问:
uniqueness of system of PDE using Lax Milgram
Consider the following question.

(5) Consider the system

(1)

$$
\begin{array}
-\Delta u(x)-v(x)&=&f(x) && x \in \Omega \\
u(x)-\Delta v(x) & = & g(x) && x \in \Omega \\
u(x) & = & 0 && x \in \partial \Omega \\
v(x) & = & 0 && x \in \partial \Omega
\end{array}
$$
where $\Omega$ is a smooth, bounded domain.
(a) Derive a weak formulation of system (1), using suitable test functions for each equation. Define a bilinear form $B((u, v),(w, q))$ such that the weak formulation is equivalent to

(2)
$$
B((u, v),(w, q))=(f, w)+(g, q)
$$
(b) What are the appropriate function spaces for $u$ and $v$ in Eq. (2)?
(c) Use the Lax-Milgram Theorem to show that the Eq. (2) has a unique solution.

transcribed from this screenshot

I know that $u,v\in H_0^1(\Omega)$ but not sure how to apply Lax Milgram here.  Finding the norm of $(u,v)$ just doesn't make sense to me. $B((u,v),(u,v))$ just isn't looking right either since I multiplied the first equation by $w$ and the second one by $q$ and each equation has an expression involving both a function of $u$ and $v$.

How would Lax Milgram be applied in this problem? What does continuity and coercivity look like

答:
Creation of the weak formulation is a standard exercise. We assume $\Omega \subset \mathbb R^n$.

Indeed, suppose that $$
\Delta u(x) - v(x) = f(x) \quad ,\quad \forall \ x \in \Omega \tag{1}
$$

Let $\phi$ be a function in $\Omega$ whose regularity will be decided based on desired manipulations. Then, assuming that $(1)$ holds, we multiply by $\phi$  (note : while writing integrals, I omit the variable $x$ and the notation $dx$ to focus on the integrand. This is common practice in PDE):
$$
\phi\Delta u - \phi v = \phi f \implies \int_{\Omega} \phi \Delta u - \int_{\Omega} \phi v
 = \int_{\Omega} \phi f$$

Let's deal with just the first term : $\int_{\Omega} \phi \Delta u$, the rest will remain as they are. We write by extension by zero (allow $\phi$ to be zero on $\partial \Omega$, so the function remains integrable) :
$$
\int_{\Omega} \phi \Delta u = \int_{\mathbb R^n} \phi \Delta u
$$

Under suitable conditions, Fubini's theorem  and integration-by-parts apply and we can write :
$$
\int_{\mathbb R^n} \phi \Delta u  = \int_{\mathbb R^{n-1}} \left[\int_{\mathbb R} \phi \Delta u \right] = -\int_{\mathbb R^{n-1}} \int_{\mathbb R} \nabla \phi \nabla u = -\int_{\mathbb R^n} \nabla \phi \nabla u = -\int_{\Omega} \nabla \phi \nabla u
$$

where the second equality follows from integration-by-parts and the fact that $\phi$ is compactly supported hence the derivatives of it vanish at infinity.

Which means we land up with the weak formulation for this particular equation :
$$
-\int_{\Omega} \nabla \phi \nabla u - \int_{\Omega} \phi v = \int_{\Omega} \phi f
$$

The conditions are quite self-evident : we require (weakly) $\nabla u , \nabla \phi$ to exist and be integrable , and furthermore vanish on the boundary. It is only natural to then call for $\phi,u \in H_0^1(\Omega)$.

The other equation is solved similarly, and results in
$$
\int_{\Omega} \nabla \phi \nabla v + \int_{\Omega} \phi u = \int_{\Omega} \phi g
$$

for all $\phi \in H_0^1(\Omega)$. Note that $v \in H_0^1(\Omega)$ from here as well.

To create our bilinear form is now extremely clear : we must "group" the two equations together. Define :
$$
B((\phi,\phi'),(u,v)) = \int_{\Omega} -\nabla \phi \nabla u - \int_{\Omega} \phi v + \int_{\Omega} \nabla \phi' \nabla v + \int_{\Omega} \phi' u
$$

so $B$ is a bilinear form on $H_0^1(\Omega) \times H_0^1(\Omega)$. Define :
$$
J : H_0^1(\Omega) \times H_0^1(\Omega) \to \mathbb R \quad ; \quad J(\phi,\phi') = \int_{\Omega} \phi f + \int_{\Omega} \phi' g
$$

this is a well defined linear functional. Our equation then becomes : find $(u,v)$ such that $$
B((\phi,\phi'),(u,v)) = J(\phi,\phi')
$$

for all $(\phi,\phi') \in H_0^1(\Omega)\times H_0^1(\Omega)$. We are firmly in Lax-Milgram territory ; the LHS is a bilinear form in $H_0^1(\Omega) \times H_0^1(\Omega)$ and the RHS is a linear functional on the same space.

To apply the Lax-Milgram theorem, whose direct consequence is the existence and uniqueness of $u,v$ satisfying the required equation (as desired), we must prove that $B$ is continuous and coercive, and $J$ is continuous.

Continuity of either one is fairly obvious :

$$
|B((\phi,\phi'),(u,v))| = \left|\int_{\Omega} -\nabla \phi \nabla u - \int_{\Omega} \phi v + \int_{\Omega} \nabla \phi' \nabla v + \int_{\Omega} \phi' u \right| \\
\leq \int_{\Omega} |\nabla \phi| |\nabla u| + \int_{\Omega} |\phi| |v| + \int_{\Omega} |\nabla \phi'| |\nabla v| + \int_{\Omega} |\phi'||u| \\
\leq \|\phi\|\|u\| + \|\phi\|\|v\| + \|\phi'\|\|u\| + \|\phi'\|\|v\| \\
\leq (\|(\phi,\phi')\|_{H_0^1 \times H_0^1}) (\|(u,v)\|_{H_0^1 \times H_0^1})
$$

where on the third line all norms are in $H_0^1$. I leave you to figure out the case for $J$.

The coercivity of $B$ follows from Poincare's inequality. Indeed, by Poincare's inequality, which can be used since $\Omega$ is bounded, we know that there exists a positive constant $C>0$ such that $\int_{\Omega} \nabla h  \nabla h \geq C \|h\|^2_{H_0^1}$ for every $h \in H_0^1(\Omega)$.

This helps massively, since :
$$
B((u,v),(u,v)) = \int_{\Omega} \nabla u \nabla u + \int_{\Omega} v \cdot v + \int_{\Omega} \nabla v \nabla v + \int_{\Omega} u \cdot u \geq C(\|u\|_{H_0^1} + \|v\|_{H_0^1}) = C (\|(u,v)\|_{H_0^1 \times H_0^1})
$$

because the terms $u \cdot u , v \cdot v$ are non-negative, and for the rest the Poincare inequality apply (note : I'm putting the dots to indicate that it is a dot product : it was one earlier as well between the $\phi$s and $u,v$ but because they were different variables I did not use the dot)

Thus , we may use Lax-Milgram and complete the problem.


问:
$\int_0^\infty |2 \arctan\sqrt{x} - \pi {x^p}/(1+x^p)| \mathrm dx $
Let $a(x) = 2 \arctan\sqrt{x}$ and $b(x) = \pi {x^p}/(1+x^p)$. I'm trying to evaluate:

$$\int_0^\infty |a(x) - b(x)| \mathrm dx $$

I've figured out I need the integrals:

$$\int_0^1 a(x)-b(x) dx = \left[A(x)-B(x)\right]_0^1$$ and
$$\int_1^\infty a(x)-b(x) dx = \left[A(x)-B(x)\right]_1^\infty$$

The first of which is finite, but the second may or may not be. $A(x)$ has an expression in terms of elementary functions:

$$A(x) = (1+x)\arctan{\sqrt{x}} - \sqrt{x}$$

But $B$ is not so nice. Wolfram Alpha gives a result in terms of the hypergeometric function:

$$B(x) = x\left(1-{}_2F_1\left[1,\frac{1}{p};1+\frac{1}{p}; -x^p\right]\right)$$

Evaluation of some things are easy $A(0)=B(0)=0$, $A(1)=\pi/2 - 1$ and $B(1)$ has an expression.

So, basically I need to know if

$$\lim_{x\rightarrow\infty} A(x)-B(x)$$

converges, and if so, to what?

答:
Using the Taylor series for $\arctan x$ at $x=\infty$, we find
$$A(x) = \frac{\pi x}{2}-2 \sqrt{x} +o(1).$$

As you mention, the $B(x)$ term is more ornery. Our limit depends fundamentally on the sign of $p$, which forces us to address $B(x)$ in cases.  Our first two are easy:

If $p=0$, then $B(x)=\pi x/2$ exactly, hence
$$A(x)-B(x) = -2\sqrt{x}+o(1).$$

If $p<0$, we compute
$$\lim_{x \to \infty} \,_2F_1\left(1,\frac{1}{p},1+\frac{1}{p};-x^p\right)=\,_2F_1\left(1,\frac{1}{p},1+\frac{1}{p};0\right)=1,$$
in which we make use of the series definition for the hypergeometric function.  It follows that $B(x)=o(x)$, hence $A(x)-B(x)=\pi x/2 +o(x)$.

If $p>0$, our task is considerably harder.  Nevertheless, we claim that $B(x) \sim x$ as $x \to \infty$, i.e. that
$$\lim_{x \to \infty}\,_2 F_1\left(1,\frac{1}{p},1+\frac{1}{p};-x^p\right)=\lim_{x \to \infty} \,_2F_1\left(1,\frac{1}{p},1+\frac{1}{p};-x\right)=0.$$
For $p>1$, write
$$\lim_{x \to \infty} \,_2F_1\left(1,\frac{1}{p},1+\frac{1}{p};-x\right)=\lim_{x \to \infty}(1+x)^{-1/p}\,_2F_1\left(\frac{1}{p},\frac{1}{p},1+\frac{1}{p};\frac{x}{1+x}\right)$$
by Euler's hypergeometric transformations.  Our hypothesis $p>1$ allows us to apply Gauss's theorem, and we find
$$_2F_1\left(\frac{1}{p},\frac{1}{p},1+\frac{1}{p},1\right)=\frac{\pi}{p} \csc\left(\frac{\pi}{p}\right).$$
Therefore our claim holds because $(1+x)^{-1/p} \to 0$ as $x \to \infty$. If $p=1$ exactly, then
$$\lim_{x \to \infty} \,_2 F_1\left(1,1,2;-x\right)=\lim_{x \to \infty} \frac{\log(1+x)}{x}=0,$$
and our claim still holds.  Finally, given $p \in (0,1)$, Euler's hypergeometric relations give
$$\lim_{x \to \infty} \,_2F_1\left(1,\frac{1}{p},1+\frac{1}{p};-x\right)=\lim_{x \to \infty} (1+x)^{-1} \,_2F_1\left(1,1,1+\frac{1}{p};\frac{x}{1+x}\right).$$
Fix $t \in \mathbb{R}$.  By definition,
$$_2 F_1(1,1,t,x)=\sum_{k=0}^\infty \binom{-t}{k}^{-1}x^k,$$
which converges at $x=1$ provided $t>2$.  In particular, setting $t=1+1/p$, we obtain a convergent sum at $x=1$, so that the value $_2F_1(1,1,1+1/p;1)$ is finite by Abel's theorem.  It follows that
$$\lim_{x \to \infty} (1+x)^{-1} \,_2F_1\left(1,1,1+\frac{1}{p};\frac{x}{1+x}\right)=0,$$
again in accordance with our claim $B(x) \sim x$.  Thus $p >0$ implies
$$A(x)-B(x) = \left(\frac{\pi}{2} -1 \right)x + o(1).$$

In conclusion, the limit
$$\lim_{x \to \infty} A(x)-B(x)$$
diverges, for any choice of $p \in \mathbb{R}$.

Note: the weaker result $A(x) \sim B(x)$ holds if and only if $p=0$.

ContentLicense: CC BY-SA 3.0


问:
What condition guarantees that every sequence in a set converges in that set?
I was wondering whether there exists a set, say $A$ such that every sequence $a_n \in A$ converges to some point in $A$. I found out that any set which contains more than one element cannot satisfy such a criteria, since if {${a_1,a_2}$} be such an $A$, then immediately I can construct a sequence $a_n=a_1,a_2,a_1,a_2,...$ which diverges. Hence I came to a conclusion that a singleton set is the only condition that satisfies such a criteria. Are there any more? If yes, what must be the condition?

答:
Of course, the question is trivial in the realm of T1-spaces. However without separation, I think, it's a nice exercise, namely:

Let $X$ be a topological space.
The following are equivalent:

Each sequence in $X$ converges.

$X$ is ultraconnected (i.e. two non-empty closed subsets have non-empty intersection) and $X$ is countably quasi-compact (i.e. "countably compact without T2").

Each countable family of non-empty closed subsets has non-empty intersection or $X$ is empty.

Proof:
$\it{1 \Rightarrow 2}$:
$X$ is ultraconnected:
Let $A, B$ be non-empty closed subsets of $X$. Pick $a \in A, b \in B$. Then the sequence $a,b,a,b, ...$ converges to some $x$. It is easy to see that $x \in A \cap B$.

In particular, each sequence has an accumulation point. Therefore, $X$ is countably quasi-compact.

$\it{2 \Rightarrow 3}$:
Let $X$ be non-empty and $\mathfrak{F}$ be a countable family of non-empty closed subsets of $X$. By ultraconnectness, $\mathfrak{F}$ has finite intersection property. Hence $\bigcap \mathfrak{F} \neq \emptyset$.

$\it{3 \Rightarrow 1}$:
Let $(x_n)_{n \in \mathbb{N}}$ be a sequence in $X$.
Then there exists a $x \in \bigcap_{n \in \mathbb{N}}\overline{\{x_n\}}$. It it easily seen that
$(x_n)_{n \in \mathbb{N}} \rightarrow x$.

Remarks:

If $X$ contains only finitely many open subsets, in particular, if $X$ is finite, 2. reduces to "$X$ is ultraconnected."

For a direct proof of $\it{1 \Rightarrow 3}$
let $(F_n)_{n \in \mathbb{N}}$ be a countable family of non-empty closed subsets of $X$. For each $n \in \mathbb{N}$ choose an $x_n \in F_n$ and consider the sequence
$x_1, x_1, x_2, x_1, x_2, x_3, ...$

ContentLicense: CC BY-SA 4.0


问:
"Drawing a Picture" of $\operatorname{Spec}(\mathbb{Z})$
Atiyah/Macdonald's commutative algebra book asks the reader to draw pictures of the prime spectrum of $\mathbb{Z}$ in exercise 1.16. I worked through it on my own, figured out what the space looks like as a set and where the open/closed sets are, but when I did some research to check my work I found the following picture:

What exactly is the squiggly line supposed to represent? is it meant to be the ideal $(0)$? That makes a vague bit of sense to me, as the closure of (0) is the entire space, but it contradicts the fact that (0) is a single point in the prime spectrum. Another source said to draw $\operatorname{Spec}(\mathbb{Z})$ as a curve passing through the ideals $(2),(3),(5),\ldots$ and a separate point off to the side to represent $(0)$, but again I don't get how drawing a curve helps clarify the structure of the space. Why not draw it as a discrete set of points and leave out the curve?

Edit: Here's another guess I just came up with: If we put the finite-complement topology on the real line and then create a quotient space by identifying every point that is not a prime positive integer as one point and call it (0), do we get $\operatorname{Spec}(\mathbb{Z})$? If this is the case it would definitely help justify the drawing.

答:
I really like Yuri Manin's way of describing how points and generic points build up affine schemes. In this sense curves in $\mathbb{C}[T_1, C_2]$ have closed points but also generic points that sit on a "lower level". These generic points in turn sit above the generic point $(0)$.

For $\operatorname{Spec} \mathbb{Z}$ it would be points sitting on a plane with the generic point below them, pretty much the two lower levels of the picture below.

$\hskip2in$


问:
Why is $\frac{n-1}{n}$ always in its simplest form?
So I was just messing around on my calculator and I noticed that for every positive integer the formula $\frac{n-1}{n}$ is always a fraction in its simplest form im not quite sure why this happens is there a case where this is not true?

答: Using Bezout's Lemma, we know that $\text{gcd}(n,n-1)=1$ because $n-(n-1)=1$. This means they have no common prime factors, so you can't find any natural number $k\neq 1$ such that $k\mid n$ and $k\mid(n-1)$, hence the fraction $\frac{n-1}{n}$ is always in its simplest form.


问:
Proving the following sequence is bounded
I have to prove the following sequence is bounded (Or prove otherwise)
$$
a_{n+1} = a_n + \frac{1}{a_n}
$$
When $a_1$ is any real positive number.

I have tried using induction. But I'm not sure if what I'm getting to is correct.

My take:

Assume $a_1$>1 (if not, then in the next step $a_2$ will be - and the proof is the same).

First we show for base case:
$$
a_2 = a_1 + \frac{1}{a_1} < 2a_1
$$

Assume by induction that this holds for $a_n$:
$$
a_n < 2a_1
$$

Here is where I got stuck. Any algebra manipulation I tried to do left me with a leftover of which I can't seem to get rid of, or an inequality which doesn't do any considerable progress.

For instance:
from the above we get:
$$
\frac{1}{a_n} > \frac{1}{2a_1}
$$
$$
a_n + \frac{1}{a_n} > a_n + \frac{1}{2a_1}
$$
Which doesn't help.
Perhaps this sequence isn't bounded?

I'd appreciate any help at all (preferably a nudge) :)

答:
First, prove inductively that $a_n > 0$ for every $n$. You deduce that
$$a_{n+1}-a_n = \frac{1}{a_n} > 0$$

so the sequence is increasing.

So either it is unbounded, either it converges. If it converged to a limit $l \in \mathbb{R}$, you would get
$$l=l+\frac{1}{l}$$

so $$\frac{1}{l}=0$$

which is impossible. So the sequence is unbounded.


问:
Notation in Fundamental Groups in Topological Spaces
I was reading about Fundamental Groups in Topological Spaces at WikiPedia. In the middle of the text, they define the following set $\pi_1(X,x_0):$

I guess I don't understand the meaning of "$\text{/homotopy}$". What does this mean?

答:
If you have a set $S$ and an equivalence relation $\sim$ on $S$, then
$$S / \sim$$
denotes the set of all equivalence classes with respect to $\sim $. In your question we have $S =$ set of all loops $\gamma : [0,1] \to X$ based at $x_0$. You have the relation $\gamma$ is homotopic to $\gamma'$ which is an equivalence relation on $S$. If you write $\gamma \simeq \gamma'$ for this equivalence relation, then
$$\pi_1(X,x_0) = S/\simeq \phantom{.}= S/\text{homotopy} .$$


问:
Bell curve and normal distribution and the empirical rule
This is an assignment on the 68-95-97 rule - or rather a close idea. Specifically we go for 68-90-95, but you'll see.

Let's assume we have a normal distribution $N(\mu,\sigma^2)$ and draw a sample of 100 elements from it, $X_1,X_2,...X_{100}$. We now define;
$$M(t) = E\bigg[\sum_{i=1}^{100} \textbf{1}_{[-t,t]}(X_i-\mu)\bigg]$$

Where the indicator function $\textbf{1}_{[-t,t]}$ is;
$$\textbf{1}_{[-t,t]}(x) = \begin{cases}
1&\text{if } x\in[-t,t]\\
0&\text{if } x\notin[-t,t]
\end{cases}$$

That is, the function $M(t)$ is the expected number of samples in the interval $\mu-t$ to $\mu+t$.

Now; to the question: How does one show the desired results;
$M(\sigma)\approx 68, M(\frac{5}{3}\sigma)\approx 90, M(2\sigma)\approx 95$?

p.s. I am familiar with the derivation that "simply" uses the CDF to calculate the given values, but it seems to me that this assignment wants to do it without the CDF?

答:
Any solution will essentially say something like $$M_n(t) = \sum\limits_{i=1}^n \mathbb P(-t \le X_i-\mu \le t)\\= n \left(\mathbb P\left(\frac{X_i-\mu}{\sigma} \le \frac{t}{\sigma} \right)-\mathbb P\left(\frac{X_i-\mu}{\sigma} \le \frac{-t}{\sigma} \right)\right) \\= n\left( \Phi\left( \frac{t}{\sigma}\right) - \Phi\left( \frac{-t}{\sigma}\right)\right)\\= n\left( 2\Phi\left( \frac{t}{\sigma}\right) - 1\right)$$
so you cannot escape the normal cumulative distribution function.  For example with $n=100$ and $t=\sigma$ you get $M_n(\sigma)=100\left( 2\Phi\left( 1\right) - 1\right)\approx 68.27$


答:
Suppose $\|f\|_1 \leq 1$. By C-S inequality we have $|Tf(t)| \leq  t$. Hence, $\|Tf\|_2\leq \frac1  {\sqrt 3}$. This value is attained when $f\equiv 1$ so $\|T\|=\frac 1 {\sqrt3 }$.

Note that $|\int_0^{t} f(s)ds| \leq \sqrt {\int_0^{t} 1^{2}ds} {\sqrt {\int_0^{t} f^{2}(s)ds}}\leq t$

Title: Norm of integral operator from $(C[0,1], \|\|_1)$ to $(C[0,1], \|\|_2).$

Body:
Let $\|\|_1$ be sup-norm and $\|f\|_2=(\int_0^1|f(t)|^2dt)^{\frac{1}{2}}$. Define linear operator $T$ from  $(C[0,1], \|\|_1)$ to $(C[0,1], \|\|_2)$ as $$T(f)(t)=\int_0^tf(s)ds$$  Then norm of $T$ is

$1.$ $1$.

$2.$ $\frac{1}{2}.$

$3.$ $\frac{1}{\sqrt{3}}.$

$4.$ none .

I am trying it like this

$$\|T(f)(t)\|=(\int_0^1|\int_0^tf(s)ds|^2)^{\frac{1}{2}}\leq (\int_0^1\int_0^t\|f\|^2_1ds)^{\frac{1}{2}}=\|f\|_1(\int_0^1tdt )^{\frac{1}{2}}=\frac{1}{\sqrt2}\|f\|_1$$ please help me to solve this problem . Thank you .


问:
Evaluate the lateral surface defined by a set rotated 360 degree around the z axis
Let $D=\left\{(y,z): \sqrt{y} \leq z \leq 2-\sqrt{y} \right\}$. Rotate $D$ $360$ degree around the $z$ axis generating the volume $V_z$ and evaluate the lateral surface of $V_z$, $|\partial V_z|$.

For evaluating the lateral surface of $V_z$, I considered these two curves:

$\gamma_1(t) = (t,\sqrt{t})$, $0 \leq t \leq 1$;

$\gamma_2(t) = (t,2- \sqrt{t})$, $0 \leq t \leq 1$.

So I continued by doing the integral $$|\partial V_z| = 2\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt +2\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt = 4\pi\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}dt.$$

Any tips on how to solve this integral? Is it possible to evaluate the surface with a different approach?

答:
The surface area of a solid of revolution around the $z$-axis is given by
$$S = 2\pi\int_{a}^{b} f(z) \sqrt{1 + f'(z)^{2}}\, dz.$$
In your case we have two functions: $f_1(z)=z^2$ (from $z=\sqrt{y}$) over the interval $[0,1]$ and $f_2(z)=(2-z)^2$ (from $z=2-\sqrt{y}$) over the interval $[1,2]$. Therefore
$$|\partial V_z|=2\pi\int_{0}^{1} f_1(z) \sqrt{1 + f_1'(z)^{2}}\, dz+2\pi\int_{1}^{2} f_2(z) \sqrt{1 + f_2'(z)^{2}}\, dz$$
Note that, by symmetry, the two integrals are equal. Moreover
$$\int_{0}^{1} f_1(z) \sqrt{1 + f_1'(z)^{2}}\, dz=\int_{0}^{1} z^2 \sqrt{1 + 4z^2}\, dz=\int_{0}^{\text{arcsinh}(2)}\cosh^2(t)\sinh^2(t)\,dt$$
where we applied the substitution $2z=\sinh(t)$. Can you take it from here? You may also take a look at How to evaluate integral: $\int x^2\sqrt{x^2+1}\;dx.$

P.S. Your approach is correct and after letting $t=z^2$ we find the same integral
$$\int_{0}^{1}t\sqrt{1+\frac{1}{4t}}\,dt=\int_{0}^{1} z^2 \sqrt{1 + 4z^2}\, dz.$$
I don't see an easier way to find this area.


问:
Prove that the algebraic expression is greater than zero
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...

答:
I considered parsing the polynomial as  $ \ (x^8 - x^5) \ + \ (x^2 -x + 1) \ $ .  Restricting techniques to ones that would likely appear in junior high school and high school curricula, we can show that $ \ x^2 - x + 1 \ $ is never equal to zero since its discriminant is $ \ -3 \ $ . So its value can never change sign; since it equals $ \ 1 \ $ for  $ \ x \ = \ 0 \ $ , it is always positive.  We can factor

$ \ (x^8 - x^5)  \ = \ x^5  ·  (x^3 - 1)  \ = \ x^5  ·  (x - 1)  ·  (x^2  +  x  + 1)  \ . \ \ \  $ [using "difference of two cubes"]

By an argument similar to the above, $ \ x^2 + x + 1 \ $ is also always positive.  The product of the first two factors is positive for  $ \ x \ > \ 1 $ and for  $ \ x \ < \ 0 \ $ , hence the complete polynomial is positive in these intervals.  It is equal to $ \ 1 \ $ at  $ \ x \ = \ 0 \ $ and  $ \ x \ = \ 1 \ $ , so it remains to establish the proposed inequality for  $ \ 0 \ < \ x \ < \ 1 \ $ .

Using "completion of the square", we write our polynomial as  $ \ (x \ - \ \frac{1}{2}  )^2 \ + \ \frac{3}{4} \ - \ (x^5 \ - \ x^8) \ $ . From what we know about power-functions, we can say that $ \ x^5 \ - \ x^8 \ > \ 0 $ in the interval  $ \ 0 \ < \ x \ < \ 1 \ $ , but I don't think there's a convincing argument using the available means to show that  $ \ 0 \ < \ x^5 \ - \ x^8 \ < \ \frac{3}{4} \ $ .  [We know that calculus will give us a maximum value for $ \ x^5 \ - \ x^8 \  $ of $ \ ( \frac{5}{8} )^{5/3}  ·  \frac{3}{8} \ \approx \ 0.171 $ at $ \ x_{min} \ = \ (\frac{5}{8} )^{1/3} \ \approx \ 0.855 \ $ . ]

Instead we may exploit symmetry and separate the polynomial into "even" and "odd" components as $ \ (x^8  +  x^2  + 1) \  - \ (x^5 + x ) \ . $ [This is a variation on what user143137 does.]  For  the first polynomial, $ \ x^8  +  x^2  + 1 \ \geq \ 1 \ $ for all real numbers (the graphs of power functions should be familiar at this level).  The second polynomial, $ \ x^5 + x  \ $ , is negative for $ \ x \ < \ 0 \ $ , so we only need to be concerned about $ \ 0 \ < \ x \ < 1 \ $ .  In that interval, we have  $ 1 \ < \ x^8  +  x^2  + 1 \ < \ 3 \ $ and $ \ 0 \ < \ x^5 + x  \ < 2 \ $ .  Each of these polynomials increases "monotonically" (although we might not use that word at this level), so we can conclude that $ \ (x^8  +  x^2  + 1) \  - \ (x^5 + x ) \ > \ 0 \ . $


问:
A function $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfies $p(x) ≥ \alpha$ and $(1 + \beta)[p'(x)]^2\leq p''(x)p(x).$
The following is exercise IV.3.1. from Analysis I by Amann and Escher.

Suppose that $\alpha, \beta, R > 0$ and $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfy $p(x) ≥ \alpha$ and $$(1 + \beta)[p'(x)]^2\leq p''(x)p(x),\quad x\geq0.$$ Show that $R < \infty$ and $p(x)\to\infty$ as $x\to R−$. (Hint: The function $p^{-\beta}$ is concave. Use a tangent line to $p^{-\beta}$ to provide a lower bound for $p$ (see Application 3.9(e)).

I think I found a counterexample. Let $\alpha:=\beta:=1$ and $p:[0,\infty)\to\mathbb{R}$ with $$p(x):=\frac{x+2}{x+1},\quad x\geq0.$$ Then $p'(x)=-1/(1+x)^2$, $p''(x)=2/(1+x)^3$, so $$(1 + \beta)[p'(x)]^2=\frac{2}{(1+x)^4}=\frac{2}{(1+x)^3}\frac{1}{1+x}\leq\frac{2}{(1+x)^3}\frac{2+x}{1+x}=p''(x)p(x)$$ for all $x\geq0$. Now $p(x)\geq1$, $R=\infty$ and $p(x)\to1$ as $x\to R-$.

Maybe I'm missing something. The question has been asked here but there are no answers yet.

答:
I wrote an email to the author of this series and the following is the screenshot of the email I received:

Since in email system, it can not show Latex, I will type it down for convenience:

$0\leq (1+\beta)p'(x)\leq p''(x)p(x)$ for $0\leq x\leq R$.
In addition, you have to add “unless p is constant” at the end of the assertion.
Now you should be able to do the exercise.

Although the question was proposed three years ago, I hope this would help you somehow.


问:
Prove that exist finitely many positive integers $n$ satisfying $\tau (n)=a$ and $n|\phi (n)+\sigma (n)$
Given a fixed positive integer $a\geq 9$. Prove there exist finitely many positive integers $n$, satisfying:

$\tau (n)=a$

$n|\varphi (n)+\sigma (n)$

My ideal is if i write the factorization $ n = \prod_{i = 1}^{m}p_i^{a_i}, $ then since $ \prod_{i = 1}^{m}(a_i + 1) = a, $ there are only a finite number of suitable $ (m, (a_1, a_2, \dots, a_m)) \in \mathbb{Z}^{+} \times \mathbb{N}^m. $ Therefore, we can fix a $ (m, (a_1, a_2, \dots, a_m)) \in \mathbb{Z}^{+} \times \mathbb{N}^m $.
For the remainder of the proof, it suffices to show there are a finite number of sets of distinct primes $ (p_1, p_2, \dots, p_m) $ such that $ n = \prod_{i = 1}^{m}p_i^{a_i} $ and $ n \vert \phi(n) + \sigma(n)$. But how can I show there are finite $ (p_1, p_2, \dots, p_m) $?

答:
Continue it like the following:
Assume for the sake of contradiction there are infinitely many such sets and let $ T $ be the set of these sets. Pick an arbitrary set $S\in T$. It's clear that $ \phi(n) = \prod_{i = 1}^{m}(p_i^{a_i - 1}(p_i - 1)) $ and $ \sigma(n) = \prod_{i = 1}^{m}\frac{p_i^{a_i + 1} - 1}{p_i - 1} $ therefore $ \frac{\phi(n) + \sigma(n)}{n} =  \prod_{i = 1}^{m}\frac{p_i - 1}{p_i} + \prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right). $ Clearly these pimes have an upper bound (we'll call it $L$) because otherwise we would have $ 1 < \frac{\phi(n) + \sigma(n)}{n} < 2. $ Therefore it's clear that there are an infinite number of sets in $ T $ that all have the same prime $ p<L. $ Now going through the same procedure over and over for all sets, we have that every element of a set $ S\in T $ is bounded by some number so $ T $ is finite and our proof is complete.

P.S. If you're wondering about the part I said that if the prime factors don't have an upper bound, $\frac{\phi(n) + \sigma(n)}{n} < 2$ , here's the proof:

As we said, $ \frac{\phi(n) + \sigma(n)}{n} =  \prod_{i = 1}^{m}\frac{p_i - 1}{p_i} + \prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right). $ Now clearly $\prod_{i = 1}^{m}\frac{p_i - 1}{p_i}$ is always less than one so assume that $\prod_{i = 1}^{m}\frac{p_i - 1}{p_i}=1-s$.
On ther other hand $\prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right)$ can be arbitrarily close to one if $p$ doesn't have an upper bound so we'll choose one of the prime factors such that $\prod_{i = 1}^{m}\left(1 + \frac{1}{p_i} + \dots + \frac{1}{p_i^{a_i}}\right)<1+s$. Then we'll have $\frac{\phi(n) + \sigma(n)}{n} < 2$ and the proof is complete.


问:
Bayesian Network, Sprinkler Example
In reference to the wet grass / sprinkler Bayesian network problem at this site:
http://www.cs.ubc.ca/~murphyk/Bayes/bnintro.html

Pr(S=1 | W=1) has been determined as 0.430.

Could someone please help me expand the expression (numerator and denominator) that yields 0.430?

As a try, to begin with ...

Pr(S=1 | W=1)
= P(S=1, W=1) / P(W=1)
= [P(S=1, W=1 | R=1)P(R=1) + P(S=1, W=1 | R=0)P(R=0) ] / P(W=1)

and then, R depends on C; there is more to denominator for sure.

Help is appreciated!

答:
I do not have the privileges to add a comment to @user732478 (https://math.stackexchange.com/users/732478/user732478) so I thought I would add it via a post.

I have been looking into the same problem as the OP.

The probability P(W=1) is 0.64710 as stated on the web-site is definitely the correct value, contrary to the suggestion from the answer referenced above. I thought it would be helpful to anyone else that came across this answer to present the workings for this.

We start with the expression for the factored joint probability as stated in a previous answer. We need to sum over the different probabilities to get the marginal probability $P(W)$ which we will then set to 1 (i.e. True).

$$P(W=1) = \sum _{c,r,s} P(C=c)P(S=s|C=c)P(R=r|C=c)P(W=1|S=s, R=r)$$

There are 3 binary elements, and so 8 possible combinations for $C,R,S$. I don't show the workings for each case rather the result of the product from the equation above, but with the respective values from the tables for the three random variables $C,R,S$. For brevity I will show $P_{CRS}$ as $P_{TTT} \dots P_{FFF}$ as follows:

$P_{TTT} = 0.03960, P_{TTF} = 0.3240, P_{TFT} = 0.009, P_{TFF} = 0.0, P_{FTT} = 0.04950, P_{FTF} = 0.0450, P_{FFT} = 0.180, P_{FFF} = 0.0$

From this we sum to get $P(W=1) = 0.6471$. This is the value quoted on the web site, when using this we arrive at the correct value for $P(S=1|W=1)$ as required in the original question.

For a check, I have also used pgmpy and calculated exactly the same answer.

Title: Bayesian Network, Sprinkler Example


问:
What is predicate lifting?
I have tried and tried again to understand coalgebraic logic (generalization of modal logics) but I can not understand the notions "predicate lifting" and "relation lifting". Are there explanations and illustrations for those notions, is this notion used outside coalgebraic logics community?

Chapter 6 of http://www.cs.ru.nl/B.Jacobs/CLG/JacobsCoalgebraIntro.pdf is about predicate lifting, but the notion seems to be so remote from everything that I know. Can someone, please, reformulate this notion in more understandable terms?

答:
Before going fully general with fibrations, one can understand predicate liftings entirely in sets. I'm writing $2^X$ for the set of predicates on $X$, that is, characteristic functions $X\to 2$ assigning every element of $X$ to its truth-value (i.e. whether the predicate holds or not).

In this sense, a predicate lifting for a functor $F: \mathsf{Set}\to\mathsf{Set}$ is a natural transformation $\lambda_X: 2^X \to 2^{FX}$, that "lifts" predicates from the set-level ($2^X$) to a predicate on functor elements ($2^{FX}$). If you have a predicate $P\in 2^X$, then $\lambda_X(P)$ is a predicate on $FX$. In coalgebraic modal logics, the  predicate liftings provide the semantics to modal operators. For example, the modal formula $\Diamond\phi$ means that there is some successor (or "next world") satisfying $\phi$. If $P$ captures the semantics of $\phi$, then a predicate lifting $\lambda$ for $\Diamond$ would send $P$ to $\lambda_X(P)\in 2^{FX}$ and $\lambda_X(P)$ intuitively has those successor structures $s\in FX$ in which at least one successor satisfies $P$ (i.e. $\phi$).

If  $\mathcal{P}$ denotes the (covariant) power set functor, then you can implement the predicate lifting for $\Diamond$ as
$$
\lambda_X(P) = \{ s \in \mathcal{P}X : \exists x\in s\colon x\in P\}
$$
Different modal operators have different semantics of course, and for $\Box\phi$ (all successor worlds satisfy $\phi$) one would define a predicate lifting as follows:
$$
\lambda_X'(P) = \{ s\in \mathcal{P}X : \forall x\in s\colon x\in P\}
$$
Instead of $\mathcal{P}$ you can put in other functors $F$ of course (with adjusted definitions of $\lambda_X$ and $\lambda_X'$).

Such a predicate lifting $\lambda_X$ 'lifts' the functor $F: \mathsf{Set}\to  \mathsf{Set}$ to a functor $\bar F: \mathsf{Pred}\to  \mathsf{Pred}$ on the category of predicates, which is a fibration over $\mathsf{Set}$. In the text you linked it seems that another generalization is done: $2^{(-)} = hom(-,2)$ gives rise to a fibration, and so you can generalize the truth values $2$ or even the base category.

Title: What is predicate lifting?


问:
Proof that $AA^T$ is invertible if $A$ has independent rows
While looking for the answer for my question I came across this post. It may be a silly idea, but if $A^{t}$ has independent rows can I just transpose it and get $A$ with independent columns and proceed as is shown in the linked solution? Or it is not working this way?

答:
Suppose $A$ is $m\times n$ and that $AA^T$ is not invertible. Then there exists $v\ne0$ (an $m\times1$ column vector) such that
$$
AA^Tv=0
$$
This implies $v^TAA^Tv=0$, hence $(A^Tv)(A^Tv)=0$, so $A^Tv=0$. Thus $v^TA=0$ and therefore $A$ doesn't have linearly independent rows.


答:
Since $\arctan u+\arctan1/u=\pi/2$ for every positive $u$,
$$
a(x)-b(x)=\frac{\pi}{1+x^p}-2\arctan\frac1{\sqrt{x}}.
$$
Let us look at the behaviour of $a(x)-b(x)$ when $x\to+\infty$::

If $p\gt1/2$, the $\arctan$ term is predominant and  $\arctan1/\sqrt{x}\sim1/\sqrt{x}$ which is not integrable, hence $a-b$ is not integrable.

If $p\lt1/2$, the $\pi/(1+x^p)$ term is predominant and $\pi/(1+x^p)\sim\pi/x^p$ when $p\gt0$ which is not integrable, hence $a-b$ is not integrable (when $p\leqslant0$, $\pi/(1+x^p)$ converges to a nonzero limit hence $a-b$ is not integrable either).

If $p=1/2$, both terms are equivalent to multiples of $1/\sqrt{x}$ and $a(x)-b(x)\sim(\pi-2)/\sqrt{x}$ hence $a-b$ is not integrable.

Finally, the function $a-b$ is never integrable at $+\infty$.

Tools used:

$\arctan u+\arctan1/u=\pi/2$ for $u\gt0$

$u/(1+u)=1-1/(1+u)$ for every $u\ne-1$

$1/(1+u)\sim1/u$ when $u\to\infty$

$\arctan u\sim u$ when $u\to0$

Integrability or nonintegrability of $u\mapsto u^a$ at $+\infty$, for various exponents $a$

NOT W|A and NO hypergeometric function

Title: $\int_0^\infty |2 \arctan\sqrt{x} - \pi {x^p}/(1+x^p)| \mathrm dx $

Body:
Let $a(x) = 2 \arctan\sqrt{x}$ and $b(x) = \pi {x^p}/(1+x^p)$. I'm trying to evaluate:

$$\int_0^\infty |a(x) - b(x)| \mathrm dx $$

I've figured out I need the integrals:

$$\int_0^1 a(x)-b(x) dx = \left[A(x)-B(x)\right]_0^1$$ and
$$\int_1^\infty a(x)-b(x) dx = \left[A(x)-B(x)\right]_1^\infty$$

The first of which is finite, but the second may or may not be. $A(x)$ has an expression in terms of elementary functions:

$$A(x) = (1+x)\arctan{\sqrt{x}} - \sqrt{x}$$

But $B$ is not so nice. Wolfram Alpha gives a result in terms of the hypergeometric function:

$$B(x) = x\left(1-{}_2F_1\left[1,\frac{1}{p};1+\frac{1}{p}; -x^p\right]\right)$$

Evaluation of some things are easy $A(0)=B(0)=0$, $A(1)=\pi/2 - 1$ and $B(1)$ has an expression.

So, basically I need to know if

$$\lim_{x\rightarrow\infty} A(x)-B(x)$$

converges, and if so, to what?


问:
Does the initial polynomial of a Sturm sequence need to be normalized?
I've written an algorithm that generates a Sturm sequence and uses that sequence to determine the number of real roots over an interval.

My algorithm is working, but I made no precaution to normalize my polynomial i.e. to ensure that the leading coefficient is 1.

I've looked around for information on whether it needs to be normalized or not but without anything concrete - although all examples I see make use of a normalized polynomial.

Does the initial polynomial need to be normalized or not? If so, why is my algorithm still working?

答: Nope, you can multiply a polynomial by a constant factor (even negative) without influencing the roots. All derived polynomials will be multiplied accordingly.</p>


问:
Proving global uniqueness of IVP
Assume that $\Omega \subset \mathbb{C}, 0 \in \Omega$ and $F$ analytic. I wish to prove that if I have an IVP $d\omega/dt = -F(\omega), w(0) = \alpha$ with solution defined on $[0,\infty)$ then for each $\alpha \in \Omega$ the solution is unique in $\Omega$. How would I go about doing this? I have a hint to consider the supremem of $t > 0$ such that two solutions are equal on [0,t) and if $t_0 < \infty$ they must be identically equal.

First I considered two solutions in an interval near $0$, as by Existence-Uniqueness the two solutions must be equal near 0. I am not sure how to continue with the supremum argument though

答:
You are half there. If global uniqueness does not hold and you have the two different solutions $w_1,w_2$ to the given IVP, then they must split apart somewhere. So you have a last point $t_0$ where both solutions are equal. This implies they are different for any $t>t_0$, at least for some interval starting at $t_0$.

But that contradicts local uniqueness for the IVP with IC $(t_0,w_0)$ where $w_0=w_1(t_0)=w_2(t_0)$.

Title: Proving global uniqueness of IVP


问:
Coproduct of $C_2$ and $C_3$ in $\mathsf{Grp}$
I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:

As far as I can tell, I need to do the following:

Describe the projections $\pi_2 : C_2 \to G$ and $\pi_3 : C_3 \to G$.

Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $\varphi_2 : C_2 \to A$ and $\varphi_3 : C_3 \to A$, there exists a unique group homomorphism $\sigma : G \to A$.

Giving a more concrete description of the cyclic groups, take $C_2 = \{0, 1\}$ and $C_3 = \{0, 1, 2\}$. Then the projections are $\pi_2(k) = x^k$ where $k=0,1$, and $\pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).

Beyond this, it is not clear to me what I should do. Any help would be appreciated!

答:
I don't know if you are satisfied with the above two brilliant and suggestive answers given by @Kevin Arlin and @drhab. So I will elaborate on  @drhab's answer, using your notations.

You can look at $G$ as a group that has strings like $xyxy^2xy$, $y^2xyxy$, or $xyxy^2$ as elements. Take $xyxy^2$ as an example.

The condition "$\sigma:G\to A$ is a group homomorphism" forces that
$$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2$$

Besides, the condition "the diagram commutes such that $\sigma \circ \pi_2 = \varphi_2$ and $\sigma \circ \pi_3 = \varphi_3$" and the fact that $\pi_2(1)=x$ and $\pi_3(1)=y$  forces the above map becomes

$$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2 = \sigma \circ \pi_2(1)\cdot  \sigma \circ \pi_3(1) \cdot \sigma \circ \pi_2(1)  \cdot (\sigma \circ \pi_3(1))^2 = \varphi_2(1)\varphi_3(1)\varphi_2(1)\varphi_3(1)^2$$

In fact, the map $\sigma: G\to A$ is completely prescribed by the above 2 "forces". So the map $\sigma: G\to A$ is the unique group homomorphism.

The point behind the proof is just what @Kevin Arlin said in his answer: the group homomorphism $\sigma: G\to$ any set $A$ is completely determined by how the homomorphism act on the group generators (Meanwhile, it must satisfy all relations. See this post). Your functions $\pi_2$ and $\pi_3$ are "injections", they identify $[1]$ in $C_2$ with $x$ in $G$, $[1]$ in $C_3$ with $y$ in $G$. After this identification, $\varphi_2$ and $\varphi_3$ actually specify how $\sigma$ should act on generators $\{x,y\}$ in $G$. So $\sigma$ has no choice, as has been proved above.


问:
Is there still any hope that the GCH could be equivalent to some large cardinal axiom?
Is there still any hope that the GCH could be equivalent to some large cardinal axiom?

Even a simple yes or not answer will be fine. Thanks!!

答:
No.

The term "large cardinals" has no explicit and well-defined meaning, but it's general meaning is an additional axiom which is strictly stronger in consistency than $\sf ZFC$, that is some $\varphi$ such that $\sf ZFC+\varphi$ can prove the consistency of $\sf ZFC$.

On the other hand, $\sf ZFC+GCH$ is equiconsistent with $\sf ZFC$ so it is not a large cardinal axiom per se.

(To slightly elaborate on the above meaning of no well-defined meaning, $0^\#$ is considered a large cardinal axiom, but it really just says that a certain real number exists; or a Jonsson cardinal which is not even an inaccessible one, but its existence implies that inaccessible cardinals do exist. Both these, and more, are considered large cardinal axioms despite not being directly related to inaccessible cardinals and so.)


问:
Neural ODE definition of derivative $\frac{d L}{dz(t)}$ (adjoint)
The authors of Neural Ordinary Differential Equations (NeuRIPS 2018 best paper award), propose to model machine learning problems with an ODE

$$ \dot z(t) = f(t, z(t), \theta) \quad{\text{s.t.}}\quad z(t_0)=z_0$$

where $f$ is e.g. a neural network with parameters $\theta$. The Gradient with respect to a loss function $L$ is computed via the adjoint method:

$$\begin{aligned}
\frac{d L}{d\theta} = -\int_{t_1}^{t_0}a(t)^T\frac{\partial f}{\partial \theta} d t
\quad\text{where}\quad
\dot a(t) = -a(t)^T\frac{\partial f}{\partial z},\,
a(t_1)=\frac{dL}{dz(t_1)}
\end{aligned}$$

I have a problem with interpreting the adjoint state $a(t)$. The paper claims that the adjoint state is given by $a(t)=\frac{dL}{dz(t)}$. However, let's consider the simple example$\,\dot z = w z,\, z(t_0)=z_0$ with solution $\hat z(t) := z^*(t;t_0, z_0, w) = e^{w (t-t_0)}z_0$ and loss function $L(w)=\frac{1}{2}|\hat z(t_1) - z_1|^2$. Then, solving the adjoint system yields:

$$\left.\begin{aligned}
\dot a(t) &= - a(t)^T \tfrac{\partial f}{\partial z} = -w \cdot  a(t)
\\ a(t_1) &= \frac{d L }{d z(t_1)} = \hat z(t_1) - z_1
\end{aligned}
\right\rbrace\implies a(t) = e^{-w(t-t_1)}(\hat z(t_1) - z_1)
$$

I veried that this is indeed correct, integrating $-\int_{t_1}^{t_0} a(t)\frac{\partial f}{\partial\theta}dt$ gives the correct derivative $\frac{dL}{dw}$.

Plugging back the solution $\hat z$ of the ODE yields $a(t)=e^{-w(t-t_1)}(e^{w (t_1-t_0)}z_0 - z_1)$ (I think I got all the signs correct here.)

However, I do not see how to mechanistically form the derivative $\frac{d L}{dz(t)}$ directly and arrive at the same result.

If we interpret $\frac{dL}{dz(t)}$ as $\frac{d}{dz}(\frac{1}{2}|z-z_1|^2)\big|_{z=\hat z(t)}$ we get $\hat z(t)-z_1$, which is unequal to $a(t)$. So how is $\frac{dL}{dz(t)}$ formally defined, such that it ends up equal ? The paper never explains this.

Secondly, I would like to know if there is a more direct way to derive the adjoint equation. I am aware of the author's derivation in the appendix as well as the derivation via Pontryagin's maximum principle. The first one appears unmotived whilst the second one for some reason requires the consideration of an optimization problem, when all we want to do is compute a derivative.

答:
As a first result to resolve the conundrum. Let $\hat z(t)=z^*(t, t_0, z_0, \theta)$, with fixed initial condition $t_0, z_0$, then it appears that the (in)correct interpretation of the term is given by:

$$\begin{alignedat}{3}
\text{incorrect:}&&  \frac{dL}{dz(t)} &\overset{\text{def}}{=} \frac{d}{d\hat z(t)}L\big(\underbrace{z^{*}(t;t_0,z_0,\theta)}_{=\hat z(t)}\big) &&\overset{\text{def}}{=}\frac{\partial}{\partial z}L(z)\Big|_{z=\hat z(t)}
\\ \text{correct:}&& \frac{dL}{dz(t)} &\overset{\text{def}}{=} \frac{d}{d\hat z(t)}L\big(z^{*}(t_1;t,\hat z(t),\theta\big) &&\overset{\text{def}}{=} \frac{\partial}{\partial z}L\big(z^*(t_1; t,z, \theta)\big)\Big|_{z=\hat  z(t)}
\end{alignedat}$$

We can observe that in the example scenario

$$\begin{aligned}
\frac{\partial}{\partial z}L(z)\Big|_{z=\hat z(t)}
&= \frac{\partial}{\partial z}\Big[\frac{1}{2}\|z-z_1\|_2^2\Big]\Bigg|_{z=\hat z(t)}
\\&= (z-z_1)\Big|_{z=\hat z(t)}
\\&= \hat z(t)-z_1   = e^{w(t_-t_0)}z_0 - z_1{\huge\color{red}{↯}}
\\
\frac{\partial}{\partial z}L(z^*(t_1; t,z, \theta))\Big|_{z=\hat z(t)}
&=\frac{\partial}{\partial z}\Big[\frac{1}{2}\|e^{w(t_1-t)}z-z_1\|_2^2\Big]\Bigg|_{z=\hat z(t)}
\\&= e^{w(t_1-t)}(e^{w(t_1-t)}z-z_1)\Big|_{z=\hat z(t)}
\\&= e^{w(t_1-t)}(e^{w(t_1-t)}\hat z(t)-z_1)
\\&=e^{w(t_1-t)}(e^{w(t_1-t)}e^{w(t-t_0)}z_0-z_1)
\\&=e^{w(t_1-t)}(e^{w(t_1-t_0)}z_0-z_1)
\\&=e^{w(t_1-t)}(\hat z(t_1)-z_1) {\huge\color{green}{\checkmark}}
\end{aligned}$$


问:
knowing that $F(x)=\int^{x^3}_0\cos({t^2})dt$, how to disprove these statements?
First statement: $F(x)=F(-x)$ for every x.

I don't understand why this is false, thinking about the area under the graph of $\cos(t^2)$, it would be equal from both sides and symmetric according to $x=0$.

Second statement: $F(1)=5$.

I have no idea how to disprove this, I tried with taking derivative $F'(x)=\cos(x^6)3x^2$, but it's positive in $(0,1]$, so $F(0)=0$, and it can reach $F(1)=5$, knowing that $F(x)$ is increasing since the derivative is positive.

 Any help and explanation are really appreciated.
Thanks in advance.

答:
(1) An easy calculation shows your function is odd and not even:
$$F(-x)= \int_0^{-x^3} \cos(t^2)dt =-\int_0^{x^3}\cos(t^2) dt = -F(x)$$
where the middle equality follows because $t \mapsto \cos(t^2)$ is an even function.

(2) $F(1)=5$ is impossible because
$$|F(1)| \leq \int_0^{1} |\cos(t^2)|dt \le 1$$


问:
global sections of ideal sheaf quotient $\mathcal{I}/\mathcal{I^{2}}$
Let $X=\mathbb{P}_{k}^{2}$, $P$ a point in $X$ with its ideal sheaf $\mathcal{I}$. Then I would like to know what is $H^{0}(X, \mathcal{I}/\mathcal{I}^{2})$.

My first thought was to consider the exact sequence
$$0\to \mathcal{I}^{2}\to \mathcal{I}\to \mathcal{I}/\mathcal{I}^{2}\to 0.$$
Taking the long exact sequence,
$$0\to H^{0}(X,\mathcal{I}^{2}) \to H^{0}(X,\mathcal{I})\to H^{0}(X,\mathcal{I}/\mathcal{I}^{2}) \to H^{1}(X,\mathcal{I}^{2})\to \cdots.$$
The first two $H^{0}$ are zero since they're powers of the ideal sheaf, then makes the fourth arrow an injective map. Hence what I only know is that dimension is bounded up by $h^{1}(X,\mathcal{I}^{2})$. Then what should I do next?

答:
Tabes Bridges' comment pretty much answers the question but I want to elaborate on some of the facts. Let's first consider the affine situation. Take an affine scheme $\operatorname{Spec} A$ and pick your closed point $\operatorname{Spec} k \to \operatorname{Spec} A$ corresponding to a maximal ideal $\mathfrak{m}$. Associated to such a point is a quasicoherent sheaf $\mathcal{I}$ of ideals, which is just the quasicoherent sheaf corresponding to $\mathfrak{m}$.

Now the quotient $\mathcal{I} / \mathcal{I}^2$ of a quasicoherent sheaf of ideals is often called the conormal sheaf because, geometrically, it identifies with the dual of the normal bundle of the closed subscheme. As the closed subscheme is just a point, this is geometrically the same as the cotangent space at that point, and indeed $\mathcal{I} / \mathcal{I}^2$ is also the cotangent sheaf at the closed point $[\mathfrak{m}]$.

When we say 'at the closed point' what do we mean? I claim that $\mathcal{I} / \mathcal{I}^2$ is the skyscraper sheaf supported on the closed point $[\mathfrak{m}]$ with value $\mathfrak{m} / \mathfrak{m}^2$. I don't actually know what the best way of seeing this (and I'd be happy if someone can remark on this), though the following commutative algebra lemma suffices in many cases:

Lemma. Let $M$ be a finitely generated $A$-module, and let $I$ be an ideal of $A$. Then $\operatorname{Supp}(M / IM) = V(I) \cap \operatorname{Supp} M$.

Now, turning to the global situation, nothing really changes. All the behaviour of the cotangent bundle should be 'local', i.e. concentrated around a small open neighbourhood of whatever closed point you consider. Therefore, it remains the case when $X$ is not affine that $\mathcal{I} / \mathcal{I}^2$ will be a skyscraper sheaf. In particular, the global sections may be identified with $\mathfrak{m} / \mathfrak{m}^2$ after choosing an open neighbourhood $\operatorname{Spec} A$ of $P$ such that $P$ identifies with $[\mathfrak{m}]$.

Remark: The module $\mathfrak{m} / \mathfrak{m}^2$ in fact has the structure of a vector space over the field $k = A / \mathfrak{m}$. Let's consider the example $X = \mathbb{P}^2_k$ that you're interested in, and assume that $k$ is algebraically closed*. Locally, this variety looks like $\operatorname{Spec} k[x,y]$, and a closed point corresponds to a maximal ideal $(x - a,y - b)$. A direct computation shows that $\mathfrak{m} / \mathfrak{m}^2$ must then be $2$-dimensional as a $k$-vector space. This is in line with the geometric interpretation: $\mathbb{P}^2$, as say a manifold, is $2$-dimensional after all.

*: Without this assumption there are additional maximal ideals, and the observation needs to be corrected. Thanks to Georges Elencwajg for noting this.


Title: knowing that $F(x)=\int^{x^3}_0\cos({t^2})dt$, how to disprove these statements?

Body:
First statement: $F(x)=F(-x)$ for every x.

I don't understand why this is false, thinking about the area under the graph of $\cos(t^2)$, it would be equal from both sides and symmetric according to $x=0$.

Second statement: $F(1)=5$.

I have no idea how to disprove this, I tried with taking derivative $F'(x)=\cos(x^6)3x^2$, but it's positive in $(0,1]$, so $F(0)=0$, and it can reach $F(1)=5$, knowing that $F(x)$ is increasing since the derivative is positive.

 Any help and explanation are really appreciated.
Thanks in advance.

Body:
1.- For $x>0$, $-x^3<0$, so $$F(-x)=\int_0^{-x^3}\cos(t^2)dt=-\int_{-x^3}^0\cos(t^2)dt \overset{(*)}{=}-\int_0^{x^3}\cos(t^2)dt=-F(x),$$ where in (*) we have used that $\cos(t^2)$ is even, so $\int_0^a=\int_{-a}^0$

The same occur when $x<0$.

2.- $\cos t^2\le 1$, so $F(1)=\int_0^1\cos t^2\le\int_0^1 dt= 1$ and it cannot be 5.


问:
Integral involving Airy function
I met the following integral when I was reading a paper:
$$\int_0^\infty Ai(y)dy=\frac{1}{3},$$
where
$$Ai(y)=\frac{1}{\pi}\int_0^\infty \cos(\alpha y+\frac{\alpha^3}{3})d\alpha.$$

The paper adopted one asymptotic result of the Airy function,
$$\int_0^x Ai(y)dy \sim \frac{1}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\exp{\left(-\frac{2}{3}x^\frac{3}{2}\right)}, \quad \text{for large x}.$$
You may find this result on Eq.(10.4.82), Handbook of Mathematical Functions, Abramowitz and Stegun. A similar result is given in Eq.(10.4.83),
$$\int_0^x Ai(-y)dy \sim \frac{2}{3}-\frac{1}{2}\pi^{-\frac{1}{2}}x^{-\frac{3}{4}}\cos{\left(\frac{2}{3}x^\frac{3}{2}+\frac{\pi}{4}\right)}, \quad \text{for large x}.$$

However, I was wondering whether we can see the first result by calculating the integrals directly, and
$$\int_0^\infty Ai(-y)dy=\frac{2}{3}$$
as well.

Any advice or references are appreciated.

答:
You can obtain it using the integral representation
$$
\operatorname{Ai}(y) = \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3} - \frac{{y^3 }}{{3t^3 }}} \right)\,\mathrm{d}t} ,\quad |\arg y|<\tfrac{\pi}{6}
$$
(cf. $(9.5.6)$). Indeed, by the Fubini theorem and the Gauss multiplication theorem for the gamma function, it is found that your integral is
\begin{align*}
& \frac{{\sqrt 3 }}{{2\pi }}\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{y^3 }}{{3t^3 }}} \right)\,\mathrm{d}y} \,\mathrm{d}t} \\ &\mathop  = \limits^{x = y^3 /(3t^3 )} \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }} \int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t \int_0^{ + \infty } \mathrm{e}^{ - x} x^{1/3 - 1} \,\mathrm{d}x \,\mathrm{d}t} \\ & = \frac{1}{{2\pi }}\frac{1}{{3^{1/6} }}\Gamma\! \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\exp \left( { - \frac{{t^3 }}{3}} \right)t\,\mathrm{d}t}
\\ &\mathop  = \limits^{s = t^3 /3} \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma\! \left( {\frac{1}{3}} \right)\int_0^{ + \infty } {\mathrm{e}^{ - s} s^{2/3 - 1} \,\mathrm{d}s} \\ & = \frac{1}{{2\pi }}\frac{1}{{\sqrt 3 }}\Gamma\! \left( {\frac{1}{3}} \right)\Gamma\! \left( {\frac{2}{3}} \right) = \frac{1}{3}.
\end{align*}
For a more general result, see $(9.10.17)$. It can be derived the same way.

Title: Integral involving Airy function


问:
Smallest hexagonal lattice with no more than one point per unit Cartesian cell, as a function of rotation angle?
This answer to How to randomly but evenly distribute nodes on a plane introduces Bridson's Algorithm for Poisson-disc sampling (original paper: Fast Poisson Disk Sampling in Arbitrary Dimensions (also here) by Robert Bridson, University of British Columbia) Also see this and animation and explanation here.

I'm seeding the initial field with some regions with points on periodic lattices which will be both square and hexagonal, and of arbitrary rotation.

Bridson's method is characterized by a minimum distance $r$ that new points can be generated

For the algorithm to go fast, one chooses a square grid of side $r/\sqrt{2}$ which is small enough that there will always be either zero or one point in each cell. This way when a new point is introduced:

it must be introduced in a currently empty cell

only eight adjacent cells must be checked to evaluate the minimum distance to all existing points.

When I seed the space with preexisting array regions, I need to make sure that they never put more than one point in any cell.

Question: What are the constraints on lattice spacing $a$ and rotation angle $\theta$ for a hexagonal lattice with unit vectors:

$$\mathbf{a_1} =  a\cos \left(\theta \right) \ \mathbf{\hat{x}} + a\sin \left(\theta \right) \ \mathbf{\hat{y}}$$

$$\mathbf{a_2} =  a\cos \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{x}} + a\sin \left(\theta +\frac{\pi}{3} \right) \ \mathbf{\hat{y}}$$

such that there will never be more than one point in a unit square cartesian grid with any origin? (In other words, no matter how you translate the origin of the hexagonal lattice with respect to the cartesian grid.)

Rough example of application, for background only. Left: real space, right: reciprocal space (Fourier transform).

答:
The two furthest points in a square are a pair of opposite vertices, i.e. the end points of a diagonal. A square with diagonal $r$ has side $r/\sqrt{2}$.

This ensures that two points with distance of at least $r$ will be in two different gridcells. Note that if you have a cell with a point, you do need to check more than just the 8 immediately neighbouring cells to be sure that there is no other point within a distance of $r$. In fact you need to check 20 cells - a 5x5 square without its corner cells and without the centre cell itself.

The two furthest points in a regular hexagon are a pair of opposite vertices, i.e. the end points of a diagonal through the centre. A regular hexagon with diameter $r$ has side $r/2$.

This side length is enough to ensure that two points of distance $r$ or more will land in different cells, but as with the squares checking the immediate neighbouring cells is not sufficient to ensure that minimum distance. You'd need to check the 6 immediate neighbours, and also the next layer of 12 neighbours.


问:
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.

According to Munkres Topology:
Definition: A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\beta$.
A space that has a countable basis at each of its points is said to be first countable.

I can also proceed indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) dt$ where $f$ is measurable. But I don't know how to prove that there exists such a function. Any help please?

答:
Simply take $f$ to be the function that is constant $0$ (but any function will do).

Suppose, for a contradiction, that $f$ has a countable local base $\{U_n\mid n \in \Bbb N\}$ in $\{0,1\}^{\Bbb R}$ (in the product topology).

For every $n$ we can find a basic product open subset of the form $B_n:=\prod_{r \in \Bbb R} O_r$, where each $O_r$ is open in $\{0,1\}$, and such that there is a finite subset $F_n \subseteq \Bbb R$ such that for all $r \notin F_n$ we know that $O_r=\{0,1\}$ and such that $$f \in B_n=\prod_{r \in \Bbb R} O_r \subseteq U_n\tag{1}$$

We thus know also that for $r \in F_n$ we have $O_r =\{0\}$ (the only non-trivial (open) subset that contains $0$ in $\{0,1\}$), although we won't use it. We could also have written the basic open subset as $\bigcap_{n \in F_r} \pi_r^{-1}[O_r]$, if that's more familiar to you.

So for each $n$ we have such a basic open subset that depends essentially only on a finite subset $F_n$ of $\Bbb R$ that sits inside $U_n$. Now, $\bigcup_{n \in \Bbb N} F_n$ is at most countable and so I can pick some real number $r_0 \notin \bigcup_{n \in \Bbb N} F_n$. Then define $O= \pi_{r_0}^{-1}[\{0\}] = \{g \in \{0,1\}^{\Bbb R}\mid g(r_0)=0\}$, which is an open subset of the product (it's basic open even) that contains $f$, the constantly $0$ function.

Because the $\{U_n\mid n \in \Bbb N\}$ form a local base at $f$ by assumption, we have that form some $m \in \Bbb N$:

$$U_m \subseteq O\tag{2}$$

But now define $g: \Bbb R \to \{0,1\}$ by $g(r_0)=1$ and $g(r)=0$ for all $r \neq r_0$.

Then in particular $g(r)$ is $0$ for all $r \in F_m$ (as $r_0 \notin F_m$ by how it was chosen!) and so $$g \in B_m$$ and so $$g \in U_m \subseteq O$$ by $(1)$ and $(2)$. But this contradicts that $g \notin O$ (as $g(r_0)= 1 \neq 0$)!  This contradiction shows that our initial assumption that $f$ has a countable local base could not have been not correct. QED.

Note finally that we can do this proof at any point of $\{0,1\}^{\Bbb R}$, so nowhere there is a local countable base and we can even further generalise this (if you know simple cardinal arithmetic) that there cannot even be a local base of size $< |\Bbb R|$, which is a stronger fact in general.


问:
If $\alpha$ is an acute angle, show that $\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$
If $\alpha$ is an acute angle, show that $\displaystyle  \int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} = \frac{\alpha}{2\sin{\alpha}}.$

My attempt:

Write $x^2+2x\cos{\alpha}+1 = (x+\cos{\alpha})^2+1-\cos^2{\alpha} = (x+\cos{\alpha})^2+\sin^2{\alpha}$, we have:

$\displaystyle  \begin{aligned}\int_0^1 \frac{dx}{x^2+2x\cos{\alpha}+1} & = \int_0^1 \frac{dx}{(x+\cos{\alpha})^2+\sin^2{\alpha}} = \bigg[\frac{1}{\sin{\alpha}}\tan^{-1}\left(\frac{x+\cos{\alpha}}{\sin{\alpha}}\right)\bigg]_{x=0}^1 \\ & = \frac{1}{\sin{\alpha}}\bigg[\color{blue}{\tan^{-1}\left(\frac{1+\cos{\alpha}}{\sin{\alpha}}\right)-\tan^{-1}\left(\frac{\cos{\alpha}}{\sin{\alpha}}\right)}\bigg] \end{aligned}$

I'm not sure, however, how the blue bit reduces to $\frac{1}{2}\alpha$. Any hints/suggestions? Thanks.

答:
Or you can use this: $$\tan^{-1}{x} - \tan^{-1}{y} = \tan^{-1}\biggl(\frac{x-y}{1+xy}\biggr)$$

I am doing just the calculation part. We have
\begin{align*} \frac{1}{1+ \frac{\cos\alpha+\cos^{2}\alpha}{\sin^{2}\alpha}}\times \frac{1 + \cos\alpha}{\sin\alpha} - \frac{\cos\alpha}{\sin\alpha} &= \frac{1}{\sin\alpha} \times \frac{\sin^{2}\alpha}{\sin^{2}\alpha + \cos^{2}\alpha + \cos\alpha} \\ &= \frac{\sin\alpha}{1 + \cos \alpha} \\ &=\frac{ \displaystyle 2 \sin\frac{\alpha}{2}\cdot \cos\frac{\alpha}{2}}{2 \cos^{2}\frac{\alpha}{2}} \\ &=\tan\frac{\alpha}{2}
\end{align*}

Now take the inverse and then get the answer.


问:
Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodule.
I saw the following result somewhere with its proof:

Result: Let $M$ be an $R$-module with a minimal generating set $S$, where $|S|\geq 2$. Then $M$ has at least two maximal submodules.

Proof: Let $x_1, x_2$ be two distinct element in $S$. Since $S$ is a minimal generating of $M$ we infer that $x_1R\nsubseteq x_2R$ and $x_2R\nsubseteq x_1R$. Hence by Zorn's Lemma there are two distinct maximal submodules, $N, P$ say, with $x_1\in N, x_2\notin N$ and $x_2\in P, x_1\notin P$.

I have doubt in the last line of the proof. My doubt: why $N$ and $P$ are maximal submodules of $M$? They are just maximal member of a class of submodules. Please explain it.

答:
The result is true even if the minimal generating set $S$ of $M$ is infinite.

Suppose $x\in S$. Then the submodule of $M$ generated by $S\setminus\{x\}$ does not contain $x$ (as otherwise $S$ is not a minimal generating set). The union of a chain of submodules that do not contain $x$ does not contain $x$. So by Zorn's Lemma there is a submodule $N$ that contains $S\setminus\{x\}$, maximal subject to not containing $x$. Any submodule properly containing $N$ must contain $S$ and so must be equal to $M$. So $N$ is a maximal submodule of $M$.

So if $x_1$ and $x_2$ are distinct elements of $S$ then there is a maximal submodule containing $S\setminus\{x_1\}$ but not $x_1$, and another maximal submodule containing $S\setminus\{x_2\}$ but not $x_2$.


问:
Finding bounds on $\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}$
I am supposed to approximate the value of $\lim_{n\to \infty}\sum_{k=0}^{n}1/\sqrt{n^2+k}$. If it were $k^2$, then it could be rewritten as an integral, which would come out to be $\ln(\sqrt{2}+1)$.

$$\lim_{n\to \infty}\sum_{k=0}^{n}\frac{1}{\sqrt{n^2+k}}\lt \lim_{n\to \infty}\frac{n+1}{n}=1$$

So the limiting sum should lie in the interval $(0,1)$. Is this correct? Thanks.

答:
Hint
$$\dfrac{1}{\sqrt{n^2 + n}} \le \dfrac{1}{\sqrt{n^2 + k}}\le \dfrac{1}{\sqrt{n^2}}$$
for $0 \le k \le n$. Sum both sides from $0$ to $n$ and see that the sums are equal, in the limit.

The limit will turn out to be $1$. Your assertion that the limiting sum is in $(0, 1)$ is incorrect. Note that even if you have $a_n < b_n$ for convergent sequences, all you can conclude is that $\lim a_n \le \lim b_n$. That is, the strict inequality cannot be passed on to the limit.


问:
Vector field Axioms
Let V be the set of all functions

f: $\mathbb{R}$ $\to$ $\mathbb{R}$

such that the n-th derivative "$f^n$" of $f$ exists for all integers $n$.

We define a non standard addition ⊕ on V by setting $f$ ⊕ $g$ = $f'$ + $g'$ where $f'$ and $g'$ are the derivatives of $f$ and $g$ respectively.

V together with non-standard additon, scalar multiplication and $0$ is not a vector space.

Which vector space axioms does V not satisfy?

答:
Assuming the multiplication with scalar is defined in the obvious way. We would need the identity $(\lambda+\mu)f = \lambda f \oplus\mu f$, but in general
$$\lambda f \oplus \mu f = \lambda f' + \mu f' \neq  \lambda f + \mu f. $$

Title: Vector field Axioms


问:
Euler–Lagrange equation: Three functions of single variable with higher derivatives
Let $f_{i} \colon [0,L] \to \mathbb{R}$ be three times differentiable. I need to consider a functional $S$ of the following form:
\begin{equation*}
S(f_{1},f_{2},f_{3}) = \int_{0}^{L}L(t, f_{1},f_{2},f_{3},f_{1}',f_{2}',f_{3}',f_{1}'',f_{2}'',f_{3}'',f_{1}''',f_{2}''',f_{3}''')\,dt,
\end{equation*}
where an apostrophe denotes differentiation with respect to the variable $t$.

I would like to write down the corresponding Euler–Lagrange equation(s). I tried to derive the result from the more general case treated on Wikipedia, but I got rather confused by the index notation. Can anybody help?

答:
Applying single function of single variable with higher derivatives case from the page to each function $f_i$ separately and you get
$$
\frac{\partial L}{\partial f_i}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial f_i'}+\frac{\mathrm{d}^2}{\mathrm{d}t^2}\frac{\partial L}{\partial f_i''}-\frac{\mathrm{d}^3}{\mathrm{d}t^3}\frac{\partial L}{\partial f_i'''}=0.
$$


问:
Would the proof change if connected subspace became connected subset?
"Let A be  a  connected  subspace  of X.  If A⊂B⊂cl(A),  then B is also connected"

I did this proof a few times before, but I realized the first time I did it, I misread subspace as subset but the answer was the same. I was wondering if the wordings did change to "Let A be a connected subset" instead of subspace, are there any changes to these proofs? Because if a subset is connected, its induced subspace topology is also connected. Now I am curious about problems where if we changed the wording from subspace to subset, the proof has to change a lot. I can't think of any where that is the case.

答:
If we consider a subset  $A$ of a topological space $X$, we always consider it to have the subspace topology it inherits from $X$ if we talk about its properties (like compact, connected etc.). This is often implicit and not always mentioned.

There is no difference between a connected subset, or a connected subspace etc. But it's good to be aware of the fact that it's common practice.


问:
$ \mathfrak{m}_{a}=\left\langle X-a^{2}, Y-a^{3}\right\rangle $ is maximal ideal
I'm attempting the following exercise.

Let $K$ be a field, and let $R$ be the ring $K[X, Y] /\left\langle X^{3}-Y^{2}\right\rangle .$ For any element $a$ of $K,$ show that the ideal
$$
\mathfrak{m}_{a}=\left\langle X-a^{2}, Y-a^{3}\right\rangle
$$
of $R$ is a maximal ideal, with $R / \mathfrak{m}_{a}$ isomorphic to $K$. Moreover, show that if $I$ is the ideal of $R$ generated by $X$ and $Y,$ then $I / m_{a} I$ is a two-dimensional $K$ -vector space for $a=0$.

Here is my proof attempt:

Proof: Let $\phi: R \to K$ such that $\phi$ sends $X,Y,k$ to $a^2,a^3,k$ for $k \in K$. This is a ring homomorphism, and also a surjection. I claim that $\ker(\phi) = \left\langle X-a^{2}, Y-a^{3}\right\rangle$; an element of $\left\langle X-a^{2}, Y-a^{3}\right\rangle$ looks like $P_1 (X-a^{2}) + P_2(Y-a^{3})$ where $P_1, P_2 \in R$, and it is clear that $\ker(\phi)\supset \left\langle X-a^{2}, Y-a^{3}\right\rangle$. Now, let $P(X,Y) \in R$ such that $\phi(P) = P(a^2,a^3) = 0$. Now, note that $P(X,Y)$ can be re-written as a finite sum of variables of the form $(X-a^2),(Y-a^3),k' $ for a constant $k'$, i.e. $P(X,Y) = P(X-a^2+a^2,Y-a^3+a^3) = Q(X-a^2,Y-a^3)$. $ P(a^2,a^3) = 0$ implies that the constant factor of $Q$, $Q(0,0)$ is $0$. Thus, $P(X,Y) = Q(X-a^2,Y-a^3) \in \left\langle X-a^{2}, Y-a^{3}\right\rangle$. By the first isomorphism theorem, we have that $R / \mathfrak{m}_{a} \cong K$.

Here is the that bit I'm even more unsure about: Let $a=0$. Then, $I = \mathfrak{m}_0 = \left\langle X, Y\right\rangle$, and we simply calculate $I/II$. $II = \left\langle X, Y\right\rangle\left\langle X, Y\right\rangle$ is the ideal generated by the finite sum $\sum_{i=1}^n P_iQ_i$ where $P_i,Q_i \in \left\langle X, Y\right\rangle$ are polynomials with degree greater than $0$, with no constant term, in $R$. So every monomial of any element of $II$ will have degree $2$; in fact it is easy to see that any monomial with degree greater than $1$ is an element of $II$. Thus $I/II$ is composed of monomials of degree $1$ in $R$, of which $X,Y$ is a $K$-basis. So $I/II$ is two-dimensional.

Is this a valid proof?

答:
Honestly, the solution is a bit hard to read. Paragraphs would certainly help. As for the bit you are unsure about, the ideas are right but the conclusions are not obvious to me, I'd be more explicit (and what follows is my attempt at doing so).

As you say, the map $e \colon k[x,y] \to k$ mapping $x$ to $a^2$ and $y$ to $a^3$ is surjective. I claim that $\ker e = \langle x-a^2,y-a^3\rangle$. Indeed, take $g$ such that $e(g) = 0$. Since $y-a^3$ is a degree $1$ monic polynomial in $k[x][y] = k[x,y]$, we can write
$$g(x,y) = (y-a^3)q(x,y) + r(x)$$
and now dividing $r$ by $x-a^2$ in $k[x]$ gives
$$g = (y-a^3)q+(x-a^2)q'+r''.$$
Applying $e$ shows that $r'' = 0$, hence $g \in
\langle x-a^2,y-a^3\rangle$, the other inclusion is a direct computation.

Thus $k[x,y]/\langle x-a^2,y-a^2\rangle \simeq k$. Since $\mathfrak m_a$ is the projection of the ideal we're dividing by in $R$, by the third isomorphism theorem
we see that

$$R/\mathfrak m_a \simeq k[x,y]/\langle x-a^2,y-a^2\rangle \simeq k,$$

as desired.

The geometric perspective is that $R = k[x,y]/(x^3-y^2)$ is the coordinate ring of the variety in $\Bbb A^2 = k^2$ defined by the equation $x^3 = y^2$. This is a curve, as the image shows:

Points on a variety are associated to certain maximal ideals of its coordinate ring. In fact, when $k$ is algeraically closed there is a bijective correspondence: this is Hilbert's Nullstelensatz. Since $(a^2,a^3)$ satisfies $(a^2)^3 = (a^3)^2$, it is a point lying on the curve, and its associated maximal ideal is $\mathfrak m_a$.

Now, you may notice that the curve is spikey at $(0,0)$. Formally, the curve at that point is not smooth. I will not get into the details, but one way to measure the smoothness of a curve at a point is by calculating the dimension of its tangent space, and checking whether it coincides with the dimension of the variety.

This may be very familiar to us in the differential setting: if $M$ is a smooth $n$-dimensional manifold, at each point the tangent space is an $n$-dimensional vector space.

Our varierty is a curve, this means that its dimension is $1$ (I won't formalize this). Hence, to see that it is not smooth at the point $(0,0)$, we should see that the tangent space is not $1$-dimensional. The tangent space at a point $p$ is defined as the dual of
$$
\mathfrak m_p/\mathfrak m_p^2
$$
where $\mathfrak m_p$ is the maximal ideal associated to this point (I don't want to get into this as it is not directly related to your question, but this seemingly contrived definition has geometric meaning). My point is that in this case the tangent space is the dual of
$$
\mathfrak m_0/\mathfrak m_0, \qquad \mathfrak m_0 = \langle x,y\rangle
$$
and what you are being asked to do is to prove that it has dimension $2$. This is an algebraic way of detecting the spikeyness of the drawing at $(0,0)$.

As for the actual computations, indeed: an element of $\langle x,y\rangle^2$ is a sum of products of the form $QP$ with $Q = xq+yq', P = xp+yp'$. Hence $QP \in \langle X^2, Y^2, XY\rangle$.  Since $x^2,y^2,xy \in \mathfrak m_0^2$, we have the other inclusion and $\mathfrak m_0^2 = \langle x^2,y^2,xy\rangle$.

As you point out, this tells us that every element of $\mathfrak m_0/\mathfrak m_0^2$ is of the form $[a+bx+cy]$ for some $a,b,c \in k$. Hence $\{1,x,y\}$ generates the quotient as a $k$-vector space. Linear independence stems from the fact that if $[a+bx+cy] = 0$, then $a+bx+cy \in \mathfrak m_0^2 = \langle x^2,y^2,xy\rangle$ and there exist $p,q,r$ for which

$$
a+bx+cy = x^2 p + y^2 q + xy r.
$$

By degree considerations, this forces us to have $p = q = r = a = b = c = 0$.


问:
Prove or disprove: if $f:[0, \infty] \to [0, \infty]$ is continuous and $f(x)\le \sqrt {x} +10$ in $[0,\infty]$...

Prove or disprove: if $f:[0, \infty] \to [0, \infty]$ is continuous and $f(x)\le \sqrt {x} +10$ in $[0,\infty]$
then there exists $c\ge 0$ such that $f(c)=c$.

I'm trying to prove this using the intermediate value theorem since $f$ is continuous, but It's not working, I'm trying to find $f(x_1)\le c$, and $f(x_2)\ge c$, so we get that there exists $c$ between them such that $f(c)=c$.

But I'm having trouble with the upper bound $f(x_2)\ge c$, I'm starting to think this statement is wrong, but I know that I might just be messing up and not getting the idea.

I would appreciate any help, thanks in advance.

答:
Hint

One can try to solve the equation

$$\sqrt x+10=x$$

on $(0,\infty )$.


问:
Morphism of covering
I have a question about some hypothesis in a classical theorem about coverings.

The theorem is the following:

Let $B$ be a connected base space, and $X$ and $Y$ be coverings of $B$ and $g, f: X\to Y$ two $B$-morphisms. If there exists $b\in B$ such that $f_{|X_b} = g_{|X_b}$ then $f=g$.

This is easy to prove is $X$ is connected. If you don't assume $X$ connected, then it is also easy to prove if you assume $B$ locally connected, as in that case every connected component of $X$ is open and closed, and therefore maps surjectively onto $B$ and thus, every connected component of $X$ contains a point of the fiber $X_b$.

This is of course a standard assumption.

But the reference I'm looking at (Douady, Algèbres et théorie galoisiennes) does not make any assumption (other than $B$ connected) and simply states that "every connected component of $X$ contains a point in the fiber $X_b$".

And I fail to see how to prove that whithout the local connectedness assumption, am I missing something?

答:
This is only a partial answer: It is true if $B$ is path connected.

Let $P$ be a path component of $X$. Note that in general the components of $X$ do not coincide with its path components; but if we prove that $P \cap X_b \ne \emptyset$ for all $b$, then also $C \cap X_b \ne \emptyset$ for all components $C$ and all $b$. Pick $x_0 \in P$ and let $b_0 = p(x_0)$, where $p : X \to B$ denotes our covering projection. Then $x_0 \in C \cap X_{b_0}$. Now let $b  \in B$. Choose a path $u$ in $B$ such that $u(0) = b_0$ and $u(1) = b$. Lift $u$ to a path $\tilde u$ in $X$ such that $\tilde u(0) = x_0$. Since $A  =\tilde u([0,1])$ is path connected, we get $A \subset P$. But $\tilde u(1) \in A \cap X_b \subset P \cap X_b$.

Title: Morphism of covering


答:
Define $g(x) = f(x) - x$ for $x \in [0, \infty)$. We wish to show that this has a root. Suppose not.

Then, $g$ can not change sign. Noting that $g(x) = f(x) \ge 0$, we conclude that $g > 0$ on $[0, \infty)$.

Note that the equation $x - \sqrt{x} - 10 = 0$ has a real root, call it $c$. Then, $$0 < g(x) = f(c) - c \le \sqrt{c} + 10 - c = 0.$$
A contradiction.

Edit: The above proof actually shows that there is a root in $[0, c \approx 13.7]$.

Title: Prove or disprove: if $f:[0, \infty] \to [0, \infty]$ is continuous and $f(x)\le \sqrt {x} +10$ in $[0,\infty]$...

Body:
Prove or disprove: if $f:[0, \infty] \to [0, \infty]$ is continuous and $f(x)\le \sqrt {x} +10$ in $[0,\infty]$
then there exists $c\ge 0$ such that $f(c)=c$.

I'm trying to prove this using the intermediate value theorem since $f$ is continuous, but It's not working, I'm trying to find $f(x_1)\le c$, and $f(x_2)\ge c$, so we get that there exists $c$ between them such that $f(c)=c$.

But I'm having trouble with the upper bound $f(x_2)\ge c$, I'm starting to think this statement is wrong, but I know that I might just be messing up and not getting the idea.

I would appreciate any help, thanks in advance.


答: Hint: Let $g(x)=f(x)-x, g(0)=f(0)\geq 0$ and $\lim_{x\to+\infty}g(x)=-\infty$ IVT implies that there exists $x$ such that $g(x)=0$.


问:
Dividing over algebraic closure
Let $M\models T$ and $C\subset B\subset M$ be finite subsets. Let $a\in acl(C)$. How can we show that $tp(a/B)$ does not divide over $C$.

Here is my attempt: Suppose $tp(a/B)$  divides over $C$. Then there is a $C$-indiscernible sequence $(\bar{b}_i)_{i<\omega}$ such that $\bar{b}_0=\bar{b}$ where $\bar{b}$ is an enumeration of $B$, and $\{\phi(x,\bar{b}_i) : i<\omega\}$ is $k$-inconsistent for some $k<\omega$. Since $a\in acl(C)$, there is finitely many copies of $a$ over $C$ which is a contradiction !!!

答:
Yes, this is correct. Let's make that last sentence precise, for completeness sake.

Let $\{a_1, \ldots, a_n\}$ be the conjugates of $a$ over $C$, with $a_1 = a$. Since $\models \phi(a, \bar{b})$ we have that for every $i < \omega$ there is $1 \leq j_i \leq n$ such that $\models \phi(a_{j_i}, \bar{b}_i)$. So by the pigeonhole principle there are $1 \leq j \leq n$ and an infinite $I \subseteq \omega$ such that $\phi(a_j, \bar{b}_i)$ for all $i \in I$. This contradicts $k$-inconsistency.

Title: Dividing over algebraic closure


问:
Solving differential equation of the form $u_x=u^n$
This is my second week of learning PDE's.
I have two equations,

$u_t=u^2$ and $y_t=3y^{2/3}$.

I know how to solve $u_x=x$ but I can't figure out what to to with those.

If you could give me a helping hand on how to solve the following PDE's and give a method of solving similar to them.

I'm still a bit confused because it is a new subject, please be respectful.

答:
Since differentiation is performed with respect to one variable only (say, we can call it $x$, or $t$), the above-mentionned "PDEs" are actually ordinary differential equations of the form $$
y' = a y^b \, .
$$
These are Bernoulli differential equations which resolution can be tackled by change of variable. Resolution is straightforward if $b =1$, with solutions of the form $y(x) = ke^{ax}$ where $k$ is an integration constant. Otherwise, setting $u = y^{1-b}$ gives
$$
u' = (1-b) y' y^{-b} = a(1-b) \, ,
$$
with solutions of the form $u = a(1-b)x+c$. Finally, by setting $c = k^{1-b}$, solutions read
$$
y(x) = \left\lbrace
\begin{aligned}
&ke^{ax} & & \quad\text{if}\quad b=1\\
&\left[a (1-b) x + k^{1-b}\right]^\frac1{1-b} & &\quad \text{otherwise}
\end{aligned}\right.
$$
where $k$ is an arbitrary constant.


问:
What is the intuition behind a low-rank covariance matrix?

Let $X, Y$ be random vectors.  Let $K = \text{Cov}(X,Y) = E[XY^T] - E[X]E[Y]^T$ be the covariance matrix of $X$ and $Y$.  Assume $K$ is low rank.

I'm trying to come up with simple intuitive examples about what having a low rank covariance matrix means but I'm having trouble.  I understand that a low rank matrix means most of the column vectors are linearly dependent on other column vectors, and I understand that the covariance matrix shows the variance relationships between each random variable.  But from here I'm having trouble coming up with an intuitive explanation or example of where this would be useful.

答:
My intuition behind low-rank covariance is that $X$ and $Y$ live in a lower-dimensional manifold.

You can visualize this in 2D by starting with a full-rank covariance and progressively bringing one of the eigenvalues to 0. You will observe that samples will all be contained in a 1D line.


问:
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.</p>

答:
I will prove the result by starting with the right hand side of the identity:

$$\begin{align}\sin^2A-\sin^2B&=(\sin A+\sin B )(\sin A-\sin B)\\
&=(2\sin\frac{A+B}{2}\cos\frac{A-B}{2})(2\sin\frac{A-B}{2}\cos\frac{A+B}{2})\\
&=(2\sin\frac{A+B}{2}\cos\frac{A+B}{2})(2\sin\frac{A-B}{2}\cos\frac{A-B}{2})\\
&=\sin(A+B)\sin(A-B)
\end{align}$$
as required,  on using the sum to product formulae in the second line of working and the double angle formulae in  the final line. I hope you found that interesting :)


问:
Deducing the equation using the trigonometric addition theorem
I've been in the trouble to deduce the right term from the left term.

$a\cos(\theta)+b\sin(\theta)=\sqrt{a^2+b^2}*\sin(\theta+\phi)$

$\tan(\phi)=\frac{a}{b}$

The textbook says that the right term can be gained using the trigonometric addition theorem.However I have no idea what should I do for next.

Can anyone tell me some hint(s)? so that I can deduce it on my own.

答:
Hint:

$\tan\phi=\dfrac ab$

$$\dfrac{\sin\phi}a=\dfrac{\cos\phi}b=\pm\sqrt{\dfrac?{a^2+b^2}}$$

Now replace the values of $a,b$


答:
HINT

Begin by assuming that
$$a\cos\theta+b\sin\theta$$ can be written in the form
$$R\sin(x+\phi)$$
for some constants $R$ and $\phi$. Expand the second expression using the angle addition formula, form some equations, and you should have it.

I realize this is a bit vague, but you have asked for a hint. If you need any more help I will be happy to elaborate :)

Title: Deducing the equation using the trigonometric addition theorem

Body:
I've been in the trouble to deduce the right term from the left term.

$a\cos(\theta)+b\sin(\theta)=\sqrt{a^2+b^2}*\sin(\theta+\phi)$

$\tan(\phi)=\frac{a}{b}$

The textbook says that the right term can be gained using the trigonometric addition theorem.However I have no idea what should I do for next.

Can anyone tell me some hint(s)? so that I can deduce it on my own.


问:
If $A,B\in \mathbb{C}^{n,n}$ are similar, does it imply that $\det(A+3I)=\det(B+3I)$?
If $A,B\in \mathbb{C}^{n,n}$ are similar, does it imply that
$\det(A+3I)=\det(B+3I)$?
If so, why? I know that $\det(A)=\det(B)$.

Body: Suppose $PAP^{-1}=B$. Then $P(A+3I)P^{-1}=PAP^{-1}+3PIP^{-1}=B+3I$.


问:
How do you define definition symbol :=?
How is "$:=$" defined formally and why?  "$\iff$", "$=$", ...? </p>

答:
I do not know of any formal meaning of the definition symbol $:=$.  I have not seen it used in mathematical logic (although I'm sure someone somewhere has used it in some formal system.)

In ordinary mathematics, the definition symbol has an informal meaning, namely to signal that the current sentence serves as a definition of what follows the symbol.  Sometimes you can substitute $=$ for $:=$ and sometimes you can substitute $\iff$, so long as you add some words to make it clear that you are defining something.

For example, the sentence "$f(x) := x^2+1$" means "we define the function $f$ by $f(x) = x^2+1$" as opposed to "$f(x)$ turns out to equal $x^2+1$ even though the function $f$ was defined previously in some other way."  And the sentence "$P(x) := x > 1$" could be used to mean "we define the property $P$ by $P(x) \iff x>1$.

I think it is usually better to use words rather than the $:=$ symbol to make it clear that you are defining something.

Title: How do you define definition symbol :=?


答:
Hint:

Integrate both members of

$$u^{-n}\,du=dt.$$

Title: Solving differential equation of the form $u_x=u^n$

Body:
This is my second week of learning PDE's.
I have two equations,

$u_t=u^2$ and $y_t=3y^{2/3}$.

I know how to solve $u_x=x$ but I can't figure out what to to with those.

If you could give me a helping hand on how to solve the following PDE's and give a method of solving similar to them.

I'm still a bit confused because it is a new subject, please be respectful.


问:
First derivative of a quadratic function
I have the following question about the derivative.

Find out the quadratic function, which takes the value $41$ at $x=-2$ and the value $20$ at $x=5$ and is minimized at $x=2$ . Calculate the minimum value of this function.

If the function is minimized at $x=2$, then $f'(2) = 0$.

My question is: the derivative of the quadratic is a line tangent to the function or the slope of the function for a given $x$? Then the derivative equals $0$ because it is minimized at the vertex and at the vertex there is no slope?

This is not really about the question above but i just want to be sure if that is the case.

答:
Just start with $f(x)=a(x-2)^2+c$.  From $f(-2)=41$ and $f(5)= 20$ you'll easily derive $a=3$ and $c=-7$.

Concerning the slope: According to Archimedes the slope of the straight line between the points $(x_1,f(x_1)$ and $(x_2,f(x_2))$ of a parabola equals the slope of the tangent at the mean of $x_1$ and $x_2$, that is $f'\bigl((x_1+x_2)/2\bigr)$.

Hence the slope at the vertex is zero due to the symmetry of the parabola.


问:
Splitting Field of $t^{50} + 1$ over $Z_{101}$
I don't know how to deal with this exercise:

Find the splitting field of $t^{50} + 1$ over $Z_{101}$.

Firstly, I don't know how to check the irreducibility in this case, but neither how to go on in order to find such field.

Thank you for your help.

答:
By Fermat's little theorem every non-zero $x\in\Bbb Z_{101}$ is a zero of $g=t^{100}-1$. As there are $100$ different such elements and $g$ has degree $100$ it follows that $$g=\prod_{0\ne x\in\Bbb Z_{101}}(t-x)$$
As $(t^{50}-1)(t^{50}+1)=g$ it follows that also $t^{50}+1$ splits into linear factors over $\Bbb Z_{101}$. Hence the splitting field of $t^{50}+1$ is already $\Bbb Z_{101}$.


问:
if $2a_n+b_n$ and $a_n+3b_n$, both converge to a finite limit, then do $a_n,b_n$ also converge to a final limit?
What I did: if if $lim_{n\to \infty}2a_n+b_n=L_1$ and $lim_{n\to \infty}a_n+3b_n=L_2$.

Then $lim_{n\to \infty}(2a_n+b_n-2(a_n+3b_n))=L_1-2L_2=lim_{n\to\infty}(-2b_n)$.

But here I'm wondering, I know that If I know that $c_n$,$d_n$, converge then $lim_{n\to\infty}c_n*d_n$ converges.

Here I know that $lim(-2)=-2$, but I don't know that $lim_{n\to\infty}b_n$ converges, it's what I'm trying to prove, so my question is:
 Is it valid to say that if $lim_{n\to\infty}(-2b_n)=L_1-2L_2$, then $lim_{n\to\infty}b_n=\frac {L_1-2L_2}{-2}$?

Body:
Hint :

$$a_n = \frac{1}{5}\left[3(2a_n+b_n)-(a_n+3b_n)\right]$$
$$b_n=-\frac{1}{5}\left[(2a_n+b_n)-2(a_n+3b_n)\right]$$


答:
$$\tan \phi=\frac{a}{b}$$
Thus,
$$\cos \phi=\frac{b}{\sqrt{a^2+b^2}}$$
$$\sin \phi=\frac{a}{\sqrt{a^2+b^2}}$$
L.H.S
$$a\cos \theta +b\sin \theta=\sqrt{a^2+b^2}\frac{\Bigl(a\cos \theta +b \sin \theta\Bigr)}{\sqrt{a^2+b^2}}$$
$$=\sqrt{a^2+b^2}\Bigl(\frac{a}{\sqrt{a^2+b^2}}\cos \theta +\frac{b}{\sqrt{a^2+b^2}} \sin \theta\Bigr)$$

I've solved most of it, you just have to apply the identity now. Hope you've got it.


问:
Existence of two subspaces in vector space $V$
Let $V$ and $W$ be finite dimensional complex vector spaces and let $f: V\xrightarrow[]{}W$ and $g: W\xrightarrow[]{}V$ be linear maps. Suppose that $f(g(w))=w$ holds for any $w\in W$. Show that there exist subspaces $V_0, V_1$ of $V$ that satisfy all of the following three conditions.

$V= V_0 \oplus V_1$ that is $V$ is the direct sum of $V_0$ and $V_1$

For any $v\in V_0$, $g(f(v))=0$

For any $v\in V_1$, $g(f(v))=v$

I tried to think so much to prove the existence but could not get a good idea, here we need to prove the existence of these two subspaces $V_0$ and $V_1$. Does that mean we need to check the three conditions(not empty, closure under addition and scalar multiplication) only? How do we use the given conditions to prove that.

I know from 1 - $V_0 \cap V_1 = \{0\}$

How do we contribute all the things? I would appreciate if anyone could show me the right path

答:
Hint

$g$ is bijective $W\to g(W)$ with inverse function $f|_{g(W)}:g(W)\to W$.


问:
If $p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k$ are non constant polynomial.
$p(z,w)=a_0(z)+a_1(z)w+\dots +a_k(z)w^k$ where $a_i(z)$ are non constant polynomials in complex variables with  $k\ge 1$.

I need know if $$\{(z,w):p(z,w)=0\}$$

which of these are true or false:

$1$. bounded with empty interior

$2$. unbounded with empty interior

$3$. bounded with non empty interior

$4$. unbounded with non empty interior

I think only $1$ is true as the set consists of roots of a polynomial in two variables with some specific degree.

Am I right?

答:
$z^{k+1}+z^kw+...+z^2w^{k-1}+zw^k$ has unbounded zero set. So 1 and 3 are ruled out.

Also, if the interior were not empty, there would've existed $U\times V$ in $\mathbb C^2$ where $p(z,w)$ is identically zero. Fix $z$ in $U$ such that it is not a root of $a_k$. You get a non constant polynomial in one variable $w$ vanishing on $V$, a contradiction. So the interior must be empty.

I only eliminated boundedness possibility.
To prove it is unbounded explicitly, just put $z=N$ large enough that $a_k(z)$ is not zero. For each $n\ge N$, we have $p(n,w_n)=0$ for some $w_n$ because $p(n,w)$ is a polynomial in one variable and it is non-constant.


问:
Linear recursion with convergent coefficients
Suppose we have a linear recurrence relation

$$x_n = \alpha_n x_{n-1}+ \beta_n x_{n-2} + y_n,$$

where

$(\alpha_n) \to \alpha >0$,

$(\beta_n) \to \beta >0$,

$(y_n) \to y > 0$, and

the limiting auxiliary equation, $m^2 - \alpha m - \beta$, has roots in $(-1,1)$

(all limits are as $n \to \infty$).

My question:

Does this imply that the sequence $(x_n)_{n=1}^\infty$ converges?

This comes from a convergent example. I am simply wondering if the above information suffices to prove it, or if one needs control on the rates of convergence, say.

答:
The answer to your question is YES ; in fact, I show below that $(x_n)$ always converges to $l=\frac{y}{1-(\alpha+\beta)}$ (notice that $1-(\alpha+\beta)\neq 0$, otherwise $1$ would be a root of $m^2-\alpha m -\beta$).

First, let us see what happens when we adapt the method in the answer to the linked question.

The scalar recursion of order 2 is equivalent to the following recursion
of order 1 for vectors via $v_n=(x_n,x_{n+1})^T$.
\begin{equation}\tag{1}\label{one}
v_{n+1}=A_nv_n+C_n,\ \ A_n=\left(\begin{matrix}0&1\\ \beta_{n+2}& \alpha_{n+2}\end{matrix}\right)
\ C_n=\left(\begin{matrix}0\\y_{n+2}\end{matrix}\right)
\end{equation}
If $\lambda,\mu$ are the roots of $m^2-\alpha m-\beta$ which satisfy $|\lambda|,|\mu|<1$ by assumption  then there is a constant matrix $T$ such that
$$\newcommand{\eps}{\varepsilon}T^{-1}A T= \left(\begin{matrix}\lambda&\eps\\0&\mu\end{matrix}\right)=:D
\mbox{ for }A=\lim A_n=\left(\begin{matrix}0&1\\ \beta & \alpha \end{matrix}\right)$$
where $\eps=0$ unless $\lambda=\mu$  and the eigenvalue $\lambda$ of $A$ has algebraic
multiplicity 1; in this case $\eps\neq0$ and it can be chosen such that
$|\lambda|+|\eps|<1$.
Putting $v_n=Tw_n$ transforms (\ref{one}) to
$$w_{n+1}=(D+F_n)w_n+ C'_n,$$
where $F_n$ is a matrix of elements tending to 0 and $C'_n=T^{-1}C_n$ is a column matrix tending to a finite limit when $n\to \infty$.

Using the maximum norm for vectors and the corresponding matrix norm, we find
$$||w_{n+1}||\leq\big(\max(|\lambda|+|\eps|,|\mu|)+||F_n||\big)||w_n||+||C'_n||.$$

By assumption, the maximum is smaller than 1 and $||F_n||$ tends to 0. Also $(||C'_n||)$ is bounded. Therefore there
exist some $0<M<1$, some $C\gt 0$ and some integer $N$ such that the big parenthesis is $\leq M$, and $||C'_n|| \leq C$ for $n\geq N$. This means that
$$||w_{n+1}||\leq M||w_{n}||+C\mbox{ for }n\geq N$$
and hence $||w_n||\leq ||w_N|| M^{n-N} + C(1+M+M^2+\ldots+M^{n-N}) \leq ||w_N|| M^{n-N} + \frac{C}{1-M}$ for $n\geq N$ :

$$
||w_n||\leq ||w_N|| M^{n-N} + \frac{C}{1-M} \tag{2}\label{two}
$$
Therefore $w_n$ is bounded and therefore also $v_n$ and $x_n$.

Now, if we put $x'_n=x_n-l$, we have

$$
x'_{n}=\alpha_n x'_{n-1}+ \beta_n x'_{n-2} + y'_n
$$

where $y'_n=y_n-(1-\alpha_n-\beta_n)l$ tends to zero. Redoing all our preceding computations with $(x'_n)$ in place of $(x_n)$, we obtain a sequence $||w'_n||$ satisfying an analogue of $\eqref{2}$ where the constant $C$ can be arbitrarily small when $N$ is large enough. This shows that $(w'_n)$ converges to zero, and hence that $(x_n)$ converges to $l$.


答:
We have that  $fg= id_W$. Now show that $(gf)^2=gf.$ Hence $gf:V \to V$ is a projection.

Show that $V_0:=ker(gf)$ and $V_1:= im(gf)$ have the desired properties.

Title: Existence of two subspaces in vector space $V$

Body:
Let $V$ and $W$ be finite dimensional complex vector spaces and let $f: V\xrightarrow[]{}W$ and $g: W\xrightarrow[]{}V$ be linear maps. Suppose that $f(g(w))=w$ holds for any $w\in W$. Show that there exist subspaces $V_0, V_1$ of $V$ that satisfy all of the following three conditions.

$V= V_0 \oplus V_1$ that is $V$ is the direct sum of $V_0$ and $V_1$

For any $v\in V_0$, $g(f(v))=0$

For any $v\in V_1$, $g(f(v))=v$

I tried to think so much to prove the existence but could not get a good idea, here we need to prove the existence of these two subspaces $V_0$ and $V_1$. Does that mean we need to check the three conditions(not empty, closure under addition and scalar multiplication) only? How do we use the given conditions to prove that.

I know from 1 - $V_0 \cap V_1 = \{0\}$

How do we contribute all the things? I would appreciate if anyone could show me the right path


答:
Just define $V_0,V_1$ as in 2 and 3, i.e. \begin{align*}
V_0&=\ker g\circ f = \{v\in V\mid g(f(v))=0\}\\
V_1&=\ker (g\circ f-id_V)=\{v\in V\mid g(f(v))=v\}
\end{align*}
Now check that these subspaces satisfy $V=V_0\oplus V_1$. Hint: $$v=v-g(f(v))+g(f(v))$$


问:
knowing that $F(x)=\int^{x^3}_0\cos({t^2})dt$, how to disprove these statements?
First statement: $F(x)=F(-x)$ for every x.

I don't understand why this is false, thinking about the area under the graph of $\cos(t^2)$, it would be equal from both sides and symmetric according to $x=0$.

Second statement: $F(1)=5$.

I have no idea how to disprove this, I tried with taking derivative $F'(x)=\cos(x^6)3x^2$, but it's positive in $(0,1]$, so $F(0)=0$, and it can reach $F(1)=5$, knowing that $F(x)$ is increasing since the derivative is positive.

 Any help and explanation are really appreciated.
Thanks in advance.

答:
Partial answer:

0)Assume $F(x) =F(-x)$, i. e. $F$ is even.

$F$ is differentiable:

1)$ F'(x) =-F'(-x);$

We get

$F'(x) =3x^2\cos ((x^3)^2)$, and

$-F'(-x) = -3(-x)^2\cos ((-x)^3)^2).$

$F$ is not even.


问:
How to compute the indefinite integral $\int \frac u{u+1}\,\mathrm du$?
How do you compute $$ \int \frac{u}{u+1}\,\mathrm du$$</p>

答:
Hint:

Integrate by parts.

$$\int\dfrac{u}{u+1}du=u \int\dfrac{1}{u+1}du-\int  \log(u+1)du$$

By the series of comments with Did,

$$\int \log u du= \log u\int1 du-\int u \cdot \dfrac{1}{u}du= (u\log u-u)$$


问:
Prove that the algebraic expression is greater than zero
Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.

It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.

My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq  x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...

答:
The AM-GM says
$$
\frac12x^8+\frac12x^2\ge\left(x^8\cdot x^2\right)^{1/2}=|x|^5\tag1
$$
and
$$
\frac12x^2+\frac12\ge\left(x^2\cdot1\right)^{1/2}=|x|\tag2
$$
Therefore,
$$
\begin{align}
x^8-x^5+x^2-x+1
&=\frac12x^8+\overbrace{\left(\frac12x^8+\frac12x^2-x^5\right)}^{\text{(1)}\implies\ge0}+\overbrace{\left(\frac12x^2+\frac12-x\right)}^{\text{(2)}\implies\ge0}+\frac12\tag{3a}\\
&\ge\frac12x^8+\frac12\tag{3b}\\[6pt]
&\gt0\tag{3c}
\end{align}
$$


答:
Let $V_1=\ker f$ and let $V_0=g(W)$.

If $v\in V_1\cap V_0$, then $v=g(w)$ for some $w\in W$ and $f(v)=0$. But$$0=f(v)=f\bigl(g(w)\bigr)=w$$and therefore $v=g(w)=0$. On the other hand, if $v\in V$, then $v=v-g\bigl(f(v)\bigr)+g\bigl(f(v)\bigr)$. Clearly, $g\bigl(f(v)\bigr)\in V_1$. Furthrmore, $f\left(v-g\bigl(f(v)\bigr)\right)=f(v)-f(v)=0$, and therefore $V=V_0\bigoplus V_1$.

If $v\in V_0$, $f(v)=0$ and therefore $g\bigl(f(v)\bigr)=0$.

If $v\in V_1$, $v=g(w)$ for some $w\in W$ and therefore$$g\left(f\bigl(v)\right)=g\left(f\bigl(g(w)\bigr)\right)=g(w)=v.$$


问:
Min/max dice problem

Suppose Anne throws two fair dice and the faces show $4$ and $3$. It is your turn and Anne states that if both dice show a number greater than $4$ and $3$, you will win the game. What is the probability of you winning the game? (i.e. if your outcome is $4$ and $5$ you will win, as the biggest gets paired with the biggest and smallest with the smallest).

By listing the favorable possibilities, we obtain:

$$\begin{aligned}
&(4,5)&\qquad (4,6) &\qquad (5,6) &\qquad (5,5)&\\
&(5,4)&\qquad (6,4) &\qquad (6,5) &\qquad (6,6)&
\end{aligned}$$

so the proability must be $8/36$ or $2/9$. However, I am interested in knowing why the following way does not work.
Let $X_1\sim\text{Unif}\,\{1,6\}$ and $X_2\sim\text{Unif}\,\{1,6\}$. I need

$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\max\{X_1,X_2\}>4)P(\min\{X_1,X_2\}>3)\\
&=\left[1-P(\max\{X_1,X_2\}\leq4)\right]P(X_1>3,X_2>3)\\
&=\left[1-P(X_1\leq4,X_2\leq4)\right]P(X_1>3)P(X_2>3)\\
&=\left[1-P(X_1\leq4)P(X_2\leq4)\right]\left[1-P(X_1\leq3)\right]\left[1-P(X_2\leq3)\right]\\
&=\left(1-\frac{4}{6}\cdot\frac{4}{6}\right)\left(1-\frac36\right)\left(1-\frac36\right)\\
&=\left[1-\left(\frac{4}{6}\right)^2\right]\left(1-\frac36\right)^2=\frac5{36}.
\end{aligned}
$$

What is wrong with this approach?

Another approach I considered is,
$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\})\\
&=P(\{X_1>4,X_2>3\})+P(\{X_1>3\,X_2>4\})\\ &=P(X_1>4)P(X_2>3)+P(X_1>3)P(X_2>4)\\
&=\frac26\cdot\frac36+\frac36\cdot\frac26=\frac13.
\end{aligned}$$

Why are the above different? Is it not true that $\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3 \implies \{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\}$?

答:
The problem is the first line of your approach. $P(A,B)=P(A)P(B)$ is only true if the events $A$ and $B$ are independent. Here the events $\max\{X_1,X_2\}>4$ and $\min\{X_1,X_2\}>3$ are not independent, since if we know the maximum is small that reduces the range of values the minimum can take.

Similarly in the second approach you use $P(C\cup D)=P(C)+P(D)$. This only works if the events $C$ and $D$ are disjoint, which they are not in this case.

Title: Min/max dice problem


答:
Your second approach is right, but formula you have used only works for two disjoint events. In general, for two events A and B, we have
$$P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$$

Title: Min/max dice problem

Body:
Suppose Anne throws two fair dice and the faces show $4$ and $3$. It is your turn and Anne states that if both dice show a number greater than $4$ and $3$, you will win the game. What is the probability of you winning the game? (i.e. if your outcome is $4$ and $5$ you will win, as the biggest gets paired with the biggest and smallest with the smallest).

By listing the favorable possibilities, we obtain:

$$\begin{aligned}
&(4,5)&\qquad (4,6) &\qquad (5,6) &\qquad (5,5)&\\
&(5,4)&\qquad (6,4) &\qquad (6,5) &\qquad (6,6)&
\end{aligned}$$

so the proability must be $8/36$ or $2/9$. However, I am interested in knowing why the following way does not work.
Let $X_1\sim\text{Unif}\,\{1,6\}$ and $X_2\sim\text{Unif}\,\{1,6\}$. I need

$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\max\{X_1,X_2\}>4)P(\min\{X_1,X_2\}>3)\\
&=\left[1-P(\max\{X_1,X_2\}\leq4)\right]P(X_1>3,X_2>3)\\
&=\left[1-P(X_1\leq4,X_2\leq4)\right]P(X_1>3)P(X_2>3)\\
&=\left[1-P(X_1\leq4)P(X_2\leq4)\right]\left[1-P(X_1\leq3)\right]\left[1-P(X_2\leq3)\right]\\
&=\left(1-\frac{4}{6}\cdot\frac{4}{6}\right)\left(1-\frac36\right)\left(1-\frac36\right)\\
&=\left[1-\left(\frac{4}{6}\right)^2\right]\left(1-\frac36\right)^2=\frac5{36}.
\end{aligned}
$$

What is wrong with this approach?

Another approach I considered is,
$$\begin{aligned}
P(\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3)&=P(\{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\})\\
&=P(\{X_1>4,X_2>3\})+P(\{X_1>3\,X_2>4\})\\ &=P(X_1>4)P(X_2>3)+P(X_1>3)P(X_2>4)\\
&=\frac26\cdot\frac36+\frac36\cdot\frac26=\frac13.
\end{aligned}$$

Why are the above different? Is it not true that $\max\{X_1,X_2\}>4,\min\{X_1,X_2\}>3 \implies \{X_1>4,X_2>3\}\cup\{X_1>3\,X_2>4\}$?