int\limits_\varepsilon ^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
$$
How can I do it? DX!

答:
$$I(x)=\int_0^\pi\frac{\sin(xt)}{t}dt$$
$$I'(x)=\int_0^\pi\cos(xt)dt$$
now $u=xt,du=xdt\Rightarrow dt=\frac{du}{x}$ and so:
$$I'(x)=\int_0^{\pi x}\frac{\cos(u)}{x}du=\left[\frac{\sin(u)}{x}\right]_{u=0}^{\pi x}$$
$$I'(x)=\frac{\sin(\pi x)}{x}$$
and it is clear that $I'(x)$ is continuous for $x>0$ and then try taking the limit:
$$\lim_{x\to0^+}\frac{\sin(\pi x)}{x}=\pi$$
so the derivative exists for all $x\in\mathbb{R}$. One other thing that you have to look at is the value of $I(0)$. we can do this:
$$I(0)=\lim_{x\to 0}\int_0^\pi\frac{\sin(xt)}{t}dt=\int_0^\pi\lim_{x\to0}\frac{\sin(xt)}{t}dt=\int_0^\pi\lim_{x\to0}\frac{xt}{t}dt=0$$
and so $I(x)\in C(\mathbb{R})$


答:
Yes, it is true.

$$\sum_i\sum_j a_ib_j=\sum_i a_i\sum_j b_j=\left(\sum_jb_j\right)\sum_ia_i=\left(\sum_i a_i\right)\left(\sum_j b_j\right)$$

You first take $a_i$ out as common factor which is independent of $j$, then take $\sum b_j$ out as common factor which is independent of $i$.


问:
Left-ratio and right-ratio in (not necessarily commutative) field
I am reading Birkhoff–Neumann (1936) about (not necessarily commutative) field $F$.

The authors use terms right-ratio and left-ratio in section 13.

Right-ratio is denoted as $[x_1, x_2, ... x_n]_r\,$  $x_i \in F$ where not all elements $x_i=0$. Similary left-ratio is $[y_1, y_2, ... y_n]_l$.

Two right-ratios $[x_1, ... x_n]_r\, ,[\xi_1, ... \xi_n]_r\,$  are "equal" iff $\exists z\in F:\xi_i=x_iz,\ i=1, ... n$

But in section 14, they also use term ennuples: not right or left ratios. We can define vector-operations and inner-product for ennuples.

I don't understand to this term (ennuples) in not commutative field and their connection to ratios. Any example (of left/right-ratios with ennuples) will be helpful.

Birkhoff, Garrett; von Neumann, John
“The logic of quantum mechanics.”
Annals of Mathemathics (2nd series) 37 (1936), no. 4, 823–843.
MR1503312
JFM62.1061.04
DOI:10.2307/1968621

答:
An "ennuple" just sounds like an element of $F^n$, and "right-ratio" just sounds like an equivalence class of ennuples under the equivalence relation $x\sim y$ iff $\exists z\in F^\ast$ such that $xz=y$. (Here I'm using the $^\ast$ to denote the nonzero elements of the ring $F$).

If we're talking about a division ring $D$, then an ennuple $x\in D^n$ would be equivalent to the members of the set of ennuples $xD^\ast$, and so together they all form a single right-ratio.

For commutative fields, both of these things would just boil down to the same thing: the subspace spanned by $x$. But in general I guess there is no guarantee they are the same subspace.

<s>The "ratio" aspect probably is a reference to the following example. If you think about pairs of nonzero integers (from $\Bbb Z^\ast$), then you can have ennuples with $n=2$: $\Bbb Z^\ast\times\Bbb Z^\ast$, and their left/right ratios would be a model of $\Bbb Q^\ast$. (For example, $2(1,2)=(2,4)$ evidences that $\frac{1}{2}=\frac{2}{4}$ as ratios.</s> Actually I suddenly realized that the relation in this case would not be symmetric! But I still think it gives a little idea of the distinction between the ennuples and equivalence classes.


答:
a. Let $v\in S\cap U$. Then $v=v+0\in S+T$, as $v\in S$ and $0\in T$. Hence $v\in S+T$ and $v\in U$, so $v\in (S+T)\cap U$. The second inclusion of a. follows by renaming the spaces.

b. Let $v\in (S\cap U)+(T\cap U)$. Then $v=x+y$, where $x\in S\cap U$ and $y\in T\cap U$. Notice that in particular, we have $x,y\in U$, so also $x+y\in U$ (as $U$ is a vector space). As $x+y=v$, we obtain $v\in U$. Also, w clearly have $v\in S+T$ (as $v=x+y$ and $x\in S$ and $y\in T$). Hence we conclude $v\in (S+T)\cap U$. This proves the desired inclusion.

This is of course quite unreadable because of all the details, but that's how one does it rigorously. :-)

Title: Show that these inclusions are correct

Body:
Let say that S,T and U are subspaces of V. How can I show these inclusions?

a) $S\cap U\subseteq(S+T)\cap U$ and $T\cap U\subseteq(S+T)\cap U \\$

b) $(S\cap U)+( T\cap U) \subseteq(S+T)\cap U$

I have no idea how to start. Any help?


问:
I have derived a new formula for π. What should I do with it?
I mean I know that I should publish it somewhere but have no idea about how to write it in a professional manner and where to publish it. Can anyone guide me?</p>

答:
What would be correct is $$\pi = \sqrt{1+\sum\limits_{n=1}^\infty \frac{6\times 2^n-n^2}{2^n \times n^2}}$$ which is close to what you said in your comment

As @lulu notes, this is essentially $\sum\limits_{n=1}^{\infty}\frac {1}{n^2}=\frac {\pi^2}6$ and is not novel - it is called the Basel problem

Your expression converges slowly to $\pi$: using the first ten terms gives about $3.049522$ while the first hundred gives $3.132077$


问:
A proper subspace $W$ of $\mathbb{R^n}$ is closed?
Let $W$ be a proper subspace of $\mathbb{R^n}$ equipped with euclidean metric $d$ defined by $d(x,y)= (\sum_{i=1}^{n}(x_i-y_i)^2)^{1/2}$. Let $W$ be a proper subspace of $\mathbb{R^n}$. Then is it closed?

Attempt: Since $W$ is a proper subspace of $\mathbb{R^n} \implies \dim(W) \leq n-1$. Now if $\dim(W)= 0$ then clearly $W$ is closed so we let $\dim(W) = k$, where $1\leq k\leq n-1$. So $W \cong \mathbb{R^k}$ and since $\mathbb{R^k}$ is complete under the above mentioned metric,we can see that any sequence in $\mathbb{R^k}$ converges to a point in $\mathbb{R^k}$ so we can thought of $W$ as $\mathbb{R^k}$ so $W$ is closed.

Is my approach correct? and If I'm wrong, give me hints.

答: What you did is correct, but I think that it is more natural to do it as follows. Let $\{e_1,\ldots,e_k\}$ be a basis of $W$ and extend it to a basis $\{e_1,\ldots,e_n\}$ of $\Bbb R^n$. Consider the linear map $F\colon\Bbb R^n\longrightarrow\Bbb R^{n-k}$ such that $F(e_{k+1})=(1,0,0,\ldots,0)$, $F(e_{k+2})=(0,1,0,\ldots,0)$ and so on…; besides, $F(w)=0$ for each $w\in W$. Then $W=F^{-1}(\{0\})$ and therefore, since $F$ is continuous and $\{0\}$ is closed, $W$ is closed.


问:
Trigonometry Word Problem--Not sure if correct
From point $A$ the angle of elevation to the top of a newly constructed building is $17.2$ deg. From point $B$ which is $153$ meters closer to the building the angle of elevation at the top of the building is $25$ deg. Solve the triangle.

smaller triangle: $\angle B = 155^\circ\,; \angle C = 7.8^\circ\,; b = 476.4\mathrm{m}\,; a = 333.3\mathrm{m}$

bigger triangle: $\angle C = 65^\circ\,; a = 333.3 \times h\,;b = 140.9\mathrm{m}\,; c = 302.2\mathrm{m}$

Height = $140.9 \mathrm{m}$ ????

答:
Let the base of the smaller triangle be $b$. Then the base of the bigger triangle is $b+153$. Let the height of the building be $h$. Then we have
$$\frac{h}{b}=\tan(25^\circ); \quad \frac{h}{b+153}=\tan(17.2^\circ).$$
It follows that
$$b\tan(25^\circ)=(b+153)\tan(17.2^\circ).$$
It follows that
$$b=\frac{153\tan(17.2^\circ)}{\tan(25^\circ) -\tan(17.2^\circ)}.$$
The calculator gives that $b\approx 302.1347$. It follows that the height $b\tan(25^\circ)$ is approximately $140.8877$. These numbers are consistent with numbers that you obtained. </p>


问:
Mutually exclusive and Independent Events
I often see that events are described as either mutually exclusive and independent events. My first question is, do mutually exclusive events simply refer to dependent events?

To add to the confusion, I also learned that mutually exclusive events can be independent event in a special case when probability is zero i.e. $P(A) = 0$ or $P(B) = 0$ .

But an event should either be independent or dependent and not both?

答:
By definition, two events $A$ and $B$ are mutually exclusive whenever $A\cap B=\varnothing$. In that case, one has $\mathbb P(A\cap B)=0$. Some people also say that $A$ and $B$ are mutually exclusive whenever $\mathbb P(A\cap B)=0$. (The second condition is weaker, but it doesn't make a difference in practice.)

Now, two events $A$ and $B$ are called independent, if $\mathbb P(A\cap B)=\mathbb P(A)\cdot\mathbb P(B)$.

Can you now see in which cases you have that $A$ and $B$ are mutually exclusive and independent?


答:
I agree with José's answer. Two additional comments:

You don't actually need $W$ to be a proper subspace: this also works if $W = \mathbb{R}^n$,

Note that the words complete and closed are used differently: the vector space $W$ can be called complete, but we cannot call $W$ closed unless we really see it as a subspace of $\mathbb{R}^n$. (I'm saying this because your last sentence seems to indicate that you think of closedness as a property of a space rather than a property of a subspace.)

Title: A proper subspace $W$ of $\mathbb{R^n}$ is closed?

Body:
Let $W$ be a proper subspace of $\mathbb{R^n}$ equipped with euclidean metric $d$ defined by $d(x,y)= (\sum_{i=1}^{n}(x_i-y_i)^2)^{1/2}$. Let $W$ be a proper subspace of $\mathbb{R^n}$. Then is it closed?

Attempt: Since $W$ is a proper subspace of $\mathbb{R^n} \implies \dim(W) \leq n-1$. Now if $\dim(W)= 0$ then clearly $W$ is closed so we let $\dim(W) = k$, where $1\leq k\leq n-1$. So $W \cong \mathbb{R^k}$ and since $\mathbb{R^k}$ is complete under the above mentioned metric,we can see that any sequence in $\mathbb{R^k}$ converges to a point in $\mathbb{R^k}$ so we can thought of $W$ as $\mathbb{R^k}$ so $W$ is closed.

Is my approach correct? and If I'm wrong, give me hints.


问:
How to prove that $DK=DL$

In $\triangle ABC$ the cevians $AD$,$BE$ and $CF$ are concurrent at the point $O$. $DL\parallel  EF$ and $BE$ intersects $DL$ at $K$. How to prove that $DK = DL$.

I applied menelaus theorem twice in the $\triangle KEL$ and got a relation $\frac{DL^2} {DK^2}=\frac{BE} {BK}\cdot \frac{CL} {CE} \cdot \frac{AL} {AE} \cdot \frac{EO} {KO}$ and also  tried to use the ratio of similar triangles but no result. any ideas  on utilizing the ratio of similar triangles.

答: You may want to start like this: Since $EF \parallel DL$, we have $\dfrac{DL}{JE}=\dfrac{DA}{AJ}$ and $\dfrac{DK}{JE} = \dfrac{DO}{OJ}$. The problem reduces to showing that $\dfrac{DA}{AJ}=\dfrac{DO}{OJ}$.

Title: How to prove that $DK=DL$


问:
Divisibility by $2020$
The problem is to prove that given $11$ natural numbers $a_1,...,a_{11}$ then there exist $11$ numbers $b_1,...,b_{11} \in \{0,1,-1\}$, not all $0$'s, such that $2020$ divides $a_1b_1+...+a_{11}b_{11}$. I know how to prove it for $12$ numbers: by pigeonhole, since $2^{12} >4040$, there exists two sequences of $b_{i} \in \{1,-1\}$ such that $a_1b_1+...+a_{12}b_{12}$ gives the same remainder mod $4040$. Take the difference and we get that $4040$ divides $a_1c_1+...+a_{12}c_{12}$ for $c_i \in {0,2,-2}$. Divide each $c_i$ by $2$ and get the result. But how to do it for $11$ numbers? If it wasn't 2020, but any odd number less then $2048$ I could do the same with $\{0, 1/2, -1/2\}$ but it does not seem to work here.

答: Note that there are $2^{11} > 2020$ subsets of the $a_i$'s. Thus by pigeonhole, there exists two different subsets with the same sum mod $2020$. Subtracting them gives us a sum of the form $a_1b_1+...+a_{11}b_{11}$ with $b_i \in \{0,1,-1\}$ that is divisible by $2020$, as desired.

Title: Divisibility by $2020$


问:
Three-dimensional gaussian integral with $(x + y + z)^2$ term
I need to analytically perform the following integral:

$$\mathcal{I} = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty e^{\frac{-r^2}{2}}(x + y + z)^2 dx dy dz$$

I know the standard way to perform multi-dimensional gaussian integrals is to split them up and calculate each dimension seperately, but I do not know how to deal with $(x + y + z)^2$ term.

Body: Let $X,\,Y,\,Z$ be three $N(0,\,1)$ IIDs so $X+Y+Z\sim N(0,\,3)$. Then$$\mathcal{I}=(2\pi)^{3/2}\Bbb E(X+Y+Z)^2=3(2\pi)^{3/2}.$$


问:
Archimedes Method To Find The Area Under A Curve
I'am reading Tom Apostol's Calculus volume-1 text (page 3 and 4),where he talks about calculating the area under a curve which eventually leads to the concept of the definite integral.In the below figure he chooses an arbitrary point on the base and denote's it's distance from $0$ by $x$.

How can the vertical distance be $x^2$?In particular,if the length of the base itself is $b$,then how come the altitude is $b^2$?

He then winds up by concluding that the area $A = \frac{b^3}{3}$ by considering approximations from above and below.

I have a problem in understanding his derivation due to unfamiliarity with mathematical induction and other rigour.Is there an informal way (i mean less rigour) of computing the area? I'll be very happy if i understand this cause i want to think and solve this problem the archimedes way!!

I'am aware of possible duplicate thread however this is purely based on apostol's text.I even went through this Area under a curve is an integral but couldn't understand it.

答:
$x\cdot x^2=x^3$, thus, the area of a rectangle.
Archimedes tells us that the area of the hyperbola is two thirds of the triangle formed by its Tangents, thus
$$\frac{x^3}{2 (2/3)} = \frac{x^3}{3}.$$


问:
Examples of properties that hold almost everywhere, but such that explicit examples are not known
In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties.
More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have
  $$
\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}
$$
  almost surely.
  Note: The result fails for all rationals.

答:
The set of normal numbers in base $b$ has full Lebesgue measure, for each integer $b\geqslant2$. Hence the set of numbers simultaneously normal in every base has full Lebesgue measure. Examples of numbers normal in a given base $b$ are easy to exhibit, à la Champernowne or à la Copeland-Erdős.

But no number is known to be normal in two given different bases simultaneously (for example in base $b=2$ and in base $b=10$).


问:
Is there any integral for the Golden Ratio?
I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:

$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$

$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$

$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$

Is there an interesting integral* (or some series) whose result is simply $\phi$?

* Interesting integral means that things like

$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$

are not a good answer to my question.

答:
Connecting three well-known constants together:
$$\int_{-\infty}^{+\infty}\sin^2\left(\frac{x}{\phi+\frac{1}{\phi}}\right)\cdot \frac{\mathrm dx}{4x^4+5x^2}=\frac{\pi}{e}\cdot\frac{1}{\left(\phi+\frac{1}{\phi}\right)^3}$$


问:
Ambiguity in orbit space notation
For an action of a group $G$ on a topological space $Y$ where each element of $G$ corresponds to a homeomorphism of $Y$, Hatcher defines the orbit space $Y/G$ of the action to be the quotient of $Y$ by the equivalence relation of being in the same orbit of the action. This confuses me. The notation $Y/G$ mentions the space and the group, but not the action. From visualising, I’m pretty sure two $Y/G$’s would not even have to be homeomorphic. Consider the subspace $X$ of $\mathbb{R}^2$ given by the union of all lines $x=z$ and $y=z$ with $z\in\mathbb{Z}$. Here are two actions of $\mathbb{Z}$ on $X$. In the first, a generator maps to the function $x+(1,0)$, in the second it maps to $x+(1,1)$. The quotient by the first action is a line  with a bunch of circles glued to it at a discrete set of points, whereas the quotient by the second is a bunch of circles indexed by the integers with the North pole of one glued to the South pole of the next. These are homotopy equivalent but not homeomorphic. So why the notation?

答:
You are right, a (left) group action is a function $\alpha : G \times Y \to Y$ with suitable properties, thus it would be correct to write $Y/\alpha$. However, it is common use (or, if you want, common abuse of notation) to write it in the form $Y/G$.
Unfortunately one cannot say much more than that.

Title: Ambiguity in orbit space notation


问:
Is $\nu(E) := \int_E f \, d\mu$ a measure on $\Omega$
Let $(\Omega, \mathcal{A},\mu)$ be a measure space. and $f:\Omega \rightarrow \bar{\mathbb{R}}^+_0$ a measurable function.

Show that $\nu(E) := \int_E f \, d\mu$ is a measure on $\Omega$

I thought about it like that...

A measure must satisfy 2 conditions.

$\nu(\emptyset)=0$

$\nu(\bigcup_{i=1}^{\infty} A_i) = \sum_{i=1}^{\infty} \nu(A_i)$ with $A_i\in\mathcal{A}$ disjoint.

The first condition looks pretty easy. Since $f$ is measurable it can be described by a step function $f = \sum_{k=1}^{\infty} \alpha_k \mathcal{X}_{E_k}$ (with $\mathcal{X}$ being the indicator function and $\alpha_i$ being the function value in that point)

Furthermore, $\int_E f \, d\mu= \sum_{k=1}^{\infty} \alpha_k \mu(E)$ with $E\in \mathcal{A}$

$$\nu(\emptyset) = \int_\emptyset f\, d\mu=\sum_{k=1}^{\infty} \alpha_k \mu(\emptyset)=\sum_{k=1}^{\infty} \alpha_k * 0=0$$

So 1) seems true to me.

I've thought about the second condition too, but I'm not sure about it.

$$\nu(\bigcup_{i=1}^{\infty}A_k)=\int_{\bigcup_{i=1}^{\infty}A_k} f\, d\mu = \sum_{k=1}^{\infty} \alpha_k \mu(\bigcup_{i=1}^{\infty}A_k)=\sum_{k=1}^{\infty} \alpha_k \sum_{i=1}^{\infty} \mu(A_i) $$
$$=\sum_{k=1}^{\infty} \sum_{i=1}^{\infty}  \alpha_k\mu(A_i) =\sum_{k=1}^{\infty} \int_{A_i}  f\,d\mu = \sum_{k=1}^{\infty} \nu(A_i) $$

So 2) must be true too and 1)+2) $\Rightarrow$ that $\nu$ is a measure on $\Omega$

Have I made a mistake? Hoping for helpful inputs, thanks in advance =)

EDIT:

For the 1st:

Since $f$ is positive and measurable, there exists a sequence of simple functions $f_n(x) \nearrow f(x)$ for all $x$.

Because $f_n$ simple $\Rightarrow f_n(x)= \sum_{k=1}^n\alpha_i\mathcal{1}$. Let $(f_n)_{n\in\mathbb{N}}$ be the monotone function sequence, consisting of simple functions with $\lim_{n\rightarrow \infty} f_n = f$

$$\nu(\emptyset) = \int_\emptyset f\, d\mu=\int_\emptyset \lim_{n\rightarrow \infty}f_n\, d\mu\overset{MCT}{=}\lim_{n\rightarrow \infty}\int_\emptyset f_n\, d\mu=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \alpha_k \mu(\emptyset\cap E_k)=\lim_{n\rightarrow \infty}\sum_{k=1}^{n} 0 = 0$$

答:
Hint

Both doesn't seem correct.

For the first one

If $f$ is simple, then $$f(x)=\sum_{k=1}^n \alpha _k\boldsymbol 1_{E_k},$$
(the sum is finite). Then $$\mu(\emptyset)=\sum_{k=1}^n\alpha _k\mu(\emptyset\cap E_k)=\sum_{k=1}^n\alpha _k\mu(\emptyset)=0.$$

If $f$ is positive, there is a sequence of simple function s.t. $f_n(x)\nearrow f(x)$ for all $x$, then using Monotone convergence theorem will do the work.

If $f$ is measurable, then $f=f^+-f^-$ where $f^+(x)=\max\{f(x),0\}$ and $f^-(x)=-\min\{f(x),0\}$ are both positive, and try to apply the previous step.

For the second one

Still doesn't work. Let $\{A_i\}_{i\in\mathbb N}$ disjoints and measurable. If $f$ is simple, i.e. $$f(x)=\sum_{i=1}^n\alpha _i\boldsymbol 1_{E_i},$$
where $E_i$ are measurable, then
\begin{align*}
\int_{\bigcup_{j=1}^\infty A_i}f\,\mathrm d \mu=\sum_{i=1}^n\alpha _i\mu\left(E_i\cap \bigcup_{j=1}^\infty A_i\right)&=\sum_{i=1}^n\alpha _i\mu\left( \bigcup_{j=1}^\infty E_j\cap A_i\right)\\
&=\lim_{m\to \infty }\sum_{i=1}^n\alpha _i\mu\left(\bigcup_{j=1}^mE_i\cap A_j\right).
\end{align*}

I let you conclude from here.

When $f\geq 0$ and $f$ measurable, apply the same recipe than the first one.


问:
Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?
WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:

$$
 \lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) =  \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
=  \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1  \\
= 1?
$$

I wonder where I'm wrong...

答:
$$\lim_{x \to 0} \left(x\cdot \sin\left(\dfrac{1}{x}\right)\right) = \lim_{\large\color{blue}{\bf x\to 0}}
 \left(\frac{\sin\left(\dfrac{1}{x}\right)}{\frac 1x}\right) =
 \lim_{\large\color{blue}{\bf x \to \pm\infty}} \left(\frac{\sin x}{x}\right) = 0 \neq 1$$</p>


问:
To prove an apparently obvious statement: if $A_1\subseteq A_2 \subseteq ... \subseteq A_n$, then $\bigcup_{i=1}^n A_i = A_n$

To prove:$$\text{If } A_1\subseteq A_2 \subseteq ... \subseteq A_n\text{ ,then } \bigcup_{i=1}^n A_i = A_n$$ using the axioms of ZFC Set Theory.

Honestly, this statement is very obvious, but I do not want to take that for granted. How can I prove this from the axioms, especially the axiom of unions?

I'm fairly new to ZFC Set Theory, so I don't know where to start. I do know that the Axiom of Unions (and Specification) gives rise to the definition of $\bigcup_{i=1}^n A_i$ for sets $A_1,...,A_n$ though. The uniqueness follows from the Axiom of Extensionality.

My thoughts:

We already know that the union exists, and it is defined such that for every element $x\in \bigcup_{i=1}^n A_i$, there exists at least one set $A_j$ such that $x\in A_j$. What's next?

Thank you!

Edit:

Induction works if the chain is finite or countably infinite. What if the chain length is uncountable? Does this result still hold, and if yes, how do we prove it?

Edit 2:

So now I wish to prove (or disprove) the following using the axioms of ZFC Set Theory:

Consider an infinite set $A$, and an index set $I$ such that $$A_1\subseteq A_2\subseteq ... \subseteq A_i \subseteq ... \subseteq A$$
where $i\in I$. The index set $I$ is uncountable. We have $$\bigcup_{i\in I}A_i = A$$

答:
To be completely formal, you'd could first prove by induction that $A_i\subseteq A_n$ for every $i\leq n$. Then prove that $\bigcup_{i=1}^n A_i\subseteq A_n$ and that $A_n\subseteq\bigcup_{i=1}^n A_i$. This last one should be trivial by definition of the union.

As for the first one, if $x\in \bigcup_{i=1}^nA_i$, then $x\in A_j$ for some $j$, and then we know (from the proof by induction) that $x\in A_n$.


问:
How can I find all the solutions of $\sin^5x+\cos^3x=1$

Find all the solutions of $$\sin^5x+\cos^3x=1$$

Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.

答:
$\sin^5x\le \sin^2x, \cos^3x\le \cos^2x$, and $\sin^2x+\cos^2x=1$, so $\sin^5x+\cos^3x$
can be equal to 1 if and only if $$\sin^5x=\sin^2x$$ and $$\cos^3x=\cos^2x$$, i.e.
$$\sin^2x(1-\sin^3x)=0$$ and $$\cos^2x(1-\cos x)=0$$
The first equation has solution: $x=k\pi$ or $x=\pi/2+2m\pi, k,m $ integers.

The second equation has solution: $x=2n\pi$ or $x=\pi/2+p\pi, n,p$ integers.

Therefore the solution is either:
(i) $x=k\pi$ and $x=2n\pi$, or equivalently, $x=2n\pi$ for some integer $n$, or

(ii) $x=k\pi$ and $x=\pi/2+p\pi$ (impossible), or

(iii) $x=\pi/2+2m\pi$ and $x=2n\pi$ (impossible) or

(iv) $x=\pi/2+2m\pi$ and $x=\pi/2+p\pi$, or equivalently, $x=\pi/2+2m\pi$ for some integer $p$.

In conclusion, $x=2n\pi$ or $x=\pi/2+2n\pi$, $n$ integers, are the solutions.


问:
How can you show that $I=J=\phi^{-2}$
How can you show that $I=J=\phi^{-2}$

Where $\phi=\frac{\sqrt{5}+1}{2}$
$$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

$$J=\int_{-\infty}^{+\infty}\frac{\sin^2(4x\ln\phi)}{4x^4+5x^2+1}\mathrm dx$$

I don't think there are such formulas to represent $\sin^2(a)+\sin^2(b)=\sin^2(c)$

$$I=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{4x^4+5x^2+1}\mathrm dx=\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(3x\ln\phi)}{(4x^2+1)(x^2+1)}\mathrm dx$$

$$I=\int_{-\infty}^{+\infty}\frac{4\sin^2(x\ln\phi)+4\sin^2(3x\ln\phi)}{3(4x^2+1)}\mathrm dx-\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)+\sin^2(x\ln\phi)}{3(x^2+1)}\mathrm dx$$

In general we got evalaute,
$$K=\int_{-\infty}^{+\infty}\frac{\sin^2(ax\ln\phi)}{bx^2+c}\mathrm dx$$

$\sin^2(a)=\frac{1-\cos(2a)}{2}$

$$K=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{1-\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx=\frac{\pi}{2b}\sqrt{\frac{b}{c}}-\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\cos(2ax\ln\phi)}{bx^2+c}\mathrm dx$$

答:
You can use this result
$$ \int_{-\infty}^\infty \frac{\cos x}{x^2+a^2}dx=\frac{e^{-a}}{a}\pi$$
to evaluate
$$ \int_{-\infty}^\infty \frac{\cos (ax)}{b^2x^2+c^2}dx=\frac1a\int_{-\infty}^\infty \frac{\cos (x)}{\frac{b^2}{a^2}x^2+c^2}dx=\frac{a}{b^2}\int_{-\infty}^\infty \frac{\cos (x)}{x^2+\frac{a^2c^2}{b^2}}dx=\frac{a}{b^2}\frac{e^{-\frac{ac}{b}}}{\frac{ac}{b}}\pi=\frac{\pi}{bc}e^{-\frac{ac}{b}} $$
For example
\begin{eqnarray}
&&\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{4x^4+5x^2+1}\mathrm dx\\
&=&-\frac13\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{x^2+1}\mathrm dx+\frac43\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{4x^2+1}\mathrm dx.
\end{eqnarray}
But
$$\int_{-\infty}^{+\infty}\frac{\sin^2(x\ln\phi)}{x^2+1}\mathrm dx=\frac12\int_{-\infty}^{+\infty}\frac{1}{x^2+1}\mathrm dx-\frac12\int_{-\infty}^{+\infty}\frac{\cos(2x\ln\phi)}{x^2+1}\mathrm dx=\frac12\pi-\frac12\cdot\pi e^{-2\ln\phi}=\frac12\pi-\frac12\pi\frac{1}{\phi^2}$$
and you can use the same way to handle the rest.

Title: How can you show that $I=J=\phi^{-2}$


问:
Evaluating $\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$
How to calculate that integral? I have no idea what substitution to use. I think polar coordinates will be needed here.

$$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}{ {{1}\over{2\pi}} e^{{{-1}\over{2}}(4x^2-2xy+3y^2)} ~\mathrm dx \mathrm dy}$$

答:
Hint

First prove that your integral converges. Then $$\frac{1}{2\pi}\iint_{\mathbb R^2}e^{-\frac{1}{2}(4x^2-2xy+3y^2)}\,\mathrm d x\,\mathrm d y=\lim_{n\to \infty }\iint_{\mathcal B(0,n)}e^{-\frac{1}{2}(4x^2-2xy+3y^2)}\,\mathrm d x\,\mathrm d y,$$
where $\mathcal B(0,n)=\{(x,y)\mid x^2+y^2\leq n\}$. Using polar coordinates allow you to conclude.


问:
Min value of $a\sec \theta +b\cos \theta$
Min value of$$a\sec \theta +b\cos \theta$$ for $$\theta \in (-90^\circ,90^\circ) $$

I have trouble in approaching this problem .Could only a method/hint be given ? I would to try it myself.

答:
Try finding the derivative of the expression, and equate it to $0$.
Like Henry pointed out, this would be easier to do in radians, as handling derivatives in degrees can get cluttered.


答:
$$
\lim_{x\to0} \frac{\sin \frac1x}{\frac1x} = \lim_{u\to\pm\infty} \frac{\sin u}{u} = 0\qquad\text{where }u=\frac1x.
$$</p>


问:
Prove that $ 2^r +2$ is sum of 2 square number, when r is a prime number not equal to 2
I was doing a question. Suddenly I got stuck at this last part of the problem. It was to prove
$ 2^r +2 = a^2 +b^2$ where $r \neq 2$, r is a prime and $ a \neq b$. Also $r^2 -1$ is a mersenne prime. I tried to use fermat's little theorum, but to no avail. Thank you.

PS note: The problem I was solving was BMO2021 Q6.

答: The question is wrong. If $r=11$, $2^{11}-7 = 2041$ is not a square number.


问:
Show $\frac{1}{\sum_{1}^{n}a^2}\leq \frac{n}{\left ( \sum_{1}^{n}a \right )^2}$
I want to show that $$\frac{1}{\sum_{1}^{n}a^2}\leq \frac{n}{\left ( \sum_{1}^{n}a \right )^2}$$

Until now, I only have $$\left (\sum_{1}^{n}a  \right )^2\leq \left (\sum_{1}^{n}a^2 \right )n$$

Please, any idea on how to procede?

答:
This is straightforward from Cauchy Schwarz inequality:
$$(a_1^2+a_2^2+...+a_n^2)(1^2+1^2+...+1^2) \ge (a_1+a_2+...+a_n)^2$$


问:
Evaluate $\int x \cos x^2 dx$
Hay I have hit this in my book

Evaluate $\int x \cos x^2 dx$.

I got $x^2 \sin(x^2) / 2 $

But I used a online calculator to check it and it is giving me $\sin(x^2)/2 $

Where dose my X go?

Title: Evaluate $\int x \cos x^2 dx$

Body: $\int x \cos(x^2) dx=\int \frac{1}{2}\cos(x^2) d(x^2)=\frac{1}{2}\sin(x^2)+C$


问:
Area of revolution of the following function, stuck on 1D integral.
Given $\gamma(t) = (t-t^2,1-t^3), t \in [0,1]$, I want to calculate the area of the revolution of the solid when $\gamma$ rotates on the $y$-axis.

Now, given I can call $x = x(t)$ and $y = y(t)$, I have the following result:

$$ A = 2\pi\int_0^1 x(t)\sqrt{(x(t)')^2 + (y(t)')^2)}\, dt $$

I do all the calculations and I get:
$$A = 2\pi\int_0^1 (t-t^2)\sqrt{1 -4t + 4t^2 + 9t^4}\,dt $$
But now I have no idea how to solve this. I tried splitting it and using substitution but I am getting nowhere.

Any tips?

答:
You have the square root of a quartic with no repeated roots. The answer can only be written in terms of elliptic integrals – but I am going to derive such an answer anyway. All elliptic integrals and functions in this answer follow Mathematica/mpmath conventions.$\newcommand{sn}{\operatorname{sn}}\newcommand{cn}{\operatorname{cn}}$

I use Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Physicists. There are formulas in that book for integrals involving the square roots of general quartics with all complex roots – of which the one here, $9t^4+4t^2-4t+1$, is an example – but they are very unwieldy. So I am going to "soften up" the integral first by making a substitution that eliminates the cubic and linear terms of $P$.

Such a substitution is of the linear fractional type, $u=\frac{t+k}{lt+1}$, $t=\frac{k-u}{lu-1}$. When I put the $t$ expression into $P$ and solve for $k$ and $l$ that make the cubic and linear terms of the resulting expression in $u$ (after multiplying by $(lu-1)^4$) zero, I get one solution as $k=0,l=-1$. So let $u=\frac t{1-t}$:
$$A=2\pi\int_0^\infty\frac{u\sqrt{10u^4-2u^2+1}}{(u+1)^6}\,du=2\pi B$$
$$B=\int_0^\infty\frac{u(10u^4-2u^2+1)}{(u+1)^6\sqrt{10u^4-2u^2+1}}\,du$$
So many powers of $u+1$ in the denominator would make it a pain if I used the tables now. Thus I perform a partial fraction decomposition of the rational function in $B$'s integrand…
$$B=\int_0^\infty\left(\frac{10}{u+1}-\frac{50}{(u+1)^2}+\frac{98}{(u+1)^3}-\frac{94}{(u+1)^4}+\frac{45}{(u+1)^5}-\frac9{(u+1)^6}\right)\frac1{\sqrt{10u^4-2u^2+1}}\,du$$
…then reduce the powers using B&F 250.02: let $P$ be a quartic, $p$ a non-root of $P$ and $n>1$, then
$$I_{-n,p}=\int_a^b\frac1{(t-p)^{-n}\sqrt P(t)}\,dt=\frac1{2(1-n)P(p)}\left( \left[\frac{2\sqrt{P(t)}}{(t-p)^{n-1}}\right]_a^b + \sum_{j=1}^4\frac{2m-2-j}{j!}P^{(j)}(p)I_{j-n,p}\right)$$
This reduces $I_{-n,p}$ to a rational linear combination of $1,I_{-1,p},I_{0,p},I_{1,p},I_{2,p}$, and in this case (with $a=0,b=\infty,P(u)=10u^4-2u^2+1$)
$$B=\frac16 I_{-1,-1} - \frac{14}9 I_{0,-1} + \frac{898}{243} I_{1,-1} + \frac{449}{243}\left( - I_{2,-1} + \frac{\sqrt{10u^4-2u^2+1}}{10(u+1)} \right) - \frac{373}{1215} + \frac{77\sqrt{10}}{540}$$
Now let $J_n=\int_0^\infty\frac{t^n}{\sqrt{P(t)}}\,dt$. When rewritten in terms of $J_n$, $B$ has no $J_1$ term:
$$B=\frac16 I_{-1,-1} + \frac{71}{243} J_0 + \frac{449}{243}\left( - J_2 + \frac{\sqrt{10u^4-2u^2+1}}{10(u+1)} \right) - \frac{373}{1215} + \frac{77\sqrt{10}}{540}$$
I have bracketed two terms together because they are infinite when standing alone but finite when combined. Now work out $J_2$ using B&F 225.09 and 361.61:
$$J_2=\frac{10^{-3/4}}2((F(\varphi,m)-2E(\varphi,m)) + \frac{u\sqrt{10u^4-2u^2+1}}{10u^2+\sqrt{10}}$$
$$\varphi=2\tan^{-1}10^{1/4}u,m=\frac12+\frac1{2\sqrt{10}}$$
The sum of the algebraic part of $J_2$ and the other term in the bracket tends to $-\frac1{\sqrt{10}}$ as $u\to\infty$; $\varphi$ tends to $\pi$ in the same limit, so the integrals become complete. Thus the bracketed expression is
$$(\cdot)=-10^{-3/4}(K(m)-2E(m))-\frac1{\sqrt{10}}$$
I can work out $J_0$ similarly using B&F 266.00; it's $10^{-1/4}K(m)$.

To work out $I_{-1,-1}$ use B&F 266.09. Letting $a=10^{-1/4}$:
$$I_{-1,-1}=\frac a2\int_0^{2K(m)}\left(\frac{a\sn u}{1-\cn u}+1\right)^{-1}\,du=\frac a2\int_0^{2K(m)}\frac{-a\cn u}{2a+(a^2+1)\sn u}\,du+\frac a2\int_0^{2K(m)}\frac{a+\sn u}{2a+(a^2+1)\sn u}\,du$$
Of these last two integrals, the $\cn$-containing one evaluates to zero. This is because $\sn$ is even and $\cn$ is odd about $K(m)$, which the integration interval is symmetric over. The same properties allow us to halve the interval of the other integral. Hence
$$I_{-1,-1}=\frac a2\int_0^{K(m)}\frac{1+(1/a)\sn u}{1+(a^2+1)/(2a)\sn u}\,du$$
Now I would apply B&F 361.69 here, but it seems incorrect and there is no errata for it. Using Mathematica (in Try It Online) I get
$$I_{-1,-1}=\frac a2\left(cK(m)+(1-c)(Z + \Pi(n,m))\right)$$
$$c=\frac{2(10-\sqrt{10})}9,Z=\frac{\sqrt{440+161\sqrt{10}}}{27}\log(-3-\sqrt{10}),n=\frac12+\frac{11\sqrt{10}}{40}$$
Finally we have
$$\boxed{A=2\pi(C_1K(m)+C_2E(m)+C_3(Z+\Pi(n,m))+C_4)}$$
$$C_1=10^{-1/4}\cdot\frac{580-247\sqrt{10}}{1215}\quad C_2=10^{-1/4}\cdot\frac{449\sqrt{10}}{1215}\quad C_3=10^{-1/4}\cdot\frac{2\sqrt{10}-11}{108}$$
$$C_4=-\frac{373}{1215}-\frac{41\sqrt{10}}{972}\quad m=\frac12+\frac1{2\sqrt{10}}\quad n=\frac12+\frac{11}{4\sqrt{10}}$$
$$Z=\frac{\sqrt{440+161\sqrt{10}}}{27}\log(-3-\sqrt{10})$$


Title: Prove that $ 2^r +2$ is sum of 2 square number, when r is a prime number not equal to 2

Body:
I was doing a question. Suddenly I got stuck at this last part of the problem. It was to prove
$ 2^r +2 = a^2 +b^2$ where $r \neq 2$, r is a prime and $ a \neq b$. Also $r^2 -1$ is a mersenne prime. I tried to use fermat's little theorum, but to no avail. Thank you.

PS note: The problem I was solving was BMO2021 Q6.

Body:
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:

In[1]:=Clear["Global`*"];
n = 2;
ParallelTable[
  If[TrueQ[2^r + 2 == a^2 + b^2 && a != b], {r, a, b}, Nothing], {r,
   2, 10^n}, {a, 0, 10^n}, {b, 0, 10^n}] //. {} -> Nothing

Running the code gives:

Out[1]={{{{3, 1, 3}}, {{3, 3, 1}}}, {{{5, 3, 5}}, {{5, 5, 3}}}, {{{7, 3,
    11}}, {{7, 7, 9}}, {{7, 9, 7}}, {{7, 11, 3}}}, {{{9, 15,
    17}}, {{9, 17, 15}}}, {{{11, 5, 45}}, {{11, 23, 39}}, {{11, 31,
    33}}, {{11, 33, 31}}, {{11, 39, 23}}, {{11, 45, 5}}}, {{{13, 25,
    87}}, {{13, 63, 65}}, {{13, 65, 63}}, {{13, 87, 25}}}}

So, we can see that you're not right! Because, for example, when $\text{r}=9\notin\mathbb{P}$ we get $2^9+2=15^2+17^2$.


答:
By Euler's reflection formula, it follows that

$$
\int_0^\infty{x^{s-1}\over1+x}\mathrm dx={\pi s\over\sin(\pi s)}\tag1
$$

Accordingly, we can find an $s$ such that $\sin(\pi s)$ can be associated with $\phi$. As it turns out, we do have some special angle that allows us to do so.

By observing the geometric properties of this triangle, we can deduce the following relationship

$$
\triangle ABC\sim\triangle BDA\cong\triangle DBA
$$

which implies

$$
{BC\over AB}={AB\over BD}
$$

Now, due to the properties of isosceles triangles, we get

$$
AB=AD=CD\Rightarrow BC=BD+CD=BD+AB
$$

Thus, we obtain

$$
1+{BD\over AB}={AB\over BD}
$$

To convenience the derivation, set $AB=1,BD=y$ so that the above identity becomes

$$
1+y=\frac1y\Rightarrow y^2+y-1=0\Rightarrow y={-1+\sqrt5\over2}=\frac1\phi
$$

Again, by the properties of isosceles, we deduce

$$
CE={1+y\over2}
$$

As a result, we obtain $\cos36^\circ$ from its definition:

$$
\cos36^\circ={CE\over CD}={CE\over AB}={1+y\over2}={1+\sqrt5\over4}=\frac\phi2
$$

Now, due to the conversion that

$$
90^\circ-36^\circ=54^\circ={3\pi\over10}
$$

we obtain

$$
\sin\left(3\pi\over10\right)=\frac\phi2
$$

Therefore, setting $s=3/10$ in (1), we obtain

$$
\fbox{$\Large\int_0^\infty{x^{-7/10}\over1+x}\mathrm dx={2\pi\over\phi}$}
$$

Title: Is there any integral for the Golden Ratio?

Body:
I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:

$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$

$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$

$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$

Is there an interesting integral* (or some series) whose result is simply $\phi$?

* Interesting integral means that things like

$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$

are not a good answer to my question.


问:
proof regarding isomorphic graphs
I am working on a graph theory proof but facing problems regarding how to approach to prove it

Consider the following statement: Given a set of n graphs, {G1, G2, · · · , Gn}, some of the pair of graphs are isomorphic, yet some pair are not. There will be even number of these graphs which are isomorphic to an odd number of graphs. Prove or disprove the statement.

As for what I can figure out, if the even number of graphs are isomorphic to the odd ones, then vice versa will also be true, disproving the statement. But we are not sure if that would be the only thing.

Pardon if any mistakes made. Thanks in advance.

答:
Consider the graph $G$ whose vertices are the graph $G_1, G_2, ..., G_n$ with two vertices adjacent if and only if they are isomorphic.  The degree of a vertex is the number of graphs (other than itself) to which the vertex is isomorphic.  By the handshaking lemma, there are an even number of vertices of odd degree.

The statement as given is false, since it overlooks the fact that each graph is isomorphic to itself.

Title: proof regarding isomorphic graphs


问:
Is this integral incomplete gamma function?
I'm trying to solve the following integral:
$$\int_{0}^{x} t^{s-1}e^t dt$$
where, $0 \leq x \leq 1, 0 \leq t \leq 1, s >0$. Although I know the following integral is an incomplete gamma function
$$\gamma(s, x) = \int_{0}^{x} t^{s-1}e^{-t} dt$$
Can someone please explain me if we can solve the other integral?

答:
What do you mean by "solve"?

The first integral diverges as
$x \to \infty$ and the second converges.

Both can be computed by putting in the power series for
$e^t$ and integrating term by term.
The first integral behaves better computationally
since all the terms are positive
so there is no cancellation.

You can get  increasingly accurate estimates
of the first integral
by repeated integration by parts,
just as for the second.


答: Almost all continuous functions are nowhere differentiable for the classical Wiener measure (see Wikipedia for example), but it's not easy to construct a counterexample. The set of nowhere differentiable functions is also comeager (for a more topology-oriented point of view).</p>

Title: Examples of properties that hold almost everywhere, but such that explicit examples are not known

Body:
In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties.
More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have
  $$
\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}
$$
  almost surely.
  Note: The result fails for all rationals.


问:
Prove $\sum\limits_{k=1}^n \binom{n}{k}k^2=2^{n-2}n(n+1)$ by combinatorial argument.
Prove $\sum\limits_{k=1}^n {n \choose k}k^2=2^{n-2}n(n+1)$ by combinatorial argument.

答: In how many ways can you form a subcommittee among $n$ people and appoint two roles within it, possibly to the same person? What if you choose who gets the roles first, and then finish the subcommittee? For the latter, simplify $n2^{n-1}+n(n-1)2^{n-2}$ to separately count the cases where the roles do or don't have the same recipient.


问:
Non-linear differential equaton for complex valued function
I try to solve the following equation,
$$\dot{q}=i\beta q(t)+\frac{\lambda}{2}q(t)\left[1-|q(t)|^2\right],$$
where $\beta$ & $\lambda$ are real constants. My best idea was to represent the function $q(t)$ as $q(t)=u(t)+iv(t)$, which brings me to the system,
$$\begin{cases} \dot{u}(t)=-\beta v(t)+\lambda u(t)\left[1-u(t)^2-v(t)^2\right]/2, \\ \dot{v}(t)=\beta u(t)+v(t)[1-u(t)^2-v(t)^2]/2\end{cases}.$$
In my opinion this transformation does not give any benifit. Then, I try to just use Wolfram Mathematica to see a solution. Wolfram  gives the following result
$$C-\frac{t}{2}=\int_{1}^{w(t)}d\xi\frac{1}{\xi\left(\lambda|\xi|^2/2-\lambda-2i\omega\right)}.$$
To be honest, I have no idea how I can derive the same result. Can anyone gives some explanation or tips what should I do to solve this equation?

答:
The Wolfram solution is just the execution of the separation of variables.

One can also get a reduction in complexity from the translation into polar coordinates $q=re^{i\phi}$,
$$
\dot r+ir\dot\phi=iβr+\fracλ2r[1−r^2].
$$
This can again be separated into real and imaginary part to give the equations
\begin{align}
\dot r&=\fracλ2r[1−r^2]\\
\dot\phi&=β
\end{align}
The first is a Bernoulli equation, the second can be easily integrated.


问:
Computing a line integral along a parabola

Compute the line integral $\int_{\gamma}F\cdot ds$, when $F(x,y) = (x,y^2)$ and $\gamma \subset \mathbf{R}^2$ is the curve of the parabola $x=3y^2$ starting at $(0,0)$ and ending at $(3,1)$.

I tried to parameterize $\gamma$ as follows $\gamma(t)=(3t^2,t)$ and then computing

$$\int_{0}^{3}F(\gamma(t)) \|\gamma'(t)\| \ dt = \int_{0}^{3}F(3t^2,t) \|(6t,1)\| \ dt = \int_{0}^{3}(3t^2,t^2) \cdot(36t^2,1) \ dt$$

but I'm quite certain this is not the correct integral. What might I be missing here?

答:
The vector field $F(x,y) = (x,y^2)$ is conservative as it is gradient of,

$f(x,y) = \displaystyle (\frac{x^2}{2} + \frac{y^3}{3}) $

So your line integral is path independent and can simply be between the end points which are $(0, 0)$ and $(3, 1)$.

$\displaystyle \int_C \vec{F} \cdot dS = \int_C \nabla f \cdot dS = f(3,1) - f(0, 0)$

On your parametrization, it should be

$\displaystyle \int_{0}^{1} (3t^2,t^2) \cdot (6t,1) \ dt = \int_{0}^{1} (18t^3 + t^2) \ dt = \frac{29}{6}$


问:
Elementary methods to evaluate $\int_{0}^{2\pi}\exp(\sin \theta)\cos(\cos \theta)\mathrm d\theta$
I have seen similar integrals being attacked by complex analysis. I am wondering whether there are elementary methods to solve such integrals.

$$\int_{0}^{2\pi}\exp(\sin\theta)\cos(\cos\theta)\mathrm d\theta$$

Any hints are appreciated. Thanks.

答:
This is inspired by a deleted solution of a similar question (Calculating the following integral using complex analysis: $\int_{0}^{\pi}e^{a\cos(\theta)}\cos(a\sin(\theta))\, d\theta$), containing no details, though:

Let $$I(a)=\int^{2\pi}_0 e^{a \sin \theta} \cos(a \cos \theta)\,d\theta.$$ Then,
$$\frac{d}{d a}I(a) = \int^{2\pi}_0 e^{a \sin \theta} (\sin \theta\cos(a \cos \theta) - \cos \theta\sin(a \cos \theta))\,d\theta.\tag{1}$$ Since
$$\frac{d}{d\theta}(e^{a \sin \theta} \sin(a \cos \theta)) = -a e^{a \sin \theta} (\sin \theta\cos(a \cos \theta) - \cos \theta\sin(a \cos \theta)),$$
the RHS of (1) is $0$ for $a>0$, i.e. $I(a)$ is a constant, implying
$$I(a)=\lim_{a\to0^+}I(a)=2\pi.\tag{2}$$
Differentiation under the integral sign and the limit in (2) can be justified using the standard ("elementary") methods of real anaysis, like dominated convergence theorem etc.

I guess the solution was deleted because the question explicitly asked for an answer using complex analysis. Let's face it: to distinguish between "elementary" and complex analysis was dubious (and mere tradition) already 100 years ago, it's absolutely ludicrous in the 21st century.


问:
Fractions in binary?
How would you write a fraction in binary numbers; for example, $1/4$, which is $.25$? I know how to write binary of whole numbers such as 6, being $110$, but how would one write fractions?</p>

Title: Fractions in binary?

Body:
Not sure how helpful this is but whenever I work with binary I always use a table.

So if you want 0.25 or 15.75 (base 10) in binary:

\begin{array}{r|rrrr|rrrr}
& 8 & 4 & 2 & 1 & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} \\
& 8 & 4 & 2 & 1 & 0.5 & 0.25 & 0.125 & 0.0625 \\
\hline
0.25 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
15.75 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0
\end{array}

Gives 0.01 and 1111.11 respectively.

And another way:

15.375 to binary for example:

\begin{align}
15 - 8 & = 7, & \quad 7 \ge 0 & \to 1 \\
7 - 4 & = 3, & 3 \ge 0 & \to 1 \\
3 - 2 & = 1, & 1 \ge 0 & \to 1 \\
1 - 1 & = 0, & 0 \ge 0 & \to 1 \\
\\
0.375 - 0.5 & = -0.125, & -0.125 \lt 0 & \to 0 \\
0.375 - 0.25 & = 0.125, & 0.125 \ge 0 & \to 1 \\
0.125 - 0.125 & = 0, & 0 \ge 0 & \to 1
\end{align}

Gives 1111.011.


答:
$\sum_{i=1}^n \binom{n}{k}k^2$ is the way to choose $k$-subset $T$ of $n$-set $S$, then choose $(a,b)$ with $a,b\in T$.

It is equivalent to

choose $a=b\in S_n$ and then choose other elements in $T$, which is $n\cdot 2^{n-1}$.

choose the ordered pair $a\neq b\in S_n$ and then choose other elements in $T$, which is $n\cdot (n-1)\cdot 2^{n-2}$.

Then we see $n\cdot 2^{n-1}+n\cdot (n-1)\cdot 2^{n-2}=2^{n-2}\cdot n(n+1)$.

Title: Prove $\sum\limits_{k=1}^n \binom{n}{k}k^2=2^{n-2}n(n+1)$ by combinatorial argument.

Body: Prove $\sum\limits_{k=1}^n {n \choose k}k^2=2^{n-2}n(n+1)$ by combinatorial argument.


问:
How to prove this using Ito's lemma?
I am stuck at the following formula which I need to prove using Ito's calculus.
$$\frac{1}{T}\int_0^T W_t^2 dt=\frac{T}{2}+\int_0^T 2W_t \left(1-\frac{t}{T}\right) dW_t$$

How to prove this?

Title: How to prove this using Ito's lemma?

Body:
Using the hint provided by @Surb set $f(x,t)=x^2(1-\frac{t}{T})$. By Ito's lemma:

$$df(x,t)=2W_t(1-\frac{t}{T})dW_t+(1-\frac{t}{T})dt-W_t^2/Tdt$$

or

$$W_T^2\times 0-W_0=\int_0^{T}2W_t(1-\frac{t}{T})dW_t+\int_0^{T}(1-\frac{t}{T})dt-\int_0^{T}W_t^2/Tdt$$

or

$$0=\int_0^{T}2W_t(1-\frac{t}{T})dW_t+(T-0.5T)-\int_0^{T}W_t^2/Tdt$$

Hence

$$\frac{1}{T}\int_0^{T}W_t^2dt=\frac{T}{2}+\int_0^{T}2W_t(1-\frac{t}{T})dW_t$$


问:
Are midpoints always the best estimate for the area under the curve using a finite number of rectangles?
Are there any scenarios where midpoints would produce a result which is less accurate than left-endpoints or right endpoints?

If so, how do I recognize when each rule would apply the most accurate result?

答:
Obviously in the limit there is no such scenario for integrable functions, since the requirement for integrability is that all choices for test points converge to the same limit.

But it is easy to construct specific cases where the midpoint is inferior. For example, estimating the integral on $[0,1]$ using equal intervals of width $\frac 1{2^n}$ for some fixed $n$, where the integrand is
$$f(t) = \begin{cases}1,& t = \dfrac{2k+1}{2^{n+1}}, k \in \Bbb N, k < 2^k\\0,&\text{otherwise}\end{cases}$$

So while you cannot force midpoints to always be worse than the endpoints, it is certainly the case you cannot expect any particular midpoint calculation to not be worse than the corresponding endpoint calculation. Or vice versa.


问:
Calculate $\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$
Please help me to calculate the $$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}$$.

My attemp is:
$\lim_{x\to1}\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}\cdot\frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{\sqrt[3]{1+7x}-\sqrt{1+3x}}\cdot\frac{\sqrt[3]{1+7x}+\sqrt{1+3x}}{\sqrt[3]{1+7x}+\sqrt{1+3x}}$, but I do not know if I did well, can anyone help me

答:
A little overkill but. You can use solving limits by substituions in here.
$$\begin{align}L &=\lim_{x\rightarrow 1} \frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}\\ &= \lim_{x\rightarrow 1} \frac{\sqrt[3]{1+7x}-\sqrt{1+3x}}{x-1}\cdot\frac{\sqrt[3]{1+7x}+\sqrt{1+3x}}{\sqrt[3]{1+7x}+\sqrt{1+3x}}\\&= \lim_{x\rightarrow 1}\frac{1}{4}\frac{\sqrt[3]{\left(1+7x\right)^{2}}-\left(1+3x\right)}{\left(x-1\right)}\end{align}$$

Take:
$$x=\frac{u^3-1}{7}$$
$$\text{As }x \rightarrow1 \text{, } u \rightarrow 2 $$
Then,
$$\begin{align}L&=-\frac{1}{4}\lim_{u \rightarrow 2} \frac{3u^{3}-7u^{2}+4}{u^{3}-8}\end{align}$$

We got a $P(x)/Q(x)$ form which you solve by removing the factors or by using L'Hopitals


问:
Bernoulli-measure of infinite word.
The Bernoulli measure $\mu$ is defined on finite blocks, that is for a block

$[y_1,y_2,\ldots,y_n]m:=\{x=(x_i)_{i\in\mathbb{Z}}: x_m=y_1, x_{m+1}=y_2,\ldots,x_{m+n}=y_n\}$

we have
$$
\mu([y_1,y_2,\ldots,y_n]m) = \prod_{j=1}^np_j,
$$
where $p_j$ is the probability of $y_j$.

I am a bit confused what the probability of an infinite word is, say, of
$$
x=(\ldots,y_1,y_1,y_1,\ldots)
$$
which has bi-infinitely many entries $y_1$.

Does this have measure zero?

Surely, one has
$$
\mu(x)=\mu\left(\bigcap_{i=-\infty}^{\infty}[y_1]_i\right)
$$

but what is the right hand side?

答: I'm assuming that there is no index $j$ with $p_j = 1$. Then, because of $\bigcap_{j \in \mathbb Z} [y_1]_j \subset \bigcap_{j = -m}^m [y_1]_j$ for any $m \in \mathbb N$, we have $$\mu \left( \bigcap_{j \in \mathbb Z} [y_1]_j \right) \leq \mu \left( \bigcap_{j = -m}^m [y_1]_j \right) = p_1^{2m+1} \xrightarrow{m\to \infty} 0 .$$

Title: Bernoulli-measure of infinite word.


问:
Which topics of mathematics should I study?
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.

答:
Here is a suggestion for an explicit first step. In my (not the most qualified) opinion, real math is grounded in rigor. Typically the first step is real analysis. You can get a sense what constitutes a proof, etc.

These notes, almost verbatim, of lectures by Fields Medal (~ Noble Prize in math) Vaughan Jones are an outstanding, very readable, self-contained entry point. They are his own treatment and are a master presenting the material:

https://sites.google.com/site/math104sp2011/lecture-notes

They start from the beginning so you can go right in.


问:
Prove that $\int\limits_0^\pi {\frac{{\sin \left( {xt} \right)}} {t} \mathrm dt} $ is continuous
How can I prove that this function is continuous? $$
f\left( x \right) = \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
$$
Some hint?
Don´t consider the zero in the endpoint of the integration zone, just take it as a limit $$
f\left( x \right) = \mathop {\lim }\limits_{\varepsilon ^ +   \to 0} \int\limits_\varepsilon ^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
$$
How can I do it? DX!

答:
$f\left( x \right)
= \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{t} \mathrm dt}
= \int\limits_0^\pi  {\frac{{\sin \left( {xt} \right)}}
{xt} x\mathrm dt}
= \int\limits_0^{x\pi}  {\frac{{\sin \left( {t} \right)}}
{t} \mathrm dt}
$

and the integral of a continuous function is continuous.


问:
$E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$. Prove $\sigma = 1_E$

Let $E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$, then $\sigma = 1_E$.

I think that the way is to see that $K$ and the elements $a_i$ are in the field fixed by $\sigma$, $C_E(\sigma)=\left\lbrace e\in E \vert \sigma (e)=e \right\rbrace$.

By definition of Galois Group, since $\sigma \in Gal (E/K)$, then $\sigma(k)=k \,\, \forall k\in K$ hence $K\subset C_E(\sigma)$ and since $\sigma (a_i)=a_i$, all the elements $a_i$ are in $C_E(\sigma)$.

How can I use that for prove that $\sigma$ is the identity?

答: Every element in E may be written as $c_0+\sum_{i=1}^n c_ia_i$ with $c_i\in K$, so $\sigma(c_0+\sum_{i=1}^n c_ia_i)=\sigma(c_0)+\sum_{i=1}^n\sigma(c_i)\sigma(a_i)=c_0+\sum_{i=1}^n c_ia_i$.


问:
Stationary shocks and contact discontinuities in conservation laws
1)What is a stationary shock or contact discontinuity?
I mean If I have $u_r<u_l$, then the entropy condition is satisfied and there exists a shock curve.
And if $u_r>u_l$, then the entropy condition is not satisfied unless I separated them by a rarefaction wave  $x/t$.
But when does a stationary shock exist?

2)Why in the scalar case the conservation law problem has a unique entropy solution, if it exists?

答:
These are different concepts. Consider Burgers' equation $u_t + uu_x=0$, and a discontinuity with left/right states $u_l$, $u_r$ propagating at the speed (Rankine-Hugoniot) $$
s =\tfrac12(u_l+u_r) \, .
$$ $\blacksquare$ A stationary shock is a discontinuity located at constant position. Thus, the shock speed $s$ vanishes. For such a solution to be admissible, we need to check the Lax entropy condition $$
u_l > s > u_r
$$
which leads to the condition $u_l = -u_r > 0 $. These stationary shocks are a particular case of shock waves. $\blacksquare$ A contact discontinuity has its shock speed $s$ given by the Rankine-Hugoniot condition, and the characteristic speeds $u_l$, $u_r$ on both sides of the discontinuity are equal. Thus, the Lax entropy condition is not satisfied, so that contact discontinuities are not admissible in the case of Burgers' equation. However, such solutions can be obtained in the case of the linear advection equation $u_t+cu_x=0$, for which the characteristic speed $c$ is the same on both sides of the discontinuity.

That's a theorem, see related literature where the proof is given.


问:
term wise product of power series converges
Suppose we have non-negative valued sequences $(a_n)_{n=1}^\infty$, $(b_n)_{n=1}^\infty$.
We know that $\sum_{i=1}^\infty a_n =\infty$ diverges, but $\sum_{i=1}^\infty a_{n}b_{n}=L$ converges to some real number $L$.

In this case, what can we say about the sequence $b_n$?

What if we add in extra condition that $a_n\to 0$ monotonically?

答:
We can say nothing about a convergence of $b_n$.

Obviously, $b_n$ may be convergent, because the case $b_n = 0 $ is possible.

Fix $a_n$. Let us construct  divergent $b_n \in \{ 0, 1 \}$ such that $\sum_n a_n b_n$ converges.

As $a_n \to 0$ and $a_n \ge 0$, we may take subsequence $n_k$ such that $a_{n_k} < \frac{1}{k^2}$ and $n_{k+1} > n_{k} + 2$.  Put $b_m = 1$ if $m = n_k$ for some $k$ and $b_m = 0$ otherwise. Hence $\liminf_n b_n = 0$, $\limsup_n b_n =1$ and $b_n$ is divergent. Moreover, we have $\sum_{n} a_n b_n = \sum_{k} a_{n_k} \cdot 1 \le \sum_{k} \frac{1}{2^k} < \infty$. Thus $\sum_{n} a_n b_n $ converges, q.e.d.


问:
why is the Column space of a Matrix and its reduced row echelon form different?
Why is the Column space of a matrix and its reduced row echelon form different? If for a 3X3 matrix A the first two columns are independent and the third column is dependent, then the first two columns span the column space of A and since it was obtained by reversing the row operations that led to the reduced row echelon form(R) which were linear operations, the same columns span the column space of the R.</p>

答:
Elementary row operations are given by invertible linear transformations, and thus preserve dependency relations of the columns.

On the other hand, applying a linear transformation can change the column space itself.


问:
Examples of properties that hold almost everywhere, but such that explicit examples are not known
In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties.
More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have
  $$
\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}
$$
  almost surely.
  Note: The result fails for all rationals.

答:
Almost all real numbers are undefineable. But by definition we can't exhibit an example.

(Somewhat) more precisely: Fix a symbol set $\Sigma$. The set of finite length sequences of symbols from $\Sigma$ is countable, hence the set of finite length sequences which define a unique real number is countable. Hence there are only countably many such real numbers, so the set of defineable real numbers has measure $0$.


答:
Let $H$ be a cyclic subgroup of $\operatorname{Gal}(E/K)$ generated by $\sigma$, i.e. $H = \langle \sigma \rangle$. Let us consider the fixed field by $H$, i.e. $E^H$. As you have mentioned, every element of $K$ as well as all $a_i$ are fixed by $\sigma$, thus $E \subset E^H$.

On the other hand, clearly $E^H \subset E$. So we have equality $E^H = E$. Now apply the Galois correspondence, which says that there is a bijection between subgroups $H$ of $\operatorname{Gal}(E/F)$ to intermediate subfields of $E/F$ by considering the operatorname $H \mapsto E^H$.

Clearly the trivial subgroup $\{e\}$ satisfies $E^{\{e\}} = E$. So by the correspondence, $H = \{e\}$ is trivial, i.e. $\sigma = e$.

Title: $E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$. Prove $\sigma = 1_E$

Body:
Let $E=K(a_1,...,a_n)$ and $\sigma \in Gal(E/K)$ such that $\sigma (a_i)=a_i$ for all $i$, then $\sigma = 1_E$.

I think that the way is to see that $K$ and the elements $a_i$ are in the field fixed by $\sigma$, $C_E(\sigma)=\left\lbrace e\in E \vert \sigma (e)=e \right\rbrace$.

By definition of Galois Group, since $\sigma \in Gal (E/K)$, then $\sigma(k)=k \,\, \forall k\in K$ hence $K\subset C_E(\sigma)$ and since $\sigma (a_i)=a_i$, all the elements $a_i$ are in $C_E(\sigma)$.

How can I use that for prove that $\sigma$ is the identity?


问:
Proof verification: Suppose $f: D \rightarrow \mathbb{R}$ is monotone. Prove that if $f(D)$ is an interval, then $f$ is continuous
I am working through a theorem in Advanced Calculus, Fitzpatrick. Ch3, Theorem 3.23

Suppose $f: D \rightarrow \mathbb{R}$ is monotone. Prove that if $f(D)$ is an interval, then $f$ is continuous

I solved it a different way than the book, I am looking for confirmation that my method is valid. If not, where did I go wrong?

My attempt
I will just do the case when $f$ is monotone increasing.

Assume $f(D)$ is an interval. Our goal is to prove $$\forall x \in D \quad \{x_n\} \text { converges to } x \rightarrow \{f(x_n)\} \text{ converges to } f(x)$$

Let $x$ be arbitrary element of $D$ and assume there exists a sequence $\{x_n\} \subset D$ that converges to $x$. There exists a subsequence of $\{x_n\}$, call it $\{x_{n_k}\}$, that is monotone increasing. It can be shown that since $\{x_n\}$ converges to $x$, then $\{x_{n_k}\}$ converges to $sup \{x_{n_k}\} = x$.

Since $f$ is monotonic increasing, $\{f(x_{n_k})\}$ is monotonic increasing. Note that this implies $\forall k \in \mathbb{N}, f(x_{n_k}) \leq f(x)$, since $x$ is the supremum of the subsequence. Therefore $\{f(x_{n_k})\}$ is a bounded monotonic sequence, which converges to $sup  \{f(x_{n_k})\} = f(x)$.

Therefore we have shown $\{f(x_{n_k})\}$ converges to $f(x)$ for arbitrary $x \in D$.

答:
That proof cannot be correct because you did not use the fact that $f(D)$ is an interval (and the statement is surely wrong without that condition).

It is correct that $f(x_{n_k})$ increasing implies that $f(x_{n_k}) \to \sup  \{f(x_{n_k})\}$, but $f(x_{n_k}) \le f(x)$ for all $k$ implies only that the supremum is $\le f(x)$. You have not shown that the limit is equal to $f(x)$.

Another (minor) error is that an arbitrary sequence converging to $x$ does not necessarily have an increasing subsequence.

You can use your approach to show that $f$ has a left and a right limit at $x$. Then assume that these are different, i.e.
$$
 \lim_{y \to x-} f(x) < \lim_{y \to x+} f(x) \, .
$$
Now show that $f$ does not take values between those limits, contradicting the assumption that $f(D)$ is an interval.


问:
Books like Stephen Abbott's Understanding Analysis
I'm not sure if this question has been asked before. I'm interested in finding textbooks that are "like" Stephen Abbott's Understanding Analysis. The distinctive feature about this textbook is that it motivates the reasoning involved in the proofs prior to presenting the proofs, making the construction behind the proofs seem all the more reasonable.

I'm hoping if other people can list textbooks similar to Abbott's book in other fields of study, algebra, complex analysis etc. The books need not be at a level comparable to that of Abbott's books; the books can be of intermediate-advanced difficulty as well.

答:
This is a late answer ($3$ years late!), but I thought I'd put my two cents in just in case somebody ever stumbles across this question.

One book that matches your description perfectly (and, I'd argue, is even more successful at what it does than Abbott's) is Michael Sipser's Introduction to the Theory of Computation. Instead of assuming the definition-theorem-proof structure standard in mathematical literature, Sipser splits his proofs into two parts: a "Proof Idea", which explains the motivation behind each proof and fleshes out the details, followed by the condensed Proof to provide the rigorous formalism.

Exercises are similarly segmented: each chapter is followed by an "Exercises" section, providing the student with an opportunity to practice basic computations and apply the learned definitions to some elementary results, and a "Problems" section, which compiles more challenging results. As is standard, some important theorems that the main text couldn't accommodate are relegated to the Problems section. The book is perfectly linear and the level of understanding demanded throughout is very even.

Perhaps the only piece of critism I can think of for this book is that it isn't advanced enough. But (in case of persisting interest) this is easily rectified by following it up with a more advanced text.


问:
Proper notation when converting ratios to fractions
When converting ratios to fractions which is the correct way of doing it? I've seen conflicting information that says the the denominator should be the total while other sources say the denominator can just be the second term.

For instance, if we have 3 apples, and 4 oranges, the ratio of apples to oranges can be written as:

$3:4$

When writing as a fraction, we can either write:

$\frac{3}{4}$

or

$\frac{3}{3+4}$ = $\frac{3}{7}$

Further when we convert this to a percent, I have seen this converted to either

$75\%$

which would mean there are 75% as many apples as oranges.

But it might also be written as

$43\%$

where the number of apples is 43% of the total items.

Are both fine as long as you are clear what you are trying to represent?

答: Yes, as long as you know that $3/7$ is apples in a whole basket(lets say) and $3/4$ is relative, ie there are 3 apples for 4 oranges and that when you need to use which.


问:
Prove that $ 2^r +2$ is sum of 2 square number, when r is a prime number not equal to 2
I was doing a question. Suddenly I got stuck at this last part of the problem. It was to prove
$ 2^r +2 = a^2 +b^2$ where $r \neq 2$, r is a prime and $ a \neq b$. Also $r^2 -1$ is a mersenne prime. I tried to use fermat's little theorum, but to no avail. Thank you.

PS note: The problem I was solving was BMO2021 Q6.


问:
Equivalence of the Lebesgue integral and the Henstock–Kurzweil integral on nonnegative real functions
Let $f:[a,b]\to[0,\infty)$ ($\mathbb{R}\ni a0$, there exists a simple function $\omega=\sum_{j=1}^n z_j\mathbf{1}_{E_j}$ (where $n\in\mathbb{Z}_+$, the $\{z_j\}_{j=1}^n$ are distinct non-negative numbers, and the $\{E_j\}_{j=1}^n$ are Lebesgue measurable sets that form a partition of $[a,b]$) such that $0\leq\omega\leq f$ and

(1a) $f$ is Lebesgue measurable and $\sum_{j=1}^n z_j\mu(E_j)\in( c-\varepsilon,c]$, where $\mu$ is the Lebesgue measure, and the analogous sum of any non-negative simple function dominated by $f$ does not exceed $c$;

OR

(1b) $f$ is Lebesgue measurable and $\sum_{j=1}^n z_j\mu(E_j)> \varepsilon$.

(2) For any $\varepsilon>0$, there exists a “gauge” function $\delta_{\varepsilon}:[a,b]\to(0,\infty)$ such that for any $(x_j,t_j)_{j=1}^n\subset[a,b]$ ($n\in\mathbb{Z}_+$) satisfying

$a<x_1<\ldots<x_{n-1}<x_n=b$,

$t_j\in[x_j,x_{j-1}]$ for all $j\in\{1,\ldots,n\}$ (where $x_0\equiv a$), and

$x_j-x_{j-1}<\delta_{\varepsilon}(t_j)$ for all $j\in\{1,\ldots,n\}$,

then

(2a) $\left|\sum_{j=1}^n f(t_j)(x_j-x_{j-1})-c\right|<\varepsilon$ for all $j\in\{1,\ldots,n\}$;

OR

(2b) $\sum_{j=1}^n f(t_j)(x_j-x_{j-1})>\varepsilon$ for all $j\in\{1,\ldots,n\}$.

I want to show that $(1a)\Longleftrightarrow(2a)$ and $(1b)\Longleftrightarrow(2b)$.

The tricky parts are as follows:

For a given Lebesgue measurable function with integral $c$ (or $\infty$), how can one construct the desired gauge function so that the Henstock–Kurzweil integral is $c$ (or does not exist, respectively)?

For a given Henstock–Kurzweil integrable function whose integral is finite or does not exist because it would be unbounded, how can one prove that it is Lebesgue measurable?

For a given Henstock–Kurzweil integrable function, how can one construct the desired step function?

I skimmed through some of the relevant literature, but I failed to find any satisfying and not-too-abstruse explanation. (Disclaimer: I'm a newbie in HK-integration with some background in measure theory.) Thank you very much for your help in advance.

Note: (1a) basically states that $\int_{[a,b]}f(x)\,\mathrm{d}\mu(x)=c$ and (1b) that $\int_{[a,b]}f(x)\,\mathrm{d}\mu(x)=\infty.$

答:
in abstract set up even an abolutely henstock kurzweil integrable function may not be measurable it all depends on whether1 is a measurable function.
on euclidean spaces every henstock kurzweil integrable function is measurable. you can refer bartle. forvreal line.the proof depends on the second fundamnental theoe oof calculus which says that the primitive is differentiable almost everywhere.
the theorem depends on vitalis lemma or austins lemma and bartles approach can be generalized to higher dimensions but  we do not get primitive is differentiable but only symmetric derivative exists almost everywhere. please refer fremlins notes.
however measurability can be deduced.
a direct proof that positive lebesgue integrable function si henstock kurzweil integrable is there in lee peng yee rydolf vyborny integral aneasy approach .
it relies on mct and dct.
howebber hk integral is far more general . it does not require integrator to be even of bounded variation for stiltjes integral . measure theory is incapable of encompassing the generality
An absolutely  hk integrable Banach valued mapping on an interval [a, b] may not be measurable. though the evaluation theorem that is first fundamental theorem of calculus holds.
Henstock has built of a theory of variation an anlaogue of outer measurre as an alternative to measure
but the drux of the thory is not relying on notion of measurability   and save fromm establishing measurability in applications. As Borel remarkedand criticized Lebesgue that he has bothere in intricate abstraction and dig the mountain to find a mole.
today the basic role pof lebesgue thory is limited to proability theory .
in practice we need  close sets open sets which are all measurable</p>


问:
Is this integral incomplete gamma function?
I'm trying to solve the following integral:
$$\int_{0}^{x} t^{s-1}e^t dt$$
where, $0 \leq x \leq 1, 0 \leq t \leq 1, s >0$. Although I know the following integral is an incomplete gamma function
$$\gamma(s, x) = \int_{0}^{x} t^{s-1}e^{-t} dt$$
Can someone please explain me if we can solve the other integral?

Body:
Make $t=-u$
$$I=\int_{0}^{x} t^{s-1}e^t dt=-\int_{0}^{-x}e^{-u} (-u)^{s-1}\,du=(-1)^{-s} (\Gamma (s)-\Gamma (s,-x))$$


问:
Show that this is true for a Riemann Integrable function
$f \ RI \ on \ [a,b]$

$inf_{[a,b]}(f) = m_1 \leq f \leq m_2 = Sup_{[a,b]}(f)$

Then,

$$m_1(b - a) \leq \int_{a}^{b}f(x)dx \leq m_2(b - a)$$

I think I just need some clarification on this. Why is this not an obvious result? I feel like this is trivial and there is nothing to show. We know that the definition of the upper and lower sums are the infimum and supremum times the size of the interval.

By defininig the inf and sup as $m_1$ and $m_2$, I feel like $m_1(b - a)$ and $m_2(b - a)$ are literally $L(f,P)$ and $U(f,P)$.

Then, since $U(f) = Inf_{P} (U(f,P))$ and $I(f) = Sup_{P} (I(f,P))$,

it is a known result that $U(f) \leq U(f,P)$ and $I(f) \geq I(f,P)$

Since we also know that for the integral to exist, $\int_{a}^{b} f(x) dx = U(f) = I(f)$. So,

$$m_1(b - a) = I(f,P) \leq I(f) = \int_{a}^{b}f(x)dx = U(f) \leq U(f,P)) = m_2(b - a)$$

Is this actually the entire proof or am I missing something? It's basically a one-step problem right now, so I'm not entirely confident I'm doing it right. Can anyone confirm\deny this for me? Thanks!

答: You wrote that “$m_1(b - a)$ and $m_2(b - a)$ are literally $L(f,P)$ and $U(f,P)$”, but this doesn't make sense unless you tell is which partition $P$ is. It turns out that what you wrote is actually true if $P=\{a,b\}$. And it follows from this that, indeed\begin{align}m_1(b-a)&=L(f,\{a,b\})\\&\leqslant\int_a^bf(x)\,\mathrm dx\\&\leqslant U(f,\{a,b\})\\&=m_2(b-a).\end{align}


问:
Proof for connected simple undirected graph with more than one vertex
Given a connected simple undirected graph G with more than one vertex, I am asked to show that there is at least one vertex $v$ that fulfills
$$\frac{1}{deg(v)} \sum_{w \in N(v)} deg(w) \geq \frac{2 |E|}{|V|},$$
where $N(v)$ is the neighborhood of vertex v, $|E|$ is the number of edges and $|V|$ is the number of vertices.

I have tried this by induction over $|E|$ or $|V|$ with the base case of $|E|=1$ being trivial, but having trouble to show that it holds when adding an edge or a vertex. Also I suspect it might help to use $\sum_{v \in V} deg(w) = 2 |E|$, but I don't understand how to quantify the $\frac{1}{deg(v)} \sum_{w \in N(v)} deg(w)$ part. Another idea would be to show that the average of the L.H.S. over all vertices is $\geq \frac{2 |E|}{|V|}$, from which the result would follow, but I am unable to show this as well.

Any help or hints would be appreciated, I am new to graph theory.

答:
Suppose, for a contradiction, that for all vertices that inequality is true. Summing over all vertices $v \in V$. We get

$$\sum _{v \in V} \sum_{w \in N(v)} \deg(w) < 2 \frac{|E|}{|V|} \sum_{v \in V} \deg(v) = \frac{1}{|V|} \left(\sum_{v \in V} \deg(v) \right)^2$$

where the last equality is by the handshake lemma.

Note that on the LHS each $\deg (w)$ will be counted once for each of its neighbours, so it will be counted $\deg(w)$ times. Hence, we have the inequality

$$\sum_{v \in V} \deg(v) ^2 < \frac{1}{|V|} \left( \sum_{v \in V} \deg(v) \right)^2$$

But this is false by the Cauchy Schwartz inequality applied to the vectors $(1,...,1)$ and $(\deg v_1,..., \deg_{|V|})$.


问:
The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$.

The lengths of the sides of a triangle are given as $|a|^2 + |b|^2 = 5|c|^2$ , where $|k|$ denotes the absolute value of $k$. Prove that the medians drawn from $A$ and $B$ are perpendicular.

What I Tried: Here is a picture :-

I would normally solve these kind of problems using elementary geometry, but I wasn't able to do it this time, simply because $(a,b,c)$ can take many values and there is probably some kind of construction or some clever proof which I am missing. One idea I thought of is that the medians divide the side lengths into $2$ equal parts, and the centroid itself divides the medians in $2:1$ ratio, and I probably have to show somehow, that Pythagoras Theorem on either of the $3$ triangles work. I took an example in Geogebra by drawing an accurate diagram with $a = 12$ , $b = 18.66$ nearly, and then $c = 10$ . This actually showed that the medians from $A$ and $B$ are really perpendicular, and I just drew a picture of that taken from Geogebra.

Now given $(a,b,c)$ , is there any way to prove this? I did not find any elementary methods like angle-chasing , pythagoras theorem , similarity and areas , etc. , but I did not find any clever use of them till now. Maybe some extra construction is required?

Can anyone help me? Thank You.

答:
Given four points $W,X,Y,Z$, let $P_{WXYZ} = WX^2 - XY^2 + YZ^2 - ZW^2$. Then we can prove that:

Claim 1. $P_{WXYZ}$ is affine-linear in each of the four points. (That is, as one of the points, say, $X$, moves along a straight line at a constant rate, the value of $P_{WXYZ}$ changes linearly.)

Claim 2. When $W,X,Y,Z$ are all distinct, $P_{WXYZ}=0$ if and only if $WY \perp XZ$.

So let's compute $P_{ADEB}$. By Claim 1 applied to point $D$ (which is halfway between $A$ and $C$), $P_{ADEB} = \frac12 P_{AAEB} + \frac12 P_{ACEB}$. By Claim 1 applied to point $E$ (which is halfway between $B$ and $C$), we get
$$
    P_{ADEB} = \frac14 P_{AABB} + \frac14 P_{AACB} + \frac14 P_{ACBB} + \frac14 P_{ACCB}.
$$
Since we only want to know if $P_{ADEB}=0$, we can forget about the $\frac14$'s and only expand out these four expressions in terms of $AB^2, BC^2, CA^2$.

We get
$$
   (AA^2-AB^2+BB^2-BA^2) + (AA^2-AC^2+CB^2-BA^2) + (AC^2-CB^2+BB^2-BA^2) + (AC^2-CC^2+CB^2-BA^2)
$$
which simplifies to $AC^2 + BC^2 - 5AB^2$. We are given that this is $0$, therefore $P_{ADEB}=0$ and $AE \perp BD$ by Claim 2.