问:
Convergence of flows of uniformly converging vector fields
My questions is about the flows of certain "well behaved" sequence of vector fields. Let $X^n: U \subset R^n \rightarrow R^n$ be a sequence of smooth vector fields which

i. Are equicontinous and equibounded

ii. $X^n$(x) converges fastly and uniformly (with "speed" $||X^{n+k}(p)-X^k(p)||< r^{k}$ for r<1 in $C^{0}$ manner) to some Holder contiunous vector field X(x)

My question is, does the flows of $X^n$ starting at a point $x \in M$ converge to a flow of X (for bounded time if needed). Now I know that flows of X need not be unique since it is just continuous but it will have flows by Peano curve theorem. And I can only prove the convergence if I have that $X^n$ are equlipschitz which is not the case because the limit is not Lipschitz for certain. What I can do is for bounded time, only find a subsequence among the integral curves which converge to an integral curve of X using Arzela Ascoli theorems (you can also show integral curves are equi-lipschitz for bounded time).

It is hard to give a counter example to this too because the vector fields satisfy many properties.

(p.s: I will give points to the first person to give a counter example to give a necessary condition or to point me in the right direction or otherwise if not I will just pick one best answer).

(p.s2: I know that if the convergence is in $C^1$ topology for instance you can show this through gronwall inequality (for fixed time). But all the references I could find or all the proofs I could give is always requires the derivatives to be uniformly Lipschitz or the limit to be Lipschitz or convergence to be in $C^1$ or higher topology)

Thanks

答:
One-dimensional counterexample. Let $U=(-1,1)$, $X(x)=x^{1/3}$,
$$X^n(x)=(x^2+\epsilon_n^2)^{-1/3}(x+(-1)^n \epsilon_n) \tag1$$
where you can choose $\epsilon_n>0$ converging to $0$ as fast as you want. Consider the solution curve of $X^n$ beginning at $0$, denoted $x_n(t)$. If $n$ is even, then $x_n'(0)$ is positive, which makes $x_n$ increasing for all times (solutions of autonomous equations don't turn back). If $n$ is odd, then $x_n'(0)<0$ and $x_n$ is decreasing for all times.

Thus, the only way for the sequence $x_n(1)$ to have a limit is to converge to $0$. But in fact, $x_n(1)>1/3$ for even $n$. Indeed, assuming $n$ is even, the value  of $ x_n(1)$  is determined from the equation $$\int_0^{x_n(1)} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx=1 \tag2$$
Since
$$\begin{split} &\int_0^{1/3} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx \\ &=
\int_0^{\epsilon_n} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx+ \int_{\epsilon_n}^{1/3} (x^2+\epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx
\\ &\le \int_0^{\epsilon_n} (2 \epsilon_n^2)^{1/3}(x+ \epsilon_n)^{-1} \,dx+ \int_{0}^{1/3} (2x^2 )^{1/3}(x)^{-1} \,dx \\& = (2\epsilon^2)^{1/3}\log 2+3^{1/3}\cdot 2^{-2/3}
 \end{split}$$
is less than $1$ for small $\epsilon$, it follows that $x_n(1)>1/3$. $\Box$

Of course, both subsequences $x_{2n}$ and $x_{2n+1}$ converge to (different) solution curves of $X$. But this is something you already know from Arzelà-Ascoli.


问:
Confidence interval for a Sum approximated with the central limit theorem
According to the central limit theorem: $S\approx N(n*\mu, n*\sigma^2)$

Normally, a confidence interval is calculated according to:

$(μ \pm 1.96*\sqrt{\frac{\sigma^2}{n}})$

Do you divide $\sigma^2$ by n in the case of an approximation of the sum?
I have seen cases, where the interval is calculated as:

$(μ \pm 1.96*\sqrt{\sigma^2})$

What is the right way to calculated the confidence interval in this case? Is the approach without n wrong or does it apply to some cases?

Thank you

答:
There seems to be some confusion about the distinction between $\mu$ and its estimate $\bar X,$ between cases where the numerical value of $\sigma$ is and is not known, and between sample size $n$ and $4.$ Perhaps the following step-by-step approach will help.

If data are a random sample from $\mathsf{Norm}(\mu, \sigma),$ where the second argument is the standard deviation, then

$$\bar X_n = \frac 1n\sum_{i=1}^n  \frac{X_i}{n}
\sim \mathsf{Norm}\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$$

Thus a 95% confidence interval (CI) for $\mu$ (when the numerical value of $\sigma$ is known) is of the form
$$\bar X_n \pm 1.96\frac{\sigma}{\sqrt{n}},$$
where 1.96 cuts probability 0.025 from the upper tail
of the (symmetrical) standard normal distribution.

Notice that the center of this CI is the sample mean $\bar X_n,$ which estimates the population mean $\mu.$

Furthermore, if $\sigma$ is unknown and estimated by the sample standard deviation
$S_n = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i=\bar X_n)^2},$ then
a 95% CI for $\mu$ is of the form
$$\bar X_n \pm t^*\frac{S_n}{\sqrt{n}},$$
where $t^*$ cuts probability 0.025 from the upper tail
of Student's t distribution (also symmetrical) with degrees of freedom $\nu = n-1.$

Finally, if you are dealing with $T_4 = \sum_{i=1}^4 X_i,$ then $T_4\sim\mathsf{Norm}(\mu_4 = 4\mu, \sigma_4 = 4\sigma).$ So if you want a 95% CI for $E(T_4) = \mu_4 = 4\mu,$ then you simply multiply by $4$ the endpoints of the corresponding CI for $\mu.$ In the case where the
numerical value of $\sigma$ is known, this would be:

$$4\bar X_n \pm 1.96\frac{4\sigma}{\sqrt{n}}.$$

In particular, if $n = 4,$ then the CI becomes

$$T_4 \pm 1.96\sqrt{4\sigma^2},$$
which agrees with @Henry's first comment,

ContentLicense: CC BY-SA 4.0


问:
$\sigma$-weak topology versus $\sigma$-strong topology on the bounded operators $B(H)$.
Let $H$ be a Hilbert space and let $T(H)$ be the traces class operators on $H$. Denote the bounded operators on $H$ by $B(H)$. We define the normal functionals on $B(H)$ by
$$B(H)_*:= \{\omega \in B(H)^*: \exists x \in T(H): \forall y \in B(H): \omega( y) = \text{Tr}(xy)\}$$
Let $B(H)_*^+$ denote the positive normal functionals.

We can introduce the $\sigma$-weak and the $\sigma$-strong topology on the space $B(H)$. The former is the locally convex topology generated by the family of seminorms
$$ B(H)\ni x \mapsto |\omega(x)|, \  \omega\in B(H)_*$$
The latter is the locally convex topology generated by the family of seminorms
$$B(H)\ni x \mapsto  \omega(x^*x)^{1/2}, \ \omega \in B(H)_*^+$$

I want to show that the $\sigma$-weak topology is weaker than the $\sigma$-strong topology (as the name suggests!). More precisely, suppose that $x_\alpha \to x$ in the $\sigma$-strong topology. Then I want to show also that $x_\alpha \to x$ in the $\sigma$-weak topology. How can I show this?

We need to show
$$\forall \omega \in B(H)_*: |\omega(x_\alpha)-\omega(x)| \to 0$$

using something about positive normal functionals but I'm unsure how to approach this.

答:
You have, for trace-class $z$ and using Cauchy-Schwarz, and writing $z=v|z|$ for its polar decomposition,
\begin{align}
|\operatorname{Tr}((x_\alpha-x)z)
&= \operatorname{Tr}(|z|^{1/2}\,(x_\alpha-x)v|z|^{1/2}\,)\\[0.3cm]
&\leq \operatorname{Tr}(|z|^{1/2}|z|^{1/2})^{1/2}\operatorname{Tr}(|z|^{1/2}v^*(x_\alpha-x)^*(x_\alpha-x)v|z|^{1/2})^{1/2}\\[0.3cm]
&=\operatorname{Tr}(|z|)^{1/2}\operatorname{Tr}(zv^*(x_\alpha-x)^*(x_\alpha-x))^{1/2}\\[0.3cm]
&\to0.
\end{align}


问:
Calculating conditional expectation of 2 vectors from dataset
So I'm having trouble understanding what I'm doing wrong here.

For context, I have some velocity components in my dataset for turbulence (simplified).

I have flattened them out so my 3 velocity vectors are A, B, C, which are all 1*N vectors now. And at each point they have a different values on a continuous spectrum (not discrete values).

So I need to find the Conditional Expectation of A conditioned on B. $E[A|B=b]$

And I have this formula to use:

$$\mu_A + \Sigma_{AB} (\Sigma_{BB})^{-1}*(b-\mu_B)$$

I'd expect if I draw a single b from the data set and ran it through that equation above, it would give me a single number for a. But my covariance matrices must not be the right dimensions because I am getting some dimension mismatch errors and even if it works it's giving me a 2x2 vector or something like that rather than a single number.

Your help/guidance would be greatly appreciated!

答:
If A, B, and C are the three components of a velocity field taken as three scalar fields, then the covariance matrices are all 1x1 like:
$\Sigma_{AB}=(A-\mu_A [1\ 1 ... 1]^T)^T (B-\mu_B[1\  1 ... 1]^T).$

So you should have got all 1x1 matrices in your expression.


问:
Solving $x^{\log(x)}=\frac{x^3}{100}$
How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?

ContentLicense: CC BY-SA 3.0

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$

Body: Since $(\log(x))^2=\log (x^{\log x})=\log (x^3/100)=3\log(x)-2$, we have $(\log(x))^2-3\log(x)+2=0$. Hence, $\log(x)=2$ and $\log(x)=1$. Therefore, $x=100$ atau $x=10$


问:
Does $M \oplus N \cong M$ imply that $N = 0$ for modules $M,N$?
Let $R$ be a commutative ring and $M,N$ be $R$-modules. Does $M \oplus N \cong M$ imply that $N = 0$?

I would think that  $M \oplus N \cong M$ implies that $(M \oplus N)/M \cong M/M = 0$, but is that true? And if yes, how would the first isomorphism induce the second one? Sorry if this question has been answered before, I could not find an answer.

Edit: And what if $M$ is assumed to be finitely generated?

答:
Assume that $M$ is finitely generated. Let $f:M \xrightarrow{\cong} M \oplus N$ be an isomorphism and let $p:M \oplus N \to M$ be the projection on the first factor.
Now $p \circ f:M \to M$ is a surjective endomorphism. It is hence an isomorphism, see here. Now as $p\circ f$ and $f$ are both isomorphisms, so is $p = (p \circ f) \circ f^{-1}$. But the kernel of $p$ is $N$, so $N=0$


问:
Show $h(\omega_1,\omega_2)=f_1(\omega_1)f_2(\omega_2)$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.
The question is from a textbook in Chinese that I use for supplementary reading.

Let $f_{i}(\omega_{i})$ be measurable on $\sigma$ -finite measure spaces $(\Omega_{i}, \mathcal{A}_{i}, \mu_{i}), i=1,2 .$ Show

$g_{i}\left(\omega_{1}, \omega_{2}\right)=f_{i}\left(\omega_{i}\right), i=1,2$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.

$h(\omega_{1}, \omega_{2})=f_{1}(\omega_{1}) f_{2}(\omega_{2}).$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.

For the first part of the question, the definition looks very similar to how random vector is defined, so I think showing

$$\{g_i(\omega_1,\omega_2)<x_i,i=1,2\}\\=\{f_i(\omega_i)<x_i,i=1,2\}\\=\{(\omega_1,\omega_2):f_i(\omega_i)<x_i,i=1,2\}\\=\{\omega_1:f_1(\omega_1)<x_1\}\times\{\omega_2:f_2(\omega_2)<x_2\}\in\mathcal{A_1\times\mathcal{A_2}}$$
would be suffice. It is the second part of the question I am unsure about.

At first I tried to show by cases, and the outcome turns out to be very nusty. I tried to show $\{0<h(\omega_1,\omega_2)<x\}$ when $x>0$ is a union of some $A_1\times A_2$ that is supposed to be in $\mathcal{A_1}\times\mathcal{A_2}$,similar for $\{h(\omega_1,\omega_2)=x\}$, when $x=0$ and $\{h(\omega_1,\omega_2)\leq x\}$ when $x<0$. The written is very massive and I don't know if doing so make sense. (it would be nice if I can find expression for $\{h(\omega_1,\omega_2)<x\}$ but I kept on confusing myself.)

However, I do know in the one-dimensional case, the product of two measurable functions is still measurable. I wonder if I could simply say $f_1(\omega_1)=g_1(\omega_1,\omega_2)$ and $f_2(\omega_2)=g_2(\omega_1,\omega_2)$ is $\mathcal{A_1}\times\mathcal{A_2}$-measurable, so will their products. The problem is when I show the first part of the question, I thought $(g_1,g_2)$ as a whole, each representing a component. They are only $\mathcal{A_1}\times\mathcal{A_2}$-measurable when considering them together. In other words, I am not sure if my reasoning makes sense or not.

If either reasonings are not correct I am not very sure how to show the second part of the question. Could someone please explain to me how should I show the second part of the question?

Also, an additional question: most of the theorems stated in the book work in the other direction: provided a function is $\mathcal{A_1}\times\mathcal{A_2}$-measurable, its section or integration over $\Omega_i$ is also measurable provided certain condition hold. Are there any theorems that state the other direction? i.e. if $f_i$ is $\mathcal{A_i}$-measurable, through operations satisfies certain condition will result in a $A_1\times A_2$-measurable, or it is rather case by case.

答:
Just do the first part separately.

1.1. Let $g_1: \Omega_1 \times \Omega_2 \rightarrow \Bbb R$  be defined by  $g_{1}\left(\omega_{1}, \omega_{2}\right)=f_{1}\left(\omega_{1}\right)$. Then, for any $x_1 \in \Bbb R$, we have
$$ g_1^{-1}((-\infty, x_1)) =\{(\omega_{1}, \omega_{2}) \in \Omega_1 \times \Omega_2 : f_1(\omega_1)< x_1\} = f_1^{-1}((-\infty, x_1)) \times \Omega_2 \in \mathcal{A}_{1} \times \mathcal{A}_{2}$$
So $g_1$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.

1.2. Let $g_2: \Omega_1 \times \Omega_2 \rightarrow \Bbb R$  be defined by  $g_2\left(\omega_{1}, \omega_{2}\right)=f_2\left(\omega_{2}\right)$. Then, for any $x_2 \in \Bbb R$, we have
$$ g_2^{-1}((-\infty, x_2)) =\{(\omega_{1}, \omega_{2}) \in \Omega_1 \times \Omega_2 : f_2(\omega_2)< x_2\} =  \Omega_1 \times f_2^{-1}((-\infty, x_2))   \in \mathcal{A}_{1} \times \mathcal{A}_{2}$$
So $g_2$ is $\mathcal{A}_{1} \times \mathcal{A}_{2}$ -measurable.

Now, for the second part, just note that
$$h(\omega_{1}, \omega_{2})=f_{1}(\omega_{1}) f_{2}(\omega_{2})=g_{1}\left(\omega_{1}, \omega_{2}\right)g_{2}\left(\omega_{1}, \omega_{2}\right)$$
and then use the fact that the product of two measurable functions (from the same domain)  is still measurable.

Remark: Regarding your additional question. Yes, there are theorem that goes from the measurability of the "component" functions to the measurability of the "combined" function, but such theorems depend of how exactly the functions will be combined into the resulting function.


问:
Is Newton-Raphson the best we can do if we only know the derivative?
I was wondering if NR is the fastest method to find a root if all we know about a function is how to evaluate it and its derivative at any point.

Since you can use the first derivative to approximate the second I was wondering if this lets you converge on a root faster (as in $\epsilon_{n+1} \approx \epsilon_n ^3$)?

答:
I came across an answer to my own question, as detailed in the linked article the following modification of Newton-Raphson does indeed only use the first derivative and converges cubically: https://www.sciencedirect.com/science/article/pii/S0377042703003911

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n - \frac{f(x_n)}{2f'(x_{n})})}$


问:
Be $K$ a normed vector space. If $A \subset K$ is compact and $p \in K$, prove that $A_p = \{x+p:x \in A\} $ is also compact.
Be $E$ a normed vector space. If $A \subset E$ is compact and $p \in E$, prove that $A_p = \{x+p:x \in A\} $ is also compact.

Hello, everyone! All right? I tried to prove this exercise, but I believe my demonstration is not correct. Could someone tell me what theorem I use to make this demonstration?

答: Given $p$, the translation $T: \>x\mapsto x+p$ of $K$ is bijective and continuous, and so is $T^{-1}: \>x\mapsto x-p$. It follows that $T$ is a homeomorphism of $K$. This implies that the image of a compact set $A$ is again compact.


问:
Integral Contour
i have an integral to solve:
$\int_{-\infty}^\infty v^2 \frac{e^{-i\omega\tau}}{\omega^2-k^2v^2}d\omega$

I got solution $\frac{\pi v}{k} \sin(kv\tau)$, but in a book the solution was $\frac{2\pi v}{k} \sin (kv\tau)$, why i got answer only half of the true answer?
Here is my work

and this

and this is what i got from some books that the answer was twice from my answer:This

did i make a mistake?

答:
There is a variant of the Residue Theorem that states if a singularity is actually on the contour instead of inside it, it's residue is counted at half value (See, for example, George Arfken's "Mathematical Methods for Physicists", Section 7.2). Setting $$f(z) = v^2\dfrac{e^{-iz\tau}}{z^2-k^2v^2}$$, the residues are
$$R(f,kv) = \lim_{z\to kv}(z - kv)f(z) = \lim_{z\to kv}\dfrac{e^{-iz\tau}}{z+kv} = \dfrac{e^{-ikv\tau}}{2kv}$$
and similarly
$$R(f,-kv) = -\dfrac{e^{ikv\tau}}{2kv}$$

So by the Residue theorem and noting that the integral around the top arc $\to 0$ as the radius $\to \infty$, the Residue theorem says

$$\begin{align}\int_{-\infty}^\infty v^2f(z)dz &= v^2\pi i(R(f,-kv)+R(f,kv))\\
&=v^2\pi i\left(\dfrac{e^{-ikv\tau}}{2kv}-\dfrac{e^{ikv\tau}}{2kv}\right)\\
&=\frac{\pi v}k\sin{kv\tau}\end{align}$$

So the answer is: you are not wrong, the book is. That quote you included from another book doesn't even make sense. It says $C$ is "chosen as a closed path" to "avoid those poles". But it is claiming an integral on the real line which must pass through those poles. I'm guessing your books are copying from a single source who made the error and the copiers, already being aware of the basic concepts, never bothered to double check it themselves (or in the case of the quote, gave a vague hand-wavy explanation without bothering to think much about it).

Title: Integral Contour


答: Let $\{U_n\}$ be an open coverage of $A_p$. Let $t: x \mapsto x+p$ be the translation map, which is a homeomorphism. So $\{t^{-1}(U_n)\}$ is an open coverage of $A$. Since $A$ is compact there is a finite subcoverage ${\{t^{-1}(U_{n_1}),...,t^{-1}(U_{n_k})\}}$. For this reason ${\{U_{n_1},...,U_{n_k}\}}$ covers $A_p$.

Title: Be $K$ a normed vector space. If $A \subset K$ is compact and $p \in K$, prove that $A_p = \{x+p:x \in A\} $ is also compact.

Body:
Be $E$ a normed vector space. If $A \subset E$ is compact and $p \in E$, prove that $A_p = \{x+p:x \in A\} $ is also compact.

Hello, everyone! All right? I tried to prove this exercise, but I believe my demonstration is not correct. Could someone tell me what theorem I use to make this demonstration?


问:
Proof regarding limits

Okay so I'm new to calculus and still trying to grasp what I can and cannot do, and was wondering if the way I showed my counter example is correct, notably the domain part..
Also I'm desperate for tips on how to prove/disprove these type of questions and I would be extremely thankful if you could show me different patterns of proofs as I want to absorb as much as I can!

答:
You did a great job :)

Writing $x \in {\Bbb R}\setminus \{x_0\}\{0\}$ is not defined.
You should have just written $x \in {\Bbb R}\setminus \{x_0\}$, where $x_0 = 0$.

Title: Proof regarding limits


答:
$x^{\log(x)}=\frac{x^3}{100}$

Taking log on both sides you get :

$log(x) log(x) = log(\frac{x^3}{100})$ = log(x) log(x) = 3logx - 2log10 = 3logx -2

$\Rightarrow (log(x))^2  = 3logx -2 $

Now putting log(x) = t

$\Rightarrow t^2=3t-2$
Now you can solve for t as this is a quadratic in t. you get (t-2)(t-1) $\Rightarrow t = 2 ; t = 1$

$\Rightarrow logx = 2  \Rightarrow x = 100 $ ; and $ logx = 1  \Rightarrow x = 10$

Title: Solving $x^{\log(x)}=\frac{x^3}{100}$

Body:
How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?


问:
3-regular graph and two-way Euler circuit
A town-planner has built an isolated city whose road network consists of $2N$ roundabouts, each connecting exactly three roads. A series of tunnels and bridges ensure that all roads in the town meet only at roundabouts. All roads are two-way, and each roundabout is oriented clockwise.
Vlad has recently passed his driving test, and is nervous about roundabouts. He starts driving from his house, and always takes the first edit at each roundabout he encounters. It turns out his journey incluldes every road in the town in both directions before he arrives back at the starting point in the starting direction. For what values of $N$ is this possible?

I have tried to turn this into an equivalent graph theory problem in which we can apply some results on Euler circuits or similar, but with no such rephrasals seem useful. Any help appreciated!

答:
wece and later Vlad provided a proof that such a tour can exist for every odd $N$. We show that there can be no such tour for even $N$.

Embed the graph in a two-dimensional surface. In order for the notion "clockwise" to be well-defined, the surface must be an orientable one. Now the number of vertices is $2N$ and the number of edges is $3N$. If a tour like the one you've described exists, then the embedding can be regarded as a map with a single face (that has $6N$ sides). But the generalization of Euler's formula,
$$
V-E+F=2-2g,
$$
must hold, where $g$ is the genus of the surface on which the graph is embedded. So we get
$$
2N-3N+1=2-2g.
$$
This is a contradiction if $N$ is even.

Added: The desired embedding is achieved by drawing the graph on a sphere with handles, which is an orientable surface. To explain this a bit more, start by drawing the graph on the sphere. There will, in general, be some crossings of edges. The graph should be drawn in accordance with the specified clockwise ordering of edges at each vertex (roundabout). To enforce this ordering, even a planar graph may sometimes need to be drawn with edge crossings.

Remove or reroute edges (without violating edge-ordering constraints) until there are no more crossings. This can be done in such a way that the graph remains connected. Now add the removed edges back, one at a time: if an edge can be drawn within a single face, do so. (The face will be divided into two faces.) If it cannot, the insertion points of the edge lie in two different faces. Cut holes in each of these faces and join the holes with a tube. In this process the faces started as two surfaces each homeomorphic to a disk and ended as a single surface homeomorphic to a cylinder. Now route the edge across the cylinder, which cuts the cylinder so it is again homeomorphic to a disk.

Once all edges have been added back, we have the desired embedding of the graph in an orientable surface. This is a 2-cell embedding, meaning all faces are homeomorphic to disks, a property that is necessary in order to apply Euler's formula.

The ideas in this sketch come from the short article,

J. H. Lindsay, An Elementary Treatment of the Imbedding of a Graph in a Surface. The American Mathematical Monthly 66(2) (1959) 117-118.

and from Jack Edmond's masters thesis

Edmonds, John Robert (1960). A combinatorial representation for oriented polyhedral surfaces. University of Maryland.

A quotation from the latter:

Theorem 2. Given a connected linear graph with an arbitrrarily specified cyclic ordering of the edges to each vertex, there exists a topologically unique, two-sided polyhedron
Whose edges and vertices are the given graph and whose clockwise edge orderings at each vertex (with respect to one of
the sides) are as specified.

These ideas have a long history, going all the way back to Lothar Heffter in the 1890s. The notion of associating an embedding with a specification of the edge orderings at each vertex of a graph now goes by the name rotation system. If you want to try out the ideas, you can verify that there are essentially three different rotation systems for $K_4$, producing three different embeddings, one spherical (genus $0$) embedding with four triangular faces and two toroidal (genus $1$) embeddings, each with two faces—either a triangle and a nonagon or a quadrilateral and an octagon.


问:
Inner products and norms
I know that every inner product induces a norm (that induces a metric), and a norm is induced by an inner product if it satisfies the parallelogram law. I know that $x^Tx$ induces $\begin{Vmatrix}
x
\end{Vmatrix}_2^2$ by definition of norm $\begin{Vmatrix}
x
\end{Vmatrix}:=\sqrt{x^Tx}$, and at the same time $\begin{Vmatrix}
x
\end{Vmatrix}_2^2$ is the norm induced by inner product $x^Tx$. I also know that $\begin{Vmatrix}
x
\end{Vmatrix}_1^2$ and $\begin{Vmatrix}
x
\end{Vmatrix}_{\infty}^2$ are not induced by inner products because it's not true that $\forall x,y\in \mathbb{R}^n$

$$\begin{Vmatrix}
x+y
\end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix}
x-y
\end{Vmatrix}_{1,\infty}^2= 2(\begin{Vmatrix}
x
\end{Vmatrix}_{1,\infty}^2+\begin{Vmatrix}
y
\end{Vmatrix}_{1,\infty}^2).$$

Then, starting from $x^TAx$ with $A=LL^T$ the Cholesky decomposition of $A$, is there a way to get $\begin{Vmatrix}
x
\end{Vmatrix}_{1}^2$ and $\begin{Vmatrix}
x
\end{Vmatrix}_{\infty}^2$? Can I still write $\begin{Vmatrix}
x
\end{Vmatrix}_{1}^2=x^Tx$ or $\begin{Vmatrix}
x
\end{Vmatrix}_{\infty}^2=x^Tx$ even if these norms are not induced by $x^Tx$?

答: No. No matter which bilinear form $(x,y)\mapsto x^T A y$ you choose, a norm not satisfying the parallelogram identity can never be induced as $\|x\|=\sqrt{x^T A x}$.

Title: Inner products and norms


问:
Alternate way to solve this algebra problem?
Working on a problem that says given the absolute value of the difference of the roots of $ax^2 + bx + c$ as $2$, what is the absolute value of the difference of the roots of $ax^2 + 6bx + 36c$?

I reasoned that the difference of roots would just be $2\sqrt{b^2 - 4ac}$. Therefore, the value of $\sqrt{b^2 - 4ac}$ is $1$. Then, I used this information to answer the question:

$2\sqrt{36b^2 - 36(4ac)} =$ ?

$2\sqrt{36(b^2 - 4ac)}$

$12\sqrt{b^2 - 4ac}$

$=12$

The question hinted that I could use Vieta's formulas to solve the problem. Is my method correct, and is there a method involving Vieta's formulas?

答: Note that the difference of roots is actually $$\frac{\sqrt{b^2-4ac}}{|a|}$$ If the roots be $p$ and $q$, then $$|p-q| =\sqrt{(p-q)^2} = \sqrt{(p+q)^2 -4pq} = \sqrt{\frac{b^2}{a^2} -4\frac ca}$$ (using Vieta’s)


问:
Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else
Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$

Is it possible to use this fact to prove something like:

$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$

Just curious.  Thanks for the help!

答:
A simpler way to see that one of $a = -b$ or $b = -c$ or $c = -a$ is true.

wlog, we can assume $abc = 1$ (why?).

So we can assume $a,b,c$ are roots of $x^3 - px^2 + qx -1 = 0$.

Your relation gives us that $q = \frac{1}{p}$ (Vieta's formulas)

Thus we get $a,b,c$ are roots of

$$x^2(x - p) + \frac{1}{p}(x - p) = 0$$

Thus the roots are $$\pm\frac{1}{p}, p$$


问:
Find an equation for the tangent plane to the surface $2xz^2+3xy-4x=7$ at the point $(1,-1,2)$, by considering a gradient of a suitable scalar field.
This is what I've done, followed by a few questions I don't undestand;

$f(x, y, z) = 2xz^2-3xy-4x=7$

Surface is at 0, So therefore we set our equation to zero $2xz^2-3xy-4x-7=0$

We then take partials and evaluate them at the point.

$F_x = x-2, F_x(1,-1,2) = 1-2=3$

$F_y = 1, F_y(1,-1,2) = 1$

$F_z = 2z, F_z(1,-1,2) = 2(2) = 4$

Now the tangent plane is $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

=$-3(x-1)+1(y+1)+4(z-1) = 0$

=$-3x+y+4z-6=0$

My questions;

I found this method of computing online and it was done for a vector. Now given this is a scalar field would it still work?

In the end we have $-3x+y+4z-6=0$, what exactly do I need to solve for? Should I bring the equation back in the original form where it was equal to a number? Or do I solve for z, x or y?

Why is surface at zero and why do we take partials?

答:
I think you made a mistake in finding the gradient vector. From there on, given you have to find the scalar equation of the plane, you can do this way -

$f(x, y, z) = 2xz^2-3xy-4x-7 = 0$ (as called out in comments, the function in the title seems wrong.

$f_x = 2z^2 - 3y - 4$

$f_y = -3x$

$f_z = 4xz$

As the gradient vector is orthogonal to the surface at a given point, it is also orthogonal (normal) to the tangent plane at that point.

Then at point $(1, -1, 2)$, normal vector to the plane will be $(7, -3, 8)$

The general equation of the plane or the scalar equation is given by,

$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$

where $(a, b, c)$ is the normal vector to the plane and $(x_0, y_0, z_0)$ is a point on the plane.

So the equation of the plane is -

$7(x-1) - 3(y+1) + 8(z-2) = 0$

i.e. $7x - 3y + 8z = 26$


问:
Find the limit function if this exist
I begin study the uniformly and pointwise convergence of functions and I have this sequence function:

$$X=\mathbb{R}\hspace{1cm}f_{n}(x) = \left\{\begin{array}{b}\ n&  \text{if }-n\leq x\leq n\\
0&\text{ if } |x|>n
\end{array}\right.$$

I need to find the limit function but I think in this case doesn't exist because throughout the sequence for some interval $[-k,k]$ this convergence to a constant function $k$ ($k \in \mathbb{N}$) but I can take some $x$ out of this interval and I can find some $\varepsilon_{0}>0$ such that:
\begin{eqnarray*}
|f_{n}(x)-f(x)|&=&K\\
&>& \varepsilon_{0}
\end{eqnarray*}

But I'm not sure this is correct, I am confusion. Can you give me some advice or hint to resolve this? Thank you.

答:
Fix $x\in\Bbb R$ and take $n_0=\lfloor |x|\rfloor +1$ (the floor of $|x|$ plus 1).

Then for any $n\ge n_0$ is clear that $|x|<n$, so $f_n(x)=n$.

This implies that $$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty} n=+\infty$$

So, the sequence is pointwise divergent.

Title: Find the limit function if this exist


问:
An elementary functional equation: $f\big(x+f(x)\big)=f(x)$
I am finding this functional equation from a past high school mathematics competition rather tricky.

Find all continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}^+$ such that:

$f\big(x+f(x)\big)=f(x)\quad \forall x\in\mathbb{R}$

and

$f(2012)=2012$.

(One must prove that the trivial constant solution is the only solution.)

答:
By induction, $f(x+nf(x))=f(x)$. Hence as you observe, each $x+nf(x)$ is injective, since
$$x+nf(x)=y+nf(y) \implies f(\cdots)=f(\cdots) \implies f(x) = f(y) \implies x + \cdot = y + \cdot$$
and therefore monotonically increasing, since $x+nf(x)\to\infty$ as $x\to\infty$.

Now for $x<y$, we have $x+nf(x)<y+nf(y)$ for all $n$ and so $f(y)-f(x)\ge0$.

But $f(x+nf(x))=f(x)$ and $f(x)>0$ implies that for all $x$ there exist points arbitrarily far away which take the same $f$ value, and hence as $f$ is non-decreasing, $f$ must be constant.


问:
Cartan subalgebra and Casimir invariants
To my understanding, if group $G$ is semisimple, $\mathfrak{g}$ is its Lie algebra, and $\Delta_T=[T,\cdot]$ is the adjoint representation, one can analyze its spectrum with $[T,U]=\lambda U$.

A particularly important value is $R=\inf_{T\in\mathfrak{g}}deg(\lambda=0)$, since there are (at most) R communing generators of $\mathfrak{g}$ and R Casimir invariants. Now,

the {$C_1,...,C_R$} commuting generators are the generators of a subalgebra called Cartan's subalgebra, which is interesting because its structure constants are zero: $[C_i,C_j]=0 \space\forall \space i,j$;

the {$H_1,...,H_R$} Casimir invariants are defined by the fact that they commute with all the generators of $\mathfrak{g}$.

My question: are $C_i$ and $H_i$ the same? Maybe under some hypotheses? It seems at least plausible, since they both are R independent operators, and they both have banal commutators.

答: No, they are not the same. The Casimir invariants $\{H_1,\ldots,H_R\}$ (in your notation) generate the center $Z(U\mathfrak{g})$ of $U\mathfrak{g}$, the universal enveloping algebra of $\mathfrak{g}$, and involve terms that are of order $\ge 2$ in $\mathfrak{g}$. The elements $\{C_1,\ldots,C_R\}$ are elements of $\mathfrak{g}$. The fact that the cardinality of these two generating sets agree is a non-trivial result.


问:
Einstein manifold
Does an Einstein manifold have to have integer $\lambda$ constant?

Why a Riemannian product between two Einstein manifoldd with different $\lambda$ constant is not Einstein?

A manifold with constant sectional curvature is Einstein, while a manifold with constant scalar curvature may not be Einstein, why?

答:
$\lambda$ can be any real number. It does not have to be an integer.

If $M=M_1\times M_2$ with $Ric_i=\lambda_ig_i$ then $Ric(X_1,X_1)=\lambda_1 g_1(X_1,X_1)$ and $Ric(X_2,X_2)=\lambda_2 g_2(X_2,X_2)$. Thus if $\lambda_1\ne \lambda_2$ then $M$ is not Einstein.

Having constant sectional curvature is more restrictive than having constant Ricci curvature. And having constant Ricci curvature is more restrictive than having constant scalar curvature. Thus: constant sectional curvature implies constant Ricci curvature (that is, Einstein) implies constant scalar curvature. But the converse converse implications are not true.

Title: Einstein manifold


问:
Help with derivation of equation of motion for a pendulum
I was reading about Lagrangian mechanics on Wikipedia (https://en.wikipedia.org/wiki/Lagrangian_mechanics) and came across the equation for motion in the "Pendulum on a movable support" section and haven't been able to figure out where the $\dot{\theta}$ in the derivative for ${x_{pend}}$ came from. Intuitively, it makes sense that there is an angular velocity $\dot{\theta}$ term but I don't know how to derive it. The equation from the article is replicated below
$${x_{pend}} = x + l \, sin\,\theta \quad \Rightarrow \quad \dot{x}_{pend} = \dot{x} + l\dot{\theta}\,cos\,\theta$$

Thank you.

答:
The key is to regard $\theta$ as a function of time. Then the second term is just an application of the chain rule
$$\frac{\delta}{\delta t}\sin(\theta(t))=\dot\theta(t)cos(\theta(t))$$


问:
Why are the empty set and the set of all real numbers both open and closed?
sorry! am not clear with these questions

why an empty set is open as well as closed?

why the set of all real numbers is open as well as closed?

答:
By definition, a set $A$ of real numbers is open when the following condition is met:

(Note that this applies equally well to the set of real numbers, just substitute $A = R$.)

$$
\hbox{$\forall x\in A, \exists\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\subset A$.}
$$

By the definition above, the set of all $x$ is no other than the set $A$ itself, and $A$ is already defined above to be open.

As for why an empty set is also open, is a rather tricky business AND DEFINITELY NOT IMMEDIATELY CLEAR why it is so as many people seem to say. First of all, it is defined rather than concluded from some logical reasoning. The reason we want an empty set to be made open is:

We want to maintain that empty set is a subset of any sets.

We want to maintain a theorem that says the union of open sets is an open set.

Now watch that the definition of an open set is the same as saying it is the union of all sets of all open balls in the set. As we want to maintain that empty set is a subset of any sets (reason number 1), we have that empty set is a subset of all sets of those open balls, thus empty set is a subset of an open set. As we also want to maintain the theorem mentioned in reason number 2 to be true, there is no other way than to define an empty set as an open set.

For the last part of the question of why the set $A$ and empty set are both also close is to remember that the definition of a close set is:

$$
\hbox{If $B \subset A$, then $A \setminus B$ is close $\iff B$ is open.}
$$

If we set $B = A$, then we have $\emptyset = A \setminus A$ in which by the definition above is close because $B = A$ is open. If we set $B = \emptyset$, then we have $A = A \setminus \emptyset$ which is close because $B = \emptyset$ is open.


答:
Your conclusion is correct, but your statement that on $[-k, k]$ the sequence converges to a constant function is not since $f_k(x)$ grows without bound as $k \to +\infty$ for every $x \in \mathbb{R}$. In fact, we can use this unbounded growth to prove the lack of convergence.

Assume $f_k$ tends to $f$ pointwise, i.e. $f_k(x) \to f(x)$ for every $x \in \mathbb{R}$. Fix $x$ and let $N$ be the smallest integer greater than $\max(|x|, f(x))$. Note that $f(x) \leq N$ and $f_k(x) = k$ for every $k > N$. Consequently,

$$
|f_k(x) - f(x)| \geq |k - N| > 1
$$

so $f_n$ does not converge to $f$ pointwise.

Since $f_n$ does not converge pointwise, it also does not converge uniformly.

Title: Find the limit function if this exist

Body:
I begin study the uniformly and pointwise convergence of functions and I have this sequence function:

$$X=\mathbb{R}\hspace{1cm}f_{n}(x) = \left\{\begin{array}{b}\ n&  \text{if }-n\leq x\leq n\\
0&\text{ if } |x|>n
\end{array}\right.$$

I need to find the limit function but I think in this case doesn't exist because throughout the sequence for some interval $[-k,k]$ this convergence to a constant function $k$ ($k \in \mathbb{N}$) but I can take some $x$ out of this interval and I can find some $\varepsilon_{0}>0$ such that:
\begin{eqnarray*}
|f_{n}(x)-f(x)|&=&K\\
&>& \varepsilon_{0}
\end{eqnarray*}

But I'm not sure this is correct, I am confusion. Can you give me some advice or hint to resolve this? Thank you.


问:
How to get eigenvectors using QR algorithm?
From everything I've heard, this matlab code ought to spit out a matrix where each row is the same. So why doesn't it?</p>
```<code>A = [ 5  2  0  0;
      3  9  4  0;
      0  9  5 -2;
      0  0 -3  4 ];
B=A;
QQQ=eye(4);

%QR algorithm
for i=1:100
    [Q,R] = qr(B);
    B=R*Q;
    QQQ = QQQ*Q;
end

(A*QQQ)./QQQ %should have constant rows, but doesn't
</code>```

答:
First, this is not true

(A*QQQ)./QQQ %should have constant rows, but doesn't

You basically state that
$$
A Q = Q \Lambda,
$$
where $Q$ is an orthogonal matrix and $\Lambda = \operatorname{diag} (\lambda_1, \dots, \lambda_n)$. This means that $A$ is ortogonaly similar to a diagonal matrix, but it is true only for normal matrices which is not the case.

After each iteration the QR algorithm the following relation holds:
$$
A Q_k = Q_k B_k
$$
But $B_k$ converge to an upper triangular matrix, not a diagonal one, as you might expect. Thus QR algorithm computes the Schur decomposition of the matrix, not its eigendecomposition.

Consider now the Schur decomposition of the original matrix:
$$
A Q = Q R
$$
It is not hard to obtain eigenvectors when the Schur decomposition is known. We need to find eigenvectors of $R$ (and the eigenvalues are already known - they are on the main diagonal of $R$).

Let $\mathbf v_i$ be the $i$-th eigenvector of $R$:
$$
R \mathbf v_i = r_{ii} \mathbf v_i \implies (R - r_{ii} I) \mathbf v_i = \mathbf 0.
$$
The corresponding $\mathbf v_i$ may be sought in form
$$
\mathbf v_i = \begin{pmatrix}
\ast\\
\ast\\
\vdots\\
\ast\\
1\\
0\\
\vdots\\
0
\end{pmatrix}
$$
with $1$ on the $i$-th row. Plugging this to the equation we obtain a triangular system for $\ast$ unknown values:
$$
R(1:i-1, 1:i-1) v_i(1:i-1) + R(1:i-1, i) - r_{ii} v_i(1:i-1) = 0
$$
Here I've used the Matlab notation for denoting submatrices. The solution of the system is
$$
v_i(1:i-1) = \left[
r_{ii} I - R(1:i-1, 1:i-1)
\right]^{-1}
R(1:i-1, i)
$$

After we have found the eigendecomposition of $R$ with $V = [\mathbf v_1, \dots, \mathbf v_n]$
$$
R V = V \operatorname{diag}(R)
$$
we can plug it into the Schur decomposition of $A$:
$$
A Q V = Q R V = Q V \operatorname{diag}(R).
$$
It is now clear that $QV$ form the eigenvectors of $A$.

Here's an updated version of the code:
https://gist.github.com/uranix/2b4bb821a0e3ffc4531bec547ea67727


Title: Alternate way to solve this algebra problem?

Body:
Working on a problem that says given the absolute value of the difference of the roots of $ax^2 + bx + c$ as $2$, what is the absolute value of the difference of the roots of $ax^2 + 6bx + 36c$?

I reasoned that the difference of roots would just be $2\sqrt{b^2 - 4ac}$. Therefore, the value of $\sqrt{b^2 - 4ac}$ is $1$. Then, I used this information to answer the question:

$2\sqrt{36b^2 - 36(4ac)} =$ ?

$2\sqrt{36(b^2 - 4ac)}$

$12\sqrt{b^2 - 4ac}$

$=12$

The question hinted that I could use Vieta's formulas to solve the problem. Is my method correct, and is there a method involving Vieta's formulas?

Body:
Note, for $ax^2+bx +c=0$
$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\frac{b^2}{a^2}-\frac{4c}a=2^2$$
Then, for $ay^2 + 6by + 36c=0$
$$(y_1-y_2)^2 = (y_1+y_2)^2-4y_1y_2=36\left(\frac{b^2}{a^2}-\frac{4c}a\right)=36\cdot 4
$$
Thus, $| y_1-y_2|= 12$.


问:
How can you solve $\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$?
$\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$

I can't use L'Hopital's Rule here. I started by substituting $\root{3}\of{x}-1$ with $t$

$\lim _{t \to 0}\frac{\ln\left(\frac{2(t+1)^3-1}{(t+1)^3}\right)}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)-3\cdot\ln{(t+1)}}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)}{t}-3\cdot\frac{\ln{(1+t)}}{t}$

As you can see I formated it so the right part will be $-3$ when I put in $0$. I don't know how to do the left part.

Thanks!

Body:
$$
\lim _{t \to 0}\frac{\ln(2(t+1)^3-1)}{t}=
\lim _{t \to 0}\frac{\ln(2(t^3+3t^2+3t+1)-1)}{t}
$$
$$
=\lim _{t \to 0}\frac{\ln(1+2t^3+6t^2+6t)}{t}
=\lim _{t \to 0}\frac{\ln(1+2t^3+6t^2+6t)}{2t^3+6t^2+6t}\cdot
\frac{2t^3+6t^2+6t}{t}
$$
$$
=1\cdot \lim_{t\to 0} \left(\frac{2t^3}{t}+\frac{6t^2}{t}+\frac{6t}{t}\right)
=1\cdot(0+0+6)=6.
$$


问:
A good book for learning mathematical trickery
I've seen several question here on what book to read to learn writing and reading proofs. This question is not about that. I've been doing that for a while, and I'm quite comfortable with proofs. I am looking for resources (books, ideally) that can teach not the concept of proofs, but rather some of the specific mathematical tricks that are commonly employed in proofs: those that mostly include clever number manipulation, ad-hoc integration techniques, numerical methods and other thing you are likely never to learn in theory-oriented books. I come mainly from applied math and engineering, and when I look at proofs from Stochastic Processes, Digital Signal Processing, Non-Linear Systems and other applied subjects, I feel like I need to learn a new method to understand every proof I read. Is there any good literature on such mathematical tricks?

答: I don't know if you're interested in inequalities, but a very nice book which teaches lots of tricks is Steele's The Cauchy–Schwarz Master Class.


问:
Munkres Topology section 50 exercise 6(e), continuous function that tends to infinity
The main problem is to prove the following theorem.

Let $X$ be a locally compact Hausdorff space with a countable basis, such that every compact subspace of $X$ has topological dimension at most $m$. Then $X$ is homeomorphic to a closed subspace of $\mathbb{R}^{2m+1}$.

Also, we have a definition.

If $f : X \to \mathbb{R}^{N}$ is a continuous map, we say $f(x) \to \infty$ as $x \to \infty$ if given $n$, there is a compact subspace $C$ of $X$ such that $\|f\|(x) > n$ (I think it means that $f(x) \notin$ 'closed $n$-ball centered at the origin') for $x \in X - C $.

And the following is part (e), where I'm stuck.

Show there exists a map $f : X \to \mathbb{R}^{N}$ such that $f(x) \to \infty$ as $x \to \infty$. [Hint: Write $X$ as the union of compact subspaces $C_n$ such that $C_n \subset$ Int $C_{n+1}$ for each $n$.]

First, I do know that locally compact second-countable space is $\sigma$-compact, so I know that there are compact subspaces $C_n$ that $C_n \subset C_{n+1}$. But how can I assure that $C_n \subset$ Int $C_{n+1}$?
Second, how can I proceed if I succeed in constructing $C_n$s that satisfy the conditions(that is, how can I make the desired function $f$)?

Edited: The 'first' part is understood. I still have the second part; suppose I have a function $f_n$ defined on $C_n$ that has value $n$ on the boundary of $C_n$. How can I extend it to $f_{n+1}$ defined on $C_{n+1}$ that equals $f_n$ on $C_n$ and has value $n+1$ on the boundary of $C_{n+1}$ so that I can eventually get a function $f$ that $f(x) \to \infty$ as $x \to \infty$?

答:
Def: A space $X$ is locally compact if each of its points has a neighbourhood base consisting of compact sets. A space $X$ is $\sigma$-compact if it can be written as union of countably many compact subsets.

Let $X$ be $\sigma$-compact and locally compact. Then $X$ can be written $X=\bigcup_\mathbb{N}U_n$, where each $U_n$ is open with $\overline U_n$ compact and satisfies $\overline U_n\subseteq U_{n+1}$.

Proof: Use $\sigma$-compactness to write $X=\bigcup_\mathbb{N}K_n$ with each $K_n$ compact. Each point of $K_1$ has a compact neighbourhood, and since $K_1$ is compact it can be covered by the interiors of finitely many such sets. Let $U_1$ be the union of these interiors so that $K_1\subset U_1$ and $\overline U_1$ is compact. Now take the compact set $\overline U_1\cup K_2$ and repeat to find a relatively compact open set $U_2$ with $\overline U_1\cup K_2\subseteq U_2$. Inductively we have a sequence of relatively compact open sets $U_1\subseteq U_2\subseteq\dots$ with $\overline U_n\subseteq U_{n+1}$ for each $n\in\mathbb{N}$. Since $K_n\subseteq U_{n+1}$ for each $n\in\mathbb{N}$ we have $X=\bigcup_\mathbb{N}K_n\subseteq\bigcup_\mathbb{N}U_n$. Thus this gives the desired exhaustion. $\square$

Of course we have a compact exhaustion of $X$ obtained by writing $X=\bigcup_\mathbb{N}\overline U_n$, but the result above is a little sharper. Notice that we haven't used any separation axioms.

To make the link with the question notice that if $X$ is locally compact and second-countable, then it is $\sigma$-compact, since any given open cover is refined by a countable cover consisting of relatively compact open sets.

We next construct an exhaustion function. Write $X=\bigcup_\mathbb{N}K_n$ with each $K_n$ compact and with $K_n\subseteq int(K_{n+1})$. We need at this stage to add the assumption that  $X$ is Hausdorff. We work inductively, beginning with the map $f_1:K_1\rightarrow \mathbb{R}$ which is constant at $1$.

Suppose we have constructed a continuous $f_n:K_n\rightarrow \mathbb{R}$ which takes the value $n$ at some point of $\partial K_n$. Since both $K_n$ and $\partial K_{n+1}=K_{n+1}\setminus int(K_{n+1})$ are closed in $K_{n+1}$, so is their union $K_n\cup \partial K_{n+1}$, and since $K_n\subseteq int(K_{n+1})$, the sets $K_n$ and $\partial K_{n+1}$ are disjoint. Thus the function $K_n\cup \partial K_{n+1}\rightarrow\mathbb{R}$ which is $f_n$ on $K_n$ and constant at $n+1$ on $\partial K_{n+1}$ is continuous. Since $K_{n+1}$ is compact Hausdorff it is normal, and hence this map has a continuous extension to a map $f_{n+1}:K_{n+1}\rightarrow \mathbb{R}$. By construction this map takes the value $n+1$ on the boundary $\partial K_{n+1}$. If we like we can replace $f_{n+1}$ by the function $\max\{f_{n+1},n+1\}$ to ensure that its maximum is achieved on the boundary, but this is not really necessary.

In this way we obtain a function $f:X=\bigcup_\mathbb{N}K_n\rightarrow\mathbb{R}$ by requiring that $f|_{K_n}f_n$. Then $f$ is well-defined because each $f_{n+k}$ extends $f_n$, and $f$ is continuous since each point of $X$ has an open neighbourhood on which it is continuous (namely some $int(K_n)$).

These seemed to be the questions you were asking about specifically. Here is another approach to the problem.

Let $X$ be locally compact Hausdorff. Then $X$ is regular Hausdorf (in fact Tychonoff). Thus if $X$ is second-countable, then it is metrisable (this is the Urysohn metrisation theorem), and in particular paracompact. Choose a countable covering $\{U_n\}_\mathbb{N}$ of $X$ by relatively compact open sets and let $\{\xi_n:X\rightarrow I\}_\mathbb{N}$ be a locally-finite partition of unity precisely subordinate to this covering. Define $f:X\rightarrow \mathbb{R}$ by
$$f(x)=\sum_\mathbb{N}n\cdot\xi_n(x).$$
Then $f$ is continuous and positive, and moreover for each $k\in\mathbb{N}$ the preimage $f^{-1}(-\infty,k]$ is compact (notice that it is contained in the compact $\bigcup^k_{i=1}\overline U_i$). If $X$ is non-compact, then $f$ is unbounded.


答:
Consider the function $F(x, y, z) = 2xz^2-3xy-4x-7.$ The level surfaces of $F$ are perpendicular to the vector field $\nabla F=(F_x,F_y,F_z).$ Thus, the equation of the tangent plane of a level surface at $(x_0,y_0,z_0)$ is given by $$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$ In your case, it is $$7(x-1)-3(y+1)+8(z-2)=0.$$

In the particular case where we have $z=f(x,y)$ then the tangent plane at $(x_0,y_0)$ is given by $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$ Just consider $F(x,y,z)=z-f(x,y)$ and apply what was said in the paragraph above.

Title: Find an equation for the tangent plane to the surface $2xz^2+3xy-4x=7$ at the point $(1,-1,2)$, by considering a gradient of a suitable scalar field.

Body:
This is what I've done, followed by a few questions I don't undestand;

$f(x, y, z) = 2xz^2-3xy-4x=7$

Surface is at 0, So therefore we set our equation to zero $2xz^2-3xy-4x-7=0$

We then take partials and evaluate them at the point.

$F_x = x-2, F_x(1,-1,2) = 1-2=3$

$F_y = 1, F_y(1,-1,2) = 1$

$F_z = 2z, F_z(1,-1,2) = 2(2) = 4$

Now the tangent plane is $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

=$-3(x-1)+1(y+1)+4(z-1) = 0$

=$-3x+y+4z-6=0$

My questions;

I found this method of computing online and it was done for a vector. Now given this is a scalar field would it still work?

In the end we have $-3x+y+4z-6=0$, what exactly do I need to solve for? Should I bring the equation back in the original form where it was equal to a number? Or do I solve for z, x or y?

Why is surface at zero and why do we take partials?


Body:
Alternative approach

Since $r_1, r_2$ are the roots of $ax^2 + bx + c = 0$,

they are also the roots of $(36a)x^2 + (36b)x + (36c) = 0.$

Since (for example) $[6r_1]^2 = 36(r_1)^2~$ and $~6(6r_1) = 36r_1$ then

$(6 r_1), (6 r_2)$

are the roots of $ax^2 + 6bx + 36c = 0$.

Since $|r_1 - r_2| = 2, ~|(6r_1) - (6 r_2)| = 12.$


问:
Use of the fact that every function is sum of an odd and an even function.
It is well know that every real variable function $f$ can be written as a sum of an odd and an even function, namely $h$ and $g$ where: $$h(x) = {f(x)-f(-x)\over 2}\;\;\;\;\;\;\;\;\;\;\;\;g(x) = {f(x)+f(-x)\over 2}$$
Now what is the use of this fact? I told that to my $\color{red}{\rm high\; school}$ students but then I don't know what to do with this fact. Is it good for a graphing or to calculate the zeroes of a function or something else...?

答:
When you are confronted with a new mathematical problem there are a few helpful rules to get ahead. One of them is: Search for the symmetries! Often a problem has in its formulation an obvious symmetry $x\leftrightarrow -x$. Such a symmetry could force the solutions to be symmetric as well.

Another example: One of the most important functions in analysis is ${\rm cis}:\ t\mapsto e^{it}$. It has a fantastic functional equation: ${\rm cis}(s+t)={\rm cis}(s)\cdot {\rm cis}(t)$ for arbitrary complex $s$, $t$. Unfortunately this function ${\rm cis}$ is complex-valued. But ${\rm cis}$ has an even part $t\mapsto\cos t$ and (up to the factor $i$) an odd part $t\mapsto\sin t$. These symmetric, resp., antisymmetric parts of ${\rm cis}$ have an immense stating in all of analysis and its applications. Similarly with the even and odd parts $\cosh$ and $\sinh$ of $x\mapsto e^x$.

I'm not sure whether it is important make a general theory of this for high school students.


问:
Sufficient condition for reducibility of polynomial $f(x,y)$
[Dual to
this question]

Suppose we have a non-constant polynomial $f(x,y)\in\mathbb{Q}[x,y]$ such that the following two conditions are satisfied:

1) For every $x_0\in\mathbb{Q}$, the polynomial $f(x_0, y)\in\mathbb{Q}[y]$ is reducible.

2) For every $y_0\in\mathbb{Q}$, the polynomial $f(x,y_0)\in\mathbb{Q}[x]$ is reducible.

My question is this:

Can we conclude from these two conditions that $f(x,y)$ is
  a reducible polynomial in $\mathbb{Q}[x,y]$?

Thanks!

答: Yes. Actually, only one of the conditions needs to hold. This follows from Hilbert's irreducibility theorem. </p>


问:
Prove that a continuous real-valued function on a closed interval in R is Riemann Integrable using only a given lemma
Prove that a continuous real-valued function on a closed interval in R is Riemann Integrable using only the following lemma:

Lemma 1: A real-valued function $f$ on the interval $[a,b]$ is integrable on $[a,b]$ iff given any $\epsilon > 0, \exists \delta > 0$ s.t. $|S_1 - S_2| < \epsilon$ whenever $S_1$ and $S_2$ are Riemann Sums for $f$ corresponding to partitions of $[a,b]$ of width less than $\delta$.

This question may already be out there. However, I haven't been able to find it because this proof can be done with other techniques. I am only allowed to use this lemma here. Can anyone explain how this would be done using only this lemma? I'm not quite seeing it at the moment. Thanks!

答:
For the reverse implication we can construct a sequence of partitions $(P_n)_n$ and Riemann sums $(S(P_n,f))_n$ where both $\|P_n\| \to 0$ and the Riemann sums form a Cauchy sequence.

Hence, there exists $I \in \mathbb{R}$ and $N \in \mathbb{N}$  such that $S(P_n,f) \to I$ as $n \to \infty$, and for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have

$$\|P_n\| < \delta, \quad |S(P_n,f) -I| < \frac{\epsilon}{2} $$

In the above, we could have taken $\delta$ such that $|S(P_1,f) - S(P_2,f)| < \frac{\epsilon}{2}$ when $\|P_1\|, \|P_2\| < \delta$.  If $P$ is any partition where $\|P\| < \delta$, then since we already have $\|P_N\| < \delta$, it follows that

$$|S(P,f) - I| \leqslant |S(P,f) - S(P_N,f)|+ |S(P_N,f) - I| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$

and $f$ must be Riemann integrable.

Construction of the sequences

There exists $\delta_1$ such that if $\|P\|, \|Q\| < \delta_1$, then $|S(P,f) - S(Q,f)| < 1$. We can assume $\delta_1 < 1$ and let $P_1$ be any partition with $\|P_1\| < \delta_1$ and $S(P_1,f)$ be any corresponding Riemann sum.

Further, there exists $\delta_2 < \min(\frac{1}{2},\delta_1)$ such that if $\|P\|, \|Q\| < \delta_2$, then $|S(P,f) - S(Q,f)| < \frac{1}{2}$. Let $P_2$ be any partition with $\|P_2\| < \min(\delta_2, \|P_1\|)$ and $S(P_2,f)$ be any corresponding Riemann sum.

Proceeding, for any $n \in \mathbb{N}$, there exists $\delta_n < \min(\frac{1}{n}, \delta_{n-1}) $ along with a partition $P_n$ and Riemann sum $S(P_n,f)$ such that if $\|P\|, \|Q\| < \delta_n$, then $|S(P,f) - S(Q,f)| < \frac{1}{n}$ and $\|P_n\| < \min( \delta_n , \|P_{n-1}\|)$.

Try to finish yourself by proving that $(S(P_n,f))$ is a Cauchy sequence.


问:
Are these solutions correct? (If the n-value is x, will the sum be y?)
$$\sum_{k=5}^{(n-1)-1\times \frac{1-(-1)^n}{2}} (n-k) \times \frac{1-(-1)^k}{2}$$

All n-values smaller or equal to the k-value are to be ignored.

if
$$n=6$$

$$ans =1 $$

if
$$n=7$$ $$ans=2$$

if
$$n=8$$
$$ans=4$$

if
$$n=9$$ $$ans=6$$

That's enough for the first one. Below is the second one.

$$\sum_{k=6}^{(n-1)-1 \times \frac{1+(-1)^n}{2}} \biggr((n-k)\times\frac{1+(-1)^k}{2}\biggl)\times2-1$$

if
$$n=7$$ $$ans=1$$
if
$$n=8$$ $$ans=3$$

if

$$n=9$$ $$ans=6$$

if
$$n=10$$ $$ans=10$$

So, are the answers correct given the n-values? I am pretty confident in these, but I can't be too sure, because I'm quite new to precalculus, my calculator is a bit basic and I've never had a precalculus class. So, if any of you could check if these answers are correct, I'd appreciate it :)

Body:
Yes they are

First sum
$$s_n=\sum _{k=5}^{(n-1)-\frac{1}{2} \left(1-(-1)^n\right)} \frac{1}{2} \left(1-(-1)^k\right) (n-k)$$

$$
\begin{array}{c|r}
 n & s_n \\
\hline
 6 & 1 \\
 7 & 2 \\
 8 & 4 \\
 9 & 6 \\
 10 & 9 \\
 11 & 12 \\
 12 & 16 \\
 13 & 20 \\
 14 & 25 \\
 15 & 30 \\
\end{array}
$$
general formula $$s_n=\frac{1}{8} \left(2 n^2-16 n+(-1)^n+31\right)$$

Second sum
$$t_n=\sum _{k=6}^{(n-1)-\frac{1}{2} \left((-1)^n+1\right)} \left(\frac{2}{2} \left(\left((-1)^k+1\right) (n-k)\right)-1\right)$$
$$
\begin{array}{c|r}
 n & t_n \\
\hline
 7 & 1 \\
 8 & 3 \\
 9 & 5 \\
 10 & 9 \\
 11 & 13 \\
 12 & 19 \\
 13 & 25 \\
 14 & 33 \\
 15 & 41 \\
\end{array}
$$
general formula
$$t_n=\frac{1}{4} \left(2 n^2-24 n+(-1)^n+75\right)$$


Title: How can you solve $\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$?

Body:
$\lim _{x \to 1}\frac{\ln\left(\frac{2x-1}{x}\right)}{\root{3}\of{x}-1}$

I can't use L'Hopital's Rule here. I started by substituting $\root{3}\of{x}-1$ with $t$

$\lim _{t \to 0}\frac{\ln\left(\frac{2(t+1)^3-1}{(t+1)^3}\right)}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)-3\cdot\ln{(t+1)}}{t} \to \lim _{t \to 0}\frac{\ln(2(t+1)^3-1)}{t}-3\cdot\frac{\ln{(1+t)}}{t}$

As you can see I formated it so the right part will be $-3$ when I put in $0$. I don't know how to do the left part.

Thanks!

Body:
$f(t)=\log (2(t+1)^3-1)$;

$f'(0)=\lim_{t \rightarrow 0} \dfrac{f(t)-f(0)}{t}=$

$\lim_{t \rightarrow 0}\dfrac{\log (2(t+1)^3-1)-\log 1}{t}$

$=\left (\dfrac{6(t+1)^2}{2(t+1)^3-1}\right )_{t=0}=6.$


问:
Uniform equicontinuity from pointwise equicontinuity over compact set
How to show that a sequence of real continuous functions over a compact set that is pointwise equicontinuous and convergent is also uniformly equicontinuous?

I thought to use Arzela-Ascoli theorem to show there is a uniformly convergent subsequence but I can only prove uniform equicontinuity from this if every subsequence is uniformly convergent.

答:
Let $\{f_i: X \to \mathbb{R}\}_{i \in I}$ be a family of real-valued continuous functions on a compact metric space $(X,d)$. Assume that this family is pointwise equicontinuous, i.e. that for every point $x \in X$, we can find for every $\epsilon > 0$ a radius $\delta_x > 0$ such that for every $i \in I$ and every $y \in X$ satisfying $d(x,y) < \delta_x$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$. We prove that this family is uniformly equicontinuous, i.e. that for every $\epsilon > 0$ we can find a $\delta > 0$ such that for every $i \in I$ and $x,y \in X$ with $d(x,y) < \delta$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$.

So fix $\epsilon > 0$ and find for each $x \in X$ a $\delta_x > 0$ such that for every $i \in I$ and every $y \in X$ satisfying $d(x,y) < \delta_x$ we have $\lvert f_i(x) - f_i(y) \rvert < \frac{1}{2}\epsilon$. Define for every $x \in X$ an open subset
\begin{aligned}
U_x := \{ y \in X \mid d(x,y) < \frac{1}{2} \delta_x\}.
\end{aligned}
Clearly we have that $x \in U_x$ and thus the open subsets $\{U_x\}_{x \in X}$ cover $X$. Since $X$ is compact, there are finitely many $x_1, \dots, x_n \in X$ such that $U_{x_1}, \dots, U_{x_n}$ cover $X$. Define
\begin{aligned}
\delta := \frac{1}{2} \min(\delta_{x_1}, \dots, \delta_{x_n}).
\end{aligned}
We show that for every $i \in I$ and for every $x,y\in X$ with $d(x,y)<\delta$ we have $\lvert f_i(x) - f_i(y) \rvert < \epsilon$. Indeed, we can find some $j \in \{1, \dots, n\}$ such that $x \in U_{x_j}$, meaning that $d(x,x_j) < \frac{1}{2}\delta_{x_j}$. Since we also have $d(x,y) < \delta \leq \frac{1}{2}\delta_{x_j}$, we find
\begin{aligned}
d(y,x_j) \leq d(y,x) + d(x,x_j) < \frac{1}{2}\delta_{x_j} + \frac{1}{2}\delta_{x_j} = \delta_{x_j}.
\end{aligned}
Therefore by the choice of $\delta_{x_j}$ we find that for all $i \in I$
\begin{aligned}
\lvert f_i(x) - f_i(x_j) \rvert < \frac{1}{2}\epsilon
\end{aligned}
and
\begin{aligned}
\lvert f_i(y) - f_i(x_j) \rvert < \frac{1}{2}\epsilon,
\end{aligned}
so in particular
\begin{aligned}
\lvert f_i(x) - f_i(y) \rvert \leq \lvert f_i(x) - f_i(x_j) \rvert + \lvert f_i(y) - f_i(x_j) \rvert < \frac{1}{2}\epsilon + \frac{1}{2}\epsilon = \epsilon.
\end{aligned}


问:
Transformation matrix of linear map on vector space whose elements are matrices
I'm starting to understand how to compute and think about the transformation matrix of a linear transformation $L:\mathbb{R}^n \to \mathbb{R}^m$. However, I'm trying to understand how this translates to a vector spaces of matrices. For example, consider the linear map $L:\mathbb{R}^{n \times m} \to \mathbb{R}^{m \times p}$, and the standard basis $\beta=\{E_{ij}|1\leq i \leq n, 1\leq j \leq m\}$ where each $E_{ij}$ contains zeros except in the $i$-th row and $j$-th column, which contains a ones. From this it follows that any matrix $Q\in \mathbb{R}^{n \times m}$ can be written as a linear combination of these basis matrices. Applying the transformation $L$ to this matrix $Q$ we obtain another matrix $L(Q)\in \mathbb{R}^{m \times p}$. In the case of vectors the transformation matrix $A$ of $L$ has columns containing the coordinates of each basis vector $L(v_i)$ w.r.t. to basis $\beta_W$. However, I'm having difficulty translating this idea to the context in which the vectors are $m\times n$ matrices.

Any help would be much appreciated. Thanks,

答:
I think the key here is you are converting your matrices into vectors by writing them as linear combinations of the basis vectors.

For example given an ordered basis (order matters for transformation matrices, you always have to keep the basis vectors in the same order!) $\beta_{3\times 3}=\{ E_{11},E_{12},E_{13},E_{21},E_{22},E_{23},E_{31},E_{32},E_{33}\}$:

$\begin{bmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{bmatrix} = a_{11}\cdot E_{11}+a_{12}\cdot E_{12}+\cdot \cdot \cdot +a_{33}\cdot E_{33} \equiv(a_{11},a_{12},...,a_{33})\in \mathbb{R}^{3\cdot 3}$

The beauty about vector spaces is you can treat whatever you are working with, in this case matrices, as a vector, so you are essentially converting your $n\times m$ matrix into a vector in $\mathbb{R}^{n\cdot m}$, you find the matrix of the linear transformation as you would for any linear transformation because you are back to familiar territory, and then that matrix allows you to input a vector (a matrix written as a linear combination of the ordered basis vectors of the domain) and you get as output another vector (another matrix written as linear combination of the ordered basis vectors of the codomain).

Hope this helps!


问:
Find $y$-Lipschitz constant
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$

I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$

I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$

now I don't know what to do.

答:
Try to find a Lipschitz constant of $x \mapsto \mathrm{e}^{-x}$, $x \ge 0$. Hint: Fundamental Theorem of Calculus.

Then, you can use this in your Lipschitz estimate.

Title: Find $y$-Lipschitz constant


问:
Showing a system of equations has no integer solutions using modular arithmetic
Quite a trivial question in comparison to the ones found on here, but here it goes:

"By applying (mod 5) to both equations, show that the system of equations below has no integer solutions:

$11x^2 + 5xy = 7 $

$9x^2 + 10y^3 = −3$   "

Now, the solution jumps straight to:

$x^2 = 2 \pmod 5$

$-x^2 = -3 \pmod 5. $

I'm struggling to figure out how exactly they get to that stage, and at that, how these last two equations imply there are no integer solutions.

答:
$11x^2+5xy=7\implies x^2\equiv2\pmod5$, because $11\equiv1\pmod5, $

$5\equiv0\pmod5$, and $7\equiv2\pmod5$.

$9x^2+10y^3=-3\implies  -x^2\equiv-3\pmod5\iff x^2\equiv 3\pmod5$,

because $9\equiv-1\pmod5$ and $ 10\equiv0\pmod5$.

There are no solutions, because   $2\equiv x^2\equiv3\pmod5$ is a contradiction.


问:
Expected number of coin flips to get three heads or three tails
We both play a game where we flip a coin. You win if 3 heads appear, I win if 3 tails appear. What is the expected number of flips for the game to end.

The heads/tails doesn't need to be consecutive. Here's my approach:

First find the expected number of flips to get three heads before game ends. This can happen in either three or four of five tosses. So the expected value is
$$\frac{1}{2^3}*3 + \frac{1}{2^4}*4C_3*4 + \frac{1}{2^5}*5C_3*5 $$

Divide this value by 2 to get the answer. Is this correct?

Many thanks.

答:
Let $P(S)$ be the probability of a given sequence and let $N(S)$ be the number of flips in that sequence. You want to find $E(N(S)) = \sum_{S} P(S)N(S)$.

There are only finitely many possible games, namely, in alphabetical order:

$HHH: P = 1/8, N = 3$

$HHTH: P = 1/16, N = 4$

$HHTTH: P = 1/32, N = 5$

$HHTTT: P = 1/32, N = 5$

$HTHH: P = 1/16, N = 4$

$HTHTH: P = 1/32, N = 5$

$HTHTT: P = 1/32, N = 5$

$HTTHH: P = 1/32, N = 5$

$HTTHT: P = 1/32, N = 5$

$HTTT: P = 1/16, N = 4$

$THHH: P = 1/16, N = 4$

$THHTH: P = 1/32, N = 5$

$THHTT: P = 1/32, N = 5$

$THTHH: P = 1/32, N = 5$

$THTHT: P = 1/32, N = 5$

$THTT: P = 1/16, N = 4$

$TTHHH: P = 1/32, N = 5$

$TTHHT: P = 1/32, N = 5$

$TTHT: P = 1/16, N = 4$

$TTT: P = 1/8, N = 3$

Grouping these by $N$ we see that:

For $N = 3$ the total $P = 1/4$

For $N = 4$ the total $P = 3/8$

For $N = 5$ the total $P = 3/8$

So $E(N) = 3 \cdot 1/4 + 4 \cdot 3/8 + 5 \cdot 3/8 = 33/8 = 4.125$


问:
Homomorphisms between fields are injective.
How would I prove this?

I know that I must show $f(a)=f(b) \Rightarrow a = b$

I also know I must use the definition of homomorphism, ie:

$f(a+b)=f(a)+f(b)$

$f(ab)=f(a)f(b)$

$f(1)=1$

I am assuming that a contradiction would be a good approach to solve this, but not quite sure on specifics.

答: The kernel of $f$ must be equal to $\{0\}$. Suppose that $f(a)=0$ for $a\ne0$; then for all $b$ one has  $f(ab)=f(a)f(b)=0$. Since, for $a\ne0$ the function $a\rightarrow ab$ is onto in a field  we have $f=0$ which is not possible because $f(1)=1$. It follows the kernel of $f$ is  $\{0\}$.


问:
$\sigma$-weak topology versus $\sigma$-strong topology on the bounded operators $B(H)$.
Let $H$ be a Hilbert space and let $T(H)$ be the traces class operators on $H$. Denote the bounded operators on $H$ by $B(H)$. We define the normal functionals on $B(H)$ by
$$B(H)_*:= \{\omega \in B(H)^*: \exists x \in T(H): \forall y \in B(H): \omega( y) = \text{Tr}(xy)\}$$
Let $B(H)_*^+$ denote the positive normal functionals.

We can introduce the $\sigma$-weak and the $\sigma$-strong topology on the space $B(H)$. The former is the locally convex topology generated by the family of seminorms
$$ B(H)\ni x \mapsto |\omega(x)|, \  \omega\in B(H)_*$$
The latter is the locally convex topology generated by the family of seminorms
$$B(H)\ni x \mapsto  \omega(x^*x)^{1/2}, \ \omega \in B(H)_*^+$$

I want to show that the $\sigma$-weak topology is weaker than the $\sigma$-strong topology (as the name suggests!). More precisely, suppose that $x_\alpha \to x$ in the $\sigma$-strong topology. Then I want to show also that $x_\alpha \to x$ in the $\sigma$-weak topology. How can I show this?

We need to show
$$\forall \omega \in B(H)_*: |\omega(x_\alpha)-\omega(x)| \to 0$$

using something about positive normal functionals but I'm unsure how to approach this.

答:
The reason this is looking so complicated is perhaps because of the fancy  definitions of the various topologies that we are using here.  As far as I can tell the "classical" definitions are:

weak topology: seminorms
$$
  x\ni B(H) \mapsto  |\langle x\xi , \eta \rangle |,
  $$
for  $\xi ,\eta \in H$.

$\sigma $-weak topology: seminorms
$$
  x\ni B(H) \mapsto  \Big|\sum_{n=1}^\infty \langle x\xi _n, \eta _n\rangle \Big|,
  \tag 1
  $$
for  $\{\xi _n\}_n,\{\eta _n\}_n\subseteq H$, with $\sum_{n=1}^\infty \|\xi _n\|\|\eta _n\|<\infty $.

strong topology: seminorms
$$
  x\ni B(H) \mapsto  \|x\xi \|,
  $$
for  $\xi \in H$.

$\sigma $-strong topology: seminorms
$$
  x\ni B(H) \mapsto  \big (\sum_{n=1}^\infty \|x\xi _n\|^2\big )^{1/2},
  $$
for  $\{\xi _n\}_n\subseteq H$, with $\sum_{n=1}^\infty \|\xi _n\|^2<\infty $.

Observing that Cauchy-Schwarz implies
$$
  \Big|\sum_{n=1}^\infty \langle x\xi _n, \eta _n\rangle \Big| \leq \big (\sum_{n=1}^\infty \|x\xi _n\|^2\big )^{1/2} \big (\sum_{n=1}^\infty \|\eta _n\|^2\big )^{1/2},
  $$
the answer becomes immediate  (it might also help to observe that, by multiplying $\xi _n$ by $\lambda $, and $\eta _n$ by $1/\lambda $ in
(1), one may assume that $\|\xi _n\| =  \|\eta _n\|$).

Of course one could also ask why are these "classical" definitions equivalent to the ones given by the OP, but
that is another story.


问:
A question about functionals and Hahn Banach theory

Let $X$ be a normed separable space.
Prove that there is a sequence $x^*_n\in X^*$ that satisfy:
$\forall 0\neq x\in X$ there is infinite $n$'s such that: $x^*_nx\neq 0$.

By a corollary of Hahn Banach which states:
If $X$ is a normed space, $x\neq0 \in X$ then there's $x^*\in X^*$ such that $||x^*||=1$ and $x^*x=||x||$.

Since $X$ is separable so I can assume that there is infinite non-zero $x\in X$ (a non-zero sequence $x_n$ in $X$) so for each such $x$ there is $x^*_i\in X^*$ such that $||x^*_i||=1$ and $x^*_ix=||x||$.Since $x\neq 0$ then $x^*_ix=||x||\neq 0$.

I am not sure if was enough coear.
Any improvements would be welcomed.

答:
Let $S$ be a countable dense subset in $X$ and for every $x \in S$ a corollary of Hahn-Banach gives a functional $x^* \in X^*$ such that $\|x^*\|=1$ and $x^*(x) = \|x\|$. I claim that $\{x^* : x \in S\}$ is your desired countable family of functionals.

Let $x \in X$, $x \ne 0$ be arbitrary. There exists a sequence $(x_n)_n$ in $S$ such that $x_n \to x$. We have

$$\|x_n\|-x_n^*(x)\le |x_n^*(x_n) - x_n^*(x)| = |x_n^*(x_n-x)| \le \|x_n^*\|\|x_n-x\| = \|x_n-x\|$$
so $$x_n^*(x) \ge \|x_n\|-\|x_n-x\|.$$
I claim that for infinitely many $n\in\Bbb{N}$ we have $\|x_n\|-\|x_n-x\| > 0$. Otherwise there exists some $n_0\in\Bbb{N}$ such that for $n \ge n_0$ we have
$$\|x_n\| \le \|x_n-x\| \to 0$$
so $x_n \to 0$. This is a contradiction since $x_n \to x$ and $x \ne 0$.

Therefore $x_n^*(x) > 0$ for infinitely many $n\in\Bbb{N}$.


答:
You could tie it to Fourier series. Not formally of course but just hint at the really neat idea that you can write a function as a series of sin (odd) and cos (even) functions.

You could also tie it to Taylor series (again not formally just a little introduction) where functions are the sums of odd powers (odd functions) and even powers (even functions). And, you could show in your examples that odd functions like sin(x) are the sums of odd powers in the Taylor series.

For actual applications, I personally have never had to use that fact to solve a problem, so I wouldn't worry about its application as a tool in High School math.

Title: Use of the fact that every function is sum of an odd and an even function.

Body:
It is well know that every real variable function $f$ can be written as a sum of an odd and an even function, namely $h$ and $g$ where: $$h(x) = {f(x)-f(-x)\over 2}\;\;\;\;\;\;\;\;\;\;\;\;g(x) = {f(x)+f(-x)\over 2}$$
Now what is the use of this fact? I told that to my $\color{red}{\rm high\; school}$ students but then I don't know what to do with this fact. Is it good for a graphing or to calculate the zeroes of a function or something else...?


问:
A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$
$$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
Several difficult integral can be handled using (1) easily. For instance we know that $$\int_{-\infty}^{\infty} \text{sech}^2 x~dx=2~~~~~~~~~~~~(2)$$ but what is interesting is that one may verify numerically that $$\int_{-\infty}^{\infty} \text{sech}^2(x-a/x)~dx=2, a>0.~~~~~~~~~(3)$$ This interesting property (1) follows from Glasser's Master Theorem. See one more interesting integral and its solutions in MSE post:s
https://en.wikipedia.org/wiki/Glasser%27s_master_theorem.

In (1) what is also interesting is that LHS is independent of $a(>0)$. I have tried proving (1) or even (3) by the ordinary rules of integration ( Real Analysis) without a success. Can someone help me?

EDIT: One more friendly (doable by hand) twin is
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$

答:
Glasser's formula (1) amounts to the statement that the image measure $\mu$ of the Lebesgue measure $\lambda$ on $\Bbb R$, under the (two-to-one) transformation $\varphi: x\mapsto x-a/x$ is also Lebesgue measure. To verify this it suffices to show that $\mu\{(A,B]\}=B-A$ for each $A<B$. Notice that if we define $\alpha_\pm:={A\pm\sqrt{A^2+4}\over 2}$ and $\beta_\pm:={B\pm\sqrt{B^2+4}\over 2}$ (so that $\varphi(\alpha_\pm) = A$ and $\varphi(\beta_\pm)=B)$ then
$$
\eqalign{
\mu\{(A,B]\}
&=\lambda\{\varphi^{-1}\{(A,B]\}\}=\lambda\{(\alpha_-,\beta_-]\}+\lambda\{(\alpha_+,\beta_+]\}\cr
&=(\beta_--\alpha_-)+(\beta_+-\alpha_+)\cr
&=B-A,\cr
}
$$
because $\beta_++\beta_-=B$ and $\alpha_++\alpha_-=A$.


问:
Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?
WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:

$$
 \lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) =  \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
=  \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1  \\
= 1?
$$

I wonder where I'm wrong...

答: Hint:  Use the definition of derivative.</p>

Body: Hint:  Use the definition of derivative.


问:
$\int_a^b\sum_{n=1}^\infty f_n(x) dx=\sum_{n=1}^\infty \int_a^b f_n(x) dx$
$\{f_n\}$ is a sequence of continuous (so also integrable) real functions on $[a,b]$ that is uniformly convergent and uniformly equicontinuous. I want to show $$\int_a^b\sum_{n=1}^\infty \frac{f_n(x)}{{\color{red}{n^2}}} dx=\sum_{n=1}^\infty \frac{1}{{\color{red}{n^2}}}\int_a^b f_n(x) dx,$$
and I cannot figure out how to do this. I think to take the limit as $n\to\infty$ and use uniform convergence to the limit $f$ which will somehow allow the sum and integrals to switch, but I cannot figure out how to do this. Help would be greatly appreciated.

Edit: In red.

答: Let $f(x) = \sum\limits_{n=1}^\infty f_n(x)$. We need $$\int_a^b f(x)dx = \lim\limits_{N \to \infty} \int_a^b \sum\limits_{n=1}^N f_n(x)dx$$ Fix $\epsilon>0$, for $N$ big enough, $|f(x)-\sum\limits_{n=1}^N f_n(x)|\leq\epsilon$. Therefore $$|\int_a^b f(x)dx -\int_a^b \sum\limits_{n=1}^N f_n(x)dx| \leq \int_a^b |f(x) - \sum\limits_{n=1}^N f_n(x)|dx \leq \epsilon (b-a)$$


问:
Term by term differentiation sum and criterion for differentiability for $f(x) =\sum_{1}^{\infty} 2^{-n} |x-1/n|$
Let $f(x) =\sum_{1}^{\infty} 2^{-n} |x-1/n|$

Compute $f'(x)$ for $x \in (1,\infty), (1/2,1)$

Prove that $f$ is not differentiable at $x=1$

For part 1 I have that since $f_n$ is differentiable at each $x\in \{1,1/2,1/3,...\}^c$ and $f$ converges for all $\mathbb{R}$ and $\sum f'_n(x)$ converges uniformly then by term by term differentiablity theorem $f'(x) = \sum f'_n(x).$

To find the sum. I'm a little unsure about this part since i got one of them to sum to 0:

If $x \in (1, \infty)$ then $x> 1/n, \forall n \in \mathbb{N}.$ Thus $f'_n(x) = 2^{-n}$ for all $n,$ which implies $f'(x) = \sum f'_n(x) = \sum_{n=1}^{\infty} 1/2^n$ which is geometric thus sum equals $\frac{1/2}{1-1/2} =1$

If $x \in (1/2, 1)$ then $x<1/n$ only for $n=1$. Thus $f'_1(x) = -2^{-1}$ and $f'_n(x) = 2^{-n}$ for $n>1.$Thus $$f'(x) = \sum_{n=1}^{\infty} f'_n(x) = -2^{-1} + \sum_{n=2}^{\infty}1/2^n = -1/2 +1/2 =0.$$Is this right?

For part 2 I'm sort of stuck. How do i show $f$ is not differentiable at $x=1$? I know that due to $|x-1/n|, f_n$ is not differentiable at $1/n, \forall n \in \mathbb{N}$. Namely $f_1$ is not differentiable at 1. So the term by term differentiability theorem fails here. But I don't think that this failing here implies $f$ is not differentiable here?

How would you show $f$ is not differentiable at $x=1$?

答:
You have correctly found $f'(x)=1$ for $x>1$ and $f'(x)=0$ for $x<1/2.$ Furthermore, since the series defining $f$ converges uniformly on $(0,2),$ $f$ is continuous on $(0,2).$ In particular, $f$ is continuous at $1.$

Suppose $x>1.$ Then by the MVT, there exists $c_x\in (1,x)$ such that

$$\frac{f(x)-f(1)}{x-1}= f'(c_x) = 1.$$

Note we needed the continuity of $f$ at $1$ to invoke MVT. Similarly, if $x\in (1/2,1),$ then

$$\frac{f(x)-f(1)}{x-1}= f'(c_x) = 0.$$

Thus the difference quotient has limit $1$ from the right and limit $0$ from the left. It follows that $f'(1)$ does not exist.


问:
What is "univalent" about univalent logic and foundations?
This is a perhaps naive question about terminology.

I know that "a bundle is said to be univalent if every other bundle is a pullback of it in at most one way (up to homotopy)" ref.

But this does not make clear why this term was chosen here.

I also wonder whether the term has further motivation, viz bivalence in logic, or the term univalence in other fields?

I would guess a term like "multi-valent" seems more fitting, considering that:

"intuitionistic logic has no interpretation as a two-valued logic, nor even as a finite-valued logic, in the familiar sense. Although intuitionistic logic retains the trivial propositions { ⊤ , ⊥ } from classical logic, each proof of a propositional formula is considered a valid propositional value, thus by Heyting's notion of propositions-as-sets, propositional formulae are (potentially non-finite) sets of their proofs." here.

答:
From nlab's article

The name univalence (due to Voevodsky) comes from the following reasoning. A fibration or bundle $p:E→B$ of some sort is commonly said to be universal if every other bundle of the same sort is a pullback of $p$ in a unique way (up to homotopy). Less commonly, a bundle is said to be versal if every other bundle is a pullback of it in some way, not necessarily unique. By contrast, a bundle is said to be univalent if every other bundle is a pullback of it in at most one way (up to homotopy). In the language of (∞,1)-category theory, a univalent bundle is an object classifier.

The univalence axiom does not literally say that anything is univalent in this sense. However, it is equivalent to saying that the canonical fibration over $\mathsf{Type}$ is univalent: every fibration with small fibers is an essentially unique pullback of this one (while those with large fibers are not, they are pullbacks of the next higher $\mathsf{Type}_1$). For a description of this equivalence, see section 4.8 of the HoTT Book (syntactically) and Gepner-Kock (semantically).

Essentially, while a category theorist could probably imagine that universes in Martin-löf type theory are 'like' object classifiers, the equality between types is too intensional for it to literally make sense. One could exhibit things as pullbacks in ways that are not provably equal (because they are merely isomorphic/equivalent, not exactly the same definition). The univalence axiom is the extensionality principle needed to make universes proper object classifiers.

Incidentally, univalence/homotopy type theory is not necessarily intuitionistic. Voevodsky's first model for type theory extended with the univalence axiom also modeled excluded middle (if my understanding is correct), and I don't think he really had any qualms about using classical axioms like that. It is just the case that many people who work with type theory, including homotopy type theory, recognize that it can be useful to not assume excluded middle/choice/etc.


问:
Geometric proof of $\sin(\theta)' = \cos(\theta)$ using infinitesimals
In his book Visual Complex Analysis, Tristran Needham uses Newton's method of infinitesimals to derive the derivative $\tan(\theta)' = 1 + \tan^2(\theta)$ by using the following picture (where $T = \tan(\theta)$):

If $d\theta$ "goes to" zero, the black triangle becomes similar to the shaded triangle which yields $\frac{dT}{L d\theta} = \frac{L}{1} \Rightarrow \frac{dT}{d\theta} = L^2 = 1 + T^2$.

If I try to apply the same logic to $\sin(\theta)$ I don't get the correct result. My thinking: use the unit hypothenuse instead of the unit base. Then $L \rightarrow 1$, $1 \rightarrow C = \cos(\theta)$, $T \rightarrow S = \sin(\theta)$, $dT \rightarrow dS$ and $L \cdot d\theta \rightarrow d\theta$. But this gives the wrong result $\frac{dS}{d\theta} = \frac{1}{C}$.

Where's my error here? Please don't simply respond by objecting that the method is not rigorous.

答:
It is the wrong picture for $\sin\theta.$

The equivalent picture would be something like...


答:
I'll go on from where you stopped at $ a^3|e^{-xy_1^2}-e^{-xy_2^2}|$. Let's note that $ 0 \leq e^{-xy_1^2}\leq 1$ because this is a monotone decreasing function of $x$, and therefore, it receives a maximum when $x=0$, such that:

$ a^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|1-e^{-xy_2^2}|.$

Let's use the fact that $ e^{-xy_2^2}\geq 0$, to conclude that:

$ a^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|1-e^{-xy_2^2}|\leq a^3\cdot 1=a^3$

Title: Find $y$-Lipschitz constant

Body:
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$

I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$

I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$

now I don't know what to do.


问:
How i can find closed form for the following sum?
$\sum_{i=1}^{n} (-1)^{i+1} \binom{n}{k}(n-k)!$

I converted to the below sum.Does anyone know of the closed form formula for this?

$\sum_{i=1}^{n}(-1)^{i+1}\frac{n!}{i!}$

Any hints or suggestions are welcomed.

答:
That’s the number of derangements of $[n]$. There isn’t a nice closed form in the usual sense, but there is one if you allow the floor function:

$$\sum_{i=1}^n(-1)^{i+1}\frac{n!}{i!}=\left\lfloor\frac{n!}e+\frac12\right\rfloor\,.$$

Equivalently, it’s the integer nearest to $\frac{n!}e$.


问:
Weird limit with/without derivation
I have no idea how can I solve the following:

$$ x^2\left(1+\frac{1}{x}\right)^{x}-e\,x^{3}\ln\left(1+\frac{1}{x}\right)$$

with $x$ tends to infinity. Thank you.

Title: Weird limit with/without derivation

Body:
I don't know if you like this or not. Noting
$$ e^t=1+t+\frac12t^2+O(t^3),\ln(1+t)=t-\frac{t^2}2+\frac{t^3}3+O(t^4). $$
Letting $t=\frac1x$ one has
\begin{eqnarray}
&&x^2(1+\frac{1}{x})^{x}-ex^{3}\ln(1+\frac{1}{x})\\
&=&\frac1{t^2}e^{\frac{\ln(1+t)}t}-\frac{e}{t^3}\ln(1+t)\\
&=&\frac1{t^2}e^{1-\frac12t+\frac13t^2+O(t^3)}-\frac{e}{t^3}\bigg(t-\frac{t^2}2+\frac{t^3}3+O(t^4)\bigg)\\
&=&\frac e{t^2}e^{-\frac12t+\frac13t^2+O(t^3)}-\frac{e}{t^3}\bigg(t-\frac{t^2}2+\frac{t^3}3+O(t^4)\bigg)\\
&=&\frac e{t^2}\bigg(1+(-\frac12t+\frac13t^2+O(t^3))+\frac12(-\frac12t+\frac13t^2+O(t^3))^2+O(t^3)\bigg)-\frac{e}{t^3}\bigg(t-\frac{t^2}2+\frac{t^3}3+O(t^4)\bigg)\\
&=&\frac e8+O(t)
\end{eqnarray}
and hence
$$ \lim_{x\to\infty}\bigg[x^2(1+\frac{1}{x})^{x}-ex^{3}\ln(1+\frac{1}{x})\bigg]=\frac e8. $$


问:
Compute the probability that $\Pr(\tau<\infty)$.
Let $B$ and $W$ be independent symmetric random walks starting at $0$ with filtration $\mathcal F_0=\{\emptyset,\Omega\}$ and $\mathcal F_n=\sigma(B_1,...,B_n,W_1,...,W_n)$.

Define $\mathcal E(B)_n=e^{B_n}(\cosh 1)^{-n}$ and $\mathcal E(W)_n=e^{W_n}(\cosh 1)^{-n}$ for all $n\ge 0$.

Also let $\tau=\inf\{n:\mathcal E(B)_n\le\frac 12\},\sigma=\inf\{n:\mathcal E(W)_n\ge 2\}$.

(a) Calculate $\Pr(\tau<\infty)$ and show that $\Pr(\sigma<\infty)\in (0,1)$.

(b) Which of the following three processes are uniformly integrable martingales:
$$\mathcal E (B)^{\sigma\wedge\tau},\mathcal E(W)^{\sigma\wedge\tau},\mathcal E (B)^{\sigma\wedge\tau}\mathcal E(W)^{\sigma\wedge\tau}.$$

We can write
\begin{align}
\Pr(\tau<\infty)&=1-\Pr(\mathcal E(B)_n>\frac 12\text{ for all }n).
\end{align}
But then how to proceed. I have no idea why we are trying to compute $\mathcal E(B)_n$ and neither do I know how to compute this probability. Can someone give me a hint? Thank you!

Edit:

I have a rough idea but it still does not work. We need to involve $\Pr(\tau<\infty)$.

Firstly, it is easy to show that $\mathcal E(B)_n,\mathcal E(W)_n$ are martingales. Secondly, it is easy to see that $\tau,\sigma$ are stopping times with respect to the filtration $\mathcal \{F_n\}_{n\in\mathbb Z_{>0}} $ . So I want to use the Optimal Stopping Theorem. Note that
\begin{align}
\mathbb E[\mathcal E(B)_\tau]&=\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}].
\end{align}
By Fatou's lemma, $$\mathbb E[\lim_{n\to\infty}\mathcal E(B)_{n\wedge\tau}]\le\liminf_{n\to\infty}\mathbb E[\mathcal E(B)_{n\wedge\tau}].$$
Since $\mathcal E(B)_n$ is a martingale, we have
$$ \mathbb E[\mathcal E(B)_{n\wedge\tau}]=\mathbb E[\mathcal E(B)_{0\wedge\tau}]=1. $$

So $1\ge \mathbb E[\mathcal E(B)_\tau].$

Write $\mathbb E[\mathcal E(B)_\tau]=\mathbb E[\mathcal E(B)_\tau 1_{\tau<\infty}+\mathcal E(B)_\tau 1_{\tau=\infty}]=\mathbb E[\mathcal E(B)_\tau;\tau<\infty]+\mathbb E[\mathcal E(B)_\tau;\tau=\infty].$

Remark: Note that we have applied Fatou's lemma, but we cannot apply the dominated convergence to get equality, since the random walk can visit any big number before it returns to some point such that $\mathcal E(B)_n\le\frac 12$.

答:
I made some progress, but still not there.

Let $X_n=B_n-B_{n-1}$, and we know $\Pr(X_n=1)=\Pr(X_n=-1)=0.5\ \forall n$, and they are independent.

Let $Y_n=e^{X_n} (\cosh 1)^{-1}>0$, it is easy to see that $\mathbb E Y_n=1$ and they are also independent since $Y_n$ is the conposition composition of a continuous function(hencec Borel measurable) and $X_n$.

Note that $\mathcal E(B)_0=1$ and for all $n>0$: $\mathcal E(B)_n=e^{B_n}(\cosh 1)^{-n}=\prod_{i=1}^n[e^{X_i} (\cosh 1)^{-1}]=\prod_{i=1}^nY_i$.

By Kakutani's theorem we conclude that $\mathcal E(B)_n\to 0$ a.s.(since $\mathbb E\sqrt{Y_i}<1$(generalized mean inequality or see:Prove inequality of generalized means) and it follows that $\prod_{i=1}^\infty Y_i=0 $)

Hence $\Pr(\tau<\infty)=1$ by definition.

Now we study $\Pr(\sigma<\infty)$.

Firstly, it is easy to show that $\mathcal E(B)_n,\mathcal E(W)_n$ are martingales.

Secondly, it is easy to see that $\tau,\sigma$ are stopping times with respect to the filtration $ \{\mathcal F_n\}_{n\in\mathbb Z_{>0}} $ . So I want to use the Optimal Stopping Theorem. Note that
\begin{align}
\mathbb E[\mathcal E(W)_\sigma]&=\mathbb E[\lim_{n\to\infty}\mathcal E(W)_{n\wedge\sigma}].
\end{align}
Since $\mathcal E(W)_{n\wedge\sigma}\le 2e(\cosh 1)^{-1}\ \forall n$, by the bounded convergence theorem we have $$\mathbb E[\lim_{n\to\infty}\mathcal E(W)_{n\wedge\sigma}]=\lim_{n\to\infty}\mathbb E[\mathcal E(W)_{n\wedge\sigma}].$$
Since $\mathcal E(W)_n$ is a martingale, we have
$$ \mathbb E[\mathcal E(W)_{n\wedge\sigma}]=\mathbb E[\mathcal E(W)_{0\wedge\sigma}]=1. $$

So $1= \mathbb E[\mathcal E(W)_\sigma].$

Write $$\mathbb E[\mathcal E(W)_\sigma]=\mathbb E[\mathcal E(W)_\sigma 1_{\sigma<\infty}+\mathcal E(W)_\sigma 1_{\sigma=\infty}]=\mathbb E[\mathcal E(W)_\sigma;\sigma<\infty]+\mathbb E[\mathcal E(W)_\sigma;\sigma=\infty].$$

If $\Pr(\sigma<\infty)=1$, then $1=\mathbb E[\mathcal E(W)_\sigma; \sigma<\infty]\ge 2$. Contradiction!


问:
How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$

答:
$$\sin(x) = \sin(a) + (x-a)\cos(a) + O((x-a)^2)$$
so the argument of the limit becomes:
$$(1+(x-a)\cot(a) )^{\frac{1}{x-a}}$$
Make $x-a=\epsilon$:
$$(1+\epsilon\cot(a) )^{\frac{1}{\epsilon}}$$

and this is related to the number e, as you can see in:
https://www.math24.net/number-e/#example3

Substitute $y=\epsilon\cot(a)$:
$$(1+y )^{\frac{\cot(a)}{y}}  =  \left[(1+y)^{\frac{1}{y}}\right]^{\cot(a)}$$


问:
Subrings of $\mathbb{Q}$
Let $p$ be prime. Suppose $R$ is the set of all rational numbers of the form $\frac{m}{n}$ where $m,n$ are integers and $p$ does not divide $n$.

Clearly then $R$ is a subring of $\mathbb{Q}$.

I now want to show that if $\frac{m}{n}$ belongs to any proper ideal of $R$ then $p|m$.

Can someone point me in the right direction.

Thanks

答: Hint: If $p$ doesn't divide $m$, can you show that $m/n$ is a unit?

Title: Subrings of $\mathbb{Q}$


问:
$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum.
I need to find supremum of:

$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$

I found out that:

$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} = \frac{m^2(n^3 + 3n^2) + 2m(n^3+3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$

$$\frac{(m^2+2m)(n^3 + 3n^2)}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}= \frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}$$
And from inequality of means (AM - GM) I know that:

$$\frac{\sqrt{(m^2+2m)^2(n^3 + 3n^2)^2}}{(m^2 + 2m)^2+(n^3 + 3n^2)^2} \leq \frac{1}{2}$$

Therefore I found a limitation of my set A. Now I need a sequence of elements belonging in $A$ that would be convergent to $\frac{1}{2}$. That is the part I can't do myself.

From graph I know that if I take $m = n$ then for $n \in [1 ; \infty)$ I get decreasing function and for that function $\frac{1}{2}$ is reached by $n < 1$. A maximal value of that function for $n \in  [1 ; \infty)$ is equal to $\frac{12}{25}$ and is reached by $n = 1$. But that is not a solution.

There is no ANY fact that would legitimise value $\frac{12}{25}$ as the supremum of that set since there always might by something bigger if $n \neq m$.

What I need is a prove that in the supremum of set $A$: $m = n$ (I don't know why that would be the case) or that there is a bigger value reached for $n$ and $m$ that are not equal.

It might be important to add that I can not use integrals nor derivatives.

答:
This is not an answer, but something you could try. You could still use differentiation. Replace the function with a continuous one,
$$f(x,y)=\dfrac{x^2y^3+3x^2y^2+2xy^3+6xy^2}{(x^2+2x)^2+(y^3+3y^2)^2},$$
Calculate $\max_{(x,y)\in\mathbb{R}^2}f(x,y)$. Then calculate the values near it which correspond to integers and pick amongst them.

PS:- Please try to make the questions a little less "wordy". At least please try to keep that in mind for all future posts.


问:
Sequence of shifts in Hilbert Space
I'd like someone to check my work on this 2-part problem:

Let $(V_n )_{n \in \mathbb{N}}$  be a sequence of operators on $\ell^2(\mathbb{N})$ defined by $(V_n x)(k) = x(n + k)$. Prove that $(V_n )_{n \in \mathbb{N}}$ converges in strong operator topology and identify its limit. Prove that the sequence of
adjoints $(V^*_n )_{n \in \mathbb{N}}$ does not converge in the strong operator topology.

My approach:

Consider the limit of the square norm of the image:
$lim_{n\to \infty}||V_n x||^2 =lim_{n\to \infty} \sum_{k=n}^\infty{x_k^2} = ||x||^2 - lim_{n\to \infty}\sum_{k=1}^n{x_k^2} = 0$. Therefore $||V_nx|| \to 0$ for all $x$, and $V_n$ converges to $0$ in SOT.

For the second part, i'm quite sure that the adjoint of $V_n$ is $(V^*_nx)(k)= 0$ if $k\leq n$, and $(V^*x)(k)=x(k-n)$ otherwise. It is then enough to mention an $x$ for which the limit $V^*_nx$ does not exist. $x_0=1,0,0...$ will do, since the sequence $V^*_nx_0$ is not Cauchy and therefore not convergent. Therefore, $V_n^*$ does not converge in SOT.

any corrections and remarks are greatly appreciated, thank you!

答:
Yes, your approach is correct. Well done!
I would like to point out that

i'm quite sure that

does not sound good as a mathematical argument,
but the adjoint is correct.
Maybe you can cite a source for that claim
or briefly confirm that the adjoint is calculated correctly.

Also, the notation
$$x_0=1,0,0...$$
should be replaced with
$$x_0=(1,0,0...)$$
in my opinion.

Title: Sequence of shifts in Hilbert Space


问:
Find the best constant $c < 0$ for $c a_n = a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0$
Let's define the following recurrence relation:

$$
\begin{align}
a_0 &= 1 \\
c a_n &= a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0
\end{align}
$$

Find the best constant $c < 0, \forall n \in \mathbb{N}$ which is farest from zero such that

$$
\begin{align}
a_n &> 0, \quad \text{if n is even} \\
a_n &< 0, \quad \text{if n is odd} \\
\end{align}
$$

My guessed answer would be $-\frac{1}{4}$. But how to prove it?

My attempt so far is to reduce the recurrence relation:

$$
\begin{align}
a_n &= \sum\limits_{k=1}^n \frac{k a_{n-k}}{c} = \frac{a_{n-1}}{c} + \sum\limits_{k=2}^n \frac{(k-1) a_{n-k}}{c} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + \sum\limits_{k=3}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + a_{n-1} - \frac{c + 1}{c} a_{n-2} \\
&= \frac{2c + 1}{c} a_{n-1} - a_{n-2} \\
\end{align}
$$

If I set my guess $c = -\frac{1}{4}$ in this reduced recurrence relation then:

$$
\begin{align}
a_n = -2 a_{n-1} - a_{n-2}
\end{align}
$$

How to proceed? Or is there any other way to prove it?

答:
From the recurrent relation $a_n = \frac{2c + 1}{c} a_{n-1} - a_{n-2}$ with $a_0 = 1$ and $a_1 = 1/c$, you can obtain an explicit formula for $a_n$:
$$a_n = p_1x_1^n+p_2x_2^n$$
with $x_1, x_2$ are the roots (can be non real) of the equation : $x^2 =  \frac{2c + 1}{c} x - 1$ and $p_1,p_2$ are the solution of the system of 2 linear equations:
$$p_1 + p_2 = a_0 = 1$$
$$p_1 x_1 + p_2 x_2 = a_1 = 1/c$$

After obtaining the analytic formula of $a_n$, the problem becomes easy. You just need to study the sign of $a_n$ in case $n$ is odd or even.


问:
Determine for which $a$ it holds that $\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$
Let $a>0$.

Determine for which values of $a$ it holds that:

$$\lim_{(x,y)\to(0,0)}\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0$$

The solution indicated is $a>3$.

Passing in polar coordinates, I obtain:

$$\frac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{(\cos\theta+\sin\theta)}$$

If $a > 3$, when $\rho \to 0$, the fraction goes to $0$, and this does not depend on $\theta$.

However, this is not enough for the limit to exist and it must exist a $g(\rho)$ such that:

$$\frac{\rho^{a-3}|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}\leqslant g(\rho)\to 0 $$

But

$$\frac{|\cos\theta-\sin\theta|^{a-1}}{\cos\theta+\sin\theta}$$

is not limited, because, for example, when $\theta \to \pm \frac{3}{4}\pi$, it goes to $\mp \infty$.

Where is the mistake?

Thanks in advance.

答:
Determine for which values of $\;a\;$ it holds that:
$$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=0\;.$$

By letting $\;X=x-y\;$ and $\;Y=x+y\;,\;$ it follows that

$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;.$

First case : $\;a\le1\;.$

$\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}\;,$

but the last limit does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed

$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$

$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y>0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$

$=+\infty\cdot(+\infty)=+\infty\;,$

$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{Y|X|^{1-a}\sqrt{X^2+Y^2}}=$

$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac1{Y|X|^{1-a}}$$\cdot\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y<0\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+Y^2}}=$

$=-\infty\cdot(+\infty)=-\infty\;.$

Second case : $\;a>1\;.$

The limit $\;\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}\;$ does not exist because we get different results when we calculate the limits of the restrictions of the function to two different subsets of its domain, indeed

$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&X=0\end{align}}\dfrac0{Y\big|Y\big|}=0\;,$

$\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}=$

$=\lim\limits_{\begin{align}(X,&Y)\to(0,0)\\&Y=|X|^{a-1}\end{align}}\dfrac{\sqrt2}{\sqrt{X^2+X^{2(a-1)}}}=+\infty\;.$

Conclusion :

In any case, for any $\;a\in\mathbb{R}\;,$ the limit

$\lim\limits_{(x,y)\to(0,0)}\dfrac{|x-y|^{a-1}}{(x+y)\sqrt{x^2+y^2}}=\lim\limits_{(X,Y)\to(0,0)}\dfrac{\sqrt2\;|X|^{a-1}}{Y\sqrt{X^2+Y^2}}$

does not exist.


问:
There are at least two primes with power 1 in $k!$ for $k>11$
My question is: need to prove that $\forall \ k \geq 11, \exists$ at least 2 prime numbers with power of 1, in k! factorization. I know I missed something, and that is why I'm looking for a hint. Thanks for help!

*** I can use that $\forall \ k > 5$ there exist at least $2$ prime numbers in the next interval: $(k,2k)$.

Body:
I can suggest using Legendre's formula
$$\nu _{p}(n!)={\frac  {n-s_{p}(n)}{p-1}}$$
Now, let's consider the primes $p_1,p_2$ s.t.
$$\frac{n}{2} < p_1<p_2 < n \tag{1}$$
Then
$$n=p_1 q_1 +r_1, 0\leq r < p_1-1$$
$$n=p_2 q_2 +r_2, 0\leq r < p_2-1$$
Both $q_1=q_2=1$ (otherwise, from $(1)$, $n>2p_1+r_1>n$ and similarly $n>2p_2+r_1>n$) and $r_1>0$, $r_2>0$. As a result
$$s_{p_1}(n)=q_1+r_1\geq 2$$
$$s_{p_2}(n)=q_2+r_2\geq 2$$
And finally
$$\nu _{p_1}(n!)={\frac  {n-s_{p_1}(n)}{p_1-1}} < {\frac  {2p_1-2}{p_1-1}}=2$$
$$\nu _{p_2}(n!)={\frac  {n-s_{p_2}(n)}{p_2-1}} < {\frac  {2p_2-2}{p_2-1}}=2$$

Of course $(1)$ works ideally for even $n$'s. What about odd $n$'s? In this case, we consider
$$\frac{n+1}{2} < p_1<p_2 < n+1$$
since $n+1$ is even and can not be prime. However we may have $p_2=n$, but then
$$\nu _{p_2}(n!)={\frac  {n-s_{p_2}(n)}{p_2-1}}={\frac  {p_2-1}{p_2-1}}=1$$


答:
You can manipulate the expression, so that

$\lim\limits_{x\to a} \left(1+\dfrac{1}{\dfrac{\sin{a}}{\sin{x}-\sin{a}}}\right)^{\dfrac{\sin{a}}{\sin{x}-\sin{a}}\dfrac{\sin{x}-\sin{a}}{\sin{a}}\dfrac{1}{x-a}}$

which is
$\;\lim\limits_{x\to a}\;e^{\dfrac{\sin{x}-\sin{a}}{x-a}\dfrac{1}{\sin{a}}}=e^{\;\dfrac{\cos a}{\sin a}}$

This is how the exponent becomes $\;\cot{a}$

Edit: Maybe I skipped a few too many steps
You can achieve the upper one expression by adding and subtracting one.
So

$1+\dfrac{\sin{x}}{\sin{a}}-1=1+\dfrac{\sin{x}-\sin{a}}{\sin{a}}=1+\dfrac{1}{\dfrac{\sin{a}}{\sin{x}-\sin{a}}}$

The exponent is

$1\cdot \dfrac{1}{x-a}=\dfrac{\sin{a}}{\sin{x}-\sin{a}}\dfrac{\sin{x}-\sin{a}}{\sin{a}}\dfrac{1}{x-a}$

Title: How can I solve this limit? $\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$

Body:
$\lim _{x \to a}\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}$
Where $a$ can be any number.

I started by using $e$ like this:

$\lim _{x \to a}e^{\ln\left(\left(\frac{\sin(x)}{\sin(a)}\right)^{\frac{1}{x-a}}\right)} \to \lim _{x \to a}e^{\ln\left(\frac{\sin(x)}{\sin(a)}\right)\cdot\frac{1}{x-a}} \to \lim _{x \to a}e^{\frac{\ln\left(\sin(x)\right)-\ln\left(\sin(a)\right)}{x-a}}$

The answer should be $e^{\cot{a}}$