问:
Show that $e^{ta}=\frac1{2\pi i}\int_{\gamma_\alpha}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda$
This is the exercise 18 on page 359 of Analysis II of Amann and Escher. I'm stuck in this exercise

Suppose $a\in\Bbb C$ and $\alpha\neq\Re(a)$. Show that for $\gamma_\alpha:\Bbb R\to\Bbb C,\,s\mapsto \alpha+is$ we have $$e^{ta}=\frac1{2\pi i}\int_{\gamma_\alpha}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda\quad\text{for }t>0\tag1$$ (HINT: the Cauchy integral formula gives $$e^{ta}=\frac1{2\pi i}\int_{\partial\Bbb D(a,r)}\frac{e^{\lambda t}}{\lambda-a}\,d\lambda\quad\text{for }t\in\Bbb R\text{ and }r>0\tag2$$ Now apply the Cauchy integral theorem.)

Trying to follow the hint I tried to create a family of closed paths $\Gamma_r=[\gamma_r]+[\delta_r]$, such that $a$ doesn't belong to it bounded regions, and then use the Cauchy integral theorem, that is

$$\int_{\Gamma_r}g(\lambda)\, d\lambda=0\quad r>0$$

for $g(\lambda):=\frac{e^{\lambda t}}{\lambda-a}$, and exploit some kind of symmetry to relate the integration on the paths $[\gamma_r]$ or $[\delta_r]$ to the Cauchy integral formula. By example, without lose of generality suppose that $\alpha>\Re(a)$, then I could define

$$\gamma_r:[-r,r]\to\Bbb C,\quad s\mapsto \alpha+is\\\delta_r:[r,r+\pi]\to\Bbb C,\quad s\mapsto re^{-i(s-r-\pi/2)}\tag3$$

and try to relate the integration on the half circle defined by $\delta_r$ with the integral in the complete circle, where for suitable enough big $r$ I can use the Cauchy integral formula. However this is not easy to deal with, because it don't show a symmetry to exploit.

Maybe I'm over-complicating and the exercise want to do something different. Can someone help me?

EDIT:

It can be shown that $(1)$, as an improper integral of Riemann, converges conditionally, and after some changes of variables the question reduces to show that

$$\frac1{2\pi i}\int_{\gamma_r}\frac{e^\zeta}{\zeta}\,d\zeta=1$$

where $\gamma_r:\Bbb R\to\Bbb C,\, t\mapsto r+it$
for any chosen $r\in\Bbb R\setminus\{0\}$. Graphing the integrand we can see that it describes two non-rectifiable spirals (symmetric respect to the real axis) that converges to some point on the real axis.

答:
I'm also confused by this problem.

But I suspect that the problem is wrong. Please see the following.
$$
\begin{aligned}
e^{ta}&=\frac{1}{2\pi i}\int_{\gamma_\alpha} e^{\lambda t}(\lambda -a)^{-1}  d \lambda\\
\iff 1&=\frac{1}{2\pi i}\int_{\gamma_\alpha} \frac{e^{(\lambda-a) t}}{(\lambda -a)} d\lambda \qquad &&\eta=t(\lambda-a)\\
\iff 1&=\frac{1}{2\pi i}\int_{\gamma_\beta} \frac{e^{\eta}}{\eta} d\eta \qquad &&\gamma_\beta=\phi(\gamma_\alpha), \qquad \phi:x\mapsto t(x-a).
\end{aligned}$$
Here note that the line $\gamma_\alpha=\alpha+ i \mathbb R$ and $\gamma_\beta=t(\alpha-a)+ i \mathbb R$.

If $\gamma_\beta$ is right to $0$, then we only need to consider the left half circle $$\gamma_1: [0,\pi]\to \mathbb C,\quad t\mapsto \beta +i Re^{it}.$$
Here $\beta$ is the center of the circle on the line $\gamma_\beta$.
$$\begin{aligned}
&\int_{\gamma_1} \frac{e^{\eta}}{\eta} d\eta\\
=&\int_0^\pi\frac{e^{\beta+i Re^{it}}\cdot (-R e^{it})}{\beta+i Re^{it}}dt \to 0,\quad R\to \infty.
\end{aligned}$$
Just as @Masacroso answered.
This part is checked by computer.

But the rest is also checked by computer to be wrong.

If $\gamma$ is left to $0$, that is $Re( \alpha -a)<0$, then we need to consider the right half circle
$$\gamma_2: [0,\pi]\to \mathbb C,\quad t\mapsto \beta -i Re^{it}.$$
Here
$$\begin{aligned}
&\int_{\gamma_1} \frac{e^{\eta}}{\eta} d\eta\\
=&\int_0^\pi\frac{e^{\beta-i Re^{it}}\cdot (R e^{it})}{\beta-i Re^{it}}dt\to 2\pi i,\quad R\to \infty
\end{aligned}$$
It is also checked by computer directly that $e^{ta}\neq \frac{1}{2\pi i}\int_{\gamma_\alpha} e^{\lambda t}(\lambda -a)^{-1}  d \lambda$ when $Re(\alpha -a)<0$.


问:
Proving that $f'(z)\neq0$ when $|z|=1$ for a continuous, analytic $f$

$f(z)$ is analytic when $|z|<1$, and it's continuous when $|z|\leq 1$.

Prove that $f'(z)$ can't be $0$ when $|z|=1$. f(z) is not a constant, and |f(z0)|=max(|z|≤1)|f(z)|,for all |z0|=1

Need some help here.I had considered about the maxium modulus principle, but I didn't get a result

答: If $f$ is analytic on $|z|<1$ and continuous on $|z|\le 1$ then for $|z|<1$, $2i\pi f(z)=\int_{|s|=1} \frac{f(s)}{s-z}ds$. If $g(t)=f(e^{it})$ is differentiable with $g'(t)=0$ then $f(e^{it}) =f(1)$ so $f$ is constant.

ContentLicense: CC BY-SA 4.0


问:
Calculation of angles in regular polyhedrons
For any regular polyhedron (tetrahedron, cube, octahedron, dodecahedron, icosahedron, ...), how do I calculate an angle from the center of the regular polyhedron to the center of each face? Obviously, the angle decreases with the increasing number of faces. What is the formula, when the number of faces n is the variable?

For cube, n= 6, the angle is 90 degrees:

答:
A full outline on corner angles of polygons, dihedral angles of polyhedra, and dihedral angles of polychora can be found here: https://bendwavy.org/klitzing/explain/dihedral.htm

It also includes the asked for supplements as well as an explicite listing of all relevant cases, even the acrons.

--- rk


问:
Probability: Matching socks
$n$ left socks are in  a drawer and n right socks are organized in a line. Each sock in the drawer has only one match in the line. On the first round one sock was randomly puled out of the drawer and matched with the first sock in the line. Same goes for all the rest. what are the chances that the first match and the last one are right pairs?

I am absolutely not sure about the way i should solve this. Thanks for your help!.

答:
Probability that the first sock will match is $1/n$. Probability that the second sock won't pick the matching sock of the last sock is $(n-2)/(n-1)$. Probability that the third sock won't pick the matching sock of the last sock is $(n-3)/(n-2)$.... Probability that $(n-2)^{th}$ sock won't match the matching sock of the last one is $2/3$. Probability that $(n-1)^{th}$ sock won't match the matching sock of the last one is $1/2$.

So the final result is:

$$\frac 1n\cdot\frac{n-2}{n-1}\cdot\frac{n-3}{n-2}\cdot\frac{n-4}{n-3}\cdot\dots\cdot
\frac23\cdot\frac12=\frac1{n(n-1)}$$

The result is the same if you consider, for example, the first and the second sock.

Simpler explanation. The chance for the first sock to get the matching one is $1/n$. the last sock can be matched with any other with equal probability. So probability that the last (second, third, seventh...) sock will be paired with the matching one is $1/(n-1)$ because you have one sock less. So the total probability is the product of these two:

$$\frac1{n(n-1)}$$

Title: Probability: Matching socks


问:
solution of a sequence in a given range
Given an equation $$x^{\frac{1}{n}}=\sqrt{1-x}$$ has a unique solution $x_n$ in $[0,+\infty)$ for each $n \in\mathbb{N}$. One can see that $$x_n=1-x^{\frac{2}{n}}$$ But how do we show that the solutions are unique?

答:
Note that the domain is $x\in [0,1]$ and let $x=\sin^2 t$ to reexpress the equation as
$$\sin^{\frac2n}t = \cos t,\>\>\>\>\>t\in [0,\frac\pi2]$$
Since $\sin^{\frac2n}t $ strictly increases from 0 to 1 and $\cos t$ strictly decreases from 1 to 0, they cross each other once over $[0,\frac\pi2]$, hence a unique solution.

Title: solution of a sequence in a given range


问:
stopped UI discrete time martingale is UI
If $\{X_n\}$ is a uniformly integrable discrete time Martingale, and if $\tau$ is a (possibly $\infty$-valued) stopping time, then $\{X_{\tau \wedge n}\}$ is uniformly integrable (note I don't care that $\{X_{\tau \wedge n}\}$ is a martingale).  The only proof I've seen of this uses stuff about test functions.  Is it possible to give an elementary proof  without this test function definition of uniform integrability?

答:
Since $(X_n)$ is UI, we know there exists $X_\infty \in L^1(\mathcal F_\infty)$ such that $(X_n) \rightarrow X_\infty$ a.s. and in $L^1$.  Furthermore, the uniform integrability and optional stopping theorem guarantee that $X_{n \wedge \tau} = \mathbb{E}[X_\infty | \mathcal F_{n \wedge \tau}]$.  For the definition of uniform integrability, we want to show that $\sup_{n} \mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}]$ tends to $0$ as $R$ tends to $\infty$.  We compute

\begin{align*}
\mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] &= \mathbb{E}[|\mathbb{E}[X_\infty|\mathcal F_{n \wedge \tau}]| 1_{|X_{n \wedge \tau}| > R}] \\
&\le \mathbb{E}[\mathbb{E}[|X_\infty||\mathcal F_{n \wedge \tau}] 1_{|X_{n \wedge \tau}| > R}] \\
&= \mathbb{E}[\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}|\mathcal F_{n \wedge \tau}]] \\
&= \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]
\end{align*}

and by Markov's inequality $P(|X_{n \wedge \tau}| > R) \le \frac 1R \mathbb{E}[|X_{n \wedge \tau}|] \le \frac 1R \mathbb{E}[|X_{\infty}|]$, where the last inequality comes from the same argument as above.  Therefore we can choose $R$ such that $P(|X_{n \wedge \tau}| > R)$ is arbitrarily small for all $n$, and therefore $\mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$ can be made arbitrarily small as well.  Since $$\mathbb{E}[|X_{n \wedge \tau}| 1_{|X_{n \wedge \tau}| > R}] \le \mathbb{E}[|X_\infty|1_{|X_{n \wedge \tau}| > R}]$$ this proves the uniform integrability of $(X_{n \wedge \tau})$.


问:
How to show that $E(\lvert e^{itY_n}-e^{itc}\rvert)$ goes to $0$ where $Y_n\xrightarrow{p}c$
I'm trying to prove this statement using Levy's theorem.

If $X_n\xrightarrow{d}X$ and $Y_n\xrightarrow{p} c$ then $X_n+Y_n\xrightarrow{d} X+c$

I have for the distance of the characteristic functions, which needs to go to $0$:
$$\begin{aligned}
\left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X+c)}\right) \right\rvert
&\le \left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X_n+c)}\right)\right\rvert + \left\lvert E\left(e^{it(X_n+c)}\right) - E\left(e^{it(X+c)}\right)\right\rvert \\
&=D_1 + D_2
\end{aligned}$$
Here $D_2$ vanishes somewhat trivially because $X_n\xrightarrow d X$. However I'm struggling with $D_1$. Using the triangle inequality we have
$$\begin{aligned}
\left\lvert E\left(e^{it(X_n+Y_n)}\right) - E\left(e^{it(X_n+c)}\right)\right\rvert
&=\left\lvert E\left(e^{it(X_n+Y_n)}-e^{it(X_n+c)}\right)\right\rvert\\
&\le E\left(\left\lvert e^{it(X_n+Y_n)}-e^{it(X_n+c)}\right\rvert\right)\\
&= E\left(\left\lvert e^{it(X_n)}\right\rvert\left\lvert e^{itY_n} - e^{itc}\right\rvert\right)\\
&= E\left(\left\lvert e^{itY_n} - e^{itc}\right\rvert\right)
\end{aligned}$$

How can we show that this goes to $0$?

答:
wlog $c=0$ (just multiply by $e^{-itc}$)

Now, fix $\epsilon>0$. For $n$ sufficiently big $\mathbb{P}(|Y_n|>\epsilon) < \epsilon$.

Thus, $\mathbb{E}[|e^{itY_n}-1|] \leq 2\epsilon + |e^{it\epsilon}-1|$ for any such $n$ (why?).

This means that $\limsup\limits_{n \to \infty} \mathbb{E}[e^{itY_n}-1|] = 0$ for any $t$ (even though the convergence "depends on $t$"), but I guess that's fine.


问:
Scheduling - Logic
Here's the scenario:

A Enterprise have 6 offices across the state. Each city have their own quantity of employees and capacity of having visitors.
During 24 weeks the Company wants that all of its employees get to know the other offices.
The problem relies on city1 with the most number of employees that can leave only 5 per week. The other cities will leave only one per week.

$6$ cities total

city1 = $31$ persons

city2 = $2$ persons $+ 1$ free spot

city3 = $3$ persons $+ 1$ free spot

city4 = $4$ persons

city5 = $3$ persons $+ 1$ free spot

city6 = $2$ persons $+ 1$ free spot

Rules:

Each person has to visit all others cities (Meaning 5 cities total)

Only $1$ person leaves from each city at the time - Except city1 $\to$ 5

Each city can accept the number of people that leaves + free spot

All visitations has to occur in $24$ weeks - But only $1$ caster per week

I tried something like this:

```<code>loop 24 weeks
    loop (total person to visit (10))
        loop cities
            -> has visited? No -> visit | Yes -> Next city
            -> All visited? Next person
</code>```

But I got $41$ weeks. Almost the double asked.
Is it possible to make in the terms requested?

答:
If I understand correctly, it cannot be done in 24 weeks or even 41 weeks.  Here is a schedule that takes 43 weeks:

Here's a schedule that takes 42 weeks and tries to minimize the number of moves from one city to another:

And here's a schedule that takes 42 weeks and tries to minimize the number of weeks away from home:

Title: Scheduling - Logic


问:
Placing a circle in a plane without overlap
Given are $n$ non-overlapping unit circles in $\mathbb R^2$ and some number $r>1$. Let $C$ be one of the circles. I want to place a new circle of radius $r$ which is tangent to $C$ and additionally does not overlap any of the other circles.

How can I find a suitable center for this new circle efficiently, or at least decide whether this is even possible or not?

答:
Let the pivot circle $C$ be placed at the origin $O$.

Then among the centers of the other circles, let's choose the two
nearest to $O$, say $A,B$.

Let's draw the circle circumscribing the three point, having center $C_c$ and radius $R$.
The maximum circle that we can fit will have center $C_c$ and radius $r=R-1$.

Now, it is possible that inside the circumscribing circle there might be a fourth point $D$
which, although being farer from $O$ than $A$, will define with $O,B$ a smaller circum-circle.

In that case we will choose that, and repeat the check.


问:
Solve equation $\sqrt{4t + 1} = 3 - 3t$
Solve equation $\sqrt{4t + 1} = 3-3t$

→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$

→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$

I'm stuck at that point. Can someone tell me what I am doing wrong?

答:
For starters, squaring $\sqrt{4t+1}=3-3t$ gives $4t+1=9-18t+9t^2$. A little rearranging and we have $9t^2-22t+8=0$.

With a bit of insight we can see that this is $(9x-4)(x-2)$.

However, I'm pretty crap at factoring so lets use the quadratic formula:

$$ax^2+bx+c=0$$ implies

$$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

So we take $$x=\frac{-(-22)\pm\sqrt{22^2-4(9)(8)}}{2(9)}$$ now step-by-step we have

$$x=\frac{22\pm\sqrt{484-288}}{18}=\frac{22\pm\sqrt{196}}{18}=\frac{22\pm14}{18}
=\frac{11\pm7}{9}$$
So for "$+$" we have $x=\frac{18}{9}=2$ and for "$-$" we have $x=\frac 4 9$.

ContentLicense: CC BY-SA 3.0

Title: Solve equation $\sqrt{4t + 1} = 3 - 3t$


问:
Range of quadratic function using discriminant

Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.

Let $2x^2-2xy+y^2=a$.

Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.

We divide both side by $y^2$ and let $t=\frac{x}{y}$.

Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.

Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?

答:
Others have pointed out where you went wrong. I just want to provide an alternative proof.

$$x^2-2xy-3y^2=4=(x+y)(x-3y)$$

Denote $z=x+y, w=x-3y$, then $zw=4, x=(3z+w)/4, y=(z-w)/4$.

We want to find the range of $$2x^2-2xy+y^2=\frac{1}{16} (5w^2+14wz+13z^2)=\frac 72+\frac{1}{16} (5w^2+13z^2) \tag 1$$ under the constraint $zw=4$.

Clearly $(1)$ can be as large as you want. And if you apply AM-GM you get the minimum
$$\frac 72+\frac{1}{16} (5w^2+13z^2) \ge \frac 72 + \frac{1}{16} 2 \sqrt{5} \sqrt{13}|wz|=\frac{7+\sqrt{65}}{2}$$


问:
Prove $\int_1^{\infty}\psi(x)(x^\frac{s}{2}+ x^\frac{1-s}{2})\frac{dx}{x} $ is Entire
Define $$f(s)= \int_1^{\infty}\psi(x)(x^\frac{s}{2}+  x^\frac{1-s}{2})\frac{dx}{x} $$ where $ \psi(x)= \sum_{n=1}^{\infty}e^{-n^2\pi x}$ is the Jacobi Theta function.

Claim- $f(s)$ is Entire

My Try :-
$s=a+ib$ then $f(a+ib)=u(a,b)+iv(a,b)$ satisfy Cauchy Riemann Equations. How to proceed?

答:
$$g(s)=\int_1^\infty \psi(x) x^s dx$$ converges for all $s$.
$$\lim_{h\to 0} \frac{g(s+h)-g(s)}{h}- \int_1^\infty \psi(x) \log(x)x^s dx $$
$$=\lim_{h\to 0}  \int_1^\infty \psi(x) \int_s^{s+h}\log(x) \frac{x^z-x^s}{h} dz dx$$
$$=\lim_{h\to 0}  \int_1^\infty \psi(x) \int_s^{s+h}\log(x)\frac1{h} \int_s^z \log(x) x^u dudz dx$$

$$=\lim_{h\to 0}  \int_1^\infty \psi(x) O(h \log^2(x) x^{1+s}) dx=0$$
Thus $$g'(s)=\lim_{h\to 0} \frac{g(s+h)-g(s)}{h}=\int_1^\infty \psi(x) \log(x)x^s dx $$ and $g$ is complex differentiable, for all $s$, it is holomorphic on the whole complex plane, by the Cauchy integral formula it is entire (analytic everywhere) and everywhere equal to its Taylor series at $s_0$.


问:
For a fixed $f \in L^1(\Omega)$, find a coercive function $h$ such that $h(f) \in L^1(\Omega)$.
I am looking for a proof of the following statement :

Let $(\Omega,\mathcal{A},\mu)$ a probability space and $f \in L^1(\Omega)$.

Show that there exists a coercive function $h$ ($h$ is positive, non-decreasing, with $\underset{t \rightarrow +\infty}{\lim} \frac{h(t)}{t} = + \infty$) such that $h(f) \in L^1(\Omega)$.

If i am looking at a simple space like the space of real sequences equipped with the counting measure, it means that if a sequence of positive numbers $a_n$ verifies $$\sum_n a_n< +\infty$$
I can find $h$ coercive such that

$$\sum_n h(a_n)< +\infty$$

which i guess look quite possible by choosing $h$ such that $h(a_n)=\lambda_n a_n$ with $\lambda _n>0$ and $\lambda_n \underset{n \rightarrow +\infty}{\longrightarrow} + \infty$, but again I'm unsure how to detail that.

Any advices are welcomed.

答:
Edit: So it's not because I don't appreciate upvotes, but the solution below is wrong, I realized. Indeed, really it "proves" that any function in $L^p$ will also be in $L^{p+\delta}$ for suitably small ($f$-dependent) $\delta$. This is wrong, though.

Indeed, the mistake made is that I ended up applying monotone convergence to a decreasing net of functions and that does not fly. Take for instance the sequence $f_n(x)=\frac{1}{nx}$ on $[0,1]$. None of them are in $L^1$, but their limit sure is.

However, I did manage to find a proof.

Note that we may, without loss of generality, assume that $f$ is strictly positive. For general $f$ you can simply define $h$ that works for $f^+$ and set $\tilde{h}(x)=0$ for $x\leq 0$.

Thus,
$$
\int_{\Omega} h(f)\textrm{d}\mu=\int_{\Omega} \frac{h(f)}{f} f\textrm{d}\mu=\int_{(0,\infty)} \frac{h(x)}{x}\textrm{d}\nu,
$$
where $\nu$ is the finite measure on $\mathbb{R}$ given by the push-forward of the measure $f\textrm{d}\mu$ under $f$. Thus, writing $g(x)=\frac{h(x)}{x}$, we've rephrased the problem as: Given a finite measure $\nu$ on $(0,\infty)$, find a positive $g\in L^1(\nu)$ such that $\lim_{x\to\infty} g(x)=\infty$ and such that $g(x)x$ is non-decreasing. We can, without loss of generality, assume that $\nu$ is a probability measure and that $\nu$ has unbounded support (the bounded support case is trivial).

Define $b_0=0$ and recursively, define $b_{k+1}=\sup\{x\in (b_k,\infty)| \nu((b_k,x))\leq 2^{-(k+1)}\}$. Next, define $g(x)=\sum_{k=1}^{\infty} k1_{(b_{k-1},b_k)}$ and note that

$$
\int_{(0,\infty)} g(x)\textrm{d}x= \sum_{k=1}^{\infty} k \mu((b_{k-1},b_k))\leq \sum_{k=1}^{\infty} k 2^{-k}<\infty
$$
Furthermore, since $g$ is non-decreasing, $xg(x)$ will also be non-decreasing. Since $\nu$ has unbounded support, we have that $g$ is increasing and all in all, we get that $h(x)=xg(x)$ is a coercive function such that $\int_{\Omega} h(f)\textrm{d}\mu<\infty$.

Wrong solution:

Well, thinking of sequences does not exploit the strongest assumption that you've been given: That you're working on a probability space.

Define $h_{\alpha}(t)=(\max\{t,1\})^{\alpha},$ which is clearly coercive for any $\alpha>1$.

Note that, $\alpha\mapsto h_{\alpha}(t)$ is non-decreasing for every $t$. Applying monotone convergence, we get

$$
\int_{\Omega} \max\{f,1\} \textrm{d}\mu=\int_{\Omega} \lim_{\alpha\to 1} h_{\alpha}(f)\textrm{d}\mu=\lim_{\alpha\to 1} \int_{\Omega} h_{\alpha}(f)\textrm{d}\mu,
$$
Now, $\int_{\Omega} \max\{f,1\}\textrm{d}\mu\leq \int_{\Omega} |f|\textrm{d}\mu+\mu(\Omega)<\infty,$ so we must have that $\int_{\Omega} h_{\alpha}(f)\textrm{d}\mu<\infty$ for $\alpha$ sufficiently small.

This gives you the desired.


问:
Do I get ALL solutions for the heat equation by using the separation of variables?
Let's assume the heat equation as:
$$\frac{\partial \omega_1}{\partial t}=\nu(\frac{\partial^2 \omega_1}{\partial x_1^2}+\frac{\partial^2 \omega_1}{\partial x_2^2})$$
Using the separation of variables as $\omega_1=X_1(x_1)X_2(x_2)T(t)$ one can find a solution as:
$$\omega_1=(C_1 \cos(\sqrt{a}x_1)+C_2 \sin(\sqrt{a}x_1))(C_3 \cos(\sqrt{b}x_2)+C_4 \sin(\sqrt{b}x_2))e^{-\nu(a+b)t}$$
Let's now assume a second function $\omega_2$ which also needs to solve the heat equation, thus can be written as:
$$\omega_2 = (K_1 \cos(\sqrt{a}x_1)+ K_2 \sin(\sqrt{a}x_1))(K_3 \cos(\sqrt{b}x_2)+ K_4 \sin(\sqrt{b}x_2))e^{-\nu(a+b)t}$$
Both functions need to fulfill the boundary conditions $\omega_i(x_2=0)=0$ and $\omega_i(x_2=L)=0$ and the continuity equation $\frac{\partial \omega_1}{\partial x_1}+\frac{\partial \omega_2}{\partial x_2}=0$.
How can I assure that by using the separation of variables I get ALL the solutions needed to solve this set of conditions?

答: There are some uniqueness theorems that tell you that certain initial-boundary value problems for the heat equation cannot have more than one solution. The goal of separation of variables is only to find a solution to the problem. Once you have found a solution, then you invoke a uniqueness theorem to show that it is the only solution.


问:
Applying a Combination of Runge-Kutta and Midpoint Rule for this IVP
For a time step $h=0.5$, how can we apply a $RK2-M$ method to solve the following IVP?
\begin{cases}
y^{\prime}(t)=-1.2 y(t)+7 e^{-0.3 t}=f(t, y(t)), \quad t \in[0,1.5] \\
y(0)=3
\end{cases}
Now the problem I am facing is in the transition of this IVP into the desired scheme. I know that $K_{1}=f(t_{i},y_{i})$ but I always get lost in computing $K_{2}$ and thus my computations are wrong. Any help is much appreciated.

答:
You have
$$
K_2=f(t_i+h/2,y_i+hK_1/2)
$$
and then
$$
y_{i+1}=y_i+hK_2
$$
This should not be hard to implement.


问:
Probability Q on rolling two 8-sided dice
Let's say I roll two 8-sided dice. I win if the sums '7' and '11' show up before we see the sum '9' TWICE. What is my probability of winning?

So this is my answer and please correct me if I am wrong:

Prob of getting either 7 or 9 —> 6/64 + 6/64 = 12/64 —> 18.75%

Prob of getting 9 twice —> 8/64 x 8/64 = 1/64 —> 1.6%

But now to find the probability of winning, would it be 18.75 - 1.6 = 17.15% ? Am I right or wrong?

Thanks!

EDIT

From the answers provided below, we have three different results:

84%
49%
60%

Which one would be the correct answer?

答:
EDIT: Thanks to the comment by @robjohn it was confirmed that my code was rolling D7 not D8. I have corrected the code and the results below.

I decided to test it empirically, writing a C# code snippet to do the rolls and test whether or not I won.

A "trial" starts with rolling the dice and ends with either winning (7 and 11 both seen at least once) or losing (9 seen twice).

A "run" consists of 100,000 trials.

The experiment consisted of 30 runs.

Results:

```<code>run 1: 50616 wins on 100000 trials: 50.616% win percentage
run 2: 50674 wins on 100000 trials: 50.674% win percentage
run 3: 50653 wins on 100000 trials: 50.653% win percentage
run 4: 50781 wins on 100000 trials: 50.781% win percentage
run 5: 50967 wins on 100000 trials: 50.967% win percentage
run 6: 50512 wins on 100000 trials: 50.512% win percentage
run 7: 50758 wins on 100000 trials: 50.758% win percentage
run 8: 50549 wins on 100000 trials: 50.549% win percentage
run 9: 50740 wins on 100000 trials: 50.74% win percentage
run 10: 50905 wins on 100000 trials: 50.905% win percentage
run 11: 50253 wins on 100000 trials: 50.253% win percentage
run 12: 50578 wins on 100000 trials: 50.578% win percentage
run 13: 50592 wins on 100000 trials: 50.592% win percentage
run 14: 50699 wins on 100000 trials: 50.699% win percentage
run 15: 50771 wins on 100000 trials: 50.771% win percentage
run 16: 50557 wins on 100000 trials: 50.557% win percentage
run 17: 50454 wins on 100000 trials: 50.454% win percentage
run 18: 50826 wins on 100000 trials: 50.826% win percentage
run 19: 50724 wins on 100000 trials: 50.724% win percentage
run 20: 50677 wins on 100000 trials: 50.677% win percentage
run 21: 50576 wins on 100000 trials: 50.576% win percentage
run 22: 50599 wins on 100000 trials: 50.599% win percentage
run 23: 50694 wins on 100000 trials: 50.694% win percentage
run 24: 50740 wins on 100000 trials: 50.74% win percentage
run 25: 50720 wins on 100000 trials: 50.72% win percentage
run 26: 50934 wins on 100000 trials: 50.934% win percentage
run 27: 50539 wins on 100000 trials: 50.539% win percentage
run 28: 50825 wins on 100000 trials: 50.825% win percentage
run 29: 50738 wins on 100000 trials: 50.738% win percentage
run 30: 50488 wins on 100000 trials: 50.488% win percentage
</code>```

C# code:

```<code>Random rnd = new Random();
int rollDice() => rnd.Next(1,9) + rnd.Next(1,9);

bool doTrial() {
    int saw7 = 0;
    int saw11 = 0;
    int saw9 = 0;

    while (true) {
        var roll = rollDice();
        //Debug.Write($"{roll} ");
        switch (roll) {
            case 7: saw7++; break;
            case 9: saw9++; break;
            case 11: saw11++; break;
        }
        if (saw7>=1 && saw11>=1)
            return true;
        if (saw9>=2)
            return false;
    }
}

for (int run = 1; run <= 30; ++run)
{

    int wins = 0;
    int trials = 100000;
    for (int i = 0; i < trials; ++i)
    {
        if (doTrial()) { wins++; }
    }
    Debug.WriteLine($"run {run}: {wins} wins on {trials} trials: {(double)wins / (double)trials * 100.0}% win percentage");
}
</code>```

Body:
EDIT: Thanks to the comment by @robjohn it was confirmed that my code was rolling D7 not D8. I have corrected the code and the results below.

I decided to test it empirically, writing a C# code snippet to do the rolls and test whether or not I won.

A "trial" starts with rolling the dice and ends with either winning (7 and 11 both seen at least once) or losing (9 seen twice).

A "run" consists of 100,000 trials.

The experiment consisted of 30 runs.

Results:

run 1: 50616 wins on 100000 trials: 50.616% win percentage
run 2: 50674 wins on 100000 trials: 50.674% win percentage
run 3: 50653 wins on 100000 trials: 50.653% win percentage
run 4: 50781 wins on 100000 trials: 50.781% win percentage
run 5: 50967 wins on 100000 trials: 50.967% win percentage
run 6: 50512 wins on 100000 trials: 50.512% win percentage
run 7: 50758 wins on 100000 trials: 50.758% win percentage
run 8: 50549 wins on 100000 trials: 50.549% win percentage
run 9: 50740 wins on 100000 trials: 50.74% win percentage
run 10: 50905 wins on 100000 trials: 50.905% win percentage
run 11: 50253 wins on 100000 trials: 50.253% win percentage
run 12: 50578 wins on 100000 trials: 50.578% win percentage
run 13: 50592 wins on 100000 trials: 50.592% win percentage
run 14: 50699 wins on 100000 trials: 50.699% win percentage
run 15: 50771 wins on 100000 trials: 50.771% win percentage
run 16: 50557 wins on 100000 trials: 50.557% win percentage
run 17: 50454 wins on 100000 trials: 50.454% win percentage
run 18: 50826 wins on 100000 trials: 50.826% win percentage
run 19: 50724 wins on 100000 trials: 50.724% win percentage
run 20: 50677 wins on 100000 trials: 50.677% win percentage
run 21: 50576 wins on 100000 trials: 50.576% win percentage
run 22: 50599 wins on 100000 trials: 50.599% win percentage
run 23: 50694 wins on 100000 trials: 50.694% win percentage
run 24: 50740 wins on 100000 trials: 50.74% win percentage
run 25: 50720 wins on 100000 trials: 50.72% win percentage
run 26: 50934 wins on 100000 trials: 50.934% win percentage
run 27: 50539 wins on 100000 trials: 50.539% win percentage
run 28: 50825 wins on 100000 trials: 50.825% win percentage
run 29: 50738 wins on 100000 trials: 50.738% win percentage
run 30: 50488 wins on 100000 trials: 50.488% win percentage

C# code:

Random rnd = new Random();
int rollDice() => rnd.Next(1,9) + rnd.Next(1,9);

bool doTrial() {
    int saw7 = 0;
    int saw11 = 0;
    int saw9 = 0;

    while (true) {
        var roll = rollDice();
        //Debug.Write($"{roll} ");
        switch (roll) {
            case 7: saw7++; break;
            case 9: saw9++; break;
            case 11: saw11++; break;
        }
        if (saw7>=1 && saw11>=1)
            return true;
        if (saw9>=2)
            return false;
    }
}

for (int run = 1; run <= 30; ++run)
{

    int wins = 0;
    int trials = 100000;
    for (int i = 0; i < trials; ++i)
    {
        if (doTrial()) { wins++; }
    }
    Debug.WriteLine($"run {run}: {wins} wins on {trials} trials: {(double)wins / (double)trials * 100.0}% win percentage");
}


问:
How to prove $\ln (2^x+3^x)$ is strictly convex?
I believe (when I look at their graphs) that the following function is strictly convex
$$\ln (2^x+3^x)$$
More generally, I think
$$\ln (a^x+b^x)$$
should be strictly convex as well. Here $a, b>0$, $a\neq b$ and $\ln$ stands for natural logarit.

But I don't know how to prove (or may be disprove it)? Could anyone help me! Thank!

Edit: According to the deleted-comments, it's interesting to note that $2^x$ and $3^x$ are log-convex, and the fact that sum of log-convex is log-convex. Now, you may look at the nice answers below to see how to prove the strictly convex.

答:
Using $\alpha = \ln a$ and $\beta = \ln b$ we have
$$
 f(x) = \ln (e^{\alpha x} + e^{\beta x}) \, .
$$
Then
$$
 f'(x) = \frac{\alpha e^{\alpha x} + \beta e^{\beta x}}{e^{\alpha x} + e^{\beta x}}
$$
and
$$
f''(x) = \frac{\alpha^2 e^{\alpha x} + \beta^2 e^{\beta x}}{e^{\alpha x} + e^{\beta x}} - \left( \frac{\alpha e^{\alpha x} + \beta e^{\beta x}}{e^{\alpha x} + e^{\beta x}} \right)^2 = \frac{(\alpha - \beta)^2 e^{(\alpha + \beta)x}}{(e^{\alpha x} + e^{\beta x})^2} > 0
$$


问:
Name for matrix operation that inverses off-diagonal elements
Is there any operation that sends matrix
$$
S=\left(\begin{array}{cc} a&b\\
c& d
\end{array}\right) \to S'
=\left(\begin{array}{cc} \:\:\:a&-b\\
-c& \:\:\:d
\end{array}\right)
\quad a,b,c,d \in \mathbb{R}
$$
by setting off-diagonal elements of 2x2 matrix to inverse?
My question is: is there a name (or notation) for this kind of matrix operation?

If I would assume that $b,c \in Im$ (imaginary) then the operation obviously can be defined as conjugate $S'=(S^*)^T$. But all values are real in the given case.

Edit: $S \in SL(2, \mathbb{R}) $

答:
Let me post my own answer.

Let $S \in \mathrm{SL(2,\mathbb{R}})$, and let C be a group of matrices with two imaginary off-diagonal elements- a subgroup of $\mathrm{SL(2,\mathbb{C}})$.
Since the S can be isomorphically mapped to some element $A \in C$ using the map
$S \to \mathrm{A=T S T^{-1}}$ as following
$$
S \to A=\left(\begin{array}{cc} \sqrt{i}&0\\ 0& \sqrt{-i} \end{array}\right)  \left(\begin{array}{cc} a&b\\ c& d \end{array}\right)  \left(\begin{array}{cc} \sqrt{-i}&0\\ 0& \sqrt{i} \end{array}\right) = \left(\begin{array}{cc} \:\:\:a&ib\\ -ic& d \end{array}\right)
$$
(it is easy to see that the map is bijective, multiplicative e t c and the isomorphism).

Then after conjugating $A \to A^*$, the inverse map $\mathrm{T^{-1} (A^*) T }$ sends $A^*$ to
$$
S'=\left(\begin{array}{cc} \sqrt{-i}&0\\ 0& \sqrt{i} \end{array}\right)  \left(\begin{array}{cc} a& -ib\\ ic& \:\:\:d \end{array}\right)  \left(\begin{array}{cc} \sqrt{i}&0\\ 0& \sqrt{-i} \end{array}\right) = \left(\begin{array}{cc} \:\:\:a&-b\\ -c& \:\:\:d \end{array}\right)
$$
The operation is thereby is complex conjugation in the isomorphic group C (a subgroup of $\mathrm{SL(2,\mathbb{C}}$) with two imaginary off-diagonal elements). And since it is conjugation in the isomorphic group it is the conjugation.

PS. The same is valid for the conjugate transpose (or Hermitian transpose).


问:
Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

Construct a triangle given side length $b$, the altitude for side $c$, and the angle bisector of $B$.

So far I only found that I can find the angle at $C$ ($\gamma$) by constructing the right triangle with leg $h_c$ and hypotenuse $b$. I played a bit with the angles, but I can't really see how to use the length of the angle bisector, since there aren't many theorems helping me.

I feel like I have to use the inscribed angle theorem and find $B$ as the intersection of 2 loci, but I only found that it lies on side $c$

Any help would be appreciated. Thanks.

答:
It is not certain if such a construction is possible.

Next best option perhaps is to find important missing sides/angles analytically,  to examine their simplicity in order to incorporate them in construction

After no luck with direct construction, a numerical iteration with assumed $ (b,d,h)= (5,4,2) =$ (base,bisector length and altitude) was  attempted.

The following generating equation ( derived using standard triangle trig relations)  in unknown $a$ gives $a\approx 3.08$.. verified to be ok.

$$\dfrac{d^2}{a}=(\sqrt{b^2-h^2}+\sqrt{a^2-h^2})\cdot
  (1-\dfrac{b^2}{(\sqrt{b^2-h^2}+a+\sqrt{a^2-h^2})^2})$$

This relation may further guide the sought construction method.


问:
Simple Math Equation find sum of 4 numbers and if greater then number X reduce all 4 numbers respectively
Im not the greatest at Math but i have the following problem:

impressions = 791.

watched 100 = 500

watched 75 = 383

watched 50 = 600

watched 25 = 700

The sum of all watched fields is 2183.

Since the sum of all watched fields is greater then impressions i want to reduce the numbers in each of the 4 fields until the sum of all 4 fields equal impressions.

I have tried many methods such as finding the percentage of 791 in 2183. Then using that percentage reduce each 4 fields respectively, but i never get a flush number.

Any ideas?

Edit: Example of 1 of my many failed attemps :p

```<code>    2183 - 791 / 2183 = 0.63
    0.63

    500 X 0.63 = 315
    500 - 315 = 185.

    383 x 0.63 = 241.29
    383 - 241.29 = 141.71

    600 x 0.63 = 378
    600 - 378 = 222

    700 x 0.63 = 441
    700 - 441 = 259
</code>```

answer : 809

809 != 791 :(

OwnerDisplayName: Thomas Sultana

Body:
Let $s$ be a scaling factor defined as $s = 791/2183.$

$500\times s  = 181.17\ldots$

$383\times s = 138.77\ldots $

$600\times s = 217.40\ldots$

$700\times s = 253.64\ldots$

Add these up:

$181.17\ldots+138.77\ldots+217.40\ldots+253.64\ldots = 791$


问:
How would i add the probability of a single event being prevented by more than one prevention method combined.
I was on my Facebook and I got stumped by something that seemed easy. Basically, it was about preventing an illness.
The post essentially read that you have a 99.95% of surviving this illness and if you get the vaccine, you have an additional 95% chance of surviving. Then, it said there was a 194.95% survival chance if you had both. "That's not right," I thought, so I've been trying to figure out how to do the math for about 30 minutes now and I don't think I have the right tools and/or knowledge to do so. Please help so I can go to sleep lol</p>

答:
To make this easier I will illustrate it by using numbers of people and not by probabilities.

Imagine you have $1,0000,000$ people, you know that $999,500$ of them will survive this illness regardless if they get a vaccine or not, while $500$ of them won't survive the illness without a vaccine. You give the vaccine to these $500$ and this vaccine is $95$% effective so $475$ of them will survive (that is $500*0.95$), so in total, out of the million people you had $999,500+475=999,975$ survived so the chance of surviving this illness with a vaccin is $\frac{999,975}{1,000,000}=0.999975$ or in other words $99.9975$%


问:
Am I allowed to use this method?
Question: solve
$\frac{d^2y}{dx^2}+\frac{dy}{dx}=2x+3$.
My attempt:
The general solution to the corresponding homogeneous DE is $y_g=c_1+c_2e^{-x}$.

Now to find the particular integral, rewrite the DE as $$(D^2+D)y=2x+3$$
We have $y_p=\frac{1}{D(D+1)} (2x+3)$.

Then $y_p=\frac{1}{D+1}(x^2+3x)$.
Here's my question:

Am I allowed to use Taylor's expansion?

If this is allowed then I can find $y_p$ easily.

We would have $y_p=1-D+D^2-D^3+D^4-...(x^2+3x)$
and $y_p=x^2+3x-(2x+3)+2-0$

Title: Am I allowed to use this method?

Body:
$$(D^2+D)y=2x+3$$
It means:
$$y''+y'=2x+3$$
Not the equation you write in the first line.
$$\frac{d^2y}{dx^2}y+\frac{dy}{dx}y=2x+3$$
$$y''y+y'y=2x+3$$
And yes you have that:
$$\dfrac 1{1+D}=1-D+D^2-D^3.......$$
But you have:
$$y_p=\frac{1}{D(D+1)} (2x+3)$$
$$y_p=\left (\dfrac{1}{D}-\dfrac 1 {D+1} \right)(2x+3)$$
$$y_p=\dfrac{1}{D} (2x+3)-\dfrac 1 {D+1} (2x+3)$$
$$y_p=(x^2+3x)-\dfrac 1 {D+1} (2x+3)$$
Or as you did:
$$y_p=\dfrac 1{D+1}(x^2+3x)=x^2+3x -(2x+3)+2$$
$$\implies y_p=x^2+x-1$$


问:
Simplify binomial sum
It is well-known that for any integer $j$
$$
\sum _{i=0}^k (-1)^{i} \binom{k}{i} (j-i)^k = k!
$$
But I wonder if there's a closed form for the following sum
$$
\sum _{i=j}^k (-1)^{i} \binom{k}{i} (j-i)^k
$$

The following table gives the value for this $k = 1,\cdots, 6$ and $j = 0, \cdots k$.

```<code>1   0
2   1   0
6   5   1   0
24  23  12  1   0
120 119 93  27  1   0
720 719 662 360 58  1   0
</code>```

答:
This answer is a(nother) supplement to @MarkoRiedel's great derivation. This time we look at some details of his proof of
\begin{align*}
\color{blue}{\sum_{p=0}^{q-1} \left\langle k \atop p \right\rangle = \sum_{p=0}^{q-1} (-1)^p {k\choose p} (q-p)^k}\tag{1}
\end{align*}
which is the crucial part in his answer. The most important aspect is the usage of multidimensional residues and how to find a proper substitution for simplification. We use the classic Integral Representation and the Computation of Combinatorial Sums by G. P. Egorychev. We cite and use substitution formulas of multi-dimensional residues and obtain this way Markos derivation. I also use his  notation in order to ease comparison.

First part:

We start with the right-hand side of (1) and obtain
\begin{align*}
\color{blue}{\sum_{p=0}^{q-1}}&\color{blue}{(-1)^p \binom{k}{p}(q-p)^k}\\
&=\sum_{p=0}^\infty (-1)^p\binom{k}{p}(q-p)^k[[p\leq q-1]]\tag{2}\\
&=[w^{q-1}]\frac{1}{1-w}\sum_{p=0}^\infty(-1)^p\binom{k}{p}(q-p)^kw^p\tag{3}\\
&=[w^{q-1}]\frac{1}{1-w}\sum_{p=0}^\infty(-1)^p\binom{k}{p}k![z^k]e^{(q-p)z}w^p\tag{4}\\
&=k![z^k][w^{q-1}]\frac{e^{qz}}{1-w}\sum_{p=0}^\infty (-1)^p\binom{k}{p}\left(e^{-z}\right)^pw^p\\
&\,\,\color{blue}{=k![z^k][w^{q-1}]\frac{e^{qz}}{1-w}\left(1-\frac{w}{e^{z}}\right)^k}\tag{5}\\
\end{align*}
In (5) we have a representation with two-dimensional residues (note $[z^k]A(z)=res_{z=0}\frac{A(z)}{z^{k+1}}$) which can be considerably simplified using a proper substitution. This substitution was nicely used by Marko, but how to come to it might not be that obvious. We'll have a look at it in the Intermezzo section where we will see how to automagically derive the appropriate substitution next to the comments below.

Comments:

In (2) we use the Iverson brackets in order to get rid of the upper limit $q-1$ of the sum.

In (3) we use an equivalent to the Iverson brackets the identity
\begin{align*}
 \color{blue}{\sum_{p=0}^n a_p=\sum_{p=0}^\infty a_p[[p\leq n]]=[w^n]\frac{1}{1-w}\sum_{p=0}^\infty a_p w^p}
 \end{align*}
which is valid, since
\begin{align*}
 \color{blue}{[w^n]}&\color{blue}{\frac{1}{1-w}\sum_{p=0}^\infty a_p w^p}
=[w^n]\sum_{l=0}^\infty w^l\sum_{p=0}^\infty a_p w^p\\
&=\sum_{l=0}^n[w^{n-l}]\sum_{p=0}^\infty a_pw^p=\sum_{l=0}^n[w^{l}]\sum_{p=0}^\infty a_pw^p\\
&\,\,\color{blue}{=\sum_{l=0}^n a_l}
\end{align*}

In (4) we use the coefficient of operator $[z^k]$ and the identity $a^k=k![z^k]e^{az}$.

In (5) we apply the binomial theorem.

Intermezzo: Change of variables under the res sign

In order to find a substitution which simplifies (5) we use Rule 5 (change of variables under the res sign) from Egorychev's classic, which is stated as transformation formula for $n$-dimensional residues. Here we state it as two-dimensional identity which readily fits our needs. We have as special case $n=2$ of Rule 5:
\begin{align*}
&[z^{k_1}w^{k_2}]\left(A(z,w)\,\prod_{j=1}^2 f_{j}^{k_j}(z,w)\right)\\
&\ \ =[u^{k_1}v^{k_2}]
\left(\left.\left[A(z,w)\left|\frac{\partial(z,w)}{\partial{(u,v)}}\right|
\prod_{j=1}^2 f_{j}^{-1}(z,w)\right]
\right|_{(z,w)=g(u,v)}\right)\tag{6}
\end{align*}
where we use the substitution:
\begin{align*}
\color{blue}{u=u(z,w)=\frac{z}{f_1(z,w)}\qquad\qquad v=v(z,w)=\frac{w}{f_2(z,w)}}\tag{7}
\end{align*}

In (6) we have $A(z,w)$ a bivariate generating function, $f_j(z,w), j=1,2$ bivariate generating functions with constant term $\ne 0$ and $(z,w)=g(u,v)=(g_1(u,v), g_2(u,v))$ is the (unique) solution of (7).

In the current situation (5) we have
\begin{align*}
A(z,w)=\frac{e^z}{1-w}\qquad f_1^{k_1}(z,w)=\left(1-\frac{w}{e^z}\right)^k\qquad f_2^{k_2}(z,w)=\left(e^{z}\right)^{q-1}
\end{align*}
and use according to (7) the substitution
\begin{align*}
u(z,w)=\frac{z}{1-\frac{w}{e^z}}\qquad\qquad v(z,w)=\frac{w}{e^z}\tag{8}
\end{align*}
We put $v=w/e^z$ in $u(z,w)$ of (8) and obtain
\begin{align*}
u&=\frac{z}{1-v}\qquad \to\qquad \color{blue}{z=u(1-v)}\tag{9}
\end{align*}
We put $z=u(1-v)$ in $v(z,w)$ of (8) and obtain
\begin{align*}
v=\frac{w}{e^{u(1-v)}}\qquad \to\qquad \color{blue}{w=ve^{u(1-v)}}\tag{10}
\end{align*}
et voilà we have automagically found the substitution which was used in Marko's proof. :-)

Calculation of the Jacobian gives
\begin{align*}
\left|\frac{\partial(u,v)}{\partial{(z,w)}}\right|^{-1}&=\left|\frac{\partial(z,w)}{\partial{(u,v)}}\right|
=\begin{vmatrix}
\frac{\partial z}{\partial u}&\frac{\partial z}{\partial v}\\
\frac{\partial w}{\partial u}&\frac{\partial w}{\partial v}\\
\end{vmatrix}\\
&=\begin{vmatrix}
1-v&-u\\
(1-v)ve^{u(1-v)}&e^{u(1-v)}-uve^{u(1-v)}\\
\end{vmatrix}\\
&=e^{u(1-v)}\begin{vmatrix}
1-v&-u\\
(1-v)v&1-uv\\
\end{vmatrix}\\
&=(1-v)e^{u(1-v)}\tag{11}
\end{align*}

Second part:

We can now continue with the expression (7). We apply the substitution rule (6) by using the substitution (7) and obtain from (6) with (9) - (11):

\begin{align*}
\color{blue}{k![z^k]}&\color{blue}{[w^{q-1}]\frac{e^{z}}{1-w}\left(1-\frac{w}{e^{z}}\right)^ke^{(q-1)z}}\\
&=k![u^{k}][v^{q-1}]
\left\{\left[\frac{e^z}{1-w}\left|\frac{\partial(z,w)}{\partial{(u,v)}}\right|
\left.\left(1-\frac{w}{e^z}\right)^{-1}e^{-z}
\right]\right|_{(z,w)=g(u,v)}\right\}\\
&=k![u^{k}][v^{q-1}]
\left\{\frac{1}{1-ve^{u(1-v)}}\,(1-v)e^{u(1-v)}\,\frac{1}{1-v}\right\}\\
&=k![u^k][v^{q-1}]\frac{1}{e^{(v-1)u}-v}\\
&=k![u^k][v^{q-1}]\frac{1}{1-v}\frac{1-v}{e^{(v-1)u}-v}\\
&=k![u^k][v^{q-1}]\sum_{p=0}^\infty v^p\frac{1-v}{e^{(v-1)u}-v}\\
&=k![u^k]\sum_{p=0}^{q-1}[v^{q-1-p}]\frac{1-v}{e^{(v-1)u}-v}\\
&=k![u^k]\sum_{p=0}^{q-1}[v^{p}]\frac{1-v}{e^{(v-1)u}-v}\\
&\,\,\color{blue}{=\sum_{p=0}^{q-1} \left\langle k \atop p \right\rangle }
\end{align*}
and the claim follows.

Title: Simplify binomial sum


问:
Finding $\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$
Finding
$$\int\frac{x^2-1}{\sqrt{x^4+x^2+1}}$$
Try: I have tried it to convert it into $\displaystyle \left(x+\frac{1}{x}\right)=t$ and $\displaystyle \left(1-\frac{1}{x^2}\right)$  but nothing happen.

I felt that it must be in Elliptical Integral of first kind, second kind or in third  kind.

Can someone explain me how to write in elliptical form

答:
The other answers to this question use complex numbers. Here is a "real-only" answer.

Let us first work out
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx\qquad y\ge0$$
"But wait a minute, isn't this always infinite?" Yes, but the integral from $0$ to $y$ (the generally implied bounds of the indefinite integral) can be obtained by a simple subtraction once we work out the antiderivative, so there's no problem.

The denominator quartic can be written as $(x^2+z^2)(x^2+\overline z^2)$ where $z=e^{i\pi/6}$. Thus Byrd and Friedman 225.09 applies:
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac12\int_0^{u_1}\left(\frac{1+\operatorname{cn}u}{1-\operatorname{cn}u}-1\right)\,du=\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du$$
where $u_1=F(\varphi,m)$, $m=\frac14$ is the parameter and $\operatorname{cn}u_1=\cos\varphi=\frac{y^2-1}{y^2+1}$. B&F 361.61 provides the solution to this second integral:
$$\int_0^{u_1}\frac{\operatorname{cn}u}{1-\operatorname{cn}u}\,du=-E(\varphi,m)-\frac{\operatorname{sn}u_1\operatorname{dn}u_1}{1-\operatorname{cn}u_1}$$
We can work out $\operatorname{sn}u_1$ and $\operatorname{dn}u_1$ using the known value of $\operatorname{cn}u_1$ and the fundamental relations between Jacobian elliptic functions, yielding
$$\int_y^\infty\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-E\left(\varphi,\frac14\right)-\frac{y\sqrt{y^4+y^2+1}}{y^2+1}=J(y)$$
The integral from $0$ to $y$ is then $J(0)-J(y)$ and $J(0)=-2E(1/4)$ (a complete integral), so
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=-2E\left(\frac14\right)+E\left(\cos^{-1}\frac{y^2-1}{y^2+1},\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
The quasiperiodicity of $E$ allows us to absorb $2E(1/4)$. Applying a few trigonometric identities then allows us to simplify the amplitude, in the process extending the domain of validity to all real numbers:
$$\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=E\left(\cos^{-1}\frac{y^2-1}{y^2+1}-\pi,\frac14\right)+\frac{y\sqrt{y^4+y^2+1}}{y^2+1}$$
$$\color{red}{\int_0^y\frac{x^2-1}{\sqrt{x^4+x^2+1}}\,dx=\frac{y\sqrt{y^4+y^2+1}}{y^2+1}-E\left(2\tan^{-1}y,\frac14\right)}$$


Title: Applying a Combination of Runge-Kutta and Midpoint Rule for this IVP

Body:
For a time step $h=0.5$, how can we apply a $RK2-M$ method to solve the following IVP?
\begin{cases}
y^{\prime}(t)=-1.2 y(t)+7 e^{-0.3 t}=f(t, y(t)), \quad t \in[0,1.5] \\
y(0)=3
\end{cases}
Now the problem I am facing is in the transition of this IVP into the desired scheme. I know that $K_{1}=f(t_{i},y_{i})$ but I always get lost in computing $K_{2}$ and thus my computations are wrong. Any help is much appreciated.

Body:
As indicated by @Lutz Lehmann, the implementation is not hard at all (a matter of practicing), The full $RK2-M$ scheme is given by :
$$
\left\{\begin{array}{l}
K_{1}=f\left(t_{i}, y_{i}\right)=-1.2 y_{i}+7 e^{-0.3 t_{i}} \\
K_{2}=f\left(t_{i}+\frac{h}{2}, y_{i}+\frac{h}{2} K_{1}\right)=-1.2\left(y_{i}+\frac{h}{2} K_{1}\right)+7 e^{-0.3\left(t_{i}+\frac{h}{2}\right)} \\
y_{i+1}=y_{i}+h K_{2}
\end{array}\right.
$$
By proceeding with computing $K_{1}$, $K_{2}$, and $y_{i+1}$ for different time intervals, we are able to recursively obtain a solution for the IVP, I have constructed this table that shows the obtained values :
$$
\begin{array}{|c|c|c|c|c|c|}
\hline \mathrm{i} & t_{i} & y_{i} & K_{1}=-1.2 y_{i}+7 e^{-0.3 t_{i}} & K_{2}=-1.2\left(y_{i}+\frac{h}{2} K_{1}\right)+7 e^{-0.3\left(t_{i}+\frac{4}{2}\right)} & y_{i+1}=y_{i}+h K_{2} \\
\hline 0 & 0 & 3 & -1.2 * 3+7=3.4 & -1.2 *\left(3+\frac{3.4}{4}\right)+7 e^{-0.075} & 3+1.8742 * 0.5 \\
& & & & =1.8742 & =3.9371 \\
\hline 1 & \frac{1}{2} & 3.9371 & -1.2 * 3.9371+7 e^{-0.15} & -1.2\left(3.9371+\frac{1}{4} 1.3004\right)+7 e^{-0.3\left(\frac{4}{4}\right)} & 3.9371+0.47497 * 0.5 \\
& & & =1.3004 & =0.47497 & =4.1746 \\
\hline 2 & 1 & 4.1746 & -1.2 * 4.1746+7 e^{-0.3} & -1.2 *\left(4.1746+\frac{1}{4} 0.17621\right)+7 e^{-0.3\left(\frac{5}{4}\right)} & 4.1746-0.25136 * 0.5 \\
& & & =0.17621 & =-0.25136 & =4.0489 \\
\hline 3 & \frac{3}{2} & 4.0489 & \mathrm{x} & \mathrm{x} & \mathrm{x} \\
\hline
\end{array}
$$
The exact solution is hence :
$$
\boxed{y(t)=-\frac{43}{9}e^{-1.2t}+\frac{70}{9}e^{-0.3t}}
$$


答:
Hint:

If $ax^2+bx+c=0$, $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Title: Solve equation $\sqrt{4t + 1} = 3 - 3t$

Body:
Solve equation $\sqrt{4t + 1} = 3-3t$

→ I squared both sides and got ► $4t + 1 = 9 - 18t- 3t²$

→ I then moved the 3t² to the left side and combined like pairs and got ► $3t² + 12t - 8 = 0$

I'm stuck at that point. Can someone tell me what I am doing wrong?


问:
If a set is in a σ-algebra, then are the subsets in the σ-algebra as well?
Let $A \in F$ where $F$ is a σ-algebra. If $B \subseteq A$, is it true that $B \in F$ ?

Intuitively I would say no. A counter example would be

$\Omega = \{1,2,3,4,5\}$, $A = \{1,2,3\}$, $B=\{1,2\}$ and $F=\{\emptyset,\{1,2,3\},\{4,5\},\{1,2,3,4,5\}\}$ is a σ-algebra.

So am I correct to assume that the claim is generally not correct?

答: You have given a correct counter-example, so yes, it is indeed not correct. Note that a $\sigma$-algebra $F$ on $\Omega$ always contains $\Omega$, by definition. So if $A\in F$ and $B\subseteq A$ were to imply $B\in F$, then applying this to $A=\Omega$ and $B$ an arbitrary subset of $\Omega$ would imply that $F=\mathcal{P}(\Omega)$ is the entire powerset, but that would not be particularly interesting.


问:
$\lim_{n \to \infty} \cos(x/\sqrt{n})^n = e^{-x^2/2}$
I am trying to prove the limit in the title.

We have

$$\lim_{n \to \infty} \cos^n\left(\frac{x}{\sqrt{n}}\right) = \lim_{n \to \infty} e^{n \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right)} $$

and I guess taking the Taylor expansion around $0$ of $\cos\left(\frac{u}{\sqrt{n}}\right)$ will get us the $-\frac{1}{2}x^2$. Using continuity of $\log$, we find that

$$n \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right) =  n \log\left( \sum_{k = 0}^\infty \frac{(-1)^k}{(2k)!} \left(\frac{x^2}{n}\right)^{k}\right) $$

What can we do next?


Title: $\lim_{n \to \infty} \cos(x/\sqrt{n})^n = e^{-x^2/2}$

Body:
I am trying to prove the limit in the title.

We have

$$\lim_{n \to \infty} \cos^n\left(\frac{x}{\sqrt{n}}\right) = \lim_{n \to \infty} e^{n \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right)} $$

and I guess taking the Taylor expansion around $0$ of $\cos\left(\frac{u}{\sqrt{n}}\right)$ will get us the $-\frac{1}{2}x^2$. Using continuity of $\log$, we find that

$$n \log\left(\cos\left(\frac{x}{\sqrt{n}}\right)\right) =  n \log\left( \sum_{k = 0}^\infty \frac{(-1)^k}{(2k)!} \left(\frac{x^2}{n}\right)^{k}\right) $$

What can we do next?

Body:
Let $\dfrac x{\sqrt n}=2y\implies n=\dfrac{x^2}{4y^2}\implies y\to0$

$$\lim_{y\to0}(\cos2y)^{x^2/4y^2}=\left(\lim_{y\to0}(1-2\sin^2y)^{-\frac1{2\sin^2y}}\right)^{-\lim_{y\to0}\dfrac{2x^2\sin^2y}{4y^2}}=?$$


问:
Are there nonlinear operators that have the group property?
To be clear: What I am actually talking about is a nonlinear operator on a finitely generated vector space V  with dimension $d(V)\;\in \mathbb{N}>1$. I can think of several nonlinear operators on such a vector space but none of them have the requisite properties of a group. In particular, but not exclusively, are there any such nonlinear operator groups that meet the definition of a Lie Group.

答:
How about the following construction. Let $M \in R_{d}(G)$ be a $d$ dimensional representation of some Lie group (e.g. $SL(2,R)$ and $V$ is some $d$ dimensional vector space. Now define:
$$
f(x,M,\epsilon)= \frac{M x}{(1+\epsilon w.x)}
$$
where the vector $w$ is chosen so that $w.M= w$ for all $M\in SL(2,\mathbb{R})$. Such vectors exist, for instance, if $d=4$ and $R_4(G)=R_2(G) \otimes R_2(G)$ then $w=\begin{pmatrix}
0 \\
1\\
-1\\
0
\end{pmatrix}$ is one such vector. It is now not too difficult to convince oneself that:
$$ \begin{eqnarray}
f(x,I,0) = x \\
f(f(x,M_1,\epsilon_1),M_2,\epsilon_2) = f(x,M_1 M_2,\epsilon_1+\epsilon_2)
\end{eqnarray}$$
so that the mapping $f$ is nonlinear and satisfies the group property.


答:
Well, for $x\ne0$, $$n\ln\cos\frac x{\sqrt n}=n\cdot\frac{\ln\cos\frac x{\sqrt n}}{\cos\frac x{\sqrt n}-1}\cdot\frac{\cos\frac x{\sqrt n}-1}{\frac{x^2}n}\cdot\frac{x^2}{n}$$

Then, you can use that $\frac{\ln T}{T-1}\to 1$ as $T\to 1$ and that $\frac{\cos u-1}{u^2}\to-\frac12$ as $s\to 0$. Therefore $$\lim_{n\to\infty}n\ln\cos\frac x{\sqrt n}=\left(\lim_{n\to\infty}n\cdot \frac{x^2}n\right)\cdot\left(\lim_{n\to\infty}\frac{\ln\cos\frac x{\sqrt n}}{\cos\frac x{\sqrt n}-1}\right)\cdot\left(\lim_{n\to\infty} \frac{\cos\frac x{\sqrt n}-1}{\frac{x^2}n}\right)=-\frac{x^2}2$$

For $x=0$ the sequence is constant.


答:
Just a minor but perhaps helpful variant of Martin R's nice answer:

$$f(x)=\ln(e^{\alpha x}+e^{\beta x})=\ln e^{\alpha x}+\ln(1+e^{(\beta-\alpha)x})=ax+\ln(1+e^{(\beta-\alpha)x})$$

so

$$f'(x)=a+{(\beta-\alpha)e^{(\beta-\alpha)x}\over1+e^{(\beta-\alpha)x}}=a+(\beta-\alpha)\left(1-{1\over1+e^{(\beta-\alpha)x}} \right)=\beta-{\beta-\alpha\over1+e^{(\beta-\alpha)x}}$$

and thus

$$f''(x)={(\beta-\alpha)^2e^{(\beta-\alpha)x}\over(1+e^{(\beta-\alpha)x})^2}\gt0$$

provided $\alpha\not=\beta$.

The only advantage here, if any, is that's a little easier to see how the key positive term $(\beta-\alpha)^2$ arises.

Title: How to prove $\ln (2^x+3^x)$ is strictly convex?

Body:
I believe (when I look at their graphs) that the following function is strictly convex
$$\ln (2^x+3^x)$$
More generally, I think
$$\ln (a^x+b^x)$$
should be strictly convex as well. Here $a, b>0$, $a\neq b$ and $\ln$ stands for natural logarit.

But I don't know how to prove (or may be disprove it)? Could anyone help me! Thank!

Edit: According to the deleted-comments, it's interesting to note that $2^x$ and $3^x$ are log-convex, and the fact that sum of log-convex is log-convex. Now, you may look at the nice answers below to see how to prove the strictly convex.


问:
Can someone please explain this vector space definition to me?
Snippet taken from "Linear Algebra" by Jim Hefferon (4th edition):

1.2 Definition Suppose that $V$ and $W$ are vector spaces of dimensions $n$ and $m$
with bases $\mathrm{B}$ and $\mathrm{D},$ and that $\mathrm{h}: \mathrm{V} \rightarrow \mathrm{W}$ is a linear map. If
$$
\operatorname{Rep}_{\mathrm{D}}\left(\mathrm{h}\left(\vec{\beta}_{1}\right)\right)=\left(\begin{array}{c}
\mathrm{h}_{1,1} \\
\mathrm{~h}_{2,1} \\
\vdots \\
\mathrm{h}_{\mathrm{m}, 1}
\end{array}\right) \quad \ldots \quad \operatorname{Rep}_{\mathrm{D}}\left(\mathrm{h}\left(\vec{\beta}_{\mathrm{n}}\right)\right)=\left(\begin{array}{c}
\mathrm{h}_{1, \mathrm{n}} \\
\mathrm{h}_{2, \mathrm{n}} \\
\vdots \\
\mathrm{h}_{\mathrm{m}, \mathrm{n}}
\end{array}\right)_{\mathrm{D}}
$$
then
$$
\operatorname{Rep}_{\mathrm{B}, \mathrm{D}}(\mathrm{h})=\left(\begin{array}{cccc}
\mathrm{h}_{1,1} & \mathrm{~h}_{1,2} & \ldots & \mathrm{h}_{1, \mathrm{n}} \\
\mathrm{h}_{2,1} & \mathrm{~h}_{2,2} & \ldots & \mathrm{h}_{2, \mathrm{n}} \\
& \vdots \\
\mathrm{h}_{\mathrm{m}, 1} & \mathrm{~h}_{\mathrm{m}, 2} & \ldots & \mathrm{h}_{\mathrm{m}, \mathrm{n}}
\end{array}\right)_{\mathrm{B}, \mathrm{D}}
$$
is the matrix representation of h with respect to B, D.

Here $h$ is homomorphoism.
What I am not able to understand that if there is a linear mapping, $f:V\to W$, then according to a previous theorem in the same book, the bases of $V$ maps to bases of $W$.
Therefore, if $B$ is the bases of $V$ containing $n$ vectors $\langle B_1, B_2, \dots, B_n\rangle$, then
$\langle h(B_1), h(B_2), \dots, h(B_n)\rangle$ is the bases for $W$.

Now shouldn't in the definition of the attached screenshot $h_ij = 1$; if $i=j$ and $h_ij = 0$; if $i\neq j$? So that the final matrix have to be an identity matrix? What's going on?

(The book is available from https://hefferon.net/linearalgebra/ and this definition is on page 214. The notation $\mathrm{Rep}_B(\vec v)$ is defined on page 124.)

答:
You've not told us how this notation is defined in the book.

I take it that $B=(\beta_1,\dots, \beta_n)$ is an ordered basis of $V$ and $D=(\delta_1,\dots,\delta_m)$ is an ordered basis of $W$.

I think that if you look at the definition of
$\text{Rep}_D (v)$
you will see that the hypothesis is that for each $i$ the vector $h(\beta_i)$ can be written
$$
h(\beta_i)=h_{1,i}\delta_1+\dots +h_{m,i}\delta_m . (*)
$$

If you now look at the definition of $\text{Rep}_{B,D}(f)$ you will see that $(*)$ are exactly the equations you need to make $\text{Rep}_{B,D}(h)$ exactly what the book says.


问:
How would you approach this problem in a timely manner regarding different way to multiply
If you had no more than 1-2 minutes, how would you approach this Q:

We have four numbers (a, b, c, d), that will create the following pairs

a*b = 5

a*c = 8

a*d = 10

b*c = 16

b*d = 32

c*d = x

Find x

EDIT

Original question pic:

答:
You can calculate $abcd$ in three ways from the given numbers:

$$abcd = ad\times bc = ac\times bd = ab\times cd $$

Every number is used exactly once in the above equations, so we must be able to find two pairs of numbers with the same product. In fact, we can do it in two ways:

$$5\times 16=10\times 8=32\times cd\\
5\times 32=10\times 16=8\times cd$$

This gives us the two solutions $cd= 5/2$ and $cd=20$.

There seems to be no reason to prefer one solution more than the other. In fact, with some more time you can find out the possibilities for the four variables themselves to confirm that both solutions are valid. In each case there are two possibilities.

The set $\{2, 5/2, 4, 8\}$ gives products $\{5, 8, 10, 16, 20, 32\}$. An alternative set that gives the same products is $\{ \sqrt{10}/2, \sqrt{10}, 8\sqrt{10}/5, 2\sqrt{10} \}$.

The set $\{ 5/4, 2, 4, 8\}$ gives products $\{5/2, 5, 8, 10, 16, 32\}$. An alternative set that gives the same products is
$\{ \sqrt{5}/2, \sqrt{5}, 2\sqrt{5}, 16\sqrt{5}/5 \}$.


问:
How to prove that $\text {GL} (E)$ is an open subset of $\mathcal L (E)\ $?

Prove that if $E$ is Banach space then $\text {GL} (E)$ is an open subset of $\mathcal L (E).$

Here $\text {GL} (E)$ denotes the space of all bounded invertible linear operators on $E$ with bounded inverse and $\mathcal L (E)$ is the space of all bounded linear operators on $E.$

Let $T \in \text {GL} (E).$ We need to show that there exists an open ball $B(T, r)$ surrounding $T$ with radius $r \gt 0$ such that $B(T,r) \subseteq \text {GL} (E).$ Let $S \in B(T,r).$ If we can show that $\|I - S \|_{\text {op}} \lt 1$ then we are through. But how do I prove that? Any help will be appreciated.

Thanks in advance.

答: Hint: $S$ is necessarily invertible if $I - T^{-1}S = T^{-1}(T - S)$ has a sufficiently small operator norm.


问:
Prove $(∀x(P(x) ∨ Q(x))) → ((∃x(¬P(x))) → (∃x(Q(x))))$ using natural deduction
Hi guys I'm trying to learn natural deduction in predicate logic and I'm struggling with this proof.

$(∀x(P(x) ∨ Q(x))) → ((∃x(¬P(x))) → (∃x(Q(x))))$

I'm trying to follow my textbook but don't have many examples likes this one. Any help would be appreciated.

So far I have two assumptions:

$(∀x(P(x) ∨ Q(x)))$ and $¬((∃x(¬P(x))) → (∃x(Q(x))))$

答:
Sketch of the proof:

From$$\exists x(\lnot P(x))$$
you'll get $$\lnot P(t)$$ for some $t$.

From $$\forall x (P(x)\lor Q(x))$$ you'll get $$P(t)\lor Q(t)$$
by using Spec (specialization).

You should now be able to conclude $Q(t)$, and by using Rule $\exists$, you get $\exists x Q(x)$.

This is just an outline, and you should be able to convert this to a formal natural deduction proof. The idea is: "if for all $x$, either $P(x)$ or $Q(x)$ holds, and for particular $x=t$, $P(t)$ does not hold, then $Q(t)$ holds."

Showing $$\vdash (∀x(P(x) ∨ Q(x))) → ((∃x(¬P(x))) → (∃x(Q(x))))$$ amounts to showing $$\{∀x(P(x) ∨ Q(x)), ∃x(¬P(x))\} \vdash ∃xQ(x)$$
using deduction theorem.


问:
Is this problem about the integral $\displaystyle \int_{a}^{b}tf(t)dt$ well-posed?
Let $f\in \mathcal{C}[0,1]$ and suppose that:

$f(1)=2$.

$\displaystyle \int_{0}^{1}f(x)dx=\frac{1}{3}$.

$\displaystyle \int_{0}^{1}f'(x)dx=\frac{25}{3}$.

Find $\displaystyle \int_{0}^{1}xf(x)dx$.

My approach: Since that $f\in \mathcal{C}[0,1]$ so, the integrations by parts is well-allowed. Then, we have $$\int_{0}^{1}xf(x)dx=f(1)-\int_{0}^{1}x(f(x)+xf'(x))dx=2-\int_{0}^{1}x(f(x)+xf'(x))dx$$
since that $u=xf(x) \implies du=f(x)+xf'(x)$ and $dv=dx\implies v=x$, then $$\int_{0}^{1}xf(x)dx=\left.x^{2}f(x)\right|_{x=0}^{1}-\int_{0}^{1}x(f(x)+xf'(x))dx=2-\int_{0}^{1}x(f(x)-xf'(x))dx$$
So, we have $$2\int_{0}^{1}xf(x)dx=2-\int_{0}^{1}x^{2}f'(x)dx$$
Note that  $$\int_{0}^{1}xf'(x)dx=\int_{0}^{1}f'(x)dx-\int_{0}^{1}f(x)dx=\frac{25}{3}-\frac{1}{3}=\frac{24}{3}$$
Now, since that $$\int_{0}^{1}x^{2}f'(x)dx=\int_{0}^{1}f'(x)dx-\int_{0}^{1}2xf(x)dx$$
so, $$2\int_{0}^{1}xf(x)dx=2-\frac{25}{3}+2\int_{0}^{1}xf(x)dx$$
but it's an absurd.

New approach (corrections): Since that $f\in \mathcal{C}^{2}[0,1]$, so the integrations by parts is well-allowed. Then $$\int_{0}^{1}xf(x)dx=f(1)-\int_{0}^{1}x(f(x)+xf'(x))dx$$
So, we have that $$2\int_{0}^{1}xf(x)dx=2-\int_{0}^{1}x^{2}f'(x)dx$$
but,  $$\int xf'(x)dx=xf(x)-\int f(x)dx=xf(x)-F(x) \implies \left. \int xf'(x)dx \right|_{x=0}^{1}=\left. xf(x)-F(x)\right|_{x=0}^{1}=f(1)-F(1)+F(0)$$
now,since that $$\int_{0}^{1}x^{2}f'(x)dx=\int_{0}^{1}x(xf'(x))dx=\left.\int xf'(x)dx\right|_{x=0}^{x=1}-\int_{0}^{1}\left(\int_{0}^{x} tf'(t)dt\right)dx=2-F(1)-F(0)-\int_{0}^{1}\left(\int_{0}^{x} tf'(t)dt\right)dx$$

but I don't know how to continue from here

答:
I don't know where

$$\int_{0}^{1}xf'(x)dx=\int_{0}^{1}f'(x)dx-\int_{0}^{1}f(x)dx$$ is coming from. If using integration by part, I would say

$$\int_{0}^{1}xf'(x)dx = \left.xf(x)\right|_{x=0}^{1} - \int_{0}^{1}f(x)dx = 2-\frac{1}{3} = \frac{5}{3}$$


问:
Show that S is non-compact and deduce further that the closed unit ball in X is non-compact.
I have an exercise I am not sure how to solve
The exercise is:
Let X be an infinite dimensional normed vector space over $\mathbb{K}$ where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$. Define the unit sphere $S=\{x \in X : \|x\|=1\}$

Show that S is non-compact and deduce further that the closed unit ball in X is non-compact.

I am not sure how to show that S is non-compact, but for the second part where I have to deduce that the closed unit ball in X is non-compact (I denote closed unit ball in X by B)  I am thinking that I will show B is compact $\Rightarrow$ X is finite.

I assume that B is compact and then since S is a closed subset of B, then S will be compact and then I can use the first part to say that X may be finite. So now I have shown that B is compact $\Rightarrow$ X is finite, hence if X is infinite then B is non-compact. But I have troubles showing that S is non-compact, so I was hoping that someone could help me with that. We have already shown following:

答:
Your Problem 3, (d) implies that $S$ is not compact.

Indeed, assume that $S$ is compact and for any $x\in S$ consider $B_x=\{v\in X\ |\ \lVert x-v\rVert < 1/2\}$. Then $\{B_x\}_{x\in S}$ is an open covering of $S$ and by compactness it has to contain a finite subcover $\{B_{x_1}, \ldots, B_{x_n}\}$. But then $\{\overline{B_{x_1}},\ldots, \overline{B_{x_n}}\}$ is a closed ball covering of $S$ and none of them contains $0$, contradicting Problem 3, (d).

Finally the closed ball cannot be compact as well because $S$ is its closed subset. And a closed subset of a compact space is compact.

Also, please do you use full "finite dimensional" and "infinite dimensional" instead of "finite" and "infinite" to avoid confusion. It took me some time to figure out that you are not talking about literally finite sets.


问:
How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?
Just a little bit unsure about this question:

How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

And by 'words' I mean just a string of alphabet - does not have to be a real world in dictionary. I do know that you need to use Inclusion/Exclusion principle which says:

$|A \cup B| = |A| + |B| - |A\cap B|$

Let U be the set of all words containing 7 letters (26 letters to the power 7)
The only thing I'm not sure is what would be the sets A and B? Or do I need a third term C?

答: Hint:  I wouldn't use inclusion/exclusion at all.  First calculate how many words of precisely 4 consonants and 3 vowels you can make.  Which does "y" count as?  You need to select 4 consonants, then select 3 vowels, then select an ordering of consonants/vowels.  How many ways can you do each of these?  Then repeat for 4,5,6, and 7 vowels.</p>


问:
When speaking of neighbourhoods in complex analysis, are we always referring to circular neighbourhoods?
In Complex Analysis, does "neighbourhood" automatically mean "circular neighbourhood", or do non-circular ones exist?</p>

答: Certainly non-circular neighborhoods are allowed. Using Cauchy's Theorem it is often possible to replace an integral along a curve by an integral along a circle. But there is no need, and little sense, in limiting the discussion only to function defined on circular neighborhoods. Moreover, the concept of analytic continuation is extremely important and one often is interested in the maximal region a given function can be extended analytically to. The resulting region will often not be circular. </p>


问:
hyperbolic metric on B.
In $\mathbb{R}^{n+1}$ we defined $\langle x,y\rangle = x_1y_1+...+x_ny_n-x_{n+1}y_{n+1}$

The upper sheet of hyperboloid $H$ in $\mathbb{R}^{n+1}$ is the set $\{x \in \mathbb{R}^{n+1} /
\langle x, x\rangle = -1 $ and  $ x_{n+1} > 0 \}$.

I have the following exercise:

Let be the hyperboloid $H$ in $\mathbb{R}^{n+1}$, consider $P$ the plane $x_{n+1} = 0$ in $\mathbb{R}^{n+1}$ and $B$ the unit ball in $P$

I) If $\phi: H \to P$ is the central projection with center $-e_{n+1}$ prove that $\phi$ is a diffeomorphism, and express $\phi$ in coordinates

I did the computations and found that this map has the following expression $$\phi\left(x_1,\ldots,x_{n+1}\right) = \left(\frac{x_1}{1- x_{n+1}},\ldots,\frac{x_n}{1- x_{n+1}}, 0\right)$$
which shows it is differentiable (because so is any input). The inverse map is $$\phi^{-1}(x_1,\ldots,x_n)= (\lambda x_1,\ldots,\lambda x_n, \lambda-1 )$$ where $\lambda = \frac{2}{1- (x_1^2+\cdots+x_n^2)}$ this shows that $\phi$ is a diffeomorphism.

Now in II) since $\phi$ is an isometry I have to prove that

$\cosh(d(x,y)) = 1 + \frac{2|x-y|^2}{(1-|x|^2)(1-|y|^2)}$ $(1)$

I am stuck at this second part

I know that $\cosh(d(x,y)) = -\langle x,y\rangle$ in $H$. From here, I tried to compute $\phi(1)$ but did not manage to.

Some hint?

Thanks

答:
First, one has a riemannian manifold $(H,g)$ where $g$ is the restriction of $\langle\cdot,\cdot\rangle$ to $TH$. Second, one has a manifold $B$ and a diffeomorphism
$$ \phi : H \to B$$
allowing to define a riemannian metric on $B$, say $g_0 = \phi_*g$. By the very definition of the push-forward metric by the diffeomorphism $\phi$,
$$
\phi : (H,g) \to (B,g_0)
$$
is an isometry. Thus, if $p$ and $q$ are points in $H$,
$$
d_g(p,q) = d_{g_0}\left(\phi(p),\phi(q) \right)
$$
so one has
$$
\cosh\left(d_g(p,q)\right) = \cosh \left(d_{g_0}(\phi(p),\phi(q)) \right)
$$
Then, your exercice reduces to the computation, in one hand, of $d_g(p,q)$ in terms of the riemannian metric $g$, and in the other hand, of $d_{g_0}(\phi(p),\phi(q))$ in terms of the euclidean metric on $B$, written $|\cdot|$.

So here are possible hints:

this can be done computing the differential of $\phi$ (this is possible because you have here an exact expression for $\phi$) and thus the exact value of $g_0 = \phi_*g$ (this can be done thanks to the exact expression of $g$ and $\mathrm{d}\phi$)

this can maybe be done an easier way: compute the expression
$$
1+2\frac{|\phi(p) - \phi(q)|^2}{\left(1-|\phi(p)|^2\right)\left(1-|\phi(q)|^2\right)}
$$
thanks to the exact expression of $\phi$, and where $|\cdot|$ is the usual euclidean norm. You may have an expression in terms of the coordinates of $p$ and $q$ in $\mathbb{R}^{n+1}$ that looks like what you already computed.

Title: hyperbolic metric on B.


问:
What are the solutions to $ac+bd=1$, for $a$ and $b$, where $a,b,c,d$ are integers and $c$ and $d$ have HCF $1$?
What are the solutions to $ac+bd=1$, for $a$ and $b$, where $a,b,c,d$ are integers, and $c$ and $d$ have HCF $1$?
A few examples:
Give all the solutions for $a$ and $b$ for the following:

$$3a+5b=1 \\
3a+7b=1 \\
4a+5b=1$$

General case:
$ac+bd=1$

I was able to do the above to provide a general formula in terms of $a$ and $b$ in terms of $n$, where $n\in \{1,2,3\ldots \}$ for the first $3$ but struggled with the general case.

I'm presuming there's some sort of way to get $a$ and $b$ in terms of $c$ and $d$, but I'm not sure.

Cheers.

答:
By the identity of Bezout, (others call it a lemma and even a theorem;French use identity) if $(c,d)=m$ then
$a\mathbb Z+b\mathbb Z= m\mathbb Z$ so in the case of $m=1$ you do have $a\mathbb Z+b\mathbb Z= \mathbb Z$. There are infinitely many solutions.


问:
Norm of equivalence class in $L^\infty$
Consider the space $L^\infty(\mathbb{R})$ and its (closed) subspace $S$ of $2 \pi$ periodic functions.
Define $\lVert f \rVert_\infty = \text{ess sup}_{x \in \mathbb{R}} |f(x)|$.

Now consider the quotient space $Q=L^\infty(\mathbb{R})/S$. Define the norm of the equivalence class $[f] = \left\{ g \in L^\infty: g-f \in S \right\}$ as $\lVert [f] \rVert_{Q} = \inf \left\{ \lVert f + g \rVert_\infty: g \in S \right\} = \text{dist}\left(f, S \right)$.

Calculate $\lVert \left[ \sin\frac{x}{2} \right] \rVert_Q $.

I know that $\lVert [f] \rVert_Q \leq \lVert f \rVert_\infty $. So $\lVert \left[ \sin\frac{x}{2} \right] \rVert_Q \leq 1$. The $Q$ norm can be expressed in several ways but taking the infimum is where I struggle. How do I proceed from here? Thanks.

答:
I would guess that the best we can do is the $0$ function, simply because the graph of $\sin(x/2)$ oscillates negative and positive orientations of the same curve over the period $2\pi$. That is, $f(x) = \sin(x/2)$ satisfies $f(x + 2\pi) = -f(x)$. If we wish to uniformly approximate $f$ by a $2\pi$ period function, any attempt to fit the function to $f$ better than just the $0$ function over $[0, 2\pi]$ will make for a worse approximation over $[2\pi, 4\pi]$, and vice-versa.

Let's prove this intuition correct. Let's suppose that we can find a function $g \in S$ such that $\|f - g\|_\infty < 1$. Let
$$\alpha = \frac{1 + \|f - g\|_\infty}{2} \in (0, 1).$$
Then, there is a non-trivial interval $I \subseteq [0, 2\pi]$ on which $f(x) > \alpha$ (simply because $\alpha < 1$, $f$ is continuous, and $f$ achieves a maximum of $1$). We then have,
\begin{align*}
\|f - g\|_\infty &= \operatorname{ess sup}_{x \in \Bbb{R}} |f(x) - g(x)| \\
&\ge \operatorname{ess sup}_{x \in \Bbb{R}} (f(x) - g(x)) \\
&\ge \operatorname{ess sup}_{x \in I} (f(x) - g(x)) \\
&\ge \operatorname{ess sup}_{x \in I} (\alpha - g(x)) \\
&= \frac{1 + \|f - g\|_\infty}{2} - \operatorname{ess inf}_{x \in I} g(x),
\end{align*}
hence
$$\operatorname{ess inf}_{x \in I} g(x) \ge \frac{1 - \|f - g\|_\infty}{2} > 0.$$
On the other hand, note that $I + 2\pi$ is another non-trivial interval, and
$$x \in I + 2\pi \implies f(x) < -\alpha.$$ Then
\begin{align*}
\|f - g\|_\infty &\ge \operatorname{ess sup}_{x \in \Bbb{R}} (g(x) - f(x)) \\
&\ge \operatorname{ess sup}_{x \in I + 2\pi} (g(x) - f(x)) \\
&\ge \operatorname{ess sup}_{x \in I + 2\pi} (g(x) + \alpha) \\
&= \operatorname{ess sup}_{x \in I + 2\pi} g(x) + \frac{1 + \|f - g\|_\infty}{2},
\end{align*}
hence
$$\operatorname{ess sup}_{x \in I + 2\pi} g(x) \le -\frac{1 - \|f - g\|_\infty}{2} < 0.$$
But, because $g$ is periodic with period $2\pi$, we must have
$$0 < \operatorname{ess inf}_{x \in I} g(x) \le \operatorname{ess sup}_{x \in I} g(x) = \operatorname{ess sup}_{x \in I + 2\pi} g(x) < 0,$$
which is a contradiction. Therefore, the norm of the equivalence class of $f$ is indeed $1$.

Title: Norm of equivalence class in $L^\infty$


问:
A subgroup can be expressed as intersection of maximal subgroups in different ways
We know from the Chinese remainder theorem that if an ideal I in a commutative ring $R$ is the intersection of finite many maximal ideals $M_i$ then the quotient ring $R/I$ is isomorphic to the direct product of $R/M_i$. And one can show that the these maximal ideals are uniquely determined.

But for a group $G$, can a subgroup $H$ be expressed as the intersection of some maximal subgroups of $G$ in different ways, e.g. $H = M_1\cap M_2 = M_3 \cap M_4$ with $\{M_1,M_2\}$ different from $\{M_3,M_4\}$? And what is the key difference here compared with the ring case?

答:
The stated property does not hold for groups.

The Klein $4$-group, $C_2\times C_2$, has three maximal subgroups: $C_2\times\{e\}$, $\{e\}\times C_2$, and the diagonal subgroup. The intersection of any two of them is equal to the trivial group. So the trivial group can be expressed as an intersection of maximal subgroups in three different ways. The same is true in $S_3$, where any two of the three subgroups of order $2$ have trivial intersection.

You can even have disjoint families of maximal subgroups with the same intersection. Modifying the first example a bit, take $C_p\times C_p$. The maximal subgroups correspond to one dimensional subspaces, so they are $\{e\}\times C_p$, and all subgroups of the form $\langle (x,x^i)\rangle$ with $0\leq i\lt p$. There are $p+1$ of them, and any two intersect trivially. So for $p\geq 3$, you can find four pairwise distinct maximal subgroups $M_1$, $M_2$, $M_3$, and $M_4$, any two of which have trivial intersection.

Note that the former example uses normal subgroups, so you cannot save the proposition by using normality as an analogue of “ideals”.

(You can get examples where the intersection is not trivial by taking a group that has a nontrivial quotient isomorphic to the Klein $4$-group; for example, the quaternion group $Q_8$, and the subgroups $\langle i\rangle$, $\langle j\rangle$, and $\langle k\rangle$, any two of which intersect at $\{1,-1\}$; the nonabelian group of order $p^3$ and exponent $p$ has a central quotient isomorphic to $C_p\times C_p$, yielding more examples in the same manner.)

What is different about groups and rings? I’m not sure what it is exactly, but here is one major difference: the only ideals of a direct product of rings, $R_1\times\cdots\times R_n$ are of the form $I_1\times\cdots\times I_n$, with $I_j\triangleleft R_j$. This is not true for subgroups (or even normal subgroups) of direct products of groups. Even for groups with no nontrivial subgroups like $C_p$, there are more maximal subgroups of $C_p\times C_p$ than the coordinate subgroups.


问:
How do I solve inequalities where both sides are not equal to zero, and that use absolute values and fractions?
For example something like

$$\dfrac{|x-2|+3}{4-|2x+8|}\geq-5.$$

I understand how to solve inequalities where there is an absolute value, however in fraction form just confuses me.

答:
Hint:

When $4-|2x+8|>0$ the inequality is always satisfied (LHS is positive).

When $4-|2x+8|<0$ the inequality is satisfied if and only if $|x-2|+3\leq -5(4-|2x+8|)$ (the inequality reverses when multiplying both sides by a negative number).


问:
how to solve the minima?
Given x＞0，y＞0, 1/x ＋ 8/$y^2$ ＝1, solve the minima of $x$＋$y$?

One guy has this problem solved by Cauchy inequality.

Can anyone have another approach or?

I worked out like this:
if 1/$x$ ＝ 8/$y^2$ ＝ 1/2 hold, then $x$＝2, $y$＝4， (1/$x$)(8/$y^2$)＝1/4 ⥤ $x$$y^2$＝32 ⥤$ xy$＝8⥤ $x$＋$y$＝2+4＝6

Is there anything wrong with my approach? If yes, please advice and explain.

Thanks

答:
Let $s:=x+y$ and let us eliminate $y$.

$$\frac1x+\frac8{y^2}=1$$ becomes

$$\frac1x+\frac8{(s-x)^2}=1$$ or

$$s=x+\sqrt{\frac{8x}{x-1}}.$$

The derivative cancels for $x=2$, and $s=6$.

Title: how to solve the minima?


答:
Add +5 to both sides and you got 0 on one side.
Now add the two terms you got on the left (you may want to search about the least common multiple)

Now the inequality is asking you for what values of x your fraction is greater than or equal to 0.

The answer is: for every value of x that makes numerator and denominator either botth positive or both negative.

Title: How do I solve inequalities where both sides are not equal to zero, and that use absolute values and fractions?

Body:
For example something like

$$\dfrac{|x-2|+3}{4-|2x+8|}\geq-5.$$

I understand how to solve inequalities where there is an absolute value, however in fraction form just confuses me.


问:
Prove that $\gcd(a, b) = \gcd(a, b, ax + by)$ for any given integers $a, b, x, y$
How can I prove that $\gcd(a,b) = \gcd(a,b,ax+by)$?

I know that $\gcd(a,b)$ can be written in linear equation like this:
$$ \gcd(a,b) = ax + by $$
Does that mean that I can substitute it to $\gcd(a,b,ax+by)$?
If yes, what's next?

答:
It is special case $\,c = ax+by\,$ of the following more general

Lemma $\ \ (a,b)\mid c\,\Rightarrow\, (a,b,c) = (a,b)\ \ \ $ [here $\,(x,y,\ldots):=\gcd(x,y,\ldots)\,$]

Proof $\ \ (a,b,c) = ((a,b),c) = (a,b)(1,c/(a,b)) = (a,b)$

where we applied the gcd associative and distributive laws.

Remark $ $ More generally, as in the descent step in the Euclidean algorithm, it is easy to prove that a gcd is preserved if we mod out one argument by another, or by the gcd of some others, e.g. $$\bmod (a,b)\!:\ c\equiv \bar c\ \Rightarrow\  (a,b,c) = (a,b,\bar c)\qquad $$

The lemma is the special case $\,\bar c = 0,\,$ i.e. $\,c\equiv 0\pmod{(a,b)}$

Though you can prove the OP directly without using these laws (by essentially inlining their proofs, e.g. via Bezout), this method will soon lead to great difficulty in more complex problems without the simplification afforded by the abstracted basic laws. So you should strive to learn these fundamental properties sooner rather than later. The proofs of these laws are no more difficult than in the special cases, and the laws generalize to wider classes of rings (where Bezout-based proofs may fail).

If you know ideal theory then you will find it illuminating to compare the gcd proofs with the ideal proofs, using the view point of unimodular transformations of ideal generators (a generalization of Bezout-based proofs).


问:
How can I find all the points where two functions of multiple variables intersect at the same value?
Given two functions $f(x_1,x_2,x_3...x_n)$ and $g(x_1,x_2,x_3...x_n)$ in $n$-dimensional space, is there a straightforward way to find the set of points where $f$ and $g$ have the same output and the same input?

In 2D space, the intersection of two curves is “usually” a discrete set of points that can be solved for and manually plugged back into the equation if you're looking for points where the curves intersect at an interesting value. In 3D, the intersection of two surfaces is a curve or discrete set of curves and you can find the intersection of that curve with a plane if you want to know where the two surfaces intersect at a particular value.

For this problem, I have I have two functions of four variables. I want to find the set of coordinates where these functions both have the same input coordinates and both have a particular output value $U$, but I don't know where to begin. Are numerical methods my only hope? Or maybe there's just some standard terminology I'm missing.

For reference, here are the equations:

$P(w,x,y,z)=\frac{1}{\sqrt{wxyz}}$

$Q(w,x,y,z)=\frac{1}{\sqrt{2}}\left(\frac{1}{wy}+\frac{1}{wz}-\frac{A}{xz}\right)$

答:
There is not a general procedure that works for any functions. But there is certainly much that can be done in your particular example.

You set the functions equal to each other:
$$\frac 1{\sqrt{wxyz}} = \frac 1{\sqrt 2}\left(\frac 1{wy}+\frac 1{wz}-\frac{A}{xz}\right)$$
A change of variables makes this a little prettier:
$$\frac 1{rstu} = \frac 1{\sqrt 2}\left(\frac 1{r^2t^2}+\frac 1{r^2u^2}-\frac{A}{s^2u^2}\right)$$
Multiply through by $r^2s^2t^2u^2\sqrt 2$:
$$\sqrt 2 rstu = s^2u^2 + s^2t^2 - Ar^2t^2$$

You can write this as a quadratic equation in each of the variables in terms of the others:
$$(At^2)r^2 + (\sqrt 2 stu)r - s^2(u^2 + t^2) = 0\\
(u^2 + t^2)s^2 - (\sqrt 2 rtu)s - Ar^2t^2 = 0\\
(Ar^2 - s^2)t^2 + (\sqrt 2 rsu)t - s^2u^2 = 0\\
(s^2)u^2- (\sqrt 2 rst)u + s^2t^2 - Ar^2t^2 = 0$$
For real solutions, each of the discriminants must be non-negative:
$$2s^2t^2u^2 +4(At^2s^2)(u^2 + t^2) \ge 0\\
2r^2t^2u^2 +4(Ar^2t^2)(u^2 + t^2) \ge 0\\
2r^2s^2u^2 + 4(Ar^2-s^2)s^2u^2 \ge 0\\
2r^2s^2t^2 + 4(s^2t^2)(Ar^2-s^2)\ge 0$$
Simplifying and switching back to $x,y,z,w$, the first two conditions turn out to be the same, as are the second two conditions. So it reduces to
$$z + 2A(z + y) \ge 0\\w + 2(Aw-x) \ge 0$$

Since $y,z > 0$ already, if $A \ge 0$, the first is automatically satisfied. The second requires $$x \le \left(\frac 12 + A\right)w$$

You can choose the values of any three of $w, x, y, z$ freely, as long as those two inequalities are observed. The fourth variable will have two possible values as determined by the corresponding quadratic equation in $r,s,t,u$. For example, if you've picked positive values for $x, y, z$, and set $s = \sqrt x, t = \sqrt y, u = \sqrt z$, then the value of $r$ is found by
$$r = \dfrac{-\sqrt 2stu \pm \sqrt{2s^2t^2u^2 +4(At^2s^2)(u^2 + t^2)}}{2At^2}\\
r = \frac{su}{\sqrt 2At}\left(-1 \pm\sqrt{1 + 2A\left(1 + \frac{t^2}{u^2}\right)}\right)$$
Squaring both sides and substituting for $w,x,y,z$,
$$w = \frac {xz}{2A^2y}\left(1 \pm \sqrt{1 + 2A\left(1 + \frac yz\right)}\right)^2$$
Similar equations can be derived to find each of the variables in terms of the other three.


答:
It probably depends on the textbook you're using. Some authors write "neighbourhood" in the sense "open neighbourhood". Some do not.

Some authors might only consider open discs, but the most common convention is that an (open) neighboorhood of a point is any open set containing the point. In practice, you can almost always work with disc shaped ones though. (Every open neighbourhood of $z$ surely contains a disc centered at $z$.)

Title: When speaking of neighbourhoods in complex analysis, are we always referring to circular neighbourhoods?

Body: In Complex Analysis, does "neighbourhood" automatically mean "circular neighbourhood", or do non-circular ones exist?</p>


答:
Suppose we have an $f$ so that

$f(1)=2$

$\int_0^1f(x)\,\mathrm{d}x=\frac13$

$\int_0^1f'(x)\,\mathrm{d}x=\frac{25}3$

and, for some $I$, we have
$$
\int_0^1xf(x)\,\mathrm{d}x=I
$$

Then, $f(x)+\alpha\sin(2\pi x)$ also satisfies 1., 2., and 3., however
$$
\int_0^1x(f(x)+\alpha\sin(2\pi x))\,\mathrm{d}x=I-\frac\alpha{2\pi}
$$
Thus, we cannot determine $\int_0^1xf(x)\,\mathrm{d}x$ simply based on 1., 2., and 3.

Title: Is this problem about the integral $\displaystyle \int_{a}^{b}tf(t)dt$ well-posed?

Body:
Let $f\in \mathcal{C}[0,1]$ and suppose that:

$f(1)=2$.

$\displaystyle \int_{0}^{1}f(x)dx=\frac{1}{3}$.

$\displaystyle \int_{0}^{1}f'(x)dx=\frac{25}{3}$.

Find $\displaystyle \int_{0}^{1}xf(x)dx$.

My approach: Since that $f\in \mathcal{C}[0,1]$ so, the integrations by parts is well-allowed. Then, we have $$\int_{0}^{1}xf(x)dx=f(1)-\int_{0}^{1}x(f(x)+xf'(x))dx=2-\int_{0}^{1}x(f(x)+xf'(x))dx$$
since that $u=xf(x) \implies du=f(x)+xf'(x)$ and $dv=dx\implies v=x$, then $$\int_{0}^{1}xf(x)dx=\left.x^{2}f(x)\right|_{x=0}^{1}-\int_{0}^{1}x(f(x)+xf'(x))dx=2-\int_{0}^{1}x(f(x)-xf'(x))dx$$
So, we have $$2\int_{0}^{1}xf(x)dx=2-\int_{0}^{1}x^{2}f'(x)dx$$
Note that  $$\int_{0}^{1}xf'(x)dx=\int_{0}^{1}f'(x)dx-\int_{0}^{1}f(x)dx=\frac{25}{3}-\frac{1}{3}=\frac{24}{3}$$
Now, since that $$\int_{0}^{1}x^{2}f'(x)dx=\int_{0}^{1}f'(x)dx-\int_{0}^{1}2xf(x)dx$$
so, $$2\int_{0}^{1}xf(x)dx=2-\frac{25}{3}+2\int_{0}^{1}xf(x)dx$$
but it's an absurd.

New approach (corrections): Since that $f\in \mathcal{C}^{2}[0,1]$, so the integrations by parts is well-allowed. Then $$\int_{0}^{1}xf(x)dx=f(1)-\int_{0}^{1}x(f(x)+xf'(x))dx$$
So, we have that $$2\int_{0}^{1}xf(x)dx=2-\int_{0}^{1}x^{2}f'(x)dx$$
but,  $$\int xf'(x)dx=xf(x)-\int f(x)dx=xf(x)-F(x) \implies \left. \int xf'(x)dx \right|_{x=0}^{1}=\left. xf(x)-F(x)\right|_{x=0}^{1}=f(1)-F(1)+F(0)$$
now,since that $$\int_{0}^{1}x^{2}f'(x)dx=\int_{0}^{1}x(xf'(x))dx=\left.\int xf'(x)dx\right|_{x=0}^{x=1}-\int_{0}^{1}\left(\int_{0}^{x} tf'(t)dt\right)dx=2-F(1)-F(0)-\int_{0}^{1}\left(\int_{0}^{x} tf'(t)dt\right)dx$$

but I don't know how to continue from here


问:
Relation between chords length and radius of circle
Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly</p>

答:
Let $a'$, $b'$ be the lengths of the perpendiculars from $M$ to the chord of length $2a$, resp. $2b$. Then
$$c^2=a'^2+b'^2=(r^2-a^2)+(r^2-b^2)\ ,$$
from which we obtain
$$r=\sqrt{a^2+b^2+c^2\over2}\ .$$</p>


问:
Parametrization of circle
So here is a parametric equation for circle where

$$x=\cos t,\quad y=\sin t$$

but this only traces out a circle when $t$ goes from $0$ to $2\pi$.

If I were to trace a circle with the restriction of $0$ to $\pi$, it would be ($\cos 2t$, $\sin 2t$).

However, why is it not $(\cos \frac{t}{2},\sin \frac{t}{2})$ since essentially here we are just dividing the original $0$ to $2\pi$ bounds by $2$, giving us $\frac{t}{2}$ is between $0$ and $\pi$? (I directly substituted $t$ with $\frac{t}{2}$ into the $\cos t$ and $\sin t$, but why would that not work?)

答:
As you've pointed out, when you use the parameterization $(\cos(t),\sin(t))$ you need $0\le t \le 2\pi$.  In other words, whatever you input into cosine and sine needs to vary between $0$ and $2\pi$ in order to trace a full circle.  If you change the input to $t/2$ you still need that input to vary between $0$ and $2\pi$.  In other words, you need $$0\le \frac{t}{2}\le 2\pi \implies 0 \le t \le 4\pi.$$  With $2t$ as the input instead you still need that input to vary between $0$ and $2\pi$, which gives you $$0 \le 2t \le 2\pi \implies 0 \le t \le \pi.$$  With whatever form you chose for your input, that input has to vary between $0$ and $2\pi$ to get a whole circle the way you've been doing it, but the domain for $t$ will change depending on the choice of input.

Edit to address comment:

You have to be a little bit careful here.  If you have the parameterization $(\cos(t),\sin(t))$, then you are correct in saying that $0 \le t \le 2\pi$ is the same as saying $0 \le t/2 \le \pi$.  However, that's all for the same $t$ and the same parameterization.  So, it's fine to say that $(\cos(t),\sin(t))$ traces out a full circle as $t/2$ varies from $0$ to $\pi$ because that's the same as saying that $t$ varies from $0$ to $2\pi$, but if you want to replace $t$ with $t/2$ in the parameterization you need to actually make that change everywhere.  So,
\begin{align}(\cos(t),\sin(t))&\longrightarrow (\cos(\color{red}{t/2}),\sin(\color{red}{t/2}))\\
0\le t \le 2\pi &\longrightarrow 0 \le \color{red}{\frac{t}{2}} \le 2\pi \implies 0 \le t\le 4\pi\\
\text{or equivalently:}\quad 0 \le \frac{t}{2} \le \pi &\longrightarrow 0 \le \frac{\color{red}{t/2}}{2}\le \pi \implies 0 \le t \le 4\pi
\end{align}

Title: Parametrization of circle


问:
Evaluate the Modified Bessel function of the second kind $K_0(z)$ for $z \to 0$
Consider the Modified Bessel function of the second kind $K_0(z)$ for $z \to 0$.

Some sources (Abramowitz and Stegun, formula 9.6.8, and this page, formula 10.30.3) state that

$$K_0(z) \sim - \log z$$

That is: $K_0(z)$ is proportional to $\log(z)$. But according to Wolfram Alpa (in the "Series expansion at z = 0" section):

$$K_0(z) = \log \left( \frac{2}{\gamma z} \right) + O(z^2) = \log 2 - \log \gamma - \log z + O(z^2)$$

where $\gamma$ is the Euler-Mascheroni constant. This is a more accurate result.

How can it be obtained? I can't find a source for this.

答:
$K_0$ can be expressed in terms of $I_0$ as follows:

$$
K_0(z)=AI_0(z)+BI_0(z)\int \frac{dx}{xI_0^2(x)},
$$
where $A=\ln 2-\gamma$ and $B=-1$ (any solution of the underlying differential equation is of that form for arbitrary $A$ and $B$). Since
$$
I_0(z)=\sum_{k=0}^\infty\left(\frac{z^k}{(2k)!!}\right)^2,
$$
and $I_0(0)=1$,
\begin{align}
K_0(z)&=I_0(z)\left(\ln 2-\gamma-\ln z+\frac{z^2}{4}+\ldots\right) \\
&=\ln 2-\gamma-\ln z+O(z^2)
\end{align}
when $z\searrow 0$.


问:
How much is the probability of $\cos x \cos(4-x)<0$ for $0<x<4$?
I have this function
$$f(x)=g(x)h(x)=\cos x \cos(4-x)$$
How much is the probability of $f(x)<0$ for $0<x<4$?

答:
The rest of the answer assumes that $x$ is chosen uniformly over $[0,4]$.

$$\cos x\cos (4-x) > 0$$ for $$x\in \left(\frac{\pi}{2}, 4-\frac{\pi}{2}\right)$$
on $[0,4]$.

So the probability you require is $$1 - \frac{4-\pi}{4} = \frac{\pi}{4}\approx 0.785$$

For reference, here is the graph on $[0,4]$ -


问:
$G/N = 1$ iff $N = G$ and $G/N = G$ iff $N = 1$
Suppose that $G$ is a finite group and $N \trianglelefteq G$. Then
$$G/N \cong \{1_G\} \iff |G/N| = |\{1_G\}| \iff \frac{|G|}{|N|}= 1 \iff |G| = |N| \iff N=G \text, $$
which answers this question. Likewise, $G/N \cong G$ if and only if $N$ is trivial.

I thought these facts were true for infinite groups as well, but now I'm having doubts. If $N=G$, then $a^{-1}b \in N$ for all $a, b \in G$, so there is only one coset and $G/N \cong 1$. If $N=1$, then $a^{-1}b \in N \iff b = a$, so every coset is a singleton and $G/N \cong G$. Is the statement below true or false?

If $G$ is a group with a normal subgroup $N$, then $G/N \cong 1$ implies $N=G$, and $G/N \cong G$ implies $N=1$.

I'd appreciate seeing lots of counterexamples if it's false. And I'd like to know if the statement generalizes to rings and ideals.

Here are some ideas. Suppose that $G/N \cong G$. The isomorphism $\phi : G/N \to G$ composes with the usual map $\pi : G \to G/N$ to produce a surjective homomorphism $\phi \circ \pi : G \to G$. Notice that $\ker \phi \circ \pi = \ker \pi = N$ since $\phi(\pi(g)) = 1_G \iff \pi(g) \in \ker \phi = \{ N \} \iff \pi(g)=N \iff g \in \ker \pi$. By the correspondence theorem, there is a bijection between subgroups of $G$ containing $\ker \phi \circ \pi =N$ and subgroups of $G$. I'd like to say "hence every subgroup of $G$ contains $N$, so $N$ must be trivial," but I'm not sure if that's true.

If $G/N$ is trivial, then $N = \ker \pi = G$. I expected the two claims to be equally difficult to prove...

EDIT: Quotient ring being isomorphic to the initial ring is relevant.

答:
$G/N = 1$ implies $N = G$, because $G/N$ has only only element (which then must be $N$).

$G/N = G$ does not imply $N = 1$. You can take $G$ to be $H \times H \times \dots$, for some group $H$, and $N = H \times 1 \times 1 \times \dots$. Then $G/N$ kills one of the copies of $H$, but $G/N = 1 \times H \times H \times \dots$ is still isomorphic to $H \times H \times \dots = G$. So for infinite groups this can indeed fail.


问:
Cauchy Principal value: Question about the integration boundaries of this integral
I want to calculate the following integral:
$$\int\limits_{C_1}\frac{1-e^{iz}}{z^2} \ dz$$
in the limit $r\to0$. $C_1$ is a semi circle with radius $r$ in the complex plane ($z=re^{i\theta},0<\theta<\pi$).

Going through the textbook solution it says:
$$\int\limits_{C_1}\frac{1-e^{iz}}{z^2} \ dz=\int\limits_{\pi}^{0}\frac{1-e^{ire^{i\theta}}}{r^2e^{2i\theta}}ire^{i\theta} \ d\theta=\ ...$$

I honestly don't know why the integration would go from $\pi$ to $0$ and not the other way around?
I think this should have something to do with the direction I'm going  around $C_1$ (my integration path) but I'm not sure how to argue this the correct way.

Any help would be really appriciated, thanks!

答:
Update 1: Using
$$ e^z-1=z+\frac12z^2+\cdots $$
then you can get
\begin{eqnarray}
&&\int_{C_1}\frac{1-e^{iz}}{z^2} \ dz\\
&=&-\int_{C_1}\frac{e^{iz}-1}{z^2} \ dz\\
&=&-\int_{C_1}\bigg(\frac{i}{z}-\frac{1}{2}-\frac1{3!}iz+\cdots\bigg) \ dz.
\end{eqnarray}
It is easy to see
$$\lim_{r\to 0}\int_{C_1}\bigg(\frac{1}{2}+\frac1{3!}zi+\cdots\bigg) \ dz=0 $$
and
$$ \int_{C_1}\frac{i}{z}\ dz=\int_0^\pi\frac{i}{re^{i\theta}}ir^{i\theta}d\theta=-\pi.
$$
So
$$\lim_{r\to 0}\int\limits_{C_1}\frac{1-e^{iz}}{z^2} \ dz=\pi. $$
Update 2: You also can use the Cauchy formula. Since $C_1$ is the half unit circle, you have
$$ \int_{C_1}\frac{1-e^{iz}}{z^2} \ dz=\frac12\int_{|z|=1}\frac{1-e^{iz}}{z^2} \ dz=\frac12\cdot2\pi i\text{Res}(\frac{1-e^{iz}}{z^2},z=0)=\pi.$$


答:
The term "neighborhood" of a point comes from topology, where it is taken to mean any open set containing that point. This convention carries over into complex analysis.

Often, a circular neighborhood of a point and a more general neighborhood of a point are interchangeable in practice. If you need a circular neighborhood of a point, it suffices to find a general open set containing the point, since then you can just take a circular neighborhood contained in that open set; and if you need a general open set, of course a circular neighborhood will suffice. So topologically there really is no distinction.

Besides, many theorems in complex analysis that use circular neighborhoods can be adapted somewhat to more general regions, the idea behind the Riemann mapping theorem.


问:
finding solution of $y'=3y^{2/3}$ with $y(0)=0$ with Picard iteration
I've been asked to apply uniqueness and existence theorem and search solution with Picard iteration. But I couldnt understand picard part.

The Picard Iteration is given by:

$$\tag 2 \displaystyle  ~~y_{n+1} = y_0 + \int^x_{x_0} f(x, y_n)dx$$

It seems all solutions are equal to $0$. Because substiting $y=0$ to $f(x,y)=3y^{2/3}$ gives $f(x,y)=0$
What can I do from here?

答:
You have to check whether the assumptions of Picard's theorem are fulfilled for your IVP. Note that $y(x)=x^3$ is another solution when $x>0$.

For $x<0$ you have to define what is meant by $y^{2/3}$ when $y<0$.


答:
That $G/N$ is trivial if and only if $N=G$ holds for all groups, finite or infinite groups. For if $N\neq G$, then let $x\in G$, $x\notin N$; we will have $xN\neq eN$, so $G/N$ contains at least one nontrivial element.

That $G/N\cong G$ implies $N$ is trivial is true for finite groups, but need not be true for infinite groups. For example, take the group $G\cong \prod_{n=1}^{\infty} C_2$, the product of infinitely many copies of the cyclic group of order $2$. If $$N=C_2\times\{e\}\times\{e\}\times\cdots\times \{e\}\times\cdots,$$
then $N$ is not trivial, but $G/N\cong G$.

For another example, the Prüfer $p$-group $C_{p^{\infty}}$ has the property that for every proper normal subgroup $N$, $C_{p^{\infty}}/N\cong C_{p^{\infty}}$. And there are infinitely many proper normal subgroups.

The magic word is “Hopfian group”. A group $G$ is Hopfian if and only if every surjective morphism $f\colon G\to G$ is a bijection; in other words, if $G/N\cong G$, then $N=\{e\}$. Every finite group is Hopfian, as are many infinite groups, but not all.

The same holds for rings: $R/I$ is trivial if and only if $I=R$, by the same argument. A ring is Hopfian if and only if $R/I\cong R$ implies $I=\{e\}$. Every finite ring is Hopfian, as are other types of rings, but not every ring is Hopfian: $\prod_{i=1}^{\infty}\mathbb{Z}_2$ is an example.

In general, in any class of algebraic objects, an object is Hopfian if every surjection $\mathscr{O}\to\mathscr{O}$ must be a bijection; and it is co-Hopfian if every injection $\mathscr{O}\to\mathscr{O}$ must be a bijection.

Title: $G/N = 1$ iff $N = G$ and $G/N = G$ iff $N = 1$

Body:
Suppose that $G$ is a finite group and $N \trianglelefteq G$. Then
$$G/N \cong \{1_G\} \iff |G/N| = |\{1_G\}| \iff \frac{|G|}{|N|}= 1 \iff |G| = |N| \iff N=G \text, $$
which answers this question. Likewise, $G/N \cong G$ if and only if $N$ is trivial.

I thought these facts were true for infinite groups as well, but now I'm having doubts. If $N=G$, then $a^{-1}b \in N$ for all $a, b \in G$, so there is only one coset and $G/N \cong 1$. If $N=1$, then $a^{-1}b \in N \iff b = a$, so every coset is a singleton and $G/N \cong G$. Is the statement below true or false?

If $G$ is a group with a normal subgroup $N$, then $G/N \cong 1$ implies $N=G$, and $G/N \cong G$ implies $N=1$.

I'd appreciate seeing lots of counterexamples if it's false. And I'd like to know if the statement generalizes to rings and ideals.

Here are some ideas. Suppose that $G/N \cong G$. The isomorphism $\phi : G/N \to G$ composes with the usual map $\pi : G \to G/N$ to produce a surjective homomorphism $\phi \circ \pi : G \to G$. Notice that $\ker \phi \circ \pi = \ker \pi = N$ since $\phi(\pi(g)) = 1_G \iff \pi(g) \in \ker \phi = \{ N \} \iff \pi(g)=N \iff g \in \ker \pi$. By the correspondence theorem, there is a bijection between subgroups of $G$ containing $\ker \phi \circ \pi =N$ and subgroups of $G$. I'd like to say "hence every subgroup of $G$ contains $N$, so $N$ must be trivial," but I'm not sure if that's true.

If $G/N$ is trivial, then $N = \ker \pi = G$. I expected the two claims to be equally difficult to prove...

EDIT: Quotient ring being isomorphic to the initial ring is relevant.