问:
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain.
$f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$

I know I have to switch the $f(x)$ and the $y$:

$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$

To find the domain:

$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$

$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$

$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$

$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$

The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$

But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.

Thanks

答:
a)

The function $f$ is not injective, therefore not bijective and cannot have an inverse therefore.

But you can calculate the individual branches (i. e. partial inverses) of $f$.

b)

You didn't rearrange your equation correctly.

Let $y$ denote $f^{-1}(x)$.

$$\sin^2\left(\frac{2y+1}{3}\right)=x$$

$$\left(\sin\left(\frac{2y+1}{3}\right)\right)^2=x$$

$$\sin\left(\frac{2y+1}{3}\right)=\pm\sqrt{x}$$

$$\frac{2y+1}{3}=\arcsin(\pm\sqrt{x})$$

$$\frac{2y+1}{3}=\pm\arcsin(\sqrt{x})$$

$$y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$$


问:
Doubt on proof of characterization of polynomials
I was trying to understand this proof of the following result:

If $f \in \mathcal{C}^\infty (\mathbb{R})$ satisfies that for every $x
> \in \mathbb{R}$ there exists a $n \in \mathbb{N}$ such that
$f^{(n)}(x)=0$, then $f$ is a polynomial.

However I don't really understand how Baire's category theorem is applied. Could anyone explain how is it applied?

Now, I was thinking that if we could state that $\operatorname{int}X \not= \emptyset$ then by Baire's category theorem, $X \cap S_n \not= \emptyset$ for some $n \in \mathbb{N}$ and then we have a contradiction because if we take $x \in \operatorname{int}(X \cap S_n)$ then there exists an interval $(a,b)$ that contains $x$ such that
$$(a,b) \subset X \cap S_n$$
so $(f^{(n)})|_{(a,b)}\equiv0$ and then $f|_{(a,b)}$ is a polynomial getting a contradiction with $x \in X$.

So, does any one have any idea if it is possible to prove that $X$ has non empty interior to conclude the proof in this way?

答:
Here's how the theorem is applied:

We make sure that $X$ is a non-empty complete metric space, and $\{X\cap S_n\}$ is its countable closed cover. Here, $closed$ is to be understood with respect to $X$. From Baire's theorem, we deduce that some $X\cap S_n$ contains a non-empty open set. Here, $open$ is to be understood with respect to $X$. Open sets in $X$ are of the form $U\cap X$, where $U$ is open in $\mathbb{R}$. So we deduce that some open interval $(a, b)$ has a non-empty intersection with $X$, and $$(a, b)\cap X\subset X\cap S_n.$$

ContentLicense: CC BY-SA 4.0


问:
Contour integral of quotient of polynomials
I am asked to compute $\displaystyle A = \int_{\partial \mathcal{D}(0,1)} \frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz$.

Let $\displaystyle g(z)= \frac{8z^2-8z+1}{4z^3-8z^2+z-2}$.

I have found $A = 8\pi i$, but when checking on WolframAlpha I get $A=-\pi$. I don't know where my mistake stems from.

Step 1: I find the root of the denominator: $\{-\frac{i}{2}, \frac{i}{2},2\}$. I remark that the moduli of $\frac{i}{2}$ and $-\frac{i}{2}$ are inferior to 1, thus two singularities lie in the contour.

Step 2: I define a function $f_1$ and a contour $C_1$ centered in $z_1=-\frac{i}{2}$.
$$\forall z \in C_1, f_1(z)=(z-z_1)g(z)$$

So I have,

$$\forall z \in C_1 \backslash \{z_1\}, g(z)=\frac{f_1(z)}{z-z_1}$$

Step 3: I apply the Cauchy integral theorem:

$$\int_{C_1}g(z)dz=\int_{C_1}\frac{f_1(z)}{z-z_1}dz=2\pi i f_1(z_1)=4\pi i$$

Step 4: I apply the same reasoning and find again $4\pi i$. (Intuitively, by symmetry, it seems that both integrals should be equal. But I may be wrong).

Step 5: I sum the two results obtained at steps 4 and 5, and get $8\pi i$.

I double-checked WolframAlpha and still finds $-\pi$.

答:
Here's another way to approach it. Consider the integral

$$\int_{\partial D(0,R)}\frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz$$

for $R>2$. The integral takes on the same value everywhere, and is only one residue away from the integral we want. Parametrizing and taking the limit we get that

$$\lim_{R\to\infty}\int_{\partial D(0,R)}\frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz = \frac{8i}{4}\cdot 2\pi =4\pi i$$

The residue at $2$ is

$$\lim_{z\to 2}\:(z-2)\cdot\frac{8z^2-8z+1}{(4z^2+1)(z-2)} = 1$$

which means the original integral we wanted was $2\pi i$


答:
Actually, that integral is equal to $2\pi i$. In fact\begin{align}\operatorname{res}_{z=\pm i/2}\left(\frac{8z^2-8z+1}{4z^3-8z^2+z-2}\right)&=\lim_{z\to\pm i/2}\left(z-\left(\pm\frac i2\right)\right)\frac{8z^2-8z+1}{4z^3-8z^2+z-2}\\&=\left.\frac{8z^2-8z+1}{4(z-2)\left(z+\left(\pm\frac i2\right)\right)}\right|_{z=\pm i/2}\\&=\frac12\end{align}and therefore your integral is equal to $2\pi i\left(\frac12+\frac12\right)=2\pi i$.
Here's another approach, which doesn't use residues. If you decompose$$\frac{8z^2-8z+1}{4z^3-8z^2+z-2}$$into partial fractions, you will get$$\frac{1/2}{z-i/2}+\frac{1/2}{z+i/2}+\frac1{z-2},$$and so, by Cauchy's integral formula, that your integral is equal to $2\pi i\left(\frac12+\frac12\right)=2\pi i$.

Title: Contour integral of quotient of polynomials

Body:
I am asked to compute $\displaystyle A = \int_{\partial \mathcal{D}(0,1)} \frac{8z^2-8z+1}{4z^3-8z^2+z-2}dz$.

Let $\displaystyle g(z)= \frac{8z^2-8z+1}{4z^3-8z^2+z-2}$.

I have found $A = 8\pi i$, but when checking on WolframAlpha I get $A=-\pi$. I don't know where my mistake stems from.

Step 1: I find the root of the denominator: $\{-\frac{i}{2}, \frac{i}{2},2\}$. I remark that the moduli of $\frac{i}{2}$ and $-\frac{i}{2}$ are inferior to 1, thus two singularities lie in the contour.

Step 2: I define a function $f_1$ and a contour $C_1$ centered in $z_1=-\frac{i}{2}$.
$$\forall z \in C_1, f_1(z)=(z-z_1)g(z)$$

So I have,

$$\forall z \in C_1 \backslash \{z_1\}, g(z)=\frac{f_1(z)}{z-z_1}$$

Step 3: I apply the Cauchy integral theorem:

$$\int_{C_1}g(z)dz=\int_{C_1}\frac{f_1(z)}{z-z_1}dz=2\pi i f_1(z_1)=4\pi i$$

Step 4: I apply the same reasoning and find again $4\pi i$. (Intuitively, by symmetry, it seems that both integrals should be equal. But I may be wrong).

Step 5: I sum the two results obtained at steps 4 and 5, and get $8\pi i$.

I double-checked WolframAlpha and still finds $-\pi$.


问:
Is this combinatorial equality correct?
I am trying to prove this equality, where $P_k$ only depends on $k$.

$$p_m - \sum_{k=m+1}^n\binom{k}{m} \sum_{h=k}^{n} (-1)^{h-k}\binom{h}{k}p_{h} = \sum_{k=m}^{n} (-1)^{k-m}\binom{k}{m}p_{k}$$

But I am really stuck!
My attempts involve trying to switch the summation, trying to prove that $$p_m - \sum_{k=m+1}^n\binom{k}{m} \sum_{h=k}^{n} (-1)^{h-k}\binom{h}{k}p_{h} = \sum_{k=m}^n\binom{k}{m} \sum_{h=k}^{n} (-1)^{h-k}\binom{h}{k}p_{h}$$, but I have not been able to get any headways.

I think an identity regarding $\sum_{h=k}^{n} (-1)^{h-k}\binom{h}{k}p_{h}$ would help.

答:
$$\begin{align*}
&\sum_{k=m+1}^n\binom{k}m\sum_{h=k}^n(-1)^{h-k}\binom{h}kp_h+\sum_{k=m}^n(-1)^{k-m}\binom{k}mp_k\\
&\qquad=\sum_{k=m}^n\binom{k}m\sum_{h=k}^n(-1)^{h-k}\binom{h}kp_h\\
&\qquad=\sum_{m\le k\le h\le n}(-1)^{h-k}\binom{h}k\binom{k}mp_h\\
&\qquad=\sum_{h=m}^np_h\sum_{k=m}^h(-1)^{h-k}\binom{h}k\binom{k}m\\
&\qquad=\sum_{h=m}^np_h\binom{h}m\sum_{k=m}^h(-1)^{h-k}\binom{h-m}{k-m}\\
&\qquad=\sum_{h=m}^np_h\binom{h}m\sum_{k=0}^{h-m}(-1)^{h-m-k}\binom{h-m}k\\
&\qquad=\sum_{h=m}^np_h\binom{h}m\sum_{k=0}^{h-m}(-1)^{h-m-k}\binom{h-m}{h-m-k}\\
&\qquad=\sum_{h=m}^np_h\binom{h}m\sum_{k=0}^{h-m}(-1)^k\binom{h-m}k\\
&\qquad=\sum_{h=m}^np_h\binom{h}m(1-1)^{h-m}\\
&\qquad=p_m\,,
\end{align*}$$

since $0^0=1$ and $0^\ell=0$ for $\ell\ge 1$.

Title: Is this combinatorial equality correct?


问:
Frequency determination from Dimension analysis
the time averaged total energy, $\bar E$,
has the following $\varepsilon$ expansion in $D$ dimension:
\begin{equation}
\bar{E}=\varepsilon^{2-D}\frac{E_0}{2\lambda}+ \varepsilon^{4-D}E_1
\end{equation}
 from the above equation how can we write
\begin{equation}
\omega_{\rm m}^2=
1-\frac{1}{2\lambda}\frac{(D-2)E_0}{(4-D)E_1}\,.
\end{equation}
where $$\omega_{\rm m}^2=1-\varepsilon_{\rm m}^2,$$  and $\lambda$ is a constant factor.
m denotes that angular frequency $\omega $ is minimum.

Can someone explain me how equation (2) can be written from equation
  (1)
  If you feel problem or need more information then please comment.
  thanks in advance.
  for more info see here in equation (39 and 40)

答:
Explaining Antoine's suggestion:

Differentiate $\bar E$ with respect to $\epsilon$ to find its optimal value. You get
$(2-D)\epsilon^{1-D} E_0/(2\lambda)+(4-D)\epsilon^{3-D}E_1=0$. Factor $\epsilon^{1-D}$ from that to get
$(2-D)E_0/2\lambda+(4-D)\epsilon^2E_1=0$. Now solve for $\epsilon^2$ to get $\epsilon^2=-E_0/2\lambda E_1(2-D)/(4-D)$. Then $1-\epsilon^2=1+E_0/2\lambda E_1(2-D)/(4-D)$ which is ame as your final answer.

ContentLicense: CC BY-SA 3.0


问:
Intersection of normal extensions is a normal extension
Let's assume that we have two normal extensions $L_1$ and $L_2$ which means that $L_1=K(A_1)$ and $L_2=K(A_2)$, where $A_1$ and $A_2$ are sets of roots of proper polynomials. How can I show that intersection of $L_1$ and $L_2$ is also normal extension? Is it clear because of the fact that $L_1 \cap L_2=K(A_1 \cap A_2) $?
I don't know if it's sufficient for the proof.

答:
No, it is not even correct that $L_1\cap L_2=K(A_1\cap A_2)$.

You can prove the claim as follows: Take any irreducible Polynomial in $K[t]$ that has a root in $L_1\cap L_2$. By normality of $L_1,L_2$ it will split into linear factors over $L_1,L_2$. By unique factorization these factorizations coincide and are therefore already in $L_1\cap L_2$.


问:
Find the length of the vector f(x) = x in L[0, 1]
Find the length of the vector $f(x) = x$ in $L_2[0, 1]$

Is the solution $L=\int\sqrt{(1+(f'(x))^2dx)}$ hence $L=\sqrt{2}$ ?

Or am I completly wrong?

答:
The equation $L = \int_I \sqrt{1+{f'}^2}$ is the euclidean length of the parametrised curve $(x,f(x))_{x\in I}$. Here, it seems that the question is not about the length of the curve as a geometric object in the plane, but would rather be the norm of the vector $\mathrm{id}$ in the abstract normed linear space $\left(L^2([0,1]), \|.\|_2\right)$ where
$$
\|f\|_2 = \sqrt{\int_0^1 |f|^2}
$$
Hence, in this case
$$
\|\mathrm{id}\|_2 = \sqrt{\int_0^1 x^2 \mathrm{d}x}
$$
This will lead to another value than $\sqrt{2}$.


问:
Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$
Find surface area of a figure defined by: $x^2 + y^2 + z^2 \leq 1$ , $x^2 + y^2 \leq y $, $y + z \geq 1$.

I drew the picture and found two surfaces: $S_1$ and $S_2$,

where $S_1$ is $\{ 0 \leq y \leq 1 \land -\sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \leq x \leq \sqrt{(\frac{1}{2})^2 - (y-\frac{1}{2})^2} \} \cap y + z = 1 \} $

and

$S_2$ is $\{ x^2 + y^2 + z^2 \leq 1 \cap x^2 + y^2 \leq y \} $ .

I wonder if this is correct:

$P(S_1) = \begin{gather*}
    \iint_D \sqrt{1+ 0^2 +(-1)^2}\,dx\,dy = ... = \frac{\sqrt{2}}{4}\pi
\end{gather*}
$

and

$P(S_2) = \begin{gather*}
    \iint_D \sqrt{1+ \frac{x^2}{1-x^2 - y^2} +  \frac{y^2}{1-x^2 - y^2}} \,dx\,dy = \cdots =\pi
\end{gather*}
$

In both cases $D$ is $\{ -\frac{1}{2} \leq x \leq  \frac{1}{2} $ and $ \frac{1}{2} - \sqrt{(\frac{1}{2})^2 - x^2}  \leq y \leq  \frac{1}{2} + \sqrt{(\frac{1}{2})^2 - x^2}\}$

Could someone please say if the idea is good, and if it isn't, what other approach should I take?
I am most familiar with this one.
Thanks in advance.

**EDIT **
There was a mistake in text above, I've corrected it.

答:
In cylindrical coordinates, the cylinder $x^2 + y^2 = y \ $ is

$r^2 = r \sin \theta \implies r = \sin \theta, \ 0 \leq \theta \leq \pi$.

This is a cylinder of radius $\frac{1}{2}$ with center at $(0,\frac{1}{2})$.

Now parametrization of cylinder is

$r(\theta, z) = (r\cos\theta, r\sin\theta, z) = (\frac{1}{2} \sin 2\theta, \sin^2\theta, z)$

$|r'_\theta \times r'_z| = 1$

Also note that at the intersection of the cylinder and the plane $y +z = 1$,

$\sin^2\theta + z = 1 \implies z = \cos^2\theta$

Also the intersection of the cylinder and the sphere $ \ \sin^2\theta + z^2 = 1 \implies z = \pm \cos\theta$

So the integral to find surface area of the cylindrical surface satisfying the given condition is

$\displaystyle 2 \int_{0}^{\pi/2} \int_{\cos^2\theta}^{\cos\theta} dz \ d\theta$

You may have to find for other surfaces similarly. The integral for other surfaces based on their parametrization are given by,

For spherical surface: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \frac{r}{\sqrt{1-r^2}} \ dr \ d\theta$

For planar bottom: $\displaystyle \int_0^{\pi} \int_0^{\sin\theta} \sqrt2 \ r \ dr \ d\theta$


问:
Using Cavalieri to derive a formula for $n$-dim $L^p$ ball
let $p>0$ and $B_p^n:=\{x\in \mathbb{R}^n : |x_1|^p+\cdots + |x_n|^p\leq 1\}$. Show that $B_p^n$ is Borel-measurable with $$\lambda^n(B_p^n)=\frac{(2\Gamma(1+\frac{1}{p}))^n}{\Gamma(1+\frac{n}{p})}$$

My idea:

I have a hint saying that I have to look at the integral $\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)$. I know that I could just use power laws and fubini to achieve a gaussian integral that I could compute. But I have to use Cavalieri's principle.

I started like this: We have:$$\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)=\int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\})\, \mathrm{d}t$$

Now I have to apply Cavalieri's principle to $\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\}$.

we have $\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\}=\{x\in \mathbb{R}^n : ||x||_p\leq (\log(t))^{1/p}\}\subset \mathbb{R}\times  \mathbb{R}^{n-1}$.

For all $y\in \mathbb{R}^{n-1}$ we cut $A_y=\{x\in \mathbb{R} : ||x||_p\leq \log(t)^{1/p}\}$. Now, we have:$$\lambda^n(\{x\in \mathbb{R}^n : e^{-(|x_1|^p+\dots+|x_n|^p)}\geq t\})=\int \limits_{\mathbb{R}^{n-1}}\lambda(A_y)\, \mathrm{d}y$$ How can I compute this integral?

答:
By Cavalieri, we can write
$$\int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x)=\int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\})\, \mathrm{d}t. $$
Notice that the set we are measuring on the right is empty for $t > 1$ and is just the scaled up unit $L^p$-ball for $0 \leq t \leq 1$,
$$ \{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\} = \log(1/t)^\frac1p \cdot B_p^n. $$
Therefore, by the scaling property of the Lebesgue measure ($\lambda^n (a B) = a^n \lambda^n (B)$), and recalling a well-known integral formula for the Gamma function, we write
$$ \int \limits_{0}^{\infty}\lambda^n(\{x\in \mathbb{R}^n : ||x||_p^p \leq \log (1/t)\})\, \mathrm{d}t = \lambda^n(B_p^n) \cdot \int_0^1 \log(1/t)^\frac{n}{p} \mathrm d t = \lambda^n (B_p^n) \cdot  \Gamma(1 + n/p). $$
At this point it is reasonable to expect
$$ \int_{\mathbb{R}^n} e^{-(|x_1|^p+\dots+|x_n|^p)}\mathrm{d}\lambda^n(x) = (2 \Gamma (1 + 1/p))^n $$
which completes the derivation. This essentially is a generalization of the argument used to find the volume of the standard $L^2$-ball in $n$-dimensions, though rather than partitioning the domain of integration explicitly by $n-1$-dimensional spheres and radii we do it instead writing the spheres as level sets.


问:
Difficult integral $\int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx$
As part of breaking up a larger integral to solve, I have ended up with the following integral:
\begin{equation*} \int_{\nu+\epsilon}^{2\nu}\frac{x^{2}}{(x-\nu)^{4}+b}dx \end{equation*}
where $\nu,b>0$ are fixed constants and $\epsilon\in (0,\nu)$ is small as I want. I have tried everything I can think of (i.e. various substitutions, breaking the integral apart even more, wolframalpha, etc..) Maybe I have missed something or maybe there is some way to do this with partial fraction decomposition, but I am not sure.

I have a solution to this integral if I assume instead that $b=0$, but in this case letting $\epsilon\to 0$ makes the answer blow up and I need it to stay bounded. In any case, I would be very grateful for any help/hints in solving the integral if $b>0$.

Edit: If an explicit solution to this integral is difficult, it would suffice for my purposes to show that the integral is bounded above by a positive constant that is independent of $\nu$ and $\epsilon$.

答:
Translate the integral

$$\int_\epsilon^\nu \frac{x^2+\nu^2}{x^4+b}dx+\int_\epsilon^\nu \frac{2x\nu}{x^4+b}dx$$

The second integral evaluates to

$$\frac{\nu}{\sqrt{b}}\left(\tan^{-1}\left(\frac{\nu^2}{\sqrt{b}}\right)-\tan^{-1}\left(\frac{\epsilon^2}{\sqrt{b}}\right)\right) = \frac{\nu}{\sqrt{b}}\tan^{-1}\left(\frac{\sqrt{b}(\nu^2-\epsilon^2)}{b+\nu^2\epsilon^2}\right)$$

For the second use the substitution $x = b^{\frac{1}{4}}t$, factor the denominator as $$t^4+1 = (t^2-\sqrt{2}t+1)(t^2+\sqrt{2}t+1)$$ and decompose with partial fractions. Can you take it from here?


答:
$$\text{Let $y^2 = f(x)=\sin^2\left(\frac{2x+1}{3}\right)$}$$

So $y \in [-1,1]$ and
$y=\sin\left(\dfrac{2x+1}{3}\right)$

where $\dfrac{2x+1}{3} = 2n\pi + \theta $ and $-\pi \le \theta \le \pi$
for some $n \in \mathbb Z$.

We compute

\begin{align}
   \theta &= \arcsin y \\
   \dfrac{2x+1}{3} &= 2n \pi + \arcsin y \\
   2x &= 6n \pi - 1 + 3 \arcsin y \\
   x &= 3n \pi - \dfrac 12 + \dfrac 32 \arcsin y \\
\end{align}

Title: How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain.

Body:
$f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$

I know I have to switch the $f(x)$ and the $y$:

$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$

To find the domain:

$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$

$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$

$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$

$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$

The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$

But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.

Thanks


问:
The direct sum $\oplus$ versus the cartesian product $\times$
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form
$$
\mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t}
$$
where $\mathbb{Z}^n$ now means $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$.

From an intuition perspective, and in the sense of sets, is this more or less the same as $\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}$? (Bear in mind I am normally using these ideas in relation to homology groups.)

答:
For abelian groups, the direct sum is a special case of the categorical notion of coproduct. This means, among other things, that if $A, B, C$ are abelian groups, there is a natural isomorphism

$$\text{Hom}(A \oplus B, C) \cong \text{Hom}(A, C) \times \text{Hom}(B, C)$$

where $\text{Hom}(X, Y)$ is the set of homomorphisms $X \to Y$. This is not hard to see. The direct product is a special case of the categorical notion of product. This means, among other things, that if $A, B, C$ are abelian groups, there is a natural isomorphism

$$\text{Hom}(A, B \times C) \cong \text{Hom}(A, B) \times \text{Hom}(A, C).$$

This is also not hard to see. The product and coproduct naturally make sense for an arbitrary number of abelian groups, and then it turns out that in order to preserve the universal properties above the direct sum ends up being the subgroup of the direct product that Pete Clark describes.

There are at least three confusing things about this setup, at least to me.

The product and coproduct agree for finitely many factors. When this happens we say the category has finite biproducts. This is emphatically not true, for example, in the category of sets, where the product is the Cartesian product and the coproduct is the disjoint union.

The coproduct in the category of groups, as opposed to the category of abelian groups, is not the direct sum. It is the free product, which is why the free product is relevant to, for example, the Seifert-van Kampen theorem (which is abstractly a theorem about how taking fundamental groups preserves certain colimits, of which coproducts are an example).

It is still possible to define the direct sum of infinitely many groups (not necessarily abelian), and this construction has neither of the universal properties above.

In any case, the fact that the product and coproduct are abstractly different should hint to you that they appear in different situations, even though they look the same. You can sometimes tell the difference by asking what happens when you extend a finite construction to infinitely many things.

For example, you mentioned homology groups. Consider the wedge $X$ of countably many circles. Then I claim that $H_1(X) = \bigoplus_{i=1}^{\infty} \mathbb{Z}$ but that $H^1(X) = \prod_{i=1}^{\infty} \mathbb{Z}$.


问:
Dimensions analysis in Differential equation
Differential equation of solitary wave oscillons  is defined by,
$$ \Delta S -S +S^3=0 $$
How can we write this equation as,
\begin{equation}
\langle(\vec{\nabla}S)^2\rangle+\langle S^2\rangle-\langle S^4\rangle=0 \tag{1}
\end{equation}
where $\langle f\rangle:=\int d^Dx f(x)$. Furthermore, another virial identity
can be found
from the scaling transformation ($\vec{x}\to \mu \vec{x}$)
by extremizing the scaled ($\vec{x}\to\mu \vec{x}$)
of the action corresponding to
$ \int d^Dx[(\vec{\nabla}S)^2+S^2-S^4/2]$ ( why $\frac{-S^4}{2}$):
\begin{equation}
(D-2)\langle(\vec{\nabla}S)^2\rangle+D\langle S^2\rangle-\frac{D}{2}\langle S^4\rangle=0 \tag{2}
\end{equation}
From Eqs. (1) and (2) one immediately finds
\begin{equation}
2\langle S^2\rangle+\frac{1}{2}(D-4)\langle S^4\rangle=0\,,
\end{equation}
which equality can only be satisfied if $D<4$.
D= Refers dimension.

I didn't understand the equation (2). Can someone do elaborately?

If you have any Query then ask me please.
Thanks in advance.

I think this article will give you enough information: link check equations (21, 41, 42 43)

答:
The first part is simply multiplication by $S$, adding an integral sign,  integration by parts for the first term.

For the second part consider a variation $\hat S(x;\mu)=S(x/\mu)$. Then do a change of variables $x=\mu y$. If the action is stationary under this spatial rescaling of $S$ then $d/d\mu$ at $\mu=1$ vanishes. Since the first term scales to $\mu^{D-2}$ with 2 from the derivative and $D$ from the Jacobian factor, this derivative gives a factor of $D-2$ as shown in the expression given. The rest are just Jacobian factors.

Edit: For example,
$$ \left< (\nabla\hat  S)^2\right> =  \int d^Dx (\nabla_x S(x/\mu))^2 = \int \mu^D d^Dy (\mu^{-1} \nabla_y S(y))^2 = \mu^{D-2} \left< (\nabla S)^2\right>$$
The derivative at $\mu=1$ is exactly $(D-2) \left< (\nabla S)^2\right>$.

To clarify the first part, first multiply the equation by $-S$. Then integrate over all space. This gives the last two terms in the next line. Then for the first one, notice that $-S\nabla\cdot\nabla S$ can be integrated by parts using the divergence theorem. This shifts the $\nabla$ to the other term and changes the sign giving the shown result. If you prefer $\nabla\cdot(S\nabla S)$ can be expanded using the product rule and then the result substituted into the integral.

The factor of a half in the action arises from wanting the Euler-Lagrange equation for this action to be the given equation. The EL eqn contains $\partial L/\partial S$ so $L=S^2 + \alpha S^4$ results in a term $2S+ 4\alpha S^3$ so to get this agreeing with the eqn, up to an overall factor, choose $\alpha=-1/2$.

The deduction of (2) follows as I discuss above; rescale by $\mu$, then differentiate at $\mu=1$.


问:
Are $\mathbb Z\times \mathbb Z/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$ as groups?
Is $(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z)\simeq \mathbb Z$,

where $4\mathbb Z\times 5 \mathbb Z$ is given as a subgroup of $\mathbb Z\times \mathbb Z$?

My feelings says that I need to find well defined $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ which has a kernel $4\mathbb Z\times 5\mathbb Z$

So using first isomorphism theorem I can show the isomorphism.

I could not find the map and how can I show it in different way?

答:
By definition of the quotients:

\begin{alignat}{1}
(\mathbb Z\times \mathbb Z)/(4\mathbb Z\times 5 \mathbb Z) &= \{(m,n)+4\Bbb Z\times5\Bbb Z\mid m,n\in\Bbb Z\} \\
&= \{(m+4\Bbb Z,n+5\Bbb Z)\mid m,n\in\Bbb Z\} \\
&= \{(k,l)\mid k\in\Bbb Z/4\Bbb Z,l\in\Bbb Z/5\Bbb Z\} \\
&= \Bbb Z/4\Bbb Z \times\Bbb Z/5\Bbb Z \\
\end{alignat}

So, your quotient is precisely (not just isomorphic to) $\Bbb Z/4 \Bbb Z\times\Bbb Z/5\Bbb Z $, which is not isomorphic to $\Bbb Z$, if only for reasons of cardinality.


问:
Limit of curvilinear integral
Let $I(R)=\int_{\gamma_R}(1+x_1^2+x_2^2)^{-1/2}\cos(x_2)ds$, where $\gamma_R=\{(x_1,x_2):x_1^2+x_2^2=R^2,x_1>0\}$. Compute $\lim_{R \to +\infty}$$I(R)$. I tried to solve it in this way: I use parameterization $x_1=R\cos\phi, x_2=R\sin\phi, \phi \in (-\pi/2,\pi/2)$ and I get: \begin{equation} I(R)=\frac{R}{(1+R^2)^{1/2}}\int_{-\pi/2}^{\pi/2}\cos(R\sin\phi)d\phi
=\frac{2R}{(1+R^2)^{1/2}}\int_{0}^{\pi/2}\cos(R\sin\phi)d\phi   \end{equation}
Now for $R \to +\infty$ :
\begin{equation}\frac{2R}{(1+R^2)^{1/2}}\sim2 \end{equation}
How could I study the integral to conclude the exercise?

Title: Limit of curvilinear integral


问:
Show that Jacobi Method does not converge for $ \frac{1}{2}\lt a \lt 1 $ in the given matrix
Show that

\begin{align}
A =
\left[ {\begin{array}{cc}
1 & a & a \\
a & 1 & a \\
a & a & 1
\end{array} } \right]
\end{align}

is a symmetric positive definite for $ \frac{1}{2}\lt a \lt 1 $, but that the Jacobi Method does not converge for $\frac{1}{2}\lt a \lt 1 $.

$$\textbf{MY ATTEMPT} $$
I know that since $A$ is SDP, $det(A) \gt 0$. Using the splitting $A=D-L-U$. I have that
\begin{align} D^{-1}(L+U) = \left[ {\begin{array}{cc}
0 & -a & -a \\
-a & 0 & -a \\
-a & -a & 0
\end{array} } \right]
\end{align}
Therefore, if the $p(D^{-1}(L+U)) \lt 1$, matrix $A$ is convergent with Jacobi Method. Otherwise, it is not.

Thus, I have the following characteristic polynomial from which I intent to obtain the eigenvalues and conclude whether the matrix is convergent with Jacobi method or not.
$$ - \lambda^3 + 3a^2 \lambda - 2a^3 = 0 $$

So my questions are:

(1) Is my approach to the question correct ? If yes

(2) How do I solve for the eigenvalues from the above cubic equation.

Please I need help friends! Thanks.

答:
This looks like a viable approach. Now you can get one eigenvalue fairly easily by guess-and-check (this might be easier by thinking about when $D^{-1}(L+U)-\lambda I$ will be singular rather than looking at the characteristic polynomial), after which you can long-divide to find the other two eigenvalues.

If you are still stuck, here is a slightly more specific hint: the eigenvalue that you can find by guess-and-check involves $a$...


问:
Two inequalities with exponential function
I am interested in seeing proofs for the following two statements:
1.)
$\forall x\in \mathbb{C}, |x|\le 1:$
$$|\operatorname{exp}(x)-1|\leq\tfrac{|x|}{1-|x|}$$

2.)
$\forall x\in \mathbb{C}, |x|\leq \tfrac{1}{2}:$
$$|\operatorname{exp}(x)-1-x|\leq |x|^2$$

Maybe one can show the second one with the help of the first one?

Body:
(To follow leoli1's comment,) note that for $|z|\le 1$:
$$
|e^z-1|\le \sum_{n=1}^\infty \frac{|z|^n}{n!}\le \sum_{n=1}^\infty|z|^n=\frac{|z|}{1-|z|}
$$

and for $|z|\le\frac12$:
$$
{|e^z-1-z|}
\le \sum_{n=2}^\infty \frac{|z|^n}{n!}
\le \sum_{n=2}^\infty\frac12|{z}|^n=\frac{|z|^2}{2(1-|z|)}\le|z|^2
$$


问:
Understanding the Steiner triple system
I am learning what Steiner systems are and I stumboled upon this Wolfram Mathworld source: https://mathworld.wolfram.com/SteinerTripleSystem.html.

I understand that $k = 3$ means that there are exactly 3 points(elements of set $X$) on each block. I also understand that the first parameter means how many points there are altogether. However I do not understand what is $\lambda$ that is the third parameter of $S$ in $S (v)=S(v,k=3,\lambda = 1)$

答:
These are some of the parameters of a block design. Quoting Wikipedia:

$v$ is the number of points (elements of $X$).

$b$, not appearing in the article you're quoting, is the number of blocks: the subsets of $X$ we're taking.

$r$, also not appearing in the article, is the number of blocks containing a given point.

$k$ is the number of points in a block (for a Steiner triple system, $k=3$).

$\lambda$ is the number of blocks containing...

...any $t$ distinct points. In the case of a Steiner triple system, $\lambda=1$ and $t=2$: for any two distinct points, there is exactly one subset containing them.

Title: Understanding the Steiner triple system


问:
The subspace $C$ of the space $(X,d)$ is compact
Let:

$(X,d)$ - complete metric space.

$K_1,K_2,\dots $ - compact sets in the space $(X,d)$ (sequence $K_1,K_2,\dots$ need not be descending)

$C_n=\{x\in X: \text{dist} (x,K_n)\le \frac 1n \}$ for  $n=1,2,\dots$

$C=\bigcap\limits_{n=1}^{\infty} C_{n}\subset X$

Claim: The subspace $C$ of the space  $(X,d)$ is compact

My attempts:

We want to show that $\bigcap\limits_{n=1}^{\infty} C_{n}$ is compact. Let's establish $V=\{ \bigcup\limits_{\alpha}:\alpha \in A\}$ - cover of $\bigcap\limits_{n=1}^{\infty} C_{n}$. From this $V$ we want to choose finite subcover for $\bigcap\limits_{n=1}^{\infty} C_{n}$. I think that we can use the total limitation of space $C$ with metric $d$ truncated to $C$. However I don't have idea how I can do it.

答:
We can, for free, arrange for $C_n$ to be a decreasing sequence without changing $\cap_n C_n$. Indeed, let (new) $C_n$ be defined as
$$
C_n:=\{x \in X \colon \: d(x,K_j)\leq \frac{1}{j}, \quad \text{for all $j\leq n$.}\}
$$
Then $C_{n} \supset C_{n+1}$ and $\cap_n C_n$ is equal to the original $C$ defined in the OP.

To show that $C$ is compact it suffices to show each $C_n$ is compact. First, let's see why $C_n$ is closed. If $K$ is a subset of $X$, then the function $X \to R$ defined by $x \mapsto d(x,K)$ is continuous -- an easy triangle inequality shows that it is in fact $1$-Lipschitz. Therefore,
$$
X \ni x \mapsto (d(x,K_1),d(x,K_2),\cdots,d(x,K_n)) \in R^n
$$
is continuous. Now, $C_n$ is the pre-image under this map of a closed box in $R^n$.

TBC


问:
Does $f(x)=x^{2}\sin\left(\frac{1}{x^2}\right)$ satisfy the relation $f(x)+f(y)−2f\left(\frac{x+y}{2}\right)=O\left(\left|x−y\right|^2\right)$?
Does $f(x)=x^{2}\sin\left(\frac{1}{x^2}\right)$,  $x\in(0,1)$ satisfy the relation $f(x)+f(y)−2f\left(\frac{x+y}{2}\right)=O\left(\left|x−y\right|^2\right)$?</p>

答:
NO. Let $x=t+d, y=t-d$, then we shall consider $f(t+d)+f(t-d)-2f(t)=O(d^2)$.

Note that
$$\lim_{d\to 0}\frac{f(t+d)+f(t-d)-2f(t)}{d^2}=f''(t),$$
so basically we need $f''$ is bounded in $(0,1)$, by calculation you will find this is not the case.


问:
analytical approximation of the solution of a PDE
Is there any method to obtain an analytical approximation (probably through a composition of exponential functions) of the solution of a differential equation?? In other words, is is possible to obtain an approximate Green's function of the PDE?

the equation in question:

$$\nabla ^ 2 \phi(\mathbf{x}) = M \frac {\phi(\mathbf{x}) }{\phi(\mathbf{x}) + C}$$

Where M and C are bounded constants of the same order as $\phi$. Additionally $\phi$ is bounded between 0 and 1.
The equation is subject to Dirichlet boundary conditions (although that is not really relevant. Could be Neuman or Robin)

Given the non-linear term on the right hand side, the equation does not have an analytical solution (to my knowledge). The obtention of the solution through finite differences (or other numerical methods that do not provide a function as the solution) should be avoided since the goal is to obtain a function of $\mathbf{x}$ that approximates the solution.

My idea is to find a method similar to how the polinomial expansion of Taylor approximates a function around a point. But for the solution of the PDE.

答:
Here is an outline of a numerical approach you can try.

You can search for a power series representation in combination with the convolution theorem of Fourier analysis.

Rewrite like this
$$\nabla^2 \phi({\bf x}) \cdot (\phi({\bf x}) + C) = M \phi({\bf x})$$

Right hand side is M times power series representation of $\phi$.

Left hand side left factor is a set of linear combinations of coefficients for this. This can be easily realized as each monomial term of a multidimensional polynomial differentiates to another. Very sparsely.

Left hand side right factor is just +C on the constant term.

Now the multiplication of power series corresponds to a multidimensional convolution of coefficients. Here is where the multidimensional Fourier transform comes into play. We Fourier transform $\nabla^2 \phi({\bf x})$ and $(\phi({\bf x})+C)$ separately and multiply each coefficient. This should equal the fourier transform of $M\phi(\bf {x})$


问:
Interior and Closure of a set in a Cofinite Topology on R
Considering the Cofinite Topology on Real Line

My question is what is the Closure and Interior of [a,b] with respect to this topology.

Actually I am a beginner. If you explain the underlying concept along with the answer to me then that would be very helpful.

Thank you.

答: If $b>a$ then $[a,b]$ contains infinitely many elements, hence the only closed subset (in the cofinite topology) containing $[a,b]$ is $\Bbb R$ itself; it follows that the closure of $[a,b]$ is $\Bbb R$. Similarly for the interior. As $\Bbb R\setminus[a,b]$ contains infinitely many elements, the only open subset contained in $[a,b]$ is the empty set, hence $[a,b]^\circ=\emptyset$.


问:
Solution to differential equation with boundary conditions
While studying transport physics, I encountered a problem, where we have to solve
$r \frac{dT}{dr} = C_1$, so the solution is $$T = C_1 \ln(r) + C_2.$$

Now with the imposed boundary conditions $T(r=R_2)= T_2$ and $T(r=R_1)= T_1$, the solution should be as follows:

$$T\left(R_{1} \leq r \leq R_{2}\right)=T_{1}+\left(T_{2}-T_{1}\right) \frac{\ln \left(r / R_{1}\right)}{\ln \left(R_{2} / R_{1}\right)}$$

How do you obtain this answer? I've tried integrating from $R_1$ to $r$ and then from $r$ to $R_2$ but that didn't work fully

Body:
Given the general solution $$T = C_1 \ln(r) + C_2$$ and boundary conditions
\begin{align}
T(R_{1}) &= T_{1}\\
T(R_{2}) &=T_{2}
\end{align}
we can plug in the boundary conditions to get:
\begin{align}
T_{1} &= C_{1}\ln(R_{1}) + C_{2}\\
T_{2} &= C_{1}\ln(R_{2}) + C_{2}.
\end{align}

We now have a system of two equations with two unknowns, and there are several techniques we could use to solve it.  One such process would be to subtract $T_{1} - T_{2}$:
$$T_{1} - T_{2} = C_{1}\ln(R_{1}) - C_{1}\ln(R_{2}) \implies C_{1} = \frac{T_{1} - T_{2}}{\ln(R_{1}) - \ln(R_{2})} = \frac{T_{1} - T_{2}}{\ln(R_{1}/R_{2})}.$$  Then, we can write:
$$T_{1} = \frac{T_{1} - T_{2}}{\ln(R_{1}/R_{2})} \ln(R_{1}) + C_{2} \implies C_{2} = T_{1} -\frac{T_{1} - T_{2}}{\ln(R_{1}/R_{2})} \ln(R_{1}).$$  This gives us
\begin{align}
T &= C_{1}\ln(r) + C_{2}\\
&=\frac{T_{1} - T_{2}}{\ln(R_{1}/R_{2})}\ln(r) + T_{1} -\frac{T_{1} - T_{2}}{\ln(R_{1}/R_{2})} \ln(R_{1})\\
&=T_{1}+\frac{T_{1}-T_{2}}{\ln(R_{1}/R_{2})}\left(\ln(r) - \ln(R_{1})\right)\\
&=T_{1} + (T_{1} - T_{2})\frac{\ln(r/R_{1})}{\ln(R_{1}/R_{2})}\\
&=T_{1}+\left(T_{2}-T_{1}\right) \frac{\ln \left(r / R_{1}\right)}{\ln \left(R_{2} / R_{1}\right)}.
\end{align}


问:
Binary vector counting
What is the count of all vectors $ \hat a \in B^{11}$, for which $|\hat a|$  = 5 and $ \mathcal v( \hat a ) > 2^5$?

where

$ B^{11} $ - all binary vectors with length 11

$|\hat a|$ - the sum of all binary vector coordinates (sum of 1s)

$ \mathcal v( \hat a )$ - decimal representation of the binary vector $ \hat a $

答: Hint: Compute the number of $\hat a$ with $|\hat a|=5$ and $v(\hat a)\le 2^5$; which positions of $\hat a$ have to be zero for the latter condition to be satisfied?

Title: Binary vector counting


问:
Is $\dim U=\dim V-1$ if U is maximal with respect to inclusion?
Does a subspace U of $V$ that is maximal with respect to inclusion have dimension $\dim U=\dim V-1$?

答: Let $v_0 \in V$ with $v_0 \notin U$. Then, by maximality of $U$, $$V = \operatorname{span}(U \cup\{v_0\}) = U+\langle v_0 \rangle,$$ where $\langle v_0 \rangle$ is the subspace spanned by $v_0$. In fact, $V = U \oplus \langle v_0 \rangle$ (why?) and then $\dim V = \dim U + 1$.


问:
Prove that $\int_{y=0}^{\infty} \int_{x:g(x) > y} f(x) \,dx \,dy = \int_{x:g(x)>0} \int_{y=0} ^{y=g(x)}\,dy$
Prove that $\int_{y=0}^\infty \int_{x:g(x) > y}  f(x) \,dx \,dy = \int_{x:g(x)>0} \int_{y=0}^{y=g(x)}\,dy ~f(x) \,dx$ where $f,g$ are continuous functions

Attempt: Let $I$ = $\int_{y=0}^\infty \int_{x:g(x) > y}  f(x) \,dx \,dy $. Now, we need to interchange the order of integration. The inner integral
$$\int_{x:g(x) > y}  f(x) \,dx$$

provides us a clue with its limits confined to $\{x:g(x) > y)\} ~$ i.e $~\{x: y<g(x) \} \implies $limits of $y$ will vary from $0$ to $g(x)$. Since the outer integral is over positive $y$ and $g(x) > y \implies g(x) > 0 ~\forall~x$ in our question.

Hence, the outer limit will be over $x : g(x) > 0$.

Although I have tried to prove the question in the above manner, this does not seem rigorous and solid proof in any way. Could someone please guide me towards more convincing proof? Will my argument serve as a proof?

答:
I'm not sure if this is any help to you, but in a similar situation I was told to use Indicator-functions, so what you are doing would roughly look like:

$$\int_{y=0}^{\infty} \int_{\{g(x)>y\}} f(x) \,dx \,dy = \int_{\mathbb{R}}\int_{\mathbb{R}} \mathbb{1}_{\{0<y\}}\cdot \mathbb{1}_{\{y < g(x)\}} \cdot f(x) \,dx \,dy \\= \int_{\mathbb{R}}\int_{\mathbb{R}} \mathbb{1}_{\{0<y<g(x)\}}\cdot \mathbb{1}_{\{0 < g(x)\}} \cdot f(x) \,dx \,dy $$

The thing you have check is that the product of the indicator functions before and after the second equality are actually the same function. Now if you change the order of integration, if you can, then you can pull out the second indicator function from the now inner integral since it doesn't depend on y, so you get:

$$\int_{\mathbb{R}} \mathbb{1}_{\{0 < g(x)\}} \cdot f(x) \int_{\mathbb{R}} \mathbb{1}_{\{0< y < g(x)\}} \cdot\,dy \,dx = \int_{\{g(x)>0\}}f(x) \int_{y = 0}^{g(x)}\, dy \,dx$$

I'm not sure though if this is any more rigorous than what you did, since it boils down to the same logic.


问:
How can you know you are performing a legal operation?
There are many examples in math where you have fake proofs of something because you do something extraneous that you were not allowed to do. For example:

Let $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots = \log 2,$$ but $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{5} + \dots = \frac{3}{2}\log 2$$ (rearranging the series). In elementary calculus, we are taught that we cannot rearrange series precisely for this reason: We may change the value that the series equals. Another example would be the equation

$$\large{x^{x^{x^{x^{\dots}}}}} = 2.$$ Then we use the property that it repeats to see that $x^2 = 2,$ which implies that $x = \sqrt{2}$ for positive $x$. But then, if we wish to solve $$\large{x^{x^{x^{x^{\dots}}}}} = 4,$$ then using the same trick we see that $$x^4 = 4 \implies \sqrt[4]{x^4} = \sqrt[4]{4} \implies x = \sqrt{2}.$$ Once again, we did something we are not allowed to do.

My question is therefore: how do we know that we aren't doing something illegal when exploring mathematics? When we try to solve problems, or discover patterns or learn, how can we tell when we do something extraneous?

答:
By knowing very well the rules of the game i.e. 1) the definitions, 2) the axioms, and 3) the theorems within a given math theory. A tiny change, or addition, or even misinterpretation of any of these... and a slip may happen leading to various paradoxes, false proofs, etc.

Math is not about syntactic, mechanical manipulation of mysterious symbols; there's deep meaning behind every symbol. Make sure you know the meaning of the symbol, only then use it.

E.g. when I see this

$\large{x^{x^{x^{x^{\dots}}}}}$

it is not immediately clear to me what it means.

So I would exercise extra care if I have to manipulate this expression.


问:
Regex with at least three even digits and odd length
As the title said, I need to make a regex for a number with at least three even digits and odd length.
As an example, the number 248 should match.
Thank you!</p>

答:
Let $e$ denote any even digit and $o$ any odd digit. Let

$$E=\big((e+o)(e+o)\big)^*$$

and

$$O=(e+o)E\,.$$

Then the expression

$$\begin{align*}
&E(eee+eeEe+eEee+eEeEe+eOeOe)E\\
&\qquad+O(eee+eeEe+eEee+eEeEe+eOeOe)O\\
&\qquad+E(eeOe+eOee+eOeEe+eEeOe)O\\
&\qquad+O(eeOe+eOee+eOeEe+eEeOe)E
\end{align*}$$

should do the trick.


问:
Do the elements of sample space constitute disjoint events?
Let, we be given a sample space $\Omega=\{x_1, x_2, x_3\}$. Now, from the axioms of probability we know $P(\Omega)=1$. Also, from the definition of union of sets, we know, $\Omega=\{x_1\}\cup\{x_1\}\cup\{x_1\}$. Thus, $P(\Omega) = P(\{x_1\}\cup\{x_1\}\cup\{x_1\})=1$. Now, my question is, does this imply $$P(\{x_1\})+P(\{x_2\})+P(\{x_3\})=1?$$

Also, if yes, does this extend to sample spaces with countably infinite number of elements?

答: That depends on what the event space of your probability space is. If your event space is just $\{\{\}, \{x_1\}, \{x_2, x_3\}, \{x_1, x_2, x_3\}\}$ then it does not even make sense to talk about $P(\{x_2\})$ and $P(\{x_3\})$ since the probability measure is a function from your event space (usually denoted $\mathcal{F}$) to $[0,1]$. Now if your event space consists of all subsets of $\Omega$ then you are correct. This is because one of the axioms of a probability space is that the probability measure is countably additive. This means that if you have a countable collection of pairwise disjoint events: $\{A_{i}\}_{i=1}^{\infty }\subseteq {\mathcal {F}} $, then $P(\bigcup _{i=1}^{\infty }A_{i})=\sum _{i=1}^{\infty }P(A_{i})$.


答:
In this topology the open sets are very big, so let's try to keep this in mind when we are computating. I will denote the closure of $[a,b]$ as $\overline{[a,b]}$

Claim $\overline{[a, b]} = \mathbb{R}$, clearly $[a,b] \subset \mathbb{R}$ so let $x \in \mathbb{R} \text{ and } U \text{ an open set such that} x \in U$, now as $U$ is open $\mathbb{R}-U$ is finite and as $[a,b]$ is infinite then $ [a,b] \not\subset \mathbb{R}-U \Rightarrow\exists y \in [a,b] - (\mathbb{R}-U)$ then $y \in [a,b] \cap U \Rightarrow [a,b] \cap U \neq \emptyset$ so by definition $x \in \overline{[a,b]}$ this is true for each $x \in \mathbb{R}$ so $\mathbb{R} \subseteq \overline{[a,b]}$ therefore we conclude $\overline{[a,b]} = \mathbb{R}$

With this in mind try to calculate the interior ¿Can $[a,b]$ contain any open set?

Spoiler: int($[a,b]$) = $\emptyset$

Title: Interior and Closure of a set in a Cofinite Topology on R

Body:
Considering the Cofinite Topology on Real Line

My question is what is the Closure and Interior of [a,b] with respect to this topology.

Actually I am a beginner. If you explain the underlying concept along with the answer to me then that would be very helpful.

Thank you.


问:
-9 modulo 12 is 3. Should get -9
STEP 1: Divide Dividend (-9) from Divisor (12).

```<code>-9 / 12 = -0.75
</code>```

Where,

0 is whole part.

75 is fractional part.

STEP 2: Whole part from the result of STEP 1 to be multiplied by Divisor (12).

```<code>0 * 12 = 0
</code>```

STEP 3: Subtract result from STEP 2 from Dividend (-9).

```<code>-9-0 = -9
</code>```

Whereas, The Actual answer is 3. Where am I missing?

答:
Following your algorithm...

You get $-0.75$ from the division. You are reading that as $0$ for the integer part and $.75$ for the rest. But the next lowest integer under $-0.75$ is $-1$. So if you want to follow this algorithm, you should be using $-1$ for the integer part (and $+0.25$ for the fractional part):

$$-9-(\stackrel{\downarrow}{-1})(12)=-9+12=3$$

Title: -9 modulo 12 is 3. Should get -9


问:
Finding the middle of n points
Sorry for my possible bad English, I have a problem that I spent a bit of time on and I have been blocked on it for a couple of hours, I'll try to translate it as best as I can:

Given an integer $n\ge3$, and given $A_1,\ldots,A_n$ points on a plane, under what condition(s) can we find points $P_1,\ldots,P_n$ such that $A_1$ is the midpoint of $[P_1,P_2]$, $A_2$ is the midpoint of $[P_2,P_3]$, ... and $A_n$ is the midpoint of $[P_n,P_1]$?

Please tell me if I wasn't clear enough in my translation, and thank you very much in advance!

答:
You can always do this when $n$ is odd, but if $n$ is even you can only do this if the respective centroids (or sums) of the even $A_i$ points and the odd $A_i$ points are coincident.

Suppose we have $P_1$ defined somehow. Then
\begin{align}
P_2 &= 2A_1-P_1 \\
P_3 &= 2A_2-P_2 = 2A_2 - 2A_1 + P_1 \\
P_4 &= 2A_3-P_3 = 2A_3 - 2A_2 + 2A_1 - P_1 \\
\end{align}
and consider that we need $P_{n+1}$ as defined by continuation to be equal to $P_1$.

When $n$ is odd, we have $P_1 = 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots-2A_2+2A_1 - P_1$

and so we see that we need $P_1 = A_n - A_{n-1} + A_{n-2}-\ldots- A_2+ A_1$

When $n$ is even , we have $P_1 = 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots+2A_2-2A_1 + P_1$,

which is only possible if $ 2A_n-2A_{n-1}+ 2A_{n-2}-\ldots+2A_2-2A_1 = 0$

or as stated  $ A_n+ A_{n-2}+\ldots+A_2 = A_{n-1}+ A_{n-3}+\ldots+ A_1$

(and in this case , the location of $P_1$ is arbitrary).

Notice this result holds regardless of the dimension of space the $A_i$ points are embedded in.

Title: Finding the middle of n points


问:
Prove $(a + b) \bmod n = (a \bmod n + b \bmod n) \bmod n$
I find I am in trouble to prove:

$$(a + b) \bmod n = (a \bmod n + b \bmod n) \bmod n ?$$

Can anyone help?

答:
There is a simple proof using the following:
\begin{equation}
x\bmod{n}=y\bmod{n} \Longleftrightarrow n \mid (x-y).
\end{equation}

Let $r = a\bmod{n}$ and $s=\bmod{n}$. By the Division Algorithm, there are integers $p$ and $q$ such that $a = pn + r$ and $b = qn + s$. Adding these equations and rewriting yields $n(p+q) = (a + b) -(r + s)$. The result now follows by the fact above.


答:
The integer division of $a$ by $d$ is defined the following way:
$$a = qd + r$$
Where $d$ is divisor, $q$ is quotient, and $r$ is the remainder, for which $$0 \le r < |d|$$ Check reference here. This is an important thing to keep in mind while doing modular arithmetic and working with congruences.

For your case:
$$ -9 = 12q + r$$
The $q$ which gives the biggest $r$ which falls in the range discussed above is $q = -1$ (which is actually integer floor of the $-0.75$ you got in your algorithm). So if you decide to go your way, you should take the integer floor of the result obtained in STEP 1.
$$ -9 = 12*(-1) + 3$$

Title: -9 modulo 12 is 3. Should get -9

Body:
STEP 1: Divide Dividend (-9) from Divisor (12).

```<code>-9 / 12 = -0.75
</code>```

Where,

0 is whole part.

75 is fractional part.

STEP 2: Whole part from the result of STEP 1 to be multiplied by Divisor (12).

```<code>0 * 12 = 0
</code>```

STEP 3: Subtract result from STEP 2 from Dividend (-9).

```<code>-9-0 = -9
</code>```

Whereas, The Actual answer is 3. Where am I missing?


问:
Risk and probability: Is there always infinite risk?
If there are an infinite number of events with a cost (risks), and total Risk is calculated by the sum of all individual risks multiplied by their probability, does everything have an infinite risk associated with it?</p>

答:
No. If the events $a_n$ all have a risk of $1$, and a probability of $2^{-n}$, then the total risk is:

$$\mathrm{risk} = \sum_{n=1}^\infty \operatorname{risk}(a_n)\Pr(a_n)=\sum_{n=1}^\infty 1\cdot 2^{-n} = 1$$


问:
Are all reflections rotations?
Can someone prove to me why or why not all reflections are just rotations?

The only requirement on rotations is that it is a square matrix such that $R^T R = I$.

I tested this equation on a bunch of the reflection matrices here: https://en.wikipedia.org/wiki/Rotations_and_reflections_in_two_dimensions#:~:text=In%20geometry%2C%20two%2Ddimensional%20rotations,side%20of%20line%20L1.

And they all seem to satisfy the rotation constraint.

So now I am not sure.

答:
You are missing a requirement for a real square matrix to be a rotation. You need that $R^T R = I$ and $det(R) = 1$. The matrix $ A =\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix} $ satisfies $A^T A = I$ but $det(A) = -1$. This is the matrix that reflects across the x-axis; it is not a rotation matrix.

Title: Are all reflections rotations?


问:
Fourier cosine transform of $\arctan(1/x)/x$
Consider
$$
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx, \quad \text{for } \omega > 0,
$$
although for $\omega \in \mathbb{C}$ would be interesting. What is the above cosine transform? Does it exist?

My attempt so far: for $x \in \mathbb{R}$ we know that
$$
\arctan(x) + \arctan(1/x) = \text{sign}(x)\pi/2,
$$
and we know that
$$
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan(x)}{x} dx = \pi \Gamma(0,|\omega|),
$$
where $\Gamma$ is the upper incomplete Gamma function. Using this, then
$$
\begin{align}
\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx
&= \int_{0}^{\infty}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx + \int_{-\infty}^{0}\cos(\omega x)\frac{\arctan\left(\frac{1}{x}\right)}{x} dx \\
&= -\int_{-\infty}^{\infty}\cos(\omega x)\frac{\arctan\left(x\right)}{x} dx -\frac{\pi}{2}\int_{-\infty}^{0}\cos(\omega x)\frac{1}{x} dx + \frac{\pi}{2}\int_{0}^{\infty}\cos(\omega x)\frac{1}{x} dx \\
&= -\pi\Gamma(0,|\omega|) - \pi\int_{0}^{\infty}\cos(\omega x)\frac{1}{x} dx.
\end{align}
$$
But the integral on the right does not make sense. I assume this is not the correct approach.

答:
If I am not running completely nuts, we can attack this via fantasy and brute force. First of all notice that the integral is even, hence we can write it as

$$2\int_0^{+\infty} \cos(\omega x) \dfrac{\arctan(1/x)}{x}\ \text{d}x$$

Now we call this integral $F(\omega, x)$.

Let's now differentiate wrt $\omega$:

$$\dfrac{\partial F}{\partial\omega} = -2\int_0^{+\infty} \sin(\omega x) \arctan\left(\dfrac{1}{x}\right)\ \text{d}x$$

This integral can be computed (there are lots of ways to do this, also it is absolutely convergent), and we have:

$$\dfrac{\partial F}{\partial \omega} = -2\left(\frac{\pi  e^{-\frac{\omega}{2}} \sinh \left(\frac{\omega}{2}\right)}{\omega}\right)$$

This result holds iff $\omega > 0$ if $\omega\in\mathbb{R}$, or as I wrote in the comment, for $\Im(\omega)\leq 0$ but we can manage to extend it for every $\omega \in \mathbb{C}$.

If you master a bit of elementary calculus techniques, you can easily find that

$$\left(\frac{\pi  e^{-\frac{\omega}{2}} \sinh \left(\frac{\omega}{2}\right)}{\omega}\right) = \frac{\pi -\pi  e^{-\omega}}{2 \omega}$$

To get $F(\omega)$ we now integrate this with respect to $\omega$:

$$F(\omega) = -2\int \frac{\pi -\pi  e^{-\omega}}{2 \omega}\ \text{d}\omega$$

The integration is rather easy, and it does make a Special Function to pop out: the Exponential Integral Special Function:

$$-2\int \frac{\pi -\pi  e^{-\omega}}{2 \omega} = -\pi  (\log (\omega)-\text{Ei}(-\omega))$$

You can find more on the Exponential Integral here: https://en.wikipedia.org/wiki/Exponential_integral

P.s. In a more rigorous way, we should attack the first part with the principal value method, just to be sure. But eventually it leads to the same result.


答: It is a bit of a matter of definition. From my understanding, talking in terms of matrices, $R^T R = I$ defines an orthogonal matrix $R$. The word rotation is used, as J. W. Tanner commented, for orthogonal matrices for which det$(R) = 1$ (while $R^T R = I$ only required det$(R)^2 = 1$). When one wants to be super-specific with this, one sometimes uses the term of a proper rotation and allows for improper rotations, which only satisfy the weaker condition det$(R)^2 = 1$.

Title: Are all reflections rotations?

Body:
Can someone prove to me why or why not all reflections are just rotations?

The only requirement on rotations is that it is a square matrix such that $R^T R = I$.

I tested this equation on a bunch of the reflection matrices here: https://en.wikipedia.org/wiki/Rotations_and_reflections_in_two_dimensions#:~:text=In%20geometry%2C%20two%2Ddimensional%20rotations,side%20of%20line%20L1.

And they all seem to satisfy the rotation constraint.

So now I am not sure.


问:
Alternative characterization of differentiability at a point
My lecturer for my Analysis class provided the following definition for differentiability at a point:

Let $(V, \|\cdot\|_V)$ and $(W, \|\cdot \|_W)$ be two finite-dimensional vector spaces.
Let $f : D \to W$ be a function and let $a \in D$.
We say that $f$ is differentiable in $a$ if there exists a linear map $L_a: V \to W$ such that, if we define the error function $\mathsf{Err}_a : D \to W$ through
$$
 \mathsf{Err}_a(x) := f(x) - f(a) - L_a (x - a)
 $$
it holds that
$$
 \lim_{x \to a} \frac{\|\mathsf{Err}_a(x)\|_W}{\|x - a\|_V} = 0.
 $$
And $L_a$ is the derivative of $f$ in $a$, denoted alternatively as $(Df)_a$.

This surely differs from the definition of differentiability in a point that I have got to known: Checking whether the limit
$$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ exists.

So far I have not seen first mentioned definition for differentiability in literature, nor do I understand the intuitive idea behind it. Any help on literary sources or explanation of any other kind is greatly appreciated.

答:
If $f : D \to \mathbb R$ for some open subset $D$ of $\mathbb R$, $a \in D$, $$\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = f'(a)$$ and you consider the linear map $L_a : \mathbb R \to \mathbb R$ given by $L_a(t) =  f'(a)t$, then of course $$\lim_{x \to a} \frac{|f(x)-f(a)-L_a(x-a)|}{|x-a|} = \lim_{x \to a} \bigg| \frac{f(x)-f(a)}{x-a}-f'(a) \bigg| = 0.$$
The first one definition of derivative is called the Fréchet derivative of $f$ at $a$, and it generalizes the classical notion of derivative.


问:
how to draw an arc?
How to draw an arc, I have these values

```<code>x  // x coordinate
y // y coordinate
r // Arc radius
startAngle // Starting point on circle
endAngle // End point on circle
clockwise  // clockwise or anticlockwise
</code>```

where x,y is the center of circle. Can someone please provide me equation for drawing an arc?

答: If you are using Mathematica, Graphics[Circle[{x, y}, r, {startAngle, endAngle}]]  does that. It is always anticlockwise, so for clockwise, swap the start and end angles. </p>

Title: how to draw an arc?


问:
Application of multinominal theorem for a tricky sum
In our lecture notes, it says that one can compute the sum
$$\sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \binom{N}{k_1,\dots,k_M} \left(\binom{k_1}{2} + \dots + \binom{k_M}{2} \right)^2$$
by using
$$\binom{k}{2}^2 = \frac{k^{\underline{4}}}{4} + k^{\underline{3}} + \frac{k^{\underline{2}}}{2},$$
where we used the falling factorials
$$n^{\underline{k}} := n (n-1) \cdots (n-k+1).$$

Since this remark is just after the multinomial theorem was stated, that is
$$\sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}}  \binom{N}{k_1,\dots,k_M} x_1^{k_1} x_2^{k_2}\dots x_M^{k_M} = (x_1+x_2+\dots +x_M)^N,\qquad \binom{N}{k_1,\dots,k_M} = \frac{N!}{k_1!\cdots k_M!},$$
I guess that one should use it on this problem, but I do not know how to get there, since using the binomial theorem on the $(\ldots)^2$-term does not give me something of the form $x_1^{k_1} x_2^{k_2}\cdots x_M^{k_M}$ or I am missing something. Does anyone have any idea, how one could start?

答:
I found a solution, maybe someone will need it in the future:
$$\begin{align*}
&\sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \binom{N}{k_1,\dots,k_M} \left(\binom{k_1}{2} + \dots + \binom{k_M}{2} \right)^2 \\
&= \sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \binom{N}{k_1,\dots,k_M} \left(\sum_{i=1}^M \binom{k_i}{2}^2 + \sum_{\substack{j,m=1\\j\neq m}}^M\binom{k_j}{2}\binom{k_m}{2} \right)\\
&= \sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \frac{N!}{k_1!\cdots k_M!}\left(\sum_{i=1}^M \left(\frac{k_i^{\underline{4}}}{4} + k_i^{\underline{3}} + \frac{k_i^{\underline{2}}}{2}\right) + \sum_{\substack{j,m=1\\j\neq m}}^M \frac{k_j^{\underline{2}}}{2} \frac{k_m^{\underline{2}}}{2}\right)\\
&= \sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \sum_{i=1}^M\left( \frac{1}{4}\frac{N!}{\prod_{\substack{l=1\\l\neq i}}^M k_l!}\frac{1}{(k_i-4)!} + \frac{N!}{\prod_{\substack{l=1\\l\neq i}}^M k_l!} \frac{1}{(k_i-3)!} + \frac{1}{2}\frac{N!}{\prod_{\substack{l=1\\l\neq i}}^M k_l!} \frac{1}{(k_i-2)!}\right) \\
&\qquad+ \sum_{\substack{j,m=1\\j\neq m}}^M \left(\frac{1}{4}\frac{N!}{\prod_{\substack{l=1\\l\neq j,m}}^M k_l!}\frac{1}{(k_j-2)!} \frac{1}{(k_m-2)!}\right)\\
&= \sum_{\substack{k_1,k_2,\dots,k_M\geq0:\\k_1+\dots+k_M=N}} \sum_{i=1}^M\Bigg( \frac{N^{\underline{4}}}{4} \binom{N-4}{\prod_{\substack{l=1\\l\neq i}}^M k_l,k_i-4} + N^{\underline{3}} \binom{N-3}{\prod_{\substack{l=1\\l\neq i}}^M k_l,k_i-3} \\
&\qquad+ \frac{N^{\underline{2}}}{2}\binom{N-2}{\prod_{\substack{l=1\\l\neq i}}^M k_l,k_i-2}\Bigg) + \sum_{\substack{j,m=1\\j\neq m}}^M \frac{N^{\underline{4}}}{4} \binom{N-4}{\prod_{\substack{l=1\\l\neq j,m}}^M k_l,k_j-2, k_m-2}\\
&= M \left(\frac{N^{\underline{4}}}{4} M^{N-4} + N^{\underline{3}} M^{N-3} + \frac{N^{\underline{2}}}{2} M^{N-2}\right) + M(M-1) \frac{N^{\underline{4}}}{4} M^{N-4}\\
&= M^{N-2}N^{\underline{3}} + \frac{M^{N-1}N^{\underline{2}}}{2} + \frac{M^{N-2}N^{\underline{4}}}{4},
\end{align*}$$
where one gets the second to last equation by multinomial theorem and counting the summands.


问:
Conceptual question about the topology of the test function space $\mathcal{D}(\Omega)$
I am studying distribution theory using Walter Rudin's Functional Analysis and although I seem to have overcome most of the technical details, there is an aspect I cannot fully understand. It is related to the topology of the test funcion space $\mathcal{D}(\Omega)$.

When equipping the space with a topology, Rudin insists on the importance of having the following condition satisfied:

If a sequence $(\phi_k)$ converges in $\mathcal{D}(\Omega)$, then there is a compact set $K \subset \Omega$ such that $\text{supp } \phi_k \subset K$ for all $k\in \mathbb{N}$

As far as I understand, it is the very condition Schwartz imposed when developing the theory; in fact, I have read that he did not even specify at topology at first, but only focused on assuring that bounded sets satisfied this condition.

In fact, Rudin starts with the topology generated by the following norms

$$ \lVert\phi \rVert_N = \{ \text{max}|D^\alpha \phi|: x\in \Omega, |\alpha|\leq N \}  $$

but finds and easy counterexample that proves that Cauchy Sequences in this topology are not convergent. In case it is relevant, the example is the following: let $\Omega = \mathbb{R}$, let $\varphi$ be a test function whose support lies in $(0,1)$ and

$$ \phi_k = \sum_{i=0}^k \frac{\varphi(x-i)}{i+1} $$

He then argues that it is necessary to add more open sets (i.e., find a strictly finer topology) such that the above mentioned condition holds.

My question is: Is there a fundamental reason why this boundedness condition is needed? Why are we so interested in having a complete test function space (I understand that it is desirable to have a complete distribution space, but I cannot see why it is a requirement in the test function space)? In the following question,

Motivation for test function topologies

someone argues that this condition is necessary to verify that distribution derivative is a local operator, but it does not seem to make a lot of sense to me. In fact, this finer topology we work with renders the distribution space larger rather than smaller, so everything that works for the distribution space should work for the dual space that would result if we used this coarser topology.

Whether this is the answer or not, it is an answer of this type what I am looking for. Thanks in advance!

答:
An answer that satisfies me is the following: if we do not impose bounded sets to lie inside some $\mathcal{D}_K$ (i.e., for any bounded set $B\subset \mathcal{D}(\Omega)$, there exists a $K\subset \Omega$ such that $\text{supp}\phi \subset K$ for all $\phi$), then, firstly, some functions in $L^1_{\text{loc}}(\Omega)$ would not be in $\mathcal{D}'(\Omega)$. This should raise a first concern on the quality of our extension of the function concept (rather than an extension, we would be suggesting an alternative).

To prove this claim, I will just give an example from which the general argument can follow. I will work in $\Omega = \mathbb{R}$ and with a particular set, but, once again, the argument can be made general easily. Consider the set $B = \{\phi_k\}_{k\in\mathbb{N}} = \{\varphi(t-k+1)\}_{k\in\mathbb{N}}$, where $\varphi$ is a $C_c^\infty(\Omega)$ function such that $\text{supp}\varphi \subset (0,1)$ and $\int_{\mathbb{R}}\varphi \neq 0$. Let now $\Lambda_f$ be the functional associated to the function $f = \sum_{i=1}^\infty i^2 \chi_{(i-1,i)}$, where $\chi_S$ represents the indicator function of the set $S$.

Then,

$$ \Lambda_f(\phi) = \int_{\mathbb{R}}f\phi $$

is not a continuous functional. Indeed, we know from the general topological vector spaces theory that

$$ \text{A linear functional is continuous} \implies \text{it is bounded}$$

and hence

$$ \text{A functional is unbounded} \implies \text{it is not continuous} $$

Let us now come back to the set $B$ I mentioned before. Our hypothesis is that $B$ is bounded, but then

$$ | \Lambda_f(\phi_k)| = |\int_{(k-1,k)}k^2\varphi(x-k+1)dx| = k^2 |\int_{\mathbb{R}}\varphi| $$

Hence, the set $\Lambda_f(B)$ is clearly not bounded, so $\Lambda_f$ cannot be in the dual space.

With little effort, an argument of this kind can be done general: without the boundedness condition,we can always find a function $L^1_{\text{loc}}(\Omega)$ such that $\Lambda_f$ is not in the dual space.

For a particular example, we can show that things can get even worse. If, for example, we use the toplogy that Rudin rejects, it can be shown that all $L^1$ functions are such that the associated distribution is continuous. But then, if we equip the dual space with the weak-$\star$ topology, it can be seen that the sequence of distributions associated to the functions $\sum_{i=1}^n i^2 \chi_{(i-1,i)}$ is a Cauchy sequence. Hence, as we have shown that the distribution associated to $\sum_{i=1}^\infty i^2 \chi_{(i-1,i)}$ would not be in the dual space, we can see that the distribution space would not be complete!


问:
Proving $\{a\}$ is a also set given that $a$ is a set. Introduction to Set Theory.
So I have just started learning about set theory and I have this proposition after introducing two axioms.

Axiom 1: Axiom of extensionality: Two sets are equal if and only if they have the same elements.

Axiom 2: Axiom of pairs: If $a$ and $b$ are sets then so is $\{a,b\}.$

The proposition is that $\{a\}$ is also a set given that $a$ is a set.

Given proof: By axiom $2$, $\{a,a\}$ is a set. By axiom $1$, this is equal to $\{a\}$. QED.

However whilst reading this proof I had a doubt.

To use axiom $1$ in the proof, we need to assume $\{a\}$ is a set right? I think this because it literally says two sets are equal if and only if elements are the same and thus for this axiom to be applicable to need a 'priori' that $\{a\}$ is a set? If that is the case are we not in a circular argument?

答:
Just to tie up some of the comments: The circularity you see here actually results from the extremely awful formulation "show $\{a\}$ is a set". Because, forgetting about set theory and just thinking logically, that is meaningless if we don't already know what $\{a\}$ is, and if we already know what it is we don't need to show it's a set.

That's the way people often phrase these things, but it's informal notation; the notation $\{\dots\}$ simply is not part of "official" set theory! When people say $\{a\}$ is a set they mean that there is a set with $a$ as its one and only element.

Or more formally, $$\exists s\left(\forall x\left(x\in s\iff x=a\right)\right).$$

Similarly, "$\{a,a\}$ is a set" means $$\exists s\left(\forall x\left(x\in s\iff x=a\lor x=a\right)\right).$$

Nothing circular about that, and it's clear the two are equivalent.


问:
If $\wp(u) = \wp(v)$, then $u-v$ or $u+v$ is a period of $\wp$
The Weierstrass $\wp$ function with an associated lattice $L$ is given by the following equation for $z \notin L:$
$$
\wp(z)=\frac{1}{z^{2}}+\sum_{\omega \in L \backslash\{0\}}\left[\frac{1}{(z-\omega)^{2}}+\frac{1}{\omega^{2}}\right].
$$
What we want to show that $\wp(u) = \wp(v)$ if and only if $u-v$ or $u+v$ is a period of $\wp$.

My attempt:

($\Leftarrow$): If $u-v$ is a period of $\wp$, then $\wp((u-v)+z)=\wp(z)$. In particular, we have that
$$\wp(u)=\wp((u-v)+v)=\wp(v).$$
Similarly, If $u+v$ is a period of $\wp$, then $\wp((u+v)+z)=\wp(z)$. In particular, we have that
$$\wp(u)=\wp((u+v)-v)=\wp(-v)=\wp(v),$$ since $\wp$ is an even funciton.

($\Rightarrow$): Is there any idea to show this direction?

答: COMMENT.- The attached figure shows a fundamental parallelogram and the values that the Wierstrass function takes in it. Can you deduce from this the properties that you enunciate?


问:
Some questions about complexification of a real vector space
Could you tell me how to prove that if $f:U \rightarrow U$ is $\mathbb{R}$-linear, then:

1) $U^{\mathbb{C}}$ is a vector space over $\mathbb{R}$ (should I check all eight conditions for a vector space?)

2) $f^{\mathbb{C}}: U^{\mathbb{C}} \rightarrow U^{\mathbb{C}}$ is the only $\mathbb{C}$ - linear mapping such that $f^{\mathbb{C}}|_U = f$

$U^{\mathbb{C}} = U \times U$ - it's a vector space over $\mathbb{C}$

$(a,b) + (c,d) = (a+c, b+d)$

$(a+bi)(c,d) = (ac-bd, ad+bc)$

$f^{\mathbb{C}}(a+ib) = f(a) + if(b)$

答:
For 1): As you know that $U^{\mathbb C}$ is a $\mathbb C$-vector space, think if restriction of the multiplication $\mathbb C \times U^{\mathbb C} \to U^{\mathbb C}$ to $\mathbb R\times U^{\mathbb C}$ can violate any condition.

For 2): Suppose $\def\C{\mathbb C}$$f^\C\colon U^\C \to U^\C$ if $\C$-linear with $f^\C|_U = f$, then we have for $x,y \in U$:
\begin{align*}
  f^\C(x,y) &= f^\C\bigl((x,0) + i\cdot(y,0)\bigr)\\
            &= f^\C(x,0) + if^\C(y,0)\\
            &= \bigl(f(x),0\bigr)+i\cdot \bigl(f(y),0\bigr)\\
            &= \bigl(f(x), f(y)\bigr)
\end{align*}
so there is at most one $\C$-linear extension of $f$, to prove existence, show that $f^\C(x,y) :=\bigl(f(x), f(y)\bigr)$ is $\C$-linear.


问:
Simple question on $O(n^{-a})$
Say that

$ \mathbb{E}(\hat{T}) - T = - \frac{1}{2n}+\frac{1}{12n^2}+O\Big(\frac{1}{n^3}\Big)$

where $O(1/n^a)$ is a quantity of order $n^{-a}$ ($a>0)$. Can we say that $ \mathbb{E}(\hat{T}) - T \rightarrow 0$ as $n \rightarrow \infty$? In other words my question is if $O\Big(\frac{1}{n^3}\Big) \rightarrow 0$ as  $n \rightarrow \infty$. It would be nice also if you could provide the rationale behind the result, thanks.

答:
Yes you can say this, because by the definition of $O$ there exists some constant $C$ such that

$$\lim_{n\to\infty}O\left(\frac{1}{n^3}\right)<\lim_{n\to\infty}C\left(\frac{1}{n^3}\right)=0$$

Title: Simple question on $O(n^{-a})$


问:
Either-or condition for set of constraints
Consider the following optimization problem
\begin{equation*}
\begin{split}
\text{min} & \sum_{j\in J} c_jx_j \\
& \quad \sum_{j\in J} a_{1j}x_j \leq b_1 \\
& \quad \sum_{j\in J} a_{2j}x_j \leq b_2 \\
& \quad x_j \geq 0 \quad \forall j\in J
\end{split}
\end{equation*}
It is well known that, if we need to impose that just one of these two constraints must be satisfied, then we can use introduce a binary variable $y\in\{0,1\}$ and two upper bounds $M_1$ and $M_2$ so that
\begin{equation*}
\begin{split}
\text{min} & \sum_{j\in J} c_jx_j \\
& \quad \sum_{j\in J} a_{1j}x_j \leq b_1 + M_1y\\
& \quad \sum_{j\in J} a_{2j}x_j \leq b_2 +M_2(1-y)\\
& \quad x_j \geq 0 \quad \forall j\in J
\end{split}
\end{equation*}

Now, imagine that we wish that just one of the following sets of constraints
\begin{equation*}
\left\{\sum_{j\in J} a_{ij}x_j \leq b_i, \quad i\in I\right\}, \quad \left\{\sum_{j\in J} a_{kj}x_j \leq b_k, \quad k\in K\right\},
\end{equation*}
with $I\cap K = \emptyset$ is satisfied. How can we formulate the optimization problem?

Any help would be appreciated.

答:
You can modify your special case of $|I|=|K|=1$ to obtain the following formulation of the general case:
\begin{align}
\sum_{j\in J} a_{ij}x_j &\le b_i + M_i y &&\text{for $i\in I$}\\
\sum_{j\in J} a_{kj}x_j &\le b_k + M_k(1-y) &&\text{for $k\in K$}
\end{align}


问:
Show that $f$ is a Borel measurable function
Let $f: \mathbb{R} \to \mathbb{R} $ be a function with
$$ x \mapsto \begin{cases}
x, \quad &\text{if} \quad x \in \mathbb{Q} \cap [0,1]  \\
1-x, \quad &\text{if} \quad x \in (\mathbb{R} \backslash \mathbb{Q}) \cap [0,1] \\
0, \quad &\text{otherwise.}
\end{cases} $$
One can write $f = \mathbb{1}_{\mathbb{Q} \cap [0,1]}(x) \cdot x + \mathbb{1}_{(\mathbb{R} \backslash \mathbb{Q}) \cap [0,1]}(x) \cdot (1-x) $ and clearly this function is measurable since it is a sum of measurable functions (this way it is just fine). But, would it be also ok, if I would show that the preimages are in $\mathfrak{B}(\mathbb{R})$, i.e.
$$
f^{-1}(x) = \mathbb{Q} \cap [0,1] \in \mathfrak{B}(\mathbb{R})
$$
and in the same fashion also for $f^{-1}(x)$ and $f^{-1}(\{0\})$? Obviously those are again all contained in the Borel $\sigma$-Algebra, but I think that is not the correct way to do it, since one has to show, that
$$ \forall B \in \mathfrak{B}(\mathbb{R}): f^{-1}(B) \in \mathfrak{B}(\mathbb{R}). $$ In my opinion there could be several cases for $B \in \mathfrak{B}$ arbitary: $x \in B $, $1-x \in B$, $\{0 \} \in B$, $\emptyset = B$, $\mathbb{R} \in B$, such that for all this $B$ I have to show, that the preimages are in $\mathfrak{B}(\mathbb{R})$. Please let me know, if I am on the right path!

答:
$\newcommand{\d}{\mathfrak{B}(\mathbb R)}$
By definition, we should be checking that for every $B \in \d$, we have that $f^{-1}(B) \in \d$.

But some niceties make things easy for us, namely the following identities :

For any $f$ and arbitrary collection of disjoint sets $U_i$ we have $f^{-1}(\cup_{i \in I} U_i) = \cup_{i \in I} f^{-1}(U_i)$.

For any $f$ and set $U$ we have $f^{-1}(U^c) = [f^{-1}(U)]^c$.

Both are easily proved by definition.

A generating set for $\d$ is given by half intervals, which are the subsets of $\mathbb R$ given by $H = \{(-\infty,x] : x \in \mathbb R\}$. So elements of $H$, for example are $(-\infty,0]$, $(-\infty,3.54]$, $(-\infty,-\pi]$ and so on.

The point is, that $H$ generates $\d$ in the sense that the smallest sigma-algebra containing $H$ is $\d$. Furthermore, $H$ is what we call as a $\pi$-family : if I intersect any two members of $H$, the intersection is also in $H$. For example , the intersection of $(-\infty,1]$ and $(-\infty,2]$ is $(-\infty,1]$ which remains in $H$.

Let's take an element in $H$, say $(-\infty,x]$. Then, the set $f^{-1}((-\infty,x])$ is nothing but the set $\{y : f(y) \leq x\}$.

Lemma : Let $f$ be a function such that for every $x \in \mathbb R$, the set $\{y : f(y) \leq x\}$ is Borel. Then $f$ itself is Borel i.e. for any $B \in \d$, $f^{-1}(B)$ is Borel.

Proof : Consider the sigma-algebra $\Sigma = \{B' \in \d : f^{-1}(B') \in \d\}$. This is the set of Borel sets, which we know satisfy the conditions for now. Note that $H \subset \Sigma$.

But we also know something else about $\Sigma$ : it is closed under complements and under infinite disjoint union, thanks to the two properties we mentioned at the outset. Also it is clear that $\mathbb R \in \Sigma$.

Now, this makes $\Sigma$ what we call as a $\lambda$-system : not quite a sigma-algebra as yet, but close. Now, we invoke the Dynkin $\pi-\lambda$ theorem :

Let $S$ be a $\lambda$ system , containing a $\pi$-system $T$. Then $S$ contains $\sigma(T)$, the smallest sigma-algebra containing $T$.

That tells us, that $\Sigma$ contains $\d$, the smallest sigma-algebra containing $H$, but then $\Sigma$ is anyway contained in $\d$ as well, so we are done.

The result of all this, is that $f$ is Borel. Why? Let's prove it using the lemma. Pick $x\in \mathbb R$. Then we have some cases :

If $x < 0$ then $f(y) < x$ for all $y$, so $\{y : f(y) \leq x\}$ is empty, hence Borel.

If $x > 1$ then $f(y) > x$ for all $y$ so $\{y : f(y) \leq x\} = \mathbb R$ is Borel.

If $x \in [0,1]$, then of course $\{y : f(y) \leq x\}$ contains $[0,1]^c$, which is Borel. Apart from this, we break into cases : if $y \in [0,1] \cap \mathbb Q$ with $f(y) \leq x$ then $y \leq x$ i.e. $y \in \mathbb Q \cap [0,x]$. Similarly, if $f(y) \leq x$ with $y \in [0,1] \cap \mathbb Q^c$ then $y \geq 1-x$. This makes up $[1-x,1] \cap \mathbb Q^c $. Thus :
$$
\{y : f(y) \leq x\} = [0,1]^c \cup \left(\mathbb Q \cap [0,x]\right) \cup \left([1-x,1] \cap \mathbb Q^c\right)
$$
a union of Borel sets, hence Borel.

Thus, we get that $f$ is Borel.

Now, a common misconception is that if for every $x$, if $\color{red}{\{y : f(y) = x\}}$ is Borel, then $f$ is Borel. This is false.

The reason? Well, the family of singleton sets $\{x\}$ don't generate $\d$, for starters. Then it doesn't even form a $\pi$-family. Finally, a Borel measurable set has more regularity, in the sense that sets that usually are indexed (or decided) by infinite sequences (like the real numbers, if treated with the decimal expansion as an infinite sequence of digits) aren't Borel because they behave far too irregularly. (More on that, but in little more detail, here).

Edit : The family $H$ may look like it too is parametrized by the real numbers, but using density of rationals, you can see that a function being Borel measurable must be a result of "countable" decision i.e. decide it at the rationals and you get it everywhere.

For a counterexample, we use the Cantor set, the most fundamental set when it comes to Lebesgue measurability, along with a standard set called the non-measurable set. The Cantor set itself is closed, hence Borel, but we can play with it nicely.

In this page, there is a good explanation on what's happening :essentially, the Cantor function is a bijection, so every set of the form $\{x : f(x) = y\}$ is a singleton hence Borel. But the problem is that the non-measurable set which sits inside $[0,1]$, that causes problems.

Unfortunately, I cannot find a simpler example, because looking for something non-Borel yet Lebesgue measurable is in itself a topic that is difficult to discuss.


问:
Struggling a bit on the factoring in this induction proof
This is a simple problem in a Number Theory book I just purchase.  I've struggled with this for a bit and I thought it may be time to ask for some extra eyes.  I'm just not seeing the next step and I'm not sure why.  So, the problem, prove $\sum_{i=1}^{n} n^2 = \frac{n(n+1)(2n+1)}{6}$.  I'm constantly getting stuck at this point $(k+1)\left [ \frac{2k^2+7k+1}{6} \right ]$.  Since I'm likely getting stuck somewhere in the process, here is my work.

\begin{align}
\sum_{i=1}^{n=k+1} n^2 & =  \frac{(k)(k+1)(2k+1)}{6} + (k + 1)^2 \\
         & =  (k+1) \left [ \frac{k(2k+1)}{6} + (k+1) \right ] \\
         & =  (k+1) \left [ \frac{2k^2 + 7k + 1}{6} \right ] \\
\end{align}

Body:
\begin{align}
\sum_{i=1}^{n=k+1} n^2  & =  \sum_{i=1}^{n=k} n^2 + (k + 1)^2 \\
 & =  \frac{(k)(k+1)(2k+1)}{6} + (k + 1)^2 \tag{ind.hyp.}\\
 & =  (k+1) \left [ \frac{k(2k+1)}{6} + (k+1) \right ] \\
 & =  (k+1) \left [ \frac{2k^2 + k}{6} +\frac{6k + 6}{6} \right ] \\
 & =  (k+1) \left [ \frac{2k^2 + 7k + \color{red}{6}}{6} \right ] \\
 & =  \frac 16(k+1)  (k+2)(2k+3) \\
\end{align}and so on.


答:
You can find this definition in Rudin's book Principles of Mathematical Analysis.  The idea is that given a function $f:\mathbb{R}^n\to \mathbb{R}^m$ (dropping the abstract vector space notion) if it's differentiable at a point $p\in \mathbb{R}^n$ then the definitin you have given say's it can be well approximated by a linear map (ie a matrix).  In the usual context we first learn where $f:\mathbb{R}\to \mathbb{R}$, let's see how we can obtain the new notion.  Suppose $f$ is differentiable at $p$, then like you said the limit
$$f'(p)=\lim_{t\to p}\frac{f(t)-f(p)}{t-p}$$
exists.
This means, the difference quotient
$$\varphi(t)=\frac{f(t)-f(p)}{t-p} \qquad (t\neq p)$$
is continuous at $t=p$.
Which means for every $\epsilon>0$ there is a $\delta>0$ so that
$$\Big | \frac{f(t)-f(p)}{t-p}-f'(p)\Big |< \epsilon \qquad  \text{for} \quad |t-p|<\delta$$
We can interpret this as there existing a function say $E(p,t)$ so that
$$f(t)-f(p)-f'(p)(t-p)=E(t,p)$$ which must satisfy $\lim_{t\to p}\frac{|E(p,t)|}{|t-p|}=0$.

Now, in the context of $\mathbb{R}$ what are the linear maps?  They are in correspondence with $\mathbb{R}$ itself, ie $1\times 1$ matricies.  So all along, the derivative that you knew and loved was always a linear map on a one dimensional vector space and to re-iterate it's the linear map that best approximates the given function about a base point.

Hope this helps!

Title: Alternative characterization of differentiability at a point

Body:
My lecturer for my Analysis class provided the following definition for differentiability at a point:

Let $(V, \|\cdot\|_V)$ and $(W, \|\cdot \|_W)$ be two finite-dimensional vector spaces.
Let $f : D \to W$ be a function and let $a \in D$.
We say that $f$ is differentiable in $a$ if there exists a linear map $L_a: V \to W$ such that, if we define the error function $\mathsf{Err}_a : D \to W$ through
$$
 \mathsf{Err}_a(x) := f(x) - f(a) - L_a (x - a)
 $$
it holds that
$$
 \lim_{x \to a} \frac{\|\mathsf{Err}_a(x)\|_W}{\|x - a\|_V} = 0.
 $$
And $L_a$ is the derivative of $f$ in $a$, denoted alternatively as $(Df)_a$.

This surely differs from the definition of differentiability in a point that I have got to known: Checking whether the limit
$$\lim_{x \to a} \frac{f(x)-f(a)}{x-a}$$ exists.

So far I have not seen first mentioned definition for differentiability in literature, nor do I understand the intuitive idea behind it. Any help on literary sources or explanation of any other kind is greatly appreciated.


问:
Iteratively solving (sparse) homogeneous linear systems
Solving (sparse) non-homogeneous linear systems can be done iteratively, by using the LSQR algorithm or similar. However, in the homogeneous case, we have $Ax =0$, where we typically want to find the non-zero solutions which is generally done by demanding a linear constraint on $x$, e.g. $|x|=1$. The method usually used to solve this constrained homogeneous system is by performing a singular value decomposition, and choosing the vector corresponding to the smallest singular value.

I'm looking for an iterative equivalent of LSQR that may be applied to sparse rectangular systems. LSQR does not seem appropriate, as it's stages implicitly assume $|b|\neq0$. Is anyone aware of such a method or can think of a method to modify LSQR appropriately?

答:
Looking through this paper, it gives a number of methods. The most straight forward seems to be the following algorithm:

Start with a random solution $x_0$

Break $A$ into it's QR decomposition, where Q is of size $m\times n$, and $R$ is upper diagonal

Solve via backwards substitution:
$$R^Tw = x_{t-1}/|x_{t-1}|$$
$$R x_t = w$$

Iterate until convergence


问:
Circular reasoning on l'hôpital
I have seen many questions about circular reasoning on this website but none of them have satisfied my doubt. How do I know if I'm doing circular reasoning on a limit by doing l'hôpital? I do know that circular reasoning is when you use l'hôpital on a limit like $\lim_{x\to 0}\frac{\sin(x)}{x}$ because the defn of the derivative of sine is cosine which is obtained by applying l'hôpital? Yeah but how do I know exactly if whenever I want to apply l'hôpital, that same l'hôpital has been used to evaluate the derivative?

答:
You appear to suggest at one point that you can use L'Hopital's rule to evaluate the derivative of a function. This is not true at all, and you don't even need to use the example of
$$
\lim_{x \to 0}\frac{\sin x}{x} \, .
$$
Consider the simpler case of how would you prove that the derivative of $x^2$ is $2x$. If your answer is 'use L'Hopital', then you would be using the fact that the derivative of $x^2$ is $2x$ in order to prove that the derivative of $x^2$ is $2x$ (a big problem!). L'Hôpital's rule is there to help you evaluate more complicated limits, where it has already been established what the derivative of all of the functions are. However, even in this case, it should be not the first tool that comes to your mind. Blindly applying L'Hôpital prevents people from thinking both analytically and creatively about the behaviour of functions. There is a little to be gained by using an overpowered tool, the proof of which is not trivial.

Title: Circular reasoning on l'hôpital


问:
Urns version of Laplace's law of succession
I'm trying to prove urns version of Laplace's law of succession my professor suggested. Laplace's law states that the chance that the next trial is a success given $j$ successes out of the first $n$ is $\frac{(j+1)}{(n+2)}$. Here is how the problem states:

"If we have $n+k+1$ urns and urn $i$ has $i$ balls labeled $1$ and $n+k-i$ labeled zero. We pick an urn at random and draw $n$ balls from it without replacement say $j$ of them are ones. Show that the conclusion of Laplace's law holds for this setup. In other word, the chance that the next ball is a one is $\frac{(j+1)}{(n+2)}$."

I've proved one version of the law, which close to this version An elementary version of Laplace's Method of Succession. I tried to use similar approach but somehow my answer always in form of $k$ and I can't get rid of it. What is the intuition behind this $k$? Is it just to increase the complexity of the problem, or it has some meaning behind it?

答:
Let $I$ be the random variable indicating the label of the urn picked, $J$ the number of ones picked (after picking $n$ balls), and $X$ the event that the next ball is one. We know that
$$
P(J = j \mid I = i) = \frac{\binom{i}{j}\binom{n+k-i}{n-j}}{\binom{n + k}{n}}
= \frac{\binom{j+(i-j)}{i-j}\binom{n+k-j-(i-j)}{k-(i-j)}}{\binom{n+k}{n}}
$$
and
$$
P(X \mid I = i, J = j) =
\begin{cases}
0 & ; i \le j \\
\frac{i-j}k & ; j \le i \le k + j
\end{cases}
$$
Compute $P(I = i \mid J = j)$ from
$$
P(I = i \mid J = j) = \frac{P(I = i, J = j)}{\sum_{\tilde i}P(I = \tilde i, J = j)} =
\frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)}
$$
where we use $P(I = i) = \frac{1}{n + k + 1}$ in the last equation. Then,
\begin{align*}
P(X \mid J = j) & = \sum_{i} P(X \mid I = i, J = j) P(I = i \mid J = j) \\
& = \sum_{i=j}^{k+j} \frac{i-j}k \frac{P(J = j \mid I = i)}{\sum_{\tilde i} P(J = j \mid I = \tilde i)} \\
& = \frac{\sum_{i=j}^{k+j}(i-j)P(J = j \mid I = i)}{k \sum_{i=j}^{k+j} P(J = j \mid I = i)} \\
& = \frac{\sum_{i=j}^{k+j}(i-j)\binom{i}{j}\binom{n+k-i}{n-j}}{k \sum_{i=j}^{k+j} \binom{i}{j}\binom{n+k-i}{n-j}} \\
& = \frac{\sum_{i=0}^k i \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}}{k\sum_{i=0}^k \binom{j+i}{i}\binom{n-j+(k-i)}{k-i}}
\end{align*}
I don't know how to make this into the desired expression $\frac{j+1}{n+2}$. Putting $k = 1$ seems to work though.</p>


问:
Prove that if $\nu(A)=\int_A fd\mu$ and $\mu$ is $\sigma$-finite, then $\nu$ is $\sigma$-finite.
Let $\nu$ be the measure defined on a $\sigma$ finite measure space with measure $\mu$ and a non negative, measurable, real valued function $f$ such that $\nu=f*\mu$ so that $\nu(A)=\int_Afd\mu$.I know that $\nu$ is a measure. My attempt has been to construct a nested sequence of measurable sets $A_i$ so that $A_i\subset A_{i+1}$ and $\Sigma_{i\in \mathbb{N}}=\Omega$. Now because $\mu$ is $\sigma$- measurable $\mu(A_i)<\infty$ $\forall A_i$. Now if $\nu(A_N)=\infty$ it follows from the linearity of the integral that $\nu(A_{i>N})=\infty$. However I haven't been able to produce a contradiction from this assumption.

答:
What you have written is in general wrong with the current assumptions. Consider the measure space $(\mathbb{R}.\mathcal{B},\lambda)$ the standard measure space on the reals with the Lebesgue measure. Then consider
$$f=\begin{cases} \frac{1}{x} & \text{for }  x>0 \\
0 & \text{for } x \leq0
\end{cases},$$
which satisfies your assumptions on $f$.
Then by monotonicity of the integral $$\nu((-1,1))=\int_0^1 fd\lambda \geq \sum_{n=1}^\infty 2^n \int_{2^{-n-1}}^{2^{-n}}d\lambda = \frac12+\frac12+\frac12+\dots = \infty,$$ and due to the nesting propert of $\sigma$-finite measures there exists a large enough $N$ such that $(-1,1) \subset A_N$ and by monotonicity of measures $\nu(A_n)=\infty \quad \forall n\geq N$,
hence $\nu$ can't possibly be sigma finite.


问:
The integral $\int_{0}^{1}\ln^n(1/x)\,dx$
It is well known that
$$
\int_{0}^{1}\ln^p(1/x)\,dx=\Gamma(p+1)
$$
In particular, if $p=n$ is a positive integer, then
$$
I_n:=\int_{0}^{1}\ln^n(1/x)\,dx=(n+1)!
$$
Therefore, we obtain the following recursive relation
$$
I_{n+1}=(n+1)I_n\quad;\quad I_1=1
$$
Is there an easy way, using only elementary methods, to establish that relation?

My attempt: Using integration by parts
\begin{align}
I_{n+1}&=\left[\ln^n\left(\frac{1}{x}\right)x\left(1+\ln\left(\frac{1}{x}\right)\right)\right]_{0}^{1}\\
&\qquad-\int_{0}^{1}(n+1)\ln^n\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)x\left(1+\ln\left(\frac{1}{x}\right)\right)\,dx
\end{align}
How do I continue from here ?

Body:
$$I_{n+1}=\int_0^1 1 \log ^{n+1}\left(\frac{1}{x}\right) \, dx=\left[x \log ^{n+1}\left(\frac{1}{x}\right)\right]_0^1- \int_0^1 -\frac{x \left((n+1) \log ^n\left(\frac{1}{x}\right)\right)}{x} \, dx=$$
$$=\left[0-\underset{x\to 0}{\text{lim}}\;x \log ^{n+1}\left(\frac{1}{x}\right)\right]+(n+1)I_n=0+(n+1)I_n$$
$$I_{n+1}=(n+1)I_n$$


问:
Let $X,Y \sim Exp(\lambda)$. Prove, by Jacobian
Let $X, Y$ are independent and $ X,Y \sim Exp(\lambda)$ Prove, by Jacobian :

$ X + Y \sim Gamma(2,\lambda)$

Can anyone help me with that?

答: Let $W:=X+Y,\,Z:=\tfrac{Y}{X}$ so $X=\tfrac{W}{1+Z},\,Y=\tfrac{WZ}{1+Z}$ and$$\lambda^2e^{-\lambda(x+y)}dxdy=\lambda^2we^{-\lambda w}dw\tfrac{2z}{(1+z)^3}dz.$$


问:
What is the solution for this?
I know that the bold 1 refers to the indicator function. However, how do I interpret the subscript [0,1]? Can someone compute the integral, including steps so that I can properly understand how the bold 1 works.

$\int_{0}^{3}x\mathbf{1}_{[1,2]}(x)dx$

答:
This is an "indicator function".  $\textbf{1}_{A}(x)=\begin{cases}1&\text{if }x\in A\\0&\text{otherwise}\end{cases}$

The bracket notation is standard notation for intervals.  $[1,2]=\{x\in \Bbb R~:~1\leq x\leq 2\}$.  Square braces imply the corresponding endpoint is included while parentheses imply it is not.  These can be mixed, for instance $[3,7) = \{x\in\Bbb R~:~3\leq x<7\}$

As for your specific integral... noting that the integrand is equal to zero outside of the interval $[1,2]$ we have

$$\int\limits_0^3x\textbf{1}_{[1,2]}(x)dx = \int\limits_1^2 xdx=\left[\frac{x^2}{2}\right]^{x=2}_{x=1}=\frac{2^2}{2}-\frac{1^2}{2}=1.5$$

Title: What is the solution for this?


答:
As pointed out in the comments, you need to assume that $X$ and $Y$ are independent.  The pdf of $X$ and $Y$ is given by $f(x) = \frac{1}{\lambda} e^{-x/\lambda}$ for $x>0$.  Then the pdf of $X+Y$ is the convolution
$$
f_{X+Y}(x) = \int_{-\infty}^{\infty} f(y)f(x-y) \,dy = \int_0^x \frac{1}{\lambda}e^{-y/\lambda} \frac{1}{\lambda} e^{-(x-y)/\lambda}\ \,dy = \frac{1}{\lambda^2} \int_0^x e^{-x/\lambda} \,dy = \frac{1}{\lambda^2} x e^{-x/\lambda},
$$
which is the desired pdf.

Title: Let $X,Y \sim Exp(\lambda)$. Prove, by Jacobian

Body:
Let $X, Y$ are independent and $ X,Y \sim Exp(\lambda)$ Prove, by Jacobian :

$ X + Y \sim Gamma(2,\lambda)$

Can anyone help me with that?


问:
Is the integral $\int_{2}^{\infty} \frac{1}{\log(x^2)}\,dx$ convergent or divergent?
How can I see if this integral:
$$\int_{2}^{\infty} \frac{1}{\log(x^2)}\,dx$$ is convergent or divergent,
cause I try to resolve the integral and I saw this is a special type of integral,
can you guys help me?
PS: I know that to check if is convergent or divergent I need to do this $$\lim_{t\to\infty}\int_{2}^{t} \frac{1}{\log(x^2)}\,dx$$

答: Hint: If $x\geqslant2$, then$$\frac1{\log(x^2)}=\frac1{2\log(x)}\geqslant\frac1{2x}.$$


问:
Extension of a continuous function $f: \mathbb{Q}\to\mathbb{R}$ to a continuous function $g:\mathbb{R}\to\mathbb{R}$.
Prove or disprove that for all continuous functions $f:\mathbb{Q}\to\mathbb{R}$ there is a unique continuous function $g:\mathbb{R}\to\mathbb{R}$ with $g(x)=f(x),\ \forall x\in\mathbb{Q}$.

Now, this is a task consisting of 2 subtask and this is the latter. In the first one (I think) I have proved that this statement is true for uniformly continuous functions. Now I will try to disprove this statement.

Let us choose $x,y\in\mathbb{R}$ with $d(x,y)<\delta/3$ where $x$ is fixed. Now let us choose $x_n\to x\wedge y_n\to y,\ x_n,y_n\in\mathbb{Q}$. Thereby $d(x_n,y_n)<\delta$. In my other proof, I now used the property for uniformly continuous functions and said that $d(f(x_n),f(y_n))<\varepsilon/2$ and followed that $g$ is uniformly continuous and unique. However, I can't use that premise here, which is why I think that this statement is false as $d(f(x_n),f(y_n))$ can be anything because of $x_n$ not being fixed (and we can't apply the continuity).

How do I go from here / is this the right path in the first place?

Any help would be greatly appreciated.

答: Edit: A continuous extension will not always exist. But it's true that if a continuous extension exists, it has to be unique. The difference between two continuous extensions is zero on the rationals, so show that the only continuous extension of the zero function $0: \Bbb{Q} \to \Bbb{R}$ to all of $\Bbb{R}$ is the zero function on $\Bbb{R}$, to show that any two continuous extensions of $f: \Bbb{Q} \to \Bbb{R}$ have to agree on all of $\Bbb{R}$.


问:
Algorithmic complexity of iterative - recursive solution
I had an exam of algorithms and data structures today and I was given an excercise I wasn't prepared for. The excercise was some pseudocode that we needed to extrapolate the time complexity out of.

The entire crux of the doubt lies on a particular part of the code :

```<code>recursive_call(int n){
 if(n > 1){
    int t = 0;
    for a in 1 to n do
       t = t + (a*2) * (a*3)

    for a in 1 to 7 do
       acc = acc + recursive_call(n/3)
    return acc;
 }
  else return n-1;
}
</code>```

The first part is trivially linear : the loop is executed n times. The second part , me and the rest of my group are in disagreement.

According to some , the complexity is actually T(n) = T(n / 3) + O(n). The recursive call is not splitted in multiple calls on the same activation record , hence it's never divided , therefore T(n/3) is not multiplied by any a>1.

Personally, I think that it doesn't matter on which part of the input you're calling the recursion , as long as you are doing it multiple times , it counts. Therefore the function should be T(n) = 7T(n/3) + O(n), because the function gets called multiple times. The fact that the input is identical is irrelevant.

So we are at an impasse. Can anyone clear this up?

答:
The second variant is correct - complexity for this implementation satisfies $T(n) = 7 T(\frac{n}{3}) + O(n)$.

It's possible to optimize this implementation by making recursive call only once, but it will formally be another algorithm with different complexity.


问:
$H^1$ function with smallest seminorm
Let $\Omega\subset\mathbb R^d$ be a Lipschitz domain and $u\in H^1(\Omega)$. Find a $w\in H^1(\Omega)$ with the same boundary values but minimal seminorm on $H^1(\Omega)$.

I've read that harmonic functions have the smallest seminorm on a $H^1$-space, but still I don't quite know how to solve this problem.

答:
So we want to minimize $Q(w) := \int_\Omega |Dw|^2\, dx$, for $t \in \mathbb R$ and $v \in H^1_0(\Omega)$ (note that $w + H^1_0(\Omega)$ is the space of $H^1$-functions having the same boundary values as $w$) we have
\begin{align*}
  Q(w + tv) &= Q(w) + t \int_\Omega (Dv, Dw)\, dx + t^2Q(v)
\end{align*}
if $Q(w)$ is minimal, we therefore have
$$ 0 \le Q(w + tv) - Q(w) = t\int_\Omega (Dv, Dw)\, dx + t^2Q(v) $$
Dividing by $t > 0$ and $t < 0$ respectively, and letting $t \to 0$, we see, this can only hold iff $\int_\Omega(Dv, Dw) \, dx = 0$. Integration by parts (note that $v$ has vanishing boundary values gives $\int_\Omega v\Delta w \, dx = 0$. As $v$ was arbitrary, we must have $\Delta w = 0$. </p>

Title: $H^1$ function with smallest seminorm


答:
It is a good start that you look at the definition of improper integrals. But in general, it is not very useful when the integrand has no easy antiderivative.

What people usually do is using the comparison principle:

If $0\le f(x)\le g(x)$ on $[a,\infty)$ then
$$
\int_a^\infty g(x)\,dx<\infty \Rightarrow \int_a^\infty f(x)\,dx<\infty
$$
and
$$
\int_a^\infty f(x)\,dx \textrm{ divergent } \Rightarrow \int_a^\infty g(x)\,dx \textrm{ divergent }
$$

This principle allows you to look at some easy integrals that you have already known whether it is convergent or not; and all you need is to compare your integral with the easy one.

In your case, observe that $\frac{1}{\log(x^2)}\ge \frac{1}{2x}$ on $[2,\infty)$.

Also, to do such problems, you should know when the following integral is convergent:
$$
\int_1^\infty \frac{1}{x^p}\,dx
$$
which is assumed to be the a priori knowledge since you need some easy integral to compare with.

Title: Is the integral $\int_{2}^{\infty} \frac{1}{\log(x^2)}\,dx$ convergent or divergent?

Body:
How can I see if this integral:
$$\int_{2}^{\infty} \frac{1}{\log(x^2)}\,dx$$ is convergent or divergent,
cause I try to resolve the integral and I saw this is a special type of integral,
can you guys help me?
PS: I know that to check if is convergent or divergent I need to do this $$\lim_{t\to\infty}\int_{2}^{t} \frac{1}{\log(x^2)}\,dx$$


问:
Query about a proof of Riemann integrability of step/simple functions
Let $f:[a,b]\to\Bbb{R}$ be a step function is of the form $$f=a_11_{[a,t_1]}+\sum\limits_{i=2}^n a_{i} 1_{(t_{i-1},t_{i}]}$$ i.e. in simple words $f(x)=a_1$ for all $x\in [a,t_1]$ and $f(x)=a_{i}$ for all $x\in(t_{i-1},t_{i}]\ \forall i\ge 2$.

I write $E_1=[a,t_1],E_{i}=(t_{i-1},t_i]\ \forall i\ge 2$, then $$f=\sum\limits_{i=1}^n a_{i} 1_{E_{i}}$$
Define
$$
\mathcal{CI}(f)=\sum\limits_{i=1}^n a_{i}(t_i-t_{i-1})
$$
We have to prove that for all $\epsilon >0$, there exists $\delta>0$ such that
$$
\left|\mathcal{CI}(f)-\sum\limits_{j=1}^k f(\xi_j)(s_j-s_{j-1})\right|<\epsilon
$$
for any portion $\mathcal{P}=\{a=s_0<\cdots<s_k=b\}$ with $\lVert \mathcal{P}\rVert(:=\text{max}\{s_j-s_{j-1}|\ j=1,\ldots,n\})<\delta$ and for any choice of $\xi_j\in(s_j,s_{j-1})$

I have tried to prove it in the following manner-

Let $\mathcal{P}=\{a=s_0<\cdots<s_k=b\}$ be a partition of $[a,b]$ and choose any $\xi_j\in (s_{j-1},s_j)$. Define a step function $g:[a,b]\to\mathbb{R}$ as follows-
$$g=\sum\limits_{j=1}^k f(\xi_j)1_{F_j}\text{ where } F_1=[a,s_1],F_j=(s_{j-1},s_j]\ \forall j\ge 2$$

Then as per our notation $\mathcal{CI}(g)=\sum\limits_{j=1}^k f(\xi_j)(s_j-s_{j-1})$

We have to approximate $|\mathcal{CI}(f)-\mathcal{CI}(g)|$. I have tried in the following manner-

$\mathcal{CI}(f)=\sum\limits_{i=1}^n a_i\lambda(E_i)$

$=\sum\limits_{i=1}^n a_i\lambda(E_i\cap [a,b])$

$=\sum\limits_{i=1}^n a_i\lambda(E_i\cap\bigsqcup\limits_{j=1}^k F_j)$

$=\sum\limits_{i=1}^n\sum\limits_{j=1}^k a_i\lambda(E_i\cap F_j)$

Similarly, $\mathcal{CI}(g)=\sum\limits_{i=1}^n\sum\limits_{j=1}^k f(\xi_j)\lambda(E_i\cap F_j)$

Then $|\mathcal{CI}(f)-\mathcal{CI}(g)|=|\sum\limits_{i,j} x_i-f(\xi_j)\lambda(E_i\cap F_j)|\le \sum\limits_{i,j}(|x_i|+|f(\xi_j)|)|\lVert \mathcal{P}\rVert\le 2nk \lVert f\rVert_{\infty} \lVert P\rVert$

Where $\lVert f\rVert_\infty=\text{sup}|f|$.

Here $n,\lVert f\rVert_\infty$ is fixed, but $k$ could vary. So I am unable to give approximation $|\mathcal{CI}(f)-\mathcal{CI}(g)|<\epsilon$.

There is a hint provided in the book that is-

Show that,
$$
\left|\mathcal{CI}(f)-\sum\limits_{j=1}^k f(\xi_j)(s_j-s_{j-1})\right|<(n-1)\lVert P\rVert \underset{i}{\text{max}}{|a_i|}
$$

But I'm unable to prove that. Can anyone help me in this regard? Thanks for help in advance.

答:
Form the partition $Q$ with points $a = x_0 < x_1  < \ldots  < x_m = b$ where

$$\{x_0,x_1, \ldots x_m\} = \{t_0,t_1, \ldots, t_n\}\cup\{s_0,s_1, \ldots, s_k\}$$

Note that $Q$ has at most $n-1$ more points and subintervals than  the partition $P = (s_0,s_1,\ldots, s_k)$. The maximum difference in the number of points occurs when  $\{t_1, \ldots, t_{n-1}\}\cap\{s_1, \ldots, s_{k-1}\}= \emptyset. $

We can write $\mathcal{CI}(f)$ and  $\mathcal{CI}(g)$ as sums of the form

$$\mathcal{CI}(f) = \sum_{p=1}^m\alpha_p(x_p-x_{p-1}),\quad\mathcal{CI}(g) = \sum_{p=1}^m\beta_p(x_p-x_{p-1}),$$

where $\alpha_p, \beta_p \in \{a_0,a_1,\ldots, a_n\}$ and $\alpha_p$ and $\beta_p$ are different on at most $n-1$ new subintervals introduced by forming partition $Q$.  We must have $|\alpha_p - \beta_p| \leqslant 2M$ where $M = \max(|a_0|,|a_1|,\ldots,|a_n|)$, and, thus

$$\left|\mathcal{CI}(f)- \mathcal{CI}(g) \right|< (n-1)\cdot \|Q\|\cdot 2M < (n-1) \cdot \|P\|\cdot 2M $$

Choose a partition $P$ such that $\delta = \|P\| < \frac{\epsilon}{2(n-1)M}$ to finish.