问:
Convergence of Sturm-Liouville eigenfunction expansions at the endpoints of an interval.
Let $\{\phi_n\}_{n=0}^\infty$ be the eigenfunctions of the regular Sturm-Liouville problem
\begin{align}
 -(p\,\phi')' + q\, \phi = \lambda \, r \, \phi \quad &\textrm{for } x \in (x_1,x_2)\\
  - a_i \, \phi(x_i)  + b_i\, (p\,\phi')(x_i) = 0 \quad &\textrm{for } i=1,2.
\end{align}
Assume that $p$ and $r$ are positive and twice continuously differentiable; assume that $q$ is continuous; the coefficients $a_i,b_i$ for $i=1,2$ are real.

Let $F(x)$ be a twice continuously differentiable function on the interval $[x_1,x_2]$. Under the above conditions, I know that
\begin{equation}
  \textrm{(I)} \quad \quad  F(x) = \sum_{n=0}^\infty \left(\int_{x_1}^{x_2} F(z) \, \phi_n(z)\, r(z)\, \textrm{dz} \right)\, \phi_n(x)
\end{equation}
with point-wise equality in the open interval $(x_1,x_2)$.

My question is: to what value does the end-point $(x=x_i)$ series
\begin{equation}
\textrm{(II)} \quad \quad  \sum_{n=0}^\infty \left(\int_{x_1}^{x_2} F(z) \, \phi_n(z)\, r(z)\, \textrm{dz} \right)\, \phi_n(x_i)
\end{equation}
converge to? Is there a general closed form expression?

If $F(x)$ satisfies the same boundary conditions as the eigenfunctions $\phi_n$, then I know that the series (I) converges to $F(x)$ uniformly on the closed interval $[x_1,x_2]$ (and so I obtain point-wise equality on the closed interval).

On the other hand, if the eigenfunctions $\phi_n$ satisfy the simpler boundary conditions $\phi_n(x_i)=0$ then the endpoint series (II) must converge to zero. The series (I) must then have a finite-jump discontinuity at the end points, e.g., jumping from $\lim_{x\rightarrow x_2}F(x)$ to $0$ at $x=x_2$. However, I am interested in the more general boundary conditions above.

I am aware of closed-form expressions for the endpoint series in the case of a Fourier expansions; I am wondering whether an analogous expression exists for regular Sturm-Liouville expansions.

Any references would be greatly appreciated.

Edit: I've linked a related question here. Is there an analogous result for Sturm-Liouville series? Do we obtain point-wise  convergence to $F(x)$ on the closed interval $[x_1,x_2]$ whenever $b_1,b_2 \neq 0$?

Edit #2: The Sturm-Liouville article on the Encyclopedia of Mathematics states that, with $b_1,b_2 \neq 0$, the expansion (I) converges under the same conditions as a Cosine series for any $F\in L^1$. Presumably, from the previous edit, this would imply that we obtain point-wise convergence to $F$ on the whole interval if $F$ is differentiable and $b_1,b_2 \neq 0$. Unfortunately, I do not have access to the articles cited in the encyclopedia.

答:
In section 9 of the Introduction to Spectral Theory: Selfadjoint Ordinary Differential Operators by Levitan and Sargsjan, the authors show that for the Sturm-Liouville problem
\begin{align}
  -y'' + q\,y = \lambda \,y\\
  y'(0) - h\, y(0) = 0 \\
  y'(\pi) + H\, y(\pi) = 0
\end{align}
on $[0,\pi]$, if $h,H \neq \infty$, then the Sturm-Liouville eigenfunction expansion converges or diverges at any point in the closed interval $[0,\pi]$ according to the behaviour of the corresponding cosine series expansion.

If one of $h$ or $H$ is infinity, one must instead compare with the $sin([n+1/2]x)$ expansion. Otherwise, if both $h=\infty, H=\infty$, then one must compare with the sine series expansion.


问:
Is it possible for a recurrence relation to not have a solution?
because I am trying to find the answer in wolfarmalpa but it show " no solution exist" !

$$a_{n}=a^{3}_{n-1}a^{2}_{n-2} ,\ a_{0}=2 , a_{1}=2a$$

I wrote it like  $a(n)=a(n-1)^3*a(n-2)^2,a(0)=2,a(1)=2$

and                $a(n)=(a(n-1))^3(a(n-2))^2,a(0)=2,a(1)=2 $

答: By induction it is easy to see that for each non-negative integer $n$ we have $a_n=2^{b_n}a^{c_n}$, where $(b_n)$ and $(c_n)$ are sequences of integer numbers such that $b_0=b_1=1$, $c_0=0$, $c_1=1$ and $b_n=3b_{n-1}+2b_{n-2}$, $c_n=3c_{n-1}+2c_{n-2}$ for each $n\ge 2$. These are linear recurrence with constant coefficients. Their characteristic polynomial $\lambda^2=3\lambda+2$  has distinct roots $\lambda_{1,2}=\tfrac 12\left(9\pm\sqrt{17}\right)$, so it has a general solution $C_1\lambda_1^n+ C_2\lambda_2^n$ for some constants $C_1$ and $C_2$. These coefficients can be easily found from the respective system of linear equations. Namely, for the sequence $(b_n)$ we have $C_1+ C_2=b_0=1$ and $C_1\lambda_1+ C_2\lambda_2=b_1=1$. For the sequence $(b_n)$ we have $C_1+ C_2=c_0=0$ and $C_1\lambda_1+ C_2\lambda_2=c_1=1$.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


问:
Determining whether two free groups are conjugates
Let $A= \langle a,b \rangle$, i.e. the free group on two generators. Also, let $B=\langle b^{-1}a^{-1}bab,b^{-1}a^2b, b^2, a^2, ab^2a^{-1}, abab^{-1}a^{-1} \rangle$, $C= \langle a^{-1}b^{-1}aba,a^{-1}b^2a, a^2, b^2, ba^2b^{-1}, baba^{-1}b^{-1} \rangle$.

I'm trying to determine whether $B$ and $C$ are conjugate in $A$. From a glance, they don't look like they are but I'm not entirely sure how to formally write it. I know that the abelinations of $B$ and $C$ are the same in $C$, but that doesn't tell me whether they are actually conjugate or not.

答:
One way to solve this by hand with pencil and paper (or e-tablet and e-pencil) is to compute the Stallings graphs of $B$ and $C$.

To describe the Stallings graph, start with the rose graph $R_A$ with one vertex $V$, and with two edges that are oriented and labelled $a$ and $b$ respectively. Any finite rank subgroup $H < A$ has a unique Stallings graph, which is a finite connected graph $\Gamma_H$ without valence 1 vertices, and which is labelled by assigning to each edge an orientation and one of the letters $a$ or $b$. The topological meaning of this is that the induced graph map $f : \Gamma_H \to R_A$ (taking each oriented, labelled edge of $\Gamma_H$ to the corresponding oriented labelled edge of $R_A$) is locally injective, and that $H$ is conjugate to the image of the induced homomorphism $f_*: \pi_1(\Gamma_H,w) \to \pi_1(R_A,v) = \langle a,b \rangle$ for any vertex $w \in \Gamma_H$. The uniqueness condition means that the label preserving isomorphism class of $\Gamma_H$ is uniquely determined by the conjugacy class of $H$.

Thus, if you can compute the Stallings graphs of $B$ and $C$ then you can examine the graphs to determine whether $B$ and $C$ are conjugate.

To compute the Stallings graph of $B$, for example, start with a rank 6 rose graph having one loop for each of the 6 words listed in the generating set for $B$. Subdivide the first loop into five edges labelled $b^{-1} a^{-1} bab$; subdivide  the second loop subdivided into 4 edges labelled $b^{-1} a a b$; and subdivide similarly for the remaining four edges.

Now repeat the following step inductively: look for a vertex having two directions coming out of that vertex with identical labels, and compute the quotient graph by identifying those two edges. For example, the first loop has first edge labelled $b^{-1}$ and the second loop has first edge labelled $b^{-1}$; so identify those edges.

The number of edges will decrease strictly with each repetition, and therefore eventually you will arrive at a graph such that at each vertex, any two directions coming out of that vertex have distinct labels. The graph you end up with might have valence 1 vertices; inductively remove each such vertex and its incident edge. At the end of that process, the resulting labelled graph is the Stallings graph of $B$.

You can find a detailed description with examples in my book.


问:
Finding linear transformation matrix without much information
I have a linear transformation $F: \mathbb R^3 \to \mathbb R[t]_2$, where $\mathbb R[t]_2$ is the vector space of polynomials up to the second degree.

I also have three different bases:
$$B = \{(1,1,0),(0,1,0),(0,0,1)\}$$
$$C = \{1, 1 + t, 1 + t^2\}$$
$$D = \{1, t, t^2\}$$

The linear transformation matrix $[F]_{B,C}$ has also been given:

$$\begin{pmatrix}1&2&-1\\1&0&-1\\0&1&0\end{pmatrix}.$$

The problem asks to write the linear transformation matrix of $[F]_{B,D}$.

What I've tried so far:

$$F(1,1,0) = something = 1(1,0,0) + 1(1,1,0) + 0(1,0,1)$$
$$F(0,1,0) = something = 2(1,0,0) + 0(1,1,0) + 1(1,0,1)$$
$$F(0,0,1) = something = (-1)(1,0,0) + (-1)(1,1,0) + 0(1,0,1)$$

So this is basically my attempt to find the linear transformation formula so I can write the matrix. But I'm not sure what to put where I wrote "something" as I'm dealing with a polynomial.

I've also thought that there may be a way to find the $[F]_{B,D}$ matrix using matrices of changing basis, but I could not think how.

Can anyone give me any hint?

答:
If you denote the identity by $I:R[t]_2\rightarrow R[t]_2$, the corresponding representation from basis $C$ to basis $D$ is (just reading from the given polynomials)

$$[I]_{C,D} =\begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$

So, the matrix looked for is

$$[F]_{B,D} = [I]_{C,D}[F]_{B,C} = \begin{pmatrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}\begin{pmatrix}1&2&-1\\1&0&-1\\0&1&0\end{pmatrix}=\begin{pmatrix}2&3&-2\\1&0&-1\\0&1&0\end{pmatrix}$$

Tudo claro agora?


问:
Distribution and pdf of sum of random variables
The independent variables $X$, $Y$ have binomial distributions with parameters $m$, $p$ and $n$, $p$, respectively. Find the pdf of $X + Y$ . What distribution does $X + Y$ have?

Can somebody give me some ideas, please?

答:
Don't use any numbers of anything fancy. Use words and an understanding of what they mean. What does it mean for $X \sim B(n,p)$ ? It means $X$ takes the value of however many time an event $E$ occurred in $n$ trails where each event occurs with probability $p$. $Y$ is nearly the same thing its just we now have $m$ trails.  Hence the variable $X+Y$ is asking us: " In $n+m$ trails how many times does $E$ occur where the chance of $E$ occurring each trail is $p$ "  This is clearly another binomial distribution with $n+m$ trails and probability $p$

Hence $X+Y \sim B(n+m,p)$

FYI this isnt a "PDF" it is a "PMF" as it is discrete not continuous


问:
If $|f(x)| \leq |x|^\alpha$ then $f$ is differentiable at $0$
In an exercise I'm asked to prove the following:

Let $\alpha > 1$. Prove that if there exists a function $f: \Bbb R \to \Bbb R$ such that:

$$\forall x \in \Bbb R, \ |f(x)| \leq |x|^\alpha$$
then $f$ is differentiable at $x = 0$

I think that we are supposed to use squeeze theorem or something like that to prove that the limit exists but the absolutes values on the condition are a little bit confusing to me to work with. How can I do this?

答: Clearly, $f(0)=0$. And then, since$$\left|\frac{f(x)-f(0)}x\right|=\left|\frac{f(x)}x\right|\leqslant|x|^{\alpha-1}.$$So, since $\alpha-1>0$, $\lim_{x\to0}\frac{f(x)-f(0)}x=0$.


问:
What is the probability mass function of max(X1, X2)?
Let X1 and X2 be two independent random variables both having discrete uniform distribution over 1, 2,.., n for some positive integer n.</p>

答: Since they are independent the event $(max(X_1,X_2)\le k)=(X_1\le k)\cap (X_2\le k)$, which has a probability of $(\frac{k}{n})^2$.


问:
solving $1+\frac{1}{x} \gt 0$
In solving a larger problem, I ran into the following inequality which I must solve:

$$ 1+\frac{1}{x} \gt 0.$$

Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.

答:
note that $\frac{x+1}{x}>0$ and $(x+1)x>0$ have the same solution(or, as you mentioned, multiply them both with $x^2$)

generally speaking, bringing the denominator over the bar doesn't change the answer, but be careful that the denominator can't be zero.

Title: solving $1+\frac{1}{x} \gt 0$


问:
Evaluate $\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z}$ with residue theorem
Here is my attempt at the problem:

$\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z}$

We have that $z=0$ is a removable singularity, so we may consider that $\tan(z)=\frac{\sin(z)}{\cos(z)}$. From here we have that $\frac{\tan(z)}{z} = \frac{\sin(z)}{z\cos(z)}$. This have a pole of order 1 at $z=\frac{\pi}{2}$. We only consider this residue because its the only residue in the circle we are integrating over. So by the residue theorem we have  $\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z} = 2\pi i \operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}]$. We can evaluate this residue by the following $\operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}] = \lim_{z \to \pi/2}(z-\frac{\pi}{2})\frac{\sin(z)}{z\cos(z)}= \lim_{z \to \pi/2}(z-\frac{\pi}{2})\frac{\sin(z)}{z\cos(z)-z\cos(\frac{\pi}{2})+z\cos(\frac{\pi}{2})}$. If we pull out a $\frac{1}{z-\frac{\pi}{2}}$ on the bottom we have $\operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}]$ = $\lim_{z \to \pi/2}\frac{\sin(z)}{z\cos'(\frac{\pi}{2})} = \lim_{z \to \pi/2}\frac{\sin(z)}{-z\sin(\frac{\pi}{2})} = \frac{1}{-\frac{\pi}{2}} = -\frac{2}{\pi}.$ So by multiplying the $2\pi i$ by this we end up with a solution of $-4i$ but the integral shouldn't be negative, so where did I go wrong on here?

答: Your computations are correct: the residue at $\frac\pi2$ is $-\frac2\pi$, and therefore the integral is $-4i$. And this number is not real; therefore, it is neither positive nor negative.


问:
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z}[\sqrt{5}]$?
Consider the integral domain $\mathbb{Z}[\sqrt{5}]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z}[\sqrt{5}]$?

I do not know the answer, so any help is welcome.

Note that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z}[\sqrt{5}]$, since its norm  $N(4+\sqrt{5})=11$ is a prime number (here as usual $N(a+b\sqrt{5})=a^2-5b^2$ for every $a, b \in \mathbb{Z}$). Anyhow $\mathbb{Z}[\sqrt{5}]$ is not a unique factorization domain, as it can be easily seen from the following factorizations $4=2 \cdot 2 = (3+\sqrt{5})(3-\sqrt{5})$. So the question is not so trivial, at least for me!

答: Note that $11=(4+\sqrt{5})(4-\sqrt{5})\in\langle 4+\sqrt{5}\rangle$, and so we have the chain of isomorphisms $$\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5}\rangle}=\frac{\mathbb{Z}[\sqrt{5}]}{\langle4+\sqrt{5},11\rangle}\cong\frac{\mathbb{Z}[t]}{\langle t^2-5,4+t,11\rangle}.$$ Also, $t^2-5=11-(4-t)(4+t)$, whence $\langle t^2-5,4+t,11\rangle=\langle 4+t,11\rangle$, and thus the above ring $\mathbb{Z}[\sqrt{5}]\big/\langle 4+\sqrt{5}\rangle$ is in fact isomorphic to $$\frac{\mathbb{Z}[t]}{\langle 4+t,11\rangle}\cong\frac{(\mathbb{Z}\big/11)[t]}{\langle \bar{4}+t\rangle}.$$ Now, $\mathbb{Z}\big/11$ is a field, so $(\mathbb{Z}/11)[t]$ is a principal ideal domain, and clearly $\bar{4}+t$ is irreducible – thus prime – in $(\mathbb{Z}/11)[t]$. This means that the above ring is a domain, and so $\langle 4+\sqrt{5}\rangle$ is indeed a prime ideal of $\mathbb{Z}[\sqrt{5}]$.


问:
Converse of Napoleon's Theorem
Consider a fixed scalene triangle $ABC$ together with a point $X$ in the semiplane determined by $BC$ opposite to $A$, a point $Y$ in the semiplane determined by $AC$ opposite to $B$ and a point $Z$ in the semiplane determined by $AB$ oppocite to $C$. Suppose that $\Delta AYC$, $\Delta CXB$ and $\Delta BZA$ are similar triangles (in that order) and moreover $\Delta XYZ$ is equilateral.

Is it true that the angles of $\Delta CXB$ are $30°$, $120°$ and $30°$?

答:
Here's a brute force derivation in the complex plane ...

Without loss of generality, take the vertices of $\triangle ABC$ to be $A=0$, $B=1+0i$, $C=c$. Then we can write
$$\begin{align}
X &= B + \rho (C-B) = 1+\rho(c-1) \\
Y &= C + \rho (A-C) = c(1-\rho) \\
Z &= A + \rho (B-A) = \rho
\end{align} \tag1$$
for some complex $\rho$. (Eg, we dilate $C$ relative to $B$ by scale factor $|\rho|$ then rotate the image about $B$ by $\arg \rho$ to get $X$; likewise for other vertices, thereby erecting identically-oriented similar triangles on each edge of $\triangle ABC$.)

As $\triangle XYZ$ is equilateral, we can get $Z$ from rotating $Y$ about $X$ by $60^\circ$, so
$$Z = X + \omega(Y-X) \quad\to\quad
c (\rho + \omega(1-2\rho) ) =
\omega\rho +(\omega -1)(1-2\rho)
\tag2$$
where $\omega := \cos60^\circ\pm i\sin60^\circ$. (The $\pm$ accounts for whether the rotation is clockwise or counter-.) The curious grouping in $(2)$ anticipates invoking the observation that $\omega-1=\omega^2$, which allows us to write

$$c (\rho + \omega(1-2\rho) ) =
\omega\left(\rho +\omega(1-2\rho)\right) \quad\to\quad (c-\omega)(\rho+\omega(1-2\rho)) = 0 \tag{3}$$
Consequently, either $c = \omega$ (making $\triangle ABC$ equilateral), or else
$$\rho = \frac{\omega}{2\omega-1}=
\frac{\frac12\pm i\frac{\sqrt{3}}{2}}{\pm i\sqrt3}=\frac1{\sqrt3} (\cos30^\circ\mp i\sin30^\circ) \tag4$$
so that $\rho$ effects the exact dilation-and-rotation to erect $30^\circ$-$30^\circ$-$120^\circ$ triangles on the edges of $\triangle ABC$ via $(1)$. (The $\mp$ distinguishes between erecting the triangles externally or internally; which is which depends upon the orientation of the vertices of $\triangle ABC$.) $\square$

Title: Converse of Napoleon's Theorem


问:
How are Taylor and Maclaurin series applied in this situation?
I am a high school student that’s having a hard time figuring out what to carry out in his report. As we know that Newton’s law of cooling shows an exponential curve in other words. This question might be stupid, I haven’t started studying both series but I am really interested on carrying out something new.  Would it be possible to use both or any series into my analysis of Newton’s curve of cooling? If not in your opinion what can I include in the analysis of the curve (even most basic would be fine).

Thanks a lot.

答:
If I understand your question correctly, you want to find a power series expansion for the exponential curve in Newton's law of cooling, which is of the form
$$ae^{bt}$$
where $a$ and $b$ are constants.

The Taylor series expansion for a function $f(x)$ at the point $a$ is
$$\sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}$$
where $f^{(n)}(a)$ denotes the $n$th derivative of $f$ evaluated at the point $a$.

The Maclaurin series expansion of $f$ is just the Taylor series of $f$ where $a=0$

Let's try to apply it to our function
$$f(t)=ae^{bt}$$
We first try to find some derivatives of $f$ in some point $a$.

let's choose for example $a=0$ (The Maclaurin series expansion).
$$f^{(1)}(t)=ab\cdot e^{bt} \Rightarrow f^{(1)}(0)=ab\cdot e^{b\cdot0}=ab$$
$$f^{(2)}(t)=ab^2\cdot e^{bt} \Rightarrow f^{(2)}(0)=ab^2\cdot e^{b\cdot 0}=ab^2$$
$$f^{(3)}(t)=ab^3\cdot e^{bt} \Rightarrow f^{(3)}(0)=ab^3\cdot e^{b\cdot 0}=ab^3$$

Do you see the pattern?

In general, we can obtain that
$$f^{(n)}(0)=a\cdot b^n$$

Now we ready to put everything together!

As I mentioned before, the Tayler series of $f$ at point $0$ (Maclaurin series) is
$$f(t)=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}t^n=\sum _{n=0}^{\infty }{\frac {a\cdot b^n}{n!}}t^n=a\sum _{n=0}^{\infty }{\frac {b^n}{n!}}t^n$$

And we obtain that the Maclaurin series of $f$ is
$$f(t)=a\sum _{n=0}^{\infty }{\frac {b^n}{n!}}t^n=a(1+bt+\frac{b^2}{2}t^2+\frac{b^3}{3!}t^3+\frac{b^4}{4!}t^4...)$$

I don't know how this can help you understand Newton's law of cooling physically
But I hope it helped. (This is my first time writing an answer so please have mercy on me in the comment section).


问:
Show that $T$ is continuous in $C([a,a+h], [\gamma - R,\gamma+R])$
Suppose that $\Phi$ is a real valued continuous function on $[a,a+h] \times [\gamma - R,\gamma + R]$ where $h > 0$, $\gamma \in \mathbb{R}$, and $R > 0$. Define $T:C([a,a+h], [\gamma - R,\gamma+R]) \to C[a,a+h]$ by $$TF(x) = \gamma + \int_a^x \Phi(t,F(t)) \, dt.$$

I want to show that $T$ is continuous on $C([a,a+h], [\gamma - R,\gamma+R])$.

My attempt: Suppose $(F_n)$ is a sequence in $C([a,a+h], [\gamma - R,\gamma+R])$ that converges to $F$, then $$|TF_n(x)-TF(x)| =\left|\int_a^x \Phi(t,F_n(t)) - \Phi(t,F(t)) \, dt \right| \leq \int_a^x \left|\Phi(t,F_n(t))- \Phi(t,F(t)) \right| \, dt.$$

This is where I'm stuck. Is there any hint or suggestion?

答: Use the hypothesis $\Phi$ is continous to go one step further in the integral


答: $|TF_n(x)-TF(X)| \leq \int_a^h \left|\Phi(t,F_n(t))- \Phi(t,F(t)) \right| dt$. Given $\epsilon >0$ choose $\delta >0$ such that $|\Phi (t,x)-\Phi(t,y)|  <\epsilon $ for $|x-y|<\delta$. This is possible because $\Phi$ is uniformly continuous. Now choose $n_o0$ such that $|F_n(t)-F(t)| <\delta$ for all $t$ if $n \geq n_0$. We now get $|TF_n(x)-TF(x)| <\epsilon (h-a)$ for all $t$ is $n \geq n_0$.

Title: Show that $T$ is continuous in $C([a,a+h], [\gamma - R,\gamma+R])$

Body:
Suppose that $\Phi$ is a real valued continuous function on $[a,a+h] \times [\gamma - R,\gamma + R]$ where $h > 0$, $\gamma \in \mathbb{R}$, and $R > 0$. Define $T:C([a,a+h], [\gamma - R,\gamma+R]) \to C[a,a+h]$ by $$TF(x) = \gamma + \int_a^x \Phi(t,F(t)) \, dt.$$

I want to show that $T$ is continuous on $C([a,a+h], [\gamma - R,\gamma+R])$.

My attempt: Suppose $(F_n)$ is a sequence in $C([a,a+h], [\gamma - R,\gamma+R])$ that converges to $F$, then $$|TF_n(x)-TF(x)| =\left|\int_a^x \Phi(t,F_n(t)) - \Phi(t,F(t)) \, dt \right| \leq \int_a^x \left|\Phi(t,F_n(t))- \Phi(t,F(t)) \right| \, dt.$$

This is where I'm stuck. Is there any hint or suggestion?


答:
After reading your comment here is how to go about it when you multiply by $x^2$. So we need to solve

$$x^2 + x > 0.$$

That can be written as

$$x(x + 1) > 0.$$

Now the product of two numbers is positive when both numbers are positive or when both are negative. Therefore we have two cases to solve:

Case 1:

$x >0$ and $x + 1> 0$

Both of these inequalities are satisfied when $x > 0$.

Case 2:

$x < 0$ and $x + 1 < 0$

Both of these inequalities are satisfied when $x < -1$.

Hope that helps!

Title: solving $1+\frac{1}{x} \gt 0$

Body:
In solving a larger problem, I ran into the following inequality which I must solve:

$$ 1+\frac{1}{x} \gt 0.$$

Looking at it for a while, I found that $x\gt 0$ and $x\lt -1$ are solutions. Please how do I formally show that these are indeed the solutions.


问:
How can the two basic binary operations (addition and subtraction) be defined in set-theoretical terms?
I recently stumbled upon this interesting definition of mathematics:

Math is the study of things that can be described as sets.

I am aware that the integers and the real numbers can be defined in terms of sets, but how can we define the operations of addition and subtraction?

答:
Just to give an idea ( but I'm far from being an authority).

You may have a look at Peterseon, Theory Of Arithmetic.

One may say that : $0$ is ( by definition) the same object as $\emptyset$.

Define $S(n)$ ( the successor of number $n$) as the set : $n \cup \{n\}$.

So, for example, $S(0) = S(\emptyset)= \emptyset\cup\{\emptyset\}=\{\emptyset\}$.

Call this number: " number 1". Hence : $1 =\{\emptyset\} =\{0\}$ by definition.

Note : continuing the process , you will get this ( at first sight)  " strange" result that every natural number is the set of all its predecessors.

Then state two rules in order to define addition as a function taking as input any couple of natural numbers ( that is, any element of the cartesian product $N\times N$ ) and sending back as output an element of $N$ (meaning that addition is an operation on $N$):

(1) $n+O=n$ ( $n$ being any natural number)

(2) $n+ S(m) = S (n+m)$ ( $n$ and $m$ being any natural numbers).

Note : the rules mean that, $a+b= a$ if $a = O$ and that $a+b= S(a+ $ predecessor of $b$) if $b\neq 0$.

So for axample:

$3+2 = 3+S(1) = S(3+1) = S (3+S(0))= S S( 3+0) = SS(3) =S(4)= 5$.

( 5 being " the successor of 4" by definition).

As to substraction, onece you have defined the integers ( using an equivalence relation on the set of couples of natural numbers) , you can say that :

$n-m = n+(-m)$.


问:
Proof P by premises $(\neg P\vee Q)\to P$
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:

$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.

The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?

答:
I have to proof next question by a formal proof that when the premise is equal to:
$(\neg P\vee Q)\to P$ has the result ,$P$, I first though that I must write:

$\neg P\vee Q\to P$, as:
$\neg(\neg P\vee Q)\vee P$ and work further from there.

The problem is that I only may use Reit ,introduction and Elimination. Can someone help me?

Indeed, that is not a fundamental inference in Natural Deduction.  Well, sure, you may derive it, but it is not really that efficient a strategy in the system.

To derive a consequent from its conditional, you would either (1) derive the antecedent so to use conditional elimination, or (2) show that a contradition would be derived under an assumption of the negation of that consequent, so to use reduction to absurdity.  Strategy (1) cannot be done here, thus you should attempt strategy (2).

$$\def\fitch#1#2{~~~~\begin{array}{|l}#1\\\hline #2\end{array}}\fitch{~~1.~(\lnot p\lor q)\to p\hspace{2ex}\textsf{Premise}}{\fitch{~~2.~\lnot p\hspace{9ex}\textsf{Assumption}}{~~3.~\\~~4.~\\~~5.~\bot\hspace{10ex}\textsf{Negation Elimination (?,?)}}\\~~6.~\lnot\lnot p\hspace{11ex}\textsf{Negation Introduction (2-5)}\\~~7.~p\hspace{14ex}\textsf{Double Negation Elimination (6)}}$$

So, determine what you will be contradicting, and a way you would derive that.


问:
Finding the integral $\int_0^\pi\frac{d\theta}{(2+\cos\theta)^2}$ by complex analysis
Trying to find the integral $$\int_0^\pi\frac{d\theta}{(2+\cos\theta)^2}$$ by complex analysis.

I let $z = \exp(i\theta)$, $dz = i \exp(i\theta)d\theta$, so $ d\theta=\dfrac{dz}{iz}$. I am trying therefore to find the integral $$\frac{1}{2iz} \oint_C \frac{dz}{\left(2 + \frac{z}{2} + \frac{1}{2z}\right)^2}$$ I am unsure of which contour I should use, and how to proceed besides that. Could anyone help?

答:
You first could take a look at an easier Integral to evaluate. Let F(a) be $$\int_{0}^{\pi}\frac{1}{a+\cos\left(x\right)}dx$$ Now, in complex form, this Integral will be $$\int_{0}^{\pi}\frac{1}{a+\frac{e^{ix}+e^{-ix}}{2}}dx$$ $$\int_{0}^{\pi}\frac{1}{a+\frac{e^{ix}+e^{-ix}}{2}}dx=2\int_{0}^{\pi}\frac{1}{2a+e^{ix}+e^{-ix}}dx$$ So after we do the substitution $$z = e^{ix}$$ This integral will be equal to $$\int_{c}^{ }\frac{2}{\left(2a+z+z^{-1}\right)iz}dz=\int_{c}^{ }-\frac{2i}{z^{2}+2az+1}dz$$ after we do some substitution. Now let's evaluate this using complex analysis. Using the quadratic equation, we notice that the pole of this equation will be at $$\sqrt{a^{2}-1}-a$$ (There's also the negative version, but that is outside of the unit circle). Then, now what we need to do is just apply the Residue Theorem (As in this particular expression we don't have major "problems" we need to deal with) $$Res_{z=\sqrt{a^{2}-1}-a}=\frac{-2i}{2\sqrt{a^{2}-1}}$$ Now we just need to multiply the Residue by 2i. So our result for F(a) will be $$\frac{\pi}{\sqrt{a^{2}-1}}$$ After all that, we get $$\int_{0}^{\pi}\frac{1}{a+\cos\left(x\right)}dx = \frac{\pi}{\sqrt{a^{2}-1}}$$ That initially was our function F(a). Now we're going to use a "trick", that is the Leibniz’s rule for differentiation under the sign of integration sign. I'm not going over all of that, but you probably already know it. $$\int_{0}^{\pi}\frac{1}{a+\cos\left(x\right)}dx = \frac{\pi}{\sqrt{a^{2}-1}}$$
Differentiating both sides in terms of a, we get $$-\int_{0}^{\pi}\frac{1}{\left(a+\cos\left(x\right)\right)^{2}}dx = \frac{-a\pi}{\sqrt{\left(a^{2}-1\right)^{3}}}$$ As we have the negative sign in both sides, we can just cancel them. Leaving us with the result $$\int_{0}^{\pi}\frac{1}{\left(a+\cos\left(x\right)\right)^{2}}dx = \frac{a\pi}{\sqrt{\left(a^{2}-1\right)^{3}}}$$
This expression works for all real values in which the magnitude of a is greater than one (if the magnitude of a is equal or less than one, the integral is undefined)
The answer for your original question is $$\int_{0}^{\pi}\frac{1}{\left(2+\cos\left(x\right)\right)^{2}}dx=\frac{2\pi}{\sqrt{27}}$$
Sorry if something was not very well described, I'm not a teacher and that's the first time I write something in StackExchange. And if you felt some lacking in the citations of the techniques I used, you can just google them and it should be very easy to find some paper about it. Thank you, I hope it was helpful to someone in any way.


问:
The functional equation $f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$
I'm having trouble with this functional equation:

Find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$:
$$f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$$

I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity, So no progress. I tried $x=y=0$, $x=0$, $y=0$, $x=-f(y)$, $y=f(x)$, and even with some assumptions on bijectivity, but nothing helped.

I don't know the source, but it seems like a pretty difficult problem, maybe even impossible to solve.

答:
I do not have a solution, but I want to share with you my ideas so far. To make things every to read, we write $f^k$ for $f\circ \dots\circ f$ $k$-times.

Applying $f$ to the left hand side, we obtain the expression
$$
f\big{(} \underbrace{f(x)}_{=:x'} + f(\underbrace{x + f(y)}_{=:y'})\big{)} = f(x' + f(y')) = y' + f\left(f(x') + f^2(y')\right) - f(x'),
$$
i.e. $$
f(f(x) + f(x+f(y))) = x + f(y) + f(f^2(x) + f^2(x+f(y))) - f^2(x).
$$
But we can also write
$$
f\big{(} f(\underbrace{x}_{=:y'}) + \underbrace{f(x + f(y))}_{=:y'}\big{)} = x + f(f^2(x + f(y)) + f^2(x)) - f^2(x+f(y)).
$$
Comparing the two quantities gives
$$
f(y) = f^2(x) - f^2(x+f(y)).
$$
For $y = f(x)$, this implies
$$
f^2(x + f^2(x)) = 0
$$
for all $x\in\Bbb R$. Also, if $f$ was surjective, we could find $y$ such that $f(y) = -x$ giving $$
f^2(x) = f^2(0) - x.
$$
Combining this with the previous yields $$f^4(0) = 0.$$

(Edit: more precisely, we get
$$
f^4(x) = f^2(0)- f^2(x) = x.
$$
In particular, subjectivity implies bijectivity and gives the inverse function explicitly.)

Naturally, the last bit completely relies on surjectivity which might not be given.

That are my thoughts so far.


问:
Understanding the Proof of Maschkes's Theorem
I am currently reading "A Course in Ring Theory" by Passman, and I know other questions about this proof have been asked here before, but I really want to understand the proof in the book.

The theorem states "Let $K$ be a field and let $G$ be a finite group. Then $K[G]$ is an Artinian ring that is Wedderburn if and only if the characteristic of $K$ dose not divide $|G|$."

I understand the direction wear he shows if the characteristic of $K$ divides the order of $G$ then $K[G]$ is not Wedderburn. For the other direction the author starts off with

"Conversely, assume that $|G| = n$ is not $0$ in $K$ and consider the $n \times n$ matrix ring $M_n(K)$, where we index the rows and columns by the elements of $G$. Then the map $\theta : K[G] \rightarrow M_n(K)$ given by $\theta(g) = (\delta_{y,xy})$ is easily seen to determine a $K$-algebra embedding, since
$$\theta(g)\theta(h) = (\delta_{y,xg})(\delta_{y,xh}) = (\delta_{y,xgh}) = \theta(gh)$$
Here, of course, $g, h, x, y \in G$. Furthermore =, $\delta_{x, y} = 1$ when $x = y$ and $\delta_{x,y} = 0$ otherwise."

There is more to this direction of the proof, but My problem is I really don't understand the map $\theta$. I know $(\delta_{y,xg}) \in M_n(K)$, but know what it looks like.

答:
I guess, it writes $\theta(g)=(\delta_{y,x{\pmb g}})$, and it's meant as the matrix with the $x,y$ (or $y,x$) entry $1$ if $y=xg$ and $0$ otherwise.

In other words, the matrix $\theta(g)$ is a permutation matrix which sends the basis element $x\in G$ of the $K$-vector space $K[G]$ to the other basis element $xg$, so this is just multiplying by $g$, which is a linear map in the group algebra.

Now, $\theta$ is defined on the basis $G$ of $K[G]$, and if we choose the correct directions (actually, you might rather need multiplication from left or by $g^{-1}$), then it maps multiplication on $G$ to matrix multiplication, as stated, and obviously it extends linearly to $K[G]$.


答:
I arrive at a proof that no such funtion exist (assuming the statement is correct).

Replace $x$ with $f(x)$ in the original equation:
$$f(f(x)) - y = f(f(f(x)) + f(f(y))) - f(f(x) + f(y)).\tag{1}$$

The right hand side is symmetric in $x$ and $y$, so we get
$$f(f(x)) - y = f(f(y)) - x,$$
or $f(f(x)) + x$ is constant for all $x$. Write $c$ for this constant. Thus $$f(f(x)) = c - x.\tag{2}$$

(In particular, this shows that $f$ is both injective and surjective.)

It follows that
$$f(c - x) = f(f(f(x))) = c - f(x),\tag{3}$$
which gives $f(\frac c 2) = \frac c 2$.

Using (2), the equation (1) becomes
$$x + y + f(2c - x - y) = c + f(f(x) + f(y)).\tag{4}$$

Put $x = \frac c 2$ and $y = c$:
$$c = f(\frac c 2 + f(c)).$$

Applying $f$ again and using (2), we get $f(c) = \frac c 4$, and then $f(0) = \frac {3c}4 $ by (3).

In (4), replace $x$ with $0$ and $y$ with $x + y$:
$$x + y + f(2c - x - y) = c + f(\frac {3c}4 + f(x + y)).$$

We compare this with (4). From injectivity of $f$, we get

$$f(x + y) + \frac {3c} 4 = f(x) + f(y).\tag{5}$$

If we define $g(x) = f(x) - \frac{3c} 4$, then we have $g(x + y) = g(x) + g(y)$. This implies that $g(rc) = rg(c) = -\frac{rc}2$ for any rational number $r$.

Therefore $f(rc) = (\frac 3 4 - \frac r 2)c$ for any rational number $r$.

Applying $f$ again gives $f(f(rc)) = (\frac 3 8 + \frac r 4) c$. Together with (2), we conclude that $c = 0$.

Thus (5) becomes $f(x + y) = f(x) + f(y)$ and (2) becomes $f(f(x)) = -x$.

The original equation can then be rewritten as:
$$f(x) + f(x) + f(f(y)) = y + f(f(x)) + f(f(f(y))),$$
or:
$$2f(x) + f(y) = 2y - x.\tag{6}$$

Apply $f$ again:
$$-2x - y = 2f(y) - f(x).\tag{7}$$

We take a linear combination of (6) and (7) and get:
$$5f(x) = 5y.$$

This obviously cannot hold for all $x, y \in \Bbb R$.

Title: The functional equation $f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$

Body:
I'm having trouble with this functional equation:

Find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$:
$$f(x) + f\bigl(x+f(y)\bigr) = y + f\Bigl(f(x) + f\bigl(f(y)\bigr)\Bigr)$$

I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity, So no progress. I tried $x=y=0$, $x=0$, $y=0$, $x=-f(y)$, $y=f(x)$, and even with some assumptions on bijectivity, but nothing helped.

I don't know the source, but it seems like a pretty difficult problem, maybe even impossible to solve.


问:
properties of recursively enumerable sets
$A \times B$ is an r.e.(recursively enumerable) set,  I want to show that $A$ (or $B$) is  r.e. ($A$ and $B$ are nonempty)

I need to find a formula. I've got an idea that I should use the symbolic definition of an r.e. set. That is, writing a formula for the function that specifies $A$ or $B$, assuming a formula exists for $A \times B$. I guess I must use Gödel number somewhere.

I should've mentioned that I asked this question over there (Computer Science) but since I am more interested in mathematical arguments and looking for formulas, I'd ask it here as well.

答:
The notion of computable or c.e. is usually defined on $\omega$. To understand what $A \times B$ being computable or c.e., you should identify order pairs $(x,y)$ with the natural number under a bijective pairing function. Any of the usual standard pairing function, the projection maps $\pi_1$ and $\pi_2$ are computable.

First note that $\emptyset \times B = \emptyset$. So the result does not hold if one of the sets is empty.

Suppose $A$ and $B$ are not empty. Suppose $A \times B$ is c.e. Then there is a total computable function $f$ with range $A \times B$. Then $\pi_1 \circ f$ and $\pi_2 \circ f$ are total computable functions with range $A$ and $B$, respectively. So $A$ and $B$ are c.e.


问:
Folded n-dimensional gaussian integral
I am dealing with the following integral:
\begin{equation}
\int_0^{\infty} dx_1 \ldots \int_0^{\infty} dx_n \quad e^{-J^T x - x^T A x}
\end{equation}
Is there any general analytical solution which gives the above integral in terms of the determinant of $A$, $J$ and (I guess) the error function?

Thanks!

答:
Update: I didn't notice the domain was different. Please ignore this answer.

<s>This is standard in quantum field theory. Take a look here or any book on the topic.</s>

Title: Folded n-dimensional gaussian integral


问:
Proving that $P(B|C \cap A^{c}) = P(B|C)$
I was solving the following question:

Most mornings, Victor checks the weather report before deciding whether to carry an umbrella.
If the forecast is “rain,” the probability of actually having rain that day is 80%.
On the other hand, if the forecast is “no rain,” the probability of it actually raining is equal to 10%.

During fall and winter, the forecast is “rain” 70% of the time and during summer and spring, it is 20%.

The probability of Victor missing the morning forecast is equal to 0.2
on any day in the year. If he misses the forecast, Victor will flip a
fair coin to decide whether to carry an umbrella.

On any day of a given season, he sees the forecast, if it says “rain” he will always
carry an umbrella, and if it says “no rain,” he will not carry an
umbrella. Are the events “Victor is carrying an umbrella,” and “The
forecast is no rain” independent? Does your answer depend on the
season?

And my attempt to solve it is as following:

Let A be the event of the forecast is "rain".

Let B be the event of Victor carrying an umbrella.

Let C be the event of Victor misses the forecast.

From the statement we have that:
$$P(A) = 0.2 \space or \space 0.7 $$
$$P(B|A) = 0.8 $$
$$P(B|A^{c}) = 0.1 $$
$$p(C) = 0.2$$

So we must prove that B and A are independent events.

Since $A^{c}$ has non-zero probability we can prove independence by proving that:

$$P(B|A^{c}) = P(B)$$

From Total Probability Theorem, we can write:
$$P(B) = P(C)P(B|C) + P(C^{c})P(B|C^{c})$$
Since $P(B|C^{c}) = P(A)$ (Because if he doesn't miss the forecast then he will carry an umbrella if the forecast is "rain"), then:
$$\implies P(B) = (0.2)(0.5)+ (0.8)P(A)$$
$$\implies P(B) = 0.1 + (0.8)P(A)$$

Now Also from the Total Probability Theorem, we can write:
$$P(B|A^{c}) = P(C)P(B|C \cap A^{c}) + P(C^{c})P(B|C^{c} \cap A^{c})$$

It's obvious that $P(B|C \cap A^{c}) = P(B|C)$, and from that It's easy to complete solving the problem.

My question is how we can prove that $P(B|C \cap A^{c}) = P(B|C)$.

答:
Since I don't find a formal proof of this, I'll write the best result that I come up with, I think it's the solution and please correct me if I'm wrong.

Proving that: $P(B| C \cap A^{c}) = P(B | C)$ is tell us that the two events A and B are conditionally independent (that condition is C here).

Since B is determined by a random process If he misses the forecast, Victor will flip a fair coin to decide whether to carry an umbrella then we can say that B is independent from A since flipping a fair coin is not related by any way to A and have a fixed probability which equal to 0.5 .


问:
Prove or Disprove an Inequality Raised to the Fifth (UPDATED... Again)
I found this interesting problem:

Given two sequences of whole numbers, $a_1, a_2, a_3 \cdots a_n$ and $b_1, b_2, b_3 \cdots b_n$ with the property that: for all $i$, $a_i + b_i = n-1$ and $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i,$ let $f(x) = \sum_{i=1}^n a_i^x$ and $g(x) = \sum_{i=1}^n b_i^x$.

Given that $f(4) \ge g(4)$ prove or disprove that $f(5) \ge g(5).$

So far, I tried using the info that $f(4) \ge g(4)$ to get: $$\sum_{i=1}^n a_i^4 - b_i^4 = (a_i - b_i)(a_i + b_i)(a_i^2 + b_i^2) \ge 0.$$

Since $a_i + b_i = n-1$,we can factor it out and divide by $n-1$ to obtain: $$\sum_{i=1}^n (a_i - b_i)(a_i^2 + b_i^2) \ge 0.$$

Then, letting $a_i^2 + b_i^2 = (a_i+b_i)^2 - 2a_ib_i = (n-1)^2 - 2a_ib_i$, I got: $$(n-1)^2 \sum_{i=1}^n (a_i - b_i) - 2\sum_{i=1}^n a_ib_i(a_i - b_i) \ge 0.$$

Since $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i, \sum_{i=1}^n (a_i - b_i) = 0.$ That leaves us with: $$\sum_{i=1}^n a_ib_i(a_i - b_i) \le 0.$$

Then, I did something similar with the $f(5) \ge g(5)$ condition. Ultimately, after lots of factoring and manipulation, I am basically left to prove/disprove that: $$\sum_{i=1}^n (a_i - b_i)a_i^2b_i^2 \ge 3(n-1)^2 \sum_{i=1}^n (a_i - b_i)a_ib_i.$$

Bringing everything over to one side and factoring yields: $$\sum_{i=1}^n (a_i - b_i)a_ib_i (a_ib_i -3(n-1)^2) \ge 0.$$

Anyone know how to continue

Body:
I just wrote some simple codes to find a counterexample and I found tons:

With $a = [ 34,  1, -19, -39,  14, -13,  43]$ and $b = [-28,   5,  25,  45,  -8,  19, -37]$,
$$7265877>7115109,\text{ but }99910101<110185881.$$

With $a = [ -3,  46, -18,  42,   9, -10, -45]$ and $b = [ 9, -40,  24, -36,  -3  ,16,  51]$,
$$11811395>11408771,\text{ but }150195321<191229081.$$

With $a = [ -8, -70, -22,  42,  56,  78, -54, -26,  40]$ and $b = [ 16,  78,  30, -34, -48, -70,  62,  34, -32]$,
$$85729632>85706592,\text{ but }1514064576<1859597376.$$

Update

Here is a counterexample in $\mathbb N^*$.

a = [ 4  7 13  1 14  8 12  7  8 10  3 12  7 13 16  3  8  7]
b = [13 10  4 16  3  9  5 10  9  7 14  5 10  4  1 14  9 10]
234857 >= 234857
3093693 < 3098013

The codes.

This problem can be described using matrices.

Let $A \in M_n(\mathbb R)$ be a positive semidefinite matrix with $\operatorname {tr} A = n(n-1)/2$. Let $B = (n-1)I_n - A$ and we have then
$$\operatorname {tr} A = \operatorname {tr} B.$$
And your proposition is:
$$\operatorname {tr} A^4\ge\operatorname {tr} B^4 \overset{?}\implies \operatorname {tr} A^5\ge\operatorname {tr} B^5,$$
and the answer is no by the counterexample.


问:
Linear interpolation of points in isometric isomorphic spaces
Suppose that we have two spaces $\mathcal{F}$ and $\mathcal{H}$ and we know that $\mathcal{H}$ is isometric isomorphic to $\mathcal{F}$, so that distances and angles are preserved. Note that we are also given the mapping from one space to the other ($\Phi$).

Then, if I do a linear interpolation between two points $\mathbf{x}_1$ and $\mathbf{x}_2$ of $\mathcal{H}$ so that:
$$\mathbf{x}_\mathrm{new} = \mathbf{x}_1 + (\mathbf{x}_2 - \mathbf{x}_1)\cdot \delta $$
where $\delta$ is some constant.

Is there any way to verify that if I compute the map of $\mathbf{x}_\mathrm{new}$ from $\mathcal{H}$ to $\mathcal{F}$ it will be a linear interpolation of $\mathbf{x}_1$ and $\mathbf{x}_2$ in $\mathcal{F}$ as well?

答: Consider the line segments joining  $\mathbf x_1$ to $\mathbf x_2$ and to $\mathbf x_{\text{new}}$.  "Linear interpolation" as you described it, means that the angle between them is $0$, and the length of the latter is $\delta$ times the length of the former.  Your hypotheses say that this information is preserved by $\Phi$.</p>


问:
Numerical Analysis or Numerical Linear Algebra Book with Available Solutions?
I am a beginning graduate student in Mathematics soon and I am planning to self-study Numerical Analysis and Numerical Linear Algebra. I know there are already reference-request questions about this, but I am looking for some more specific books.

I am looking for books in the following categories:

Books with difficult problems, and solutions in the back (ideal)

Books with difficult problems

Books with solutions in the back

Books with separate solution manuals

Problem books (ideally with solutions in the back, or solution manuals)

Could you please recommend me some textbooks, and tell me in which category they are? Thank you very much, any recommendations are immensely appreciated!

答:
I would recommend this book:

Burden, Richard L., and J. Douglas Faires. Numerical Analysis. 9th ed. 2010. ISBN-13: 978-0538733519.

Since difficulty is necessarily subjective, I wouldn't attempt to answer what category it is in. However, here's a free PDF version in case you want to check out what it's like.


问:
Asymptotic approximation for $E[X_n]$ where $\{X_n\}$ converges in probability to a constant $\mu$?
I have a sequence of bounded random variables $\{X_n\}$, i.e. $|X_n| \le C$ for all $n$.

Suppose this sequence convergences in probability to a constant $\mu$ with known rate of convergence, e.g.
$$
X_n - \mu = o_p\bigg(\frac{1}{n}\bigg), \quad \quad \quad \quad \text{or} \quad \quad \quad \quad X_n - \mu = O_p\bigg(\frac{1}{n}\bigg).
$$

Suppose also that $\{X_n\}$ is uniformly integrable. Then $X_n$ converges in expectation to $\mu$. But what about the asymptotic rate of convergence of the expectation? Do the $O_p$ and/or $o_p$ rates of convergence translate into a corresponding rate of convergence for the expectation? For example, do they allows us to say something like the following: As $n \to \infty$,
$$
E[X_n] = \mu + O\bigg(\frac{1}{n}\bigg).
$$

答:
I borrow the argument of shrinking bumps (centered at zero, of width $1/\sqrt{n}$ and constant height $1$) from Convergence in probability implies boundedness in $L^1$? to answer your question.

WLOG, assume that $\mu = 0$.  Consider $((0,1],\mathcal{B}((0,1]),\text{Leb})$, and
the family of random variables $X_n(\omega) := 1_{(0,1/\sqrt{n})}(\omega)$ for $\omega \in (0,1]$.  Fix $\epsilon \in (0,1)$ (independent of $n$).  $X_n$ is an indicator function, so we write $X_n$ instead of $|X_n|$.

Check that $nX_n = O_p(1)$.  (As the choice of $\epsilon$ doesn't depend on $n$, it's also $o_p(1)$.)  Observe that the value of $X_n$ is either $0$ or $1$.

\begin{align}
P(nX_n \ge \epsilon) &= P(X_n \ge \epsilon/n) = 1/\sqrt{n} \to 0 \text{ as } n \to +\infty, \\
\text{but } E[nX_n] &= E[n1_{(0,1/\sqrt{n})}] = n (1/\sqrt{n}) = \sqrt{n} \to \infty.
\end{align}
Hence $o_p$ and $O_p$ don't translate to $o$ and $O$ in expected values respectively.


问:
Covariant functor, and left exact
I'm reading A first course of Homological Algebra by Northcott, and there is something that the author said it was straightforward. But for some reason, I just don't see the straightforwardness of it.

Definition

And let $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$; where $\mathscr{C}_\Lambda$; $\mathscr{C}_\Delta$ be the covariant functor between the category of $\Lambda-$left module, and $\Delta-$left module.

$F$ is said to be left exact iff for every exact sequence of left module: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C \rightarrow 0$, the following induced sequence is also exact:
$0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$

Property

Of course if $F$ is a left exact functor, then it'll preserve monic.

Theorem

Suppose $F:\mathscr{C}_\Lambda \to \mathscr{C}_\Delta$ is a left exact covariant functor, and that $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} C$ is exact in $\mathscr{C}_\Lambda$. Prove that $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is also exact in $\mathscr{C}_\Delta$.

Here's my proof, I'll be very glad if you guys can check my proof, as well as give me some hints to prove it in another easier way (if another way exists).

Proof

Since $F$ preserves monic, the sequence $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi)} F(C)$ is exact at $F(A)$.

$\pi \sigma = 0 \Rightarrow 0 = F(\pi \sigma) = F(\pi)F(\sigma)$

What left is to prove $\ker F(\pi) \subset \mbox{im } F(\sigma)$

Consider the exact sequence: $0 \rightarrow A \xrightarrow{\sigma} B \xrightarrow{\pi} \mbox{im }\pi' \rightarrow 0$, where $\pi'$ is the same as $\pi$ with the codomain contracted to be $\mbox{im } \pi$ since $F$ is left exact, it means that: $0 \rightarrow F(A) \xrightarrow{F(\sigma)} F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi)$ is also exact.

Consider the following chain of homomorphism: $F(B) \xrightarrow{F(\pi')} F(\mbox{im }\pi) \xrightarrow{F(i)} F(C)$

Since $i$ is monic, $F(i)$ is also monic (I think so, but is it right?)

So $\ker F(\pi) = \ker [F(i)F(\pi')] = \ker F(\pi') = \mbox{im }F(\sigma)$.

Is my proof valid? And is there any shorter proof? As the author said that it's straightforward, and I don't really think my proof can be classified to be straightforward.

Thanks guys very much,

Have a good day,

答:
I think you got to the point beginning with the sentence "Consider the exact sequence..." You can probably omit the work before that, although there is no mistake there. After that sentence, you get that the sequence is exact at $F(A)$ for free.

The mathematical leap was that you took the given and recognized that there is a relevant short exact sequence related to the exact sequence you were given, and you could apply the definition of "left exact functor" to that. To wrap things up, you showed that nothing was lost in the leap, and that you could pass back to the original sequence's maps.

Title: Covariant functor, and left exact


问:
Shifting the index of a sequence to prove convergence.
I'm currently studying sequences in Real Analysis and really struggling with answering this basic question:

Prove that if $x_n→a$, then for any fixed positive integer k the sequence $\{y_n\},y_n=x_{n+k}$ also converges to $a$

I honestly have no idea how to even start this proof. Could someone provide me with some hints? Specifically, what does $y_n=x_{n+k}$ represent in terms of sequences? Is this a case of induction?

答:
Suppose $x_n \to$ a with a $\in \mathbb{R}$. Then by the definition of limit we can write:
$\forall \varepsilon \geq 0(\exists n\in \mathbb {N} (\forall m\in \mathbb {N} (m\geq n\implies |x_{m}-a|<\varepsilon )))$.

Note that given $k \in \mathbb {N}$, n $< n+k$. So it's still true that:

$\forall \varepsilon \geq 0(\exists n\in \mathbb {N} (\forall m\in \mathbb {N} (m\geq n\implies |x_{m+k}-a|<\varepsilon )))$, by the definition of $y_n$ we get:

$\forall \varepsilon \geq 0(\exists n\in \mathbb {N} (\forall m\in \mathbb {N} (m\geq n\implies |y_{m}-a|<\varepsilon )))$. So $y_n$ converges by definition (to a).


答:
A first thing one could do is to understand intuitively why the desired property is correct. When we have a sequence $(x_n)$ converging to a number $a$, then
for all $n$ big enough, $x_n$ gets really close to $a$. Since for $k\in \mathbb{N}$, sequence $(x_{n+k})=(x_{k+1},x_{k+2},\ldots)$ just ignores the first $k$ terms of $(x_n)$, the behaviour does not change for large enough $n$ and therefore it is reasonable to suspect that it also converges to $a.$

Now, for the proof.

Heuristic. We start by stating precisely what we want to prove:

for all $\varepsilon>0$ there exists $n_0\in \mathbb{N}$ such that for all $n\geqslant n_0$, $|x_{n+k}-a|<\varepsilon$.

Since you want to prove something for all $\varepsilon>0$, let ${\color{red}\varepsilon}>0$ (be random, but fixed). Now we are looking for a suitable $n_0$ such that for all $n\geqslant n_0$, $|x_{n+k}-a|<{\color{red}\varepsilon}$. We don't have any other choice than using what we already have:  $\varepsilon>0$ and the fact that $(x_n)$ converges to $a$. Especially, if you use the convergence, you know that for all
${\color{blue}\varepsilon}>0$ there exists $n_0\in \mathbb{N}$ such that for all $n\geqslant n_0$, $|x_{n}-a|<\varepsilon$. Note that, although we used the same notation,
the ${\color{red}\varepsilon}$ at the beginning of our proof is not the same as ${\color{blue}\varepsilon}$ (since the last one comes along with quantifier for all ($\forall$)). Since you know that for all
${\color{blue}\varepsilon}>0$ something interesting happens, an interesting choice would be to use ${\color{red}\varepsilon}$ (which is positive).
Indeed, using ${\color{red}\varepsilon}>0$ there is some
$n_1\in \mathbb{N}$ such that for all $n\geqslant n_1$, $|x_{n}-a|<{\color{red}\varepsilon}$.

Ok, now we found (basically, not really found, but proved the existence of a) $n_1$ such that terms  $x_{n}$ get ${\color{red}\varepsilon}$ close to $a$ thereafter. Now it all comes to the relation of $n_1$ and $k.$
If $n_1$ is bigger than  $k$, then the terms $x_{n_1},x_{n_1+1},\ldots$ are terms of the sequence $(x_{k+1},x_{x+2},\ldots)$ and of course are ${\color{red}\varepsilon}$ close to $a$, so if we choose $n_0:=n_1-k\in \mathbb{N},$ then for $n\geqslant n_0$ one gets $n+k\geqslant n_1$ and so terms
$x_{n+k}$ are ${\color{red}\varepsilon}$ close to $a$. On the other hand, if
$n_1\leqslant k$, then for all $n\geqslant 1$ we have $n+k> n_1$ and so terms
$x_{n+k}$ are ${\color{red}\varepsilon}$ close to $a$, so in this case choosing $n_0:=1$ is good enough. Eventually, one could choose: $$n_0:=\left \{\begin {array}{lll}
n_1-k&,~ n_1>k \\
1&, n_1\leqslant  k\\
\end{array}
\right..$$

Formal presentation. (includes all the above, without comments and what instructors often write and let you discover the rest):
let $\varepsilon>0$. Since $(x_n)$ converges to $a$, there exists $n_1\in \mathbb{N}$ such that for all $n\geqslant n_1$ $|x_{n}-a|<\varepsilon$.
If $n_1>k$, then   for $n\geqslant n_1-k$, we   have $|x_{n+k}-a|<\varepsilon$ and
if $n_1\leqslant k$ then for $n\geqslant 1$, we   have $|x_{n+k}-a|<\varepsilon$.

Notes.

Choice of $n_0$ is not unique. If you chose $n_0$ to be $n_1+k$, then you are sure that for $n\geqslant n_0$, integer $n$ exceeds $k$ so $x_n$ is term of
$(x_{n+k})$ and  also integer $n$ exceeds $n_1$ so $x_n$ is $\varepsilon$ close to $a$. It is up to you to choose whatever $n_0$ you prefer (and works of course).

The above (hope was not too long) contains all the ideas you need to read and write formal proofs.

Title: Shifting the index of a sequence to prove convergence.

Body:
I'm currently studying sequences in Real Analysis and really struggling with answering this basic question:

Prove that if $x_n→a$, then for any fixed positive integer k the sequence $\{y_n\},y_n=x_{n+k}$ also converges to $a$

I honestly have no idea how to even start this proof. Could someone provide me with some hints? Specifically, what does $y_n=x_{n+k}$ represent in terms of sequences? Is this a case of induction?


问:
Best approximation to reflexive subspace - converse
Let $X$ be a normed vector space.

I know if $X$ is reflexive, then for every normed space $Y$ that $X$ is a subspace of $Y$, $X$ has a best approximation to every $y \in Y$ (which is to say there exists $x\in X$ that $d(y, X) = \|x-y\|$).

I need to prove the converse of the above fact. To be more precise, if for every normed space $Y$ that $X \subset Y$, $X$ has a best approximation to every $y \in Y$, is it true that $X$ is reflexive?

答:
$\newcommand{\dist}{\text{dist}}
\newcommand{\R}{\mathbb R}
\newcommand{\diam}{\text{diam}}
\newcommand{\and}{\quad \text{and} \quad}
$The conjecture proposed by the OP is  true:  every non-reflexive space may be embedded into a larger space $Y$ in such
a way that the minimization problem
$$
  \|y-x_0\| = \inf\{\|y-x\| : x \in X\}
  $$
admits no solution for some $y$ in $Y$.   Proving this conjecture obviously entails constructing the larger space $Y$,
so it is necessarily a somewhat long argument.

Lemma.  Let $X$ be a normed space and let $C$ be any nonempty convex subset of $X$ whose  diameter satisfies
$$
  \diam(C)\leq 2.
  $$
Then $X\oplus \R$ becomes a normed
space if equipped with the norm
$$
  \|(x,a)\| = |a| + \dist(x,aC),\quad \forall x\in X,\quad \forall a\in \R.
  $$
In addition, the correspondence  $x\mapsto (x, 0)$ defines  an isometry from $X$ onto a closed hyperplane of $X\oplus \R$.

Proof. It is obvious that
$\|(x,0)\| = \|x\|$,
so the isometry claim is clear.
We leave it for the reader to prove that
$$
  \|\lambda (x, a)\| =   |\lambda |\|(x, a)\|, \and   \|(x, a)\| =  0 \Rightarrow (x, a)=0,
  $$
and we concentrate instead in  proving  the triangle inequality
$$
  \|(x_1+x_2,a_1+a_2)\| \leq    \|(x_1,a_1))\| +  \|(x_2,a_2)\|,
  \tag 1
  $$
which we will first verify
in the special case that $a_1$ and $a_2$ have the same sign.
If both $a_1$ and $a_2$ are zero, the claim follows trivially, so we assume that $a_1+a_2\neq 0$.

Fixing any $\varepsilon >0$, choose $y_i$ in $C$, for each $i=1,2$, such that
$$
  \|x_i-a_iy_i\|<\dist(x_i,a_iC)+\varepsilon .
  $$
Then
$$
  y:= \frac{a_1}{a_1+a_2}y_1 + \frac{a_2}{a_1+a_2}y_2
  \tag 2
  $$
lies in $C$, whence
$$
  \dist(x_1+x_2,(a_1+a_2)C) \leq
  \|x_1+x_2-(a_1+a_2)y\| = $$$$ =
  \|x_1+x_2-(a_1y_1+a_2y_2)\| \leq
  \|x_1-a_1y_1\| +   \|x_2-a_2y_2\| < $$$$ <
  \dist(x_1,a_1C) + \dist(x_2,a_2C) + 2\varepsilon .
  $$
Combining this with the fact that $|a_1+a_2| \leq |a_1| +|a_2|$, we have proven (1).

Let us now prove (1) in the remaining case that $a_1$ and $a_2$ have opposite signs.   Unlike the previous case the
above argument employing (2) does not work because $y$ will not be in $C$.    It is here that the bound on the diameter of $C$ becomes relevant.

Interchanging $(x_1,a_1)$ and $(x_2,a_2)$ if
necessary, we may assume that $|a_1|\geq |a_2|$.  Furthermore, since we already know  that $\|\cdot\|$ is homogeneous, we may
multiply $(x_1,a_1)$ and $(x_2,a_2)$  by $-1/a_2$ and hence assume that $a_2=-1$.  Finally, given that $a_1$ is
positive and $|a_1|\geq 1$, we may write $a_1=1+b$, with $b\geq 0$.  Therefore the two elements we are working with may now be
represented by
$$
  (x_1,1+b)\and (-z_2,-1),
  $$
where we made a last minute  parameter change, namely $x_2=-z_2$.
Unraveling (1)  with these assumptions we see that we need to prove that
$$
  \dist(x_1-z_2,bC)\leq  2 + \dist(x_1,(1+b)C)+ \dist(z_2,C).
  \tag 3
  $$

Fixing any $\varepsilon >0$, choose $y_1$ and $y_2$ in $C$ such that
$$
  \|x_1-(1+b)y_1\|< \dist(x_1,(1+b)C) +\varepsilon ,
  $$
and
$$
  \|z_2-y_2\|< \dist(z_2,C) +\varepsilon .
  $$
Then
$$
  \dist(x_1-z_2,bC)\leq
  \|x_1-z_2 - by_1\| = $$$$ =
  \|x_1-(1+b)y_1- (z_2 -y_2) + y_1 - y_2\|\leq  $$$$ \leq
  \|x_1-(1+b)y_1 \|+\|z_2 - y_2\| + \|y_1 - y_2\|\leq  $$$$ \leq
  \dist(x_1,(1+b)C) + \dist(z_2,C) +2\varepsilon  + \diam(C).
  $$
Since $\varepsilon $ is arbitrary and $\diam(C)\leq 2$, we see that (3) is satisfied.
QED

This said, let us be given a non-reflexive space $X$.  By Smulian's Theorem (see below) choose a nested sequence $C_1
\supseteq C_2 \supseteq C_3 \supseteq \cdot \cdot \cdot \ $ of nonempty closed bounded convex subsets with empty intersection.   By
scaling everything by the same fixed factor we may assume that each $C_n$ is contained in the unit ball of $X$, and
hence, in particular, $\diam(C_n)\leq 2$.

For each $n$ let $\|\cdot\|_n$ be the norm on $X\oplus \R$ constructed in the Lemma relative to $C_n$, and define a new norm on
$X\oplus \R$ by
$$
  \|(x, a)\| =
  \sum_{n=1}^\infty   \frac{\|(x, a)\|_n}{2^n}.
  $$

Since  $\|(x, 0)\|_n = \|x\|$, for every $n$ and every $x\in  X$, and since   $\sum_{n=1}^\infty   1/2^n=1$, we see that  $\|(x,
0)\| = \|x\|$, so
$X$
also embeds into $X\oplus \R$ relative to the above norm.

We now claim that there is no point in $X$ (or rather in its canonical image in $X\oplus \R$) minimizing the distance to the
vector $u:= (0,-1)$.  To see this notice that for every $x$ in $X$ one has that
$$
  \|x-u\|_n =
  \|(x, 0)-(0, -1)\|_n =
  \|(x, 1)\|_n =
  1 + \dist(x,C_n) \geq  1,
  $$
with equality iff $x\in C_n$.  Therefore
$$
  \|x-u\| \geq
  \sum_{n=1}^\infty  \frac 1{2^n} = 1,
  $$
so we see that
$\dist(u,X)\geq 1$.
On the other hand, if $x\in  C_k$, then
$$
  \|x-u\| =
  1 +   \sum_{n=k+1}^\infty  \frac{\dist(x,C_n) }{2^n}
  $$
which can be made as close as desired to 1 if $k$ is large enough, meaning that in fact   $\dist(u,X)=1$.

The punch line is then that there is no $x$ in $X$ such that
$$
  \|x-u\| =   1 +   \sum_{n=1}^\infty  \frac{\dist(x,C_n) }{2^n} = 1,
  $$
since such an $x$ needs to be in every $C_n$, but we know that $\bigcap_nC_n$ is empty.

Not many people read Russian, so here is the statement:

Theorem (Smulian [1]) A normed space $X$ is reflexive iff every nested
sequence $C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdot \cdot \cdot \ $ of nonempty closed
bounded convex subsets of $X$ has nonempty intersection.

[11] V.L. Smulian, On the principle of inclusion in the space of the type (B), Mat. Sb. (N.S.) 5 (1939) 317-328.


问:
Why is it correct to assume $f = e^{\lambda x}$ in differential equation?
I was solving the following differential equation:

$$f'' + f = 0$$

And, as I learned in class, I assumed that $f = e^{\lambda x}$ and then found the roots of the polynomial and found the solution. But then the following came to my mind:

Why is it correct to assume that $f = e^{\lambda x}$? How do we know that by doing this we are indeed finding all the solutions of the equation and we aren't only finding the solutions that take that form? Because doing this looks like a guessing game, we try to guess the solution and then find it by correcting and fixing the values of $\lambda$.

答:
Let’s start with the first degree linear differential equation and see why the homogenous solution is exponential function.

$$
\begin{align}
0&=\frac{d}{dx}y+\lambda y\\
\\
&=e^{\lambda x}\left(\frac{d}{dx}y+\lambda y\right)\\
\\
&=e^{\lambda x}\frac{d}{dx}y+\lambda e^{\lambda x}y\\
\\
&=\frac{d}{dx}\left(e^{\lambda x}y\right)
\end{align}
$$

We see that the differentiation of $e^{\lambda x}y$ yields $0$ i.e. $e^{\lambda x}y=C$ or equivalently $y=Ce^{-\lambda x}$. Therefore, this is not a guessing game. However, they don’t teach students this and only teach students to immediately assume exponential function.

Now let’s solve your question. Recall that $\frac{d}{dx}()$ is a linear operator.

$$
\begin{align}
0&=\frac{d^{2}}{dx^{2}}y+y\\
\\
&=\left(\frac{d^{2}}{dx^{2}}()+()\right)y\\
\\
&=\left(\frac{d}{dx}()+i()\right)\left(\frac{d}{dx}y-iy\right)\\
\\
\end{align}
$$

Now we define $z=\frac{d}{dx}y-iy$ to obtain a first degree differential equation

$$
\begin{align}
0&=\frac{d}{dx}z+iz\\
\\
&=e^{ix}\frac{d}{dx}z+ie^{ix}z\\
\\
&=\frac{d}{dx}\left(e^{ix}z\right)\\
\\
\\
z&=Ce^{-ix}
\end{align}
$$

Now we substitute back to definition of $z$

$$
\begin{align}
Ce^{-ix}&=\frac{d}{dx}y-iy
\\
\\
Ce^{-i2x}&=e^{-ix}\frac{d}{dx}y-ie^{-ix}y\\
\\
&=\frac{d}{dx}\left(e^{-ix}y\right)
\end{align}
$$

Now integrate the last row and substitute $D_{1}=\frac{-C}{2i}$ to obtain

$$
\begin{align}
D_{1}e^{-i2x}+D_{2}&=e^{-ix}y\\
\\
\\
D_{1}e^{-ix}+D_{2}e^{ix}&=y
\end{align}
$$

There you go. Now if you are interested, try to solve $y’’+2y’+y=0$ this way and see if you understand why the solution is $y=D_{1}e^{-x}+D_{2}xe^{-x}$


问:
What is $\frac{(an)!}{n!}$?
How can $\frac{(an)!}{n!}$ be expressed in terms of $a!$, $n!$, $a$, $n$ (and maybe Pochhammers)?</p>

答: If you don't insist on being exact, you can use Stirling's approximation.  $$\frac{(an)!}{n!}\approx \frac {(an)^{an}e^n}{n^ne^{an}}\sqrt{a}=a^{an}n^{an-1}e^{-(an-1)}\sqrt a$$ which will be very close-within about a factor of $1+\frac 1{12n}$</p>

Title: What is $\frac{(an)!}{n!}$?


问:
Homeomorphism between $\mathbb{P}^n$ and $D^n / R$
I've to find a homeomorphism between $\mathbb{P}^n$ and $D^n / R$ with the equivalence relation $R$ on $D^n$ given by: $(x,y) \in \mathbb{R}$ iff $x=y$ or ($(x,y \in S^{n-1}$ and $x = -y$). I know that this is the same as proving that $S_n$ positive is homeomorphic to $\mathbb{P}^n$, but I don't know how to prove this.

答:
Define $i : D^n \to S^n, i(x) = (x,\sqrt{1-\lVert x \rVert^2})$. This map embeds $D^n$ as the closed upper hemisphere of $S^n$.

You know that $\mathbb P^n$ is the quotient space of $S^n$ obtained by identifying antipodal points. Let $p : S^n \to \mathbb P^n$ denote the quotient map. The map $H = p \circ i : D^n \to \mathbb P^n$ has the property $H(x) = H(y)$ if and only if $(x,y) \in R$. Moreover, for each $x \in S^n$ we have $x \in i(D^n)$ or $-x \in i(D^n)$, hence $H$ induces a continuous bijection $h : D^n/ R \to \mathbb P^n$. But $D^n/ R$ is compact (as a quotient space of a compact space) and $\mathbb P^n$ is Hausdorff, thus $h$ is a homeomorphism.


问:
Invariant subspace $U$ of direct sum $V\oplus V$ of irreducible representation $V$ is isomorphic to $V$
I am studying from Hall's book "Lie Groups, Lie Algebras, and Representations" and I'm stumped on the following question:

Suppose that $V$ is an irreducible finite-dimensional representation of a group or Lie algebra over $\mathbb{C}$, and consider the associated representation $V\oplus V$. Show that every nontrivial invariant subspace $U$ of $V\oplus V$ is isomorphic to $V$ and is of the form $$U=\{(\lambda_1v,\lambda_2v)|v\in V\}$$ for some constants $\lambda_1$ and $\lambda_2$, not both zero.

From the looks of it, I think I should be using Schur's lemma, which tells me that if $V$ is an irreducible complex representation of a group or Lie algebra and $\phi:V\to V$ is an intertwining map of $V$ with itself, then $\phi=\lambda I$ for some $\lambda\in\mathbb{C}$. An intertwining map is a linear map which commutes with the action. My understanding is that other authors just call this a homomorphism of representations.

However, I do not see how to connect the end result with Schur's lemma. Perhaps the projection maps will come in handy here? Any help is appreciated.

答:
A hint would be to write down the short exact sequence
$$0\rightarrow U \rightarrow V \oplus V \rightarrow W \rightarrow 0.$$

We now have an intertwining map (i.e. homomorphism of representations) $\phi: V \oplus V \rightarrow W$.

This map, of course, is a direct sum of two maps $\phi_1: V \rightarrow W$ and $\phi_2: V \rightarrow W$.

What does Shur's lemma say about $\phi_1$ and $\phi_2$?


问:
Does the following graph have a Hamilton circuit?

A Hamilton circuit (or path) is a path that visits each vertex exactly once (except the start/end point) and ends at the starting point.

I've stared at this for quite a while and cannot find a Hamilton circuit yet my guide says that one exists. Where is the path?

答: Here's the lexicographically first Hamilton cycle (found using a backtracking algorithm):


问:
Four Number Theorem : Let $a$, $b$, $c$, $d$ be integers such that $ab = cd$.
Let $a$, $b$, $c$, $d$ be integers such that $ab = cd$. Then there exist integers $x$, $y$, $z$, $w$ such that $xy = a$, $zw = b$, $xz = c$, $yw = d$.

My Progress: I tried playing with $\gcd(a,c)$, $\gcd(a,d)$,$\gcd(b,c)$, $\gcd(b,d)$.
But I am not able to proceed.

Please post hints rather than solution, it helps me a lot.

Thanks in advance.

答:
Firstly note that it is sufficient to prove the theorem when $a,b,c,d,x,y,z,w$ are all natural numbers. For if any of the given numbers is $0$ then the solution tuple $(x,y,z,w)$ is trivial and if there are negatives involved you can look for  $x,y,z,w$ for $\lvert a \rvert,\lvert b \rvert, \lvert c \rvert, \lvert d \rvert$ and then adjust for signs.

If $b = 1$ you can take $(x,y,z,w) = (c,d,1,1)$,
say the result holds for all $a,b,c,d$ when $1 \leq b < n$ and say $an = cd$ for some $a,c,d$.
Let $p$ be a prime divisor of $n$ then $p \vert c$ or $p \vert d$. Say $p \vert c$, then we'll have an equation of the form $am = c'd$ where $n=mp,c=pc'$ and $1 \leq m<n$ so by hypothesis there exists $(r,s,t,u)$ all naturals such that $a = rs, m = tu, c' = rt, d = su$
that gives $n = (pt)u$ and $c = r(pt)$, therefore $ (r,s,pt,u)$ is the tuple corresponding to $an = cd$, similarly one can find the tuple if $p \vert d$. This proves the theorem for natural numbers by induction.


问:
Looking for a function that approximates a parabola
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?

答:
There's no such thing as a "piecewise function". In the phrase "piecewise defined function", the word "piecewise" modifies the word "defined", not "function".  The term "piecewise" refers to how the function is defined, not the function itself.

$$
t(x) =
\begin{cases}
-x & x < 0 \\
x& x \geq 0
\end{cases}
$$

and

$t(x)=|x|$

are the same function. In the first case, the definition is given piecewise, and in the second, it isn't. You can make any piecewise definition non-piecewise with enough absolute values.  For instance, the following gives a graph along the lines of yours:

$f(x) = \frac 12(200-||x-50|-|x-150||)(100-|x-50|)$


问:
Give an example of a set A with equivalence relations R and S for which S∘R ≠ R∘S
I have to give an mere example where SR is not equal to RS.

I have tried a couple of numerical examples but am not able to figure any out.

答:
Somehow I've never come across relation composition before!

Anyway, this should work. Let $A = \{a, b, c\}$. Let $R$ equate $a, b$ and leave $c$ alone. Let $S$ equate $b, c$ and leave $a$ alone.

Then $aRSc$ but we don't have $aSRc$.

Let me know if I misunderstood the definition and I'll edit or delete my post.


问:
Why these solutions from this ODE are global?
Let consider the Cauchy's Problem : $y'(t) = y(t)(1-y(t))(1-K(t)y(t)), \ t\in J$ and  $y(0)=\frac{1}{4}$.

First if $K(t)=K\in \mathbb{R}$, I have to prove that the maximal of this problem is maximal.

Applying the Cauchy-Lipschitz's theorem, we know that there exists only one maximal solution in an open interval $I\subset J$ of the Cauchy's Problem. Here $f(t,y(t))=y(t)(1-y(t))(1-Ky(t))$ and $f\in \mathcal{C}^1$. Moreover the solution can be written as follow :

$y(t)=\frac{1}{4}+\int_{0}^{t}f(s,y(s)) \mathrm{d}s$.

Then I don't really know how to proceed. If $f$ is bounded on $I$, I know that the maximal solution is global but here we do not have lots of informations on $f$. If the interval was compact, I could have done it by contradiction using the "blow-up in finite time" theorem.

Second, same question if $K(t)=\frac{\sin(t)}{1+t}$.

Thanks in advance !

答:
If $K(t) = K\in\mathbb{R}$ is a constant, then this equation is autonomous: $f$ does not depend explicitly on $t$. In this case, it is enough to show that a solution $y(t)$ with $y(0) = \frac{1}{4}$ remains bounded in time on its maximal interval of existence. For if there exist constants $M\leq y(t) \leq N$ for all $t$, then you can just show that $f = f(y)$ has a Lipschitz constant $L$ on the set $\{y:M\leq y\leq N\}$. Since $y(t)$ never leaves this range, this Lipschitz constant is effectively global: you can extend your solution by a time $\sim \frac{1}{L}$ infinitely many times.

To see why $y(t)$ is bounded by constants, use the Picard-Lindelof theorem again: note that this equation has constant solutions $y(t) = 0$, $y(t) = 1$, and $y(t) = K$. Therefore a solution satisfying $y(0) = \frac{1}{4}$ obeys the global bounds $0 < y(t) < 1$. For otherwise, there would be some time $t_0$ for which either $y(t_0) = 0$ or $y(t_0) = 1$. But any solution satisfying $y(t_0) = 0$ for some $t_0$ must in fact be uniformly $0$, since we already know $0$ is a solution and you can show that solutions are locally unique. Since we know $y(0) = \frac{1}{4}$ it cannot be the $0$ solution. Similarly $y$ cannot be the constant solution $y(t) = 1$. Therefore the only possibility is that $y(t)$ remains bounded between $0$ and $1$.


问:
Basis for $\mathbb{R}^\mathbb{N}$ implies axiom of choice?
Let $\mathbb{R}^\mathbb{N}$ denote the vector space over $\mathbb{R}$ of sequences of real numbers, with multiplication and addition defined by component. It's well-known that though the subspace $\mathbb{R}^\infty$ of sequences with only a finite number of nonzero terms has a basis $\mathbf{e}_1 = (1, 0, 0, 0, \ldots), \mathbf{e}_2 = (0, 1, 0, 0, \ldots)$, this is not a basis of $\mathbb{R}^\mathbb{N}$ (expressing the constant sequence $(1, 1, 1, \ldots)$ would require an infinite sum $\mathbf{e}_1 + \mathbf{e}_2 + \mathbf{e}_3 + \cdots$, and infinite sums in generic vector spaces are undefined). It's also been proved that the statement that all vector spaces have a basis is equivalent to the axiom of choice.

I'm interested, though, in the specific space $\mathbb{R}^\mathbb{N}$. Has it been proved that a basis for this set requires the axiom of choice and cannot be described explicitly? This isn't a homework question or anything; I'm just curious.

答:
No single concrete set admitting a certain property would imply the axiom of choice. Period. The axiom of choice is a global statement, and statements about a set with a certain property are local (I'm not talking about a global statement, e.g. "For every set $A$, $A\times X$  can be well-ordered" implies the axiom of choice for any fixed set $X$, that's cheating).

We can always have the axiom of choice failing, as badly as we want it to fail, while the real numbers, and every set that you would ever cared about, can be well-ordered so that all vector spaces "that matter" have a basis. In other words, the axiom of choice is a global statement, so its negation is not about one set. It's about the existence of a counterexample.

(We actually don't even know if there is a field $F$ such that "All vector spaces over $F$ have a basis" implies the axiom of choice; speaking of global statements disguised as local statements.)

On the other hand, it is consistent that every set of reals has the Baire property, which implies that every linear $T\colon\Bbb{R^N\to R^N}$ is continuous. Alas, being a separable space, there can only be $2^{\aleph_0}$ continuous functions; but we can easily show that a basis of $\Bbb{R^N}$ must have size $2^{\aleph_0}$ as well, and therefore there would be $2^{2^{\aleph_0}}$ linear functions just induced by permutations of such a basis. And so, indeed, if all sets of reals have the Baire property, no basis for $\Bbb{R^N}$ can exist.


问:
Trouble in Proposition 6.7 in Atiyah MacDonald: Why is $N_i/N_{i+1}\subset M_i/M_{i+1}$
I am trying to understand the proof of Proposition 6.7 in Atiyah and Macdonald. I am stuck on the following part of the proof.

Let $(M_i)$ be a composition series for $M$. Let $N$ be a submodule of $M$. Define $N_i=N\cap M_i$. Then $N_i/N_{i+1}\subset M_i/M_{i+1}$.

This should be simple but I cannot see what I am missing. Clearly $N_i\subset M_i$. Therefore clearly also $N_i/M_{i}\subset M_i/M_{i+1}$. But when we quotient $N_i$ by $N_{i+1}$ we are quoitenting by less stuff, so its not obvious why we have an inclusion.

答:
Consider the composition

$$N_{i} \xrightarrow{u} M_{i} \xrightarrow{\pi} M_{i}/M_{i+1}$$

where $u$ denotes inclusion, and $\pi$ denotes the canonical quotient morphism. The kernel of $\pi \circ u$ is $u^{-1}(\ker(\pi))$. Since $u$ is just the inclusion map, we have $u^{-1}(\ker(\pi)) = N_{i} \cap \ker(\pi)$. The kernel of $\pi$ is $M_{i+1}$, so we conclude that the kernel of $\pi \circ u$ is exactly $N_{i} \cap M_{i+1} = N_{i+1}$, and so $\pi \circ u$ descends to an injective map $N_{i}/N_{i+1} \to M_{i}/M_{i+1}$ by the universal property of the quotient, as desired.


问:
What kind of mathematical rule was broken in here?
I just wanted to write the iterated version of:

$$ a^2 - b^2 = (a+b)(a-b)$$

As:

$$ n^x - 1 = (n^{ \sqrt{x}} +1) \color{red}{(n^{ \sqrt{x}} -1)} = \\
(n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1) \color{red}{(n^{ \sqrt{\sqrt{x}}} -1)} = \\
(n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) \color{red}{(n^{ \sqrt{\sqrt{\sqrt{x}}}} - 1)} = \dots
 $$

And so:

$$n^x - 1 = (n^{ \sqrt{x}} +1)(n^{ \sqrt{\sqrt{x}}} +1)(n^{ \sqrt{\sqrt{\sqrt{x}}}} +1) ... \Rightarrow \\n^x - 1 = \prod_{k=1}^{\infty} (n^{x^{\left (\frac{1}{2k}\right)}} + 1)$$

But when plugging $n=1$ we get:

$$ 1^x - 1 \equiv 0 = \prod_{k=1}^{\infty} (1^{x^{\left (\frac{1}{2k}\right)}}  + 1) \equiv \prod_{k=1}^{\infty} (1+1) = \prod_{k=1}^{\infty} 2 = \infty$$

Where did I go wrong? I think that I "omitted" the $n^{x^{(\frac{1}{2k})}} - 1$ part, but isn't it valid when we talk about infinite series? I would appreciate if you could clear things up.

Thank you!

答:
Should be instead $$n^x-1 = (n^{x/2}-1)(n^{x/2}+1)$$,
as in general,

$$n^x-1 \not = (n^{\sqrt{x}}-1)(n^{\sqrt{x}}+1).$$

This was your error.


问:
Derivative of Inverse of sum of matrices
Given is the function $f : \mathbb{R}^p \to \mathbb{R}$ with

$$
f(x) = q(x)^{\top} G^{-1} q(x)
$$

where $G = A + x_1 B_1 + \ldots + x_p B_p$.

The matrices $A, B_1, \ldots, B_p \in \mathbb{R}^{n \times n}$ are all symmetric positive definit.

$q: \mathbb{R}^n \to \mathbb{R}^n$ and the Jacobian $\nabla q$ is known.

Is it possible to derive a closed form for $\nabla f$? For me, the hard part is
$G^{-1}$. Any hints or suggestions are really appreciated!

Body:
$$\frac1{\Delta x}\big[(A+(x+\Delta x)B)^{-1}-(A+xB)^{-1}\big]=$$
$$\frac1{\Delta x}(A+xB)^{-1}(A+xB)\big[(A+(x+\Delta x)B)^{-1}-(A+xB)^{-1}\big]
(A+(x+\Delta x)B)
(A+(x+\Delta x)B)^{-1}$$
$$=-(A+xB)^{-1}B
(A+(x+\Delta x)B)^{-1}\rightarrow-(A+xB)^{-1}B(A+xB)^{-1}$$ as $\Delta x\rightarrow0$.

So that $\partial f/\partial x_j=2\partial q/\partial x_j^\top G^{-1} q -q^\top G^{-1}B_jG^{-1}q$.


问:
Induction proof: $2\times6\times10\times ... \times (4n-2) = (2n)!/n!$
What I'm missing to solve this problem is the algebra steps to prove the inductive step.  The base case is clear: $2 = 2!/1!$.  Suppose it's for true for $k \ge 1$.  Must show it's true for $k + 1$.  Using the hypothesis, we can write

$$2\times6\times10\times14\times18\times ... \times (4k-2) = \frac{(2k)!}{k!}.$$

Now I take the obvious step: I multiply $4(k+1) - 2$ on both sides of the equation and I get

\begin{align*}
2\times6\times10\times14\times18\times ... \times (4k-2) [4(k+1) - 2] = \frac{(2k)!}{k!} 4(k+1) - 2).
\end{align*}

Now my job must be putting the RHS into the form $(2(k+1))!/(k + 1)!$.  But I'm not able to do it.  I look at a special case such as $k = 3$.  I get $14 \times 6!/3!$ and that equals $8!/4!$, but it doesn't seem obvious how to put it in the general case.  I need some algebra help.

Body:
What you need is

$$4(k+1)-2=4k+2=2(2k+1)=\dfrac{2(2k+1)(2k+2)}{2(k+1)}=\dfrac{(2k+1)(2k+2)}{k+1}.$$

Can you take it from here?


问:
Weird non-homogeneous wave equation
I've been trying to solve the Cauchy problem \begin{align}&u_{tt}-u_{xx}+u=0~; x \in \mathbb{R},t>0 \tag{1}\\&u(x,0)=\cos x ~\text{for}~x\in \mathbb{R}\\&u_t(x,0)=\sin x~\text{for}~x\in \mathbb{R}\end{align}
I've seen the class of problems of the form

\begin{align}&u_{tt}-c^2u_{xx}=f(x,t)~; (x,t) \in \mathbb{R}\times (0,\infty)\\&u(x,0)=\varphi(x)~\text{for}~x\in \mathbb{R}\\&u_t(x,0)=\psi(x)~\text{for}~x\in \mathbb{R}\end{align}But in the problem (1) the vector field is dependent of $u$, how to solve these kind of problems? I don't think separation of variable makes sense without boundary conditions, also changing coordinates from $u$ to $w$ by $u(x,t) \to w(\xi,\eta)$ gives me : $w_{\xi\eta}=\frac{1}{4}w$ which is again hard to solve without further knowledge. Is there any other clever change of coordintes applicable? or Cauchy-Kovalevskaya Theorem could be of any help to find an explicit solution? Appreciate any suggestions!

答: This is a case of the Klein-Gordon equation. Since the initial conditions are periodic in the spatial variable, the solution should also be periodic in space. So you can instead pose this problem with the spatial domain replaced by $[0,2\pi)$ and impose periodic boundary conditions. Then you can solve via separation of variables and using Fourier series.

Title: Weird non-homogeneous wave equation


问:
Three vertices of a triangle lie on $y^2=x^3$

If ${P(a,b),Q(c,d),R(m,n)}$ are the centroid, orthocenter, circumcenter respectively of a scalene triangle with  vertices on the curve $y^2=x^3$, then find the value of $$\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$$

I know that the centroid, orthocenter and circumcenter lie on the straight line and centroid divides orthocenter and circumcenter in ratio $2:1$. Thus

$$a=\frac{2m+c}{3},b=\frac{2n+d}{3}$$ As you can see I am not able to use  $y^2=x^3$.

I think there is some trick involved!

答:
$\begin{array}{} G=(a,\sqrt{a^3}) & H=(c,\sqrt{c^3}) & O=(m,-\sqrt{m^3}) \end{array}$

$\begin{array}{} \frac{x_{H}-x_{G}}{x_{G}-x_{O}}=2 & \frac{c-a}{a-m}=2 & a=\frac{2m+c}{3} \end{array}$

$\begin{array}{} \frac{y_{H}-y_{G}}{y_{G}-y_{O}}=2 & \frac{\sqrt{c^3}-\sqrt{a^3}}{\sqrt{a^3}+\sqrt{m^3}}=2  \end{array}$

$\frac{\sqrt{c^3}-\sqrt{(\frac{2m+c}{3})^3}}{\sqrt{(\frac{2m+c}{3})^3}+\sqrt{m^3}}=2$

One solution that checks the system is:

$m=\frac{1}{2}\left( 2+\sqrt{6+\sqrt[3]{9}}-\sqrt{12-\sqrt[3]{9}}+\frac{30}{\sqrt{6+\sqrt[3]{9}}}  \right) ·c$

$\begin{array}{} m=α·c & α≈0.158873884116  \end{array}$

numerically checking the value of the expression k

$k=\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$

$k=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}-\frac{1}{\sqrt{m}}$

$\begin{array}{} m=α·c & a=\frac{1+2α}{3}  \end{array}$

$k=\left(\frac{\sqrt{3}}{\sqrt{1+2α}}+1-\frac{1}{\sqrt{α}}   \right)·\frac{1}{\sqrt{c}} ≈\frac{-2·10^{-15}}{\sqrt{c}}$

Indicating that the value of k is zero

Proof:

$k=\left(\frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}}   \right)·\frac{1}{\sqrt{c}} =0$

$\begin{array}{} \text{Solving} & \frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}}=0 & β=0.158873884146 \end{array}$

β is equal to α which is the solution of the initial system.


问:
If $\operatorname{sp}(A) \cup \operatorname{sp}(B)=\operatorname{sp}(A\cup B) \Rightarrow A\cup B$ is linearly dependent
$\newcommand{\sp}{\operatorname{sp}}$

Let $V$ be a vector space over $F$ field, and let $A,B$ be two different, disjoint, non empty sets of vectors from $V$.

If $\sp(A) \cup \sp(B)=\sp(A\cup B) \Rightarrow A\cup B$ is linearly dependent

I've started by saying that if $\sp(A) \cup \sp(B)=\sp(A\cup B)$
then $\sp(A) \subseteq \sp(B)$ or $\sp(B)\subseteq \sp(A)$.

Thus if we assume, WLOG, that $\sp(A) \subseteq \sp(B)$ and we take $v_1 \in A$ and multiply it by scalars $\alpha_1,\ldots,\alpha_k$ we'll get a vector from $B$.

Thus $ A\cup B$ is linearly dependent.

I feel like this proof is not good enough, where is it failing?

答:
We use contraposition at this proof:

suppose $A\cup B$ be linear independence then elements of A as $\{\alpha_1,..,\alpha_n\}$ with elements of B $\{\beta_1,...\beta_m\}$ will be linear inedependence so $\alpha_1+....+\alpha_n+\beta_1+....\beta_m \in span(A\cup B)$ but it is not belong to span(A)$\cup$span(B)


问:
Why is it correct to assume $f = e^{\lambda x}$ in differential equation?
I was solving the following differential equation:

$$f'' + f = 0$$

And, as I learned in class, I assumed that $f = e^{\lambda x}$ and then found the roots of the polynomial and found the solution. But then the following came to my mind:

Why is it correct to assume that $f = e^{\lambda x}$? How do we know that by doing this we are indeed finding all the solutions of the equation and we aren't only finding the solutions that take that form? Because doing this looks like a guessing game, we try to guess the solution and then find it by correcting and fixing the values of $\lambda$.

答:
$$f'' + f = 0$$
There are many ways to solve this differential equation. Here is one of them:
$$f''\cos x +f \cos x=0$$
$$f''\cos x \color{red}{ -f' \sin x +f'\sin x}+f \cos x=0$$
$$(f'\cos x)' +(f \sin x)'=0$$
After integration we get:
$$f'\cos x +f \sin x=C_1$$
That you can rewrite as:
$$\left ( \dfrac {f}{ \cos x}\right)'=\dfrac {C_1}{\cos^2 x}$$
Integrate and you are done.
$$ \dfrac {f}{ \cos x}= {C_1}{\tan x}+C_2$$
Finally:
$$ {f}(x)= {C_1}{\sin x}+C_2 \cos x$$


问:
Midpoints on Riemannian manifold with bounded curvature
Let $M$ be a Riemannian manifold with non-negative sectional curvature everywhere bounded above by some $\kappa>0$. For simplicity, assume that the injectivity radius is everywhere infinite on $M$, so the log map at any $x\in M$, denoted by $Log_x$, is defined everywhere on $M$.

Now, let $x_1,x_2$ be fixed points in $M$.

For any $x\in M$, I want to quantify the difference between :

taking the midpoint of $Log_x(x_1)$ and $Log_x(x_2)$ in the tangent space $T_xM$ and map it back to $M$, and

taking the midpoint $x^*$ of $x_1$ and $x_2$, defined as the minimizer of $d(y,x_1)^2+d(y,x_2)^2$ over $y\in M$, where $d$ is the geodesic distance.

In other words, can we bound
$$\left\|Log_x(x^*)-\frac{1}{2}\left(Log_x(x_1)+Log_x(x_2)\right)\right\|_x^2$$
by a quantity that only depends on $\kappa$ and $d(x,x^*)$ ?

And what about the case where $M$ has non-positive sectional curvature everywhere bounded below by $-\kappa$ (for some $\kappa>0$) ?

答:
For fixed $x_1,x_2$ one does get a bound for above depending only on $\kappa$ and $d(x,x^*)$. Let me only consider $x$ for which $d(x,x^*)<\frac{1}{4}d(x_1,x_2)$. For other choices of $x$, triangle inequality provides the upper bound $d(x,x^*)$ upto a constant. Now for $d(x,x^*)<\frac{1}{4}d(x_1,x_2)$, first one observes $$\frac{1}{4}d(x_1,x_2)\leq d(x,x_1)\leq \frac{3}{4}d(x_1,x_2),$$ again by triangle inequality and similarly for $d(x,x_2)$. Let $\theta_x$ be the angle between $\log_x(x_1)$ and $\log_x(x_2)$. Then $$\|\frac{1}{2}(\log_x(x_1)+\log_x(x_2))\|=\frac{1}{2}\sqrt{d(x,x_1)^2+d(x,x_2)^2+2d(x,x_1)\cdot d(x,x_2)\cos\theta_x}$$ and $\|\log_x(x^*)\|=d(x,x^*)$. One can use Toponogov's triangle comparison theorem to bound $\theta_x$ in terms of $\kappa$ and $d(x,x^*)$. Note that as $d(x,x^*)$ goes to zero, $\theta_x$ goes to $\pi$. The curvature lower bound says that the $\theta_x$ can not be too small.

Edit: In the case of strictly positive sectional curvature, note that Topogonov's theorem says that the angle between $\log_x(x_1)$ and $\log_x(x_2)$ can not be less than $\pi-\epsilon$ for any $\epsilon>0$ as $x\to x^*$ (you need one of the sides to be small, let it be the one joining $x$ and $x^*$). If the angle is greater than $\pi+\epsilon$ for some $\epsilon>0$, as $x\to x^*$, then again Arzela-Ascoli gives a contradiction to an infinite injectivity radius.


答:
As an expansion of Calvin Khor's answer, you need to generate a suitable approximation to the square pulse function, which can be done in a number of different ways.

Consider that the error function goes from -1 to 1, with a relatively rapid transition from one side to the other. Using this function, we can write
$$
f_a(x)=\frac{\text{erf}(ax)+1}2-\frac{\text{erf}(a(x-1))+1}2=\frac{\text{erf}(ax)-\text{erf}(a(x-1))}2
$$
Here is what that function looks like for $a=5$ (red), $a=50$ (blue), and $a=500$ green):

We can translate this as needed, then multiply by our required polynomial, in order to get our fit. Fitting $(1-x^2)$ leads us to
$$
f(x)\approx (1-x^2)f_a(2x+1)
$$
which, for the same values of $a$, gives

Any sequence that converges towards the square pulse will work. Instead of $\text{erf}$, you could use $\tan^{-1}$ (with appropriate rescaling), for example, or $\tanh$.

Title: Looking for a function that approximates a parabola

Body:
I have a shape that is defined by a parabola in a certain range, and a horizontal line outside of that range (see red in figure).

I am looking for a single differentiable, without absolute values, non-piecewise, and continuous function that can approximate that shape. I tried a Gaussian-like function (blue), which works well around the maximum, but is too large at the edges. Is there a way to make the blue function more like the red function? Or, is there another function that can do this?


问:
question on subgroups of prime order

Let $G$ be a group and let $\,H,\, K\,$ be subgroups of $\,G,\,$ each of order $\,p,\,$ where $\,p\,$ is prime.

  Show that either $\,H\cap K =\{e\},\,$ or $\,H=K.\,$

Is the result true if $\,p\,$ is not a prime number?   Justify your answer.

答:
I'm assuming that you want $H,K$ subgroups of $G$, rather than $H<K$ subgroups (probably just a capitalization typo!)

Hint: $H\cap K$ is a subgroup of $H$ and of $K$. What are the possible orders of subgroups of a prime order group?

After you answer that hint, you should be much more clear on the possiblities for what $H\cap K$ is.

As for the second half of the question, look at something simple like $\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$. Try looking at some obvious order 4 subgroups.

Title: question on subgroups of prime order


问:
Geometry : Incircle of isosceles triangle and Menelaus
Geometry Problem:

Let $ABC$ be an isosceles triangle with $AC = BC$. It's incircle touches $AB$ at $D$,$ BC$ at $E$. A line goes through A and intersects incircle at $F$ and $G$. Line $AB$ intersects $EF$ and $EG$ in $K$ and $L$ respectively. Prove $DK=DL$.

This problem is supposed to be solved by using menelaus. I extended some line segments and applied menelaus wherever I could but couldn't do it. Can some shed some light on it?

Edit:

My main focus is on $∆GEF$, by applying menelaus here we get $ \frac{FA}{AG} \frac{GL}{LE} \frac{EK}{FK}=1$ and also consider $∆ACD$ and $∆BCD$, we get $\frac{DN}{NC} \frac{CW}{WA} \frac{AK}{KD} = 1$ and $\frac{DP}{PC} \frac{CE}{EB} \frac{BL}{LD}=1$. I am not sure how they would be of any help.

答:
The tangent line $AH$ to the circle and $HE \>|| \>KL$ assure

$$\angle AHF = \angle HEF = \angle FKA = \alpha$$

which makes AFHK cyclic and leads to

$$\angle KHA = \angle KFA = \angle GFE = \angle CEG = \angle LEB = \beta$$

Along with $\angle HAK = \angle EBL$ and $HA = EB$, the $\triangle HKA$ and $\triangle ELB$ are congruent. Therefore $KA = LB$ and, in turn, $DK = DL$.


答:
A guess at what a solution might be does not have lead to all possible solutions.  For example, consider $y'' = 0$.  If you guess that $y(x) = e^{\lambda x}$ then $y''(x) = \lambda^2 e^{\lambda x}$, so $y'' = 0$ only when $\lambda = 0$, which means $e^{\lambda x}$ fits the equation only when $\lambda = 0$.  That leads us to the constant solution $y(x) = 1$, whose span is constant functions.  This completely misses the solutions $y(x) = ax$ for different constants $a$.

Similarly, if you guess a solution to $y'' + 2y' + y = 0$ has the form $y(x) = e^{\lambda x}$ then $y'' + 2y' + y = (\lambda^2 + 2\lambda + 1)e^{\lambda x}$ so to satisfy the ODE is equivalent to $\lambda^2 + 2\lambda + 1 = 0$, and thus $(\lambda + 1)^2 = 0$.  Therefore $\lambda = -1$, and we are led to the solution $e^{-x}$.  Its constant multiples gives us solutions $ce^{-x}$, but this is missing the solution $xe^{-x}$.

What helps you know you have genuinely found all solutions to a linear differential equation is knowing in advance the dimension of the solution space. (If the differential equation is not linear, then the solution space is not closed under linear combinations so things become much more complicated.)  It can be proved that an $n$th order constant coefficient linear ODE has an $n$-dimensional solution space, so as soon as you find $n$ linearly independent solutions, their linear combinations give you all solutions. As I showed above, guessing at formulas for solutions does not have to lead you to a full set of linearly independent solutions.

Title: Why is it correct to assume $f = e^{\lambda x}$ in differential equation?

Body:
I was solving the following differential equation:

$$f'' + f = 0$$

And, as I learned in class, I assumed that $f = e^{\lambda x}$ and then found the roots of the polynomial and found the solution. But then the following came to my mind:

Why is it correct to assume that $f = e^{\lambda x}$? How do we know that by doing this we are indeed finding all the solutions of the equation and we aren't only finding the solutions that take that form? Because doing this looks like a guessing game, we try to guess the solution and then find it by correcting and fixing the values of $\lambda$.


问:
Inequality between infinity-norm and 2-norm of a matrix
So, I am trying to show that for a matrix $A \in \mathbb{R}^{m \times n}$, $||A||_{\infty} \leq \sqrt{n} ||A||_2$.

Starting from $||A||_{\infty} = \max \limits_{||x||_{\infty}=1}||Ax||_{\infty} = \max \limits_{||x||_{\infty}=1}\Big( \max \limits_{i=1, \cdots, m}|\sum_{j=1}^n a_{ij}x_j|\Big) $, I'm trying to connect it to $||A||_2=\max \limits_{||x||_2=1}\Big( \sum_{i=1}^m (\sum_{j=1}^n a_{ij}x_j)^2   \Big)$.

I know that ${||x||_{\infty}=1}$ implies $|x_j| \leq 1$, but I cannot see how to push it toward having a $\sqrt n$ in the RHS. How can I do that? Am I on the right path at all?  Thanks.

答: Let $x = (x_1, \ldots, x_n)$. Since $|x_i| \leq \lVert x \rVert_\infty$ for all $i$, by squaring and adding we get $\lVert x \rVert_2^2 \leq n \lVert x \rVert_\infty^2$, and taking square roots gives $\lVert x \rVert_2 \leq \sqrt{n} \lVert x \rVert_\infty$. Can you take things from here?


答: Hint: $H$ and $K$ are both cyclic, so let $H$ be generated by $a$.  What if $K$ contains  $a$?  What if it doesn't?

Title: question on subgroups of prime order

Body:
Let $G$ be a group and let $\,H,\, K\,$ be subgroups of $\,G,\,$ each of order $\,p,\,$ where $\,p\,$ is prime.

  Show that either $\,H\cap K =\{e\},\,$ or $\,H=K.\,$

Is the result true if $\,p\,$ is not a prime number?   Justify your answer.


问:
Different forms of Tangent Angle Addition Identity
I am trying to derive the tangent angle addition identity:
$$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x\tan y}$$
by the argument shown here:

While I can arrive at the correct identity algebraically and using trig ratios in a geometric diagram proof, I'm confused about why I cannot substitute 1 for the numerator, i.e.
$$\tan(x+y) = \frac{1}{1-\tan x\tan y}$$ as I am using a unit square and $1 = \tan x+\tan y$. I tried graphing these to compare and evaluating values but they are definitely not equal.

Is there a general rule for making substitutions in diagram proofs like this one? If I cannot use $1$ in the numerator to represent $\tan x+\tan y$, then why am I allowed to use $1$ in the denominator?

答: @Blue pointed out that while I have defined the bottom edge of my rectangle as 1, there is nothing in the diagram to suggest that it is a square. Thus, I cannot set its height as 1 and cannot assert that $1 = tanx+tany$! I am embarrassed as this has been bothering me for several days; finding this simple solution was very satisfying. Thanks, @Blue!


问:
How does this summation work?
For $i = 1,2,\ldots,19$ we have:

$$\left(\sum_{j=1, j\neq i}^{19}\alpha_j\right)\cdot(1.5) + (\alpha_i)\cdot(3) = 1.5 \implies \left(\sum_{j=1}^{19}\alpha_j\right)(1.5) + (\alpha_i)(1.5) = 1.5$$

Does anyone know how to prove this?
Any intuition behind it?

I thought it would make sense when written as:

$$\alpha_1 = \frac 1 2 - \frac 1 2 \left(\sum_{j=2}^{19}\alpha_j\right)$$

$$ \vdots $$

$$\alpha_{19} = \frac 1 2 - \frac 1 2 \left( \sum_{j=1}^{18}\alpha_j \right)$$

$\alpha_i$ are  unknowns

But I still fail to see why.

答:
The desired claim follows quickly from the identity: $$\left(\sum_{j=1,\,j\neq i}^{19}a_j\right)+a_i=\sum_{j=1}^{19}a_j$$

Now just multiply both sides by $1.5$ and add $1.5\times a_i$ to both sides.

Title: How does this summation work?


问:
Prove that there exists integers $r,s$ such that $r^k\equiv a \pmod{p_1}$ and $s^k \equiv a \pmod{p_2}$
Given that $k$ is a positive integer and $p_1,p_2$ are co primes such that for some integers $q,a$ and $r,s$
$$q^k \equiv a \pmod{p_1p_2} \iff r^k \equiv a \pmod{p_1} \:\text{ and }\:s^k\equiv a \pmod{p_2}$$

My try:
I tried to prove from right to left first.

$$r^k=mp_1+a$$ and $$s^k=np_2+a$$

Now by Euclid's algorithm we have $m=m'p_2+r_2$ and $n=n'p_1+r_1$

Thus we get $$r^k=m'p_1p_2+p_1r_2+a$$ and $$s^k=n'p_1p_2+r_1p_2+a$$

I got stuck here.

答:
This is a special case $\,f(x) = x^k-a\, $ of the result below, which shows that if $\,m,n\in\Bbb Z\,$ are coprime and $\,f\in\Bbb Z[x]\,$ is a polynomial with integer coefs, then the roots of $\,f\bmod mn$ correspond to CRT-combining the roots of $f\bmod m\,$ and $\,f\bmod n.\,$ In particular this yields the sought existence claim:

$$ f\,\ \text{is solvable $\!\bmod mn\iff f\,$ is solvable $\!\bmod m\,$ & $\!\bmod n$}\qquad$$

Suppose that $\,f(x)\,$ is a polynomial with integer coefs and $\,m,n\,$ are coprime integers. By CRT, solving  $\,f(x)\equiv 0\pmod{\!mn}\,$ is equivalent to solving $\,f(x)\equiv 0\,$ mod $\,m\,$ and mod $\,n,\,$ and each CRT combination of a root $\,r_i\,$ mod $\,m\,$ with a root $\,s_j\,$ mod $\,n\,$ corresponds to a unique root $\,t_{ij}\bmod mn\,$ i.e.

$$\begin{eqnarray} f(x)\equiv 0\!\!\!\pmod{\!mn}&\overset{\rm CRT}\iff& \begin{array}{}f(x)\equiv 0\pmod{\! m}\\f(x)\equiv 0\pmod{\! n}\end{array} \\
&\iff&  \begin{array}{}x\equiv r_1,\ldots,r_k\pmod {\!m}\phantom{I^{I^{I^I}}}\\x\equiv s_1,\ldots,s_\ell\pmod{\! n}\end{array}\\
&\iff& \left\{ \begin{array}{}x\equiv r_i\pmod{\! m}\\x\equiv s_j\pmod {\!n}\end{array} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}^{\phantom{I^{I^{I^I}}}}\\
&\overset{\rm CRT}\iff& \left\{ x\equiv t_{i j}\!\!\pmod{\!mn} \right\}_{\begin{array}{}1\le i\le k\\ 1\le j\le\ell\end{array}}\\
\end{eqnarray}\qquad\qquad$$

You can find many concrete worked examples of this isomorpism in prior posts, e.g. below

$\,f(x) = (x\!+\!2)(x\!+\!3)\,$ and $\,m,n =
   11,17$

$f(x) = x(x-1)\,\ $ [idempotents]

$x^2=1 \pmod{\!7},\ x^3=1\pmod{\! 9},\,$ generalization where $f$ depends on the modulus.


问:
Logic - An application of combining first-order logic statements
Consider the sets $\Omega_b$ and $B_a$ which are dependent on $b$ and $a$, respectively and the function $ x:\mathbb{R} \rightarrow D $. There are three propositions:
$$
\begin{align}
&P_1 : \forall a > 0: \exists b>0 : \Omega_b \subseteq B_a\\
&P_2 : \forall b > 0: \exists T > 0 : \forall t > T : x(t) \in \Omega_b\\
&P_3 : \forall a > 0: \exists T > 0 : \forall t > T : x(t) \in B_a
\end{align}
$$

I want to show that $P_1$ and $P_2$ imply $P_3$ in formula form,  that is, $P_1 \land P_2 \Rightarrow P_3$. I believe it is true but I can't figure out how to rigorously show that.

What I tried:

From $P_1$ and axiom of choice, it is guaranteed that a choice function $c:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $\Omega_{c(a)} \subseteq B_a$ exists.

From $P_2$ together with $P_1$, we have $\forall a > 0: \exists T > 0 : \forall t > T : x(t) \in \Omega_{c(a)} \subseteq B_a$.

答:
$\def\fitch#1#2{\quad\begin{array}{|l}#1 \\ \hline #2\end{array}}$
$\newcommand\tab[1][1cm]{\hspace{#1}}$

I spent some time learning the Fitch-style system and tried applying it to my problem. The proposition seemed intuitively true, but I didn't know how to formally and unambiguously prove it.

$$
 \fitch{
  1.~\forall a > 0: \exists b>0 : \Omega_{b} \subseteq B_{a}\\
  2.~\forall b > 0: \exists T > 0 : \forall t > T : x(t) \in \Omega_{b}
 }{
  \fitch{3.~\boxed{a_{1}}~a_{1}>0}{
   4.~\exists b>0:\Omega_{b} \subseteq B_{a_{1}}~\tab[4cm]\forall\text{-Elim: 1}\\
   \fitch{5.~\boxed{b_{1}}~b_{1}>0 \land \Omega_{b_{1}}\subseteq B_{a_{1}}}{
    6.~\exists T > 0 : \forall t > T : x(t) \in \Omega_{b_{1}} \subseteq B_{a_{1}}~\tab[1cm]\forall\text{-Elim: 2}
   }\\
  7.~\exists T > 0 : \forall t > T : x(t) \in B_{a_{1}}~\tab[2cm]\exists\text{-Elim: 4, 5-6}
  }\\
 8.~\forall a > 0: \exists T > 0 : \forall t > T : x(t) \in B_{a}~\tab[1cm]\forall\text{-Intro: 3-7}
 }
$$