问:
Can someone clarify this implication
I'm reading a finance book, and I saw this implication that I don't understand. I mean where this g function come from? If someone can clarify this I would appreciate. Thanks.

If a have a function like $f(x,t)$ and the following equation

$ \frac{d f}{f}=\sigma dx \Rightarrow lnf - lng(t)=\sigma x$

答:
What you actually are looking at is the equation

$$\frac{df(x,t)}{dx}=\sigma f(x,t)$$

One common trick of solving this, is abusing the notation and writing this as

$$\frac{df}{f}=\sigma dx$$

and then integrating both sides. This is what was done in the book.
Note however that we are differentiating with respect to $x$, although the function depends on $x$ and $t$. One way of understanding this, is that you solve the equation for every possible $t$, one at a time. So imagine a fixed $t$ and solve the differential equation. You will get a constant from the integration. But for any 'other' given $t$ the constant you get, might be a different one. So if you look at the global picture, the constants depend on $t$. In other words, they form a function $g(t)$.

If you want to do a reality check and just convince yourself that the solution makes sense, rearrange the solution for $f$ and differentiate. You will see that the function $g(t)$ dies during the process.

Title: Can someone clarify this implication


问:
Boundary Flowout Theorem
I'm trying to prove Theorem 9.24 (Boundary Flowout Theorem) from John. M. Lee's Introduction to Smooth Manifolds . There, all steps are like proving Flowout Theorem , but in last one, I just can show that ${\Phi|_o}_{\delta}$ is an injective smooth immersion, not an embedding
(We can't use Proposition 4.22(d) beacause of boundary in M).

Also there is a hint in Problem 9.11  but I have no idea how to use it:

"define $\Phi$ first
in boundary coordinates and use uniqueness to glue together the local definitions.
To obtain an embedding, make sure $\delta(p)$ is no more than half of the
first time the integral curve starting at $p$ hits the boundary (if it ever does)."

答:
The theorem that you're asking how to prove is the following (rephrased from the book):

Boundary Flowout Theorem: Suppose that $M$ is a smooth manifold with nonempty boundary and that $N$ is a smooth vector field over $M$ such that $N$ is inward-pointing at any point of the boundary $\partial M$. Then there exists a smooth positive function $\delta : \partial M\rightarrow\mathbb{R}_+$ and a smooth map $\Phi : \mathcal{P}_\delta\rightarrow M$, where
$$\mathcal{P}_\delta=\left\{ \left(t,p \right) : p\in\partial M\ \ 0\le t<\delta \left(p \right)\right\},$$
such that $\Phi$ is a smooth embedding onto an open neighborhood of $\partial M$ that satisfies the following property: for any $p\in\partial M$, the curve $t\mapsto\Phi \left(t,p \right)$ for $0\le t<\delta \left(p \right)$ is the integral curve of $N$ starting at $p$.

You should follow the first half of the hint given in the book, which is done as follows:

1.) Choose boundary coordinates $ \left(U,\varphi \right)$ of $M$.

2.) Push $N$ down the image of $\varphi$ to $\mathbb{H}^n$ to get $\hat{N}=\varphi_\ast N$, and then extend that smoothly to an open subset of $\mathbb{R}^n$. Call that smoothly extended vector field ${\hat{N}}_E$ ("$E$ " for "extension").

3.) Then use the flow of ${\hat{N}}_E$ to construct a local version of $\Phi$.

4.) As a crucial step, prove that the local versions of $\Phi$ constructed as so are smooth embeddings onto open sets.

5.) Then glue these local constructions of $\Phi$ by a smooth partition of unity and prove that it satisfies the desired properties stated in the theorem.

In step 5, once you're able to prove that the constructed $\Phi$ is an injective smooth immersion, in order to prove that it's a topological embedding just note that it can be restricted to open subsets of $\mathcal{P}_\delta$ on which it's equal to the local versions of $\Phi$ that you constructed in step 3. Since you proved in step 4 that the local versions of $\Phi$ are smooth embeddings onto open sets, the same will hold for $\Phi$ on these open subsets of $\mathcal{P}_\delta$. That with the injectivity of $\Phi$ then implies that $\Phi$ is a smooth embedding onto an open subset of $M$ (which is clearly a neighborhood of $\partial M$ since $\Phi \left(0,p \right)=p$). I wrote up a full proof on my website:

https://sites.math.washington.edu/~hgrebnev/D&Writings/Z_PDF_Documents_I/Jack%20Lee%20Smooth%20Manifolds%20Notes.pdf

As is pointed out in the author's errata: the second part of the hint given in the book doesn't actually apply to this theorem because the integral curves of $N$ can never in fact hit the boundary. The reason for this is that if an integral curve of $N$ starting at a boundary point $p\in\partial M$ ever hit the boundary again, say at a point $q\in\partial M$, then $N$ would not be inward pointing at $q$.

ContentLicense: CC BY-SA 4.0

Title: Boundary Flowout Theorem


问:
Let $u_1,u_2,u_3,u_4$ be four linearly independent vectors in $R^n$. For what value of $a$ is the set linearly dependent?
Let $u_1,u_2,u_3,u_4$ be four linearly independent vectors in $R^n$. For what value of $a$ is the set {${u_1+4u_2, u_2+3u_3, u_3+3u_4, u_4+au_1}$}
linearly dependent?

My method for solving this question was to write the terms as a vector equation equal to 0, expand the equation, and collect like terms until I was left with an equation of the following form. The $c$ terms are scalars multiplying the vectors:

$$u_1(c_1+c_4a) +u_2(4c_1+c_2)+u_3(3c_2+c_3)+u_4(3c_3+c_4)$$

I then set the four $c$ expressions equal to zero, and solved between them to determine that $a=1/36$ makes the set linearly dependent.

What I don't understand is how setting the scalar terms equal to zero, which I believe indicates the trivial solution and thus linear independence, returns the value of $a$ that will cause linear dependence.

答:
Let's examine this more closely then. You wish to test the linear independence of these four vectors. They are linearly independent if the only solution for $c_1, c_2, c_3, c_4$ to the following equation:
$$c_1(u_1+4u_2) + c_2(u_2+3u_3) + c_3(u_3+3u_4) + c_4(u_4+au_1) = 0 \tag{1}$$
is $c_1 = c_2 = c_3 = c_4$. Otherwise the set is linearly dependent. As you note, $(1)$ is equivalent to
$$(c_1+c_4a)u_1 +(4c_1+c_2)u_2+(3c_2+c_3)u_3+(3c_3+c_4)u_4 = 0. \tag{2}$$
As $u_1, u_2, u_3, u_4$ are assumed to be linearly independent, we can conclude from this that
$$c_1 + c_4 a = 4c_1 + c_2 = 3c_2 + c_3 = 3c_3 + c_4 = 0. \tag{3}$$
Note also the converse holds true too. If $(3)$ holds, then so does $(2)$ rather trivially. So, in total, we have $(1) \iff (2) \iff (3)$. The linear independence of our set is equivalent to the system of equations in $(3)$ having only the solution $c_1 = c_2 = c_3 = c_4 = 0$; if there are non-trivial solutions to $(3)$, then there are non-trivial solutions to $(1)$, and if there are no non-trivial solutions to $(3)$, then the same can be said of $(1)$.

Note that now we have removed all mention of $u_1, u_2, u_3, u_4$. This is a question purely in terms of systems of linear equations now.

Next comes the solving. Rearranging (ready for substitution), we get
\begin{align*}
c_1 &= -c_4 a \\
c_2 &= -4c_1 \\
c_3 &= -3c_2 \\
c_4 &= -3c_3.
\end{align*}
Then, performing the substitutions:
$$c_4 = -3c_3 = -3(-3c_2) = 9(-4c_1) = -36(-c_4a) = 36ac_4,$$
or, rearranging,
$$(36a - 1)c_4 = 0.$$
Now, let's consider two possibilities here: either $36a - 1 = 0$ or not. If $36a - 1 \neq 0$, then we may divide both sides by $36a - 1$ (we can divide by any non-zero number), and hence the above equation implies $c_4 = 0$. From here, we may back substitute into our system to discover $c_1 = 0$, hence $c_2 = 0$, and finally $c_3 = 0$. That is, the only solution when $36a - 1 \neq 0$ is $c_1 = c_2 = c_3 = c_4 = 0$. Hence this is the only solution for $(1)$ as well, and so our vectors are linearly independent if $36a - 1 \neq 0$.

Otherwise, if $36a - 1 = 0$, i.e. if $a = 1/36$, then we suddenly get a free variable $c_4$. Let's choose a non-zero value for $c_4$, e.g. $c_4 = 1$. Then $c_1 = -c_4a = -1/36$, $c_2 = -4c_1 = 1/9$, $c_3 = -3c_2 = -1/3$, and we can verify that these values then also satisfy $c_4 = -3c_3$ as well. Thus, we have a non-trivial solution
$$(c_1, c_2, c_3, c_4) = (-1/36, 1/9, -1/3, 1),$$
which implies that $(1)$ is linearly dependent.

I hope that clarifies things.


问:
interesting Integral , alternative solution.
Show the following relation:

$$\int_{0}^{\infty} \frac{x^{29}}{(5x^2+49)^{17}} \,\mathrm dx = \frac{14!}{2\cdot 49^2 \cdot 5^{15 }\cdot 16!}.$$

I came across this intgeral on a physics forum and solved by  (1) making a substitution (2) finding a recursive formula (with integration by parts).

I would like to see a clever alternative solution to this one, since the only way I would solve such an integral is by finding a recursive formula. But I'm sure there are other ways/methods to do it (maybe more elegant) that  I want to discover.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over
\pars{5x^{2} + 49}^{17}}\,\dd x
= {14! \over 2\cdot 49^{2} \cdot 5^{15}\cdot 16!}}:\
{\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over
\pars{5x^{2} + 49}^{17}}\,\dd x}
\,\,\,\stackrel{5x^{2}/49\ \mapsto\ x}{=}\,\,\,
{1 \over 2\cdot 49^{17}}\,\pars{49 \over 5}^{15}
\int_{0}^{\infty}{x^{\color{red}{15} - 1} \over \pars{1 + x}^{17}}\,\dd x
\end{align}
It can be evaluated by means of the
Ramanujan's Master Theorem.
Namely,
\begin{align}
&{1 \over \pars{1 + x}^{17}} =
\sum_{k = 0}^{\infty}{-17 \choose k}x^{k} =
\sum_{k = 0}^{\infty}{16 + k \choose k}\pars{-1}^{k}x^{k}
\\[5mm] = &\
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{17 + k} \over 16!}{\pars{-x}^{k} \over k!}
\end{align}
and
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{x^{29} \over
\pars{5x^{2} + 49}^{17}}\,\dd x} =
{1 \over 2\cdot 49^{2} \cdot 5^{15}}\,
\Gamma\pars{\color{red}{15}}\,
{\Gamma\pars{17 - \color{red}{15}} \over 16!}
\\[5mm] = &\
{1 \over 2\cdot 49^{2} \cdot 5^{15}}\,14!\,{1! \over 16!} =
\bbx{{14! \over 2\cdot 49^{2} \cdot 5^{15}\cdot 16!}}
\approx 2.8433 \times 10^{-17} \\ &
\end{align}

ContentLicense: CC BY-SA 3.0


问:
Derivative question
I need to find the derivative of:

$$
h(x) = \int_{0}^{x^2} (1-t^2)^{1/3} \, dt
$$

Would the answer to that just be:

$$
(1-x^4)^{1/3}?
$$

答: Let $v = x^2$ then $$\frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = f(v)$$ so $$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} f(t) \mathrm{d}t = \frac{\mathrm{d}v}{\mathrm{d}x} \frac{\mathrm{d}}{\mathrm{d}v} \int_0^{v} f(t) \mathrm{d}t = 2x f(x^2).$$</p>

Title: Derivative question


问:
rounding up to nearest square
Say I have x and want to round it up to the nearest square. How might I do that in a constant time manner?

ie.

$2^2$ is 4 and $3^2$ is 9. So I want a formula whereby f(x) = 9 when x is 5, 6, 7 or 8. What it does when x = 4 or 9 doesn't really matter.

I could write a function to do this easily enough. ie.

```<code>while (sqrt(x) is not int) {
  x++;
}
</code>```

But I'd rather not do a brute force approach. Any ideas?

Thanks!

答: You can take the ceiling of $\sqrt{x}$ and square it.</p>

Title: rounding up to nearest square


问:
Random polynomial and $x^n - 1$
Suppose $f(x) = x^n - 1$ over $\mathbb{F}_2$.

Now suppose that $g(x)$ is a randomly chosen polynomial of degree $\leq n - 1$ over $\mathbb{F}_2$

What is the probability that $\gcd(f, g) = 1$?

答:
Let's generalize:

How many monic polynomials of degree $D$ divide $x^n-1$ within $\mathbb{F}_q$?

We will reduce this to the problem of counting nonnegative solutions to a linear equation.

First, write $n=q^km$ with $q\nmid m$, so $x^n-1=(x^m-1)^{q^k}$. The irreducible factors of $x^m-1$ correspond to the $\mathrm{Gal}(\mathbb{F}_q(\mu_m)/\mathbb{F}_q)$-orbits (where $\mu_m$ are the roots of $x^m=1$). We don't actually need to calculate these orbits or the corresponding factors - it suffices to determine how many factors there are of each degree.

Every $m$th root is expressible as $g^{uv}$ where $u\in(\mathbb{Z}/m\mathbb{Z})^{\times}$ and $v\mid m$. (This can be proved by using the Chinese Remainder Theorem on $\mathbb{Z}/m\mathbb{Z}$.) Note $g^{uv}$ is a primitive $(m/v)$th root of unity. The size of this orbit is the smallest $k$ for which $q^k(uv)\equiv uv$ mod $m$, which is equivalent to $(m/v)\mid (q^k-1)$ (note the Frobenius automorphism $\wp(x)=x^q$ generates the Galois group), which is $k=\mathrm{ord}_{m/v}(q)$. Setting $d=m/v$, we can say there are $\varphi_d=\phi(d)/\mathrm{ord}_d(q)$-many irreducible factors of $x^m-1$ of degree $\mathrm{ord}_d(q)$ which are minimal polynomials of $d$th roots.

We may write $g(x)$ as a product of these irreducible factors, each with multiplicity $e\le q^k$.

Therefore, the answer to the general problem equals the number tuples $\big(e_i^{(d)}\big)$ satisfying

$$ \sum_{d\mid m} \mathrm{ord}_d(q)\big(e_1^{(d)}+\cdots+e^{(d)}_{\varphi_d}\big) = D $$

with $0\le e_{i}^{(d)}\le q^k$ for all $d\mid m,1\le i\le \varphi_d$.

Title: Random polynomial and $x^n - 1$


问:
In computing eigenvalues, why does $A− \lambda I$ have to be a singular matrix?
This question is being asked not because textbooks do not provide for an explanation about this. They do but, I really can't seem to grasp it very well. I know matrices but honestly my professor has not defined what a singular matrix means.

" To be of any use, the nullspace of $A− \lambda I$ must contain vectors other than zero."

I also do not have an idea of what nullspace is. I have tried searching about it but would also like to ask for those concepts here as it might be explained better.

I get that with eigenvalues and eigenvectors, we don't want the eigenvector $v$ in $Av = \lambda v$ to be zero because that would just result into a useless solution. But I don't get where all the talk about it being singular came from. I do understand from a definition I read that a singular matrix has a determinant of zero, which led to why $|A− \lambda I| = 0$ came and from there I can do the solutions.

I have read about the derivation of the equation $|A− \lambda I| = 0$ and the only part I don't get is how it was concluded that $A− \lambda I$ has to be singular. I am aware that probably I don't have a good grasp of the definitions and if I did I would understand why it led to becoming singular. The idea is probably everywhere on the Internet, I just do not manage to get it or find a good enough reference for me to do so. But that's why I'm asking here, probably a good explanation or a reference with a good explanation will be mentioned. Thank you.

答:
Start with $Av=\lambda v$. Equivalently, we could write $Av=\lambda Iv$, since $Iv=v$. Then we can bring both terms to the left to get $Av-\lambda Iv=0$, or equivalently,
$$(A-\lambda I)v=0.$$
This means that $v$ is in the nullspace of the matrix $A-\lambda I$ (by definition, any vector $x$ such that $Mx=0$ is in the nullspace of matrix $M$). Any matrix with a non-trivial nullspace is singular.

We can also work backward. If $A-\lambda I$ is singular matrix, then we know that it must have at least one vector $v$ in its nullspace (other than the zero vector). This vector satisfies
$$(A-\lambda I)v=0\implies Av-\lambda I v=0\implies Av=\lambda Iv=\lambda v.$$
Therefore, if $A-\lambda I$ is singular, any vector in its nullspace is an eigenvector of $A$ with eigenvalue $\lambda$.


问:
Solve the recursion, $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$
Bring the following recursion relation to an explicit expression:

$$a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$$
$a_{0} = 0$, $a_1 = 1$, $a_2 = 2$

All the examples I have seen were with maximum 2 steps back ($a_{n-2}$) and I thought I know how to solve those but I'm having a hard time both with the Generating function and the Characteristic polynomial methods.

The Generating function should start from $n = 3$ but what would happen to the defined values?

For the Characteristic polynomial, Does it mean I'll have to find the roots of a polynomial from a 3rd degree?

答:
To expand one of the comments, if $A_n$ is a solution of the homogeneous part $$A_n=3A_{n-1}-3A_{n-2}+A_{n-3}$$ and $B_n$ satisfies $$B_n=3B_{n-1}-3B_{n-2}+B_{n-3}+8$$ then $kA_n+B_n$ satisfies the original recurrence.

The characteristic polynomial of the homogeneous part is $(x-1)^3$. I have given some answers to questions similar in content going into a clunky form of why the following works - generating functions give the same answer, and a sound proof - $A_n=(pn^2+qn+r)1^n$ where $p, q, r $ are arbitrary, and I've emphasised $1^n (=1)$ because it would be $2^n$ if the generating function had a factor $(x-2)^3$, and it would be a cubic in $n$ if it were $(x-\alpha)^4$ (etc).

The difficulty then arises because the inhomogeneous part $8=8\times 1^n$, and $1$ is a solution of the characteristic polynomial, so the obvious test function $B_n=constant$ won't work. In fact, because $A_n$ involves a quadratic, it is the next highest power you test: $B_n=kn^3$ to find the value of $k$.

As I say, generating functions will get you there, as will other methods, if you want to prove the method works. On the other hand, if you want to identify a solution quickly, it is useful to know which functions to test. $A_n+B_n$ has three parameters $p, q, r$ which determine the sequence, and these correspond to the three values of $a_n$ which are required to do the same. So, given the initial values (or three values of $a_m$), it is easy to show that the solution is unique. The test functions I have suggested work, so they provide the solution.

Do try it out, because the theory all hangs together.


问:
Minimizing the maximum of $|x^2 - x - k|$
Let $k$ be a real number. Also, let $M(k)$ be the maximum value of $f(x) = |x^2 - x - k|$ in $0 \le x \le 1$. What are all real numbers $k$ which minimize $M(k)$?

答:
Let $\{T_n(x)\}$ be the Chebyshev polynomials of the first kind
https://en.wikipedia.org/wiki/Chebyshev_polynomials.

Recall that for any given $n\ge 1$, among the polynomials of degree $n$ with leading coefficient $1$,
$\frac{1}{2^{n-1}}T_n(x)$ is the one of which the maximal absolute value on $[-1, 1]$
is minimal which is equal to $\frac{1}{2^{n-1}}$. see "Minimal $\infty$-norm" https://en.wikipedia.org/wiki/Chebyshev_polynomials.

We have
\begin{align}
\max_{x \in [0, 1]} |x^2 - x - k| &= \max_{y \in [-1, 1]} |(\tfrac{y+1}{2})^2 - \tfrac{y+1}{2} - k|\\
&= \frac{1}{4} \max_{y \in [-1, 1]} |y^2 - 1 - 4k|\\
&\ge \frac{1}{4} \cdot \frac{1}{2^{2-1}}\\
&= \frac{1}{8}.
\end{align}
Also, $\frac{1}{8}$ is attainable at e.g. $k = -\frac{1}{8}$ (corresponding to $y^2 - 1 - 4k = \frac{1}{2} T_2(y) = y^2 - \frac{1}{2}$). We claim that $k = -1/8$ is the unique solution to $\max_{y \in [-1, 1]} |y^2 - 1 - 4k| = \frac{1}{2}$.
Indeed, by letting $g(y) = |y^2 - 1 - 4k|$, from $g(0) \le 1/2$ and $g(1) \le 1/2$, we have $|1 + 4k|\le \frac{1}{2}$
and $|4k| \le 1/2$ which results in $k = -\frac{1}{8}$.


问:
How to write negation of statements?
How to write negation of following statements in words?

```<code>1. Any integer is either positive or negative.
2. There is a child who is loved by everyone.
3. The connector is loose or the machine is unplugged.
4. No politician cheat voters.
</code>```

And how to write negation of following statement in symbols?

```<code>5. (∀x)(∃y)(x^2 <y)
</code>```

For 3rd one my answer is "The connector is not loose and the machine is not unplugged" and for 4th one my answer is "politician cheat voters". I am not sure...Are they correct?

答:
Write the negation of each statement

Some telephone can take photographs
All houses have two stories
Some cars are hybrids
All golf course are green
Some drivers are not safe

Title: How to write negation of statements?


问:
Can someone clarify Example I.I.2 from Hardy's Course of Pure Mathematics?
"If $\lambda, m,$ and $n$ are positive rational numbers, and $m > n$, then $\lambda(m^2 − n^2), 2\lambda mn$, and $\lambda(m^2 + n^2)$ are positive rational numbers. Hence show how to determine any number of right-angled triangles the lengths of all of whose sides are rational."

What does he mean by "how to determine any number of right angled triangles"?

答:
We know that $3$ lengths can form a right angled triangle iff those lengths satisfy $a^2+b^2=c^2$.

But no matter what we choose for $m, \; n$ and $ \lambda$.
$$(2\lambda mn)^2+(\lambda(m^2-n^2))^2=4\lambda^2m^2n^2+\lambda^2m^4-2 \lambda^2m^2n^2+\lambda^2n^4$$
$$=\lambda^2m^4+2\lambda^2m^2n^2+\lambda^2n^4=(\lambda(m^2+n^2))^2$$ as we wanted.

So we can choose any number of rationals $m, \; n$ and $\lambda$ to produce any number of right-angled triangles.

A simple example of an infinite solution set is if you keep $m$ and $n$ constant, but vary $\lambda$, you will get infinite solutions as there are an infinite amount of rational numbers.


问:
Bartle's theorem 7.6: If () is Cauchy in measure, there exists a subsequence () of () such that () converges a.e.

I'm having trouble understanding this passage of Bartle's Elements of Integration, chapter 7, theorem:

Theorem 7.6: Let $(f_n)$ be a sequence of measurable real-valued functions which is Cauchy in measure. Then there exists a subsequence which is converges almost everywhere and in measure to a measurable real-valued function $f$.

I can see how it would be this way if the sets were disjoint, because then we would have $\mu(\cup_{j=k}^{\infty}E_k)=\sum_{j=k}^{\infty}\mu(E_k) $. Also I can see that $\mu(E_n\cap E_m) < 2^{-max\{n,m\}}$ but I'm failing to understand how the whole measure of $F_k$ will be less than $2^{-(k-1)}$.

If anyone could help me, I would be most grateful

答:
Use subadditivity:
$$\mu(F_k) = \mu\left(\bigcup_{j=k}^{\infty}E_j\right)
\leq \sum_{j=k}^{\infty}\mu(E_j) \leq \sum_{j=k}^{\infty}2^{-j} = 2^{-k+1}\sum_{j=1}^{\infty}2^{-j} = 2^{-k+1}$$


问:
find fixed vectors $(V = [0, 0, 1])$ new coordinates in new coordinate system
For a robotic camera I'm using, in its default position its Euler angles (roll, pitch, yaw) are 0. In that case, when I am trying to detect planes I look for planes with the normal vector $v = [0, 0, 1]$

However, lets say the camera has moved and now has some given roll, pitch and yaw.

The world is still the same and so I need to find the same planes, what is the equivalent normal vector I should look for in this new position?

Thanks

答:
You could write it as the product of three rotations

$$R = R_r R_pR_y$$

$$R = \begin{bmatrix}1 & 0 & 0 \\0 &\cos \alpha & \sin \alpha \\ 0 & -\sin\alpha & \cos\alpha\end{bmatrix}
\begin{bmatrix}\cos \beta & 0 & \sin \beta \\0 &1 & 0 \\ -\sin\beta & 0 & \cos\beta\end{bmatrix}
\begin{bmatrix}\cos \gamma & \sin \gamma & 0 \\-\sin \gamma &\cos \gamma & 0 \\ 0 & 0 & 1\end{bmatrix}$$

So this would be your rotation matrix to reach from your initial position to your final position if $\alpha, \beta , \gamma$ are roll, pitch and yaw

Can you use this to see what happens to $(0,0,1)$


问:
Motivation behind weak form and the test function?
I am not a mathematician so please bear with me. I have been reading this document to have a better understanding of FEA from the perspective of an engineer.

I am having a lot of difficulty seeing the intuition/motivation behind the weak form and why we multiply by a test function and then integrate.

Almost every text I've seen begins with declaring that we should multiply by a test function and then integrate it as if it's an axiom.

The author states that the motivation behind the weak form is the realization two vectors are equal if their inner products with some arbitrary test function is the same. Then the author continues on.

Where does this test function come from? I understand we later restrict it to be from Sobolev space for the fact that this necessitates that means the function will behave well within the domain of interest (for example by being square integrable).

Q: Why do we use the test function and integrate to find the weak form? What is the motivation/intuition? I understand everything we do afterwards, but it's the initial inception of the weak form that eludes me.

Thank you

答:
In the physical world, it is impossible to measure a function at a point. The best you can do is to measure a local average of a function over a small region near a point.

For example, we conceptualize the temperature field of the room you are sitting in as a function being defined at every point. But really, all you can do is measure the local average of the temperature near a point with a thermometer. You could manufacture more and more accurate thermometers, and the region over which you are taking the average becomes smaller and smaller. But you can never know the exact value of the temperature at a point. Temperature at a point is not even a well-defined concept, if you think about zooming in to the atomic scale.

Integrating against a test function is like taking a "measurement" of a field. By saying that
$$\int_\Omega f(x) v(x) dx = \int_\Omega g(x) v(x) dx$$
holds for all test functions $v$, we are saying that $f$ and $g$ agree for all possible measurements you could make of them. The test function $v$ is like the thermometer averaging profile from the above example.

Edit: Here's a picture I made illustrating the idea:


问:
Constructing a Hausdorff space with a given fundamental group via a matrix space
Consider the topological $\mathbb{R}$-vector space $V$ of $2 \times \infty$ matrices $A$ of the form $$A=\left [ \begin{matrix}
    a_1 & a_2 & a_3 & \cdots \\
    b_1 & b_2 & b_3 & \cdots \\
   \end{matrix} \right ]$$

where all but finitely many entries are zero. The topology is induced by a metric $d$ defined by $d(A,B)=\max_{ij}(|A_{ij}-B_{ij}|)$.

We consider in particular the subspace (topological subspace, not a vector subspace) $U$ of $V$ consisting of matrices whose columns span $\mathbb{R}^2$.

We want to prove two things. First, that $U$ is contractible (as a topological space). Second, and this is really the hard part, we want to use $U$ to construct, for any finite subgroup $G$ of $GL_{2}(\mathbb{R})$, a Hausdorff space $X_{G}$ whose fundamental group is $G$.

I'm already stuck on contractibility. A very naive approach is trying to contract things to $0$, but this doesn't work since the matrix with all entries $0$ does not have columns that span $\mathbb{R}^2$.

答:
I will be using  the notation
$$
  (u_1, u_2,u_3, \ldots ) :=   \pmatrix{a_1 & a_2 & a_3 & \ldots \cr b_1 & b_2 & b_3 & \ldots }
  $$
where each $u_j$ is the $j^{\text{th}}$ column of the matrix on the right hand side.

Here is how to construct a homotopy from the identity  map on $U$ to the shift map
$$
  S:(u_1, u_2,u_3, \ldots ) \mapsto  (0,  u_1, u_2,\ldots ).
  $$

Define
$$
  H: (t, u) \in  [0,1]\times U\mapsto v\in  U,
  $$
where
$$
  v_j = \left\{\matrix{
  (1-t)u_1, & \text{if }  j=1,  \cr
  (1-t)u_j + t u_{j-1}, & \text{if }  j>1.
}
\right.
  $$

In other words,  the $1^{\text{st}}$ coordinate of $H(t, u)$ consists of a  slowly vanishing $1^{\text{st}}$ coordinate
of $u$, while the $j^{\text{th}}$ coordinate of $H(t, u)$ runs along te segment joininig the $j^{\text{th}}$ and
$(j-1)^{\text{st}}$ coordinates of $u$.

It is evident that $H(0,\cdot)$  is the identity map, that $H(1,\cdot)$  is the shift,  so it is enough to check that
$H$ is well defined in the sense that $H(t,u)$ indeed lies in $U$ for every $(t,u)\in  [0,1]\times U$.   In order to do this we
will  verify that when
$$
  v=H(t,u),
  $$
the span of the columns of $v$, here denoted
$\text{span}(v)$, contains $\text{span}(u)$.

When $t=1$ this is obvious, so we assume that $t\in [0,1)$.  Since $v_1$ is a nonzero multiple of $u_1$, we have that $u_1\in
\text{span}(v)$.  By induction, assuming that $u_{j-1}$ lies in $\text{span}(v)$,  notice that
since
$$
  v_j=(1-t)u_j + t u_{j-1},
  $$
we have that
$$
  u_j =   \frac{v_j- t u_{j-1}}{1-t} \in  \text{span}(v).
  $$

The rest of the argument to prove that $U$ is contractible  has been explained in the comments by @freakish, and the
last part of the question follows from Allen Hatcher's "Algebraic Topology" book, also pointed out by @freakish.

Incidentally,  the argument above is essentially also due to @freakish even though they claim in a comment that it does
not work.


问:
equality of integrals without trigonometry
can someone show the equality of these two integrals without using any trigonometry;
$$ \int_{0}^{1} \frac{dt}{\sqrt{1-t^2}} = 2 \int_{0}^{1} \sqrt{1-t^2} \, dt $$

i'm working through a derivation of relationship between a circle's circumference and area and this equality isn't obvious to me. that is I can't see how to do it without trigonometric substitutions which I don't want to make since $\pi$ hasn't been defined yet...

答:
Integrate by parts.  First, show that

$$\int_0^1 dt \, \sqrt{1-t^2} = \int_0^1 dt \frac{t^2}{\sqrt{1-t^2}}$$

You do this by noting that

$$\int_0^1 dt \, \sqrt{1-t^2} = [t \sqrt{1-t^2}]_0^1 - \int_0^1 dt \, t \frac{d}{dt} \sqrt{1-t^2}$$

Then note that

$$\int_0^1 \frac{dt}{\sqrt{1-t^2}} - \int_0^1 dt \frac{t^2}{\sqrt{1-t^2}} = \int_0^1 dt \, \sqrt{1-t^2}$$

Combining these two equations, you get the stated relation.


问:
Permute columns by pre-multiplying and rows by post-multiplying?
I was looking at Gilbert Strang's lectures on Linear Algebra and noticed that in lecture 2, Elimination with Matrices, around the 40nth minute he mentions that you can use the permutation matrix,
$$P=
\begin{bmatrix}
    0       & 1 \\
    1       & 0
\end{bmatrix}
$$
so that $AP$ is a permutation of $A$'s columns and $PA$ is a permutation of $A$'s rows.

I was wondering if there exists a matrix $P'$ such that $P'A$ is a permutation of $A$'s columns and $AP'$ is a permutation of $A$'s rows.

How to prove there is no $P'$ such that $P'A=AP$ and $AP'=PA$ for all
$n\times n$ matrices?

For which matrices $A$ there is such a $P'$?

Tried the $2\times 2$ case

$L$ pre-multiplies $A$ and permutes its columns,
$$
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4
\end{bmatrix}
\begin{bmatrix}
a       & b \\
c       & d
\end{bmatrix}=\begin{bmatrix}
b       & a \\
d       & c
\end{bmatrix} \Rightarrow
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
bd-ac       & a^2-b^2 \\
d^2-c^2       & ac-bd
\end{bmatrix}=L
$$

$R$ post-multiplies $A$ and permutes its rows,

$$
\begin{bmatrix}
a       & b \\
c       & d
\end{bmatrix}
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4
\end{bmatrix}=
\begin{bmatrix}
c       & d \\
a       & b
\end{bmatrix} \Rightarrow
\begin{bmatrix}
x_1       & x_2 \\
x_3       & x_4
\end{bmatrix}=\frac{1}{ad-bc}\begin{bmatrix}
cd-ab       & d^2-b^2 \\
a^2-c^2       & ab-cd
\end{bmatrix}=R
$$

$L\neq R$ in the general case

$det(A)\neq 0$ for $R$ and $L$ to exist

if $det(A)\neq 0$ and $a=d=0$, then $P'=L=R$

Additional notes

I feel I have to clarify further. I was looking for a matrix $P'$ that behaves just like $P$ but from the opposite side, meaning $P'$ permutes columns when pre-multiplying $A$ and rows when post-multiplying $A$.

答: If there is an ($n\times n$) matrix $P’$ such that $P'A=AP$ and $AP'=PA$ then $PA^2=AP’A=A^2P$. The latter holds if $A^2$ is a linear combination of powers of $P$. Moreover, by equivalences (I) $\Leftrightarrow$ (IV) and (I) $\Leftrightarrow$(V)  of Theorem from Will Jagy’s answer, the set of matrices commuting with $P$ coincides with the set of linear combinations of powers of $P$ iff the permutation represented by $P$ is a cycle. Conversely, if $PA^2=A^2P$ and $\det A\ne 0$ we have $P'A=AP$ and $AP'=PA$ for $P’=APA^{-1}$.


问:
Help proving $\lim_{h \to 0} \int_{1}^{5} f(x+h)dx = \int_{1}^{5}f(x)dx$
Suppose a function $f$ is integrable on $[0,6]$. Prove that

$$\lim_{h \to 0} \int_{1}^{5} f(x+h)dx = \int_{1}^{5}f(x)dx$$

This seems like the easiest thing, I would normally just bring the limit inside the integral and evaluate $f(x+h)$ to $f(x)$, but its asking to prove that this is the case and I haven't even used the fact that its integrable on $[0,6]$. I tried the epsilon delta proof this way. Let $\epsilon > 0$ and assume

$$\left|\int_{1}^{5}f(x+h)dx - \int_{1}^{5}f(x)dx\right| < \epsilon$$

I tried to find a $\delta$ so that $|f(x+h)-f(x)| < \delta$, but had no luck. Any help would be appreciated.

答:
$\int_1^{5}f(x+h)dx=\int_{1+h}^{5+h}f(x)dx=\int_1^5f(x)dx-\int_1^{1+h}f(x)dx+\int_{5}^{5+h}f(x)dx$

Since $F(x)=\int_1^xf(x)dx$ is a continuous function we have that $\lim_{h\to0}\int_a^{a+h}f(x)dx=\lim_{h\to 0} F(a+h)-F(a)=0$

So when we pass to the limit we have that
$$\lim_{h\to0}\int_1^{5}f(x+h)dx=\lim_{h\to0}\int_1^5f(x)dx-\int_1^{1+h}f(x)dx+\int_{5}^{5+h}f(x)dx=\int_1^5f(x)dx$$


问:
Is the relation $\{(x^4,x):x\in\mathbb{R}\}$ a function that maps from $\mathbb{R}$ to $ \mathbb{R}$?
Is $\{(x^4,x):x\in\mathbb{R}\}$ a function that maps from $\mathbb{R}$ to $ \mathbb{R}$?

Since $x^4$ will make all the negative elements in $\mathbb{R}$ positive, does it map to $\mathbb{R}^+$ instead? However since $\mathbb{R}^+ \subseteq \mathbb{R}$ does it still map from $\mathbb{R}$ to $ \mathbb{R}$? Im so confused

答:
The range of a function just needs to be a set that includes all the function values.  You have defined a relation that has multiple cases of the same element in the first position with different elements in the second.  It is therefore not a function from $\Bbb R$ to $\Bbb R$.

If you define a function $f(x)=x^4$, that is a fine function from $\Bbb R$ to $\Bbb R$.  There is no requirement that the function values cover the range.  We define surjective functions as those that do have a value in the domain that maps to every element of the range.

If you try to define a function $f(x^4)=x$, that is not a function at all because $x^4=1$ will give pairs $(1,1)$ and $(1,-1)$.  An input value is not mapped to a unique output value, which is required of a function.


问:
Quadratic form of nonlinear differential equations
I've got nonlinear differential equations like this:
\begin{equation}
\vec{y}={\begin{pmatrix} S\\E\\I\\R \end{pmatrix}},\vec{f}(x,\vec{y})=\frac{\partial \vec{y}}{\partial t}={\begin{pmatrix}
  -10^{-6}SI\\10^{-6}SI-0.4E\\0.4E-0.5I\\0.5I
\end{pmatrix}}
\end{equation}
In class my TA tell us we can rewrite it in quadratic from:
$Y_{n \times 1}^{\prime}=f\left(x, Y_{n \times 1}\right)=B_{n \times 1} Y_{n \times 1}^{T} A_{n \times n} Y_{n \times 1}+C_{n \times n} Y_{n \times 1}$,so we can get it derivative for further calculating,But I don't have an idea how to get Matrix B and A,is there any limit or I can just write a special form which I guess out?

Body:
\begin{align}
\frac{d}{dt}\begin{bmatrix}
S\\
E\\
I\\
R
\end{bmatrix} \, = \, \begin{bmatrix}
-10^{-6}\\
\,\,\,\,\,10^{-6}\\
0\\
0
\end{bmatrix} \Big[
S \,\, E \,\, I \,\,  R \Big]\begin{bmatrix}
0 & 0 & \frac{1}{2} & 0\\
0 & 0 & 0 & 0\\
\frac{1}{2} & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix} \begin{bmatrix}
S\\
E\\
I\\
R
\end{bmatrix} \,+\, \begin{bmatrix}
0 & 0 & 0 & 0\\
0 & -0.4 & 0 & 0\\
0 & \,\,\,\,0.4 & -0.5 & 0\\
0 & 0 & \,\,\,\,0.5 & 0
\end{bmatrix} \begin{bmatrix}
S\\
E\\
I\\
R
\end{bmatrix}
\end{align}


问:
A polynomial $P(x)$ of degree $5$ with lead coefficient one,increases in the $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$
A polynomial function $P(x)$ of degree $5$ with leading coefficient one,increases in the interval $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$. Given that $P(0)=4$ and $P'(2)=0$, find the value of $P'(6)$.

I noticed that $1,2,3$ are the roots of the polynomial $P'(x)$.$P'(x)$ is a fourth degree polynomial. So

Let $$P'(x)=(x-1)(x-2)(x-3)(ax+b)$$
Now I expanded $P'(x)$ and integrated it to get $P(x)$ and use the given condition $P(0)=4$, but I am not able to calculate $a,b$.

答: The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-2 find a possible formula for P(x).</p>


问:
Can an odd perfect number be divisible by $825$?
I know that an odd perfect number cannot be divisible by $105$. I wonder if that's also the case for $825$.</p>

答:
No it cannot.

Let $n=\prod p_i^{\alpha_i}=2^{\alpha_1}3^{\alpha_2}5^{\alpha_3}\cdots$ where $p_i$ is the $i$th prime number and let $S(n)$ be the sum of the divisors of $n$.

Suppose $n$ is an odd perfect number divisible by $825=3\cdot 5^2\cdot 11$, then $S(n)=2n$ and $\alpha_1=0,\alpha_2\ge 1,\alpha_3\ge 2,\alpha_5 \ge 1$.

Since
$$
\begin{align}
S(n)& = n\left(1+\frac{1}{3}+\cdots+\frac{1}{3^{\alpha_2}}\right)
\left(1+\frac{1}{5}+\cdots+\frac{1}{5^{\alpha_3}}\right)\cdots
\\
& = n \prod_i \left(\sum_{j=0}^{\alpha_i} p_i^{-j}\right)
\end{align}$$

then we must have $\alpha_2\ge 2$, since $1+1/3 = 4/3$ but $S(n)=2n$ is not divisible by 4. Likewise $\alpha_4 \ge 2$ since $1+1/11=12/11$.

Then either $\alpha_2=2$, and since $1+1/3+1/9=13/9$ it must be that $13\mid n$. Or $\alpha_2>2$. Both cases lead to contradictions so there cannot be such an odd perfect number.

If $\alpha_2=2$ then
$$
2 = \frac{S(n)}{n} \ge \frac{13}{9} \left(1+\frac{1}{5}+\frac{1}{25}\right)
\left(1+\frac{1}{11}+\frac{1}{121}\right)\left(1+\frac{1}{13}\right) > 2
$$
which cannot be, or if $\alpha_2>2$
$$
2 = \frac{S(n)}{n} \ge \left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)
\left(1+\frac{1}{5}+\frac{1}{25}\right)
\left(1+\frac{1}{11}+\frac{1}{121}\right) > 2
$$
which also cannot be.

This is adapted from this solution mentioned in the comments to the linked question.


问:
Limiting behavior of $x + \sqrt{x} - \sqrt{x + \sqrt{x}}$
As $x$ and $y$ vary through the real numbers, how do the interval families $\left[ x, x + \sqrt{x} \right]$ and $\left[ y - \sqrt{y}, y \right]$ differ?

Not by much, it turns out: consider the two maps: $f : x \mapsto x + \sqrt{x}$, $g : y \mapsto y - \sqrt y$. The composition $g \circ f$ rapidly converges to the the identity minus a constant:

$$\lim_{x \rightarrow \infty} x - (g \circ f)(x) = \lim_{x \rightarrow \infty} x - \left( (x + \sqrt{x}) - \sqrt{x + \sqrt x)} \right) = 0.5.$$

Yet this fact is false for any exponent greater than 0.5. Indeed, for any fixed $\varepsilon > 0$:

$$\lim_{x \rightarrow \infty} = x - \left( (x + x^{0.5 + \varepsilon}) - (x + x^{0.5 + \varepsilon})^{0.5 + \varepsilon} \right) = \infty.$$

How can I understand this sudden change in behavior???

EDIT: another unusual observation. what if we add ceiling functions to both, so that $\hat{f} : \lceil x \mapsto x + \sqrt{x} \rceil $, $\lceil g : y \mapsto y - \sqrt y \rceil$. Then $g \circ f$ appears to oscillate between $-1$ and 0, though possibly converges to 0. I'd like to prove or disprove: does it always take the values either $-1$ or $0$? Is it "eventually" 0? Thanks.

答:
First,
$$x-\big(x+\sqrt{x}-\sqrt{x+\sqrt{x}}\big)=\sqrt{x}-\sqrt{x+\sqrt{x}},$$
so I'm just going to work with that.
$$\sqrt{x}-\sqrt{x+\sqrt{x}}=\sqrt{x}-\sqrt{x}\sqrt{1+x^{-1/2}}.$$
When $x>1$, we can use the binomial series
$$\sqrt{1+x^{-1/2}}=1+\frac{1}{2}x^{-1/2}-\frac{1}{8}x^{-1}+\frac{1}{16}x^{-3/2}\cdots$$
Then we get
$$\sqrt{x}-\sqrt{x}\sqrt{1+x^{-1/2}}=\sqrt{x}-\sqrt{x}-\frac{1}{2}+\frac{1}{8}x^{-1/2}-\frac{1}{16}x^{-1}\cdots.$$
The $\sqrt{x}$ terms cancel, and the terms of order $x^{-1/2}$ and smaller vanish as $x$ gets large, leaving only $-0.5$.

Now, in the other case:
$$x^{0.5+\epsilon}-(x+x^{0.5+\epsilon})^{0.5+\epsilon}=x^{0.5+\epsilon}-x^{0.5+\epsilon}(1+x^{-0.5+\epsilon})^{0.5+\epsilon}.$$
For $\epsilon<0.5$ and $x>1$, we can again use the binomial series
$$(1+x^{-0.5+\epsilon})^{0.5+\epsilon}=1+(0.5+\epsilon)x^{-0.5+\epsilon}+\cdots$$
This time, however, we get
$$x^{0.5+\epsilon}-x^{0.5+\epsilon}(1+x^{-0.5+\epsilon})^{0.5+\epsilon}=x^{0.5+\epsilon}-x^{0.5+\epsilon}+(0.5+\epsilon)x^{2\epsilon}+\cdots.$$
The higher order terms in the series will vanish for large $x$ (under the conditions we put on $\epsilon$) as you can check, but the $x^{2\epsilon}$ will blow up.


问:
Find point farthest on $(x-5)^2+(y+1)^2+(z-7)^2=69$ from point $(2, 3, -6)$
So, I have the equation of a sphere $$(x-5)^2+(y+1)^2+(z-7)^2=69$$ and a point, $P(2, 3, -6)$. I am supposed to find the point on the sphere farthest away from $P$.

I tried to figure out how to use parametric equations to represent the sphere and maximize that using the distance formula. However, they are multiple variables in this equation, and this led me to nowhere.

答:
The line connecting $P(2,3,-6)$ and the farthest point $M(a,b,c)$ is parallel to $M$’s normal surface vector $(a-5,b+1,c-7)$, i.e.
$$\frac{a-2}{a-5}=\frac{b-3}{b+1}=\frac{c+6}{c-7}$$

Substitute into $(a-5)^2+(b+1)^2+(c-7)^2=69$ to obtain the point
$$M\left(5+3\sqrt{\frac{69}{194}}, -1-4\sqrt{\frac{69}{194}}, 7+13\sqrt{\frac{69}{194}}\right)
$$


问:
Is $X - Y(Y^\top Y)^{-1}Y^\top XY(Y^\top Y)^{-1}Y^\top$ positive semi-definite or negative definite?
Let $X$ is a positive definite matrix and $Y$ is a matrix of full column rank. Is the following matrix positive semi-definite or indefinite:

$$
X - Y(Y^\top Y)^{-1}Y^\top XY(Y^\top Y)^{-1}Y^\top
$$

I tried to decompose it into a product of $(\cdot)$ and $(\cdot)^\top$, but couldn't.

答:
Let $U$ denote a matrix whose matrix form an orthonormal basis for the column space of $Y$. We have
$$
Y(Y^TY)^{-1}Y^T = UU^T,
$$
so that the matrix under consideration is
$$
M = X - UU^TXUU^T.
$$
Let $V$ denote a matrix whose columns extend the columns of $U$ to an orthonormal basis. Let $W = [U \ \ V]$. We find that $M$ is similar to
$$
W^TMW = \pmatrix{U^T\\ V^T} [X - UU^TXUU^T] \pmatrix{U & V} =
\pmatrix{0 & U^TXV\\ V^TXU & V^TXV}.
$$
With that, it is easy to see that $M$ will be positive semidefinite if and only if $U^TXV = 0$, which occurs if and only if the column space of $Y$ is an invariant subspace of $X$, which occurs if and only if $XP = PX$, where $P = Y(Y^TY)^{-1}Y^T$.

If $M$ is not positive semidefinite, then it must be indefinite.


答:
A relation $f \subseteq X \times Y$ is called a function when

$$
\begin{align}
\text{(right-unique): }&\forall x \in X: \forall y \in Y: \forall z \in Y: ((x,y) \in f \land (x,z) \in f) \Rightarrow y = z\\
\text{(left-total): }&\forall x \in X: \exists y \in Y: (x,y)\in f
\end{align}
$$
The relation you showed is left-total but not right-unique if you choose the domain $X = \mathbb{R}_+$ and the codomain $Y = \mathbb{R}$ and therefore, it is not a function. Likewise, the relation is not left-total nor right-unique if you choose $X = \mathbb{R}$ and $Y = \mathbb{R}$ and therefore, it is still not a function.

Title: Is the relation $\{(x^4,x):x\in\mathbb{R}\}$ a function that maps from $\mathbb{R}$ to $ \mathbb{R}$?

Body:
Is $\{(x^4,x):x\in\mathbb{R}\}$ a function that maps from $\mathbb{R}$ to $ \mathbb{R}$?

Since $x^4$ will make all the negative elements in $\mathbb{R}$ positive, does it map to $\mathbb{R}^+$ instead? However since $\mathbb{R}^+ \subseteq \mathbb{R}$ does it still map from $\mathbb{R}$ to $ \mathbb{R}$? Im so confused


问:
Gre combinatorics problem
In a local election there are $2$ candidates for mayor, $4$ candidates for sheriff, and $5$ candidates for dogcatcher on the ballot. In each of the three categories a voter may vote for exactly one candidate or none. How many different ways can a voter fill out the ballot?

I don't think I understand the question, but I do feel like it's a combinatorics question.

Have been working for like 8 hours on practice questions and my mind is absolutely slow at the moment. In particular, I'm not sure what it means by "In each of the three categories a voter may vote for exactly one candidate or none" then we can count the different ways from it.

Any help is highly appreciated.

答: The answer is $3\times 5\times 6 = \boxed{90}$ since the voter has to make three independent decisions. The first decision requires the voter to choose between one of three options (vote for nobody, vote for the first candidate, or vote for the second candidate). The other two cases are similar.

Title: Gre combinatorics problem


问:
rounding up to nearest square
Say I have x and want to round it up to the nearest square. How might I do that in a constant time manner?

ie.

$2^2$ is 4 and $3^2$ is 9. So I want a formula whereby f(x) = 9 when x is 5, 6, 7 or 8. What it does when x = 4 or 9 doesn't really matter.

I could write a function to do this easily enough. ie.

```<code>while (sqrt(x) is not int) {
  x++;
}
</code>```

But I'd rather not do a brute force approach. Any ideas?

Thanks!

答:
I don't know if this is quite what you're looking for, but you can use the fact that
$ \ (n+1)^2 \ = \ n^2  \ + \ (2n + 1) \ $ .   Once you reach a perfect square $ \ n^2 \ $ , the next $ \ 2n \ $ integers would be the ones that get "rounded-up".  This is at least reduces the process to a test for perfect-square and a counting loop.</p>

Title: rounding up to nearest square


问:
The Laplace transform of a function $\phi(t)$ can be written as $\hat{\phi}(s) = \cosh (2x\sqrt{s})/(\sqrt{s}\sinh\sqrt{s})$.

The Laplace transform of a function $\phi(t)$ can be written as $$\hat{\phi}(s) = \cosh (2x\sqrt{s})/(\sqrt{s}\sinh\sqrt{s}).$$ Determine $\phi(t)$ where $x$ is real.

So I know that $$\cosh(at) = \frac{s}{s^2 - a^2}$$ and I know that $$\sinh(at) = \frac{a}{s^2 - a^2}$$ but I'm stuck on how to tet to that point where I can take the inverse Laplace function and determine my answer.

答:
The inverse Laplace Transform, $\phi(t)$, is given by the Bromwich integral

$$\begin{align}
\phi(t)&=\frac1{2\pi i }\int_{c-i\infty}^{c+i\infty}\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}\,ds\\\\
\end{align}$$

The integrand has no branch point singularities since $\frac{\cosh(2xz)}{z\sinh(z)}$ is an even function of $z$.  There are simple poles of the integrand at $s=0$ and $s=-n^2\pi^2$.  Then, we can write for $t>0$

$$\begin{align}
\phi(t)&=\sum_{n=0 }^\infty \text{Res}\left(\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}, s=-n^2\pi^2\right)\\\\
&=\sum_{n=0 }^\infty\lim_{s\to -n^2\pi^2}\left((s+n^2\pi^2)\frac{\cosh(2x\sqrt s)}{\sqrt s\sinh(\sqrt{s})}e^{st}\right)\\\\
&=\sum_{n=0 }^\infty (-1)^n2\cosh(i2xn\pi)e^{-n^2\pi^2t}\\\\
&=2\sum_{n=0}^\infty (-1)^n \cos(2n\pi x)e^{-n^2\pi^2 t}
\end{align}$$


问:
How to determine if a ray intersects a line?
I'm trying to determine how you could check if a ray, given an origin and a direction, and a line, given 2 points (Not a line segment) intersect.

Ray: Origin(x,y), direction(x,y)

Line: point1(x,y), point2(x,y)

I thought about doing a case analysis, such as checking if they are parallel, check if the ray is to the left of the line then check the direction etc. but I believe there is a better way to do it involving the cross products.

I'm trying to write a program to do this check, but my issue lies with coming up with a method to perform this check. I'm kind of clueless as to doing an overly complicated case-analysis.

答:
Two lines through points $z_1,z_2$
and $z_3,z_4$ intersect at a point
($\overline{z}$ is the complex conjugate of $z$):
\begin{align}
z&=
f_{\times}(z_1,z_2,z_3,z_4)=
\frac{
(z_1-z_2)\,(\overline{z_3}\,z_4-\overline{z_4}\,z_3)
-(z_3-z_4)\,(\overline{z_1}\,z_2-\overline{z_2}\,z_1)
}{
(z_1-z_2)\,(\overline{z_3}-\overline{z_4})
-(z_3-z_4)\,(\overline{z_1}-\overline{z_2})
}
\tag{1}\label{1}
,
\end{align}
unless the denominator is zero, which is a special case
that need to be treated separately.

If we set $z_1=0$, $z_2=d$ (direction of the ray),
and the endpoints
of given segment $A=z_3,\ B=z_4$,
then  \eqref{1} simplifies to

\begin{align}
z&=
\frac{d\,(\overline{A}\,B-\overline{B}\,A)}{
d\,(\overline{A}-\overline{B})
-\overline{d}\,(A-B)}
\tag{2}\label{2}
.
\end{align}

Given that, consider two real numbers $t,\,s$ such that
conditions
\begin{align}
z&=t\cdot d
,\\
\text{along with }\quad
z&=(1-s)\cdot A+s\cdot B
\quad\text{hold}
.
\end{align}

If $t\ge0$ and $s\in[0,1]$ then $z$ is the sought point.


问:
Condition when improper Riemann integral is the same as the Lebesgue integral
Let $ -\infty < A < B \leq +\infty$, and let $f:[A,B)\to \mathbb{R}$ be almost continuous everywhere (sucht that $f_{[A,b]} \in \mathcal{R}(A,b)$ for all $A<b<B$). Then $\mathcal{R} \int_A^B f(x)dx := \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx.$

The following is a condition when the improper Riemann integral is the same as the Lebesgue integral. How do I prove this?

If $f \geq 0$, then

$$\mathcal{L} \int_{[A,B)} fd\lambda = \lim_{b \to B} \mathcal{R} \int_A^b f(x)dx \in \mathbb {\overline R}_{\ge 0}.$$

答:
Well, by definition if a function is riemann integrable then $$\int_a^bf(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$$

Notice that the function $g_n(x)=\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$ if $x\in \left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right)$ is increasing and converges to $f$ almost everywhere, also

$$\int_{[a,b]}g_{n}d\lambda=\frac{1}{n}\sum_{i=1}^n\inf\left\lbrace f(x):x\in\left[a+\frac{(i-1)(b-a)}{n},a+\frac{i(b-a)}{n}\right]\right\rbrace$$
And by the bounded convergence theorem we have that $\int_{[a,b]}fd\lambda=\lim\int_{[a.b]}g_nd\lambda=\int_a^bf(x)dx$

Now take $f:[A,B)\to\mathbb R$ and an increasing sequence $b_n\to B$
$$\int_{[A,B)}f\chi_{[A,b_n]}d\lambda=\int_{[A,b_n]}fd\lambda=\int_A^{b_n}f(x)dx$$
So by bounded convergence $\int_{[A,B)}fd\lambda=\lim\int_{[A,B)}f\chi_{[A,b_n]}d\lambda=\lim\int_A^{b_n}f(x)dx=\lim_{b\to B}\int_A^bf(x)dx$
Where the last equality holds because we are asuming that $\lim_{b\to B}\int_A^bf(x)dx$ conveges so for every sequence $b_n\to B$ $\int_A^{b_n}f(x)dx$ must have the same limit.


答:
A relation is a function over a set if, given any particular value in that set, that value appears as left-hand value in exactly one ordered pair.

In your relation, both $(16,2)$ and $(16,-2)$ appear, so the value $16$ appears as the left-side value in two different ordered pairs, which means this is not a function.


问:
Complex number sequence question
So if I have $$z_{0}=4\sqrt2 \, \left( \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) \right)$$ and $$z_{1}=2 \left( \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) \right)$$

$$z_{n}=2\sqrt{2} \, \left(\cos\left(\frac{5\pi}{4}\right) + i \sin\left(\frac{5\pi}{4} \right) \right)\, \frac{z_{n-1}}{|z_{n-2}|}$$ for $n\in \mathbb{N}$, with $n \geq2$. So the $z_{n}$ is sequence. Can I find two things,

Determine smallest positive integer $p$ such that for every positive integer $n$, |$z_{n+p}$|=|$z_{n}$|

Determine smallest positive integer $q$ such that for every positive integer $n$, $z_{n+q}$=$z_{n}$

These 2 is very similar, first one is looking for abslute value of it, and second one is not. I think at end of the sequence, there is $\frac{z_{n-1}}{|z_{n-2}|}$. This is maybe related. Anyway of using this to prove it? I am stuck.

答: Hint: Write each of the numbers in polar form; $z_n=r_ne^{i\theta_n}$. See if you can use the given recurrences to set up recurrences for $r_n$ and $\theta_n$, which should be easier to deal with.

Title: Complex number sequence question


问:
Ring Homomorphism (mapping of unit element) between a ring and an integral domain
Let R and S be two rings with unit elements. Let us assume that the unit element of R is $1_R$
and the unit element of S is $1_S$. Let $\phi$ : R → S be a ring homomorphism. Show that if S is
an integral domain, then $\phi$ ($1_R$) = $1_S$.

Note: $\phi$($1_R$) $\neq 0_S$ (Zero element in S)

I want to prove this via contradiction specifically and have proceeded as follows:

$\exists y \in S \ \ s.t.$  $ y \ \phi(1_R)  \neq \ y $

Then-

$ y \ \phi(1_R)- y  \neq  0_S $

I am stuck here and don't know how to proceed ahead with this proof using contradiction.

答:
You don't need to prove it by contradiction. Note that
$$1_R 1_R = 1_R,$$
so
$$\phi(1_R 1_R) = \phi(1_R) \phi(1_R) = \phi(1_R) = \phi(1_R) 1_S,$$
and since $S$ is an integral domain, the cancellation laws implies that $\phi(1_R) = 1_S$.


问:
Integrate ${\sec 4x}$
How do I go about doing this? I try doing it by parts, but it seems to work out wrong:

$\eqalign{
  & \int {\sec 4xdx}   \cr
  & u = \sec 4x  \cr
  & {{du} \over {dx}} = 4\sec 4x\tan 4x  \cr
  & {{dv} \over {dx}} = 1  \cr
  & v = x  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \int {4x\sec 4x\tan 4xdx}   \cr
  & \int {4x\sec 4x\tan 4xdx} :  \cr
  & u = 4x  \cr
  & {{du} \over {dx}} = 4  \cr
  & {{dv} \over {dx}} = \sec 4x\tan 4x  \cr
  & v = {1 \over 4}\sec x  \cr
  & \int {4x\sec 4x\tan 4xdx}  = x\sec 4x - \int {\sec 4x} dx  \cr
  & \int {\sec 4xdx}  = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $

I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!

EDIT: Is there an easier way to do this?

答:
Use two substitutions. The first substitution transforms the integrand into $\sec \theta$, whose evaluation was asked here. The second substitution is the Weirstrass substitution. (See comment below). In the present case the second integral becomes an easy table integral:

$$\begin{eqnarray*}
\int \sec 4x\,dx &=&\frac{1}{4}\int \sec \theta \,d\theta ,\qquad \theta =4x
\\
&=&\frac{1}{4}\int \frac{1}{\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}
\,dt,\qquad t=\tan \frac{\theta }{2} \\
&=&\frac{1}{2}\int \frac{1}{1-t^{2}}dt=\frac{1}{2}\operatorname{arctanh}t+C \\
&=&\frac{1}{2}\operatorname{arctanh}\left( \tan \frac{\theta }{2}\right) +C \\
&=&\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) +C.
\end{eqnarray*}$$

This integral can be rewritten as

$$\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) =\frac{1}{4}\ln \left\vert
\tan 2x+1\right\vert -\frac{1}{4}\ln \left\vert 1-\tan 2x\right\vert .$$

Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form

$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$

where $P,Q$  are polynomials in $\sin \theta,\cos \theta$

$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$

which converts the integrand into a rational function in $t$. We know from trigonometry (see this answer) that

$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$

Title: Integrate ${\sec 4x}$


问:
Integral of $\exp(-\frac{c_1}{x}-\frac{c_2}{x^2})$ on positive real line
Is it possible to find
$$
\int_{0}^{K}
\exp\left(-\,{c_{1} \over x} - {c_{2} \over x^{2}}\right)
\,\mathrm{d}x
$$
or at least a good approximation to it, where $c_{1}$ and $c_{2}$ are positive and $K$ is a very large number $?$.

I tried to consider the change of variables
$-u = -c_{1}/x - c_{2}/x^{2}$ but to no avail.

Note: Approximation can include known functions such as Erf, or Gamma, if necessary.

Body:
As said, series expansion could be the solution. Writing
$$\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)=1+\sum_{n=1}^p \frac {c_n}{n! \,x^n}=f(x)$$ the $c_n$'s would be
$$\left(
\begin{array}{cc}
 1 & -a \\
 2 & a^2-2 b \\
 3 & -a^3+6 a b \\
 4 & a^4-12 b a^2+12 b^2 \\
 5 & -a^5+20 b a^3-60 b^2 a \\
 6 & a^6-30 b a^4+180 b^2 a^2-120 b^3 \\
 7 & -a^7+42 b a^5-420 b^2 a^3+840 b^3 a \\
 8 & a^8-56 b a^6+840 b^2 a^4-3360 b^3 a^2+1680 b^4 \\
 9 & -a^9+72 b a^7-1512 b^2 a^5+10080 b^3 a^3-15120 b^4 a \\
 10 & a^{10}-90 b a^8+2520 b^2 a^6-25200 b^3 a^4+75600 b^4 a^2-30240 b^5
\end{array}
\right)$$ and
$$\int\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)\,dx=1-a\log(x)-\sum_{n=2}^p \frac {c_n}{(n-1)\,n! \,x^{n-1}}+ C$$

For sure, there is a  problem with a lower bound equal to $0$.


问:
standard deviation can not be $6$.
For the frequency distribution :
Variate x : $x_1 , x_2  , x_3 ,  x_ 4 ........,  x_{15}$
Frequency f : $f_1 ,  f_2  , f_3  , f_ 4 ........, f_{15}$   where $0 < x_1 < x_2  < x_3 <  x_ 4 ........<  x_{15} = 10$ and $\sum f_i >0$.

Then the standard deviation  can not be $6$.

Can anyone please help me by giving some hints? I have no idea.

答:
There is a standard result that $$\sigma_x^2 \leq \frac{R_x^2}{4}$$ where $\sigma_x^2$ is the variance of $X$ and $R_x$ is the range of $X$

Note that here $R_x = 10 - 0 = 10$

Using this result, $$\sigma_x^2 \leq \frac{10^2}{4} = 25 $$
Or, $$\sigma_x \leq 5$$

Proof:

Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i \leq n} x_i = b$

Then $R_x = b - a$

$\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$

Now $\sum_{i=1}^n (x_i - c)^2$ is least when $c = \bar{x}$

Hence $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_{i=1}^n (x_i - \frac{a+b}{2})^2 = \sum_1 (x_i - \frac{a+b}{2})^2 + \sum_2 (x_i - \frac{a+b}{2})^2$

where $\sum_1$ includes the values of $x$ less than or equal to $\frac{a+b}{2}$ and $\sum_2$ includes values of $x$ greater than $\frac{a+b}{2}$

Or, $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_1 (a - \frac{a+b}{2})^2 + \sum_2 (b - \frac{a+b}{2})^2 = \sum_1 \frac{R_x^2}{4} + \sum_2 \frac{R_x^2}{4} = n \frac{R_x^2}{4}$

Or, $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \leq \frac{R_x^2}{4}$

Title: standard deviation can not be $6$.


问:
Importance and Intuition of Polynomial Rings
I know that a polynomial ring $R[x]$ is the ring with elements consisting of polynomials with coefficients in $R$. However, this definition leaves me confused when I try to really understand the concept of polynomial rings, rather than just accept that definition for what it is. Where did the polynomials come from? Why are polynomial rings important? Is there a way to “construct” the polynomials in the context of Abstract Algebra, so that we may see why studying rings made up of polynomials is important? Are polynomial rings actually just sets of polynomials, or are the polynomials just a concrete way to represent some more abstract idea? What does the variable $x$ mean in polynomial rings? I apologize for all the questions, I’m just having trouble seeing the importance or intuition behind them.

答:
$R[x]$ can be thought of as the simplest way to add one element to the ring $R$ and still have a ring.

Suppose I have a ring $R$ and I want to add one extra element which is not already in $R$. Call that extra element $x$. Consider the set $A = R\cup \{x\}$. To make $A$ into a ring, we need to add in all sums of terms that look like $r x^k$ where $r\in R$ and $k\in \mathbb{N}$. The simplest way to define multiplication in this ring is just to let multiplication of two members of $R$ stay the same and let all the products $r x^k$ be unique elements of the new ring. This is just the polynomial ring $R[x]$.

(This is actually just a restatement of the comment by @lhf, but it is likely that OP does not know what a universal $R$-algebra is.)

The philosophy behind polynomial rings is that we are not committing ourselves to saying exactly what $x$ represents. It is just an "indeterminate" and we can manipulate it according to formal rules like $x^{a} x^{b} = x^{a+b}$. If at some point we wish to specify a value for $x$, we can do so using a ring homomorphism $R[x] \rightarrow R$ that replaces every instance of $x$ with that particular value.

Note also we can define an homomorphism from $R[x]$ to the ring of all functions from $R$ to $R$ with pointwise operations. Just send $r \in R$ to the constant function $r$ and send $x$ to the identity map on $R$. In the case $R = \mathbb{R}$, this map is an isomorphism onto its image. This is how we get the correspondence between polynomial rings and polynomials as we know them from grade school algebra.


问:
WLOG Can assume length of 1.
I am looking at a proof of the following:

$$\mathbb{P}(\sum a_iX \ge t)\le \text{exp}\bigg(\frac{-t^2}{||a||_2}\bigg)$$

They start WLOG we can assume $||a||^2_2=1$. I am trying to understand why this is the case. The only thing I can come up with is that I can rescale

$$\mathbb{P}(\sum a_iX \ge t)=\mathbb{P}(\sum \frac{a_iX}{||a||_2} \ge \frac{t}{||a||_2})=\mathbb{P}(\sum \frac{a_iX}{||a||_2} \ge k)$$

for some $k\ge 0$

Is my logic correct?

答:
No, your rescaling is wrong. You should rescale by $\|a\|_2$ (not $\|a\|_2^2$): note that otherwise it won't be homogeneous.

In more detail, note that if you prove the case for unit vectors, then indeed the general case follows:

$$\mathbb{P}\left\{\sum_i a_iX_i \ge t\right\}=\mathbb{P}\left\{\sum_i \frac{a_iX_i}{\|a\|_2} \ge \frac{t}{\|a\|_2}\right\}=\mathbb{P}(\sum_i a'_iX_i \ge t') \stackrel{(\ast)}{\leq} e^{-{t'}^2} = e^{-\left(\frac{t}{\|a\|_2}\right)^2} = e^{- \frac{t^2}{\|a\|_2^2}}$$
where for $(\ast)$ we applied the result to $a' = \frac{a}{\|a\|_2}$ (which is a unit norm vector by construction) and $t' = \frac{t}{\|a\|_2}$.

Title: WLOG Can assume length of 1.


问:
Disk method around vertical line

Set up an integral to find the volume of the solid generated by revolving the region from x=1 to x=2 about the line x = 3 using the disk method. Do not evaluate the integral. Your answer should not contain absolute value signs. (hint: the inverse function of $e^{ax}$ is $\frac{lnx}{a}$ and the inverse function of ln(ax) is $\frac{e^x}{a}$.)

Here is a graph of the problem

With a = 48, my two equations that make the region are ln(48x) (top) and $e^{-48x}$ (bottom). My initial approach was to split the area into three integrals, ln(48) to ln(96), $e^{-48}$ to ln(48), and $e^{-96}$ to $e^{-48}$. Now, I don't know what to put in the integrands.

Any help is appreciated.

答:
I figured it out myself. Here's the answer:

$$\pi\int_{\ln48}^{\ln96}\left(\left(3-\frac{e^x}{48}\right)^2-1\right)dx+\pi\int_{e^{-48}}^{\ln48}(2^2-1)dx+\pi\int_{e^{-96}}^{e^{-48}}\left(\left(3-\frac{\ln x}{48}\right)^2-1\right)dx$$

Title: Disk method around vertical line


问:
Covering map for two circles
Given a bouquet of two circles $\mathbb S^1 \vee \mathbb S^1$, I want to find a covering map $q: E \to \mathbb S^1 \vee \mathbb S^1$, whose induced homomorphism $q_* : \pi_1(E, e) \to \pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ satisfy below given $n \geq 2$:

(1) the image $q_* (\pi_1(E, e))$ inside $\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ is a free group with $n$ generators

(2) the image $q_* (\pi_1(E, e))$ is a normal subgroup of $\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e))$ such that $[\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e)) : q_* (\pi_1(E, e))] = n - 1$ ($|\pi_1(\mathbb S^1 \vee \mathbb S^1, q(e)) / q_* (\pi_1(E, e))|$)

If $n = 2$, I think I can simply take $E$ to be a bouquet of two circles and just let $q$ to be a homoemorphism. For $n > 2$, should I consider $q$ from a bouquet of $2 n$ circles? Could someone give me a walk through of this question please?

答:
Your idea of taking a bouquet on $2n$ circles unfortunately doesn't work; any covering space $X$ of a bouquet $(\bigvee_{i=1}^kS^1,\star)$ will be a directed graph where every vertex has $k$ incoming edges and $k$ outgoing edges, corresponding to the $k$ incoming and $k$ outgoing edges of $\star$. Indeed, a vertex of $X$ with degree not equal to $2k$ would have no neighborhood homeomorphic to a neighborhood of $\star$, thus preventing $X$ from being a covering space. (To see this, consider the number of connected components of a small neighborhood after removing the vertex.) Here is an alternative approach; we use the fact that $q_*$ is an injection.

Label the circles of $S^1\vee S^1$ as $a$ and $b$, and let $F=\pi_1(S^1\vee S^1)=\langle a,b\rangle$. Consider the map $\phi:F\rightarrow C_{n-1}$, where $C_{n-1}=\langle c\rangle$ is cyclic of order $n-1$, induced by $a,b\mapsto c$. By the first isomorphism theorem, the kernel of this map will be a normal subgroup of $F$ of index $n-1$; furthermore, the elements of $\ker\phi$ will be precisely those of form $a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k}$, where $\sum_{i=1}^km_i+n_i$ is divisible by $n-1$. This subgroup corresponds to the fundamental group of a graph of the following form on $n-1$ vertices. (This illustration is of the case $n=5$.)

Such a graph can be equipped with the obvious map to $S^1\vee S^1$ that takes every vertex to $\star$; this will be a covering map by the remark in my first sentence. Furthermore, such a graph has $2(n-1)$ edges and maximal trees of length $n-2$, so its fundamental group will be free on $2(n-1)-(n-2)=n$ generators, as desired.

Incidentally, this photo comes from Hatcher, a great resource to know of if you are learning algebraic topology.

Edit: Here are some answers to the questions you asked in the comments. For question 0; yes, the graph I've shown is a topological space, with topology induced by the subspace topology in $\mathbb{R}^2$, satisfying the desired properties of your problem. Alternatively, if you prefer the language of cell complexes, then a "graph" is simply a $1$-dimensional cell complex.

For question 1, do you know the definition of a cyclic group? There is a cyclic group of every finite cardinality. The answer to question 2 requires knowledge of cyclic groups, so if you're not familiar with them you should first read the wikipedia article I've linked. In any case, this is the fleshed out version of my argument. Every element of $F=\langle a,b\rangle$ is of the form $a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k}$ for some $m_i,n_i\in\mathbb{Z}$. We have $$\phi(a^{m_1}b^{n_1}\dots a^{m_k}b^{n_k})=\phi(a)^{m_1}\phi(b)^{n_1}\dots\phi(a)^{m_k}\phi(a)^{n_k}=c^{m_1}c^{n_1}\dots c^{m_k}c^{n_k}=c^{\sum_{i=1}^km_i+n_i},$$ which will equal the identity in $C_{n-1}$ if and only if $\sum_{i=1}^km_i+n_i$ is divisible by $n-1$.

The answer to question 3 makes more sense after you know the answer to question 4, so I'll answer question 4 first. The main theorem is that the fundamental group of a graph is a free group, generated by the edges of the graph minus the edges of (any) maximal tree in the graph. This is quite a core theorem on fundamental groups, and is genuinely invaluable if you want to study covering spaces of graphs like $S^1\vee S^1$. For a reference on this theorem, see section IV.4. here, and for a worked through example see example 1.22 in the Hatcher book I linked to above.

Finally, for question 4, the subgroup $\ker\phi<\langle a,b\rangle$ doesn't exactly correspond to the graph I've shown, it corresponds to the fundamental group of that graph. Using the theorem I've mentioned in the answer to question 3, it is straightforward to verify this. Before you've learned the theorem though, if you just want to see this fact intuitively, try just picking a vertex of the graph as a basepoint and drawing out loops along the edges of the graph, writing down $a$, $b$, $a^{-1}$, or $b^{-1}$ depending on which edge you've traversed and in which direction... what kind of paths do you get?

Title: Covering map for two circles


问:
Finding the distance between groups of points?
Let's say we have two groups of points : $a_1,a_2,a_3,\dots,a_n$ and $b_1,b_2,b_3,\dots,b_m$, in $2$-dimensional or $N$-dimensional space.

I have the distance between every $a$ and $b$ point, but not between the points in the group internally, i.e.

$D = |a_i,b_j|$ is known,

$D = |a_i,a_j|$ and $D = |b_i,b_j|$, unknown

My question is, is the overall distance between the two groups (I imagine centers of the groups) just the average of all the distances, OR it is calculated some other way?

答:
If you would like to find the distance between centers (or means) of the groups, you need the within-group data.

Let us simplify the problem to perhaps an unrealistic level: imagine the space is 1-dimensional. Then, all the points will be on the same line. Now imagine two points in group $a$ are at the same distance from the same point in group $b$ . It could be that those two points are exactly at the same location, or it could be that they are located on opposite sides of the group $b$ point at the same distance from it. In the former case, the center of the two points in group $a$ would be their shared location. In the latter case, their center would be on the group $b$ point. So you see, even in such a simple setting, it is not possible to determine where the center of a group of points is. Having multiple points does not invalidate this argument. And in multiple dimensions, the same problem persists. We lose information when we use absolute values for distance.


问:
Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$
Can somebody help me with this.

Compute the infinite series of the $\displaystyle{\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n}.}$

I have no idea how to start with this.  This is my first time encounter with the problem.  I'm sorry.  If you could, I will appreciate this.

What I did by the way is to expand the product.  However, is there a way to do this in a way that it does not like doing the expanding classical form?

答:
Is this acceptable or can somebody point out my mistake?

$\prod_{n=1}^{\infty} \frac{(n+1)^2}{n^2+2n} = \lim_{n \rightarrow \infty} \prod_{k=1}^n \frac{(k+1)^2}{k^2+2k} = \lim_{n\rightarrow \infty} \left[\frac{2{(2)}}{1{(3)}} \cdot \frac{{(3)}(3)}{{(2)}(4)} \cdots \frac{(n-1){(n-1)}}{(n-2){n}} \cdot \frac{{(n)}{(n)}}{{(n-1)}{(n+1)}} \cdot \frac{(n+1)(n+1)}{n(n+2)} \right] = \lim_{n \rightarrow \infty} \frac{2(n+1)}{n+2} = \lim_{n\rightarrow \infty} \frac{2\left(1 + 1/n\right)}{1 + 2/n} = 2$.


问:
Probability that a selection from a random set from a multivariable equation $z=f(x,y)$ is the maximum, given $y$.
I was watching a Veritasium video about the effect of luck and skill on success, and to test, he did the following:

He randomly generated values for luck and skill (say from 0 to 1). He then added them, such that the sum $z=0.95S+0.05L$. He did this for a large number of samples, and then found that the top 11 samples had an average luck score of 0.947.

So I wanted to investigate this analytically in the following manner:

Have an equation $z=0.95x+0.05y$ (x,y are replacements for S and L)

For a sample $S$ of n values drawn from this equation $z$, such that $x\in[0,1], y\in [0,1]$, what is the probability that the number, $s$, chosen from this sample is the maximum value of this sample, given that $y=k$. That is, find $Pr(s=S_{max}|y=k)$

Note: when drawing values from the equation,  the choices of $x$ and $y$ follow a uniform distribution from $[0,1]$..

I then want a theoretical graph where the y-axis is this probability, and the x-axis is $k$, (k ranging from 0 to 1).

Perhaps this graph/equation is given by some equation.

How do I find this graph (Analytically not numerically).

EDIT : <s> The sample size n is not a parameter in this instance, it is just for the sake of visualisation. As we are looking at expected probabilities, $n$ does not make a difference </s>

答:
We have $f_{0.95X}(x) = 1/0.95 \cdot 1_{[0,0.95]}(x)$ and $f_{0.05Y}(x)=20 \cdot 1_{[0,0.05]}(x)$. Assuming $X$ and $Y$ are independent, the pdf of $Z$ is:

$$f_Z(t) = \int_0^t f_{0.95X}(y)f_{0.05Y}(t-y) dy = \begin{cases}
\frac{20}{0.95}t & \text{if } 0 \leq t \leq 0.05 \\
\frac{1}{0.95} & \text{if } 0.05 \leq t \leq 0.95 \\
\frac{20}{0.95}(1-t) & \text{if } 0.95 \leq t \leq 1 \\
0 & \text{otherwise.} \end{cases}$$

Integrate this function to get the cdf:
$$F_Z(t) = \begin{cases}
0 & \text{if } t < 0 \\
\frac{10}{0.95}t^2 & \text{if } 0 \leq t \leq 0.05 \\
\frac{1}{40 \cdot 0.95} + \frac{t-0.05}{0.95} & \text{if } 0.05 \leq t \leq 0.95 \\
\frac{400}{19}(t-\frac{1}{2} t^2) -\frac{181}{19} & \text{if } 0.95 \leq t \leq 1 \\
1 & \text{otherwise.} \end{cases}$$

The probability that you ask for is $g(k) = P(0.05k+0.95X \geq \max\{Z_1,Z_2,\ldots,Z_{s-1}\})$. where $X$ and $Z_i$ are iid. You can compute this probability as:
$ \int_0^1P(0.05k+0.95x \geq \max\{Z_1,Z_2,\ldots,Z_{s-1}\})f_x(x) dx$, which is $ \int_0^1 F_Z(0.05k+0.95x)^{s-1} f_x(x) dx.$ I will simplify this integral if/when I have more time.


问:
Convergence of sequence
Does the following:

$$
\begin{align}
x_0 & = a \\
x_1 & = x_0 + \frac{1}{1}(x_0 + x_0(c - 1)) \\
x_2 & = x_1 + \frac{1}{2}(b + x_1(c - 1)) \\
x_3 & = x_2 + \frac{1}{3}(b + x_2(c - 1)) \\
& {}\,\vdots \\
x_n & = x_{n-1} + \frac{1}{n}(b + x_{n-1}(c - 1))
\end{align}
$$

where $x_{0}$ is switched to b in the $x_{2}$ term onwards, converge to:

$$\frac{b}{1 - c}\quad\text{?}$$

答:
First, the constant $c$ is slightly misleading. Why not take $d := 1-c$ so that: $$x_{n+1} = \left(1-\frac{d}{n}\right) x_{n} + \frac{b}{n}$$
Let $y_n := x_n - t$ for some $t$ that we'll establish in a second. Then:
$$ y_{n+1} + t = \left(1-\frac{d}{n}\right) y_{n} +  \left(1-\frac{d}{n}\right)t + \frac{b}{n}$$
which is more naturally written as:
$$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} + \frac{b-td}{n}$$
Thus, let $t := \frac{b}{d}$, so that we have:
$$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} $$
This allows you to compute $y_n$ almost explicitly:
$$ y_{n+1} = y_k\prod_{m=k}^n \left(1-\frac{d}{m}\right) $$
where $k$ is reasonably small (say, $k=3$). It is a classical theorem in analysis that because $\sum_m \frac{d}{m}$ diverges to $+\infty$, the product $\prod_{m=k}^n \left(1-\frac{d}{m}\right) $ converges to $0$ as $n \to \infty$. Thus, we have $y_n \to 0$ as $n \to \infty$.
We may now recover $x_n$ and find its limit: $x_n = y_n + t \to t = \frac{b}{1-c}$. Hence, the conjectured convergence is true. And it doesn't really matter is anything changes at the initial steps in the induction. </p>

Title: Convergence of sequence


问:
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$
I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$

I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2}  \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?

答:
Just another way to do it

$$\frac{1+x^{2}}{1+x^{4}}=\frac{1+x^{2}}{(x^2+i)(x^2-i)}=\frac 12\left( \frac{1+i}{x^2+i}+\frac{1-i}{x^2-i}\right)$$ and you could even continue with partial fractions.

Without any simplifications
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac{2 (i+1) \tan ^{-1}\left(\frac{x}{\sqrt{i}}\right)-(i-1) \left(\log
   \left(\sqrt{i}-x\right)-\log \left(\sqrt{i}+x\right)\right)}{4 \sqrt{i}}$$ Now, playing with the complex numbers,
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac{\tan ^{-1}\left(1+\sqrt{2} x\right)-\tan ^{-1}\left(1-\sqrt{2}
   x\right)}{\sqrt{2}}$$ REcombining the arctangents,
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac 1{\sqrt{2}}\tan ^{-1}\left(\frac{\sqrt{2} x}{1-x^2}\right)$$


问:
Show that if $x \not\equiv23 \pmod{101}$, then $x^{33} \not\equiv9 \pmod{101}$
Can I have help to prove that for all integers $x$, if $x \not\equiv23 \pmod{101}$, then $x^{33} \not\equiv9 \pmod{101}$?

First I start with contrapositive, but how I can work with it? I try splitting the modulus, but it is prime. Can someone help please? Thanks.

答:
We have:

$$x(x^{33})^3 \equiv 1 mod (101)$$

$$x^{33}\equiv (r^{1/3}=\frac 1 x^{\frac 13}) mod (101)$$

Due to statement we must have:

$r^{1/3}\neq 9$, or, : $\frac 1x \neq 729 $, or:$x\neq\frac 1{729}$

Now we have $x\neq 23 +101 k$.

That means $x$ may be: $x=101k+22.$

In this case for $k=7$ we get: $x=707+22=729,$ which contradicts the second condition.


问:
Finding the minimal subalgebra as certain matrice were given
Let us consider the su(4) Lie algebra which is with 15 orthogonal generators and is of rank 3. Suppose I know the fundamental representation $F_{1-15}$ (which are 4-by-4 traceless Hermitian matrices) of these 15 generators, and according to which I can also obtain the adjoint representation $A_{1-15}$ of these generators by the commutators ($A_{1-15}$ are 15-by-15 purely-imaginary and Hermitian matrices).

Here are the questions:

(1) Given a certain Hermitian matrix $B = \sum_i a_i A_i$ with $a_i$ being real numbers, whether I can always find some other two Hermitian matrices, say B' and B'', such that $\{B,B',B''\}$ construct a Cartan subalgebra of su(4)? If so, how to find out the explicit forms of B' and B''?

(2) One step further, given two Hermitian matrices $B$ and $B'$ which do not commute, i.e. $[B,B']\neq 0$, whether there must exist a Hermitian matrix $B''$ such that $\{B,B',B''\}$ can construct a subalgebra of su(4)? If so, how to find out the minimal subalgebra, and what is the explicit form of $B''$?

These two questions were written in the adjoint representation, but it seems that they do not rely on certain representations. Namely, e.g., I can replace all $A_i$ by $F_i$ in both questions.

For the question (2), I can raise some special cases that make this work. For example, if we consider the two matrices $B = \sigma^x\otimes I$ and $B' = \sigma^y\otimes I$ with $\sigma^{x,y,z}$ being the Pauli 2-by-2 matrices and $I$ being the 2-by-2 identity. And obviously, we can take $B'' = \sigma^z\otimes I$ such that $\{B,B',B''\}$ form an su(2) subalgebra in the su(4).

However, I have no idea about the more general case. I want to try this proof by first putting $B$ in certain Cartan subalgebra, and then demonstrating that $B$ and $B'$ cannot generate all the simple roots.

答:
For question (1):

A subalgebra of a (semisimple) Lie algebra is called toral if it is abelian and all its elements are $ad$-semisimple.

Lemma 1: If $\mathfrak g$ is a Lie algebra over a field over characteristic $0$, then a subalgebra $\mathfrak h$ of $\mathfrak g$ is a Cartan subalgebra if and only if it is maximal toral (i.e. it is toral, and no subalgebra which properly contains $\mathfrak h$ is toral).

Proof: See Equivalence of Two Cartan Subalgebra Definitions in Semi-Simple Lie Algebra and If X commutes with all elements of the Cartan subalgebra, then X is in the Cartan Subalgebra?. Confer also Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?.

Fact 2 (alluded to in https://math.stackexchange.com/a/3820346/96384): For $n \ge 2$, every element of the Lie algebra $\mathfrak{su}(n)$ is $ad$-semisimple. Equivalently, the only nilpotent element of $\mathfrak{su}(n)$ is zero. Equivalently, the only $ad$-diagonalisable element of $\mathfrak{su}(n)$ is zero.

(Try to show this fact and/or the equivalences by hand. Such Lie algebras are called "anisotropic". For the ground field $\mathbb R$, this is the same as what is commonly called "compact". An analogous statement on the Lie group level is in Compact semisimple Lie groups contain no nontrivial unipotent elements? )

Conclusion: For every element $0\neq x \in \mathfrak{su}(4)$ there exist elements $x', x''$ such that $x, x', x''$ span a Cartan subalgebra of $\mathfrak{su}(4)$.

Proof: The one-dimensional subalgebra spanned by $x$ is toral because of Fact 2. Every toral subalgebra is contained in a maximal toral subalgebra, i.e. (by Lemma 1) a Cartan subalgebra. We know that Cartan subalgebras of $\mathfrak{su}(4)$ are three-dimensional.

This translates to your statement for a representation which consist of whatever matrices.

I am not entirely sure what you mean in question 2, but as said in a comment, it is known that every simple Lie algebra can be generated by two elements. In particular, for each simple Lie algebra contained in $\mathfrak{su}(4)$ (of which there are plenty, of various dimensions, including of course the entire $\mathfrak{su}(4)$), one can find two elements $x,x' \in \mathfrak{su}(4)$ which generate that subalgebra (and one could then just choose a third $x''$ contained in that one). If one can even pick a third element, there are obviously many more subalgebras one could generate. I find it hard to imagine any subalgebra of $\mathfrak{su}_4$ for which one could not find three elements which generate it.

But then again, as said I am not sure if that is what you ask. In particular I don't understand what you mean by finding "the minimal subalgebra" and "the explicit form" of the third element?

Added in response to comment: In principle, it is straightforward to find elements $x', x''$ to a given $x$ as above: Because as said every non-zero element of our LA is semisimple, just compute the centralizer of $x$, take one element $x'$ from that which is not a scalar multiple of $x$. Next, compute the centralizer of $x'$, take the intersection with the centralizer of $x$ you computed before; by general theory, this intersection still contains elements linearly independent from the space spanned by $x$ and $x'$; take one of them and call it $x''$. (I am not saying the computations of centralizers are easy, but they are doable.)

Maybe you should clarify for yourself here that starting with one element $x$, you can certainly find elements $x', x''$ such that they form a basis of a CSA, but in general this will be far from unique. Even in the "most unique" case of $x$ being a regular element, i.e. it is contained in a unique CSA, if you have found such $x', x''$, you can obviously replace them by any linear combination of $x, x', x''$ as long as the resulting space is still three-dimensional.

This is e.g. the case for $B = diag(0,i,-i,0)$. (Also note that I use standard math notation, for me $\mathfrak{su}(4)$ consists of traceless skew-Hermitian matrices. To get physics notation, divide everything by $i$.) Here the only choices for $B'$ and $B''$ are two linearly independent elements of the space $\{ diag(ia, -ia,ib,-ib) : a,b \in \mathbb R \}$.

But e.g. if your first element is $B=diag(i,-i,0,0)$, you can also choose $B' = \pmatrix{0&0&0&0\\0&0&0&0\\0&0&0&1\\0&0&-1&0}$, a non-diagonal matrix, and then e.g. $B'' = diag(0,2i,-i,-i)$. (Some centralizers are easily computed.)


问:
Diffeomorphism from Inverse function theorem
I often heard that it is possible to show by using the inverse function theorem that if a function is smooth (i.e. arbitrarily often differentiable), a bijection between open sets, and has a non-singular jacobian, then it is a smooth diffeomorphism. but somehow the inverse function theorem that I know and that wikipedia seems to know, only states that if it is a continuously differentiable bijection with nonzero jacobian, then its inverse function is also continuously differentiable. But how do you get from this, to the statement that I proposed above? I don't see the implication.</p>

答:
Let $LI^m$ and $L^m$ be the set of invertible linear transformations of $\mathbb{R}^m$ in itself and the set of linear transformations of $\mathbb{R}^m$ in itself respectively. Define $\operatorname{Inv}: LI^m\to L^m$ by $\operatorname{Inv}(A)$ is the inverse matrix of $A$. First you can note that if $f$ is a function satisyfing your conditions, then $$(f^{-1})'(y)=[f'(g(y))]^{-1}$$

This implies that $(f^{-1})'=(\operatorname{Inv})\circ f'\circ f^{-1}$. Because $\operatorname{Inv},f\in C^\infty$, you can conclude from the last formula (by using chain rule) that $f^{-1}\in C^\infty$.

Update: I do not understood OP's comment in the bounty, nevertheless I give here more explanations:

Since $f'\neq 0$ we have that $\operatorname{Inv}(f'(g(y)))$ is well defined for all $y$, so the derivative of $(f^{-1})$ dependes on functions that are differentiable again. By a induction argument OP's can finish the proof. Also the proof that $\operatorname{Inv}$ is $C^\infty$ was given in the comments by @ronno. As I have wrote to @Lipschits in the comments, I know another proof of the fact that $\operatorname{Inv}$ is $C^\infty$, but it is really big and @ronnos's proof is more straightforward.


问:
Pointwise convergence of integrable functions to a real measurable function
I want to prove the following result:

If $\left( X, \mathcal{B}, \mu \right)$ is a $\sigma$-finite measure space and $f: X \rightarrow \mathbb{R}$ is a measurable function, then there is a sequence of integrable simple functions which converges pointwise to $f$.

To do so, I am given the hint as contruct $\phi_n: X \rightarrow \mathbb{R}$, as

$$\phi_n = \chi_{X_n} \left( \sum\limits_{k = 1}^{n2^n} \dfrac{k}{2^n} \chi_{f^{-1} \left[ \frac{k}{2^n}, \frac{k + 1}{2^n} \right]} + n \chi_{f^{-1} \left[ n, \infty \right)} \right).$$

Here, $X = \bigcup\limits_{n = 1}^{\infty} X_n$, where $\mu \left( X_n \right) < \infty$ for each $n \in \mathbb{N}$. Now, I can see that each $\phi_n$ is a simple function (since it can take only finitely many values and is defined on measurable sets. Also, it is integrable with the estimate

$$\int\limits_X \phi_n \mathrm{d}\mu = n \mu \left( X_n \right) \left( n 2^{n - 1} + 1 \right).$$

However, what I cannot see (directly) is that why should $\phi_n \to f$ pointwise?

答:
There are some typos in your definition of $\phi_n$. First of all, this works only for non-negative measurable functions $f$ since your $\phi_n$'s are all non-negative.  The sum starts from $k=0$.  Also, you cannot include both the end points so I will drop the right end-point in this proof.

For any $x$ note that $f(x) <n$ for $n$ suffuiceintyl large. For such $n$  there exists a unique $k$ such that $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$. This implies $k \leq n 2^{n}$. Now $\frac k {2^{n}} \leq x< \frac {k+1} {2^{n}}$ and $\phi_n(x)=\frac k {2^{n}}$. Thes two inequalities give $|f(x)-\phi_n(x)|$ is less the the length of the interval $[\frac k {2^{n}}, \frac {k+1} {2^{n}})$ which is $\frac 1 {2^{n}}$. Hence $\phi_n(x) \to f(x)$. If $f$ is also integrable then so so each $\phi_n$ since $0 \leq \phi_n \leq f$


问:
Find $99^{101103} \bmod 101$
Find $99^{101103} \bmod 101$.

The solution doesn't make any sense to me for this. They use Fermat's Little Theorem. Also used the fact that $101$ is prime, $a^{100} = 1 \text{ mod } 101$

$$99^{101103} = 99^{101100} \cdot 99^3 = (99^{100})^{1011} \cdot 99^3 \equiv 1^{1011}\cdot (-2)^3 \equiv -8 \equiv 93 \ (\text{ mod } 101\ ).$$

I don't understand how they got $(99^{100})^{1011} \rightarrow 1^{1011}$ and where did the $(-2)^3 $ come from?

答:
Recall the fact that $a^{100} \equiv 1 \pmod{101}$ (provided that $a$ is not divisible by $101$).

Let $a = 99.$

Also $99 \equiv -2 \pmod{101},$
because $-2 = 99 - 101.$

Title: Find $99^{101103} \bmod 101$


问:
Chain Rule with 1/cos(x)
In the statement of the attached snapshot it states:

$1/\cos(x)$ is made up of $1/g$ and $\cos()$:

$$f(g) = 1/g$$

How is $1/\cos(x)$ partially made up of $1/g$?

Where does $1/g$ come from?

答:
Let $g(x) = \cos x$. Then $\frac{1}{\cos x} = \frac{1}{g}$. Then
$$\left(\frac{1}{\cos x}\right)' = \left(\frac{1}{g}\right)' = -\frac{1}{g^2}g' = -\frac{1}{\cos ^2x}(-\sin x) = \frac{\sin x}{\cos^2x}$$

Title: Chain Rule with 1/cos(x)


问:
Expressing hyperelliptic integrals through series
Suppose I want to evaluate the following integral, but non-numerically:
$$\int_1^2\frac1{\sqrt{x^5-x+1}}\,dx$$
This is, of course, a hyperelliptic integral, which cannot in general be expressed in terms of elementary (or even elliptic) functions. Furthermore, it is quite well-known that the denominator polynomial here is not even solvable in radicals, but just assume we know the roots. (It is far easier to numerically find polynomial roots than to numerically integrate.)

Byrd and Friedman have this to say about the matter:

For the evaluation of hyperelliptic integrals, one must usually resort to direct numerical integration or to the use of complicated series expansions.

This second part has me intrigued. How can I express the integral above using series, hypergeometric or otherwise? Is there a general procedure to derive such a series representation?

Note that if there were just two terms in the polynomial, the integrand would be expressible as a binomial series; the integral itself would then be expressible using the $_2F_1$ function. But there are three terms here, so $_2F_1$ cannot be immediately applied.

Taylor and Padé expansions are not quite desirable either, since they require much work at extended precision and will not converge over the entire domain if truncated.


问:
Large deviations Exercise (Durret 2.7.6)
I'm studying probability theory and doing the exercises in Durret v5.

Let $X_1,\ldots, X_n$ iid with $EX_1 = 0$. Show that if $\epsilon,a>0$, then
then  $$\liminf_{n\to\infty }\frac{P(S_n \geq na)}{nP(X_1 \geq
n(a+\epsilon))}\geq 1$$

Hint: Let $F_n =\{ X_i \geq n(a+\epsilon)
> \text{for exactly one } i\leq n\}$.

Edit: I tried using the inequality $P(S_n \geq na) \geq P(S_{n-1}\geq -n\epsilon)P(X_n \geq n(a+\epsilon))$, but i don't know what to do next.

答:
Here's my answer after working on it for some time. A cleaner or shorter proof is welcome:

Let $F_n =\{X_i \geq n(a+\epsilon)\textrm{ for exactly one } i\leq n\}$ and note that $F_n = \bigsqcup_{i=1}^{n} G_i$, where $G_i$ is the event that the large element has index $i$. This gives us that:
\begin{align*}
\frac{P(S_n \geq na)}{nP(X_1\geq na)} &\geq \frac{P(F_n \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em]
&=\frac{\sum_{i=1}^{n}P(G_i \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em]
&=\frac{nP(G_1 \cap \{S_n \geq na\})}{nP(X_1\geq n(a+\epsilon))} \\[0.5em]
&=\frac{P(X_1\geq n(a+\epsilon))P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right)}{P(X_1\geq n(a+\epsilon))} \\[0.5em]
&= P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right)
\end{align*}
Since $EX_1 = 0$, then $E|X_1| <\infty$ and $nP(X_1 \geq n(a+\epsilon)) \xrightarrow{n\to\infty}0$. This gives us that:
$$P\left(\max\limits_{i<n}X_i \geq n(a+\epsilon)\right) \leq nP\left(X_1 > n(a+\epsilon)\right) \xrightarrow{n\to\infty}0. \ (*)$$
We have that $\liminf P\left(\left\{\max\limits_{i<n}X_i < n(a+\epsilon)\right\} \cap \{ S_{n-1}\geq -n\epsilon\} \right)$ is $1$ by the weak law of large numbers and $(*)$. This gives us the result.


问:
Does negative third derivative imply negative first derivative?
Does negative third derivative imply negative first derivative? For a system, the negative of the derivative of the Lyapunov function means the system is stable. How about the negative third derivative?</p>

答:
No, consider $f(x) = x-x^3$ at $x=0$.

$f'(0) = 1 > 0$, while $f'''(0) = -6 < 0$.

Perhaps you are assuming additional contraints?


答:
Yes, prove the contrapositive.  Let $x^{33}\equiv9\bmod101$.

Then $x^{-1}\equiv (x^{100})x^{-1}\equiv(x^{33})^3\equiv   9^3=729\equiv22$.

Thus $x\equiv22^{-1}\equiv23\bmod101$.

Title: Show that if $x \not\equiv23 \pmod{101}$, then $x^{33} \not\equiv9 \pmod{101}$

Body:
Can I have help to prove that for all integers $x$, if $x \not\equiv23 \pmod{101}$, then $x^{33} \not\equiv9 \pmod{101}$?

First I start with contrapositive, but how I can work with it? I try splitting the modulus, but it is prime. Can someone help please? Thanks.


问:
Can I write e(mathematical constant) in fraction?
Proof of e can be in fraction

e is 2.718281828.....
so as you can see, there is a pattern. So, I break it down into 2.7 + 0.01828 + 0.0000001828 .....

and then I use sum of infinity to get the fraction

Well, in mathematics theory , e cannot be in fraction. But I tried to use sum of infinity and proof it that it can be in fraction. Am I correct?

答:
You have proved that

$$
e' = 2.718281828(1828)...
$$

is rational. This is a special case of a more general statement that a number is rational if and only if its decimal expansion is repeating or terminating. However, $e'$ is not the same number as

$$
e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n = 2.718281828\color{red}{45904}... .
$$

In fact, it is a fairly well-known fact that $e$ is irrational (transcendental even). See for example this wikipedia article for a few proofs of this fact.

That said, your observation provides a convenient mnemonic for the first ten digits of $e$!


问:
Finding all cyclic subspaces of $\mathbf{C}^2$ and $\mathbf{R}^2$ w.r.t to rotational linear transformation
I have been studying linear algebra and came across this question that I am not quite sure how to solve.

The question begins by letting $T:\mathbf{R}^2 \rightarrow \mathbf{R}^2 $ be a rotation counter-clockwise 90 degrees and let $S: \mathbf{C}^2 \rightarrow \mathbf{C}^2$ be defined by $S(x,y) = (-y,x)$. The question asks to find all cyclic subspaces of $\mathbf{R}^2$ with respect to $T$ and all cyclic subspaces of $\mathbf{C}^2$ with respect to S.

The hint says to consider the independence or dependence of the list $x, Tx$ for $x \in \mathbf{R}^2$ and the set $y,Sy$ for $y\in\mathbf{C}^2$ but I'm not sure how to solve this problem

Any help would be greatly appreciated!

答:
For both types of linear transformations it is true that
$$T^2=-I, T^4=I$$
where $I$ is the identity transformation. This means that we only need to check whether $x,Tx$ are linearly independent, since all the other vectors generated by powers of $T$ are linearly dependent on these two.

Now to find all possible cyclic subspaces, consider the vector subspaces of $\mathbb{R^2}$ and of $\mathbb{C^2}$ one by one ranked by their dimensionality.

The zero-dimensional subspace is a cyclic subspace for both vector spaces. When $x=(0~ 0)^T$ it is obvious that $Tx=x=0$ as well and the subspace closes.

Consider a general vector in $\mathbb{R}^2$, $x=\begin{pmatrix}x_1\\x_2\end{pmatrix}$. Then
$$Tx=\begin{pmatrix}-x_2\\x_1\end{pmatrix}$$

Unless $x_1,x_2=0$ there exists no $\lambda\in \mathbb{R}$ such that $Tx+\lambda x=0$. Thus the two vectors are always linearly independent of one another and therefore they form a basis of $\mathbb{R^2}$. We conclude that for $T:\mathbb{R}^2\to\mathbb{R}^2$, the two T-cyclic subspaces are the origin and $\mathbb{R^2}$ itself.

Consider a general vector in $\mathbb{C}^2$, $x=\begin{pmatrix}z_1\\z_2\end{pmatrix}$. Then
$$Tx=\begin{pmatrix}-z_2\\z_1\end{pmatrix}$$
Suppose there exists  $\lambda\in \mathbb{C}$ such that $Tx+\lambda x=0$. This equation can be satisfied if
$$z_1=\lambda z_2~,~  z_2=-\lambda z_1$$
substituting the second equation into the 1st one and discarding the case when $z_1=z_2=0$  this leads to
$$\lambda^2+1=0$$
which indicates that $\lambda=\pm i~, ~z_1=\pm i z_2$ can satisfy the equation! We conclude that there are two additional separate one-dimensional T-cyclic subspaces in $\mathbb{C}^2$ spanned by the vectors $\begin{pmatrix}1\\i\end{pmatrix}, \begin{pmatrix}1\\-i\end{pmatrix}$, that $T$ leaves invariant. $\mathbb{C}^2$ is another obvious cyclic subspace generated by any vector other than the ones mentioned above.


问:
Questions on a subset of $C[0,1]$
Problem: Let $$\mathcal{F} =\left\{F\in C[0,1]: F(x) = \int_0^x f(t) \, dt, \,\, f\in C[0,1],\, \, ||f||_\infty \leq 1\right\}. $$
Questions:

a) Show that $\mathcal{F}$ is not closed.

b) Show that $\overline{\mathcal{F}} = \{f \in C[0,1]:f(0) = 0 , \, \, \text{Lip}(f)\leq 1\}$ (Lip represents Lipschitz constant).

I'm not quite sure how to start with these two questions. For part (a), I was trying to construct a sequence that does not converge in $\mathcal{F}$ but this seems to be difficult. For part (b), I was given a hint to find $F_n \in \mathcal{F}$ so that $F_n(\frac{k}{2^n}) = (1-\frac{1}{n})f(\frac{k}{2^n})$ but I'm not too sure how does that help with this question.

Any suggestion or hint for these two parts? Thank you.

答:
Hints: Let $g$ be any integrable function on $[0,1]$ which is not almost everywhere equal to any continuous such that $|g(x)| \leq 1$ for all $x$. (Eg: characteristic fucntion of $(0,\frac 1  2)$).  Let $f(x)=\int_0^{x} g(t)dt$. Then $|f(x)-f(y)|\leq |x-y|$. We can find  sequence $(g_n)$ of continuous functions  such that $|g_n(x)| \leq 1$ for all $n,x$ and $\int_0^{1} |g_n(x)-g(x)|dx \to 0$. Let $f_n(x)=\int_0^{x} g_n(t)dt$. Then each $f_n \in \mathcal F$ and $f_n \to f$ uniformly. Hence, $f$ belongs to the closure of $\mathcal F$ (w.r.t. the sup norm). But $f$ does not belong to $\mathcal F$.

For the second part let $f$ be any Lipschitz function with Lipschitz constant less than or equal to $1$ vanishing at $0$. Then $f$ is absolutely continuous, and we can write $f(x)=\int_0^{x} f'(t)dt$. Approximating $f'$ by continuous functions as above we see that $f$ belongs to the closure of $\mathcal F$.  The reverse inclusion is obvious.

Title: Questions on a subset of $C[0,1]$


问:
What is the extension of Bayesian Network into cyclic graph?
The wikipage of Bayesian Network says

"Formally, Bayesian networks are directed acyclic graphs whose nodes represent random variables in the Bayesian sense"

But in the model I need to build, cyclic structure of constraint is necessary. For example, A influences B, B influences C, C influences A.

What should I use then(instead of Bayesian Network)?

ps. I saw "Markov Network", but it says that the variables should have the "Markov property"(Memorylessness), which is not necessarily true in my intended application, in the sense that some variable is influenced by the history of others as well as its own.

Thank You!

Matt

答:
You might be interested in this paper on discovering cyclic causal models: http://arxiv.org/abs/1206.3273

While cycles can be introduced into directed graphical models, it makes it significantly more complicated to compute the probability of some configuration. For the graphical model $X\rightarrow Y$, if we know the marginal probability of $X$ and the conditional probability of $Y$ given $X$, then the joint probability of $X=x$ and $Y=y$ is just $Pr(X=x,Y=y) = Pr(X=x)Pr(Y=y|X=x)$, and in general any probability can be computed by a combination of multiplication and marginalization. If by contrast we have a cyclic graphical model $X \leftrightarrow Y$, then this is best viewed as a Markov chain $\ldots X\rightarrow Y \rightarrow X \rightarrow Y \ldots$ where, even if we know the transition probabilities $p(x|y)$ and $p(y|x)$, we still have to solve for the stationary state of a Markov chain to compute various probabilities. For cyclic models with larger graphs the situation becomes even more complicated.

I realize this doesn't solve your problem exactly though. Spelling out the problem you're trying to solve in more detail might help. For instance, if the influences between A, B and C are symmetric (A influences B the same way B influences A) then an undirected graphical model would make more sense. If there is a temporal aspect to the influence of one variable on another, a Markov chain would make more sense (and in some sense a cyclic causal model can be seen as a kind of Markov chain, where we are trying to reason about properties of the stationary distribution).