问:
If f is Riemann Integrable, then the lower sums tend to the integral.
The title may be a bit confusing so here's what I'm trying to establish:

Let $f: [a,b] \to \mathbb{R}$ be Riemann integrable and let $\{P_k\}_{k=1}^{\infty}$ be a sequence of partitions with $\lim_{k \to \infty} |P_k| = 0$. Prove that $\lim_{k \to \infty} L(f,P_k) = \int_{a}^{b} f(x) \ dx$.

Proof Attempt:

Let $\{T_k\}_{k=1}^{\infty}$ be a sequence of associated evaluation sets. Then, by the sequential definition of Riemann Integrability:

$$\lim_{k \to \infty} R(f,P_k,T_k) = I$$

We will attempt a direct method of proof. So, consider:

$$|L(f,P_k)-I| = |L(f,P_k)-R(f,P_k,T_k)+R(f,P_k,T_k) - I| \leq |L(f,P_k)-R(f,P_k,T_k)| + |R(f,P_k,T_k)-I|$$

It is easy to find some kind of bound on $|R(f,P_k,T_k)-I|$. So, we only worry about $|L(f,P_k)-R(f,P_k,T_k)|$. For $k \in \mathbb{N}$:

$$P_k = \{x_0 ^{(k)} = a < x_1 ^{(k)} < \ldots,x_{n_k} ^{(k)} = b \}$$

$$T_k = \{t_1 ^{(k)},t_2 ^{(k)},\ldots, t_{n_k} ^{(k)} \}$$

Then, we can write out:

$$L(f,P_k) = \sum_{i=1}^{n_k} m_{i} ^{(k)} \Delta x_i ^{(k)}$$

$$R(f,P_k,T_k) = \sum_{i=1}^{n_k} f(t_i ^{(k)}) \Delta x_i ^{(k)}$$

$$R(f,P_k,T_k)-L(f,P_k) = \sum_{i=1}^{n_k} \left(f(t_i ^{(k)})-m_i ^{(k)} \right) \Delta x_i ^{(k)}$$

$$R(f,P_k,T_k)-L(f,P_k) \leq \sum_{i=1}^{n_k} \left(f(t_i ^{(k)})-m_i ^{(k)} \right) |P_k|$$

Since $f$ is Riemann Integrable, it is bounded so it has a supremum $M$ and an infimum $m$ on the interval $[a,b]$. Hence

$$|R(f,P_k,T_k)-L(f,P_k)| \leq |P_k| \cdot \sum_{i=1}^{n_k} 2M = 2Mn_k |P_k|$$

This is where I'm quite stuck now because I don't know how to use this to bound the quantity on the left-hand side. Can someone give me a hint on how I can continue?

答:
I will point out before starting that in some sense this is the easy direction to prove because the initial definition of Riemann integrability is so a priori strong even though it is ultimately equivalent to the Darboux definition.

Anyway, you used too much brute force in passing to the global bound on $f$ in the last step. Your last RHS is actually guaranteed to not go to zero because $n_k |P_k| \geq b-a$.

Instead, you need to argue that $|f(t_i^{(k)})-m_i^{(k)}|$ can be made small by suitable choice of $t_i^{(k)}$: specifically, you really would like it to be at most  $\frac{\epsilon_k}{\Delta x_i^{(k)}}$ where $\epsilon_k$ is your favorite sequence of positive numbers going to zero. If you had that, then you would be done. But you do have that, by the definition of infimum.


问:
how to solve $\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x)$?
how to solve $\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x) $ when $\lim\limits_{x \to +\infty} f(x) = +\infty$ and $\lim\limits_{x \to +\infty} g(x) = +\infty$?

for example what are the solutions of $\lim\limits_{x \to +\infty} x \over \lim\limits_{x \to +\infty} 2x $ or $\lim\limits_{x \to +\infty} x \over \lim\limits_{x \to +\infty} x^2 $?

I think we can't use the L'hopital rule because ${\lim\limits_{x \to +\infty} f(x) \over \lim\limits_{x \to +\infty} g(x)} \neq \lim\limits_{x \to +\infty} {f(x) \over  g(x)} $ . (not sure)

So please tell me how to solve these kinds of problems and which theorems we use must use.

Thanks

答:
This is tricky territory. It is true that the limit
$$
\lim_{x \to \infty} \frac{x}{2x}
$$
both exists and is equal to $1/2$. However, we can't split the limit of a quotient into a quotient of limits. This is because
$$
\frac{\lim_{x \to \infty}x}{\lim_{x \to \infty} 2x}=\frac{\infty}{\infty} \, ,
$$
which is undefined. $\infty/\infty$ is as meaningless as $\text{yellow}/\text{blue}$. In fact, some mathematicians prefer not to write
$$
\lim_{x \to \infty}2x = \infty \, ,
$$
instead saying that 'the limit does not exist'. What those mathematicians are trying to emphasise is that infinity is not a number. In fact, saying that
$$
\lim_{x \to \infty}2x = \infty
$$
is just a shorthand for saying that as $x$ gets larger, $2x$ also gets larger and larger. To be more precise, the value of $2x$ is unbounded. We are most certainly not saying that $2x$ is approaching the number infinity.

We can use L'Hopital's rule to evaluate
$$
\lim_{x \to \infty}\frac{x}{2x}
$$
but this is needlessly heavy handed. $x/(2x)=1/2$ for all $x$, and so the limit simplifies to
$$
\lim_{x \to \infty}\frac{1}{2} = \frac{1}{2} \, .
$$
Once again, we don't care about $x=\infty$, whatever that means.

ContentLicense: CC BY-SA 4.0


问:
Coercive bilinear form on Hilbert space
I need to show the two following results. If true, it must be a simple proof but I do not seem to be able to make it work. Thank you in advance.

Consider a continuous symmetric bilinear form $B$ on a real Hilbert space $H$. Let $D$ be a closed subspace of that Hilbert space over which the form is coercive, i.e, there exists $\alpha >0 \in \mathbb{R}$ such that $B(d,d)\geq \alpha \|d\|^2$ for all $d\in D$. Take the new closed subspace of $H$ generated by $D$ and a one dimensional subspace. Assume $B$ is definite positive on the new subspace. Then I need to show that $B$ is also coercive on the new subspace.

It seems like the logical thing to do is to use the Lax-Milgram theorem but I am not exactly sure how to do this.

答:
Here is the answer to my own question. Please feel free to comment. Any suggestion is welcome:

Let $\widetilde{H}$ be the closed subspace generated by $D$ and the one-dimensional subspace of $H$. Since $B$ is positive definite, there exists a one-to-one bounded symmetric operator $L$ on $\widetilde{H}$ such that $B(u,v)=\langle Lu,\,v\rangle$. Let $y\in \widetilde{H}$ such that $\langle Ld,\,y\rangle=0$ for all $d\in D$. Then, any element $u$ of $\widetilde{H}$ has a unique representation $u=ay+d$ for some $a\in \mathbb{R}$ and $d\in D$. Let $\widetilde{L}$ be the restriction of $L$ on $D$, i.e., if $v\in \widetilde{H}$ and $Lv=a'y+d'$ for some $a'\in \mathbb{R}$ and $d'\in D$, then $\widetilde{L}v\equiv d'$. The operator $\widetilde{L}$ is such that $B(d_1,\,d_2)=\langle \widetilde{L}d_1,\,d_2\rangle$ for all $d_1,d_2 \in D$. Since $B$ is coercive on $D$, $\widetilde{L}$ is an isomorphism on $D$ by the Lax-Milgram theorem. Thus the image of $D$ under $L$ is a closed subspace of codimension 1 in $\widetilde{H}$. Since $L$ is one-to-one, it follows that $L$ is an isomorphism on $\widetilde{H}$ and thus $B$ is coercive on $\widetilde{H}$.

ContentLicense: CC BY-SA 3.0

Title: Coercive bilinear form on Hilbert space


问:
find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem
I'm trying to prove with squeeze theorem that the limit of the following series equals 1:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the left side of the inequality I did:

$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} <  \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the right side, at first I did the following:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} <  \frac{n\sqrt[n]{n}}{n}$$

But then I realized it wasn't true and that the direction of this inequality is the opposite.

Do you have any idea which series with limit 1 is bigger from the original series?

Thanks!

答:
You'll want to know some things about how big $\sqrt[n]{n}$ is. The key facts to prove are:

For $n$ a positive real number, it's increasing when $n < e$ and decreasing when $n>e$. For integers, $3^{1/3} \approx 1.44$ is the largest value, with $2^{1/2} \approx 1.41$ taking second place.

As $n \to \infty$, $\sqrt[n]{n} \to 1$. A more precise estimate of $\sqrt[n]{n}$ as $n \to \infty$ is $1 + \frac{\log n}{n}$, but we won't need it.

So we are averaging a few large terms, and many many terms close to $1$. One good way to deal with a situation like that with the squeeze theorem is to separate into two parts:
$$
   \frac1n \sum_{k=1}^n \sqrt[k]{k} = \frac1n \sum_{k=1}^{\sqrt n}\sqrt[k]k + \frac1n \sum_{k=\sqrt{n}+1}^{n}\sqrt[k]k.
$$
What can we say about these two parts?

In the first sum, we have $\sqrt n$ terms, each of which is at most $3^{1/3}$. So the sum is at most $3^{1/3} \sqrt n$, and we're dividing by $n$. This sum goes to $0$.

In the second sum, we have nearly $n$ terms, each of which is less than $\sqrt[k]{k}$ for $k = \sqrt n$. So they add up to less than $n \sqrt[k]{k}$. When we divide by $n$, we get $\sqrt[k]{k}$ where $k=\sqrt n$, and this approaches $1$ as $n \to \infty$.

(The specific cutoff of $\sqrt n$ is very flexible: any function $1 \ll f(n) \ll n$ would do.)


问:
Finding the lowest value for $a$ at which $\lim_{n\to\infty}\frac{(2n)!}{(n!)^a}$ is convergent
So I've noticed that for larger values of $a$ the function below forms a bell curve shape whereas for lower values it diverges, I would like to find an approximation for the lowest value at which this is true:
$$f_a(x)=\frac{(2x)!}{(x!)^a}$$

I have thought about using stirling's approximation:
$$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n$$
which would give me:
$$\frac{(2x)!}{(x!)^a}=\frac{\sqrt{4\pi x}\left(\frac{2x}{e}\right)^x}{\left(2\pi x\right)^{a/2}\left(\frac xe\right)^{ax}}=\sqrt{2}\left(2\pi x\right)^{1-a/2}\left(\frac xe\right)^{x(1-a)}2^x$$
$$=2^{x+(3-a)/2}\pi^{1-a/2}x^{x(1-a)+1-a/2}e^{-x(1-a)}$$
if we break this down we are really looking for the convergence of:
$$x^{x(1-a)+1-a/2}e^{-x(1-a)}$$
Then my thoughts were that the $x^x$ term seems the most important term, so the answer would just be $a\ge1$ but I feel like this is not correct. Does anyone have any thoughts? Thanks

答:
When $a=2$ the quotient is $\binom{2n}{n}$ so it goes to infinity.  Let $a = 2+\alpha$ with $\alpha>0$

The $$\begin{align}
\frac{(2n)!}{(n!)^a}&=\frac1{(n!)^\alpha}\binom{2n}{n}\\
&\sim\frac1{(n!)^\alpha}\frac{(2n)^{2n}e^{-2n}\sqrt{4\pi n}}{n^{2n}e^{-2n}2\pi n}\\
&=\frac1{(n!)^\alpha}\frac{4^n}{\sqrt{\pi n}}\\
\end{align}$$
which goes to $0$ as $n\to\infty$  To see this, take the logarithm and recall that $\log n! \sim n\log n$.

So there is a largest $a$, namely $2$, such that the sequence does not converge, but no smallest $a$ such that it does.


问:
Multi-dimensional Inverse Fourier Transform of $\cos(a\left|q\right|)$
I am trying to compute $$\int d^Dq \; e^{iq\cdot\left(r-r'\right)}\cos\left(c_L\left|t-t'\right|\left|q\right|\right),$$ that is, the $D$-dimensional inverse Fourier transform of $\cos\left(a\left|q \right|\right),\;a>0$. I am interested in $D=1,2,3$ and am unskilled with distributions. I have computed the $D=1$ case to be
$$\pi\delta\left(\left|r-r'\right|-c_L\left|t-t'\right|\right)+\pi\delta\left(\left|r-r'\right|+c_L\left|t-t'\right|\right).$$

Edit: When transformed into spherical coordinates, the integrals become:

$D=1:$
$$2\int_0^{\infty}dq\;\cos\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$
$D=2:$
$$2\pi\int_0^{\infty}dq\;q\,J_0\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$
$D=3:$
$$4\pi\int_0^{\infty}dq\;q\,\sin\left(\left|r-r'\right|q\right)\cos\left(c_L\left|t-t'\right|q\right)$$

答:
Spherical coordinates are good candidates to evaluate these integrals. The $d$-dimensional measure is given by $d^d k = k^{d-1}d^{d-1}\Omega\,dk$, where $d^{d-1}\Omega$ is the surface measure on the sphere $S^{d-1}$ and $k=|k|$. For example, in 3 dimensions, the integral would be
$$\int_0^\infty k^2 dk  \int_0^{2\pi}d\phi \int_0^\pi \sin \theta \,d\theta \,e^{ik|r-r'|\cos\theta} \cos(c_L|t-t'|k),$$
where I have chosen the $z$-axis as parallel to the $\vec{r}-\vec{r}'$ vector so that $\vec{k} \cdot (\vec{r}-\vec{r}')= k|r-r'|\cos\theta$.

Edit:
For $d=3$, your integral is
$$-2\pi\frac{d}{d|r-r'|}\int_{-\infty}^\infty \cos\left(|r-r'|q\right)\cos\left(c_L|t-t'|q\right)\,dq.$$
This is formally the same integral as for the case $d=1$, with an additional derivative. In the final result, you will have $\delta'$, the distributional derivative of $\delta$, defined by $\langle-\delta', f\rangle = \langle\delta, f'\rangle$, where $\langle g, h\rangle$ denotes the inner product $\int_{-\infty}^\infty g(x)\,h(x)\,dx$, and $f\in C^\infty_c$ is a smooth function with compact support.


答:
This is not a full answer to the question, but many answers are implying that the function $n\mapsto n^{1/n}$ is strictly increasing. This is not the case. To see this:

Let $y=x^{1/x}$. Then $\ln y=\frac 1x \ln x$ so $\frac{y'}{y}=\frac{1}{x^2}(1-\ln x)$. Since $y>0$, this implies that $y$ is increasing on $(0,e)$ and decreasing on $(e,\infty)$.

Hence, do not use upper bound of $n^{1/n}$.

Title: find limit of $\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$ with squeeze theorem

Body:
I'm trying to prove with squeeze theorem that the limit of the following series equals 1:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the left side of the inequality I did:

$$\frac{1+\sqrt{1}+\sqrt[3]{1}+...+\sqrt[n]{1}}{n} <  \frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n}$$

For the right side, at first I did the following:

$$\frac{1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n}}{n} <  \frac{n\sqrt[n]{n}}{n}$$

But then I realized it wasn't true and that the direction of this inequality is the opposite.

Do you have any idea which series with limit 1 is bigger from the original series?

Thanks!


问:
*Weak Lusin* vs *Strong Lusin*
In a book that I’m reading, the author presents two versions of Lusin’s theorem:

Let $X$ be a topological space with a regular measure $\mu$, and $Y$ to be second countable (i.e., admits a countable family $(B_i)_i$ Of open sets such that for any open set $B \subset Y$, we can represent $B$ as an union of $B_i$’s).

Weak Lusin - Under the above assumptions, if $f: X \to Y$ is measurable, then, for every $\epsilon>0$ there exists a compact set $K\subset X$ s.t $\mu(K^c) <\epsilon$ and $f \mid_K
$ is continuous.

Strong Lusin - Under the same assumptions, if
$f:X\to \mathbb R^d$ measurable, then for every $\epsilon>0$ there exists a compact set $K\subset X$ and a continuous function $g:X\to \mathbb R^d$   s.t $\mu(K^c) <\epsilon$ and $f=g$ on $K$.

The thing is, these two seem very similar to me. I can’t quite comprehend how the existence of a function $g$ coinciding with a function $f$ on $K$, differs from having a continuous function $f$ only on $K$. Is the different mainly due to the fact that exists a function $g$ continuous?
Another different seems to be the change in the space $Y$. The strong version is restricted to $\mathbb R^d$.

Can anyone clarify the difference in this two versions? Perhaps present a counterexample.

答:
To go from your Weak Lusin to Strong Lusin one needs to be able to do the following: if $K\subset X$ is compact, and $f\colon K \to Y$ is continuous, then there is a continuous extension of $f$ to whole of $X$, i.e. there is a continuous $F:X \to Y$ such that $F(x)=f(x)$ whenever $x \in K$. If this is possible then after finding the map $g$ in the Weak Version, throw out the part of $g$ outside $K$. Use this extension to find a new continuous $g$ that agrees with the old $g$ along $K$.

Now is this extension always possible? If $Y$ is a Euclidean space this extension is always possible (Evans and Gariepy measure theory and fine propeties of functions has a proof), in general not.

As for an example of how $f$ being continuous on $K$ differs from having a continuous $F$ that restricted to $K$ equals $f$, consider $E=(-1,0)\cup(0,1) \subset \mathbb{R}$ and the function that assigns $-5$ to all negative $x$ and $+5$ to all positive $x$. Now, $f:E \to \mathbb{R}$ is continuous (!). However, there does not exist any continuous $F:\mathbb{R} \to \mathbb{R}$ so that its restriction to $E$ coincides with $f$ -- you just won't be able to assign anything to $F(0)$ in a continuous way. Notice that necessarily I had to choose noncompact $E$. But my example was to show why the two notions are not the same.

Title: *Weak Lusin* vs *Strong Lusin*


问:
If there exsits a constant M such that $ \forall x\in [a,b], \forall n\geq 1,|S'_n(x)|\leq M, $ then $\sum_{n=1}^\infty a_n(x)$ converges uniformly.

Let series of functions $\sum_{n=1}^\infty a_n(x)$ converges on $[a, b]$, its partial sum are $S_n(x)$. If there exsits a constant M such that
$$
\forall x\in [a,b], \forall n\geq 1,|S'_n(x)|\leq M,
$$
then $\sum_{n=1}^\infty a_n(x)$ converges uniformly.

I tried to prove the problem by Cauchy criterion.
$$
S_{n+p}(x)-S_n(x)=S_{n+p}'(\xi_1)(x-a)+S_{n+p}(a)-S_{n}'(\xi_2)(x-a)-S_{n}(a)
$$
But it seems we are unable to to control $|S'_{n+p}-S'_n|$, so I  amstuck here.

Is my way wrong, if so, please suggest a correct way. Appreciate any help!

答:
We have $S_n(x) \to S(x)$ pointwise and $|S_n(x) - S_n(y)| \leqslant |S_n'(\xi)||x-y| \leqslant M|x-y|$ for all $x,y \in [a,b]$ and all $n \in \mathbb{N}$.

For all $\epsilon > 0$ there exists $\delta > 0$ such that if $|x-y| < \delta$, then for all $n \in\mathbb{N}$ we have $|S_n(x) - S_n(y)| < \epsilon/3$. Taking the limit as $n \to \infty$ it follows that $|S(x) - S(y)| \leqslant \epsilon/3$ as well.

Since $[a,b]$ is compact there exist finitely many points $x_1, \ldots x_m$ such that $[a,b] \subset \cup_jI(x_j,\delta),$ where $I(x_j,\delta)$ is the open interval with center $x_j$ and radius $\delta$.

By pointwise convergence, there exists $N_j \in \mathbb{N}$ such that if $n \geqslant N_j$ then $|S_n(x_j) - S(x_j)|< \epsilon/3.$ Let $N = \max_{1 \leqslant j \leqslant m}N_j.$

Given any $x \in [a,b]$ there exists $j$ such that $x \in I(x_j,\delta)$. This implies $|x - x_j| < \delta$ and if $n \geqslant N$ we have

$$|S_n(x) - S(x)| \leqslant |S_n(x) - S_n(x_j)|+|S_n(x_j) - S(x_j)|+|S(x_j) - S(x)| \\\leqslant \frac{\epsilon}{3} +  \frac{\epsilon}{3}+  \frac{\epsilon}{3} = \epsilon,$$

proving uniform convergence.


问:
Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?
It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of
\begin{align}
TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\
(v,p)&\mapsto p
\end{align}
(I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.

答:
There are two major issues here.

First of all, the only way for the union to not be disjoint and make any sense at all is if the tangent spaces are assumed to live in a common universal set, which isn't generally the case.

Secondly, even if you embed the tangent spaces in a common universe, e.g. by embedding the manifold inside some Euclidean space, then you lose crucial information by taking a regular union, since the line between tangent vectors and points becomes blurred and points living in distinct tangent spaces may get identified. For example, consider the tangent bundle of the circle embedded in $\Bbb{R}^2$, as in the following picture:

By taking a regular union, you get the subset of $\Bbb{R}^2$ consisting of all the points in red as your tangent bundle. But then all the points which lie on multiple lines "forget" which tangent space they belong to. For example, the point $(1,1)$ lies on the tangent space at $(1,0)$ as well as the tangent space at $(0,1)$. Therefore you lose the canonical surjection $\bigcup_p T_pM \to M$ which is used, among many other things, to endow the tangent bundle with a manifold structure. Disaster!

Finally, to adress the part where you mention vector fields: it is important to note that such objects are not mere set-theoretic sections of the map $\bigcup_p T_pM \to M$; they are continuous or smooth sections. And in order for this to make sense, we need a topology/smooth structure on the tangent bundle.


问:
Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$

$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$

I'm having trouble solving this equation. I've tried squaring both sides and got this

$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$

after simplifying

$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$

and that's where I got stuck.

答:
There is a formula to simplify nested radicals like that
$$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$
It is effective only if $(a^2-b)$ is a perfect square. In your example both of them have this property
$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$
$$\sqrt {11 -  \sqrt {112}} - \sqrt {8 -\sqrt {28}} = a$$
$$\sqrt {11 -  \sqrt {112}} =\sqrt{\frac{11+\sqrt{121-112}}{2}}-\sqrt{\frac{11-\sqrt{121-112}}{2}}=\sqrt{\frac{11+3}{2}}-\sqrt{\frac{11-3}{2}}=$$
$$=\sqrt 7-2$$
$$\sqrt {8 -\sqrt {28}} =\sqrt{\frac{8+6}{2}}-\sqrt{\frac{8-6}{2}}=\sqrt 7-1$$
so finally
$$\sqrt 7-2-(\sqrt 7-1)=-1$$

There is a similar formula for the sum
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$


问:
Solving $\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$.
ETA: since commenters noted I missed a sign I am correcting that.

This is one of those Mathsolutionzz instagram problems. I was curious if I did this correctly, because I looked at the comments, and the solutions others came up with are rather different (and not integers).

The problem:

$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$

So my first step was simple: square both sides and get rid of one of the radicals. We then get

$$2+\sqrt{2-\sqrt{2+x}} = x^2 \tag{1}$$
which I can rearrange to:
$$\sqrt{2-\sqrt{2+x}} = x^2-2 \tag{2}$$

Then we repeat the process, square both sides again:
$${2-\sqrt{2+x}} = (x^2-2)^2 \tag{3}$$

rearrange:
$$-\sqrt{2+x} = (x^2-2)^2-2 \tag{4}$$

rinse, repeat:
$$2+x = ((x^2-2)^2-2)^2 \tag{5}$$

and where I originally had a solution at $x=2$, because I messed up a sign, this time we get a polynomial:
$$x^8-8x^6+20x^4-16x^2-x+2 \tag{6}$$

Which is the really ugly polynomial that @Lubin noted (thanks!) and has several roots.

All that said, were I trying to find the roots by hand, I know the roots can't be negative (since this problem doesn't seem to allow for imaginary solutions and we are sticking to real numbers).

So my next go-to would be factoring out the polynomial. When I originally did this I would try factoring one side of the equation with the opposite-sign binomial, like this:

$$2+x = ((x^2-2)^2-2)^2$$
$$2+x = (x^2-2)^4-4(x^2-2)^2+4$$
$$x-2 = (x^2-2)^4-4(x^2-2)^2$$
$$x-2 = [(x^2-2)^2-2(x^2-2)][(x^2-2)^2+2(x^2-2)]$$
$$x-2 = (x^2-2)[(x^2-2)-2][(x^2-2)+2]$$
$$x-2 = (x^2-2)(x^2-4)(x^2)$$
$$x-2 = (x^2-2)(x-2)(x+2)(x^2)$$

Which at least tells me that +1, +2, -1, -2 don't work, really except insofar as x=2 makes the whole thing zero. But I am curious if this is a bad method, or simply erroneous. (I do realize that ultimately I'd probably want to move the x-2 to the RHS and expand it leaving me with the degree 8 polynomial anyway, but I thought this worth a shot).

答:
By squaring$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x,$$
you arrive at
$${2+\sqrt{2\color{red}-\sqrt{2+x}}} = x^2,$$
then
$$ 2\color{red}-\sqrt{2+x}=(x^2-2)^2$$
and
$$ 2+x=((x^2-2)^2-2)^2.$$
While $x=2$ and $x=-1$ are guessable solutions to this, they are not solutions to the original equation. The problem is that each squaring could introduce spurious solutions, i.e., $a=b$ implies $a^2=b^2$, but not vice versa. In fact, using the correct minus sign (you somehow replaced it with a $+$) leads to the same final polynomial even though the solutions cannot be the same.


问:
Prove that the sum of the radii of the circles

$ABCD$ is a cyclic quadrilateral. Prove that the sum of the radii of the circles drawn inside the triangles $\Delta ABC$ and $\Delta CDA$ is equal to the sum of the radii of the circles drawn inside the triangles $\Delta BCD$, $\Delta DAB$.

I came up with the idea that I can use the Japanese theorem to prove that the distance between the centers is equal - they make a rectangle - but I can't solve it. Please help.

答:
Hint. Use Carnot's Theorem: Given a triangle $\Delta ABC$, let $O$ denote its circumcenter, $R$ its circumradius, and $r$ its inradius. Let $O_1,O_2,O_3$ be furthermore the orthogonal projections of $O$ onto $BC, CA, AB$ respectively. We then have $$OO_1+OO_2+OO_3=R+r$$ Notice: The segment $OO_i$ is taken to be negative if $OO_i$ lies completely outside $\Delta ABC$ and positive otherwise.

Here, $\color{blue}{OO_2}$ would be negative, while $\color{red}{OO_1, OO_3}$ are positive. For convenience, let $AB=:c, BC=:a, CA=:b$. Notice that $OO_3BO_1$ is a cyclic quadrilateral since $\angle BO_3O+\angle OO_1B=90^\circ+90^\circ=180^\circ$, and hence, you can use Ptolemy's Theorem in order to infer $$\begin{align*}OB\cdot O_1O_3&=OO_3\cdot BO_1+O_3B\cdot OO_1\\\iff R\cdot \frac{b}2&=OO_3\cdot \frac{a}2+\frac{c}2\cdot OO_1\end{align*}$$ Analogously, you will obtain \begin{cases}R\cdot a=OO_3\cdot b+OO_2\cdot c\\
R\cdot b=OO_1\cdot c+OO_3\cdot a\\
R\cdot c=OO_2\cdot a+OO_1\cdot b \end{cases}

Add these up and consider the well-known equation $$r\cdot (a+b+c)=2\cdot [\Delta ABC]=OO_1\cdot a+OO_2\cdot b+OO_3\cdot c$$ (do you see now why it is important to take $OO_2$ to be negative?). The first part is just a consequence of dividing $\Delta ABC$ into three triangles with the incenter as a vertex. The second part is trivial.
$$\begin{align*}R\cdot (a+b+c)&=OO_1\cdot (b+c)+OO_2\cdot (c+a)+OO_3\cdot (a+b)\\ R\cdot (a+b+c)+r\cdot (a+b+c)&=OO_1\cdot (a+b+c)+OO_2\cdot (a+b+c)+OO_3\cdot (a+b+c)\\\iff R+r&=OO_1+OO_2+OO_3\end{align*}$$

Now, going back to your problem, it is rather easy to finish once we have this gem :)

 (I'll be referring to the image.) Notice that using Carnot's Theorem twice, once for $\Delta ABD$ and again for $\Delta BCD$, we obtain $$R+r_1=OO_1+OO_5+OO_4\qquad \text{and}\qquad R+r_2=OO_2+OO_3+OO_5$$ Notice that $OO_5$ is negative for $\Delta ABD$ and positive for $\Delta BCD$. Thus, if you add these two equations, you'll get $$r_1+r_2=OO_1+OO_2+OO_3+OO_4-2R$$ It is easy to see that this expression will be identical when referring to $r_3+r_4$.


问:
In $\int \frac{1\;dt}{t}=\ln|t|+C$, why do we write $|t|$?
$$\int \frac{1\;dt}{t}=\ln|t|+C$$

Why do we write $|t|$?

答:
The function $\frac{1}{t}$ is integrable on any bounded interval $I \subset \mathbb{R}\setminus\{0\},$ i.e. also on for example $(-3,-2).$ However, $\ln t$ is not defined for negative $t$ so the right hand side in the following is undefined:
$$\int_{-3}^{-2} \frac{1}{t} dt = \big[\ln t \big]_{-3}^{-2}.$$

What is then a suitable primitive function for negative values of $t$? Assume that $-\infty < a < b < 0.$ Then, making the substitution $t=-s,$ with $s>0,$ we get
$$
\int_a^b \frac{1}{t} dt
= \int_{-a}^{-b} \frac{1}{-s} (-ds)
= \int_{-a}^{-b} \frac{1}{s} ds
= \big[\ln s \big]_{-a}^{-b}
= \big[\ln (-t) \big]_{a}^{b}
.
$$

Thus, for $t>0$ a primitive function to $\frac{1}{t}$ is given by $\ln t$ while for $t<0$ a primitive function is given by $\ln(-t).$ We can write this as
$$
F(t) = \begin{cases}
\ln t & \text{for $t>0$} \\
\ln(-t) & \text{for $t<0$}
\end{cases}
$$
but we can merge these as $\ln|t|.$


答:
The value you're looking for is $a = 2$.

Indeed, one has, according to Stirling's formula:

$$
\frac{(2n)!}{(n!)^a} \quad \underset{n \to +\infty}{\Large{\sim}} \quad \frac{ \sqrt{2 \pi 2 n} (2n)^{2n} e^{-2n} }{ \left( \sqrt{2 \pi n} (n)^{n} e^{-n} \right)^a } = \sqrt{2} \sqrt{2 \pi}^{1-a} n^{ (2-a)n + \frac{1-a}{2} } e^{(a-2)n} 2^{2n}
$$

which you can rewrite as $C \frac{n^{f(n)}}{K^n}$ with $ C :=  \sqrt{2} \sqrt{2 \pi}^{1-a}$, $K = 4e^{a-2} $ and $f(n) = (2-a)n + \frac{1-a}{2}$.

Then:

if $a < 2$, one has $\frac{(2n)!}{(n!)^a} \underset{n \to +\infty}{\longrightarrow} +\infty$.

if $a = 2$, the limit is also $+\infty$ because of the term $4^n$

if $a > 2$, the limit is $0$.

Title: Finding the lowest value for $a$ at which $\lim_{n\to\infty}\frac{(2n)!}{(n!)^a}$ is convergent

Body:
So I've noticed that for larger values of $a$ the function below forms a bell curve shape whereas for lower values it diverges, I would like to find an approximation for the lowest value at which this is true:
$$f_a(x)=\frac{(2x)!}{(x!)^a}$$

I have thought about using stirling's approximation:
$$n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n$$
which would give me:
$$\frac{(2x)!}{(x!)^a}=\frac{\sqrt{4\pi x}\left(\frac{2x}{e}\right)^x}{\left(2\pi x\right)^{a/2}\left(\frac xe\right)^{ax}}=\sqrt{2}\left(2\pi x\right)^{1-a/2}\left(\frac xe\right)^{x(1-a)}2^x$$
$$=2^{x+(3-a)/2}\pi^{1-a/2}x^{x(1-a)+1-a/2}e^{-x(1-a)}$$
if we break this down we are really looking for the convergence of:
$$x^{x(1-a)+1-a/2}e^{-x(1-a)}$$
Then my thoughts were that the $x^x$ term seems the most important term, so the answer would just be $a\ge1$ but I feel like this is not correct. Does anyone have any thoughts? Thanks


问:
Roots of a cubic equation
I have the following equation:

$s^3+as+b=0$

Now I want the values for a and b for which the given equation has the following complex roots:

$c \pm di$

I don't really care about the remaining root.

Any ideas? I have Matlab available, but I don't know how I can use it to solve this problem.

Title: Roots of a cubic equation

Body:
The polynomial $s^3+as+b$ must be a multiple of $(s-c-di)(s-c+di)=s^2-2cs+(c^2+d^2)$. This implies (with $e$ as third root)
$$s^3+as+b=(s-e)(s^2-2cs+(c^2+d^2)\\=s^3-(e+2c)s^2+(2ec+c^2+d^2)s+e(c^2+d^2),$$
hence $e=-2c$ and then $a=2ec+c^2+d^2=-3c^2+d^2$ and $b=e(c^3+d^2)=-2c(c^2+d^2)$.


答:
One may combine the following two facts:

1. If $a_n\to a,\,$ then $\,\frac{1}{n}(a_1+\cdots+a_n)\to a$.

2. $\sqrt[n]{n}\to 1$.

Another way to show it is the following:
$$
\sqrt[2k]{k}=1+a_k\Longrightarrow \sqrt{k}=(1+a_k)^{k}\ge 1+ka_k
\Longrightarrow 0\le a_k<\frac{1}{\sqrt{k}}
$$
and hence
$$
1<\sqrt[n]{n}=(1+a_n)^2=1+2a_n+a_n^2<1+\frac{2}{\sqrt{n}}+\frac{1}{n}\le
1+\frac{3}{\sqrt{n}}
$$
and thus
$$
1<\frac{1}{n}(1+\sqrt{2}+\cdots+\sqrt[n]{n})<1+\frac{3}{n}\left(1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\right) \\ <1+\frac{3}{n}\cdot (2\sqrt{n}+1)\to 1.
$$
It remains to show that
$$
1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}<2\sqrt{n}+1
$$
which can be easily done inductively.


问:
Two different results using two different ways to solve limit, how come
So im learning about limits from tutorials and at the same time im solving the examples myself. Im getting solutions that sometimes differ from the explainer's.

Lets take this limit and the way it is solved on the video:

$$\lim_{x \to -2} \dfrac { (x^3 - x^2 - 6x)}{(x^2+2x)}$$
$$\lim_{x \to -2}\dfrac {x(x^2 - x - 6)}{x(x+2)}$$
$$\lim_{x \to -2} \dfrac {(x+2)(x-3)}{(x+2)}$$
$$\lim_{x \to -2} (x-3)=-5$$

If i try solving it by taking the highest power as per the rule,

$$\lim_{x \to -2} \dfrac {(x^3 - x^2 - 6x)}{(x^2+2x)}$$

$$\lim_{x \to -2} \dfrac {x^3}{x^2}$$

$$\lim_{x \to -2} x=-2$$

or is the rule only for limits at infinity? Its a bit unclear to me.

答:
The first result is correct. You can get a confirmation by l'Hospital rule since you limit is of the form $0/0$.

Taking the derivatives of numerator and denominator, we find

$$\lim_{x\to -2}\frac{3x^2-2x-6}{2x+2}=\frac{10  }{-2}=-5$$


问:
6 Districts and 6 Robberies
The question - A city with 6 districts has 6 robberies in a particular week. Assume the robberies are located randomly, with all possibilities for which robbery occurred where equally likely. What is the probability that some district had more than 1 robbery?

Solution says there are 6^6 possible configuration for which robbery occurred where (i.e. the denominator). I got stuck when I thought each district could have had 0 robbery or 1 or 2 ... 6 robberies. District A having more than one robberies would imply some other district having zero robbery. How do you arrive at the figure of 6^6 then?

Solution also says the problem is identical to throwing a die 6 times and finding the probability of having a different face show up each time. But again, the probability of a face showing up in one throw is independent of the face which showed up on the previous throw, but in the case of a district, a robbery happening in the first district would imply none of the other cities could have 6 robberies since the total is fixed unlike in the case of the die.

答:
You are misunderstanding the meaning of the $6^6$ term. Imagine ordering the location of the robberies by the time they occurred. One possible list of districts is "A, A, B, D, A, E." Another is "E, A, B, D, C." There are $6^6$ possible outcomes since there are six possible districts for the first robbery, six districts for the second robbery, and so on.

For the numerator, you want to count sequences where each district appears exactly once, such as "E, A, B, D, C."

Title: 6 Districts and 6 Robberies


答:
Observe that $$\lim_{x\to \infty} \frac{x^3-x^2-6x}{x^3}=\lim_{x\to\infty} 1-\frac 1x-\frac{6}{x^2}=1$$

This limit being $1$ is why you can substitute out the two functions in limits. In general:

$$\lim_{x\to \alpha}g(f(x))= \lim_{x\to\alpha}g(f_\alpha(x))\iff \lim_{x\to\alpha}\frac{f(x)}{f_\alpha(x)}=1$$

What this means in a graphical sense is that both $f$ and $f_\alpha$ converge at the same rate to the same value when $x$ approaches $\alpha$.

In your example, we note that $$\lim_{x\to -2}\frac{x^3-x^2-6x}{x^3}=1-(-\frac12)-(\frac32)=0$$
and so this condition is failed. Hence we can't substitute the expressions in our limit.

The first method is exactly the way to do it, because given $P(x)=Q(x)\frac{x-\beta}{x-\beta}$, we have that $$\lim_{x\to\beta} P(x)=\lim_{x\to\beta} Q(x)$$

Title: Two different results using two different ways to solve limit, how come

Body:
So im learning about limits from tutorials and at the same time im solving the examples myself. Im getting solutions that sometimes differ from the explainer's.

Lets take this limit and the way it is solved on the video:

$$\lim_{x \to -2} \dfrac { (x^3 - x^2 - 6x)}{(x^2+2x)}$$
$$\lim_{x \to -2}\dfrac {x(x^2 - x - 6)}{x(x+2)}$$
$$\lim_{x \to -2} \dfrac {(x+2)(x-3)}{(x+2)}$$
$$\lim_{x \to -2} (x-3)=-5$$

If i try solving it by taking the highest power as per the rule,

$$\lim_{x \to -2} \dfrac {(x^3 - x^2 - 6x)}{(x^2+2x)}$$

$$\lim_{x \to -2} \dfrac {x^3}{x^2}$$

$$\lim_{x \to -2} x=-2$$

or is the rule only for limits at infinity? Its a bit unclear to me.


问:
How to find a pair of possible solutions (x, y) of this system of equations?
Let $(x, y)$ be a pair of possible solutions of this system of equations:
$x + y = 7$;
$x^2 + y^2 = 25$.

This exercise is all about finding $xy$ product.
there are four possible solutions,

(a) 15

(b) 12

(c) 8

(d) 6

Here's how I've solved it:
I've rewritten the second equation as "$x + y = 5$", since each member was squared.

Now, I've rewritten again this equation as "$y = 5 - x$".
I've substituted and I've got: "$x + 5 - x = 7$", but x cancels out.  Therefore I can say this is not the way to solve it.

I try another one: I've rewritten the first equation as "$y = 7 - x$". Then, I've substituted, and I've got: "$x^2 + (7-x)^2 = 25$"
The last step is: $x^2 = -12$, which isn't solvable using real numbers. So, it's not solvable using this method.

If you know an alternative method, let me know.

答:
Since you asked for alternative methods, here is another one:

$$(x+y)^2 = x^2+y^2+2xy\Leftrightarrow $$ $$xy=\frac 12\left(\underbrace{(x+y)^2}_{=7^2}-\underbrace{(x^2+y^2)}_{=25}\right)=12$$


问:
proof verification on problem with uniformly continuity and $dist(A,B)$
Let $(X,d)$ and $(Y,p)$ and $f:X\rightarrow Y$ uniformly continuous. Prove that for every $y_1,y_2\in f(x)
 $ with $y_1\neq y_2$ its true that $$dist( f^{-1}(y_1), f^{-1}(y_1))>0$$
where $dist(A,B)=inf\left \{ d(x,y): x\in A,y\in B\right \}$ .

Because $f$ is uniformly continuous $X$ is compact and Subjective so for every  $y\in f(x)\exists \: x:f(x)=y\Rightarrow f^{-1}(y)=x$ is enough to show that $dist(x_1,x_2)>0$ and then is enough to show that $d(x_1,x_2)>0$

let's suppose $d(x_1,x_2)=0$ from the axioms of a metric space $\Leftrightarrow x_1=x_2$

$\Rightarrow  f^{-1}(y_1)=f^{-1}(y_2)\Rightarrow y_1=y_2$ which is a contradiction from our hypothesis.

答:
Suppose that $\operatorname{dist}(f^{-1}(\{y_1\}),f^{-1}(\{y_2\})) = 0$, and let $\varepsilon>0$. Then, because $f$ is uniformly continuous, there is a $\delta>0$ such that for any $x_1,x_2 \in X$ with $d(x_1,x_2)<\delta$ we have $\rho(f(x_1),f(x_2)<\varepsilon$.

Now, since $\operatorname{dist}(f^{-1}(\{y_1\}),f^{-1}(\{y_2\}) = 0$ we can find $z_1 \in f^{-1}(\{y_1\})$ and $z_2 \in f^{-1}(\{y_2\})$ such that $d(z_1,z_2) < \delta$. Thus, $\rho(y_1,y_2) = \rho(f(z_1),f(z_2)) < \varepsilon$.

Hence, we showed that for any $\varepsilon>0$ we have $\rho(y_1,y_2) < \varepsilon$, which implies that $\rho(y_1,y_2) = 0$.


答:
Suppose $M \subset \mathbb{R}^n$ is a submanifold. Then one may define its tangent bundle to be the union of all the tangent vector spaces  $T_pM$ to $M$ at points $p$. The problem is what is to be understood as "union" here.

Suppose you define it to be the union as subsets of $\mathbb{R}^n$.
For example, if $M = \mathbb{R} \subset \mathbb{R}$, then at each point the tangent space is $\mathbb{R}$, so all the tangent subsets are equal to the same subspace of $\mathbb{R}$, namely $\mathbb{R}$, and so is their union.

Now suppose $M = \mathbb{S}^1 \subset \mathbb{R}^2$. Then any vector line $D$ can be seen as the tangent space of a point of the circle. So here, the usual union as subsets of $\mathbb{R}^2$ will be the union of all vector lines of $\mathbb{R}^2$, which is $\mathbb{R}^2$.

These constructions are "extrinsic" as they depend not only on $M$ but also on the ambient space $\mathbb{R}^n$. For example, if one says $M \subset \mathbb{R}^n \simeq \mathbb{R}^n\times\{0\} \subset \mathbb{R}^m$, then one can define a "different $TM$", depending on the viewpoint (even if they will be isomorphic).

But in these two examples, if you take a random element of the union, you don't know at which point it is tangent. You lose a lot of geometric sense.

The idea to avoid this is to take a disjoint union, namely
$$TM = \bigcup_{p\in M} \{p\}\times T_pM.$$
An element of this union is of the form $(p,v)$ with $v \in T_pM$, so each element has in its construction more data than in the previous example.

For the first example, this construction gives $T\mathbb{R} = \bigcup_{p\in\mathbb{R}} \mathbb{R} = \mathbb{R}\times \mathbb{R}$, and any tangent vector is of the form $(x,t)$ where $t$ is tangent to $x$.

For the circle, it gives $T\mathbb{S}^1 = \bigcup_{\theta \in \mathbb{S}^1} \{\theta\} \times T_{\theta}\mathbb{S}^1$, etc.

For an abstract manifold, there is no "ambient space", so the usual union of the tangent spaces cannot be defined as an union of subspaces of the same fixed set. Therefore, it would have been a bad construction as we would not have been able to extend it. But the disjoint-union allows us to define, for a general manifold $M$ that is not embedded in a Euclidean space,
$$T_pM = \bigcup_{p\in M}\{p\}\times T_pM,$$
where $T_pM$ is an intrinsic notion in $M$, depending only on the differential structure.

In addition, this construction shows that there is a natural structure of fiber-bundle on the tangent space $TM$ (this is a more general concept), and this construction automatically gives a smooth function $\pi : TM \mapsto M$ that is just the projection $(p,v) \mapsto p$.

If one wants to define a vector field by the idea that "at each point $p$ it has a tangent vector to $p$", then it can be rigorously defined using this construction as a continuous map $X : M \to TM$ such that $X(p) = (p,v_p)$. This is equivalent to saying that $X$ is a (continuous) section of $\pi$, that is $\pi\circ X = \mathrm{id}_M$.
Usually we require the vector field to be smooth, which means $X$ is smooth (a smooth section).

Edit: It is a common problem for geometers, when they have to give a talk to non-specialists, to give a clear presentation while the audience has no idea of the principal objects we use, such as manifolds, vector bundles, etc. My experience is this: do not lose time giving overly complicated definitions if what is really important is the geometric sense. Just say that a manifold is a geometric notion that can extend the definition of surfaces, etc. Define tangent vectors visually. Say that a tangent field is a field of tangent vectors without talking about bundles. Same for covectors. If you have to talk about operators in bundles, just talk about how they act on vectors. You will gain a lot of time and the audience will probably understand many more things than if you had given some overly complicated rigorous statements.

Title: Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?

Body:
It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of
\begin{align}
TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\
(v,p)&\mapsto p
\end{align}
(I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.


答:
As shown in this answer, the Binomial Theorem says that for $n\ge1$,
$$
\begin{align}
1\le n^{1/n}
&\le1+\sqrt{\frac2n}\tag{1a}\\
&\le1+\frac{2\sqrt2}{\sqrt{n}+\sqrt{n-1}}\tag{1b}\\[3pt]
&=1+2\sqrt2\left(\sqrt{n}-\sqrt{n-1}\right)\tag{1c}
\end{align}
$$
Thus,
$$
\frac nn\le\frac1n\sum_{k=1}^nk^{1/k}\le\frac1n\left[n+2\sqrt2\sum_{k=1}^n\left(\sqrt{k}-\sqrt{k-1}\right)\right]\tag2
$$
and, because the sum on the right side of $(2)$ telescopes, we have
$$
1\le\frac1n\sum_{k=1}^nk^{1/k}\le1+\frac{2\sqrt2}{\sqrt{n}}\tag3
$$
to which we can apply the Squeeze Theorem.


问:
Functoriality of the Fundamental group
The fundamental group is a functor from the category of pointed topological spaces to the category of groups.

Therefore every base-point preserving continuous function $f$ between pointed topological spaces induces a homomorphism $f_*$ between the fundamental groups. This is done by composing the loops with $f$, which is well-defined, because homotopy is also preserved under $f$.

Can we switch this around?

Every group is the fundamental group of a CW-complex, which can be constructed according to how many generators and relations the group has.

Can a continuous function be constructed for every homomorphism such that the continuous function induces the homomorphism? If the fundamental group functor is 'surjective', one has a pre-image at least.

How do you go from the algebraic to the topological with the morphisms?  I have no idea.

答:
Let $B : \mathbf{Grp} \to \mathbf{Top}_*$ be the functor obtained by defining $B G$ to be the geometric realisation of the nerve of $G$ (considered as a 1-object category), i.e. the simplicial set
$$\cdots \mathrel{\lower{0.5ex}{\begin{array}{c} \smash{\to} \\ \smash{\to} \\ \smash{\to} \\ \smash{\to} \end{array}}} G \times G \mathrel{\lower{0.5ex}{\begin{array}{c} \smash{\to} \\ \smash{\to} \\ \smash{\to} \end{array}}} G \rightrightarrows 1$$
where the degeneracies maps insert the unit element at the appropriate location and the face maps compose adjacent pairs of elements.

It is well-known that $B G$ is a $K (G, 1)$ Eilenberg–MacLane space, i.e. $B G$ is a path-connected topological space such that $\pi_1 (B G, *) \cong G$ and $\pi_n (B G, *) = 1$ for all $n > 1$. Moreover, the isomorphism $\pi_1 (B G, *) \cong G$ is induced by the obvious correspondence: send each element of $G$ to the loop in $B G$ that realises the corresponding 1-simplex in the nerve. It follows that $\pi_1 \circ B$ is naturally isomorphic to $\mathrm{id}_{\mathbf{Grp}}$ as a functor.

Title: Functoriality of the Fundamental group


问:
Localization by a prime ideal
Let $A$ be a commutative ring with identy and let $P\subseteq A$ a prime ideal. We consider $$A_P:=\bigg\{\frac{a}{s}\:\bigg|\; a\in A, s\in A\setminus P\bigg\}.$$

I must prove that $$\frac{a}{s}\;\text{is a unit in}\; A_P\iff a\in A\setminus P$$

$(\Rightarrow)$ If $\frac{a}{b}\in A_P$ is a unit, exists $\frac{b}{t}\in A_P$ such that $$\frac{a}{s}\frac{b}{t}=1\implies (ab,st)\equiv (1,1),$$ therefore exists $u\in A\setminus P$ such that $u(ab-st)=0$, then $$a(ub)=ust\in A\setminus P$$, since $P$ is prime $a\in A\setminus P$.

for the other implication?

could someone suggest me something? I tried, but I can't.

Thanks!

答: The other direction is even easier: the point is that we can write $a$ into the denominator, and we have $\dfrac sa\dfrac as=1$.

Title: Localization by a prime ideal


问:
Relation between 1-norm of matrix and Euclidean norm of vectors
Suppose I have column vectors $u, v$ in a Hilbert space. I can define rank-1 matrices $uu^*$ and $vv^*$, where $^*$ denotes the transpose conjugate. If it is the case that

$$\|uu^* - vv^*\|_1 \leq \varepsilon,$$

where $\|\cdot \|_1$ is the trace norm or nuclear norm or Schatten 1-norm defined as $\|A\|_1 = \text{Tr}(\sqrt{A^*A})$, can one say something about an upper bound for

$$\|u - v\|?$$

Here $\|\cdot \|$ is the Euclidean 2-norm for vectors.

In my case, I have the additional knowledge that $uu^*$ and $vv^*$ are positive semidefinite with unit trace if that helps with the proof.

答:
The most you can say is $\|u-v\|\le 2$.

First of all, notice that $Tr(uu^*)=1\implies \|u\|=1$, and the same holds for $v$. As a consequence,
$$\|u-v\|\le \|u\| + \|v\| = 2. $$

Take now $v = -u$. We have that
$$
vv^* =uu^* \implies \|vv^*-uu^*\| = 0 \le \varepsilon
$$
for any $\varepsilon>0$, but
$$
\|u-v\| = \|2u\| =  2.
$$


问:
Why can we do limit manipulation on one limit but not the other?
For example:

$$ \lim_{x \rightarrow 1} (x)^{\frac{1}{x-1}} \neq (\lim_{x \rightarrow 1} (x))^{\frac{1}{x-1}} \equiv 1$$

But when we talk about euler limits (the example may not be accurate, but there are a lot of limits that do this):

$$ \lim_{x \rightarrow 1} (1+ \frac{1}{x-1})^{x^2-1} = \lim_{x \rightarrow 1} (1+ \frac{1}{x-1})^{(x+1)(x-1)} = e^{ \lim_{x \rightarrow 1}x+1} = e^2$$

Meaning that we calculate the base of the power, getting that number $e$ and then proceeding to calculate the exponent power $e^{\lim ~ \dots}$

What is the actual difference here? Why in the first one I can't calculate the base of the power and in euler limits I can? I really don't get it.. Thanks!

答:
In the first case, the limit is of the form $1^\infty$ which is indeterminate.  In the second case, you're skipping a step.  We have $$\lim_{x \rightarrow 1} \left(1+ \frac{1}{x-1}\right)^{(x+1)(x-1)}=
\lim_{x \rightarrow 1}\left(\left(1+ \frac{1}{x-1}\right)^{x-1}\right)^{x+1}$$ and the limit of the base is $e$, the limit of the exponent is $2$, and there's no indeterminacy.


答:
Well, thanks to Hagen and to you, the polynomial that would have the solution(s) among its roots is the octic
$$
x^8-8x^6+20x^4-16x^2-x-2\,.
$$
Rational Root Theorem tells you where to look, $\pm1$ and $\pm2$, if you want rational numbers. Of these, only $+2$ and $-1$ are roots, but neither of these is a solution to your problem. Looking at the Desmos graph, I see that there are eight, count’em eight, real roots: four positive, four negative. The four negative ones are excluded by the statement of the problem, and $2$ doesn’t work. I guess it remains to check the other three.

Title: Solving $\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$.

Body:
ETA: since commenters noted I missed a sign I am correcting that.

This is one of those Mathsolutionzz instagram problems. I was curious if I did this correctly, because I looked at the comments, and the solutions others came up with are rather different (and not integers).

The problem:

$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$

So my first step was simple: square both sides and get rid of one of the radicals. We then get

$$2+\sqrt{2-\sqrt{2+x}} = x^2 \tag{1}$$
which I can rearrange to:
$$\sqrt{2-\sqrt{2+x}} = x^2-2 \tag{2}$$

Then we repeat the process, square both sides again:
$${2-\sqrt{2+x}} = (x^2-2)^2 \tag{3}$$

rearrange:
$$-\sqrt{2+x} = (x^2-2)^2-2 \tag{4}$$

rinse, repeat:
$$2+x = ((x^2-2)^2-2)^2 \tag{5}$$

and where I originally had a solution at $x=2$, because I messed up a sign, this time we get a polynomial:
$$x^8-8x^6+20x^4-16x^2-x+2 \tag{6}$$

Which is the really ugly polynomial that @Lubin noted (thanks!) and has several roots.

All that said, were I trying to find the roots by hand, I know the roots can't be negative (since this problem doesn't seem to allow for imaginary solutions and we are sticking to real numbers).

So my next go-to would be factoring out the polynomial. When I originally did this I would try factoring one side of the equation with the opposite-sign binomial, like this:

$$2+x = ((x^2-2)^2-2)^2$$
$$2+x = (x^2-2)^4-4(x^2-2)^2+4$$
$$x-2 = (x^2-2)^4-4(x^2-2)^2$$
$$x-2 = [(x^2-2)^2-2(x^2-2)][(x^2-2)^2+2(x^2-2)]$$
$$x-2 = (x^2-2)[(x^2-2)-2][(x^2-2)+2]$$
$$x-2 = (x^2-2)(x^2-4)(x^2)$$
$$x-2 = (x^2-2)(x-2)(x+2)(x^2)$$

Which at least tells me that +1, +2, -1, -2 don't work, really except insofar as x=2 makes the whole thing zero. But I am curious if this is a bad method, or simply erroneous. (I do realize that ultimately I'd probably want to move the x-2 to the RHS and expand it leaving me with the degree 8 polynomial anyway, but I thought this worth a shot).


问:
Finding the constants of an exponential equation
Find $A$ and $k$ if $f(x)=Axe^{-kx}$ has a critical point at $x=10$, and at that critical point, $f(x)=12$.

I'm entirely lost in this question. I can find the general derivative of $Axe^{-kx}$ pretty easily, but solving gets difficult fast. I'd at least like a hint first to see what direction to even move in.

(The problem)

Body:
$$f'(x)=Ae^{-k x} (1- k x)$$
we know that $f(x)$ has a critical point at $x=10$ thus
$$f'(10)=0\to Ae^{-10k}(1-10k)=0$$
which means $$1-10k=0\to k=\frac{1}{10}$$
Furthermore we know that $f(10)=12$ so
$$f(x)=A x e^{-k x}\to f(10)=10Ae^{-\frac{10}{10}}=12$$
and finally
$$k=\frac{1}{10};\;A=\frac{6}{5}e$$


问:
Product of functions with mean zero has mean zero?
Let $f,g:[0,L] \subset \mathbb{R} \longrightarrow \mathbb{R} $ be a differentiable and periodics function, with period $L>0$.

Question 1. If $f$ and $g$ has mean zero, that is,
$$ \int_0^L f(x) \; dx=\int_0^L g(x) \; dx=0$$
then
$$\int_0^L f(x)\cdot g(x) \; dx=0?$$

I know that $f'$ and $g'$ has mean zero, since
$$\int_0^L f'(x) \; dx=f(L)-f(0)=0 \quad \text{and} \quad \int_0^L g'(x) \; dx=g(L)-g(0)=0. \tag{1}$$

In this question, the only idea I had was to use the fact $(1)$ and integration by parts, but did not get anything.

Question 2. If there exists a differentiable periodic function $\varphi:[0,L] \subset \mathbb{R} \longrightarrow \mathbb{R} $, with period $L>0$, mean zero and derivative $\varphi '$ differentiable , such that
$$\int_0^L \varphi''(x)f(x) \; dx + \int_0^L \varphi'(x) g(x)\; dx=0, $$
then $f$ and $g$ has mean zero?

In the last question, I don't have ideas. But it does not seem to me to be valid in general.

答:
For the first question, consider the case of $f(x) = g(x) = \sin(x)$. Then the integral from $0$ to $2\pi$ would be $$\int_0^{2\pi}\sin^2(x)dx = \pi$$

So it doesn't have to equal $0$, meaning that the first statement is not necessarily true.

For the second question, let $f(x) = g(x) = \sin(x)+1$. Let $\phi(x) = \cos(x)-\sin(x)$. The sum of the integrals would then be $$\int_{0}^{2\pi}\left(\sin\left(x\right)-\cos\left(x\right)\right)\left(\sin\left(x\right)+1\right)dx+\int_{0}^{2\pi}\left(-\sin\left(x\right)-\cos\left(x\right)\right)\left(\sin\left(x\right)+1\right)dx = 0$$
but the integral from $0$ to $2\pi$ of $f(x)$ or $g(x)$ is not $0$, so the second statement is also false.


问:
Weak Konig's Lemma and Sequential Compactness
May I ask how is "Sequential Compactness" related to "Weal Konig's Lemma"? If we have a bounded set in real number space, how do we prove that every infinite sequence in this set has an infinite subsequence such that this subsequence converges to some point in the set via weak Konig's lemma, which states that if a binary tree has arbitrarily long full finite paths, then it as an infinite path?</p>

答:
Let $\langle x_n:n\in\omega\rangle$ be a bounded sequence in $\Bbb R$, and let $X=\{x_n:n\in\omega\}$. The result is trivial if $X$ is finite, so assume that $X$ is infinite. $X$ is bounded, there is a closed interval $I_{\langle\rangle}$ such that $X\subseteq I_{\langle\rangle}$. Given an interval  $I_\sigma$ for some finite sequence $\sigma$ of zeroes and ones, bisect $I_\sigma$, let $I_{\sigma^\frown 0}$ be the left half, and let $I_{\sigma^\frown 1}$ be the right half. Let

$$\mathscr{I}=\big\{I_\sigma:\sigma\in{^{<\omega}\{0,1\}}\big\}\,;$$

$\langle\mathscr{I},\supseteq\rangle$ is a binary tree of height $\omega$. Let $\mathscr{I}_0=\{I_\sigma\in\mathscr{I}:I\cap X\text{ is infinite}\}$. It’s straightforward to verify that $\mathscr{I}_0$ is a subtree of $\mathscr{I}$ with arbitrarily long finite branches, so it has an infinite branch, from which a convergent subsequence is easily constructed.


答:
Substituting $x$ with $2\cos\theta$ will suffice to get the result to your question $$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$ will be $\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}} = 2\cos\theta$

On LHS by Half angle cosine formula

$\sqrt{2+\sqrt{2-2\cos\frac{\theta}{2}}}$ = $\sqrt{2+2\sin\frac{\theta}{4}}$ = $\sqrt{2+2\cos(\frac{\pi}{2}-\frac{\theta}{4}})$ = $2\cos(\frac{\pi}{4}-\frac{\theta}{8})$

Now $\theta = \frac{\pi}{4}-\frac{\theta}{8}$ on further simplification we can get $\theta$ = $\frac{2\pi}{9}$ which is $40°$

Therefore $$x = 2\cos\frac{2\pi}{9} \qquad \text{or}\qquad 2\cos40°.$$


答:
This is exactly the same as the birthday problem.  But instead of determining the probability that 2 kids have the same birthday you're determining the inverse -- that no 2 robberies occur in the same district.

You could do it easily by computing the likely hood that no district had 2 robbers.  And then subtract from 1.

1st robber can pick any district (probability = 1 = 6/6).
2nd robber must pick one of the remaining 5 districts (5/6).
3rd robber must pick one of the remaining 4 districts (4/6).
4th robber (3/6).
5th robber (2/6).
6th robber (1/6).

So the answer is:
$$1- \frac{6!}{6^6} = 0.0154$$

Title: 6 Districts and 6 Robberies

Body:
The question - A city with 6 districts has 6 robberies in a particular week. Assume the robberies are located randomly, with all possibilities for which robbery occurred where equally likely. What is the probability that some district had more than 1 robbery?

Solution says there are 6^6 possible configuration for which robbery occurred where (i.e. the denominator). I got stuck when I thought each district could have had 0 robbery or 1 or 2 ... 6 robberies. District A having more than one robberies would imply some other district having zero robbery. How do you arrive at the figure of 6^6 then?

Solution also says the problem is identical to throwing a die 6 times and finding the probability of having a different face show up each time. But again, the probability of a face showing up in one throw is independent of the face which showed up on the previous throw, but in the case of a district, a robbery happening in the first district would imply none of the other cities could have 6 robberies since the total is fixed unlike in the case of the die.


问:
Show that for $a \ne 0, b \in \Bbb{R}$ that $\int_{\Bbb{R}}f(ax+b)d\lambda(x) = \frac{1}{|a|}\int_{\Bbb{R}}f(x)d\lambda(x)$
Show that for $a \ne 0, b \in \Bbb R$ that $\int_{\Bbb R}f(ax+b)d\lambda(x) = \frac1{|a|}\int_{\Bbb R}f(x)d\lambda(x)$ where $\lambda$ is the Lebesgue measure. (Note that both sides are well defined and equal or neither are well-defined).

I have this question for an assignment, but I don't really know where to start. Do I approximate $f$ by simple functions and try to show the assertion for simple functions? I don't need a full answer, but if someone could point me in the right direction or help me get started, that would be great :)

答:
Hint

Suppose $f=\boldsymbol 1_{A}$ where $A$ is measurable. Then, $$\int_{\mathbb R}\boldsymbol 1_A(ax+b)\,\mathrm d \lambda (x)=\int_{\mathbb R}\boldsymbol 1_{\frac{A-b}{a}}(x)\mathrm d \lambda (x)=\lambda \left(\frac{A-b}{a}\right)=\frac{1}{|a|}\lambda (A)=\frac{1}{|a|}\int_{\mathbb R}f(x)\,\mathrm d \lambda (x).$$

Now, try to see what happen when $f$ is simple, i.e. $f(x)=\sum_{i=1}^n\alpha _i\boldsymbol 1_{A_i}$ for $\alpha _i\in\mathbb R$ and $A_i$ measurable. Then use the fact that any positive function is the increasing limit of a sequence of simple function, i.e. there is $(f_n)$ of simple function s.t. $f_n\nearrow f$. And finally, if $f$ is measurable, then $f=f^+-f^-$ where $f^+$ and $f^-$ are positive function (namely, $f^+=f(x)\vee 0$ and $f^-(x)=-(f(x)\wedge 0)$.


问:
Present value of deferred life annuity-due
Kimberly is told that she can receive a 250,000 death benefit from her husband's life insurance in annual installments of 25,000 at the beginning of each year for 11 years and a final payment of 16,265 at the beginning of the 12th year.

Alternatively, Kimberly may receive annual installments of 13,000 at the beginning of each year for life, with a certain period of 10 years.

Calculate the present value of a 10-year deferred life annuity-due of one dollar per annum at Kimberly's issue age.

Okay so ultimately, I figured the present value of a 10-year deferred life annuity-due would be the discounted rate of 10 years times the present value of a perpetuity due....

To get the discount rate I used the first payment option of an 11-year annuity due with annual payment of 25,000 plus a final payment of 16,265 at time t=11 since the annuity-due begins payment at time t=0, this means that at time t=11, the final payment will be made thus,

$$ 250000=25000\require{enclose} \ddot a_{\enclose{actuarial}{11}{i}} + 16265v^{11} $$

this formula generates interest rate i=3% thus, the discount rate d = .03/1.03 = .029 or 2.9%

Now, for perpetuity-due it is simply payment over discount which in this case = 13,000 / .029 = $448,275.86.

However, because it is deferred for 10 years and we only want it in terms of 1 dollar per annum, this becomes,

$$v^{10} * \frac {1}{.029}\\ \\ = (1.03)^{-10}*34.483 \\ \\= 25.6584 $$

However, my answer sheet gives a different value without explaining the solution  so I have no clue how to interpret this question.. can somebody please explain this for me?

答:
There's no perpetuity here.  Kimberley is offered a life annuity.

From the first option, we can compute the interest rate.  In the second option, Kimberley is offered a life annuity with period certain, also worth $250,000$.  A life annuity with a $10$-year period certain is the sum of a a ten-year annuity certain, and a 10-year deferred life annuity.  Since ww know the interest rate, we can compute the value of the annuity certain, and then subtraction gives the value of the life annuity.  We just have to divide by $13,000$ to get the answer to the question posed.


问:
Laurent Series and residue of $\frac{z}{(z-1)(z-3)}$ around z = 3
As mentionned in the title, I'd like to get the function's Laurent series and after its residue, I have tried to separate the two denominators to get a partial fraction but I still have a z at numerator that's bothering me ...

So far : $$
        \frac{z}{(z-1)(z-3)} = \frac{-z}{2(z-1)} + \frac{3z}{(z-3)}
$$

I can obviously try $\omega = z-3$ and change my variable but I don't feel it'll will work ... Because I will still have this z on top of my fraction ...

Could you help me ?
Thank you very much ;)

答: $z=3$ is a simple pole so $Res(f)_{z=3}=\lim_{z\to 3} f(z)={3\over 2}$</p>


问:
How to interpret Gauss formula - number of irreducible monic polynomials
Say I want to find the number of monic irreducible polynomials in $\mathbb{F}_{3}[X]$ of degree 2, by using the following formula (Gauss formula?). How does the formula work?

There is no explanation of it in my book or examples with it, and I honestly doesn't understand how the sum works.
What are we summing over? Like what index are we summing over, where does the sum start and end? What do they mean by d divides n written in the lower index?

答:
In your case, $n = 2$, $p = 3$. The expression
$$ \sum_{d|n} $$
means that your index set consists precisely of the divisors of $n$. In this case, the index set corresponds to $d = 1$ and $d = 2$. So the formula in your case is just
$$ N_2 = \frac{1}{2}\left(\mu(2) 3^1 + \mu(1) 3^2\right)  = \frac{1}{2}\left(-3 + 9\right) = 3. $$

If instead $n = 4$, then the terms of your sum would be over $d = 1$, $d = 2$, and $d =4$. So the summation has three terms. If $n = 10$, the terms of the sum would be $d = 1$, $d = 2$, $d = 5$, and $d = 10$.

As for why the formula works, this link may be useful.


问:
$\mathbb{R}^{\mathbb{N}}$ Hilbert space with uncountable orthonormal basis?
Let's say we have a Hilbert space of sequences $u_n\in\mathbb{R}^{\mathbb{N}}$, equipped with some inner product $(\cdot,\cdot)$. Let's say there is a Hermitian operator $M$ acting on this space, with a continuous spectrum of eigenvalues $\lambda\in\mathbb{R}$. Because $M$ is Hermitian, its eigenvectors are automatically orthogonal to each other $(a^{\lambda},a^{\lambda'})=0$ for $\lambda\neq\lambda'$. The eigenvectors are assumed to be in the Hilbert space (i.e. have finite norm).

Owing to the existence of this Hermitian operator $M$, this Hilbert space has an uncountable orthonormal basis. I don't see any obstacles for such a scenario to occur, but it seems counter-intuitive for a sequence $u_n$ to have finite norm while also being orthogonal to an uncountable number of sequences.

Does this scenario make sense? Or is there something about $M$ and the Hilbert space which makes it impossible?

Edit: The reason I ask this question is that I seem to have come across a scenario in which this does in fact occur. The Hilbert space in question has the unusual inner product
$$(u,v)=\sum_{n=0}^{\infty}\frac{(2n+1)!!}{(2n)!!}u_nv_n.$$
And the Hermitian operator is
$$M_{nm}=(4n+3)\delta_{nm}+2n\delta_{n-1m}+(2n+3)\delta_{n+1m}.$$
You can check that it is Hermitian with respect to this inner product by noting that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric. This operator can be written in the form $M=A+A^{\dagger}+3$, where
$$A_{nm}=2n\delta_{nm}+(2n+3)\delta_{n+1m}.$$
These operators obey the commutator
$$[A,A^{\dagger}]=2M.$$
This means in particular that
$$e^{cA}Me^{-cA}=e^{2c}M$$

Which implies that if $v$ is an eigenvector of $M$ with eigenvalue $\lambda$, then $e^{-cA}v$ is an eigenvector with eigenvalue $e^{2c}\lambda$, for any constant $c$. Thus providing a continuous spectrum of eigenvalues.

To find an eigenvector $v$, we may solve the recursive relation associated with the equation $Mv=\lambda v$, and find that there are normalizeable solutions for real $\lambda<0$, with asymptotic behavior
$$v_n\sim(-1)^n\frac{1}{\sqrt{n}}e^{-\sqrt{2n|\lambda|}},\quad n\to\infty$$

Hermiticity of $M$

Since this has been questioned a couple of times I will show that $M$ is Hermitian with respect to this inner product. As I mentioned it is equivalent to showing that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric.

$$\frac{(2n+1)!!}{(2n)!!}M_{nm}=\frac{(2n+1)!!}{(2n)!!}(4n+3)\delta_{nm}+\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}+\frac{(2n+3)!!}{(2n)!!}\delta_{n+1m}$$

This is symmetric since

$$\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}=\frac{(2m+3)!!}{(2m)!!}\delta_{m+1n}.$$

Reviewing the definition of a Hermitian operator, it presumes it is a bounded operator, which we have found that this operator is not (assuming the rest of my calculations are correct). Perhaps the statement about orthogonality of eigenvectors is not true when the "Hermitian" operator is unbounded?

答: An example of such a Hilbert space is $\ell^2=\{ x_n| \sum x_n^2<\infty\}$. This is a complete inner product space with inner product $\langle x_n,y_n\rangle=\sum x_ny_n$. We can automatically see that the sequences with a one in a single entry create an orthonormal basis for $\ell^2$, i.e. $B=(e^i_n)$ where $e_n^i=\delta_{ni}$. This implies that there cannot be an uncountable basis for such a vector space and hence such a hermitian operator cannot exist. I don't believe we can construct another sequence Hilbert space since the norm must be compatible with the inner product.


问:
Probability density functions and cumulative distribution functions: open or closed intervals?
In Statistics, the probability density function, $f_X(x)$ and the cumulative distribution function, $F_X(x)$, of a real-valued random variable $X$ are said to have the following meanings:

$$f_X(x)=\Pr[X=x] \space\space;\space\space F_X(x)=\Pr[X\leq x]$$

And they are related by

$$f_{X}(x)=\frac{d}{d x} F_{X}(x)  \space\space;\space\space  F_X(x)=\int_{-\infty}^{x} f_{X}(u) du$$

However,  I have found in various sources different criteria for including or not including the values of the endpoints of the interval. When do we consider a closed interval and an open interval? Which of the following options would be the correct one?

$F_X(x)$ equals $\Pr[X < x]$ or $\Pr[X\leq x]$?

$F_X(b)-F_X(a)$ equals $\Pr[a<X< b]$, $\Pr[a\leq X< b]$, $\Pr[a< X \leq b]$ or $\Pr[a \leq X \leq b]$?

$f(x)dx$ equals $\Pr\big[X\in(x,x+dx)\big]$, $\Pr\big[X\in (x,x+dx]\big]$, $\Pr\big[X\in[x,x+dx)\big]$ or $\Pr\big[X\in[x,x+dx]\big]$?

答:
In short : You have said that $f$ is a density i.e it relates to a continuous distribution so whether you include the end points or not it wont matter.

Some more :

Firstly if your distribution is continuous it doesn't matter since it will take any one discrete value with probability $0$.

However it is standard to include the interval (and matters if your distribution is discrete or mixed), i.e $F_X(x)=P(X\leq x)$ - which answers your first question. In fact this also answers your second question since it implies $F_X(b)-F_X(a)=P(a<X\leq b)$.

For your third question since $f$ is written as a density it does not matter. This is similar to taking a Riemann integral of some function over $[a,b]$ or $(a,b)$ its the same.


问:
Confusion over real functions as a complex Fourier series
I am a bit confused with the following, I mean I do understand what these notes say, but does that mean that for real functions that the sum is only from 0 to  infinity rather than negative infinity to positive? Are terms doubling up? And do complex conjugates need to be taken?

答:
He is just breaking 3.22 into two sums in the following way
$$
S(x)=\frac{a_0}{2}+ \sum_{n=1}^\infty \frac{a_n}{2}\left(e^{\frac{in\pi x}{l}}+e^{\frac{-in\pi x}{l}} \right)+\frac{-ib_n}{2}\left(e^{\frac{in\pi x}{l}}-e^{\frac{-in\pi x}{l}} \right)
$$
$$
S(x)=\frac{a_0}{2}+ \sum_{n=1}^\infty \left(\frac{a_n}{2}-\frac{ib_n}{2}  \right)e^{\frac{in\pi x}{l}}+\sum_{n=1}^\infty \left(\frac{a_n}{2}+\frac{ib_n}{2}  \right)e^{-\frac{in\pi x}{l}}.
$$
Now, on the right most sum let $n \to -n$ and you have
$$
S(x)=\frac{a_0}{2}+ \sum_{n=1}^\infty \left(\frac{a_n}{2}-\frac{ib_n}{2}  \right)e^{\frac{in\pi x}{l}}+\sum_{n=-1}^{-\infty} \left(\frac{a_{-n}}{2}+\frac{ib_{-n}}{2}  \right)e^{\frac{in\pi x}{l}}.
$$
Here I think there is a typo in your book and the $c_n$ should be defined as follows
$$
\begin{cases}
c_0=\frac{a_0}{2}, \\
c_n=\frac{a_n}{2}-\frac{ib_n}{2}, \ n>0,\\
c_n=\frac{a_{-n}}{2}-\frac{ib_{-n}}{2}, n<0.
\end{cases}
$$
Notice that as he defined you would have to know the values $a_n, b_n$ for negative $n$ and in fact he defined $c_n$ in two different ways for positive values of $n$. Which makes no sense. But using the definition I give above notice that
$$
S(x)=c_0+ \sum_{n=1}^\infty c_n  e^{\frac{in\pi x}{l}}+\sum_{n=-1}^{-\infty} c_n e^{\frac{in\pi x}{l}}=\sum_{n=-\infty}^\infty c_n  e^{\frac{in\pi x}{l}}.
$$


问:
Prove this function is integrable over an interval
Let $f:[0,2] \to R$ defined as $f(x)= \begin{cases} 1 & x \neq 1 \\ 0 & x = 1 \end{cases}$

Prove that $f$ is integrable in $[0,2]$ and obtain its integral.

Well since the function it's not continuous nor monotonic, then I think this must be proved by definition, but I don't know how to find the partitions, upper and lower sum, etc.

Body:
hint

Let $ \epsilon>0 $ small enough.

Consider the partage
$$P=(0,1-\frac{\epsilon}{3},1+\frac{\epsilon}{3},2)$$

then

$$U(f,P)-L(f,P)=$$
$$(1-1)(\frac{1-\epsilon}{3}-0)+$$
$$(1-0)(1+\frac{\epsilon}{3}-(1-\frac{\epsilon}{3}))+(1-1)(2-(1+\frac{\epsilon}{3}))=$$
$$\frac{2\epsilon}{3}<\epsilon$$

Its integral is given by

$$\int_0^1f+\int_1^2f=$$
$$\lim_{n\to+\infty}(\int_0^{1-\frac 1n}dx+\int_{1+\frac 1n}^2dx)=$$
$$\lim_{n\to+\infty}(1-\frac 1n)+(2-1-\frac 1n)=$$
$$\lim_{n\to+\infty}(2-\frac 2n)=2$$m


问:
Show that$ f(U)$ is open in $R^n$
Suppose $f : {R}^n → {R}^n$ is a continuously differentiable function such that $f '(x)$ is injective for every $x \in {R}^n$. Show that $f(U)$ is open in ${R}^n$, for every open set $U \subset {R}^n$.

So I'm trying to work off this question I found (Let $g:\mathbb R^n\rightarrow \mathbb R$ continuous and injective. Show that $g(U)$ is open if $U\subset \mathbb R^n$ open.) and I was wondering if that would work or is there a better more clean way?

答:
Suppose $x^*\in U$ and let $y^*=f(x^*)$. We are given that $f'(x^*)$ is invertible and so the inverse function theorem gives us open sets $V,W$ such that $x^* \in V, y^* \in W$, $W=f(V)$ and a $C^1$ function $h:W \to V$ such that $h(f(x)) = x$ for all $x \in V$.

Let $O \subset U \cap V$ be open and $x^* \in O$ and $P = f(O)$. Since $f(O) = h^{-1}(O)$ and $h$ is continuous ($C^1$ in fact) we see that $P$ is open and hence $y^* \in P \subset f(U)$ and so $f(U)$ is open.

Title: Show that$ f(U)$ is open in $R^n$


问:
in a graph with minimum degree 1, get the expected value of the degree of a uniform random neighbor of a uniformly random vertex
I came across this exercise in one of my old notes in class:

Consider a simple undirected graph $G=(V,E)$ on $n$ vertices such that the minimum degree is 1. Let $x$ be a uniformly random vertex and $y$ be a uniformly random neighbor of $x$. Let $Z$ be the degree of $y$. Show that $$\mathbb{E}[Z] = \frac{1}{|V|} \sum_{ab \in E} \left(\frac{deg(a)}{deg(b)}+ \frac{deg(b)}{deg(a)} \right).$$

I'm not too sure how to get the given expression for $\mathbb{E}[Z]$, my gut feeling is that it is the sum (over all vertices) of the product of the probability of obtaining a vertex and the probability of obtaining the neighbors of the vertex and the degree of the neighbor. How do we get the expected value to be that sum (over the edges) of ratios of degrees of the incident vertices?

答: For every edge $ab$, there is a $\frac1{|V|} \cdot \frac1{\deg a}$ probability of starting at $a$ and going to $b$, which results in $Z = \deg b$. There is also a $\frac1{|V|} \cdot \frac1{\deg b}$ probability of starting at $b$ and going to $a$, which results in $Z = \deg a$.


问:
Will $B$ form an abelian group? Yes/No

Is the following statement is true/false ?

Given $A$ is an abelian  group and  $B$ is any group, suppose  $f: A \to B $ is homomorphism and  $f$ is not onto; then  $B$ is a abelian group

My attempt : I think  true . Take element  $a_1, a_2 \in  A$ and it is given that $A$ is abelian so $a_1a_2=a_2a_1$

since $f$ is homomorphism  then it implies  $ f(a_1 a_2)=f(a_1)f(a_2) = f(a_2)f(a_1)= f(a_2a_1)$

so $f(a_1)f(a_2) = f(a_2)f(a_1)$ where $f(a_1) , f(a_2) \in B $

This implies $B$ is abelian group

答: This is clearly false: take $B$ to be any non-abelian group and $A$ to be any abelian group. Then there is a map $f: A \to B$ sending all elements of $A$ to the identity element of $B$, which is clearly a non-surjective homomorphism.

Title: Will $B$ form an abelian group? Yes/No


问:
Heun's method: Why is it of second order?
I want to get an intuitive understanding of Heun's method and why it has order of consistency 2, i.e. the error lies is in $O(h^{3})$ if $h$ is the stepwidth., I have read it everywhere, my professor says it without proving it. All the resources I have seen so far on the web either omit the proof as trivial, and the only one that does the proof does it wrong. I am at a loss. Can someone explain it to me in as much detail as possible?

答:
I am assuming you mean the explicit method given by
\begin{align}
&k_1 = f(t_n, y_n) \\
&k_2 = f(t_n + h, y_n + hk_1) \\
&y_{n + 1} = y_n + \frac{h}{2}\left(k_1 + k_2\right)
\end{align}
To analyze truncation error, assume $y_n = y(t_n)$.
From Euler's method we know
$$y_n + hk_1 = y(t_n + h) + O(h^2)$$
Therefore by Taylor expanding the y argument to the first degree
$$k_2 = f(t_n + h, y(t_n + h) + O(h^2)) = f(t_n + h, y(t_n + h)) + O(h^2)$$
Hence Heun's method is almost the trapezoid method for integration applied to $f$ to approximate $y(t_n + h) - y_n = \int_{t_n}^{t_n + h}f(t, y(t))\,dt$ except that it differs from it by $\frac{h}{2}O(h^2) = O(h^3)$. The trapezoid method is known to be $O(h^3)$ locally, so the error of Heun's method is $O(h^3)$ since it differs from the trapezoid method by $O(h^3)$.

Alternatively, Taylor expanding Heun's method and comparing it to the Taylor expansion of $y(t_n + h)$ will give you the result. This strategy can give the conditions for a Runge Kutta method to achieve a desired order.

For example for two stage methods of order 2:

Assume a method of the form
\begin{align}
&k_1 = f(t_n, y_n) \\
&k_2 = f(t_n + c_2h, y_n + ha_{21}k_1) \\
&y_{n + 1} = y_n + h(b_1k_1 + b_2k_2) \\
\end{align}
Then Taylor expanding $k_2$ to $O(h^2)$ gives
\begin{align}
k_2 &= f + f_{t}c_2h + f_{y}ha_{21}k_1 + O(h^2) \\
&= f + (c_{2}f_{t} + a_{21}f_{y}f)h + O(h^2).
\end{align}
where the arguments $(t_n, y_n)$ are omitted.
Hence
\begin{align}
y_{n + 1} = y_{n} + (b_1 + b_2)fh + (b_{2}c_{2}f_{t} + b_{2}a_{21}f_{y}f)h^2 + O(h^3).
\end{align}
On the other hand,
\begin{align}
y(t_n + h) = y_{n} + y'(t_n)h + \frac{y''(t_n)}{2}h^2 + O(h^3).
\end{align}
Since $y'(t) = f(t, y(t))$, we get
\begin{align}
&y'(t_n) = f \\
&y''(t_n) = f_t \cdot 1 + f_yy' = f_t + f_{y}f.
\end{align}
Therefore
$$y(t_n + h) = y_n + fh + \frac{f_t + f_yf}{2}h^2 + O(h^3)$$
Matching terms gives
\begin{align}
&b_1 + b_2 = 1 \\
&b_2c_2 = \frac{1}{2} \\
&b_2a_{21} = \frac{1}{2}.
\end{align}
Heun's method is $c_2 = 1$, $b_1 = b_2 = \frac{1}{2}$, $a_{21} = 1$.


问:
$L^p$ integrability condition
I am trying to prove an equivalent condition for functions belonging in $L^p$. It's convergence of $\sum_{n\in\mathbb{Z}}2^{np}m(|f|>2^n)$. the one direction is an easy application of Chebyshev's inequality. However I am having trouble with the reverse direction. that is whenever the series converges, then $\|f\|_p<+\infty$.

答:
Let $ A_n = \{ |f| > 2^n \} $ so the sequence of sets $ A_n $ is decreasing.  Define $ B_n = A_n \backslash A_{n+1} $.  On $ B_n $ we have $ 2^n < |f| \leq 2^{n+1} $.  Define $ \psi $ by $ \psi(x) = 2^{n+1} $ on each $ B_n $, i.e.
$$
\psi = \sum_n {2^{n+1} \chi_{B_n}}
$$.  Now $ \psi $ dominates $ |f| $ and we have
$$
\int {|f|^p} \leq \int \psi^p = 2^p \sum_n {2^{np}|B_n|} \leq 2^p \sum_n {2^{np} \mu A_n} < \infty
$$
as desired.

Title: $L^p$ integrability condition


问:
Prove inequality transformation
I want to show that the inequality

(1) $\int_{0}^{\infty}\vert h(t,x) \vert \varphi(x)\leq e^{-\mu(t-y_{0})} \int_{0}^{\infty}\vert h(y_{0},x)\vert \varphi (x) dx < \infty$

with

(2) $  h(t,x)=\tilde{n}(t,x)-\int_{0}^{\infty}n^{0}(y)\phi(y)dyN(x)$

is equal to

(3) $\int_{0}^{\infty}\vert mN(x)-\tilde{n}(t,x)\vert \varphi(x)dx\leq C_{0}e^{-\mu(t-y_{0})}\int_{0}^{\infty}\vert mN(x)-\tilde{n}(y_{0},x)\vert \varphi(x)dx dx$

with

$m=\int_{0}^{\infty}n^{0}(x)\phi(x)dx.$

My idea was just du insert (2) into (1), but then I don't know where the $C_0$ and the second dx gets from. And I'm not sure if the second dx is a mistake or there on purpose.

答:
Given that
$$m=\int_{0}^{\infty}n^{0}(x)\phi(x)\ \mathrm dx$$
We have
$$h(t,x)=\tilde{n}(t,x)-mN(x)$$
For the left hand side, we have
$$\int_{0}^{\infty}\left|h(t,x)\right|\varphi(x)\ \mathrm dx$$
$$=\int_{0}^{\infty}\left|\tilde{n}(t,x)-mN(x)\right|\varphi(x)\ \mathrm dx$$
For the right hand side, we have
$$e^{-\mu\left(t-y_0\right)}\int_{0}^{\infty}\left|h(y_0,x)\right|\varphi(x)\ \mathrm dx$$
$$=e^{-\mu\left(t-y_0\right)}\int_{0}^{\infty}\left|\tilde{n}(y_0,x)-mN(x)\right|\varphi(x)\ \mathrm dx$$
The extra $\mathrm dx$ on the right hand side of your post is a typo and $C_0$ is most likely a positive constant. Without more information, we cannot tell where $C_0$ actually comes from. You were pretty much on the right track.

Title: Prove inequality transformation


问:
Finding asymptotes to general functions
I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:

Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functions.

From pre-calculus I recall finding asymptotes of rational functions, but how do I find vertical, horizontal, and oblique asymptotes of a more general function (such as an exponential or square root, etc.)?

答:
In general, the Taylor Series is the last thing you should be looking at.

In fact, the Taylor Series was not looked at; they calculated the Laurent Series. The point is that:

$$\operatorname{e}^y = 1 + y + \frac{1}{2}y^2 + \frac{1}{3!}y^3 + \cdots $$

Replacing $y$ with $2/x$ gives:

\begin{array}{ccc}
\operatorname{e}^{2/x} &=& 1 + \frac{2}{x} + \frac{1}{2}\left(\frac{2}{x}\right)^2 + \frac{1}{3!}\left(\frac{2}{x}\right)^3 + \cdots \\ \\
&=& 1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots
\end{array}

The actual function under consideration was $x\operatorname{e}^{2/x}+1$ and so we have:

\begin{array}{ccc}
x\operatorname{e}^{2/x}+1 &=& x\left(1 + \frac{2}{x} + \frac{2}{x^2} + \frac{4}{3x^3} + \cdots\right) + 1 \\ \\
&=& \left(x + 2 + \frac{2}{x} + \frac{4}{3x^2} + \cdots\right) + 1 \\ \\
&=& x + 3 + \frac{2}{x} + \frac{4}{3x^2} + \cdots
\end{array}

The point here is that as $x$ gets larger, anything with an $x$ in the denominator gets smaller: $1/x \to 0$ as $x \to \infty$. So as $x \to \infty$ we have $2/x \to 0$, $4/3x^2 \to 0$, etc. All of "the tail" gets smaller and smaller very quickly. The only terms that we are left with are the $x$ and the $3$.

There is a general definition for an asymptote. They don't even need to be lines.

Title: Finding asymptotes to general functions


问:
use power series expansion to prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$
prove $e^{z_1+z_2}=e^{z_1}e^{z_2}$

hint: If $\sum_{n=0}^{\infty} a_n \sum_{n=0}^{\infty} b_n = \sum_{n=0}^{\infty} c_n$ then

$ c_n=\sum_{k=0}^{n} a_k b_{n-k} $ provided that $ \sum_{n=0}^{\infty} a_n , \sum_{n=0}^{\infty} b_n$  are convergent

using power series expansion: $e^{z}= \sum_{n=0}^{\infty} {z^n\over n!}$

$e^{z_1}= \sum_{n=0}^{\infty} {z_1^n\over n!}$ and $e^{z_2}= \sum_{n=0}^{\infty} {z_2^n\over n!}$

$e^{z_1}e^{z_2}= \sum_{n=0}^{\infty} {z_1^n\over n!} \sum_{n=0}^{\infty} {z_2^n\over n!}=\sum_{k=0}^{n} {z_1^k\over k!}  {z_2^{n-k}\over {n-k}!} = e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}$ [Using binomial thm.]

Now, $e^{z_1+z_2}= \sum_{n=0}^{\infty} {(z_1+z_2)^n\over n!}=e^{z_1}e^{z_2}$
but iam not sure of the binomial thm part

答:
We obtain
\begin{align*}
\color{blue}{e^{z_1}e^{z_2}}&=\sum_{k=0}^\infty\frac{z_1^k}{k!}\sum_{l=0}^{\infty}\frac{z_2^l}{l!}\\
&=\sum_{n=0}^\infty\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{z_1^kz_2^l}{k!l!}\tag{1}\\
&=\sum_{n=0}^\infty\sum_{k=0}^n\frac{z_1^kz_2^{n-k}}{k!(n-k)!}\tag{2}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{n!}{k!(n-k)!}z_1^kz_2^{n-k}\right)\frac{1}{n!}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}z_1^kz_2^{n-k}\right)\frac{1}{n!}\\
&=\sum_{n=0}^\infty\frac{(z_1+z_2)^{n}}{n!}\tag{3}\\
&\,\,\color{blue}{=e^{z_1+z_2}}
\end{align*}
and the claim follows.

Comment:

In (1) we change the order of summation by introducing $n$ and summing up according to increasing $n$.

In (2) we set $l=n-k$.

In (3) we apply the binomial theorem.


问:
Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$
If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$

So $x+y+z=0$

We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.

答:
Hint

Define
$$
x={a-b\over b-c}\quad,\quad y={b-c\over c-a}\quad,\quad z={c-a\over a-b}
$$
and try to minimize $x^4+y^4+z^4$ constrained to $xyz=1$.

Alternatively, you can also define
$$
x=\ln\left|{a-b\over b-c}\right|\quad,\quad y=\ln\left|{b-c\over c-a}\right|\quad,\quad z=\ln\left|{c-a\over a-b}\right|
$$
and minimize $e^{4x}+e^{4y}+e^{4z}$ constrained to $x+y+z=0$.

Edit

Thanks to @RiverLi's comment below, also three more constraints should be included:
$$
{y(x+1)+1=0
\\
z(y+1)+1=0
\\
x(z+1)+1=0
}
$$


问:
Number of solutions of the differential equation: $(1-t)y\frac{dy}{dt} = 1 - y^2, y(1/2) = 2 $
\begin{array}{lcl} (1-t)y\frac{dy}{dt} = 1 - y^2 \\ y(1/2) = 2 \end{array}

Given this  differential equation, I was able to prove that there are multiple solutions that pass through the given point, by joining the solutions $y = \pm \sqrt{1 + 12(1-t)^2}$ and the solutions $ y = \pm 1 $, but not infinite, as it was stated in the exercise.
Is the plot with the red line a solution, too?

Could someone explain me what is wrong with my answer?

Thanks in advance

答:
$$\begin{array}{lcl} (1-t)y\dfrac{dy}{dt} = 1 - y^2 \\ y\left( \dfrac 12 \right) = 2 \end{array}$$
And we also have:
$$\frac{ydy}{1-y^2} =\dfrac {dt}{ 1 -t}$$
$$\frac{dy^2}{y^2-1} =\dfrac {2dt}{ t-1}$$
$$\ln  |{y^2-1}| =2\ln |{ t-1}|+C$$
$$ (y^2-1) =K ( t-1)^2$$
$$ y^2 =1+K ( t-1)^2$$
And $K=12$ from initial condition $(y(1/2) = 2 )$. As @Raffaele commented the only solution is:
$$ y =\sqrt {1+12 ( t-1)^2}$$
Your solution looks good to me. I see no mistakes.

$y=\pm 1$ are both solutions of the DE. But they don't satisfy the initial condition.


问:
Product of functions with mean zero has mean zero?
Let $f,g:[0,L] \subset \mathbb{R} \longrightarrow \mathbb{R} $ be a differentiable and periodics function, with period $L>0$.

Question 1. If $f$ and $g$ has mean zero, that is,
$$ \int_0^L f(x) \; dx=\int_0^L g(x) \; dx=0$$
then
$$\int_0^L f(x)\cdot g(x) \; dx=0?$$

I know that $f'$ and $g'$ has mean zero, since
$$\int_0^L f'(x) \; dx=f(L)-f(0)=0 \quad \text{and} \quad \int_0^L g'(x) \; dx=g(L)-g(0)=0. \tag{1}$$

In this question, the only idea I had was to use the fact $(1)$ and integration by parts, but did not get anything.

Question 2. If there exists a differentiable periodic function $\varphi:[0,L] \subset \mathbb{R} \longrightarrow \mathbb{R} $, with period $L>0$, mean zero and derivative $\varphi '$ differentiable , such that
$$\int_0^L \varphi''(x)f(x) \; dx + \int_0^L \varphi'(x) g(x)\; dx=0, $$
then $f$ and $g$ has mean zero?

In the last question, I don't have ideas. But it does not seem to me to be valid in general.

Body:
Another counter example:

$$\int_0^2{(x-1)}{dx} = \int_0^2{(1-x)}dx = 0$$

But:

$$\int_0^2{(x-1)(1-x)}dx = -\int_0^2{(x^2+2x+1)}dx  = -8/3 -4 - 2$$


问:
Prove that the eigenvalues of skew-Hermitian matrices are purely imaginary
I would like some help on proving that the eigenvalues of skew-Hermitian matrices are all pure imaginary. I have gotten started on it, but am getting stuck.

Attempt at proof: $Av=\lambda v \implies A \bar{v}=\bar{\lambda}\bar{v}.$

Also, $v^TA^T=\lambda v^T \implies v^TA^T\bar{v}=\lambda v^T \bar{v} \implies ?$

Should I conjugate both sides next?

答:
Consider the Euclidean inner product of $x,y$ $\in$ Complex vectors $C^n$    defined as : $$\langle x,y \rangle = x^Hy$$
Where $x^H$ is the hermitian transpose of x.We also know $\overline{\lambda}$ is eigenvalue of $x^H$: $$Ax = \lambda x$$
$$(Ax)^H = (\lambda x)^H$$
$$x^H A^H = \overline{\lambda} x^H$$
Now:
$$\langle Ax,x \rangle = (\lambda x)^H x = \overline{\lambda} x^H x$$
Also :
$$\langle Ax,x \rangle = (Ax)^H x = x^H A^H x = x^H(A^Hx)$$
Since we know $A^H = -A$:
$$x^H(A^Hx) = x^H(-Ax) = - x^H(\lambda x) = -\lambda x^H x$$
Therefore:
$$-\lambda =  \overline{\lambda}$$
And $\lambda$ is purely imaginray.


问:
Questions about homotopy lifting theorem.
I’m reading the following material and would like to ask some clarification.

Theorem B8.5 (Homotopy lifting). Let $p: \tilde{X} \rightarrow X$ be a covering, and $H: Y \times I \rightarrow X$ be a homotopy between maps $f_{0}, f_{1}: Y \rightarrow X .$ Given any lift $\tilde{f}_{0}: Y \rightarrow \tilde{X}$ of $f_{0},$ there is $a$
unique lift $\tilde{H}: Y \times X \rightarrow \tilde{X}$ of $H$ agreeing with $\tilde{f}_{0}$ on $Y \times\{0\}$

Proof. For each $y \in Y,$ there is a unique lift of the path $H(y, \cdot): I \rightarrow X$ starting at $\tilde{f}_{0}(y) .$ Collecting these lifts, we get a possibly discontinuous function $\tilde{H}: Y \times I \rightarrow \tilde{X} ;$ we just need to show that this $\tilde{H}$ is continuous. Fixing any $y \in Y$ it suffices to find a neighborhood $N \ni y$ such that $\tilde{H}$ is continuous on $N \times I$.

Consider a cover $\left\{U_{\alpha}\right\}$ of $X$ by trivializing neighborhoods for the covering $p: \tilde{X} \rightarrow X .$ The pre-images $H^{-1} U_{\alpha}$ cover $Y \times I .$ By the corollary to the Lebesgue lemma, we can find a neighborhood $N$ of $y$ and a partition $\left\{t_{i}\right\}$ of $I$ such that each $H\left(N \times\left[t_{i-1}, t_{i}\right]\right)$ is contained in a trivializing neighborhood that we call $U_{i}$.

Start with $\tilde{H}=\tilde{f}_{0}$ on $N \times\left\{t_{0}\right\}$ and proceed by induction on $i=1, \ldots, n .$ If we know $\tilde{H}$ on $N \times\left\{t_{i-1}\right\},$ then we look at the trivialization of the covering over $U_{i} .$ On $N \times\left\{t_{i-1}\right\},$ the lift $\tilde{H}$ is given by a continuous map to the fiber $F ;$ since the interval $\left[t_{i-1}, t_{i}\right]$ is connected, the only way to extend the lift to $N \times\left[t_{i-1}, t_{i}\right]$ is to make it constant in $t .$ Thus $\tilde{H}$ is continuous on $N \times\left[t_{i-1}, t_{i}\right],$ and in particular on $N \times\left\{t_{i}\right\}$, which is what we need to continue the induction. Once we reach $t_{n}=1,$ we see that $\tilde{H}$ is continuous on all of $N \times I$ as claimed.

Here’s the statement of the corollary to Lebesgue’s lemma:

Corollary B7.3. Suppose $Y$ is a topological space, $\left\{U_{\alpha}\right\}$ is an open cover of $Y \times I$, and $y \in Y$ is any point. Then there exists a neighborhood $N$ с $Y$ of $y$ and a finite partition $0=t_{0}<t_{1}<\cdots<t_{n}=1$ of I such that each $N \times\left[t_{i-1}, t_{i}\right]$
is contained in some $U_{\alpha}$.

My questions are as follows:

Is it absolutely necessary that the interval $I$ in $N \times I$ is partitioned into several intervals $[t_{i-1},t_i]$? Shouldn’t $H(N \times I)$ already fit into some $U_\alpha$? The reason for this question is because I’m picturing the pre-images $H^{-1}U_\alpha$ as $U_\alpha \times I$ (so they are like “cylinders”, and $N \times I$ would just fit into the intersection of some of those).

I’m not clear about the last part of the argument, in particular this statement:

On $N \times\left\{t_{i-1}\right\},$ the lift $\tilde{H}$ is given by a continuous map to the fiber $F ;$ since the interval $\left[t_{i-1}, t_{i}\right]$ is connected, the only way to extend the lift to $N \times\left[t_{i-1}, t_{i}\right]$ is to make it constant in $t .$

It’d be great if someone could explain. (Also as a side note, another argument is that the lift is given by a locally constant function to $F$, hence the conclusion. But in an earlier part of the text it specifies that the function to $F$ is locally constant if the fiber is discrete. How does it extend to this setting, where the fiber is $I$?)

答:
Here is a simple example which exhibits that the answers to your questions in 1 are "no".

Let $Y = I$, and let $U_\alpha$, for each point $\alpha = (y,t) \in Y \times I$, be the box neighborhood
$$U_\alpha = (I \times I) \cap \bigl((y-.001,y+.001) \times (t-.001,t+.001)\bigr)
$$
You can see, for example, that its going to take a few hundred thousand separate elements of the collection $\{U_\alpha\}$ just to cover $Y \times I = I \times I$, because each $U_\alpha$ has area $\le .002 \times .002 = .000004$ and $I \times I$ has area $1$.

Now pick any $y \in Y$, and any neighborhood $N \subset Y$ of $y$. Then there does not exist any $\alpha$ such that $y \times I \subset U_\alpha$, and therefore there also does not exist any $\alpha$ such that $N \times I \subset U_\alpha$. I fact, you can see that its going to at least a five hundred separate $U_\alpha$'s in order to cover just $y \times I$, because each $U_\alpha$ has height $\le .003$ whereas $y \times I$ has height $1$. So it will also take at least five hundred separate $U_\alpha$'s to cover $N \times I$.

Perhaps what sent you astray is in thinking that $H^{-1} U_\alpha$ has the form $U_\alpha \times I \subset Y \times I$. That doesn't even make sense, $U_\alpha$ is a subset of $X$, not of $Y$. (I also worry that the clashing uses of $U_\alpha$ in Theorem B8.5 and Corollary B7.3 could be confusing).

Regarding 2, the phrase "...is to make it constant in $t$" is ambiguous: What is it? To make this clearer, you should read that phrase as "is to make the lift to the fiber constant in $t$". Keep in mind, the fiber $F$ that we are working with in this passage of the proof is the one in the formula $p^{-1}(U_\alpha) \approx U_\alpha \times F$ which comes from the definition of a "trivializing neighborhood". What we are trying to construct is the continuous function $\tilde H : N \times [t_{i-1},t_i] \to U_\alpha \times F$. Since $F$ is discrete, every continuous function to $F$ from a connected space (such as $[t_{i-1},t_i]$) is constant.


问:
Showing subset is dense
Let $X$ be finite discrete metric space. Since it is finite and discrete every map from $X$ to real numbers will be bounded and continous. Let $C$ be the set of such functions. I want to show that $C$ is seperable where $C$ is metric space with respect to the norm $||f|| = \sup{f}$.

I consider the set $B =\{f :  \text{the image of $f$ is the subset of rational numbers}\}$

I feel that $C$ has same cardinality as $\mathbb{R^n}$ and $B$ has same cardinality as $\mathbb{Q^n}$ where $n$ is cardinality of $X$. Since $\mathbb{Q^n}$ is seperable in $\mathbb{R^n}$, is it enough to prove that these sets have same cardinal numbers as mentioned above and conclude that $B$ is seperable in $C$?

Or

How can I show that any function which is not in $B$ is limit point of $B$?

答:
In fact $C$ is homeomorphic to $\Bbb R^n$. If $X=\{x_1,\ldots,x_n\}$, let

$$\varphi:C\to\Bbb R^n:f\mapsto\langle f(x_1),\ldots,f(x_n)\rangle\,;$$

then $\varphi$ is a homeomorphism, and since $\Bbb Q^n$ is dense in $\Bbb R^n$, $\varphi^{-1}\left[\Bbb Q^n\right]$ is a countable dense subset of $C$. The metric induced by the norm on $C$ even translates nicely to $\Bbb R^n$: if

$$B(f,\epsilon)=\{g\in C:\|g-f\|<\epsilon\}$$

is the open $\epsilon$-ball centred at $f$ in $C$, then

$$\varphi[B(f,\epsilon)]=\prod_{k=1}^n\big(f(x_k)-\epsilon,f(x_k)+\epsilon\big)\,,$$

the open box of side length $2\epsilon$ centred at $\varphi(f)$.

Title: Showing subset is dense


问:
What is the upper bound for edges that are not going in or out of node 1 in this graph?
I have the following graph with the following problem:

We want to obtain a Hamiltonian cycle by supplying all nodes with 1 unit of flow. Initially the node 1 will send 4 units of flow through one and only one of its edge to the next node. The next node consumes 1 unit of flow and sends the rest through one and only one of its edge to the next node and so on until we arrive back at node 1. Node 1 will then consume the last unit of flow.

The upper bound of flow that you can send through one of the edges going out of node 1 is 4 units and the upper bound of flow that comes back through one of the edge to node 1 is 1 unit. I want to know what would be the upper bound of flow that can be sent on edges that are not connected to node 1.

答: An upper bound is $3$, more generally $n-1$ for a graph with $n$ nodes.


问:
Why second order logic is more expressive that the first order logic?
As far as I understand, the second order logic is more expressive than the first order logic because it can make statements about predicates. However, I do not understand why this problem (impossibility to make statements about predicates) cannot be solved within the first order logic.

Let's assume that we have a first order logic system with some predicates. For example we can have the following statements in it:

```<code>CapitalOf(Berlin, Germany)
HasBorderWith(Germany, France)
CapitalOf(Paris, France)
Country(Germany)
City(Paris)
</code>```

So, in the above example we have the following predicates: CapitalOf, HasBorderWith, City, Country. Now, let's assume that we want / need to make some statements about predicates. FOL syntax does not allow it. But we can reformulate all our statements in the following way:

```<code>Relation(CapitalOf, Berlin, Germany)
Relation(HasBorderWith, Germany, France)
Relation(CapitalOf, Paris, France)
Class(Country, Germany)
Class(City, Paris)
</code>```

Now, all the predicates (CapitalOf, HasBorderWith, City, Country) have the same status as entities (Germany, France, Berlin, Paris). So, what can us stop now from making statements about the predicates using FOL?

答:
You're misunderstanding what second-order logic does. I think part of this is due to the use of the term "predicate," which is unfortunate since it has connotations which are not relevant here; it's better to describe second-order logic as allowing quantification over all subsets of, relations on, and fuctions on the domain. Crucially we're quantifying over all such second-order objects, not just - for example - the definable ones, or the ones specifically named in the language of the structure itself.

Here's a specific example of a second-order sentence with no first-order equivalent:

Every injection of the domain to itself is a surjection.

In symbols:

$\forall F[(\forall x,y(F(x)=F(y)\rightarrow x=y))\rightarrow (\forall x\exists y(F(y)=x))].$

This sentence is true in a given structure exactly when that structure is finite. However, by the compactness theorem for first-order logic there is no first-order sentence which does this: if $\varphi$ is a first-order sentence with arbitrarily large finite models, then $\varphi$ has an infinite model.

Here's another example. Suppose my language has two unary predicate symbols $U$ and $V$. The statement "$U$ and $V$ hold of the same number of elements" is second-order expressible ("There exists a function which restricts to a bijection from $U$ to $V$") but - again via compactness - is not first-order expressible.


问:
Laurent Series and residue of $\frac{z}{(z-1)(z-3)}$ around z = 3
As mentionned in the title, I'd like to get the function's Laurent series and after its residue, I have tried to separate the two denominators to get a partial fraction but I still have a z at numerator that's bothering me ...

So far : $$
        \frac{z}{(z-1)(z-3)} = \frac{-z}{2(z-1)} + \frac{3z}{(z-3)}
$$

I can obviously try $\omega = z-3$ and change my variable but I don't feel it'll will work ... Because I will still have this z on top of my fraction ...

Could you help me ?
Thank you very much ;)

Body:
Hint:
$$
\frac{3z}{z - 3} = \frac{3(z - 3) + 9}{z - 3}
$$
\begin{align*}
\frac{-z}{2(z - 1)} &= \frac{1}{2}\frac{-(z - 3) - 3}{(z - 3) + 2}\\
&= -\frac{1}{2}\frac{z - 3}{(z - 3) + 2} - \frac{1}{2}\frac{3}{(z - 3) + 2}\\
&= -\frac{z - 3}{4}\cdot\frac{1}{(z - 3)/2 + 1} - \frac{3}{4}\cdot\frac{1}{(z - 3)/2 + 1}
\end{align*}
I think you should be able to expand these with the help of well-known power series...


问:
How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?
I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem?

$$
\lim_{n \to +\infty} n^{\frac{1}{n}}
$$

答:
Applying the Binomial Theorem, we can say
$$
\begin{align}
\left(1+\sqrt{\frac2n}\right)^n
&\ge\color{#C00}{1}+\overbrace{\binom{n}{1}\sqrt{\frac2n}}^{\sqrt{2n}}+\overbrace{\ \binom{n}{2}\frac2n\ }^{\color{#C00}{n-1}}\tag{1a}\\[6pt]
&\ge\color{#C00}{n}\tag{1b}
\end{align}
$$
Therefore,
$$
1\le n^{1/n}\le1+\sqrt{\frac2n}\tag2
$$
to which we can apply the Squeeze Theorem.