问:
Prove$\sum_{|\alpha|<1, f(\alpha)=0}\log \frac{1}{|\alpha|}= \sum_{\Re(\rho)>1/2}\log\ |\frac{\rho}{1-\rho}|$
Let
$$ f(z)= (s-1)\zeta(s) $$
where $s=\frac{1}{1-z}$ and $\zeta$ denotes the Riemann Zeta function.

Prove that,$$\sum_{|\alpha|<1, f(\alpha)=0}\log \frac{1}{|\alpha|}= \sum_{\Re(\rho)>1/2}\log\ \left|\frac{\rho}{1-\rho}\right|$$
I found this in this paper of Balazard Saias and Yor-.

My try- I have understood the whole paper but this one line i could not get.

答:
This is very elementary math as below:

Let: $\rho=\frac{1}{1-\alpha}$

Then $1/\rho=1-\alpha$, so $\alpha=\frac{\rho-1}{\rho}$, so $1/\alpha=\frac{\rho}{\rho-1}$ so $|1/\alpha|=|\frac{\rho}{\rho-1}|=|\frac{\rho}{1-\rho}|$
Since $f(\alpha)=0$ corresponds to $\zeta(\rho)=0$ and of course $|\alpha| <1$ corresponds to $\Re \rho > 1/2$ the result should be clear now.

As an aside, the above is part of the proof of the Balazard Saias Yor criterion (which itself is part of a large number of integral criteria equivalences to RH, but it looks simpler so it has more appeal to amateurs) to prove/disprove the Riemann Hypothesis using the formula proved in the mentioned paper:

$\int_{\Re s =1/2}\frac{\log |\zeta(s)|}{|s|^2}d|s|=2 \pi\sum_{\Re(\rho)>1/2, \zeta(\rho)=0}\log\ \left|\frac{\rho}{1-\rho}\right| $

so in particular the integral is zero iff RH is true


问:
Does a semicircle have 2 right angles?
If you have a line passing through the middle of a circle, does it create a right angle at the intersection of the line and curve?

More generally, is it valid to define an angle created between a line and a curve? Is the tangent to the curve at the point of intersection a valid interpretation (I.e a semi circle has 2 right angles)

I saw it in a debate thread and it got me curious now.

答:
As you identify in your question, the real point of contention is the definition of angle. As the other answers have indicated, if your definition includes angles at the intersection of two curves then a semicircle certainly has two right angles.

However, as the controversy on Twitter (a few days ago at time of posting, started because this question was in someone's daughter's primary school homework) indicates, angle as commonly used is often restricted to the intersections of (straight) lines. Under this definition there is no angle at the vertices of a semicircle.

Which definition is more common is an empirical question (and the consensus seems to be that the wider definition is common among professional mathematicians, while the stricter definition might be something a layperson would describe), but to my mind the fact that there exist differing definitions makes your question ill-formed. I think the educational value is to expose the fact that we should be careful with how we think about our definition.

ContentLicense: CC BY-SA 4.0

Title: Does a semicircle have 2 right angles?


问:
How to prove that $\lim\limits_{x \to 0 }\;x^{-a}e^{\frac{-1}{x^{2}}} =0$ for all a?
I'd like your help with this:

I tried using L'Hôpital's Rule and all kinds of arithmetic to prove that
$$\lim_{x \to 0 }\left(x^{-a}e^{\left(\frac{-1}{x^{2}}\right)}\right) = 0$$ for every $a$, and it didn't work.

($a=0$ is trivial)

Any hints?

Thank you.

答:
Actually, this result you hope for is false if $a$ is allowed to be an arbitrary real number (but it only fails for stupid reasons).  If $a = -\frac{1}{2}$, for example, then

$$\lim_{x \to 0^{-}} \frac{\sqrt{x}}{e^{\frac{1}{x^2}}} $$

does not exist.

If one is restricted to integer values of $a$, then the result does as was previously discussed.  Also, the limit $\displaystyle \lim_{x \to 0} |x|^{-a} e^{- \frac{1}{x^2}}$ always exists for all $a$.

ContentLicense: CC BY-SA 3.0


问:
How to prove that two parametrizations of a surface M give the same tangent plane
If you have 2 parametrizations of a surface $M$, $x$ and $y$, such that for some point $P \in M$, $x(u_0,v_0)=y(s_0,t_0)=P$. How do you show that the tangent plane of $M$ at $P$ are the same? That is how do I show that $\text{Span}(x_u,x_v)=\text{Span}(y_s,y_t)$.

答:
Ted Shifrin's answer can be generalized:

Let $A$ be an affine space with a real and normed translation space $V$ and let $x\colon U\subset\mathbf{R}^n\to A$ and $y\colon U'\subset\mathbf{R}^n\to A$ be injective maps defined on open subsets of $\mathbf{R}^n$ such that the transition maps $t:=y^{-1}\circ x\colon U\cap U'\to U\cap U'$ and $t^{-1}=x^{-1}\circ y\colon U\cap U'\to U\cap U'$ are differentiable in $a\in U\cap U'$ and $b:=y^{-1}(x(a))\in U\cap U'$. Finally, we require that $x$ and $y$ are differentiable in $a$ and $b$, respectively*.

In this case, $Dt(a)\colon\mathbf{R}^n\to\mathbf{R}^n$ and $Dt^{-1}(b)\colon\mathbf{R}^n\to\mathbf{R}^n$ are isomorphisms inverse to each other** and according to the chain rule,
\begin{equation}
Dy(b)=Dx(a)\circ Dt^{-1}(b)
\end{equation}
and
\begin{equation}
Dx(a)=Dy(b)\circ Dt(a).
\end{equation}
Thus, $Dx(a)(\mathbf{R}^n)=Dy(b)(\mathbf{R}^n)=:S\subset V$.

In your notation,
\begin{equation}
x_i=\lim\nolimits_{\delta\to 0}\frac{x(a_1,\ldots,a_{i-1},a_i+\delta,a_{i+1},\ldots,a_n)-x(a_1,\ldots,a_n)}{\delta}=Dx(a)(e_i)
\end{equation}
and $\text{span}\{x_1,\ldots,x_n\}=S$.*** Thus, we just proved the result you asked for.

Remark: $(x_1,\ldots,x_n)$ is a basis of $S$ if and only if $Dx(a)$ is injective and because of the two centered equations, $Dx(a)$ is injective if and only if $Dy(b)$ is injective.

*If $X$ and $Y$ are affine spaces with real and normed translation spaces $V$ and $W$, a map $f\colon D\subset X\to Y$ is called differentiable in $p\in D$, if $p$ is an inner point of $D$, $f$ is continuous in $p$ and a linear map $A\colon V\to W$ with the following property exsits: For each $0<\epsilon$, there is a $0<\delta$ such that
\begin{equation}
\|f(p+v)-f(p)-A(v)\|\leq\epsilon\|v\|
\end{equation}
for all $v\in V$ with $\|v\|<\delta$. If $f$ is differentiable in $p$, there is a unique linear map with this property, called the derivative of $Df(p)$ of $f$ in $p$.

**You can prove this by using the chain rule and the fact that a map $f\colon A\to B$ is injective if and only if there is a map $g\colon B\to A$ such that $g\circ f=\text{id}_A$ and surjective if and only if there is a map $g\colon B\to A$ such that $f\circ g=\text{id}_B$.

*** If $f\colon D\subset\mathbf{R}^n\to A$ is differentiable in $a=(a_1,\ldots,a_n)\in D$, the partial derivatives
\begin{equation}
\partial_i f(a)=\lim\nolimits_{\delta\to 0}\frac{x(a_1,\ldots,a_{i-1},a_i+\delta,a_{i+1},\ldots,a_n)-x(a_1,\ldots,a_n)}{\delta}\in V
\end{equation}
exist for all $i\in\{1,\ldots,n\}$ and
\begin{equation}
Df(a)(x_1,\ldots,x_n)=\sum_{i=1}^n x_i\cdot \partial_i f(a).
\end{equation}


问:
Showing equivalence of expressions for the joint probability of two events
If I have random variable $v$ distributed by CDF $G$ on $[\underline{v},\bar{v}]$, I know that $Pr(v\geq a-x \hspace{1mm} and \hspace{1mm} v\geq a)=Pr(v\geq a-x | v  \geq a)Pr(v\geq a)= Pr(v\geq a)$, where $x$ is non-negative. But I cannot seem to get the same answer if instead I write $Pr(v\geq a | v\geq a-x)Pr(v\geq a-x)$. These two should clearly be equivalent. Thank you.

答:
If $x\ge 0$, then clearly
$$
\{v\ge a-x\}\cap\{v\ge a\}=\{v\ge a\},
$$
and so, the probability of these events is the same.

If you, nevertheless, want to use conditioning,
$$
\mathsf{P}(v\ge a\mid v\ge a-x)=\frac{\mathsf{P}(v\ge a)}{\mathsf{P}(v\ge a-x)},
$$
assuming that the denominator is positive, so that you get the same result.


问:
trigonometric equation containing sin(x) and x
I have a problem solving the following trigonometric equation and would be happy for any help.

$$\sin(2x) = 3,5x$$

What I tried is:

$$2\sin(x)\cos(x) = 3,5x\\
 2\sin(x)(1-\sin(x)) = 3,5x\\
 2\sin(x)-2\sin^2(x) = 3,5x\\
 \sin(x)-\sin^2(x) = 1,75x$$

And now I don't know what to do ...

答:
There is only one solution: $x=0$.

Note that if $|x|>\frac27$, then it is impossible for this equality to hold since $|3.5x|>1\geq|\sin(2x)|$.

Now use the Taylor expansion of $\sin$ to show that there is no other solution in $\left(\frac{-2}7,\frac27\right)$


问:
Derangement problem but with functions instead of permutations, using inclusion-exclusion.
How many functions $f : [n] \to [n]$ are there with no fixed points, i.e. such that $f(i) \ne i$ for all $i$? Use inclusion-exclusion to solve it.

答: For each $i$ there are exactly $n - 1$ non-fixed point values for $f(i)$, so $(n -1)^n$ functions in all.


问:
Double sigma formula with nested index
I am trying to obtain a formula for:

$G = m_1m_2 + m_1m_3 + m_2m_3$ using sigma notation.

I've come up with:
$\sum_{i=1}^{3} \sum_{j=2}^{3} m_im_j = \sum_{i=1}^{3}m_im_2 + m_im_3 = m_1m_2 + m_1m_3 + m_2m_2 + m_2m_3 +m_3m_2 + m_3m_3$. My question is, how do I exclude the values $m_2m_2, m_3m_2, m_3m_3$ in order to obtain $G$? That is, how should I exclude the values where $i \geq j$? Also, is there any way that I can express the above sum "without" the double sigma?

答:
A common notation is
\begin{align*}
\color{blue}{G }= m_1m_2 + m_1m_3 + m_2m_3\color{blue}{=\sum_{1\leq i<j\leq 3}m_im_j}
\end{align*}

Title: Double sigma formula with nested index


问:
Multiplication/Cayley tables for the Dihedral Groups
I am currently doing a group theory problem, which asks for the multiplication table of the dihedral group $D_4$. Having looked up the answer online, I do not understand how some of the elements arose. For example, the elements on the columns and rows are $e,a,a^2,a^3,b,ba,ba^2,ba^3$. I understand how those elements came about, but I don't understand why e.g. $ab,a^2b,a^3b$ are not considered here.  Can anyone help?

答:
What you are doing, implicitly, is expressing the group $D_4$ with a presentation. A presentation of a group is made of two parts: a set of generators, and a set of relations.

Generators are commonly denoted with lowercase letters: $a,b,c,...$

Relations are expressions that generators satisfy: for instance, normally you have some relations with a single generator $a^4=e$, $b^2=e$, etc. , and then relations that tell you how the generators combine, such as $ab=ba$ (commutativity), or more complex ones like $ab=cba$.

Given a presentation, it is not easy in general to understand what group you are looking at: this is because the group is defined as all words in the generators (think $abba$) modulo the relations (meaning that if $ab=ba$, $abba=abab=baba$ etc.). So... you can have words that look different, but are actually the same when you apply the relations.

In your case, $D_4$ is given by $$\langle a,b \; |\; a^4=b^2=e, ab=ba^{-1} \text{ (or } b^{-1}ab=a^{-1})\rangle $$

So, for example:

$$ab = ba^{-1} = ba^3$$
$$a^2b = a(ba^{-1}) = ba^{-1}a^{-1} = ba^2$$

and, in general, whatever expression you consider can be reduced to one of the eight elements given by your book.


问:
Proving that if $y=\operatorname{arcsec}(x)$ then $\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$
I'm trying to prove this formula however, I cannot seem to figure out how to single out the $x$ and remove its power. I would very much appreciate your help towards this.

$$y=\operatorname{arcsec}(x) \implies\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$$

Here's my working out, my reasoning is wrong somewhere, although I cannot find where I went wrong:

$$y = \operatorname{arcsec}; \quad\sec(y)=x$$

Then taking the derivative of both sides:

$$\sec(y)\tan(y)\frac{dy}{dx}=1; \frac{dy}{dx}=\frac{1}{\sec(y)\tan(y)}=\frac{1}{\sqrt{1-x^2}(x^2+1)}$$

How do I simplify the last equation?

答:
What you can show is that
$$
\frac{d}{{dx}}\sec ^{ - 1} (x) = \frac{1}{{x\sqrt {x^2  - 1} }}
$$
for $x>0$. Indeed, by the formula for the derivative of the inverse function, we find
\begin{align*}
\frac{1}{{\left[ {\frac{d}{{dt}}\sec t} \right]_{t = \sec ^{ - 1} (x)} }} & = \left[ {\frac{{\cos ^2 t}}{{\sin t}}} \right]_{t = \sec ^{ - 1} (x)}  = \left[ {\frac{1}{{\left| {\sec t} \right|\sqrt {\sec ^2 t - 1} }}} \right]_{t = \sec ^{ - 1} (x)} \\& = \frac{1}{{\left| x \right|\sqrt {x^2  - 1} }} = \frac{1}{{x\sqrt {x^2  - 1} }}
\end{align*}
for $x>0$.


答: Assuming that you are trying to show that for $x>0$ $$\frac{d}{dx} \operatorname{arcsec}(x) = \frac{1}{x\sqrt{x^2-1}},$$ we can proceed as follows.  Just as you did, we start with $$y = \operatorname{arcsec}(x) \implies \sec(y) = x.\tag{1}$$  This then gives us $$\frac{dy}{dx}=\frac{1}{\sec(y)\tan(y)} = \frac{1}{\sec(\operatorname{arcsec}(x))\tan(\operatorname{arcsec}(x))}.$$  Then, we have $\sec(\operatorname{arcsec}(x)) = x$, and for $\tan(\operatorname{arcsec}(x))$ we consider the Pythagorean identity $$\tan^{2}(y) = \sec^{2}(y) - 1 \implies \tan(y) = \sqrt{\sec^{2}(y) - 1}.$$  Now, using $(1)$, we can write $$\tan(\operatorname{arcsec}(x)) = \sqrt{\sec^{2}(\operatorname{arcsec}(x)) - 1} = \sqrt{x^{2} - 1}.$$  All together, then, we get $$\frac{d}{dx} \operatorname{arcsec}(x) = \frac{1}{x\sqrt{x^2-1}}.$$

Title: Proving that if $y=\operatorname{arcsec}(x)$ then $\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$

Body:
I'm trying to prove this formula however, I cannot seem to figure out how to single out the $x$ and remove its power. I would very much appreciate your help towards this.

$$y=\operatorname{arcsec}(x) \implies\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$$

Here's my working out, my reasoning is wrong somewhere, although I cannot find where I went wrong:

$$y = \operatorname{arcsec}; \quad\sec(y)=x$$

Then taking the derivative of both sides:

$$\sec(y)\tan(y)\frac{dy}{dx}=1; \frac{dy}{dx}=\frac{1}{\sec(y)\tan(y)}=\frac{1}{\sqrt{1-x^2}(x^2+1)}$$

How do I simplify the last equation?


问:
Integer-valued polynomials divisible by 2 or 3
I would like some help with the following problem:

Suppose that $p \in \operatorname{Int}(\mathbb{Z}) = \{p \in \mathbb{Q}[X] : p(\mathbb{Z}) \subseteq \mathbb{Z}\}$ is such that for all $z \in \mathbb{Z}$, either $2$ divides $p(z)$ or $3$ divides $p(z)$.
Show that the quantifiers may be reversed, i.e., that either for all $z \in \mathbb{Z}$ we have $2$ divides $p(z)$, or for all $z \in \mathbb{Z}$ we have $3$ divides $p(z)$.

I thought of using the fact that $\{\binom{X}{k} \mid k \in \mathbb{N}\}$ is a $\mathbb{Z}$-basis of $\operatorname{Int}(\mathbb{Z})$, where $$\binom{X}{k} = \frac{X (X - 1) \cdots (X - k + 1)}{k!}$$ (in particular $\binom{X}{0} = 1$).
Therefore, for a fixed $p \in \operatorname{Int}(\mathbb{Z})$, I can write $$p = \sum_{i=1}^k a_i \binom{X}{i}$$
for some $k \in \mathbb{N}$ and $a_1, \ldots, a_k \in \mathbb{Z}$.

I tried to show the contrapositive:
suppose that there is $z_2, z_3 \in \mathbb{Z}$ such that $2 \nmid p(z_2)$ and $3 \nmid p(z_3)$.
Then there is $1 \leq i, j \leq k$ such that $2 \nmid a_i$ and $3 \nmid a_j$.
But then I don't quite know how to find a $z$ with both $2 \nmid p(z)$ and $3 \nmid p(z)$,
just as I don't see how "for all $z \in \mathbb{Z}$, either $2$ divides $p(z)$ or $3$ divides $p(z)$" relates to the coefficients of $p(z)$.

Is my approach correct, and how should I proceed?

答:
Let $q$ be an integer valued polynomial and let $x,y$ be integers. Then $x-y$ divides $q(x)-q(y)$.

Well, we wish this was true. It is not : take $P_2(x) = \frac{x(x-1)}{2}$ then $P_2(4) - P_2(2) = 6-1 = 5$ is  not a multiple of $4-2$.

Let's imagine it was the case. Then, we could do the following : let $a_1$ be such that $2$ doesn't divide $p(a_1)$ and $a_2$ be such that $3$ doesn't divide $p(a_2)$. We'd know that for all integers $k,l$, we have $p(a_1+2k) \equiv p(a_1) \pmod{2}$ and $p(a_2 + 3l) \equiv p(a_2) \pmod{3}$. In particular, for any $k,l$ integers, $p(a_1+2k)$ is never a multiple of $2$ and $p(a_2+3l)$ is never a multiple of $3$. However, because $2$ and $3$ are co prime we can find $k,l$ integers such that $a_2-a_1 = 2k-3l$. That is , $a_1+2k = a_2+3l = K$, say. But then $p(K)$ is not a multiple of $2$ or $3$, a contradiction.

Now, the wish we want is not granted by the mathematical gods. But we can do something very interesting, that makes the above argument a possibility. I state a lemma in general, which I then exploit to show the answer. First a definition.

Let $p$ be a prime and $K$ an integer. Then $\nu_p(K)$ is the largest power of $p$ which divides $K$ i.e. such that $p^{\nu_p(K)}$ divides $K$ but $p^{\nu_p(K)+1}$ does not divide $K$. We refer to this as the $p$-adic valuation.

Now, the $p$-adic valuation is non-negative. However, one can also easily verify that $\nu_p(a+b) \geq \min\{\nu_p(a),\nu_p(b)\}$, and $\nu_p(ab) = \nu_p(a)+\nu_p(b)$ quite easily.

Lemma[Cahen-Chabert] : Let $f$ be an integer valued polynomial, with $\deg(f) < p^h$ for some positive integer $h$. Then
$$
\nu_p(f(b)-f(a)) \geq \nu_p(b-a) -h+1
$$
for all integers $a,b$.

Proof : Replacing $f(x)$ by $x \to f(a+x) - f(a)$ we can assume that $a=0$ and $f(0) = 0$. Thus it is enough to show that $\nu_p(f(b)) \geq \nu_p(b) - h+1$. We know that $f_n(x) = \binom xn$ forms a basis of integer valued polynomials, so we can verify that this holds for $f_n$ with $n<p^h$.

Well, for $n=0,1$ it is pretty obvious. For $n \geq 2$, write :
$$
f_n(x) = \frac xn g(x)
$$
where $g(x) = \frac{\prod_{i=1}^{n-1}(x-i)}{(n-1)!}$. Now, for any $b$ we get $nf_n(b) = bg(b)$ and therefore $$
\nu_p(f_n(b)) = \nu_p(b) + \nu_p(g(b)) - \nu_p(n)
$$

but $g$ is a well known integer polynomial, since the product of any $n$ consecutive integers is a multiple of $n!$ (this can be shown in multiple ways). Also, $n<p^h$ so $\nu_p(n) \leq h-1$. Therefore, the result follows.

So we have the Cahen-Chabert lemma. This plays a very important role in the proposed proof.

Indeed, let $p$ be an integer valued polynomial such that $p(z)$ is either a multiple of $2$ or $3$ for every $z$. Then we know that there exist $D,D' > 1$ such that $\deg p < 2^D$ and $\deg p < 3^{D'}$.

Now, let $a_1$ be any integer. Then consider $a_1 + k2^{D}$. Then let $a = a_1$ and $b = a_1 + k2^D$ in the Cahen-Chabert lemma. We get :
$$
\nu_2(f(a_1+k2^D) - f(a_1)) \geq \nu_2(k2^D) - D+1 \geq 1
$$

Therefore, $f(a_1+k2^D) - f(a_1)$ is a multiple of $2$ for all integers $k$.

In similar fashion, for all $a_2$, we know that $f(a_2 + l3^{D'}) - f(a_2)$ is a multiple of $3$ for all integers $l$.

Now, can you figure out the rest, using the argument we wished was the case?

Note : I call this lemma the Cahen-Chabert lemma : in truth, I do not know where one can find it. It does seem to be a useful result, though, judging on this question at least.

I'd like to think there are generalizations. For example, one can probably replace $2,3$ by any pair of coprime integers. I also think one can replace a pair of coprime integers by any set of integers with gcd $1$ (for example, an integer valued polynomial such at every $z$ we have $p(z)$ is a multiple of $6,15$ or $10$ then $p(z)$ is divisible by one of them for all $z$).

These can be tried as exercises. I will elaborate upon requirement the way to do these, but roughly it's about playing with the $h$ and $p$ in the Cahen-Chabert lemma.


问:
Lower bound for $\prod_{j=0}^{k}(1-a_{k-j})$
I am trying to find if

$$\prod_{j=0}^{k}(1-a_{k-j})$$

can be lower bounded by some constant $c$, with $0 \leq a_n \leq 1$. I have seen in other post that

$$\prod \limits_{j=0}^{k}(1-a_{k-j}) \geq1- \sum\limits_{j=k}^{n}a_{k-j}$$

but I do not know how to proceed. Given $\lim_{m\to \infty} a_m \to 0$ can we find such a lower bound $c$?

答:
We cannot bound this product with some $c>0$. For example, let us take a sequence that satisfy the given conditions:
$$a_n = \begin{cases} 2, & n = 0 \\ 1/2, & n = \overline{1,k} \\ 0, & n > k \end{cases}$$
Definitely, $a_n \to 0$ and $a_n \ge 0$. However,
$$\prod_{j=0}^{k}(1-a_{k-j}) = \prod_{j=0}^k(1-a_j) = (1-2)\left(1 - \frac{1}{2}\right)^k < 0$$
Note that by choosing $a_0$ appropriately, we can make this product be less than any real number.

Therefore, with no constraints on $a_n$, we cannot lower bound the product.

EDIT: After the additional condition $0 \le a_n \le 1$, we have
$$1-a_0 \ge 1-1 = 0 \\
1-a_1 \ge 1-1 = 0 \\
... \\
1-a_k \ge 1-1 = 0 $$
Multiplying all inequalities we get $$\prod_{j=0}^k(1-a_j) \ge 0$$


问:
Another Topology on the Prime Spectrum of a Ring
Let's fix a commutative ring $R$. We'll write $X = \operatorname{Spec}R$ for the set of prime ideals of $R$. Finally, let's write $\beta X$ for the Stone-Čech compactification of $X$.

We can define a function $\lim  : \beta X \to X$ using the following recipe:
$$ f \in \lim \mu \iff \mu(\{p \in \operatorname{Spec} R : f \in p \}) = 1$$
Indeed, $\lim \mu$ is prime by Łoś's theorem, or an easy direct verification.

I believe this function actually endows $X$ the structure of a $\beta$-algebra.  In other words, this endows $X$ with a compact Hausdorff topology. In particular, it must be distinct from the Zariski topology.

My questions: does this topology have a name? Is there somewhere I could read about this?

答:
This is the constructible topology, the topology generated by both the Zariski open sets and the basic Zariski closed sets (i.e., the closed sets defined by principal ideals).  Another way to say it is that it is the topology generated by declaring the distinguished Zariski open subsets $D(f)$ to be clopen, rather than just open.  Or, it is what you get by considering $\operatorname{Spec} R$ as a subset of $\mathcal{P}(X)\cong \{0,1\}^X$ and giving it the product topology, where you put the discrete topology on $\{0,1\}$.  (This description is immediate from your description of what limits look like, since you are just taking pointwise limits in $\{0,1\}^X$.)

It can also be described purely topologically from the Zariski topology: it is the topology generated the sets that are compact and open in the Zariski topology, together with their complements.  This topology is useful in the abstract theory of spectral spaces (spaces that are homeomorphic to the spectrum of a ring with the Zariski topology); see https://stacks.math.columbia.edu/tag/08YF for instance.  It is also of interest because the constructible sets (in the sense of algebraic geometry) are exactly the clopen subsets in the constructible topology (so the constructible topology makes $\operatorname{Spec} R$ the Stone space of the Boolean algebra of constructible sets).


问:
Homology group of $X$
Can I say that the homology group  $ H_n(X)$  it gives us the information of how many holes the topological space $X$ has ? And for an example to say that the $H_n(\bigtriangleup^{n} )=0$ because n simplexis dont have holes and compare it to $H_n(X )=1$ where $X$ here is a torus which has one hole? Meaning if $ H_n(X)\neq H_n(Y) $ then $X$ can't be isomorphic to $Y$ ?

答:
The latter is true: if $X$ is homeomorphic to $Y$, then $H_n(X)$ is isomorphic to $H_n(Y)$ for all $n$, because the construction of the $H_n$ is a functor, so transforms continuous functions to morphisms (of groups) in a way that preserves composition and identities. This is why by the contrapositive we can say that $H_n(X) \neq H_n(Y)$ for some $n$ implies $X \not\simeq Y$, and this is one of the applications of homology (and of other functors like homotopy groups and cohomology etc.).

What $H_n(X)$ "measures" exactly is hard to say, others might have more to say on that. I've always found it somewhat "magical" (you take a large quotient of maps and voila we get $\Bbb Z$...), though it is inspired by simplicial subdivisions in of surfaces in  $\Bbb R^n$ etc.

Title: Homology group of $X$


问:
Probability and Combinatorics (samples)

There are only $10$ polar bears left in the wilderness and they are slowly dying of starvation. We know for certain that 4 of them are pregnant, the rest are either male or not pregnant. We wish to save as many of the pregnant mother bears as possible. However, the high cost of search and rescue only allows us to catch randomly $5$ of these $10$ bears in the wilderness.

Find the probability that at least $1$ pregnant mother bear is caught in this sample of $5$ bears.

My idea is that by finding the combination of arranging the 10 bears into the sample, this can be used to calculate the probability. For example, this would be $_{10}C_{5} = 30240$. However, how would I calculate the number of arrangements of the four bears that will go into the sample? I considered $_{5}C_{4}$, yet this would be illogical as it only includes the number of arrangements of the four bears in the sample, but does not consider the arrangements in the sample not selected. Is my working correct?

答:
The first part is correct. You will need $_{10}C_5$, as this is the total number of combinations of picking 5 bears out of 10. But the second part starts to go awry. We are not interested in $_5C_{\text{anything}}$ because it looks like the 5 is referring to what you already selected. This is generally not the way to go. Instead, you need to find the probability that 1 pregnant mother is in the sample, 2, 3, or all 4. This is $\frac{_4C_1*_6C_4+_4C_2*_6C_3+_4C_3*_6C_2+_4C_4*_6C_1}{_{10}C_5}$, where each term in the numerator represents choosing however many pregnant mother bears times choosing however many male or not pregnant mother bears. Divide this all by the number of ways to choose 5 bears out of 10.

It is easier to think of the complement, one minus the probability of selecting no pregnant mother bears. $1-\frac{_6C_5}{_{10}C_5}$. This is the same as adding up all the terms above.

Title: Probability and Combinatorics (samples)


问:
Probabilistic recursion
I have the following recursion:

$$p_{t+1} = \begin{cases} 1 \text{ with probability } 1-p_t \\ \alpha p_t \text{ with probability } p_t \end{cases}$$

for $0<\alpha<1$. Numerical simulations show that $\mathbb{E}[p_t]$ converges to a steady-state value as $t\rightarrow \infty$, irrespective of the initial condition, only dependent on $\alpha$. I'm not sure how to approach solving it in closed-form. It seems like it should be possible.

I have tried solving it by imposing a self-consistent condition, i.e.

$$\mathbb{E}[p_{\infty}] = \begin{cases} 1 \text{ with probability } 1-\mathbb{E}[p_{\infty}] \\ \alpha \mathbb{E}[p_{\infty}] \text{ with probability } \mathbb{E}[p_{\infty}] \end{cases}$$

I'm not sure if this or the next step is valid, I can neither justify it nor state exactly what's going wrong (but something is wrong, as the answer I get does not match my simuluations). Setting $\mathbb{E}[p_{\infty}] = 1\cdot(1-\mathbb{E}[p_{\infty}]) + \alpha \mathbb{E}[p_{\infty}] \cdot \mathbb{E}[p_{\infty}]$ and solving for $\mathbb{E}[p_{\infty}]$ gives a value which is different from simulations.

Can someone provide a solution or a resource for solving such probabilistic recurrences, or let me know where I'm going wrong and how to proceed?

答:
An observation that might prove fruitful for you, too much for a comment, but likely not the answer you're looking for. Anyway ...

It seems your interest is in the long-time behavior of $p_t$. When your process starts, there is considerable flexibility in how $p_t$ evolves. However, that changes the first time you hit the "set $p_t = 1$" situation. The next step after will necessarily yield $p_{t+1} = \alpha$, and the one following $p_{t+2}$ being set either to $\alpha^2$ or back to 1, etc. So after hitting that $p_t = 1$ reset button, all the values of $p_{t'}$ you generate are discrete and of the form
$1, \alpha, \alpha^2 \alpha^3, \dots$.

This helps you understand the later process, which you might imagine as having states
$S_0, S_1, S_2, S_3, \dots$. State $S_k$ has value $\alpha^k$, and you can write your process in terms of the transition matrix
\begin{equation}
T =
\begin{bmatrix}
           0 &         1 &         0 &         0 &         0 &      0 &  \cdots \\
  1 - \alpha &         0 &    \alpha &         0 &         0 &      0 &  \cdots \\
1 - \alpha^2 &         0 &         0 &  \alpha^2 &         0 &      0 &  \cdots \\
1 - \alpha^3 &         0 &         0 &         0 &  \alpha^3 &      0 &  \cdots \\
1 - \alpha^4 &         0 &         0 &         0 &         0 &   \alpha^4 & \cdots \\
1 - \alpha^5 &         0 &         0 &         0 &         0 &      0 & \cdots \\
\vdots & \vdots& \vdots & \vdots & \vdots & \vdots &
\end{bmatrix}
\end{equation}
The Perron eigenvector $[\pi_0, \pi_1, \pi_2, \dots]$ of $T$ gives you the distribution of the steady state, and the expected value of $p_t$ tending to
$p_\infty = \pi_0 + \pi_1\alpha + \pi_2\alpha^2 + \cdots$.

Title: Probabilistic recursion


问:
Hopf-Rinow Theorem for Riemannian Manifolds with Boundary
I am a little rusty on my Riemannian geometry. In addressing a problem in PDE's I came across a situation that I cannot reconcile with the Hopf-Rinow Theorem. If $\Omega \subset \mathbb{R}^n$ is a bounded, open set with smooth boundary, then $\mathbb{R}^n - \Omega$ is a Riemannian manifold with smooth boundary.  Since $\mathbb{R}^n - \Omega$ is closed in $\mathbb{R}^n$, it follows that $\mathbb{R}^n - \Omega$ is a complete metric space. However, the Hopf-Rinow Theorem seems to indicate that $\mathbb{R}^n - \Omega$ (endowed with the usual Euclidean metric) is not a complete metric space since not all geodesics $\gamma$ are defined for all time. Am I missing something here? Do the hypotheses of the Hopf-Rinow theorem have to be altered to accommodate manifolds with boundary?</p>

答: Hopf-Rinow concerns, indeed, Riemannian manifolds with no boundary.


问:
Division in inequality
I have the problem
$(x-3)*(x+3) \leq x*(3+x)$

and with the following steps

$(x-3)*(x+3) \leq x*(3+x)$

$x^2-9 \leq 3x+x^2$

$-9 \leq 3x$

$-3 \leq x$.

However, my first thought was not to go this way but rather divide by $(3+x)$, which yields

$x-3 \leq x$.

Obviously this does not work. My first thought was that it is not allowed to divide by $(3+x)$ since this contains the solution -3 for x, which would equal to 0 $(-3+3) = 0$.

Question: Why does this not work? Is my thought (it does not work since it would include a division with 0) correct?

Thank you for your help!

答: You can't divide by $x+3$ if $x=-3$. Furthermore, if $x<-3$, then $x+3<0$. So, after dividing by $x+3$, you will get $x-3\geqslant x$ in this case (which is impossible, of course).

Title: Division in inequality


问:
Showing a set is dense in $l^1$
I have been studying functional analysis and I came across this question: Show that $$F := \{(s_n)\in l^1 : (ns_n)\notin l^{\infty}\}$$ is dense in $l_1$.

So here is my approach and I would like some feedback if possible: $(e_n) \in l^1$ (a sequence with unit value in the nth position) is certainly part of this set as $(ne_n)$ is unbounded. We know already that the span of $(e_n)$ is dense in $l^1$. Every other sequence in F can be represented by the span of $(e_n)$. Hence, $(e_n)$ is dense in F itself. Hence, F is dense in $l^1$.

Is this correct? I am sorry that I can be very vague in structuring my arguments (I am new to mathematics) and was wondering if anyone could help me with structuring this better or correct me if I am wrong somewhere.

答:
Given $\epsilon>0$ and a sequence $(s_n)\in\ell^1$.

Let $u_n:=\begin{cases}\frac{m}{2^m}&n=2^m\\0&o/w\end{cases}$; and let $\epsilon_n:=\frac{\epsilon u_n}{4}$.

Then $(\epsilon_n)\in\ell^1$ with $\|(\epsilon_n)\|_1=\frac{\epsilon}{2}$, since $\sum_nu_n=\sum_m\frac{m}{2^m}=2$.

Consider $b_n:=s_n\pm\epsilon_n$, with the sign chosen to be the same as that of $s_n$;

then $(b_n)\in\ell^1$ since $\|(b_n)\|_1\le\|(s_n)\|_1+\|(\epsilon_n)\|_1<\infty$;

and $(nb_n)\notin\ell^\infty$ since $|nb_n|=|ns_n|+n\epsilon_n\ge n\epsilon_n$, and $nu_n=\begin{cases}m&n=2^m\\0&o/w\end{cases}$ is unbounded.

Hence $$\|(s_n)-(b_n)\|_1=\|(\epsilon_n)\|_1=\frac{\epsilon}{2}$$

This means that every ball around $(s_n)\in\ell^1$ has a sequence $(b_n)\in F$ within it.

Title: Showing a set is dense in $l^1$


问:
Is a an integer to the power of an irrational number irrational?
This seemed like a very obvious thing. I was playing around with exponents and logarithms, and I just started wondering, "if $a$$\in$$\mathbb Z$, and $r$$\in$$\mathbb R \setminus \mathbb{Q}$, is $a^r$ irrational?"

As I thought about this, it was almost obivous, but I could not prove it. Is there a proof to the above statement? Namely, is every integer to the power of a real number a real number that is also irrational? (This is not true for real numbers though. Namely, real numbers to the power of real numbers could be rational, i.e., $\sqrt 2^{2log_22}$=$1$ which is an integer let alone a rational). How would also one go about proving such a statement?

EDIT: I have found a mistake. I meant to say if $a^r$ is not an integer, then it is real. (If $a$$\in$$\mathbb Z$, and $r$$\in$$\mathbb R$, and if $a^r$ is not an integer, is then $a^r$ irrational?)

答:
[I am assuming you mean an integer to the power of an irrational number.] What about $3 = 2^{\log_2 3}$? We show that $\log_2 3$ is indeed irrational; if
$2^{\frac{m}{n}}=3$ for some integers $n,m \ge 1$ then raising both sides to the $n$-th power gives $2^m = 3^n$ for some integers $m,n$. Which is impossible as $2^m$ is even and $3^n$ is odd.


答:
I think it is easiest to prove this directly. Fix some $\varepsilon>0$ and some $(x_n)\in\ell^1$. We know there must exist some $K\in\mathbb N$ such that $\sum_{n=K}^\infty|x_n|<\varepsilon/2$. Now note that there exists an infinite set $A\subset \mathbb N$ so that
$$\sum_{n\in A} 1/\sqrt n<\varepsilon/2.$$
Now define some $(s_n)$ by $s_n=x_n$ for all $n\leq K$. For all $n\geq K$ let
$$s_n=\begin{cases}1/\sqrt n &  \text{if}~ n\in A\\ 0 & \text{if}~n\notin A.\end{cases}$$
Clearly $(s_n)\in\ell^1\cap F$, and $\|(s_n)-(x_n)\|_1<\varepsilon$. As $(x_n)$ and $\varepsilon$ are arbitrary it follows that indeed $F$ is dense in $\ell^1$.

Title: Showing a set is dense in $l^1$

Body:
I have been studying functional analysis and I came across this question: Show that $$F := \{(s_n)\in l^1 : (ns_n)\notin l^{\infty}\}$$ is dense in $l_1$.

So here is my approach and I would like some feedback if possible: $(e_n) \in l^1$ (a sequence with unit value in the nth position) is certainly part of this set as $(ne_n)$ is unbounded. We know already that the span of $(e_n)$ is dense in $l^1$. Every other sequence in F can be represented by the span of $(e_n)$. Hence, $(e_n)$ is dense in F itself. Hence, F is dense in $l^1$.

Is this correct? I am sorry that I can be very vague in structuring my arguments (I am new to mathematics) and was wondering if anyone could help me with structuring this better or correct me if I am wrong somewhere.


问:
Find the sum of all 5 digit numbers formed using $\{0,1,2,3,4\}$
Given solution

The sum of all the numbers is
$10 \times 24 \times 10^4 + 10 \times 18 \times 10^3 + 10 \times 18 \times 10^2 + 10 \times 18 \times 10 + 10 \times 18 \times 1. $

My work

Clearly, the number of 5 digit numbers that can be formed is
$$4\times 5\times 5 \times 5 \times 5$$
$$=2500$$

The first 5 digit number is $10000$

So sum of the numbers should be
$$S=\frac {1250}{2} (20000 + 2499)$$
$$=14061875$$

But given answer is $2599980$

Where am I going wrong?

答:
Trying to collate comments:

First, it is important to understand what the question is asking.  The problem wants you to find the sum of all numbers in the set $\{x~:~x~\text{is a five digit number with nonrepeating digits whose digits all come from the set }\{0,1,2,3,4\}\}$.

That is to say, it is asking for you to evaluate $$10234 + 10243 + 10324 + 10342 + 10423 + 10432 + 12034 + 12043 + 12304 + 12340 + 13024 + \dots + 43012 + 43021 + 43201 + 43210$$

This is a sum of several five-digit numbers.  You ask "what is wrong with my solution."  What is wrong is that you attempted to use a formula regarding a sum of terms in an arithmetic progression however the terms in this sum are not in an arithmetic progression.

"I don't know whether the question is asking for sum of digits or sum of numbers formed by the digits"

The question is asking for sum of numbers formed by the digits, however through some clever reorganization of our thought process we can accomplish this by looking at digits individually.

Next, some groundwork reminding you how to add...

Recall that if we want to add two numbers, for example $12340 + 31420$ we can do this by adding the respective positions together...  That is to say, $\color{red}{1}\color{blue}{2}\color{green}{3}\color{purple}{4}0 + \color{red}{3}\color{blue}{1}\color{green}{4}\color{purple}{2}0$ we add the red digits together $\color{red}{1}+\color{red}{3}=\color{red}{4}$, we add the blue digits together, $\color{blue}{2}+\color{blue}{1}=\color{blue}{3}$ etc...  Our result is $\color{red}{4}\color{blue}{3}\color{green}{7}\color{purple}{6}0$

Formally, what happened here was that $12340$ can be interpreted as $1\cdot 10^4 + 2\cdot 10^3 + 3\cdot 10^2 + 4\cdot 10^1 + 0\cdot 10^0$.  Similarly $31420$ can be broken up.  We combined like terms, having added all of the terms who were multiples of $10^4$ together, adding all those who were multiples of $10^3$ together, etc...

The same principle works when adding many numbers together... however we might normally want to keep track of "carry-over digits" like how $\color{red}{1}\color{blue}{8}+\color{red}{1}\color{blue}{8} = 36$ and not $\color{red}{2}\color{blue}{16}$.  So, we can't just concatenate the results of the separate additions... but that doesn't mean that the way of thinking of this is flawed... here we do indeed have $18+18 = (1\cdot 10 + 8) + (1\cdot 10 + 8) = (1 + 1)\cdot 10 + (8 +8)$, that is still perfectly correct.

"Your answer seems to calculate sum of digits (from my perspective)"  While yes, we did add digits individually, we did so keeping track of which digit place they originated from by keeping the $10^4$ and $10^3$ and such terms with them as well...

Now... we take this idea but apply it on a much grander scale.

In our overall grand sum, there are as mentioned several five digit numbers.  How many?  $4\cdot 4!$, seen by routine application of counting principles.  (Pick the first digit, noting it may not be zero.  Then pick the second digit noting it can't be what was picked for the first, then pick the third digit, etc... for $4\cdot 4\cdot 3\cdot 2\cdot 1$ total outcomes)

Now... we organize our thoughts as alluded to by thinking of $10234$ as $1\cdot 10^4 + 0\cdot 10^3 + 2\cdot 10^2 + 3\cdot 10^1 + 4$, we think of $10243$ as $1\cdot 10^4 + 0\cdot 10^3 + 2\cdot 10^2 + 4\cdot 10^1 + 3$ etc...

Now... we may freely rearrange terms in the sum, moving all of the $10^4$ terms together, then moving all of the $10^3$ terms together, all of the $10^2$ terms together, and so on and we evaluate each of these separately.

Factoring out each of the $10^4$'s $10^3$'s etc... after rearranging and sorting within each parenthetical phrase we now have a sum which looks like:

$$(1+1+1+\dots+2+2+\dots + 3+3+\dots +4+4+\dots)\cdot 10^4 + (0+0+\dots+1+1+\dots+2+2+\dots+3+3+\dots+4+4+\dots)\cdot 10^3 + \dots + (0+0+\dots+1+1+\dots+2+2+\dots+3+3+\dots+4+4+\dots)$$

The challenge here is that it is difficult at a glance to evaluate the sums of these parentheses without more information... but we have more information.  Specifically, we know there are $4\cdot 4!$ terms in each of these sums.  Next, we know that $1$'s, $2$'s, $3$'s, and $4$'s each occur the same amount of times as each other within each parentheses... so in that first parentheses a quarter of the numbers being added are $1$, a quarter are $2$, etc...

So, we have the first parenthetical term is  $(\underbrace{1+1+\dots}_{4!~\text{terms}}+\underbrace{2+2+\dots}_{4!~\text{terms}}+\dots)\cdot 10^4$ which simplifies to $4!\cdot (1+2+3+4)\cdot 10^4$

Now, the rest of the parenthetical phrases are all similar to each other.  Like before, there are some number of $1$'s, $2$'s, $3$'s and $4$'s and due to symmetry we know there will be just as many $1$'s as there are $4$'s etc...

However the $0$'s are there in the mix as well.  We count how many times $0$ appears as the $1000$'s digit in the $5$-digit numbers we are counting.  By counting principles, we pick the first digit, the second digit we are forcing to be zero, we then pick the third digit, fourth digit, and fifth digit for a total of $4!$ different numbers with $0$ in the second place.

As such, it follows from symmetry that $3\cdot 3!$ different numbers have a $1$ in the second place, similarly for $2$'s, $3$'s etc...

So, the second parenthetical phrase in the grand sum is:

$$(\underbrace{0+0+0+\dots}_{4!~\text{terms}}+\underbrace{1+1+\dots}_{3\cdot 3!~\text{terms}}+\underbrace{2+2+\dots}_{3\cdot 3!~\text{terms}}+\dots)\cdot 10^3$$

which simplifies as

$$(4!\cdot 0 + 3\cdot 3! \cdot 1 + 3\cdot 3!\cdot 2 + \dots) \cdot 10^3 = 3\cdot 3!\cdot (1+2+3+4)\cdot 10^3$$

The same analysis holds for the hundreds digits, tens digits, and ones digits.

We have as a final grand total then:

$$10234 + 10243 + 10324 + 10342 + 10423 + 10432 + 12034 + 12043 + 12304 + 12340 + 13024 + \dots + 43012 + 43021 + 43201 + 43210$$

$$=4!\cdot (1+2+3+4)\cdot 10^4 + 3\cdot 3!\cdot (1+2+3+4)\cdot 10^3 + 3\cdot 3!\cdot (1+2+3+4)\cdot 10^2 + 3\cdot 3!\cdot (1+2+3+4)\cdot 10^1 + 3\cdot 3!\cdot (1+2+3+4)$$

$$=24\cdot 10\cdot 10^4 + 18\cdot 10\cdot 1111$$

$$=2599980$$


问:
How to determine the center of symmetry of a function?
I'm learning about functions and I came to a question that I'm struggling at:

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

The question is:

Show that the graph of $f$, has a center of symmetry to be determined in an orthonormal system.

What is the best approach to solve this question?

答:
Let's discover where the center of symmetry could be

Knowing that the map has a vertical asymptote for $x = -1$, the abscissa of the center of symmetry is $-1$. Let's find the ordinate. We have

$$\begin{aligned}f(x) &= \frac{x^2 -2}{x+1} = \frac{\left(x + 1\right)^2-2}{x + 1}\\
&=\frac{\left(x + 1\right)^2-2x-3}{x + 1}\\
&=x-1-\frac{1}{x+1}
\end{aligned}$$

Therefore $y = x-1$ is another asymptote of the map. The center of symmetric if it exists lie at the intersection of both asymptotes and therefore has for coordinates $C= (-1,-2)$.

Let's prove that $C$ is indeed a center of symmetry

For this, we need to notice that

$$-4-f(-2-x) = f(x).$$

Which I leave to the reader.

In general, for $(a,b)$ to be a center of symmetry you need to prove that $2b−f(2a−x)=f(x)$.


问:
Isolated Points and Topological Transitivity
I'm looking for a hint on this question:

Suppose $X$ is a metric space with at least one isolated point and $T:X\to X$ is a topologically transitive dynamical system. Show that $X$ is necessarily finite, and $X=O(x)$ for any point $x\in X$.

Any help is welcome. Thanks in advance.

答:
Michael provides a counterexample to the statement:

Suppose $X$ is a metric space with at least one isolated point and $T\colon X\to X$ is a topologically transitive dynamical system where $T$ is not necessarily continuous. Show that $X$ is necessarily finite, and $X=O(x)$ for any point $x\in X$.

But judging by the context of the exercise I assume that $T$ is supposed to be continuous, since this is the standard assumption for a dynamical system when a topology is involved. So I will provide a proof of the statement if $T$ continuous. Moreover, I will use the following definition of transitivity

A topological dynamical system $T: X\to X$ is called topologically transitive if for every pair $U, V\subseteq X$ of open sets there exists a $k\in \mathbb{Z}_{\geq 0}$ such that $T^k(U) \cap V \neq \emptyset$.

As Micheal mentioned, it is enough to show that an isolated point is a periodic point. Let us do this now.
Let $x$ be an isolated point of $X$ and set
$$
\mathcal{X} = \{y\in X\mid \exists k\geq 0\colon T^k(y) = x\}.
$$
Note that by transitivity $\mathcal{X}$ is dense in $X$. Let $\mathcal{C}$ denote the set of all connected components of $X$ with non-empty interior and $\mathcal{D}$ the set of all connected components of $X$ with empty interior. Since each $C\in \mathcal{C}$ is open we can use density of $\mathcal{X}$ to find a $y_C\in C\cap \mathcal{X}$ so that there exists a $k\geq 0$ such that $T^k(y_C) = x$. Now $T$ is continuous, so $T^k$ is continuous and since continuous functions map connected sets on connected sets we must have that $T^k(C) = \{x\}$. Note that we can apply the same argument for any $D\in \mathcal{D}$ with $D\cap\mathcal{X}\neq \emptyset$. So we can partition $X = A\cup B$ where
$$
A = \{y\in X\mid \exists C\in\mathcal{C}\colon y\in C\text{ or }\exists D\in\mathcal{D}\colon D\cap\mathcal{X}\neq 0 \text{ and }y\in D\}
$$
and $B = A^c$. Observe that both $A$ and $B$ are $T$-invariant and $B$ has empty interior (by transitivity). Thus we have two dynamical systems $T|_A$ and $T|_B$ which we can study separately. By the above connected-component-argument we conclude that $x$ is in the orbit of every element in $A$. In particular, we can find a $k\geq 0$ for $T(x)$ such that $T^{k+1}(x) = x$, so $x$ is a periodic point of $T|_A$. Since $B$ has non-empty interior, it cannot be open, in particular for every $\epsilon > 0$ there must be an element in $A$ in an $\epsilon$-neighborhood of $B$, otherwise this $\epsilon$-neighborhood of $B$ is $B$ itself. So finiteness of $A$ implies that $B$ is empty. From this it follows that $X$ is indeed finite and $O(x) = X$ for all $x\in X$.


问:
Find the number of positive integer pairs $(x,y)$ such that $\frac 1x + \frac 1y=\frac{1}{2007}$ such that $x<y$
$$2007(x+y) =xy$$
$$\implies (x-2007)(y-2007)=2007^2$$

Now the values so $x$ and $y$ should depend on the factors of $2007^2$, which is something I am not very sure about

How should I proceed?

答:
First off, I assume this is a middle/high school contest problem from the year 2007.  If so, then every student going into the contest should have memorized the prime factorization of 2007 (as long as their sponsor knew this unsaid rule).

That said, if you know some basic divisibility rules, you can simplify this on the spot.  For instance, the sum of the digits of 2007 is 9, so 2007 is divisible by 9.  Doing so:
$$\frac{2007}{9} = 223$$
which no longer has a digit sum divisible by 3.  We can check primes up through $\lfloor \sqrt{223}\rfloor = 14$ for the remaining divisibility to prove to ourselves that $223$ is a prime."

Therefore, we find $2007 = 3^2 \cdot 223$ is the prime factorization, $2007^2 = 3^4 \cdot 223^2$ is its prime factorization.  Therefore $2007^2$ has $(4+1)(2+1) = 15$ positive integral divisors.

One divisor is $2007$, resulting in $2007^2 = 2007 \cdot 2007$ which in turn would result in $x=y$ so we toss that, leaving us with $14$.

By symmetry, half of these will result in $x < y$ and half will result in $y<x$, so the answer is $\frac{14}{2} = 7$.


答:
Actually I think you can solve this problem by dividing by $x+3,$ provided that you handle the sign of this term correctly and also take care of the special case
$x + 3 = 0.$

Consider three cases: $x + 3 = 0,$ $x + 3 > 0$, $x + 3 < 0.$

Case $x + 3 = 0$:

In this case $(x-3)(x+3) = 0 = x(3+x),$ so it is certainly the case that
$(x-3)(x+3) \leq x(3+x)$. In this case $x = -3,$ so $x = -3$ is one solution.

Case $x + 3 > 0$:

In this case, multiplying or dividing by $x + 3$ on each side of an inequality leaves the inequality intact, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \leq x$.
But $x-3 \leq x$ for all $x$, so every $x$ such that $x + 3 > 0$ is a solution.
In other words, every $x > -3$ is a solution.

Case $x + 3 < 0$:

In this case, multiplying or dividing by $x + 3$ on each side of an inequality reverses the direction of the inequality, that is, $(x-3)(x+3) \leq x(3+x)$ if and only if $x-3 \geq x$.
But there is no $x$ such that $x-3 \geq x$ for all $x$, so this case does not find any solutions.

Conclusion: The only solutions are the ones found in the first two cases, namely, $x$ must satisfy either $x = -3$ or $x > -3.$ A more succinct way of expressing this is that $x \geq -3.$

Title: Division in inequality

Body:
I have the problem
$(x-3)*(x+3) \leq x*(3+x)$

and with the following steps

$(x-3)*(x+3) \leq x*(3+x)$

$x^2-9 \leq 3x+x^2$

$-9 \leq 3x$

$-3 \leq x$.

However, my first thought was not to go this way but rather divide by $(3+x)$, which yields

$x-3 \leq x$.

Obviously this does not work. My first thought was that it is not allowed to divide by $(3+x)$ since this contains the solution -3 for x, which would equal to 0 $(-3+3) = 0$.

Question: Why does this not work? Is my thought (it does not work since it would include a division with 0) correct?

Thank you for your help!


问:
Evaluation of "concentration of measure function" at zero.
Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). We define the concentration function of the triple $X,d$ and $\mu$ as follows:

$\alpha_{(X,d,\mu)}(r)=\sup\{1-\mu(A_r):A\subset X\text{ is measurable and }\mu(A)\geq 1/2\}$, $r\geq0$,

where we define $A_r=\{x\in X:d(x,A)<r\}$ (EDIT: I adopt the convention $A_0=A$).

I am interested in the value of the following (sparing you the elementary details that leads to the equality below):

$\alpha_{(X,d,\mu)}(0)=\sup\{\mu(B):B\subset X\text{ is measurable and }\mu(B)\leq 1/2\}$

If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$,
then it clearly is the case that $\alpha_{(X,d,\mu)}(0)=1/2$.
If there is no set with measure $1/2$, then the situation is not so clear.
I can easily find examples where there is no set of measure $1/2$ and $\alpha_{(X,d,\mu)}(0)$ is not $1/2$ (Dirac probability measure comes to mind), but I am unable to either prove (by finding an example ideally) or disprove that there exists a probability measure on some metric space such that no set has a measure of $1/2$ but $\alpha_{(X,d,\mu)}(0)=1/2$.

答: It turns out that the image of any finite measure is closed (see here), which implies that $1/2$ (or any other constant $\alpha\in[0,1]$ for that matter) cannot be a limit point of the image of $\mu$ without being itself in the image of $\mu$.</p>


问:
If you roll a fair 6-sided die and then flip a fair coin that number of times, what is the probability that you will get at least two heads?
My idea is to use disjoint events and calculating the probability of getting at least two heads for each number rolled. For example, if I roll a 3, I would calculate the probability with the expression $(\frac{1}{6}) (\frac{1}{2})^3 \binom{3}{2} + (\frac{1}{6}) (\frac{1}{2})^3\binom{3}{3})= \frac{1}{12}$ and then add up the probabilities of getting at least two for each rolls, since the events are disjoint, summing to $\frac{67}{128}$. Is this a valid solution? Is there a better approach to solving this problem?

答:
It is valid. You find that P(at least 2 heads|die=1) = 0, P(at least 2 heads|die=2)=1/4, P(at least 2 heads|die=3)=1/2, P(at least 2 heads|die=4)=11/16, P(at least 2 heads|die=5)=13/16, and P(at least 2 heads|die=6)=57/64. Then 1/6*(0+1/4+1/2+11/16+13/16+57/64)=67/128.

There is a way you can numerically approximate the answer and that is to use simulation. You can write code to run 10000 rolls of the die to calculate the probability that you get at least 2 heads. Then do this 100 times, and on each iteration gets a probability of getting at least 2 heads. The mean of these 100 probabilities is a cross-validated probability of 0.523544. We can check that $\frac{67}{128}\approx0.5234375$, which is very close.


问:
Continuity and inverse continuity in the homemorphism of $\mathbb{C}\mathbb{P}^1$ and $S^2$
In the proof of the homeomorphism between $\mathbb{C}\mathbb{P}^1$ and $S^2$. I am following the following approach:

Let $X=\{ [z_0:z_1] \in \mathbb{C}\mathbb{P}^1  \; | \; z_1 \neq 0 \} $. Show that $\phi : X \rightarrow \mathbb{C} $ s.t. $[z_0:z_1] \mapsto z_0/z_1 $ is a homemorphism. Once I prove the homeormorphism, then using the stereographic projection and one-point compactification, I can show that homeomorphism between $\mathbb{C}\mathbb{P}^1$ and $S^2$.

My attempt: $\phi$ is surjective since, any $z \in \mathbb{C}$ can be mapped by $[z:1] \in \mathbb{C}\mathbb{P}^1 $.

$\phi$ is injective since, $z_0/z_1 =z_2/z_3 \Rightarrow [z_0:z_1] = [z_2:z_3] $.

Doubt: How do I show that $\phi$ is continuous map and also an open map or continuous and inverse is also continuous to show the homeomorphism?

答: Note that $\phi$ is well-defined : $(z_0,z_1) \sim (z_2,z_3)$ under the equivalence relation implies $\phi([z_0,z_1]) = \phi([z_2,z_3]) = \frac{z_0}{z_1}$, so the image of $\phi$ does not depend on the representative pair. Also a map from a quotient (here $q:\Bbb C \times \Bbb C^\ast \to \Bbb{CP}^1$, sending $(z_0,z_1)$ to $[z_0,z_1]$, induces the quotient topology) is continuous iff the composition with $q$ is continuous (standard universal property for quotient maps) and here $\phi \circ q$ equals the map $(u,v) \to \frac{u}{v}$ which is continuous. ($\Bbb C$ is a topological group..). That the inverse is continuous is also clear, as the map $z \to (z,1) \in \Bbb C \times \Bbb C^\ast$ is also continuous..)


问:
Which kinds of rational functions of one variable have an inverse relation that contains a branch that is a rational function?
Let's consider the rational functions whose numerator and denominator of the function term are coprime.

Which kinds of rational functions of one variable have an inverse relation that contains a branch that is a rational function?

Which kinds of polynomial functions of one variable have an inverse relation that contains a branch that is a rational function?

I assume the degree of the numerator and the degree of the denominator of the function term has to be less than or equal to $1$. Some calculations with algebraic equations with undetermined parameters as coefficients seem to show that. But I'm not sure.

答:
If $R$ is a rational function and $S$ a branch of $R^{-1}$ in an open set $U \subset \Bbb C$ then $S(R(z)) = z$ in $U$. If $S$ is also a rational function then it follows that $S(R(z)) = z$ globally (as meromorphic functions).

It follows that $S$ is injective and therefore has degree one. Then $R$ has degree one as well.

So the only rational functions with a (local) rational branch of the inverse are rational functions of degree one (which are the Möbius transformations).


问:
How to obtain $\frac{\pi^2}{6}$ from $\displaystyle\int_0^1 \frac{-\ln(1-x)}{x} \, dx $
\begin{align}
& \int_0^1 \frac{-\ln(1-x)}{x} \, dx = \int_0^1 1+\frac{x}{2}+\frac{x^2}{3} + \frac{x^3}{4}+\cdots \, dx \\[8pt]
= {} &\left[1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4} + \cdots \right]_0^1 = 1+\frac{1}{2}+\frac{1^2}{3} + \frac{1^3}{4}+\cdots-1
\end{align}

I think I did something wrong, I thought I could just plug $x=1$ into the series, but it is wrong, it is a harmonic series truncated at 1. The correct solution should be the Basel problem $1+\dfrac{1}{2^2}+\dfrac{1}{3^2} + \dfrac{1}{4^2} + \cdots = \dfrac{\pi^2}{6}$

Basically, result just claim that the above integral is equal to $\displaystyle\int_0^1 -\log(1-x) \, dx$, which is impossible.

Also, is there a generalization of the integral for $\zeta(4), \zeta(6), \zeta(8)$. I guess they are related to the dilogarithm function or even polylogarithm function, am I correct?

答: $-\ln(1-x)=\sum_{n=1}^\infty \frac{x^n}{n}$ for $x \in [0,1)$, so after dividing by $x$ you get $\sum_{n=1}^\infty \frac{x^{n-1}}{n}$ which is the series you wrote inside the integral. But then you didn't do the term by term integration, which is how the squares are introduced in the denominator to make the connection to $\zeta(2)$.


问:
Convergence of the alternating $p$-series $\sum_{n\ge1} \frac{(-1)^{n-1}}{n^p}$
Assuming $p>0$, we consider the $p$-series $$
1-\frac{1}{2^p}+\frac{1}{3^p}-\frac{1}{4^p}+\cdots
$$
For what $p$-values is the series convergent? For what $p$-values is the series absolutely convergent? Which $p$-values will make it is conditionally convergent?</p>

答: we will use the criteria for convergence of alternating series which involves following two conditions 1) (an) must be non-increasing sequence of positive terms 2) limit of an should be 0 as n tends to infinity(https://i.stack.imgur.com/on0xd.jpg)</p>


答:
Dividing both sides of the inequality
$$
(x-3)(x+3) \leq x(x+3)
$$
by $x+3$ is problematic for two reasons. The first is that it can lead to division by zero, as you have already noted. The second is that when you multiply or divide an inequality by a negative number, you have to switch the direction in which the inequality is written. Hence, if $x+3$ is smaller than $0$, then you have to flip the $\leq$ to $\geq$. If you insist upon dividing by $x+3$, this approach can be made to work if you split the problem into cases.

Case 1: if $x+3>0$, then dividing both sides by $x+3$ does not switch the direction of the inequality. Hence,
$$
(x-3) \leq x \\
-3 \leq 0 \, .
$$
The final statement is true, and so for $x+3>0$, the inequality must hold.

Case 2: if $x+3=0$, then we can't divide both sides by $x+3$. But notice that if $x+3=0$, then
$$
(x-3)(0) \leq x(0)
$$
which is the same as $0 \leq 0$. Hence, the inequality is true for $x+3=0$.

Case 3: if $x+3<0$, then we have to switch the direction of the inequality, and we get that
$$
x-3 \geq x \\
-3 \geq 0
$$
The above inequality has no solutions, and so the inequality is not true for $x+3<0$. Putting everything together, we find that $x+3$ must be greater than or equal to $0$ for the inequality to be true. The solution is thus $x\geq -3$.

In general, if $a \geq b$, then $-a \leq -b$. Hence, if we multiply both sides by $-1$, we have to switch the direction of the inequality. This is because
\begin{align}
& a \geq b \\
\implies & 0 \geq b - a \\
\implies & -b \geq -a \\
\implies & -a \leq -b \, .
\end{align}

Title: Division in inequality


问:
Proving that $x^2 + 4$ is not divisible by $3$
I need to show the following:

For any integer $x$, $x^2 + 4$ is not divisible by $3$.

I was trying proof by contraposition, but I do not believe that is the most efficient way to go about this.  Can anybody point me in the right direction, please?

Body: Let $x=3k+a$, where $a=-1,0,1$. Then $$x^2+4=(3k+a)^2+4=3(2ak+3k^2)+(a^2+4)$$ So $3\mid x^2+4\iff 3\mid a^2+4$. Checking the values $a=-1,0,1$ we see that $3\not\mid a^2+4$.


问:
Real Mathematical Analysis Prelim problem 4.60
I am stuck on the following problem from Pugh's book:

Does there exist a continuous function $$f: [0,1] \rightarrow \mathbb{R}$$ such that $$\int_0^1 xf(x)\,dx  = 1$$ and $$\int_0^1 x^n f(x)\,dx  = 0$$ for $$n = 0,2,3,\ldots$$

The progress I made so far is that, for any polynomial $p$, we must have $p'(0) = \int_0^1 p(x)f(x)\,dx$. Also from Cauchy-Schwarz we can deduce that $\int_0^1 f^2(x) \, dx \geq 3$. We also know that the polynomials are dense in the space of continuous functions, but I don't know how to use this fact to solve the problem.

Thank you

答: Presume such a function $f$ existed. Then the function $g(x)=x^2f(x)$ would be another continuous function on $[0,1]$ such that $\int_0^1 x^n g(x)dx=0$ for all $n=0,1,2\ldots$. Now, that would imply that $\int_0^1p(x)g(x)dx=0$ for every real polynomial $p$, and so, due to density of all polynomials, you would have $g(x)=0$. Thus, $f(x)=\frac{g(x)}{x^2}=0$ for $x\ne 0$. Continuity gives us $f(x)=0$ for all $x\in[0,1]$ - which is impossible, because $\int_0^1xf(x)dx=1\ne 0$.


问:
Big-O: Prove $2^n$ is $O(n!)$
I am a little stuck trying to prove that $2^n$ is $O(n!)$. Obviously, I can tell in a few ways that this is the case. For one, based on Big-$O$ hierarchy, the exponential is beneath the factorial in terms of complexity. I also substituted $C = 2$ and tried several values of $n$, all following the Big-$O$ definition of $f(n) \le cg(n)$.

However, my professor says neither of these is a sufficient justification and I should be able to use the definition alone.

How can I do this? I know in the past it has been advantageous for me to compare the functions by converting one into a form that resembles the other, but I just can't figure out how to relate the factorial and $2^n$ functions.

Can anyone point me in the right direction?

答:
Can you show $$\lim_{n\to\infty}\frac{2^n}{n!}=0\text{ ?}$$

Hint $$\frac{2^n}{n!}=\frac{2}{1}\frac{2}{2}\frac{2}{3}\cdots\frac{2}{n-1}\frac{2}{n}$$

so as soon as $n\geq 4$, this is

$$\frac{2^n}{n!}\leq \frac{2}{1}\frac{2}{2}\cdot1\cdots 1\cdot \frac{2}{n}=\frac{4}{n}$$

Then choose, say, $\epsilon =1$..

ADD Let $a_n=2^n,b_n=n!$. Then showing $$a_n\leq C\cdot b_n$$ eventually is equivalent to showing $$\log a_n\leq \log C+\log b_n$$ eventually, that is $$n\log 2\leq \log C+\log n!$$

In the left hand side, we sum $\log 2$ $n$ times, while on the right hand side we sum $\log 2,\log 3,\dots$ and so on. This is essentially the idea I gave before: in one, the product is fixed: one always multiplies by $2$; while in the other the factors grow without bound.

Title: Big-O: Prove $2^n$ is $O(n!)$


问:
Big-O notation for algorithm
I'm working in an algorithm and I have achieved to know that its cost is $O \Big(\prod_{i = 1}^{N} i \cdot \log i \Big)$. I don't know how to simplify my cost even more.

Can anyone help me? Thank you so much

答:
Hint:
$$
O\left(\prod_{i = 2}^{i \leq N}i\log i\right) = O\left(N!\hspace{1mm}\log(2)\log(3)\dots\log(N)\right)
$$
The complexity shows that this algorithm is not by any means efficient.

Title: Big-O notation for algorithm


问:
Min Max diameter of subgraphs from a given graph.also given the number of subgraphs.
Given a graph and the number of groups we want to divide the graph into, Find the best way to divide the graph, such the max diameter of all the groups is minimum

The graph is undirected, the number of nodes in each of groups(or sub-graphs) may not necessarily be the same.

答:
This is NP-complete, by reduction of the clique cover problem : graphs of diameter 1 are cliques, and deciding whether there's a partition of a given graph into $k$ cliques is NP-complete.

If instead of diameter you were looking for radius, it would be NP-complete by reduction of the dominating set problem (where the radius = 1).


问:
Tips and tricks for proofs involving algebraic / equation manipulation
I recently completed a solid chunk of my first truly formal math book on Set Theory and decided to move on to Michael Spivak's Calculus (as I heard it was a good place to begin one's journey into Analysis). For the purpose of self-studying, I purchased the solution manual, as well.

After completing the first two chapters and going through all practice problems, I am now painfully aware that I am not particularly great at proofs involving clever equation manipulations. As I progressed through my aforementioned set theory book, I felt like the number of "proof flavors" (different strategies one could invoke to tackle the claim) were quite manageable. Figuring out which path to pursue  seemed to be categorically recognizable.

However, when it comes to proofs involving equation manipulation, I literally feel like I am staring into a sea of hundreds of possible permutations (in the English sense) that one might use to arrive at the desired conclusion. From what I have seen so far, the strategies largely seem to involve the invocation of Field Axioms (e.g. creative usages of $0$ and $1$ through the existence of additive inverses and  multiplicative inverses, respectively) and factoring (that often times feels quite contrived).

I was wondering if there were any general tips and tricks as to how one might improve his/her ability in recognizing the type of equation manipulation that a given proof is suggesting (yelling) to carry out. I am assuming the seasoned mathsters have some sort of pattern recognition that goes off in their heads that allows for efficient determination of starting points.

I apologize if this question is a little too broad. I recognize that every proof may necessarily be "special", so in the absence of a specific example, perhaps this question is unanswerable. But here's hoping. Book recommendations are also greatly appreciated! Cheers~

答:
I'm about halfway through Spivak's Calculus myself.

The specific problem you mention (Prove $\sqrt 6-\sqrt 2 - \sqrt3$ is irrational) is a tricky one. I certainly wasn't able to do it myself. I ended up having to look at the published solution. Coming back to it today, several months later though, I managed it.

(Note: the next part of that problem ("prove that $\sqrt2$ + $\sqrt[3]{2}$ is irrational") is marred by a hint ("start by working out the first 6 powers of this number") that's meant to push you towards the published solution, which relies on linear algebra and is certainly outside the scope the book. Usually Spivak is loath to rely on things without justification. This is an unusual lapse, made more baffling by the existence of a much more appropriate solution.  Compounding things, I think there's also a typo in the table in his solution. There are mistakes in the text, ranging from typos to more serious problems. When you think there might be a mistake, MSE can be very helpful.)

More generally, many of the problems in the book are challenging. Often, I have to look up the solution, at least for a hint.

The difficulty is also somewhat front-loaded. The first 4 Chapters which in 3rd edition are "Basic Properties of Numbers", "Numbers of Various Sorts", "Functions", and "Graphs" (with 3 appendices), were in some ways more challenging for me than anything that's come since.

Once you hit the first chapter on "Limits", things settle into a groove.

Not that it's "easy" from that point on; the problems remain challenging throughout. Just that, I think you'll find from limits on, it gets really fun.

Don't be discouraged if you aren't able to "ace" the problems. Many of the early problems are introducing you to concepts and techniques you will use later on. The full import of some things doesn't become clear until many chapters later.

You can learn most of Spivak's clever tricks in Spivak, as opposed to trying to learn them elsewhere ahead of time just in case they are in Spivak.

Certainly look outside of Spivak if you need to brush up on something specific, but I don't know if it's worth it trying to just generally get better at clever arithematic tricks before moving on with the text.

Remain diligent about doing the problems. If you make sure you understand them, even the ones you need help with, you will get a lot out of it.

Go back periodically and try re-working problems that initially gave you a hard time.

The book is a tough, glorious, frustrating, edifying beast. Stick with it at least until the end of the "Limits" chapter. I think at that point you'll be happy you did.


答:
You seem to be assuming that as you have $2500$ five digit numbers and the first one is $10000$ that the numbers will be all the numbers from $10000$ to $12499$.  But you aren't doing all the numbers.  You are only doing the numbers with digits $0,1,2,3,4$ and you skipping the numbers with any digits between $5$-$9$.

This is certainly not and arithmetic progression.

But if we designate the $k$th number as $a_k$ and the five digits as $b_{k,0},b_{k,1},...etc.$ then $a_k = 10000b_{k,4} + 1000b_{k,3} + 100b_{k,2} + 10b_{k,1} + b_{k,0}$.

And $\sum a_k = 10000\sum b_{k,4} + 1000\sum b_{k,3} +... etc.$

So it a matter of just calculating the sums of the digits.

For any number with a first digit of $w$ (where $w$ can by $1,2,3,4$) there we $5^4$ numbers begining with $w$.  So  $10000\sum b_{k,4} = 100000(5^4)\cdot 1 + 10000(5^4)\cdot 2 + ... etc= 1000(5^4)(1+2+3+4)$.$.

And for any number with $j$th digit, $j$ not the first digit, there will be $4\cdot 5^3$ numbers with that same digit.

SO hopefully it will be clear that

$\sum a_k = $

$10000\sum b_{k,4} + 1000\sum b_{k,3} +100\sum b_{k,2} + 10\sum b_{k,1} + \sum b_{k,0} =$

$10000\cdot 5^4 (1+2+3+4) + 1000\cdot 4\times 5^3 (0+1+2+3+4) + 100\cdot 4\times 5^3 (0+1+2+3+4)+10\cdot 4\times 5^3 (0+1+2+3+4) + 4\times 5^3 (0+1+2+3+4)=$

$5^4\cdot 100000\cdot 10 + 4\cdot 5^3\cdot 10(1000 + 100 + 10 + 1)=$

$62500000 + 5000\times 1111 = $

$62500000 + 5555000= 68055000$

Which.... doesn't match the answer at all.

.....

So I'm thinking the are asking for numbers with no repetitions.  They use all $0,1,2,3,4$ and each exactly once.

In that case:

There are $4!$ numbers that begin with $1,2,3$ or $4$.  And so if you add only what those numbers contribute to the sum, that should be:  $10^4 \times 4! \times (1+2+3+4)$.

And for ever other digit... well, if that digit is $0$ there are $4!$ options for the other $4$ digits.  But if that digit isn't $0$ there are $3$ options for the first digit which can not be $0$ and $3!$ for the remaining digits.  So if we add only what the $k$th digit contributes to the sume we get $10^k (4!\times 0 + 3\cdot 3!(1+2+3+4)) = 10^k(3\cdot 3!(1+2+3+4))$.

So the sum is $(10^4 \cdot 4! + 3\cdot 3!\cdot 1111)(1+2+3+4)=$

$(10^4\cdot 24 + 18\times 1111)\cdot 10 = $

$2400000 + 199980= 2599980$

Title: Find the sum of all 5 digit numbers formed using $\{0,1,2,3,4\}$

Body:
Given solution

The sum of all the numbers is
$10 \times 24 \times 10^4 + 10 \times 18 \times 10^3 + 10 \times 18 \times 10^2 + 10 \times 18 \times 10 + 10 \times 18 \times 1. $

My work

Clearly, the number of 5 digit numbers that can be formed is
$$4\times 5\times 5 \times 5 \times 5$$
$$=2500$$

The first 5 digit number is $10000$

So sum of the numbers should be
$$S=\frac {1250}{2} (20000 + 2499)$$
$$=14061875$$

But given answer is $2599980$

Where am I going wrong?


答:
Let $q_n\in\mathbb{R}[X]$ such that $\|f-q_n\|_{\infty}\rightarrow 0$ and let $p_n=q_n+\frac{(1-x)^{m_n}}{m_n}q_n'(0)$ where $m_n=\lfloor|q_n'(0)|+1\rfloor^2+n+1$, then $p_n'(0)=0$ for all $n$ so that $\int_0^1 p_n(x)f(x)dx=0$ for all $n$. Moreover, $\|f-p_n\|_{\infty}\leqslant\|f-q_n\|_{\infty}+\frac{|q_n'(0)|}{m_n}\rightarrow 0$ thus, $$ \int_0^1 f(x)p_n(x)dx\rightarrow\int_0^1 f(x)^2dx $$
and therefore $\int_0^1 f(x)^2dx=0$ which means that $f=0$, $f$ being continuous. This contradicts the fact that $\int_0^1 xf(x)dx=1$.

Title: Real Mathematical Analysis Prelim problem 4.60

Body:
I am stuck on the following problem from Pugh's book:

Does there exist a continuous function $$f: [0,1] \rightarrow \mathbb{R}$$ such that $$\int_0^1 xf(x)\,dx  = 1$$ and $$\int_0^1 x^n f(x)\,dx  = 0$$ for $$n = 0,2,3,\ldots$$

The progress I made so far is that, for any polynomial $p$, we must have $p'(0) = \int_0^1 p(x)f(x)\,dx$. Also from Cauchy-Schwarz we can deduce that $\int_0^1 f^2(x) \, dx \geq 3$. We also know that the polynomials are dense in the space of continuous functions, but I don't know how to use this fact to solve the problem.

Thank you


问:
Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$
I hit a wall on this question. Below are my steps

$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$

Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$

$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$

Use partial fraction decomposition,
$$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$

Solve for $A$ and $B$:

$$\begin{cases}
A=\frac{1}{-2a^{2}}
\\
B=\frac{1}{2a^{2}}
\end{cases}$$

Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$
Factoring out $a$ yields
$$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$

Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.

My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$

Feedback says my answer is wrong. Where did I mess up?

答:
$$I=\int\frac{-7x}{x^4-a^4}dx=-7\int\frac{x}{x^4-a^4}dx$$
now for ease lets let $u=x^2\Rightarrow dx=\frac{du}{2x}$ and $b=a^2$ so:
$$I=-\frac72\int\frac{du}{u^2-b^2}$$
now let $u=bv\Rightarrow du=b\,dv$ so:
$$I=-\frac{7b}{2}\int\frac{dv}{b^2(v^2-1)}=-\frac{7}{2b}\int\frac{dv}{v^2-1}$$
now this is a standard integral that you can solve with PFD or using a substitution


问:
Is mathematics one big tautology?
Is mathematics one big tautology? Let me put the question in clearer terms:

Mathematics is a deductive system; it works by starting with arbitrary axioms, and deriving therefrom "new" properties through the process of deduction. As such, it would seem that we are simply creating a string of equivalences; each property can be traced back logically to the axioms. It must be that way, that's how deductive systems work!

If that be the case, then in what sense are we introducing novel or new ideas? It would seem that everything is simply equivalent to the fundamental set of axioms that we choose to start with. Is there a precise step in mathematical derivation that we can isolate as going beyond pure logic? If so, then how does this fit in with the fact that mathematics is deductive? Must we change our view of mathematics as purely deductive? And if not, then is there a way to reconcile the feeling of creativity in mathematics with the fact that it boils down to pure logic?

I'm trying to figure out the true nature of what's going on here.

答:
I think the answer you might be looking for is that truth within a mathematical model is an axiom or combination of axioms, augmented with a number of definitions. There can be no other truths. Whether math represents truth in general is discussed in

http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=truth;action=display;num=1268924603;start=

Title: Is mathematics one big tautology?


问:
Minimizing the surface area with a constant volume
I want to minimize the surface area of a rectangular prism, with a constant volume. The dimensions of the prism is $25 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$. If it is "flattened out" [with the top face not included], then the dimensions of the "flattened out" prism is $41\text{ cm} \times 24 \text{ cm}$.

Then, $l = 41 - 2c$, $w = 24 - 2c$, and $h = c$. When I use the formula to get the surface area and differentiate with respect to $c$, all $c$ terms vanish. What did I do wrong?

The process that I need to do is differentiation. I think it will be a minimum if $l = w = h$ using AM-GM, but this will not be counted.

Edit: I noticed that the only values of $c$ satisfying $c(41 - 2c)(24 - 2c) = 1600$ are $\left\{8, \frac{49 - \sqrt{1601}}{2}, \frac{49 + \sqrt{1601}}{2}\right\}$. Since $0 \leqslant c \leqslant 12$, then the only values are the first two in the list. When I substituted the second value to the formula for the surface area, it is approximately $1675.906\,\mathrm{cm}^{2}$. Then, the minimum surface area must be $728\,\mathrm{cm}^{2}$. Is this correct?

Edit [March 03, 2021]: $c$, $24 - 2c$, and $41 - 2c$ is now ignored. Dimensions are now free to take any form satisfying the constraints, accepted answer.

答:
If the question is to find dimensions of a rectangular box, with constant volume of $1600$, that minimizes the surface area of the box. The box is closed on top.

Surface area $S = 2l w + 2 l h + 2 w h$

$V = 1600 = lwh \implies h = \frac{1600}{wl}$

So, $S = 2lw + \frac{3200}{w} + \frac{3200}{l}$.

Now take partial derivative wrt $w$ and $l$ and equate to zero to find $w, l, h$ that minimizes surface area.

$\displaystyle S'_w = 2l - \frac{3200}{w^2} = 0 \implies l = \frac{1600}{w^2}$

$\displaystyle S'_l = 2w - \frac{3200}{l^2} = 0 \implies w = \frac{1600}{l^2}$

Solving both, $\displaystyle l = w = 4 \sqrt[3]{25}$.
Then use this to find value of $h$ and you should get a cube.


问:
Rational rank one decomposition of symmetric positive semidefinite integer matrices

Problem: Given an $n\times n$ symmetric positive semidefinite (PSD) $\color{blue}{\textbf{integer}}$ matrix $Q$ with $\mathrm{Rank}(Q) = r$, find
$\color{blue}{\textbf{integer}}$ vectors $u_i, i=1, \cdots, r$ and
positive $\color{blue}{\textbf{integer}}$ pairs $(c_i, d_i), i=1, \cdots, r$ such that $$Q = \frac{c_1}{d_1} u_1u_1^\mathsf{T} +
\frac{c_2}{d_2} c_2u_2 u_2^\mathsf{T} + \cdots + \frac{c_r}{d_r} u_r u_r^\mathsf{T}.$$

Background:

The problem arises when we express a non-negative polynomial as a sum of squares (SOS) of polynomials.

See e.g. https://web.stanford.edu/class/ee364b/lectures/sos_slides.pdf.

When we get $f(x) = Z(x)^\mathsf{T}Q Z(x)$ (see page 8 in the link above), we hope to do rational rank one decomposition of $Q$
such that we can express $f(x)$ as sum of squares of polynomials that is $f(x) = \sum_{i=1}^r [f_i(x)]^2$.

Also see my answers for the following questions:

https://mathoverflow.net/questions/193753/this-inequality-why-cant-solve-it-by-now-only-four-variables-inequality,

https://mathoverflow.net/questions/239243/an-inequality-concerning-lagranges-identity

Discussion:

This task is easy by the following fact:

Fact 1: Let $B$ be an $n\times n$ real symmetric positive semidefinite matrix. Let $u\in \mathbb{R}^n$ is a vector
such that $u^\mathsf{T}Bu \ne 0$, then
\begin{align}
\mathrm{Rank}\left(B - \frac{1}{u^\mathsf{T}Bu} Buu^\mathsf{T}B\right) &= \mathrm{Rank}(B) - 1, \tag{1}\\
B - \frac{1}{u^\mathsf{T}Bu} Buu^\mathsf{T}B &\succeq 0. \tag{2}
\end{align}
For (1), see Theorem 1.1 in [1]. (2) is obvious.

According to Fact 1, we have the following process:

First we choose any integer vector $v_1$ such that $v_1^\mathsf{T}Qv_1 \ne 0$,
let $Q_1 = Q - \frac{1}{v_1^\mathsf{T}Qv_1} Qv_1v_1^\mathsf{T}Q$.

Second, choose any integer vector $v_2$ such that $v_2^\mathsf{T}Q_1v_2 \ne 0$,
let $Q_2 = Q_1 - \frac{1}{v_2^\mathsf{T}Q_1v_2}Q_1v_2v_2^\mathsf{T}Q_1$.

$\cdots\cdots$

Finally we get $Q = \frac{1}{v_1^\mathsf{T}Qv_1} Qv_1v_1^\mathsf{T}Q + \frac{1}{v_2^\mathsf{T}Q_1v_2}
Q_1v_2v_2^\mathsf{T}Q_1 + \cdots$.

Example 1:

$$Q = \left(\begin{array}{rrr} 16 & -17 & 12\\ -17 & 23 & -6\\ 12 & -6 & 19 \end{array}\right),$$
$v_1 = [1, 0, 0]^\mathsf{T}$,
$$Q_1 = \left(\begin{array}{rrr} 0 & 0 & 0\\ 0 & 79/16 & 27/4\\ 0 & 27/4 & 10 \end{array}\right),$$
$v_2 = [0, 1, 0]^\mathsf{T}$,
$$Q_2 = \left(\begin{array}{rrr} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 61/79 \end{array}\right).$$
Terminate since $\mathrm{Rank}(Q_2) = 1$. We have
$$Q = \frac{1}{16} u_1u_1^\mathsf{T} + \frac{1}{1264}u_2u_2^\mathsf{T} + \frac{61}{79}u_3u_3^\mathsf{T}$$
where
$u_1 = [16, -17, 12]^\mathsf{T}, u_2 = [0, 79, 108]^\mathsf{T},
u_3 = [0, 0, 1]^\mathsf{T}$.

Example 2: For the same $Q$ as Example 1, we choose different $v_1, v_2$.

$$Q = \left(\begin{array}{rrr} 16 & -17 & 12\\ -17 & 23 & -6\\ 12 & -6 & 19 \end{array}\right),$$
$v_1 = [2, 1, -1]^\mathsf{T}$,
$$Q_1 = \left(\begin{array}{rrr} 23/2 & -19/2 & 27/2\\ -19/2 & 21/2 & -17/2\\ 27/2 & -17/2 & 37/2 \end{array}\right),$$
$v_2 = [1, 0, -1]^\mathsf{T}$,
$$Q_2 = \left(\begin{array}{rrr} 61/6 & -61/6 & 61/6\\ -61/6 & 61/6 & -61/6\\ 61/6 & -61/6 & 61/6 \end{array}\right).$$
Terminate since $\mathrm{Rank}(Q_2) = 1$. We have
$$Q = \frac{1}{2} u_1u_1^\mathsf{T} + \frac{1}{3}u_2u_2^\mathsf{T} + \frac{61}{6}u_3u_3^\mathsf{T}$$
where
$u_1 = [3, -5, 1]^\mathsf{T}, u_2 = [-2, -1, -5]^\mathsf{T},
u_3 = [1, -1, 1]^\mathsf{T}$.

Example 3: See https://mathoverflow.net/questions/239243/an-inequality-concerning-lagranges-identity

$$Q = \left(\begin{array}{rrrr} 7942 & -4138 & 917 & -4721\\ -4138 & 7934 & -4717 & 921\\ 917 & -4717 & 6000 & -2200\\ -4721 & 921 & -2200 & 6000 \end{array}\right),$$
$v_1 = [1, 0, 0, 0]^\mathsf{T}$,
$$Q_1 = \left(\begin{array}{rrrr} 0 & 0 & 0 & 0\\ 0 & 22944392/3971 & -16833934/3971 & -6110458/3971\\ 0 & -16833934/3971 & 46811111/7942 & -13143243/7942
\\ 0 & -6110458/3971 & -13143243/7942 & 25364159/7942 \end{array}\right),$$
$v_2 = [0, 1, 0, 0]^\mathsf{T}$,
$$Q_2 = \left(\begin{array}{rrrr} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 7984289425/2868049 & -7984289425/2868049
\\ 0 & 0 & -7984289425/2868049 & 7984289425/2868049 \end{array}\right).$$
Terminate since $\mathrm{Rank}(Q_2) = 1$. We have
$$Q = \frac{1}{7942} u_1u_1^\mathsf{T} + \frac{1}{22778045158}u_2u_2^\mathsf{T}
+ \frac{7984289425}{2868049}u_3u_3^\mathsf{T}$$
where
$u_1 = [7942, -4138, 917, -4721]^\mathsf{T}, u_2 = [0, 11472196, -8416967, -3055229]^\mathsf{T}$ and
$u_3 = [0, 0, 1, -1]^\mathsf{T}$.

Question:

From Examples 1-2, we see that the numbers ($c_i$'s, $d_i$'s, elements of $u_i$'s, in absolute value) in Example 2 is smaller than that in Example 1
which is due to different choice of $v_1, v_2$. We say the rank one decomposition in Example 2 is better.

From Example 3, we see that the numbers ($c_i$'s, $d_i$'s, elements of $u_i$'s, in absolute value) can be quite large (the biggest one is $22778045158$)
even when the numbers in $Q$ is not large (the biggest one is $7942$). Is there another $v_1, v_2$ such that
the numbers are much smaller?

One can imagine that when $n$ is large and $\mathrm{Rank}(Q)$ is large (e.g., $n>100$, I have ever encountered), the numbers may be very very large if we randomly choose $v_1, v_2, \cdots$.

Is there algorithms to find the better solution by nicely choosing $v_1, v_2, \cdots$?

Any comments and solutions are welcome and appreciated.

Reference

[1] Moody T. Chu, Robert E. Funderlic and Gene H. Golub, “A Rank-One Reduction Formula and its Applications to Matrix Factorizations”,
SIAM Review, Vol. 37, No. 4 (Dec., 1995), pp. 512-530.

答:
Beginning with rational symmetric $H,$   we may calculate a nonsingular rational $L$ such that $H = L D L^T$  with $D$ diagonal rational. Taking the diagonal elements $d_i$ of $D,$  we know $D = \sum d_i e_i e_i^T,$  where $e_i$ is the column vector with a single $1$ at position $i$ and otherwise $0.$  It follows that
$$ H = \sum  d_i L e_i e_i^T L^T = \sum d_i (Le_i) (Le_i)^T$$

There are two basic methods for diagonalizing a quadratic form. One may repeatedly complete the square, as in many quadratics forms books. Students seem to have trouble with that, although it has always seemed easy to me. The opposite method is suitable for computers, so I will put in output from my program.

for now, see reference for linear algebra books that teach reverse Hermite method for symmetric matrices

The matrix I call $L$ above becomes $L=Q^T$  in the output below

$$ \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 \frac{ 17 }{ 16 }  & 1 & 0 \\
 -  \frac{ 174 }{ 79 }  &  -  \frac{ 108 }{ 79 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 &  - 17 & 12 \\
 - 17 & 23 &  - 6 \\
12 &  - 6 & 19 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  \frac{ 17 }{ 16 }  &  -  \frac{ 174 }{ 79 }  \\
0 & 1 &  -  \frac{ 108 }{ 79 }  \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 &  \frac{ 79 }{ 16 }  & 0 \\
0 & 0 &  \frac{ 61 }{ 79 }  \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
 -  \frac{ 17 }{ 16 }  & 1 & 0 \\
 \frac{ 3 }{ 4 }  &  \frac{ 108 }{ 79 }  & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 &  \frac{ 79 }{ 16 }  & 0 \\
0 & 0 &  \frac{ 61 }{ 79 }  \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 &  -  \frac{ 17 }{ 16 }  &  \frac{ 3 }{ 4 }  \\
0 & 1 &  \frac{ 108 }{ 79 }  \\
0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrr}
16 &  - 17 & 12 \\
 - 17 & 23 &  - 6 \\
12 &  - 6 & 19 \\
\end{array}
\right)
$$

$$ \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc $$

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 \frac{ 2069 }{ 3971 }  & 1 & 0 & 0 \\
 \frac{ 3060867 }{ 11472196 }  &  \frac{ 8416967 }{ 11472196 }  & 1 & 0 \\
1 & 1 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
7942 &  - 4138 & 917 &  - 4721 \\
 - 4138 & 7934 &  - 4717 & 921 \\
917 &  - 4717 & 6000 &  - 2200 \\
 - 4721 & 921 &  - 2200 & 6000 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  \frac{ 2069 }{ 3971 }  &  \frac{ 3060867 }{ 11472196 }  & 1 \\
0 & 1 &  \frac{ 8416967 }{ 11472196 }  & 1 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
7942 & 0 & 0 & 0 \\
0 &  \frac{ 22944392 }{ 3971 }  & 0 & 0 \\
0 & 0 &  \frac{ 7984289425 }{ 2868049 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 -  \frac{ 2069 }{ 3971 }  & 1 & 0 & 0 \\
 \frac{ 917 }{ 7942 }  &  -  \frac{ 8416967 }{ 11472196 }  & 1 & 0 \\
 -  \frac{ 4721 }{ 7942 }  &  -  \frac{ 3055229 }{ 11472196 }  &  - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
7942 & 0 & 0 & 0 \\
0 &  \frac{ 22944392 }{ 3971 }  & 0 & 0 \\
0 & 0 &  \frac{ 7984289425 }{ 2868049 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  -  \frac{ 2069 }{ 3971 }  &  \frac{ 917 }{ 7942 }  &  -  \frac{ 4721 }{ 7942 }  \\
0 & 1 &  -  \frac{ 8416967 }{ 11472196 }  &  -  \frac{ 3055229 }{ 11472196 }  \\
0 & 0 & 1 &  - 1 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
7942 &  - 4138 & 917 &  - 4721 \\
 - 4138 & 7934 &  - 4717 & 921 \\
917 &  - 4717 & 6000 &  - 2200 \\
 - 4721 & 921 &  - 2200 & 6000 \\
\end{array}
\right)
$$

$$ \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc $$


问:
On the integrals of two complex valued functions which get closer
Suppose $f,g$ are complex valued continuous functions defined on positive real axis.  Suppose for all $\epsilon>0$, there is $M$ such that $|f(x)-g(x)|<\epsilon$ for $x>M$ and it is given that $\left|\int_0^\infty f(x)dx\right|<c_1<\infty$.  Is it then true that $|\int_0^\infty g(x)dx|<c_2<\infty$ for some constant $c_2$?

答: I guess the answer is no because consider for instance $g(x)=f(x)+1/x$.  Then given $\epsilon>0$ there is $M$ such that $1/x<\epsilon$ for sure.  But $|\int_0^\infty g(x)dx|$ is not finite.


问:
equivalent conditions to check if given two systems of linear equations are equivalent.
I am reading Linear Algebra by Hoffman and Kunze. The definition for equivalent system is

Two systems of linear equations are equivalent if each equation in each system is a linear combination of the equations in the other system.

While going through one of the exercise, it was asked to check if given two systems are equivalent. I find it bit lengthy to check if each equation from each system can be written as linear combination of other equations from other system. So I look for some tricks and read few related questions asked here. After spending some time, I came to know about following conditions which were scattered here and there. I wanted to have them in one place and check if it's correct.

For two homogenous systems of equations, following statements are equivalent :

Two systems are equivalent.

If $A$ and $B$ are matrices of coefficients of respective systems, $A$ and $B$ are row equivalent.

$AX=0$ and $BX=0$ have same set of solutions.

$A$ and $B$ have same row reduced echelon form.

For two non-homogenous systems of equations, say $AX=Z_1$ and $BX=Z_2$, following statements are equivalent :

Two systems are equivalent.

If $A'$ and $B'$ are respective augmented matrices, then $A'$ and $B'$ are row equivalent.

$AX=Z_1$ and $BX=Z_2$ have same (non-empty)set of solutions.

$A'$ and $B'$ have same row reduced echelon form.

Have I listed correctly. Thanks.

答:
Yes, you are correct.

References below are to your text Linear Algebra.

For homogeneous systems,

1 and 2 imply each other by the definitions of linear combination and equivalent on page 4, elementary row operations on page 6, and row-equivalent on page 7,

1 implies 3 by Theorem 1.1 (Chapter 1, Theorem 1),

2 and 4 imply each other by the first corollary to Theorem 2.11 and the fact that row-equivalence is an equivalence relation (page 7),

2 implies 3 by Theorem 1.3, and

3 implies 4 by the discussion after the proof of Theorem 1.5 and the fact that row-equivalence is an equivalence relation.

Those items are overkill as a proof, but I included more than necessary for completeness.

For non-homogeneous systems,

1 and 2 imply each other by the definitions of linear combination, equivalent, elementary row operations, and row-equivalent,

1 implies 3 by Theorem 1.1,

2 and 4 imply each other by the first corollary to Theorem 2.11 and the fact that row-equivalence is an equivalence relation, and

3 implies 4 by the discussion after Theorem 1.7 and the fact that row-equivalence is an equivalence relation.


问:
perfect right-triangles that are not super-perfect
I'm seeking some insight on the answer to this problem from Project Euler.

Consider the right angled triangle with sides $a=7, b=24$, and $c=25$.

The area of this triangle is $84$, which is divisible by the perfect numbers $6$ and $28$.

Moreover, it is a primitive right angled triangle as gcd$(a,b)=1$ and gcd$(b,c)=1$.

Also, c is a perfect square.

We will call a right angled triangle perfect if

```<code>1. it is a primitive right angled triangle
2. its hypotenuse is a perfect square
</code>```

We will call a right angled triangle super-perfect if

```<code>1. it is a perfect right angled triangle and
2. its area is a multiple of the perfect numbers 6 and 28
</code>```

How many perfect right-angled triangles with $c \le n$ exist that are not super-perfect?

The answer turns out to be 0.

This outcome is not immediately intuitive to me.

Is there some property of Pythagorean triples that leads to this conclusion?

For the given case, it is evident that the area is a multiple of the perfect numbers 6 and 28, so we see that this perfect right-angled triangle is super-perfect, but how can we generalize this property for all perfect right-angled triangles?

Any insight is much appreciated.
Thanks

答: 33,56,65 and the rest are not perfect triangles. The hypotenuse is not a perfect square in any of them</p>


问:
Python coding with numpy sympy
New to python, and I am working on a problem with the ultimate goal of being able to evaluate

$$\frac{\sqrt{x}\bigg[\prod_{k=0}^{n-1}(\sqrt{x}+2^k)- \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)\bigg]}{\prod_{k=0}^{n-1}(\sqrt{x}+2^k)+ \prod_{k=0}^{n-1}(-\sqrt{x}+2^k)}$$

at whichever value of $x, n$ I choose.

First, I probably need to expand the function: how should I go about doing so? Perhaps sympy?
Then, how would I evaluate the sympy function?

答:
With sympy it's even easier (and more precise):

```<code>from sympy.abc import x, k, n
from sympy import Product, sqrt, simplify
nm = sqrt(x) * (Product(sqrt(x) + 2**k, (k, 0, n-1)) - Product(-sqrt(x) + 2**k, (k, 0, n-1)))
dn = (Product(sqrt(x) + 2**k, (k, 0, n-1)) + Product(-sqrt(x) + 2**k, (k, 0, n-1)))
r = nm / dn
# assign values, e.g., x=2, n=30 and simplify the expression r
r = r.subs({x:2, n:30})
simplify(r.doit(), full=True)
</code>```

The above code will produce the output as the following rational number:

$\displaystyle \frac{15027793389823468799645258066452193394332497033326437886769757423624751185248020668383259645787227100364629147062572788303205899}{10480566136263082489017135219769116375568335802982506381129986143601518100707743247130303530922557209156825461255161211247319516}$

We may want to approximate the output rational number obtained with a float, using the following code:

```<code>Float(simplify(r.doit(), full=True), 20) # with precision of 20 digits
# 1.4338722922444847313
</code>```

Title: Python coding with numpy sympy

Body:
With sympy it's even easier (and more precise):

from sympy.abc import x, k, n
from sympy import Product, sqrt, simplify
nm = sqrt(x) * (Product(sqrt(x) + 2**k, (k, 0, n-1)) - Product(-sqrt(x) + 2**k, (k, 0, n-1)))
dn = (Product(sqrt(x) + 2**k, (k, 0, n-1)) + Product(-sqrt(x) + 2**k, (k, 0, n-1)))
r = nm / dn
# assign values, e.g., x=2, n=30 and simplify the expression r
r = r.subs({x:2, n:30})
simplify(r.doit(), full=True)

The above code will produce the output as the following rational number:

$\displaystyle \frac{15027793389823468799645258066452193394332497033326437886769757423624751185248020668383259645787227100364629147062572788303205899}{10480566136263082489017135219769116375568335802982506381129986143601518100707743247130303530922557209156825461255161211247319516}$

We may want to approximate the output rational number obtained with a float, using the following code:

Float(simplify(r.doit(), full=True), 20) # with precision of 20 digits
# 1.4338722922444847313


Title: Big-O: Prove $2^n$ is $O(n!)$

Body:
I am a little stuck trying to prove that $2^n$ is $O(n!)$. Obviously, I can tell in a few ways that this is the case. For one, based on Big-$O$ hierarchy, the exponential is beneath the factorial in terms of complexity. I also substituted $C = 2$ and tried several values of $n$, all following the Big-$O$ definition of $f(n) \le cg(n)$.

However, my professor says neither of these is a sufficient justification and I should be able to use the definition alone.

How can I do this? I know in the past it has been advantageous for me to compare the functions by converting one into a form that resembles the other, but I just can't figure out how to relate the factorial and $2^n$ functions.

Can anyone point me in the right direction?

Body:
HINT:

$$\frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}=\frac{2^{n+1}n!}{2^n(n+1)!}=\frac2{n+1}$$


问:
Powers of primitive roots.
Let  $w= \cos\frac{2k\pi}{n} + i \sin\frac{2k\pi}{n}$ be a primitive $n^{\text{th}}$ root of unity, ie, $w^n=1$ and $w^m \neq 1$ for $m \leq n$. Then the powers
$$1, w, w^2, \ldots, w^{n-1}$$
are all the $n^{\text{th}}$ roots of unity.

Comments: I tried by use the fact that if $w_0 = (\cos\frac{2\pi}{n} + i \sin\frac{2\pi}{n})$ then the $n^{\text{th}}$ roots of unity are $1, w_0, w_0^2, \ldots, w_0^{n-1}$. This case is the case in that $k=1$ in the problem. The way forward would be this using $\mod{n}$? Or there would be one where I would not need the result I commented.

答:
Here is roadmap:

$1, w, w^2, \ldots, w^{n-1}$ are $n$th roots of unity.

$1, w, w^2, \ldots, w^{n-1}$ are all different. This is the crucial point.

There are exactly $n$ complex $n$th roots of unity.

Title: Powers of primitive roots.


问:
points matching minimizing sum of square distances
I have $2$ sets of $n$ points in $\mathbb{R}^d$ $A$ and $B$.

I am trying to find a bijective function $f:[n]\rightarrow[n]$ such that
$$\sum_{i=1}^n{||A_i-B_{f(i)}}||_2^2$$

is minimized, i.e., to find a pairing between the sets that minimize the sum of square distances.

I thought of using the Hungarian Method which takes $O\left(n^3\right)$ time but is there any way to do it faster? I was looking for papers that try to do it faster but all I could find is papers that talks about minimizing sum of distances instead of square distances.

Is there any way to do it faster than $O\left(n^3\right)$ and either get an optimal solution or a solution with a provable cost of at most $(1+\epsilon)$OPT, where OPT is the optimal cost?

答:
What you're looking for is called the $2$-Wasserstein distance between your sets of points.

The Hungarian method (also named (Khun-)Munkres, depending on the paper) has long been used to solve this exactly, but auction methods converge really fast.

Moreover, in $\mathbb R^d$ with low dimension, you can even use kd-tree methods to make local requests and accelerate the computation. You can see this in a paper (here the sets of points $A$ and $B$ are in a space a little more complicated than $\mathbb R^2$, but you could forget about that and the paper is quite clear). The complexity is hard to estimate because it depends on the heuristic on which the auction method is based, but the paper regress their method's speed to $O(n^{1.6})$. In dimension $d > 2$ it might be a little less efficient but it's better than $O(n^3)$ anyway.


问:
Solving first order non-linear ODE with logarithm
I want to solve for $t \in \mathbb{R}, u'(t)=-u(t)\ln \lvert u(t) \rvert$.

I defined two cases: $\mathbb{R^*_+}$ and $\mathbb{R^*_-}$.

For $\mathbb{R^*_+}$:

$$\frac{du}{u}=-\ln(u(t))dt$$

And by integrating, it follows that:

$$\ln\lvert u \rvert = -u\ln(u) + u$$

But I'm stuck here, I don't know what to do next.

答:
The above solution is absolutely correct.  A slightly quicker way is to substitute
$$
u(t) = e^{f(t)}
$$
producing the differential equation:
$$
f'=-f
$$
eventually leading to the same answer.


问:
The polynomial subspace
Let $A$ be a set of 6 polynomials in $\mathbb{R}_5[x]$ over $\mathbb{R}$ field,
assume $sp(A) = \mathbb{R}_5[x]$ which of the following is true?

1. It might be that $A$ holds exactly 4 polynomials from second degree.

2. It might be that $A$ holds exactly 4 polynomials from first degree.

I think the two argument are equivalent, and in order to span $\mathbb{R}_5[x]$
a set must have $6$ vectors, and since 4 of the vectors are in 2nd or 3rd degree it means there will always be an unreachable vector and thus the groups described in 1. and 2. cannot span $\mathbb{R}_5[x]$.

答:
In short:

$$A:=\,\{x^4\,,\,x^4+1\,,\,x^4+x\,,\,x^4+x^2\,,\,x^3\,,\,x^5\}\;\implies \,\text{Span}(A)=\Bbb R_5[x]$$

and $\,A\,$ contains four quartic polynomials (check the above vectors are line. independent) .

OTOH, if $\,A\,$ contains four linear polynomials then it automatically is a linearly dependent set, since all these four linear pol's are elements of $\;\Bbb R_1[x]\;$ , which is a two-dimensional space, so that $\,\text{Span}(A)\;$ cannot be the whole $\,\Bbb R_5[x]\,$

Title: The polynomial subspace


问:
How to determine the center of symmetry of a function?
I'm learning about functions and I came to a question that I'm struggling at:

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

The question is:

Show that the graph of $f$, has a center of symmetry to be determined in an orthonormal system.

What is the best approach to solve this question?

答:
Well, it is quite obviously $(-1,-2)$ if you draw a graph.

Notice that it lies in the intersection of both asymptots: $x=-1$ and $y=x-1$.

 A general rule simply involves shifting the function. Proving that $(a,b)$ is  a center of symmetry for $f(x)$ is the same as proving that the function
 $$g(x):=f(x+a)-b$$
 is symmetric around the origin. This means you have to prove $g(-x)=-g(x)$.

 So, in your case $g(x) ={x^2-1\over x}$ which is clearly odd.


答: Since $\prod_{i=2}^Nf(i)=\exp\sum_i\ln f(i)$ is approximately proportional to $\exp\int_c^N\ln f(x)dx$,$$\prod_{i=2}^Ni\ln i\in O\left(\exp\int(\ln N+\ln\ln N)dN\right)=O(\exp(N\ln\ln N-\operatorname{li}(N)+N\ln N-N))$$or thereabouts. The logarithm is asymptotic to $\tfrac12N^2\ln N$.

Title: Big-O notation for algorithm

Body:
I'm working in an algorithm and I have achieved to know that its cost is $O \Big(\prod_{i = 1}^{N} i \cdot \log i \Big)$. I don't know how to simplify my cost even more.

Can anyone help me? Thank you so much