问:
"Minimal" definition of Wiener process
Almost everywhere (for example on Wiki) the Wiener process is defined as a process $W(t)$ such that:

$W(0)=0 $;

It has independent increments;

The increments in an interval $\Delta t$ are gaussian with expected value 0 and variance $D\Delta t$;

The process is sample continuous;

Is this definition minimal? Or could the 4) be dropped?

My idea is trying to prove the fact that the following  holds $\forall \epsilon>0$:

$$lim_{\Delta t \rightarrow0}P(|\Delta W|> \epsilon)=0$$

where $\Delta W$ is the increment of the process in $\Delta t$. This should be a sufficient condition for sample continuity (sometimes it is called Lindeberg condition) that is valid for Markov processes according to Feller "Handbook of stochastic methods". Of course the definition of the Wiener process entails that the process is Markov because it has independent increments.

答: The first $3$ conditions are not quite enough to show that $W$ has continuous sample paths, but they will ensure that there exists a modification that does have sample paths by Kolmogorov's continuity theorem.  For example, let $W$ be a standard Brownian motion and define $X(t) = W(t)1_{W(t) \ne 1}$.  For any fixed $t$ we have that $X(t) = W(t)$ a.s. so the first $3$ conditions will hold for $X$ ($X(0)=0$, $X(t)-X(s)$ is independent of $X(u)-X(v)$ for any $v < u < s < t$, and $X(t)-X(s) \stackrel{d}{=}W(t)-W(s) \sim N(0,t-s)$), but $X(t)$ does not have continuous sample paths.

Title: "Minimal" definition of Wiener process


问:
What is the distance between the line with Cartesian equation $x - 4 = y - 5 = z$ and the line with Cartesian equation $x - 2 = y - 5 = z - 1$?
What is the distance between the line with Cartesian equation $x - 4 = y - 5 = z$ and the line with Cartesian equation $x - 2 = y - 5 = z - 1$?

I am not sure how to do it. I think the lines are parallel with the same direction but I'm not sure.

答:
HINT

The vector equation of the first line can be written as
\begin{align*}
L_{1}(a) = (a,a+1,a-4) = (0,1,-4) + a(1,1,1)
\end{align*}

On the other hand, the vector equation of the second line can be written as
\begin{align*}
L_{2}(b) = (b,b+3,b-1) = (0,3,-1) + b(1,1,1)
\end{align*}

Consequently the lines $L_{1}$ and $L_{2}$ are parallel.

In order to calculate the distance between them, let us take $(0,3,-1) - (0,1,-4) = (0,2,3)$.

Hence the distance $d$ can be expressed as
\begin{align*}
d = \left\|(0,2,3) - \frac{1}{3}\langle(0,2,3),(1,1,1)\rangle(1,1,1)\right\|
\end{align*}

Can you take it from here?

ContentLicense: CC BY-SA 4.0


问:
how to get brownian bridges maximum and minimum's joint distribution
I'm wondering how to get brownian bridge's maximum and minimum's joint distribution.Brownian bridge's definition is as below.

$B_t = W_t - tW_1$ for $t \in [0,1]$, W is wiener process

How to get it? In fact, I didn't get maximum's distribution of brownian bridge. I cannot catch any idea.. I tried to get this distribution via conditional probability, that fix $W_1$, but it is still hard one. Can you help me?

edit) I think that it is concerned with joint distribution of brownian motion's maximum and minimum, and W(1). Then how to attain this one?

答:
Let $M$ and $m$ be the Max and min of $B_t$. Let us find the joint cdf of $M$ and $-m$, defined by
$$
P(\{M< a\}\cap \{-m< b\}),
$$
In fact, the argument you can use is exactly the continuous version of the discrete problem here: A generalization of reflection. Namely, Brownian motion is the limit of a discrete random walk, and there is an exact formula for the number of random walks of a given length who stay between two horizontal lines, using the reflection principle.

Applying that logic, we get that
$$
P(M<a,-m<b)
=\frac1{P(W_1=0)}\left(
\sum_{k\in \mathbb Z} P\big(W_1=2k(a+b)\big) -
\sum_{k\in \mathbb Z} P\big(W_1=2k(a+b)+2a\big)
\right)
$$
Note that this does not quite make sense, as all the probabilities are zero, but you have to understand this as a limit. Anyways, what you get is
$$
P(M<a,m<-b)=\sum_k\exp(-2k^2(a+b)^2)-\sum_k\exp\left(-2\big(k(a+b)+a\big)^2\right)
$$
This gives the joint cdf of $M$ and $-m$.


问:
Showing measure of a finite intersection is positive and a best possible bound

Suppose that $E_1 \cdots E_n \subset \left]0,1\right[$ are Lebesgue measurable sets and suppose that $\sum_{k=1}^n m(E_k)>n-1$. Show that $m\left(\bigcap_{k=1}^n E_k \right) > 0$. Show that the bound $n-1$ is best possible.

Here is what I managed to do: Assume for a contradiction that $m\left(\bigcap_{k=1}^n E_k\right)=0$, then
\begin{align}
    m\left(\left]0,1\right[\setminus \bigcap_{k=1}^n E_k\right)=m \left(\left]0,1\right[ \cap \left[\bigcap_{k=1}^n E_k \right]^c \right) & \le \sum_{k=1}^n m\left(\left]0,1\right[\setminus E_k \right)=n-\sum_{k=1}^n m(E_k) \nonumber
\end{align}
Then we have
\begin{align}
    m\left(\left]0,1\right[\setminus \bigcap_{k=1}^n E_k \right) = 1 \le n-\sum_{k=1}^n m(E_k) \implies \sum_{k=1}^n m(E_k) \le n-1 \nonumber
\end{align}
Which is a contradictions and thus $m \left(\bigcap_{k=1}^n E_k \right)>0$.

But how do I show that the bound on the sum which is $(n-1)$, is the best possible bound?

答: We can write this statement as follows: $m(\bigcap_{i=1}^{n}E_i)=0$ implies $\sum\limits_{i=1}^{n} m(E_i) \leq n-1$. If $n-1$ is not the best bound then there exists $t <n-1$ such that $m(\bigcap_{i=1}^{n}E_i)=0$ implies $\sum\limits_{i=1}^{n} m(E_i) \leq t$. To get a contradiction to this take $E_i=(0, \frac  s n), 1\leq i \leq {n-1}$  and $E_n=(\frac s n, 1)$ where $s=\frac {n(t-1)} {n-2}$.


问:
Finding the number of $2017$-digit numbers with leading digit $2$
Let $N$ be the number of $2017$-digit numbers such that the leading digit is $2$ and there are an odd number of $9$'s. Find the remainder when $N$ is divided by $1000.$

I first jumped to the thought of using some form of casework on the number of $9$'s and trying to find a pattern. Because it is given that the leading digit is $2,$ there are $2016$ digits where you wish to place the $9$'s and the other digits. If there is a singular $9,$ then there are $$\binom{2016}{1} \cdot 9^{2015}$$ possible numbers; if there are three $9$'s, then there are $$\binom{2016}{3} \cdot 9^{2013}$$ possible numbers, and so on. Therefore, I came to the conclusion that $$N = \sum_{n = 0}^{1007} \binom{2016}{2n + 1} \cdot 9^{2015 - 2n}.$$ However, I am unsure how to calculate this large sum, and I was also wondering if there was a more efficient way to do this problem.

EDIT: According to Wolfram Alpha, the answer is $672,$ although I was looking for a method to do this without calculators.

Body:
You've already got
$$N = \sum_{n = 0}^{1007} \binom{2016}{2n + 1} \cdot 9^{2015 - 2n}$$

By the binomial theorem, we have
$$(1+x)^m=\sum_{n=0}^{\lfloor m/2\rfloor}\binom{m}{2n}x^{2n}+\sum_{n=0}^{\lfloor(m-1)/2\rfloor}\binom{m}{2n+1}x^{2n+1}\tag1$$
and
$$(1-x)^m=\sum_{n=0}^{\lfloor m/2\rfloor}\binom{m}{2n}x^{2n}-\sum_{n=0}^{\lfloor (m-1)/2\rfloor}\binom{m}{2n+1}x^{2n+1}\tag2$$

Subtracting $(2)$ from $(1)$, we get

$$\sum_{n=0}^{\lfloor (m-1)/2\rfloor}\binom{m}{2n+1}x^{2n+1}=\frac 12\bigg((1+x)^m-(1-x)^m\bigg)$$

So, when $m=2016$ and $x=\dfrac 19$, we obtain
$$\sum_{n=0}^{\lfloor (2016-1)/2\rfloor}\binom{2016}{2n+1}\cdot 9^{-2n-1}=\frac 12\bigg(\bigg(1+\frac 19\bigg)^{2016}-\bigg(1-\frac 19\bigg)^{2016}\bigg)$$

Multiplying the both sides by $9^{2016}$, we have

$$\begin{align}N=\sum_{n=0}^{1007}\binom{2016}{2n+1}\cdot 9^{2015-2n}&=\frac{9^{2016}}{2}\bigg(\bigg(1+\frac 19\bigg)^{2016}-\bigg(1-\frac 19\bigg)^{2016}\bigg)
\\\\&=\frac{1}{2}\bigg(10^{2016}-8^{2016}\bigg)
\\\\&=5\cdot 10^{2015}-4\cdot 8^{2015}\end{align}$$

Using Euler's theorem,
$$2^{\varphi(5^3)}\equiv 2^{100}\equiv 1\pmod{5^3}\implies 2^{103}\equiv 8\pmod{1000}$$
so we have
$$8^{2015}\equiv (2^{103})^{58}\cdot 2^{71}\equiv 8^{58}\cdot 2^{71}\equiv (2^{103})^2\cdot (2^{10})^{3}\cdot 2^9\equiv 64\cdot 824\cdot 512\equiv 832\pmod{1000}$$

Therefore, we finally get
$$N=\sum_{n=0}^{1007}\binom{2016}{2n+1}\cdot 9^{2015-2n}\equiv 0-4\cdot 832\equiv \color{red}{672}\pmod{1000}$$


问:
This is regarding vector space in linear algebra...
Let $V$ be the set of all positive real numbers.On $V$ define the addition and the scalar multiplication in the following manner.

addition:  $x+y = xy$

scalar multiplication : $rx= x^r$

Prove or disprove whether $V$ is a vector space.

my answer

chose $1,2$ belongs to $V$

$1+2=3$ not equal to $1\cdot2= 2$

hence $V$ is not closed under addition ,hence $V$ is not a vector space... is the answer simple as that??? please help me friends...it will be very useful for my  Linear Algebra exam...

答:
Every field is a vector space (of dimension 1) over itself, whith addition given by the additive structure of the field, and scalar multiplication given by the multiplicative structure. In your case, there is simply a change of notation: the sum is denoted as a product, and consistently the product $rx$ becomes a power $x^r$.

For the other question you raised in the comments, consider that the matrix associated to the map sending $u_i$'s to $v_i$'s is of the form (e.g. take $n=3$)
$$\left(
\begin{array}{ccc}
1&0&0\\
1&1&0\\
1&1&1
\end{array}
\right)
\left(
\begin{array}{c}
u_1\\
u_2\\
u_3
\end{array}
\right)
=
\left(
\begin{array}{c}
u_1\\
u_1+u_2\\
u_1+u_2+u_3
\end{array}
\right)$$

The matrix is invertible ( $\det=1$), hence the map also is, so it is a linear bijective map, and you should know that such a map sends a basis in a basis

ContentLicense: CC BY-SA 3.0


问:
Geometric series word problem: Number of games played
If a tournament consists of 64 teams playing sudden death games until 1 team is declared the winner. How many games, in total, are played during this tournament?

My approach:
Teams: 64, 63, 62, ....., 1

Two teams play 1 game, therefore, all together 32 games. But where is the GP???

答:
As you said in your post, two teams play 1 game. Then, if 64 teams will play, only 32 teams will win that game. For the second game, only 16 teams will remain.

Notice the pattern that the number of teams is divided by 2 every round.

In this case, for only 1 team to win, we need to divide 64 by 64 or $2^{6}$. Since this calculation follows the pattern, then we need 6 matches for only 1 team to win.

Edit:

The geometric sequence in this problem is $a_{n} = \frac{1}{2}a_{n-1}$ or $a_{n} = a_{0}\left(\frac{1}{2}\right)^{r}$. It is given that $a_{0} = 64$ and we need to solve for $r$ such that $a_{n} = 1$.

A solution of this problem is like the following:
\begin{align*}a_{n} &\;=\; a_{0}\left(\frac{1}{2}\right)^{r} \\ 1 &\;=\; (64)\left(\frac{1}{2^{r}}\right) \\ 2^{r} &\;=\; 64 \\ 2^{r} &\;=\; 2^{6} \\ r &\;=\; 6\end{align*}

Therefore, 6 games are needed.


问:
Question about terminology regarding nets

Show that $U$ is an open subset of a topological space $X$ if and only if for every point $z\in U$ each net converging to $z$ is co-finitely contained in $U$.

Is this question as stated correct? IMO, I think it's supposed to say eventually in $U$ rather than co-finitely contained in $U$.

答:
No, it’s not. Let the index set be $\Bbb Z$ in the usual order and define the net $f:\Bbb Z \to \Bbb R$ be defined by $f(n)=\min(0,n)$. Then $f$ converges to $0$ but infinitely many terms of the net (for $n<0$) are outside the open neighbourhood $(-1,1)$ of $0$.

It’s indeed more customary and correct to say that the net is eventually in the open set. Tails need not be cofinite as it is the case for sequences.


问:
Efficient way to model this "peaking filter" function.
CONTEXT

I'm working on an image processing algorithm, which uses pixel lightness values to determine how much the pixel should be darkened.

A small range of lightness values should be "selected" by the function, with a smooth transition to the unselected ones. The desired behaviour is visually described below:

```<code>X is the pixel lightness
Y is the multiplier to be applied to the pixel lightness later on
</code>```

I should be able to control 3 parameters:

A: Filter center

B: Some kind of Bandwidth

C: Gain

WHERE I AM NOW

At first, I've tried using a simple parabola and clipping y values above 1:

$y = B(x-A)^2+(1-C)$

However, the harsh clipped transition adds noise to the final image. I need a smoother transition.

So, I gave it another shot with two very expensive tanh curves, reflected by the filter center:

$y = \left\{
 \begin{array}{ll}
  1-\Big[tanh(B(x-A)+B')+1\Big](C/2)  & \mbox{if } x \leq A \\
  1-\Big[tanh(B(A-x)+B')+1\Big](C/2)  & \mbox{if } x > A \\
 \end{array}
\right.$

The results are better, however since I'll be processing 1080p videos with this algorithm, I don't think calculating tanh for every single pixel is a good thing.

THE FINAL QUESTION

Is there a way to model this behaviour without expensive operations such as tanh?

答:
If a polynomial is suitable, then consider

$$
f(x)=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4
$$

Solving for the $c_n$, using the conditions:

$$
f'(A)=f'(A+B/2)=f'(A-B/2)=f(A+B/2)=0 \quad; \quad f(A)=C
$$

Using your notation: $A$ is position of center, $B$ is width, $C$ is the height. This is just for the peak: you need to subtract it from 1 to get the 'upside down version'. The coefficients are

$$
c_0=\frac{C}{B^4}(B^4-8B^2A^2+16A^4) \\
c_1=\frac{16C}{B^4}(B^2A-4A^3) \\
c_2=-\frac{8C}{B^4}(B^2-12A^2) \\
c_3=-\frac{64CA}{B^4} \\
c_4=\frac{16C}{B^4}
$$

Here is a plot with $A=1$, $B=0.1$, $C=2$

Of course we are only interested in the graph between $A-B/2 \ < x < \ A+B/2$. It's tangent to the horizontal at the boundary.


问:
Show that the quotient norms fail to be a norm in the quotient space $\ell^{\infty}/c_{00}$.
Let $(V, \|\ \cdot\ \|)$ be a normed space over $\mathbb{R}$ or $\mathbb{C}$ and let $W\subset V$ be a linear subspace. We define the following equivalence relation on $W$: $$x\sim y\iff x-y\in W.$$ Then, we can define the quotient space $$V/W=V/\sim=\{[x]:x\in V\},\ \ \text{where}\ \ [x]\ \ \text{is the equivalence class}.$$ And we define the quotient "norm" to be $$\|[x]\|_{V/W}=\inf_{y\in W}\|x-y\|.$$

It is known that $\|\ \cdot\ \|_{V/W}$ is a semi-norm, and when $W$ is not closed, it may fail to be a norm since it may fail the property that $$\|[x]\|_{V/W}=0\iff [x]=[0].$$

One counterexample has been given here: Concrete counterexample for norms on quotient spaces, but it seems that the definition of the quotient in this post is a little bit different than mine.

Therefore, I tried to come up with an example by considering $c_{00}$ as a subspace of $\ell^{\infty}$ (as long as the elements in $c_{00}$ comes from $\ell^{\infty}$, the subspace property is clear).

Firstly note that $c_{00}$ is not closed in $\ell^{\infty}$. Indeed, for each $n\in\mathbb{N}$, set $$x_{n}:=\Bigg(1, \dfrac{1}{2},\dfrac{1}{3},\cdots,\dfrac{1}{n},0,0,\cdots\Bigg)\ \ \text{and}\ \ x=\Bigg(1, \dfrac{1}{2}, \dfrac{1}{3},\cdots\Bigg).$$ Then $(x_{n})_{n=1}^{\infty}\subset c_{00}$ and $\|x-x_{n}\|_{\infty}=\frac{1}{n+1}\longrightarrow 0$, and thus $x_{n}\longrightarrow x$ in $\ell^{\infty}$, but $x\notin c_{00}$.

However, I cannot come up with a sequence  $x\in c_{00}$ (so that $[x]=0$) but $\|[x]\|_{\ell^{\infty}/c_{00}}\neq 0.$

Any idea? Thank you!

答:
You have $x \in \ell^\infty \setminus c_{00}$ (and consequently $[x] \neq 0$), so if $\|\cdot\|_{V/W}$ were a norm, then we would expect $\|[x]\|_{V/W} > 0$. However,
$$0 \le \inf_{y \in c_{00}} \|x - y\| \le \|x - x_n\| \to 0,$$
as $n \to \infty$. By squeeze theorem, we have
$$0 = \inf_{y \in c_{00}} \|x - y\| = \|[x]\|_{V/W},$$
proving $\|\cdot\|_{V/W}$ is not a norm after all.


问:
About mulit-variate Fourier series
If $f(x,y)$ is $2\pi$ periodic with respect to $x$ and $2\pi$ periodic with respect to $y$ respectively, then can I write $$ f(x,y) = \sum_{j,k \in \mathbb Z} c_{jk} e^{ijx} e^{iky}$$ where $$ c_{jk} = \frac{1}{4\pi^2}\int_0^{2 \pi} \int_0^{2 \pi} f(x,y) e^{-ijx} e^{-iky} dxdy \;?$$ I am wondering the sufficient condition to write in this way. Is the condition $f(x + 2\pi, \cdot) = f(x, \cdot)$ and $f(\cdot, y + 2 \pi) = f(\cdot, y)$ sufficient? Or do I need $f(x + 2 \pi, y+2\pi) = f(x,y)$ ?</p>

答:
The conditions that $f$ is $2\pi$ periodic in the $x$ and $y$ directions separately imply together that $f(x + 2\pi, y + 2\pi) = f(x,y)$. Periodicity in both directions is needed.

However, there is also another condition that is required for the above to be true, which is that $f$ must be square-integrable; that is, $ \int_0^{2\pi} \int_0^{2\pi} |f(x,y)|^2 \, dx \, dy$ must be defined and finite.

Title: About mulit-variate Fourier series


问:
find analytic function whose real part is given

Let $u(x,y)$ be a harmonic polynomial of real coefficients and $$f(z)=2u(\frac{z}{2}, \frac{z}{2i})-u(0,0)$$
  prove that for any $z\in \mathbb{C}$, $\text{Re} f(z)=u(x,y)$.

I tried to use the condition $u_{xx}+u_{yy}=0$ to construct harmonic conjugate $v$ then deduce $f$, but it comes out a lot of integrals, and I don't know how to simplify. A similar question goes here where $u$ is degree one wrt $y$ but here it's more general. Thanks for any help.

答:
Since $u(x, y) = u\left(\frac{z+\bar z}{2}, \frac{z - \bar z}{2}\right) = P(z, \bar z)$, we can define $P(z, \bar z)$ to be a harmonic polynomial, such that
$\frac{\partial^2 P}{\partial z\partial \bar z}$ is zero. Therefore every monomial in $P$ can't be the form $z^i\bar z^j$ otherwize the Laplace is not zero.

Write in the form of $P(z, \bar z) = c_0 + \sum_{k = 1}^n c_kz^k + d_k \bar z^k$ and thus $$u(x, y) = c_0 +\sum_{k=1}^n c_k(x+iy)^k + d_k(x-iy)^k.$$

From that the coefficients of $u$ are real numbers, we have $c_k = d_k$, and $$u(x, y) = c_0 +\sum_{k=1}^n c_k(x+iy)^k + c_k(x-iy)^k, c_i\in \mathbb R.$$

From definition of $f$, we have $f(z) = 2u(z/2, z/2i)-c_0=c_0+2 \sum_{k=1}^nc_kz^k$.

Notice that $\text{Re}(2z^k)=z^k+\bar z^k$ we can easily get $\text{Re}(f(z)) = u(x,y)$.


问:
Bayes' Theorem with 3 parameters
Can you please help me figure out how the following equation was derived using Bayes' Theorem? The left-hand side of the equation is the posterior distribution.

$$f(x,y,z\mid D)=\frac{f(D\mid x,z)f(x\mid y)f(y)f(z)}{f(D)}$$

I started by directly applying Bayes' Theorem but I seem to not able to go forward that arrives at the equation above.

$$f(x,y,z\mid D)=\frac{f(D\mid x,y,z)f(x,y,z)}{f(D)}$$

Any help is appreciated. Thank you!

Title: Bayes' Theorem with 3 parameters

Body:
The equation as written is not correct.

Here is a concrete example.  Let $D$ be the event that $X + Y + Z$ is even, where $X, Y, Z$ have the joint probability function
$$\begin{array}{c|c|c|c}
x & y & z & D & \Pr[(X,Y,Z) = (x,y,z)] \\
\hline
0 & 0 & 0 & \text{even} & 1/3 \\
0 & 0 & 1 & \text{odd} & 1/5 \\
0 & 1 & 0 & \text{odd} & 1/7 \\
0 & 1 & 1 & \text{even} & 1/8 \\
1 & 0 & 0 & \text{odd} & 1/10 \\
1 & 0 & 1 & \text{even} & 1/15 \\
1 & 1 & 0 & \text{even} & 1/32 \\
1 & 1 & 1 & \text{odd} & 1/1120 \\
\end{array}$$
There is nothing special about the fractions I chose except that they happen to be Egyptian.  Then we compute
$$\Pr[D] = 1/3 + 1/8 + 1/15 + 1/32 = 89/160,$$
$$\Pr[(X,Y,Z) = (1,0,1) \mid D] = \frac{1/15}{89/160} = 32/267,$$
$$\Pr[Y = 0] = 7/10,$$
$$\Pr[Z = 1] = 1319/3360,$$
$$\Pr[X = 1 \mid Y = 0] = \frac{1/10 + 1/15}{7/10} = 5/21,$$
$$\Pr[D \mid X = Z = 1] = \frac{1/15}{1/15 + 1/1120} = 224/227.$$
Hence the RHS is
$$\frac{(224/227)(5/21)(7/10)(1319/3360)}{89/160} = \frac{21104}{181827} \ne \frac{32}{267}.$$


问:
$2^{nd}$ countable implies separable
I was looking at this proof that $2^{nd}$ countable implies separable. Could someone clarify if my understanding is correct?

Suppose $X$ is $2^{nd}$ countable having countable basis $\beta=\{B_i|i \in \mathbb{N}\}$. For each $B_i \neq \varnothing$ let $b_i$ be an element of $B_i$ and let $Z$ be the set of all such $b_i$. Then $Z$ is countable. If $V \in \tau$ then $B_i \subset V$ for some $i$ and so $b_i \in V$ thus $V \cap Z \neq \varnothing$. So $Z$ is dense in $Z$ and $X$ is separable.

So in this proof, they basically take a countable basis, choose one point $b_i$ in each basis element in the collection, and form the set $Z$ of all of these individual points coming from exactly one basis element? Then $V \in \tau \implies B_i \subset V$ for some $i$. And this follows from the definition of a basis condition that for each open set $U \in \tau$ and $x \in U$, there is a $B \in \beta$ such that $x \in B  \subset U$? So each open set contains at least one of the $b_i's$ and so each open set intersects $Z$?

答:
Yes, this is in essence the argument. Minor nitpick: we need a non-empty open subset $U$ and so there is some $x\in U$ and only then can we say that there is a basic open $B_i$ such that $x \in B_i \subseteq U$ and then $b_i \in U\cap Z \neq \emptyset$ shows $U$ intersects $Z$ so $Z$ is dense.

That $Z$ is (at most) countable might need an appeal to a theorem too, plus the (unavoidable) use of Countable Choice is swept under the rug somewhat. At least acknowledge the fact.

Title: $2^{nd}$ countable implies separable


问:
Simplest proof that $\sum 1/p$ grows like $\log \log x$
I am seeking recommendations for a simple proof that $\sum_{p<x} 1/p$ grows like $\log (\log (x))$.

By simple I mean suitable for secondary school or 1st year university students - not those who are already trained in mathematics.

Ideally the proof would start from Euler's product formula for the Riemann zeta function, but this is not necessary.

答:
Without the knowledge of partial summation, we just use the traditional summation by parts:

Let $\pi(x)$ denote the number of prime numbers less than or equal to $x$, so for all $n\in\mathbb Z^{>0}$

$$
\pi(n)-\pi(n-1)=
\begin{cases}
1 & n\text{ is prime} \\
0 & \text{otherwise}
\end{cases}
$$

Applying this to our problem, we have

$$
\begin{aligned}
\sum_{p\le N}\frac1p
&=\sum_{n=1}^N{\pi(n)-\pi(n-1)\over n}={\pi(N)\over N}+\sum_{n=1}^{N-1}{\pi(n)\over n}-\sum_{n=1}^N{\pi(n-1)\over n} \\
&={\pi(N)\over N}+\sum_{n=2}^{N-1}{\pi(n)\over n}-\sum_{n=2}^{N-1}{\pi(n)\over n+1}={\pi(N)\over N}-\sum_{n=2}^{N-1}\pi(n)\left[{1\over n+1}-\frac1n\right] \\
&={\pi(N)\over N}+\sum_{n=2}^{N-1}\pi(n)\int_n^{n+1}{\mathrm dt\over t^2}={\pi(N)\over N}+\int_2^N{\pi(t)\over t^2}\mathrm dt
\end{aligned}
$$

Let $N=\lfloor x\rfloor$, then

$$
\int_N^x{\pi(t)\over t^2}\mathrm dt=\pi(N)\int_N^x{\mathrm dt\over t^2}={\pi(N)\over N}-{\pi(x)\over x}
$$

As a result, the above formula applies to all $x\in\mathbb R_{>0}$:

$$
\sum_{p\le x}\frac1p={\pi(x)\over x}+\int_2^x{\pi(t)\over t^2}\mathrm dt
$$

By the prime number theorem, we know that there exists a positive constant $K$ such that

$$
\left|\pi(x)-{x\over\log x}\right|\le{Kx\over\log^2x}
$$

As a result, using big O notation, the above thing becomes

$$
\begin{aligned}
\sum_{p\le x}\frac1p
&={1\over\log x}+\int_2^x{\mathrm dt\over t\log t}+\int_2^x\mathcal O\left(1\over t\log^2t\right)\mathrm dt+\mathcal O\left(1\over\log^2 x\right) \\
&=\log\log x+\mathcal O(1)
\end{aligned}
$$

That we can immediately conclude the error term is of $\mathcal O(1)$ is mainly because it is evident that the other terms would not exceed some constant.

P.S. Indeed a stronger result Mertens' theorem which states

$$
\sum_{p\le x}\frac1p=\log\log x+B_1+\mathcal O\left(1\over\log x\right)
$$

can be proven using elementary methods without prime number theorem, but it will require more technical details.


问:
Can a set have a complement in intuitionistic ZF?
Does IZF (ZF formulated in intuitionistic logic) prove that for any set $a$, $\{x: x \notin a \}$ does not exist?

答:
Consider the rank function $\operatorname{rank}$ defined by
$$\operatorname{rank} a = \bigcup\{\operatorname{rank}x + 1 \mid x\in a\}.$$

Then every element of $a$ has a rank smaller than $\operatorname{rank} a$, but you can show that any ordinal $\alpha$ has rank $\alpha$, so any ordinal greater than $\operatorname{rank} a$ is not a member of $a$.

Now I claim that the class $C=\{\beta\in\mathrm{Ord}\mid \alpha\in\beta\}$ is a proper class for any $\alpha$. Assume that $C$ is a set, then $\gamma:=\bigcup C$ is also an ordinal since $C$ is a set of ordinals. Then $\gamma\in C$, which is a contradiction since it implies $\gamma\in C\subseteq \gamma$.

Here are some remarks on my proof:

$\mathsf{CZF}^-$ (Constructive ZF without Subset Collection) suffices for the proof.

In fact, we can derive more on $C=\{\beta\in\mathrm{Ord}\mid \alpha\in\beta\}$.

Define a proper class $A$ is inexhaustible if for every subset $x\subseteq A$ of $A$ there is $y\in A$ such that $y\notin x$. (That is, $A$ is greater than every subset of it.) Classically, every proper class is inexhaustible, but $\mathsf{IZF}$ need not prove it.

We can see that if every set is subcountable (i.e., every set is an image of a subset of $\omega$), then $\mathcal{P}(1)$ is a proper class, but not an inexhaustible class. (It follows from that $\mathcal{P}(1)$ is a double complement of $2$ and no double complement of a set is inexhaustible.) However, the statement 'every set is subcountable' is not consistent with $\mathsf{IZF}$, although it is compatible with $\mathsf{CZF}$.

A slight modification of the previous proof shows $C$ is inexhaustible: let $x\subseteq C$, and take $\gamma:=\bigcup x$. Then we have $\gamma\in C$ but $\gamma\notin x$.


答:
The direction vector of both lines is $(1,1,1)$ and the unit direction vector

$\hat{v}$ = $\frac{1}{\sqrt3}(1,1,1)$.

Now take a point $A$ on line $1$ and $B$ on line $2$. The easiest from the equation of lines are $A(4, 5, 0)$ and $B(2, 5, 1)$.

Given both lines are parallel, the distance $d$ between the lines is then simply $|\vec{AB} \times \hat{v}|$ (this is because if the angles between vectors is $\theta$, cross product will have the magnitude $|AB \sin \theta|$ which is the perpendicular distance between the lines).

So $d = \frac{1}{\sqrt3}|(-2, 0, 1) \times (1, 1, 1)|$

Body:
What is the distance between the line with Cartesian equation $x - 4 = y - 5 = z$ and the line with Cartesian equation $x - 2 = y - 5 = z - 1$?

I am not sure how to do it. I think the lines are parallel with the same direction but I'm not sure.


问:
How to prove the equality of these sets?
I'm quite new to set theory (and to constructing proofs in general). To prove the equality of $A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$, would you just prove that  $A\cup(B\cap C)\subseteq (A\cup B)\cap(A\cup C)$ and that $(A\cup B)\cap (A\cup C)\subseteq A\cup(B\cap C)$? If so, how? Thanks in advance!

答:
prove the equality of $A\cup(B\cap C)=(A\cup B)\cap(A\cup C)$

Alternative approach

Prove that LHS and RHS match in the Venn Diagram analog to a truth table chart.

For general problems of this type, with complications, if all else fails, you can consider cases.  Here, there are three pertinent sets, $A,B,$ and $C$.  Therefore, there are 8 possibilities, re a specific element either is or is not in each of $A,B,$ and $C.$

Therefore, all you have to do is construct the appropriate chart.

$$\begin{array}
{|c|c|c|c|c|c|c|c|}
\hline
A & B & C & B\cap C & LHS & A\cup B & A\cup C & RHS \\
\hline
t & t & t & t & t & t & t & t \\
\hline
t & t & f & f & t & t & t & t \\
\hline
t & f & t & f & t & t & t & t \\
\hline
t & f & f & f & t & t & t & t \\
\hline
f & t & t & t & t & t & t & t \\
\hline
f & t & f & f & f & t & f & f \\
\hline
f & f & t & f & f & f & t & f \\
\hline
f & f & f & f & f & f & f & f \\
\hline
\end{array}$$


问:
Show that $V/W$ is a Banach space when $V$ is a Banach space and $W$ is a closed linear subspace.
A similar question and partially complete proof have been posted here: How to show that quotient space $X/Y$ is complete when $X$ is Banach space, and $Y$ is a closed subspace of $X$?, and in fact my question comes from this post.

Let $V$ be a normed space and $W\subset V$ be a linear subspace. I have proved the following:

Lemma 1: The followings are equivalent:

$V$ is a Banach space;

For every sequence $\{x_{k}\}\subset V$ such that $\sum_{k=1}^{\infty}\|x_{k}\|<\infty$ the series $\sum_{k=1}^{\infty}x_{k}$ is convergent, i.e. there exists $y\in V$ such that $$y=\lim_{M\rightarrow\infty}S_{M}\ \ \ \ S_{M}:=\sum_{k=1}^{M}x_{k}.$$

Basically, it shows that $V$ is a Banach space $\iff$ absolutely convergence implying convergence. And following the post above, I want to use this lemma, to prove the following statement:

If $V$ is a Banach space and $W$ is closed, then $(V/W, \|\ \cdot\ \|_{V/W})$ is a Banach space, where $$\|[x]\|_{V/W}:=\inf_{y\in W}\|x-y\|_{V},\ \ \text{and}\ \ [x]\ \ \text{is the equivalence class}.$$

I have proved the most of it:

Let $([x_{n}])_{n=1}^{\infty}$ be a sequence in $V/W$ such that $\sum_{n=1}^{\infty}\|[x_{n}]\|_{V/W}<\infty$.  We want to show that $\sum_{n=1}^{\infty}[x_{n}]$ converges to some $[x]\in V/W$, and then Lemma 1 concludes the proof.

Recall the definition of the quotient norm, for each $n\in\mathbb{Z}_{\geq 1}$, we have $\|[x_{n}]\|_{V/W}=\inf\{\|x_{n}-y\|_{V}, y\in W\}.$ It then follows from the definition of infimum that, for each $n\in\mathbb{Z}_{\geq 1}$,  there exists $y_{n}\in W$ so that $$\|x_{n}-y_{n}\|_{V}\leq \|[x_{n}]\|_{V/W}+\dfrac{1}{2^{n}}.$$

It then follows that $\sum_{n=1}^{\infty}\|x_{n}-y_{n}\|_{V}<\infty.$

But $(x_{n}-y_{n})$ is a sequence in $V$ and $V$ is a Banach space, by Lemma 1, we know that the series $\sum_{n=1}^{\infty}x_{n}-y_{n}$ converges to some $x\in V$.

But then how could I conclude that $\sum_{n=1}^{\infty}[x_{n}]$ converges to $[x]$ in $V/W$? I tried to evaluate $$\Bigg\|\sum_{n=1}^{\infty}[x_{n}]-[x]\Bigg\|_{V/W}=\inf_{y\in W}\Bigg\|\sum_{n=1}^{\infty}(x_{n}-x)-y\Bigg\|_{V}$$ But I don't know how to use the above result.

Thank you for any help!!

答:
You are almost done:

Recall that $|| z+W ||_{V/W} \leq ||z|| $ for all $z \in V$ (i prefer to use the notation $z+W$ for the coset), thus

$$||\Big (\sum_{k=1}^n (x_k-y_k) -x \Big) +W  ||_{V/W} \leq || \sum_{k=1}^n (x_k-y_k)-x||_V \xrightarrow{n \to \infty} 0.$$
Notice that the LHS is equal to
$$ ||     \Big(\sum_{k=1}^n  x_k  -x +W \Big)  -  \Big (\sum_{k=1}^n  y_k +W \Big )||_{V/W}  = ||  \sum_{k=1}^n  x_k    -  x +W ||_{V/W}, \  \text { since } \ \sum_{k=1}^n y_k \in W .$$
So, $ \sum_{k=1}^n x_k +W  \xrightarrow{n\to \infty} x$+W.


问:
Regularized Incomplete Beta Function
I try to solve this property of Regularized Incomplete Beta Function. How can you solve a statement :

$I_x (a,a) = 1 - \frac{1}{2} I_{4x(1-x)}(a,\frac{1}{2})  $ where $1/2 \leq x \leq 1$.

Some definitions :
$I_x (a,b) = \frac{B(x;a,b)}{B(a,b)}$ with B(x;a,b) is incomplete Beta function and B(a,b) is Beta function.

I tried to use substitution rule with z = 1 - x. But it doesn't go any further.

Could someone give me idea more please ?

答:
Using the definitions you wrote
$$\text{lhs}=\frac{B_x(a,a)}{B(a,a)}$$
$$\text{rhs}=1-\frac{B_{4 (1-x) x}\left(a,\frac{1}{2}\right)}{2 B\left(a,\frac{1}{2}\right)}$$ Now, for $ \frac 12 \leq x \leq 1$
$$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-3} (4 (1-x)-4 x) ((1-x) x)^{a-1}}{\sqrt{1-4 (1-x) x}
   B\left(a,\frac{1}{2}\right)}$$
Since $ \frac 12 \leq x \leq 1$, the last term simplify and write
$$\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$
$$\frac d {dx}\text{[lhs-rhs]}=\frac{(1-x)^{a-1} x^{a-1}}{B(a,a)}-\frac{2^{2 a-1} ((1-x) x)^{a-1}}{B\left(a,\frac{1}{2}\right)}$$ Divide by $(1-x)^{a-1} x^{a-1}$ to get
$$\frac{\frac d {dx}\text{[lhs-rhs]} } {(1-x)^{a-1} x^{a-1} }=\frac{1}{B(a,a)}-\frac{2^{2 a-1}}{B\left(a,\frac{1}{2}\right)}=0$$ by the properties of the beta function.

So, for $ \frac 12 \leq x \leq 1$, $\text{[lhs-rhs]}$ is a constant. For $x=1$ for example
$$\text{lhs}=\frac{B_1(a,a)}{B(a,a)}=1 \qquad \text{and} \qquad \text{rhs}=1-\frac{\Gamma \left(a+\frac{1}{2}\right) B_0\left(a,\frac{1}{2}\right)}{2
   \sqrt{\pi } \Gamma (a)}=1$$ Then, the identity.

Title: Regularized Incomplete Beta Function


问:
Does anything precede zeration in the hyperoperators
"Hyperoperator" family of operators is defined recursively like so:
$$
a\langle{0}\rangle{b} = a+b\\
a\langle{1}\rangle{b} = ab = a+a+a+a... \; \text{(w/ b a's)}\\
a\langle{2}\rangle{b} = a^{b}= aaaa... \; \text{(w/ b a's)}\\
...
$$
We can also extend the hyperoperators to "-1st operator" using the method described in this old post:
$$
a\langle{-1}\rangle{b} = a@b = \{a\neq{b}:\text{max}(a,b)+1,a=b:a+2\}\\
$$
This operator (called zeration) satisfies the property that $a+b = a\langle-1\rangle{a\langle-1\rangle}a...\text{(w/ b a's)}\\$.

My question is: Can we extend hyperoperators below zeration indefinitely while preserving the property I described above?

答:
Your "zeration" definition is unusual. It is usualally defined as increment by one operation:

$$a\langle0\rangle b=b+1$$
Its repeating $b\langle0\rangle b$ for $c$ times is addition $c+b$.
Increment is not commutative and does not depend on the first argument.

Moreover, Since $\Gamma(1)=\Gamma(2)=1$, addition and multiplication are both commutative, but since $\Gamma(0)$ goes to infinity, this means that zeration does not depend on the first argument at all. And this is true for any hyperoperation of negative integer order as well. Because the Gamma function has poles there.


问:
Sum of elements of $A \cap(B \cup C)=400 \times 274$. Find the closed form of $C$.
Question:

$A=\{x: x=3$ digit natural number $\}$

$B=\{x: x=9 k+2 ; k \in N\}$

${C}=\{{x}: {x}=9 {k}+I ; {k} \in {Z}\}, $ $0 \leq {I}<9$

Sum of elements of $A \cap(B \cup C)=400 \times 274 .$
Find the value of $I$

To be honest, I don't know how to start without making an assumption.

$$A \cap(B \cup C)=(A \cap B)\cup (A \cap C)$$
Assuming that $(A \cap B)$ and $(A \cap C)$ doesn't have any element common, one can find the sum of elements of $(A \cap B)$ and $(A \cap C)$ which would be equal to $400 \times 274 $.

Sum of elements of $(A \cap B)$:
$$\sum_{k=11}^{110}(9k+2)=54450+200=54650$$

Sum of elements of $(A \cap C)$:

Elements of $C$ are $$\cdots\cdots90+I,99+I,108+I\cdots990+I,999+I\cdots\cdots$$
Elements of $A$ are $$100,101,102\cdots 999$$
So possible elements of $(A \cap C)$ [By keeping $0 \leq {I}<9$ in mind] must be:$$99+I,108+I\cdots990+I\tag1\label{eq1}$$
So
$$\sum_{k=11}^{110}(9k+I)=54450+100I$$
Then after adding and solving $$I=5$$
And by luck $9k+2$ and $9k+5$ does not have any element common from $k=11$ to $k=110$.

How can this be solved without making that risky assumption? Any alternate.

Seeing range in $\eqref{eq1}$ and guessing the elements of $(A \cap C)$ worked well in this question but there is a single value of $I$ so that won't work every time.

答: It is not by luck that those two sets do not have any elements in common. Assume $9k+5=9\ell +2$, where $k,\ell$ are integers. Then doing some algebra, we see $9(\ell-k)=3$. Do you see the problem here? This should help you make the assumption a lot less risky.


问:
Mersenne primes before computers
On the Wikipedia page there is an ordered list of Mersenne primes and the dates they were discovered. The largest such primes and most recent discoveries were made with the help computers. But the largest Mersenne prime discovered without computer aid is $2^{127}-1$ by $\acute{\text{E}}\text{douard Lucas}$ in 1876, which has more than $38$ digits.

$2^{127}-1 = 170141183460469231731687303715884105727$.

How can one prove that such a large number is prime without the help of a computer?

Food for thought:

$2^{127}-1$ where $127= 128 -1=2^7-1$ where $7=8-1 =2^3-1$ where $3=4-1=2^2-1$.

答:
We sometimes underestimate the capability and perseverance that mathematicians had in those times, the age of no calculator, just calamo charta.

Edouard Lucas discovered that $M_{127}$ was prime. However, contrary to what may be popularly thought, he did not use the Lucas-Lehmer test!

Indeed, Lucas only knew that his test which he had devised in 1856, was sufficient, and that too only for certain values of the exponent. The Lucas-Lehmer test was not fully proved until Lehmer came along in 1930 and proved it.

Indeed, the result used by Lucas was one that he supposed true : it was not given a rigorous proof until 1913 by Carmichael, but was considered a true statement at the time of Lucas' computation.

Let $F_k$ denote the $k$th Fibonacci number, and let $n \in \mathbb N$. Then $n$ is a prime number if :

$n \equiv \pm 3 \pmod{10}$.

$n | F_{n+1}$.

$n \nmid F_m$ for all $1 \leq m \leq n$ which are divisors of $n+1$.

and similarly $n$ is a prime number if :

$n \equiv \pm 1 \pmod{10}$.

$n | F_{n-1}$.

$n \nmid F_m$ for all $1 \leq m \leq n-2$ which are divisors of $n-1$.

Since $2^{127}-1 \equiv 7 \pmod{10}$ (which does not require great computation capacity!) we can apply the first test. Indeed, note that $n+1 = 2^{127}$, so all(!) we have to verify is that $2^{127}-1$ divides $F_{2^{127}}$ but not $F_{2^m}$ for any $1\leq m \leq 127$.

Which Lucas did, using ideas that eventually led to the Lucas-Lehmer test : in other words, the Fibonacci and Lucas numbers satisfy the relation $F_{2n} = L_nF_n$ and $L_{2n} = L_n^2 -2 (-1)^n$ so one can get $(F_{2n},L_{2n})$ from $(F_{n},L_n)$! Furthermore, these respect modulo rules as well, and thus the computation is made simpler.

It also helped that Lucas had his own work to fall back on : he already had prime factorizations of the first sixty Fibonacci numbers , all his hard work.

Prior to Lucas' work, different methods were used to show primality (and non-primality!). Let's take $M_{31} = 2^{31}-1$. This was not shown prime by the Lucas-Lemher test. Indeed, it came with a startling observation by Euler :

If a prime $p$ divides $M_{31}$ then $p \equiv 1,63 \pmod{248}$.

I would like to know how he proved this particular assertion, but all I can say is that it comes down to primes dividing forms of certain kinds. For a simple exercise, try to prove that if a prime $p$ divides a number of the form $x^2 - 2y^2$ then $p \equiv \pm 1 \pmod{8}$.

Following this observation, Euler proceeded to roughly compute $\sqrt{M_{31}} \approx 46340$ and then used a prime table to find all primes which were congruent to $1$ or $63$ modulo $248$ and see if they divided $M_{31}$. No such prime was found. $M_{31}$ is prime!

An extension of this observation was made by Landry : Let $n \in \mathbb N$ and suppose it is known that every factor of $n$ is of the form $ax+b$ for some (known) integers $a,b$, then from $n = (ax_1+b)(ax_2+b)$  he saw that there existed $q,h,r$ integers such that :
$$
x_1x_2 = q-bh \\
x_1+x_2 = r+ah\\
h = \frac{q-x_1(r-x_1)}{ax_1+b}
$$

using which he was able to test for existence of $q,h,r$ for various values of $x_1$. He proved the existence of a $B>0$, such that it is enough to have no solutions to the first two equations for $x_1 > B$ and no solution to the third equation for $x_1 \leq B$ for $n$ to be prime. If there is a solution at some stage, then $n$ is not prime!

Landry used this to factorize $2^n \pm 1$ for all $n \leq 64$ completely (with the apparent exception of four of them  near $64$), and also obtained the largest prime number known then :
$$
\frac{2^{53}+1}{321} = 2805980762433
$$

They do work very very hard in computation, though. Lucas confirmed that $M_{67}$ was composite, but no prime factor was found till $1903$ when Frank Nelson Cole took , after "three years of Sundays", to the blackboard in a famous "no speech" seminar , proceeded to square $2$ repeatedly and find $$
2^{67}-1 = 147573952589676412927
$$

and on the other side of the board, proceeded to multiply : $$
\quad \quad \ \ \ \quad \quad \quad 193707721 \\
\quad \quad \quad\times 761838257287 \\
-----------\\
147573952589676412927 \\
-----------
$$

in high school fashion. He said nothing , and cue rapturous applause. And why not!

Title: Mersenne primes before computers


问:
Locally injective graph homomorphism into a tree
$\DeclareMathOperator{\star}{\mathcal{star}}$

Let $X$ be a connected graph, $T$ be a tree and $\phi: X \rightarrow T$ be a locally injective graph homomorphism. Prove that $\phi$ is injective!

Here, a graph means a pair of sets $X = (X^0,X^1)$ with three maps:

$\alpha: X^1 \rightarrow X^0$

$\omega: X^1 \rightarrow X^0$

$^\bar{}: X^1 \rightarrow X^1$

such that for any $e \in X^1$ the following holds

$\overline{\overline{e}} = e \neq \overline{e}$

$\alpha(e) = \omega(\overline{e})$

A graph homomorphism $\phi:X \rightarrow Y$ is a map from $X^0 \cup X^1$ to $Y^0 \cup Y^1$ such that

$\forall v \in X^0 \ \phi(v) \in Y^0$

$\forall e \in X^1 \ \phi(e) \in Y^1$

$\forall e \in X^1 \ \phi(\alpha(e)) = \alpha(\phi(e))$

$\forall e \in X^1 \ \phi(\omega(e)) = \omega(\phi(e))$

$\forall e \in X^1 \ \phi(\overline{e}) = \overline{\phi(e)}$

A graph homomorphism $\phi: X \rightarrow Y$ is locally injective if $\phi \restriction \star (x)$ is injetive $\forall x \in X^0$, where $\star(x) := \{e \in X^1 \ | \ \alpha(e) = x \}$.

A path between two vertices $v,\tilde{v} \in X^0$ is a sequence of edges $(e_1,e_2, \ldots, e_n) \in (X^1)^n$ such that

$\alpha(e_1) = v$, $\omega(e_n) = \tilde{v}$

$\forall 1 \le i < n \ \omega(e_i) = \alpha(e_{i+1})$

A graph $X$ is connected if for any pair of vertices $(v,\tilde{v}) \in (X^0)^2$ there is a path from $v$ to $\tilde{v}$.

I've tried to prove it via induction,but somehow seem to miss the property leading to my desired contradiction if I assume, $\phi$ was not injective. What goes wrong if I look at a path from $v$ to $\tilde{v}$ with $\phi(v) = \phi(\tilde{v})$ where either $v \neq \tilde{v}$ or the related edges are not equal?

答: A cycle is a path from $v$ to $v$. Trees are connected graphs without (simple) cycles. You have built a cycle in $T$ but it is not necessarily simple. Try to get a simple (i.e. with all intermediate vertices different) cycle starting from the cycle you have.</p>


问:
Geometric series word problem: Number of games played
If a tournament consists of 64 teams playing sudden death games until 1 team is declared the winner. How many games, in total, are played during this tournament?

My approach:
Teams: 64, 63, 62, ....., 1

Two teams play 1 game, therefore, all together 32 games. But where is the GP???

答:
This is the old tennis ball question.  How many tennis balls are needed if the loser takes the tennis ball home as a consolation prize.

Each match eliminates 1 team.  Of the 64 teams, 63 need to be eliminated so that a winning team can be declared.

The actual answer for $N$ teams will be $(N-1)$ matches, where if a team is given a bye from one round to the next, that does not count as a match played.

Addendum

Responding to Simran's Comment/Question

$1 + 2 + 4 + 8 + \cdots + 2^{(n-1)} = 2^n - 1.$


问:
Scalar Multiple times $n \times n$ Identity Matrix Proof
How to prove
$\det(kA) = k^{n}$ where $A$ is the $n \times n$ Identity Matrix using induction.

I started off using the base case, forming my inductive hypothesis and trying to factor out $k^{n}$ times $k$ for the $(n + 1)$-th case by using the definition of the determinant but I am completely lost on the technicalities and notation.

答:
The rule to calculate the determinant of a matrix by cofactor expansion along the first row:

$$\begin{vmatrix}
a_{11} & a_{12} & \cdots& a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \ddots &\vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{vmatrix}$$
$$= (-1)^{1+1}\cdot a_{11}\cdot |A_{11}| + (-1)^{1+2}\cdot a_{12}\cdot |A_{12}| + \cdots + (-1)^{1+n}\cdot a_{1n} \cdot |A_{1n}|$$

where $|A_{ij}|$ is the determinant of the matrix by crossing out the $i$-th row and $j$-th column. For example $|A_{11}| = \text{(crossing out the first row and first column)}=$

$$\begin{vmatrix}
 a_{22} & a_{23} &\cdots & a_{2n}\\
 a_{32} & a_{33} &\cdots & a_{3n}\\
 \vdots & \vdots & \ddots &\vdots \\
 a_{n2} & a_{n3}& \cdots & a_{nn}
\end{vmatrix}_{(n-1)\times(n-1)}$$

But now given the matrix:

$$\begin{vmatrix}
k & 0 & \cdots& 0\\
0 &k & \cdots & 0\\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & k
\end{vmatrix}_{n\times n}$$

This make our life a whole lot easier because if you notice in the above formula for the determinant by cofactor expansion along the first row, the coefficients depend on the matrix entries, and in this case $a_{12},...,a_{1n} =0$.

So

$\begin{vmatrix}
k & 0 & \cdots& 0\\
0 &k & \cdots & 0\\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & k
\end{vmatrix}_{n\times n} =$
$ k\cdot \begin{vmatrix}
k & 0 & \cdots& 0\\
0 &k & \cdots & 0\\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & k
\end{vmatrix}_{(n-1)\times (n-1)} =$

$ k^2\cdot \begin{vmatrix}
k & 0 & \cdots& 0\\
0 &k & \cdots & 0\\
\vdots & \vdots & \ddots &\vdots \\
0 & 0 & \cdots & k
\end{vmatrix}_{(n-2)\times (n-2)} =$
$\cdots =k^{n-2}\cdot \begin{vmatrix}
k &0\\
0 & k
\end{vmatrix} = k^{n-2}\cdot (k^2 -0) = k^n$

Basically once you get comfortable with determinants you'll notice that for diagonal matrices (matrices with nonzero entries only in the diagonal, $a_{11},a_{22},...,a_{nn} \not=0$, the rest all zero), the determinant is just the diagonal entries multiplied, in this case they are all $k$ and the matrix is $n\times n$ so there are $n$ diagonal entries, hence $k^n$.


问:
Convergence of metric sequence to 0

Let $(x_n)$ be a sequence in $M$, and $l ∈ M$. Prove that $(x_n)$ converges to $l$ if and only if the sequence of real numbers $(d(x_n,l))$ converges to $0$.

I'm able to prove right to left but not left to right. I've tried using contradiction but I'm not sure how to set up an appropriate radius

Side question: How does one include MathJax formatting in one's question?

答: The assertion $\lim_{n\to\infty}x_n=l$ means$$(\forall\varepsilon>0)(\exists N\in\Bbb N):n\geqslant N\implies d(x_n,l)<\varepsilon\tag1$$and the assertion $\lim_{n\to\infty}d(x_n,l)=0$ means$$(\forall\varepsilon>0)(\exists N\in\Bbb N):n\geqslant N\implies|d(x_n,l)-0|<\varepsilon.\tag2$$Since $|d(x_n,l)-0|=d(x_n,l)$, $(1)$ and $(2)$ are the same assertion.

Title: Convergence of metric sequence to 0


答:
Left to right:

Given that $x_n \to l$ and therefore $\forall \epsilon \gt 0, \exists N_\epsilon \in \mathbb N: n\ge N_\epsilon\implies d(x_n,l)\lt \epsilon\implies |d(x_n,l)-0|\lt \epsilon $

Title: Convergence of metric sequence to 0

Body:
Let $(x_n)$ be a sequence in $M$, and $l ∈ M$. Prove that $(x_n)$ converges to $l$ if and only if the sequence of real numbers $(d(x_n,l))$ converges to $0$.

I'm able to prove right to left but not left to right. I've tried using contradiction but I'm not sure how to set up an appropriate radius

Side question: How does one include MathJax formatting in one's question?


问:
Why does $(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1,\;k\ge2\;\implies\;\text{isPrime}(k)$
Let $k$ be a integer such that $k\ge2$

Why does
$$(k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor = 1$$
only when $k$ is prime?

Example:

$$\pi(n) = \sum _{k=4}^n \left((k-2)!-k \left\lfloor \frac{k!}{(k-1) k^2}\right\rfloor \right),\;n\ge4$$

where $k=4,$ since:

$$\pi(4)\quad=\quad(4-2)!-4 \left\lfloor \frac{4!}{(4-1) 4^2}\right\rfloor = 2$$

I've tried to evaluate it in different forms, and I am probably just overlooking something obvious; So if anyone has any information in regard to this, please share.

答:
If $k$ is not a prime, then the equation is $0$ modulo $k$. (We have $(k-2)! \equiv 0 \pmod{k}$.)

Conversely if $k$ is a prime, then by Wilson's theorem $(k-2)! \equiv 1 \pmod{k}$.

Edit: (To complete the argument.) If $k$ is a prime, then $(k-2)! = 1 + l k$ for some $l \ge 0$. Equivalently $l = \frac{(k-2)! - 1}{k}$ and we must show that $l = \left\lfloor \frac{k!}{(k-1)k^2} \right\rfloor$. But this is true since

$$\left\lfloor \frac{k!}{(k-1)k^2} \right\rfloor = \left\lfloor \frac{(k-2)!}{k} \right\rfloor$$
and the greatest number below $(k-2)!$ that is divisible by $k$ is $(k-2)! - 1$.


问:
Basic question on Bregman divergences and strong convexity
I have seen the following claim in several research papers:

Let $p \in (1,2]$, and consider the function $\Phi(x)=\frac{1}{2}\|x\|^2_p$ (where $x\in\mathbb{R}^d$). Then the corresponding Bregman divergence $D_\Phi$ is $(p-1)$-strongly convex w.r.t the $\ell_p$ norm: $$D_\Phi(y,x) \geq \frac{p-1}{2} \|x-y\|^2_p$$ for all $x,y\in\mathbb{R}^d$.

Typically, no other detail is given, besides "cf. Ball et al. (1994)." I cannot see how that follows immediately from that paper, however. Is it obvious?

答:
I believe this is just the fact that for continuous convex functions midpoint strong convexity is equivalent to strong convexity.

For instance, suppose we define a function $g \colon \mathbf{R}^n \to \mathbf{R}$to be $\lambda$-midpoint strongly convex w.r.t. $\|\cdot\|$ provided that:
$$
(\star) \qquad g\Big(\frac{u + v}{2}\Big) \leq \frac{g(u) + g(v)}{2}
 - \frac{\lambda}{8} \|u - v\|^2, \quad \mbox{for all}~u, v \in \mathbf{R}^n.$$
This definition is clearly equivalent to:
$$
\frac{g(x + y) + g(x - y)}{2} \geq g(x) + \frac{\lambda}{2} \|y\|^2, \quad
\mbox{for all}~x, y \in \mathbf{R}^n.
$$
Above we simply use the homogeneity of the norm. Proposition 3 in this jargon says that the function $f_p \colon u \mapsto \tfrac{1}{2} \|u\|_p^2$ is $(p-1)$-midpoint strongly convex w.r.t. $\|\cdot\|_p$.

Therefore, to establish your claim, it suffices to verify that $(\star)$ (i.e., midpoint strong convexity) is equivalent to strong convexity.


问:
Coercivity of inverse operator
Suppose that $A: \mathbb{R}^n \to \mathbb{R}^{n}$ satisfies:
$$
m \|u\|^{2}_{2} \leq \langle u,Au\rangle \leq M\|u\|_{2}^{2} \text{ for all } u \in \mathbb{R}^{n}
$$
but is not necessarily symmetric. By coercivity we have that $A^{-1}$ exists and is bounded:
$$
\|A^{-1}u\|_{2} \leq \frac{1}{m}\|u\|_2
$$
(I think this is just a simple version of the Lax-Milgram theorem). Can one say anything about the coercivity of $A^{-1}$? Specifically, if $A$ were symmetric and diagonalizable then one could say that:
$$
\|A^{-1}u\|_{2} \geq \frac{1}{M}\|u\|_2
$$
Is this true when $A$ is non-symmetric?

答: Just replace $u$ by $A^{-1}v$ in the first chain of inequalities. Then use that $A^{-T}A^{-1}$ is positive definite.

Title: Coercivity of inverse operator


答: I'm assuming $(M,d)$ is a metric space and will use the continuity of $d$. Since the sequence $\{(x_n, l)\}_n$ in $M^2$ converges to $(l,l)$, so $d(x_n,l)$ will converge  to $d(l,l)=0$.

Title: Convergence of metric sequence to 0


问:
Given $x_{0}$ and $x_{1}$, is there a formula to find $x_n = d\,(2\,x_{n-1} - x_{n-2})$ at a specified $n$?
The question covers the main problem I want to solve: $d$ can be any scalar constant (ideally in between 0 and 1, if that makes the problem easier), as well as $x_{0}$ and $x_{1}$. I tried figuring this out on my own, but if the answer is there it is eluding me (this uses an almost identical sequence, but I have no idea what difference equations are to know if that's relevant). This was inspired after learning about the Hugo Elias algorithm for efficiently generating water ripples, if that helps at all. I would be grateful for any help regarding this problem!

As a side note, I may not know the correct terms for everything, so please feel free to let me know or modify this question with more accurate terminology!

答:
Yes, there is a closed form for $x_n$.

Consider the sequence $\mathcal F(a,b,f_0,f_1)=(f_n)_{n\ge0}$ with $$f_{n+2}=af_{n+1}+bf_n\qquad n\ge0,\tag1$$
and $a,b,f_0,f_1$ known constants. Then consider the generating function
$$f(x)=\sum_{n\ge0}f_nx^n.$$
Multiply $(1)$ by $x^{n+2}$ and sum over $n\ge0$:
$$\sum_{n\ge2}f_nx^n=ax\sum_{n\ge1}f_nx^n+bx^2\sum_{n\ge0}f_nx^n,$$
which is
$$f(x)-f_1x-f_0=ax(f(x)-f_0)+bx^2f(x).$$
Solving for $f$, we get
$$f(x)=\frac{(f_1-af_0)x+f_0}{1-ax-bx^2}.$$
Then we can write $$\frac{(f_1-af_0)x+f_0}{1-ax-bx^2}=\frac{u_1}{1-v_1x}+\frac{u_2}{1-v_2x},$$
for the constants $u_i,v_i$ such that
$$\begin{align}
u_1v_2+u_2v_1&=af_0-f_1,\qquad &u_1+u_2&=f_0,\\
v_1+v_2&=a,\qquad &v_1v_2&=-b.
\end{align}$$
This system of equations has the solution
$$\begin{bmatrix}
v_1 \\
v_2 \\
u_1 \\
u_2 \\
\end{bmatrix}
=
\begin{bmatrix}
\tfrac{a+\sqrt{D}}{2}\\
\tfrac{a-\sqrt{D}}{2}\\
\tfrac{1}{\sqrt{D}}\left(f_1+\tfrac{-a+\sqrt{D}}{2}f_0\right)\\
\tfrac{1}{\sqrt{D}}\left(-f_1+\tfrac{a+\sqrt{D}}{2}f_0\right)
\end{bmatrix},\qquad D=a^2+4b.\tag2$$
We have
$$\sum_{n\ge0}f_nx^n=\frac{(f_1-af_0)x+f_0}{1-ax-bx^2}=\frac{u_1}{1-v_1x}+\frac{u_2}{1-v_2x}=u_1\sum_{n\ge0}(v_1x)^n+u_2\sum_{n\ge0}(v_2x)^n,$$
that is
$$\sum_{n\ge0}f_nx^n=\sum_{n\ge0}(u_1v_1^n+u_2v_2^n)x^n,$$
so that $$f_n=u_1v_1^n+u_2v_2^n.\tag3$$

Since you are given $x_0,x_1$, we have $x_n=f_n\in\mathcal F(2d,-d,x_0,x_1)$.
$$\begin{bmatrix}
v_1 \\
v_2 \\
u_1 \\
u_2 \\
\end{bmatrix}
=
\begin{bmatrix}
d+\sqrt{d(d-1)}\\
d-\sqrt{d(d-1)}\\
\tfrac{1}{2\sqrt{d(d-1)}}\left(x_1+(-d+\sqrt{d(d-1)})x_0\right)\\
\tfrac{1}{2\sqrt{d(d-1)}}\left(-x_1+(d+\sqrt{d(d-1)})x_0\right)
\end{bmatrix},$$
so that
$$\begin{align}
x_n=\tfrac{1}{2\sqrt{d(d-1)}}\Bigg\{&\left(x_1+\left(-d+\sqrt{d(d-1)}\right)x_0\right)\left(d+\sqrt{d(d-1)}\right)^n\\
&+\left(-x_1+\left(d+\sqrt{d(d-1)}\right)x_0\right)\left(d-\sqrt{d(d-1)}\right)^n\Bigg\}.
\end{align}$$


问:
Modular arithmetic $a=bq+r$
How do I show that if $r$ is the remainder when $a$ is divided by $b$ (e.g. $a = bq + r$) then $a\equiv r (\mod b)$?

I assume I'm supposed to use the division algorithm and/or the quotient remainder theorem.

Thank you.

答:
Definition of congruence modulo $b$:

$a\equiv r\mod b$ iff $b\mid a-r$ (divisibility) iff there exists $q$ such that $bq = a-r$.

Now if $a=bq+r$, then $b\mid a-r$ and so $a\equiv r\mod b$.

Title: Modular arithmetic $a=bq+r$


问:
Triangle with integral side lengths and $\angle A=3\angle B$

$ABC$ is a triangle with integral side lengths. Given that $\angle A=3\angle B$, find the minimum possible perimeter of $ABC$.

I got this problem from an old book (which did not provide even a hint). I can think of some approaches, but all of them result in complicated Diophantine equations that would not be solvable without the help of a computer. Any suggestions?

答:
Giving it a try using elementary techniques.

In $\triangle ABC$, let $B=\theta$, $A=3\theta$, $C=\pi-4\theta$. By sine-rule
$$\frac{a}{\sin 3\theta}=\frac{b}{\sin \theta}=\frac{c}{\sin 4\theta}$$

so that $$a=b(3-4\sin^2\theta) \quad , \quad c=b(4\cos \theta \cos 2\theta)$$

On further simplification,
$$a=b(4\cos^2\theta-1) \quad , \quad c=b \cdot 4\cos \theta (2\cos^2 \theta-1)$$

$c>0$ places bound on $\cos \theta$ : $\cos \theta > 1/\sqrt{2} \Rightarrow \theta < 45^{\circ}$ which makes sense since $45^{\circ}+3\cdot 45^{\circ}=180^{\circ}$

We may choose any suitable value of $\cos \theta > 1/\sqrt{2} \approx 0.7071$.

Taking $\cos \theta = 3/4=0.75$, yields $a/b=5/4$, $c/b=3/8$. So one class of triangles would be multiples of
$$(a,b,c)=(10,8,3)$$

Taking $\cos \theta = 5/6$, yields $a/b=16/9$, $c/b=35/27$. So another class of triangles would be multiples of
$$(a,b,c)=(48,27,35)$$

and so on.


问:
What the range of $r$ and $\theta$?
If $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac{y}{x}$ for $0\leq x\leq 1$ and $0\leq y\leq 1$, what the range of $r$ and $\theta$?

I think since $0\leq x\leq 1$ and $0\leq y\leq 1$ then $0\leq r\leq 1$.

since $0\leq x\leq 1$ and $0\leq y\leq 1$, then $\frac{y}{x}>0$ so $0\leq \theta\leq \frac{\pi}{2}$.

Does this answer correct? (Or maybe $r$ depend on $\theta$?)

答: Draw a picture. You will  see that $r$ goes from $0$ to $\frac 1 {\cos \theta}$ for $0 \leq \theta \leq \frac {\pi}4$ and  $r$ goes from $0$ to $\frac 1 {\sin \theta}$ for $\frac {\pi} 4 \leq \theta \leq \frac {\pi} 2$.

Title: What the range of $r$ and $\theta$?


问:
Show that $\{0,1\}^\mathbb{R}$ is separable
Show that $\{0,1\}^\mathbb{R}$ is separable. I was trying to think about how to show this. First I considered the subset $S=\{f \in\{0,1\}^\mathbb{R}:f(a)=0,a \in \mathbb{Q}^{c}\}$. However I was told that although this set is dense it is uncountable being of cardinality $2^{|\mathbb{N}|}$. I was given a hint to use the bijective correspondence with elements in this set, and elements in the power set of $\mathbb{R}$ and consider the collection of all finite subsets of $\mathbb{R}$. So in this case I should consider the subset $S=\{f\in\{0,1\}^\mathbb{R}:f(a)=0,\text{for all but finitely many }a\}$. This set is supposedly countable and dense.

Now I am rethinking this. I see the set of all finite subsets of $\mathbb{N}$ is countable, so maybe I was told to consider the set $S=\{f \in\{0,1\}^\mathbb{R}:f(a)=0 \ \text{for all but finitely many}\ a \in \mathbb{N}\}$? If I do this, to prove this subset is countable, would I just need to prove the equivalent statement that the set of all finite subsets of $\mathbb{N}$ is countable?

Once the set is proved to be countable, the next thing to prove is that the closure of the set is equal to $\{0,1\}^\mathbb{R}$. That the set is dense. I will try to give this an attempt.

Let $x \in \{0,1\}^{\mathbb{R}}$. Let $\prod U_{\alpha}$ be an open set containing $x$. Then $U_\alpha=\{0,1\}$ for all but finitely many $\alpha$. Now I am struggling to figure out what to do if $f(a)=1$ for only finitely many irrationals or non whole numbers. Maybe I should go back to considering the second set? The set of functions that are zero at each coordinate but finitely many? If this is the case then again  Let $x \in \{0,1\}^{\mathbb{R}}$. Let $\prod U_{\alpha}$ be any open set containing $x$. The $U_\alpha=\{0,1\}$ for all but finitely many alpha. So certainly it contains a point in $S$ since it contains any function $0$ everywhere but finitely many coordinates. What would be a better way to show this?Is the set of all finite subsets of $\mathbb{R}$ actually countable?

Unfortunately I am hitting a brick wall with how to actually come up with all of the math I need to show this set is dense. How should I try to show any open set containing $x$ must contain some elements in $S$?

答:
Your first set $S_1 = \{f \in \{0,1\}^{\Bbb R}\mid \forall \alpha \in \Bbb Q^c: f(\alpha)=0\}$ is uncountable (in the power-set representation it is the set of all subsets of $\Bbb Q$, really) and not dense, but closed. The argument is essentially the same as I gave in my earlier reply: if $A \notin S_1$ (in the powerset view, so $A \subseteq \Bbb R$) then $\exists \alpha \in \Bbb Q^c \cap A$ and then $[\{\alpha\};\emptyset]$ is a basic neighbourhood of $A$ that misses $S_1$. Here I use basic open sets of $P(\Bbb R)$ of the form $$[F;G]=\{A \subseteq \Bbb R\mid F \subseteq A, A \cap G = \emptyset\}$$ where $F,G$ are disjoint (WLOG) finite subsets of $\Bbb R$.

Your second attempt, $S_2 = \{A \subseteq \Bbb R\mid A \text{ finite }\}$ ( I only use the powerset view in the remainder) is a better attempt: it is at least dense, but still uncountable as already the subset $\{\{x\}: x \in \Bbb R\}$ is uncountable (size continuum). Denseness is clear: if $[F;G]$ is basic open as above, then $F \in S_2 \cap [F;G]$, so every basic open set intersects $S_2$.

To show that we indeed have a countable dense subset: let $D$ be the set of finite disjoint unions of open intervals of the form $(q,q’)$ where $q,q’ \in \Bbb Q, q < q’$. Elementary set theory tells us that $D$ is countable indeed and if $[F;G]$ is basic open, define $H = F \cup G = \{h_0,h_1,\ldots, h_n\}$ (in increasing order) and pick $q_0 < q_1 < \ldots q_n < q_{n+1}$, all rationals, such that $q_0 < h_0 < q_1 < h_1 < q_2 < \ldots < h_n < q_{n+1}$ and define $A$ to be the union of the $(q_i, q_{i+1})$ that contain all the $h_i$ that lie in $F$. It’s then clear that $A \in D \cap [F;G]$ and denseness has been shown.


问:
Elasticity function of a given demand function
$$
q=D(x)=\frac{300}{x}\ \ \ \ \ \ \ \  \text { (given ) }
$$
as we know
$$
\begin{array}{l}
E(x)=\dfrac{-\frac{d q}{d x}}{\frac{q}{x}} \\
\dfrac{d q}{d x}=\dfrac{-300}{x^{2}}
\end{array}
$$
$$
E(x)=-\left(-\frac{300}{x^{2}} \times \frac{x \times x}{300}\right)=1
$$

The demand function is given
$D(x) = 300/x$
See where am i going wrong
I used
$E(x)= ( -{\rm d}q/{\rm d}x)/(q/x)$
${\rm d}q/{\rm d}x= -300/x^2$
$E(x)= 1$

Body:
The formula is $$\epsilon=|\frac{dQ/Q}{dP/P}|=|\frac{dQ}{dP}\frac{P}{Q}|$$

Now

$$Q=300/P$$
$$\frac{dQ}{dP}=-300/P^2$$
$$\epsilon=|\frac{dQ}{dP}\frac{P}{Q}|=|-300/P^2\times P/Q|=|-300/PQ|=|-1|=1$$


问:
Finding zeros of a function
I need to find when $f(x)=0$, where:
\begin{equation}
f(x)=1-\dfrac{k}{x}-\dfrac{k}{3}\dfrac{e^{-ax}}{x}+\dfrac{4k}{3}\dfrac{e^{-bx}}{x}
\end{equation}
Here, $k$, $a$ and $b$ are positive constants. I think this problem analytically is impossible, but I really don't know. I've tried to use Mathematica but it doesn't give any answer and I'm pretty desperate.

Thank you so much.

答:
A priori, I suppose that $x=0$ cannot be an interesting zero  of function
$$f(x)=1-\dfrac{k}{x}-\dfrac{k}{3}\dfrac{e^{-ax}}{x}+\dfrac{4k}{3}\dfrac{e^{-bx}}{x}$$ So, consider
$$g(x)=x\,f(x)=x-k-\dfrac{k}{3}e^{-ax}+\dfrac{4k}{3}e^{-bx}$$

We have
$$g(0)=0 \qquad g'(0)=1+\frac{1}{3} k (a-4 b)\qquad g''(0)=-\frac{1}{3} k \left(a^2-4 b^2\right)$$ So, if $g'(0)<0$ and $g''(0) >0$, there is a chance to find something.

Expanding $g(x)$ as a Taylor series around $x=0$, we have
$$g(x)=x \left(\frac{1}{3} k (a-4 b)+1\right)-\frac{1}{6} x^2 \left(k \left(a^2-4
   b^2\right)\right)+\frac{1}{18} k x^3 \left(a^3-4 b^3\right)-\frac{1}{72} x^4
   \left(k \left(a^4-4 b^4\right)\right)+O\left(x^5\right)$$ Dividing by $x$ and using series reversion
$$x=t+\frac{\left(a^3-4 b^3\right)}{3 \left(a^2-4 b^2\right)}t^2 +\frac{ \left(5
   a^6+12 a^4 b^2-64 a^3 b^3+12 a^2 b^4+80 b^6\right)}{36 \left(a^2-4
   b^2\right)^2}t^3+O\left(t^{4}\right)$$ where $t=\frac{2k( a -4 b) +6}{(a^2 -4 b^2) k}$

For the case $a=1,b=2,k=1$ considered by @GEdgar, the above gives as an estimate
$$x=\frac{5605384}{6834375}=0.820175$$ which is not fantastic. But, starting with this estimate, Newton method will work like a charm (one single iteration for $8$ exact significant figures !).
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 0.82017507 \\
 1 & 0.92826114
\end{array}
\right)$$

Edit

Just for the fun of it, I built the series expansion up to $O\left(x^{15}\right)$ (this is just ridiculous !) and used series reversion (I shall not report any formula here since they are monsters).

For the worked case, the estimate is
$$x=\frac{116953055942942882914859816106777623661064}{126941147255319208068560063838958740234375}$$ which is $0.921317$.

Title: Finding zeros of a function


问:
Discrete metric space with more than one element not connected
Prove that a metric space with the discrete metric and more than one element is not connected.

Attempt:Let $X$ be a discrete metric space with more than one element. If $A$ is any nonempty proper subset of $X$, $A$ and $X-A$ is a separation of $X$, since all subsets of discrete metric spaces are open.

I realize this proof is the same proof that spaces with the discrete topology are not connected, is it appropriate to prove this result for metric spaces like this?

答:
Connectedness is a topological property, it does not really depend on the metric, just on its induced topology (defined in terms of open and closed sets etc.). The discrete metric induces the discrete topology.

So I'd say it's totally appropriate. Just let it stand.


问:
Does a curve $C \subset X$ that contains a singularity of $X$ pick up extra multiplicities?
Suppose $X$ is a variety over a algebraically closed field, and $C \subset X$ is a curve, that contains a singularity of $x$ of $X$. If $D$ is a Cartierdivisor, whoose support $|D|$ contains $x$, is it true that $D|_C$ vanishes at $x$ with multiplicity $\geq 2$?

答:
No. For example, let $X \subset \mathbb{P}^3$ be a quadratic cone, let $C \subset X$ be its ruling, and let $D = H \cap X$, where $H \subset \mathbb{P}^3$ is a hyperplane passing through the vertex of the cone and transverse to $C$. Then
$$
D \vert_C = (H \vert_X) \vert_C = H\vert_C
$$
has multiplicity 1.


问:
Borel/ Lebesgue on $\chi_D$ for $D = \{(x,x):x \in E \}$
Let $E \subset [0,1]$.

Define $D = \{(x,x):x \in E \}$.

Let $f= \chi_D$.

a) Are $f_x, f^y$ Borel for all $x,y \in [0,1]$?

b) Is it true that $f$ is Borel iff $E$ is Borel?

c) If we have both $\mu$ and $\nu$ being the Lebesgue measure, is it true that $f$ is $\mu \times \nu $- measurable iff $E$ is Lebesgue measurable?

My thinking:

a)Yes, since $f_x, f^y$ can only be $0$ or $1$.

b) Yes, since $f$ is Borel iff $D$ is Borel.

c) Yes, since $f$ is Lebesgue measurable iff D is.

Could someone let me know if I have the above correct? Thank you!

Thanks!

答:
Your answers are not corrcet.

For fixed $x \in E$ we have $f_x(y)=1$ if $y=x$ and $0$ otherwise.

For fixed $x \notin E$ we have $f_x(y)=0$ for all $y$.

This makes $f_x$ Borel, hence also Lebesgue measurable always. [No condition on $E$].

Similarly, $f^{y}$ is Borel, hence also Lebesgue measurable always.

Now can write down the answers to all parts of the question.


问:
Let p be a prime and k a positive integer such that $a^k$mod p = a mod p for all integers a. Prove that p - 1 divides k - 1.

Let p be a prime and k a positive integer such that $a^k$mod p = a mod
  p for all integers a. Prove that p - 1 divides k - 1.

I think I need to use Fermat's Little Theorem, and I can get  $a^k$mod p =  $a^p$mod p, and I think it follows that  $a^{k-1}$mod p = $a^{p-1}$mod p = 1, but from there I'm stumped.

I think it follows that $a^{k-1}$ = m$a^{p-1}$, but why couldn't m be a fraction, i.e. k - 1 divides p - 1 but not the other way around?

答:
Another proof would be to consider the group $U(p)$ of multiplicative units in $\mathbb Z / p \mathbb Z$.  For $p$ prime the order of $U(p)$ is $p-1$.  Since $p$ is prime, $U(p)$ is cyclic (see Keith Conrad's exposition) and so it has elements of all orders $d$ dividing $p-1$.

If $a$ is such an element of order $d$ then our hypothesis is that $a^{k-1} \equiv 1 \mod p$, and so $|a|=d$ divides $k-1$.  So any divisor of $p-1$ is a divisor of $k-1$, hence $p-1$ divides $k-1$.

This avoids Galois theory, unless of course you used Galois theory to prove that $U(p)$ is cyclic.  But that can be avoided because all we really needed was that $U(p)$ has an element of order $q$ for each prime $q$ dividing $p-1$, which is guaranteed by Cauchy's theorem.


问:
Converse of a theorem regarding vector potentials
It is easily seen by substitution that if
$\vec {B} = \text {curl} \vec {A}$ , then $\text {div} \vec {B}=0$. Can we prove the converse as well?

Let $\vec {B}=(B_x,B_y,B_z)$. Then,

$$
\frac {\partial B_x}{\partial x}+\frac {\partial B_y}{\partial y}+\frac {\partial B_z}{\partial z}=0
$$

This equation is valid for,
$$
\frac {\partial B_x}{\partial x}=Z-Y, \frac {\partial B_y}{\partial y}=X-Z, \frac {\partial B_z}{\partial z}=Y-X,\ \ \ \ \ \ \ \ \ \ \ (1)
$$

Where $X,Y,Z$ are arbitrary quantities whatsoever. Now, $$
X=\frac {\partial B_y}{\partial y}+Z=-\frac {\partial B_z}{\partial z}+Y
$$
This implies that $X$ can be expressed as a $2^{\text {nd}}$ order partial derivative of some quantity, say $A_x$, since the denominator has both $\partial y$ and $\partial z$.
Hence,
$$
X= \frac {\partial ^2 A_x }{\partial y\ \partial z},Y= \frac {\partial ^2 A_y }{\partial z\ \partial x}, Z= \frac {\partial ^2 A_z }{\partial x\ \partial y}
$$
Solving the set (1) of equations, we have,
$$
B_x= \left( \frac {\partial A_z}{\partial y}-\frac {\partial A_y}{\partial z} \right),...\text{and 2 similar expressions.}
$$

This is my line of thinking - I have made use of the 'form' of the differential-fractions. Yet I feel there are logical holes in the above outline. I need some suggestions, and references as well for a proper proof of the theorem. Thanks!

答:
There are several questions about this already on this site. The answer is "it depends". It depends on the domain of $B$. If for example, the domain of $B$ is all of $\Bbb{R}^3$ or say an open ball, or more generally any star-shaped open set (I'm sure you can allow for more general situations as well) then yes the converse is true (look up Poincare's lemma for a proof).

BUT, we need some sort of restriction on the domain of the function because otherwise it is not true, as seen by the following counterexample: let $U=\Bbb{R}^3\setminus\{0\}$ and define $\mathbf{B}:U\to\Bbb{R}^3$ as $\mathbf{B}(\mathbf{r})=\frac{\mathbf{r}}{r^2}$. Then, $\mathbf{B}$ is divergence free, but doesn't admit a vector potential since the integral over the unit sphere is $4\pi\neq 0$ (if $\mathbf{B}$ did admit a vector potential on $U$ then by Stokes' theorem the integral must be $0$ since the unit sphere has no boundary, but it is not, hence there is no vector potential on $U$).


问:
Sum of elements of $A \cap(B \cup C)=400 \times 274$. Find the closed form of $C$.
Question:

$A=\{x: x=3$ digit natural number $\}$

$B=\{x: x=9 k+2 ; k \in N\}$

${C}=\{{x}: {x}=9 {k}+I ; {k} \in {Z}\}, $ $0 \leq {I}<9$

Sum of elements of $A \cap(B \cup C)=400 \times 274 .$
Find the value of $I$

To be honest, I don't know how to start without making an assumption.

$$A \cap(B \cup C)=(A \cap B)\cup (A \cap C)$$
Assuming that $(A \cap B)$ and $(A \cap C)$ doesn't have any element common, one can find the sum of elements of $(A \cap B)$ and $(A \cap C)$ which would be equal to $400 \times 274 $.

Sum of elements of $(A \cap B)$:
$$\sum_{k=11}^{110}(9k+2)=54450+200=54650$$

Sum of elements of $(A \cap C)$:

Elements of $C$ are $$\cdots\cdots90+I,99+I,108+I\cdots990+I,999+I\cdots\cdots$$
Elements of $A$ are $$100,101,102\cdots 999$$
So possible elements of $(A \cap C)$ [By keeping $0 \leq {I}<9$ in mind] must be:$$99+I,108+I\cdots990+I\tag1\label{eq1}$$
So
$$\sum_{k=11}^{110}(9k+I)=54450+100I$$
Then after adding and solving $$I=5$$
And by luck $9k+2$ and $9k+5$ does not have any element common from $k=11$ to $k=110$.

How can this be solved without making that risky assumption? Any alternate.

Seeing range in $\eqref{eq1}$ and guessing the elements of $(A \cap C)$ worked well in this question but there is a single value of $I$ so that won't work every time.

答:
In words, $A \cap(B \cup C)$ contains all three-digits numbers which leave remainder $2$ and $I$ on division by $9$. For example,

$$A \cap B = \{ 99+2, \ldots, 990+2 \} $$

Clearly $A \cap B$ has $110-11+1=100$ elements whose sum is
$$100 \cdot \frac{101+992}{2}=200 \cdot \frac{1093}{4}=200 \times 273.25$$

It is clear that $I \neq 2$ and $A \cap B$ and $A \cap C$ are disjoint. We observe there would be $100$ terms in $A\cap C$ too. The numbers have been chosen conveniently. The total sum can be written as
$$\sum (9k+2)+(9k+I)=\sum 2(9k+2) + (9k+I) - (9k+2)$$
$$=2\sum (9k+2) + \sum (I-2)$$
$$=400 \times 273.25 + 100(I-2)$$

Therefore, $$100 (I-2) = 400 \, (274-273.25) \Rightarrow \boxed{I=5}$$


问:
$3\int_{0}^{1}(f'(x))^2dx \geq (2\int_{0}^{1}f(x)dx)^2 \impliedby 2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$
Let $f : \mathbb{R} \to \mathbb{R} $ be a differentiable function.  Suppose that $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$

Show that $$3\int_{0}^{1}(f'(x))^2 \,\mathrm dx \geq (2\int_{0}^{1}f(x)\,\mathrm dx)^2$$

Body:
by integrating by parts,
$$\int_0^{\frac{1}{2}}g_1(x)f'(x)\,\mathrm dx=g_1(x)f(x)\bigg|_0^{\frac{1}{2}}-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx \tag{1}$$
$$\int_{\frac{1}{2}}^1g_2(x)f'(x)\,\mathrm dx=g_2(x)f(x)\bigg|_{\frac{1}{2}}^1-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx \tag{2}$$
Since $2\int_{0}^{\frac{1}{2}}f(x)\,\mathrm dx=\int_{\frac{1}{2}}^{1}f(x) \,\mathrm dx$, we let $g_1'(x)=a_1\neq0$ and $g_2'(x)=b_1\neq0$.
$$g(x)=\begin{cases}
g_1(x)=a_1x+a_0,x\in[0,\tfrac{1}{2})\\
g_2(x)=b_1x+b_0,x\in(\tfrac{1}{2},1]
\end{cases}$$
Let
$g_1(x)f(x)\bigg|_0^{\frac{1}{2}}+g_2(x)f(x)\bigg|_{\frac{1}{2}}^1=0$, we get
$$\begin{cases}
g_1(0)=g_2(1)=0\\
g_1(\tfrac{1}{2})=g_2(\tfrac{1}{2})
\end{cases} \Rightarrow\quad \begin{cases}
a_0=b_1+b_0=0\\
\tfrac{1}{2}a_1+a_0=\tfrac{1}{2}b_1+b_0
\end{cases}$$
Hence, we get $a_0=0,a_1=a_1,b_0=-a_1,b_1=a_1$,
$$g(x)=\begin{cases}
g_1(x)=a_1x,x\in[0,\tfrac{1}{2})\\
g_2(x)=-a_1x+a_1,x\in(\tfrac{1}{2},1]
\end{cases}$$
from $(1)+(2)$, we have
\begin{align*}
\int_0^1g(x)f'(x)\,\mathrm dx&=-\int_0^{\frac{1}{2}}g_1'(x)f(x)\,\mathrm dx-\int_{\frac{1}{2}}^1g_2'(x)f(x)\,\mathrm dx\\
&=-\frac{a}{3}\int_0^1f(x)\,\mathrm dx+\frac{2a}{3}\int_0^1f(x)\,\mathrm dx=\frac{a}{3}\int_0^1f(x)\,\mathrm dx
\end{align*}
by Cauchy-Schwarz inequality, we get
$$\left(\frac{a}{3}\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \int_0^1g^2(x)\,\mathrm dx\int_0^1(f'(x))^2\,\mathrm dx$$
and
$$\int_0^1g^2(x)\,\mathrm dx=\frac{a_1^2}{24}+\frac{a_1^2}{24}=\frac{a_1^2}{12}$$
Therefore,
$$\left(\int_0^1f(x)\,\mathrm dx\right)^2\leqslant \frac{3}{4}\int_0^1(f'(x))^2\,\mathrm dx$$


问:
Is minimal polynomial of field of rational numbers and field of real numbers are same? Why?
Is minimal polynomial of field of  rational numbers and field of real numbers are same?    Why?</p>

答:
Depends.

The minimal polynomial of the imaginary unit $i\in\Bbb C$ is $x^2+1$, which is a polynomial over the rationals and the reals.

But the minimal polynomial of $\sqrt 2\in\Bbb R$ is $x^2-2$ over the rationals and $x-\sqrt 2$ over the reals.


问:
Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent
Show that series $$\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$$

is divergent . What inequality can we use here. i tried various method but none of these give any result. Any hint please .

答: Take the exponential of the denominator. It can be seen that $\log\log(n) n$ is smaller than $e^n$ for large enough $n$. Since $x \mapsto e^x$ is monotonically increasing, this means the deonminator is smaller than $n$. The series can then be compared with the harmonic series.


问:
Classifying the critical points when second derivative test fails.
$$f(x, y) = x^4 + y^4 - x^3$$

\begin{align*}
\frac{\partial f}{\partial y} &= 4y^3 = 0 \implies y = 0\\
    \frac{\partial f}{\partial x} &= 4x^3 - 3x^2 = 0 \implies \text{ either } x = 0 \text{ or } x = \frac{3}{4}
\end{align*}
Hence, $\left(0, 0 \right) $ and $ \left( \frac{3}{4}, 0 \right) $ are critical points. The Hessian is $H(x, y) = \begin{bmatrix}
12 x^2 - 6x & 0\\
0 & 12y^2
\end{bmatrix}$.

$H(0, 0) = \textbf{0}$ so no conclusion.

$H\left ( \frac{3}{4}, 0 \right ) = \begin{bmatrix}
\frac{9}{4} & 0\\
0 & 0
\end{bmatrix}$ hence the test is inconclusive.

Are they any other ways I can find the nature of critical points: local min, local max or saddle?

答:
Second derivatives are of limited help, particular with functions of two or more variables with power-functions with exponents larger than $ \ 2 \ . $  When the Hessian is just giving you zero at a point, it is often useful to go back to the properties of the function to see what issues are arising.

In the $ \ xz-$ plane, the function is $ \ f(0, y) \ =  \ y^4 \ , $ which you know is always "concave upward".  What causes a problem for the Hessian is that this is a curve with a very "flat" minimum, so using a second derivative doesn't probe behavior at the origin "deeply" enough (you have the same difficulty in single-variable calculus).  But in the neighborhood of $ \ ( 0,0) \ , $ we see that  $ \ f_{yy} \ = \ 12y^2 \ \ge \ 0 \ . $

In the $ \ yz-$ plane, however, we have  $ \ f(x, 0) \ =  \ x^4 \ - \ x^3 \ , $ for which there is an inflection point at the origin along the $ \ x-$ direction.  With $ \ f_{xx} \ = \ 6x · (2x - 1) \ , $ the concavity switches from positive for $ \ x \ < \ 0 \ $ to negative for $ \ 0 \ < \ x \ < \ \frac12 \ , $ then back to positive for $ \ x \ > \ \frac12 \ . $  So close to the origin, the sign of the curvature of the surface changes:  the surface has a "ripple in the bottom" and there is a small region over which $ \ f(x, y) \ < \ 0 \ . $  Hence, the origin is a saddle point.

This is not something that will be picked up by simply evaluating the Hessian at $ \ (0,0) \ ; $ it would be better to investigate the determinant function  $ \ 72x · (2x - 1) · y^2 \ . $

[The other point that you found, $ \ \left( \frac34 \ , \ 0 \right) \ , $ is sitting in a neighborhood where $ \ f_{xx} \ $ and $ \ f_{yy} \ $ are both positive, so the surface curvature is positive there and you presumably found this to be the location of the absolute minimum for the function.]


问:
Simplest proof that $\sum 1/p$ grows like $\log \log x$
I am seeking recommendations for a simple proof that $\sum_{p<x} 1/p$ grows like $\log (\log (x))$.

By simple I mean suitable for secondary school or 1st year university students - not those who are already trained in mathematics.

Ideally the proof would start from Euler's product formula for the Riemann zeta function, but this is not necessary.

答:
$$(1+O(\frac1{\log n}))n\log n = \log n! = \sum_{p^k \le n} \lfloor n/p^k \rfloor\log p $$ $$= (1+O(\frac1{\log n}))\sum_{p \le n} n \frac{\log p}{p}\tag{1}$$
Followed by a partial summation
$$\sum_{p\le n} \frac1p = \frac{\sum_{p\le n} \frac{\log p}p}{\log n}+\sum_{m \le n-1} (\sum_{p\le m} \frac{\log p}p) (\frac1{\log m}-\frac1{\log (m+1)})$$
$$ =  \frac{(1+O(\frac1{\log n}))\log n}{\log n}+\sum_{m\le n-1}(1+O(\frac1{\log m})) \log m  \frac{1+O(\frac1{m})}{m\log^2 m}$$ $$=\log \log n+C+O(\frac1{\log n})$$


问:
normal bundle of level set
Let $M$ be a Riemannian manifold and $S \subset M$ a regular level set of a smooth function $f:M\rightarrow \mathbb{R}^k$. How can I show that the normal bundle of $S$ is trivial?

If $k=1$ then $\text{grad}f$ is a global frame for $NS$ but I am not sure how to show it formally for the general case. Thanks!

答: If $S=f^{-1}(x)$ where $f:M\to N$ and $x$ is a regular value then $f_*:T_yM\to T_xN$ is (by definition of regular value) surjective, $T_yS$ is the kernel of $f_*$, hence $f_*$ is an isomorphism of $(T_yS)^\perp\to T_xN$, i.e. (letting $y$ vary) it gives an isomorphism $(TS)^\perp\cong S\times  T_xN$. </p>

Title: normal bundle of level set


问:
Is infinite a infinite or finite
Long back I have watched a documentary based on a mathematician named Cantor. According to that documentary, Cantor claimed that infinite does not exists, it is only finite. Is that True?</p>

答: I didn't see the documentary so can't say for sure, but possibly what it said is that Cantor claimed that INFINITE NUMBERS don't exist.  Cantor's vitriolic opposition both to infinitesimals and infinite numbers is a well-known fact; see for example this recent text. In fact, Cantor called infinitesimals "the cholera bacillus of mathematics".  His other epithets for infinitesimals include "paper numbers" and "an abomination", see for example this.</p>

Title: Is infinite a infinite or finite


问:
Can this differential equation be solved manually?
$ x\;\ddot{y} - 2\dot{x} + 9y + \frac{2}{3} = 0$

I tried making use of wolfram alpha but it was unable to provide me with a solution.
I also wanted to apply the Laplace transform method, however I am not sure how to apply it given the first term of the equation.

答: Note that if $y_0(t)$ is fixed, the equation is an ODE of 1° order without initial condition, and if $x_0(t)$ is fixed, the equation is an ODE of 2° order without initial condition.


答: By Cauchy condensation test, the given series converges if and only if $\sum_{n=1}^\infty a_n$ does, where $$a_n=\frac{2^n}{(n\log 2)^{\log(n\log 2)}}.$$ But $a_n\not\to 0$ (and in fact $a_n\to\infty$) as $n\to\infty$. Hence, the given series diverges.

Title: Show that series $\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$ is divergent

Body:
Show that series $$\sum_{n=2}^{\infty} \frac{1}{(\log(n))^{\log(\log(n))}}$$

is divergent . What inequality can we use here. i tried various method but none of these give any result. Any hint please .


问:
Net areas of definite integral
Given
$$
\int_{-4}^{2} f(x) dx = 5\\
\int_{-4}^{-2} f(x) dx = A\\
\int_{-2}^0 f(x) dx = B\\
\int_{0}^{2} f(x) dx = C
$$
$A$ and $C$ are areas below the $x$-axis. By interpreting definite integrals as net areas, how do I find the value of $\int_{-4}^{2} (f(x) - 2x - 3) dx$?

I tried making $-2x - 3 = g(x)$, then finding the net area of $g(x)$ from $2$ to $-4$ (which is $18.5$) then adding them together. So my final answer is $5+18.5$ units. I am not sure if I am doing it right because they have different regions above and below the $x$-axis.

答:
Your method is correct but your interpretation and computation of the integral seems to be wrong. Also it looks like there's more information available than you need which may be confusing. Given the problem just use what you need from what's available (a general approach).

Splitting the integral as you did, $$\int_{-4}^2f(x)-\int_{-4}^{2}(2x+3)dx=5-\left[x^2+3x\right]_{-4}^2=5-\left((4+6)-(16-12)\right)=-1.$$

From how you asked your question, make sure you read your integrals correctly, e.g. $\int_{-4}^2$ is the integral from $-4$ to $2$, so $$\int_{-4}^2f(x)dx=[F(x)]_{-4}^2=F(2)-F(-4).$$ This is probably where your misunderstanding lies.

Title: Net areas of definite integral


问:
An apparent dichotomy in the calculation of a limit with indeterminacy $ 0^0 $
$$L=\lim_{x \to 0}(x+2x^2+3x^3+4x^4+5x^5)^\frac{1}{\log(4x+3x^2+2x^3+x^4)}$$

答:
By taking logarithms on both members we got
$$\log L= \lim_{x \to 0}\frac{\log(x+2x^2+3x^3+4x^4+5x^5)}{\log(4x+3x^2+2x^3+x^4)},$$
which is of the indeterminate form $\infty/\infty$. By applying successively L'Hôpital's rule, I got $L=e^{5/4}$ as result. However correct is $L=e$. I was unable to find my mistake. Thanks for helping!


问:
Integral with respect to Probability Distribution
I want to integrate the following, but am not sure exactly how to do so. I'm working with a probability distribution $F(x)$ (i.e. $f(x)$ is a density function):

$$\int_0^\infty \int_t ^\infty x^k dF(x) dt$$

I know I can write this as:
$$\int_0^\infty \int_t ^\infty x^k f(x) dx dt$$
But I fail to see how this helps with simplification. I'm tempted to somehow call the inner $E[x^k]$, but that seems incorrect, as we're integrating from $t$ to $\infty$. Would really appreciate some help with this one!

答: Interchanging the order of integration you get $\int_0^{\infty} \int_0^{x} dt x^{k}f(x)dx=\int_0^{\infty} x^{k+1} f(x)dx$. If your random variable $X$ is positive this is just $EX^{k+1}$.


答:
Note that over the field $\mathbb{R}$ of the real numbers irreducible polynomials have degree $\leq2$.

On the other hand over the field $\mathbb{Q}$ of rational numbers you can find irreducible polynomials of any degree, meaning that the degree of irreducible polynomials is not bounded. For instance, if $p$ is a prime number the polynomial
$$
1+X+X^2+\cdots+X^{p-1}
$$
is notoriously irreducible.

This answers your questions because irreducible polynomials are the minimal polynomials of their roots.

Body: Is minimal polynomial of field of  rational numbers and field of real numbers are same?    Why?</p>


问:
Compute $\sum_{n=1}^{+\infty} (-1)^n \frac{a_n}{n}$
For $n \in \mathbb{N}^\star$, define $a_n := \frac{1 \times 3 \times \cdots \times (2n-1)}{2 \times 4 \times \cdots \times (2n)}$ and $f(x) = \sum_{n=1}^{+\infty}  \frac{a_n}{n} x^{2n}$.

One easily gets by induction that $\forall n \in \mathbb{N}^\star, a_n \leq \frac{1}{\sqrt{2n+1}}$ and this helps us show that $f$ is continuous on $[-1, 1]$.

Let $S(x) = \sum_{n=1}^{+\infty} a_n x^{2n}$. One shows that :

$$
\forall x \in (-1, 1), S(x) = \frac{1}{\sqrt{1 - x^2}} - 1
$$

How do I use that to compute $\sum_{n=1}^{+\infty} (-1)^n \frac{a_n}{n}$ ? I see it is $f(i)$, but I have no idea how to use this and $S$ to get a clean expression. Thanks for the help.

答:
Observe that
$$
\sum_{n=1}^\infty\frac {a_n}nx^{2n}=f(x)=2\int_0^x\frac {S (t)}t dt=2\log\frac2 {1+\sqrt {1-x^2}}.\tag1
$$
Substituting into (1) $x=i$ one obtains:
$$
\sum_{n=1}^\infty (-1)^n\frac {a_n}n=f (i)=\log (12-8\sqrt2).\tag2
$$


问:
Find an integrable $g(x,y) \ge |e^{-xy}\sin x|$
I want to use Fubini theorem on $$\int_0^{A} \int_0^{\infty} e^{-xy}\sin x dy dx=\int_0^{\infty} \int_0^{A}e^{-xy}\sin x dx dy$$

Must verify that $\int_M |f|d(\mu \times \nu) < \infty$. I'm using the Lebesgue theorem, so far I've come up with $g(x,y)=e^{-y}$ but am not sure whether it's correct.

My argument is that if $x\in (0,1)$ then the $\sin x$ part is going to ensure that the inequality holds.

答:
You can take $g(x,y)=e^{-xy}|\sin x|$ in order to get an integrable function on $[0,A]\times [0,+\infty)$. Indeed,
$$\int_0^{+\infty}e^{-xy}dy=\frac 1x$$
and the integral $\int_0^A\frac{|\sin x|}{x}dx$ is convergent.

With $g(x,y)$, we wouldn't get a function which majorize $e^{-xy}\sin x $ everywhere (but only on for $y\geqslant 1$), which is not a problem.


问:
Correct the Fourier transform of a shear-distorted image? ($x_{new} = x + \alpha y$)
I have some old STM images that appears to be sheared by about 5 degrees; in other words as scanning progressed from top to bottom the scanning area slowly drifted left-to-right, by the end the scanning was shifted to the side by say 5% up to 10% of the length of the scan.

$$x_{new} = x + \alpha y$$

where $|\alpha| < 0.1$ and the original data is within the square: $-1 <= x, y <= 1$.

The straightforward way to fix this would be to re-interpolate the original image to de-shear it, then crop off the triangular bits along the sides so that the FT is performed on a rectangular area of all good data, and I will probably do that.

In fact, since I'm using a 2D Hamming window I might even be able to completely ignore the blank triangular areas that the de-shearing produces (replace unknown data with image median).

Question: But I am curious if there is a transform that I can apply (again by re-interpolation) to the result of the Fourier transform that will produce an equivalent result without de-shearing the original image. I'm only interested in the central are of the FT so I don't care about lost edges.

I have a hunch that the answer is yes and it will turn out to be a similar type of linear transform or shearing of the FT.

In fact it looks like it's very simple and this is what's happening

$$\omega_{y,new} = \omega_{y} + \beta \omega_{x}$$

with $\beta = \alpha$

when I make some plots, but I don't know how to go about showing this to be true.

Related:

Separating scaling, rotation and sheer coefficients correctly in this case

答:
Your intuition was correct:

Call the original image $I(x,y)$  the sheared image is $I(x+\alpha y, y)$.
I will use the Fourier integral and not the sum, but up to discretization, the argument should stand. Define the Fourier integral of $I$:
$$\hat I(\omega_x,\omega_y) = \mathcal F \circ  I(x,y) = \iint dx dy e^{i(\omega_x x+\omega_y y)}I(x,y) $$
Then
$$\iint dx dy e^{i(\omega_x x+\omega_y y)}I(x+\alpha y,y)=\iint dx dye^{i(\omega_x (x +\alpha y)+(\omega_y-\omega_x\alpha) y)}I(x+\alpha y,y)$$
$$=\iint du dv e^{i(\omega_x u+(\omega_y-\omega_x \alpha) v)} I(u,v)=\hat I(\omega_x ,\omega_y-\omega_x\alpha)$$  Thus
$$\mathcal F \circ  I(x+\alpha y,y) = \hat I(\omega_x, \omega_y -\omega_x\alpha)$$


问:
Setting up integration within a truncated octahedron
I am trying to set up a triple integral over a region bounded by a truncated octahedron (see https://en.wikipedia.org/wiki/Truncated_octahedron). This surface is described by the intersection of 14 surfaces, given by the equations
$$|x|+ |y| + |z| = \frac{3}{2}$$
$$|x| = 1$$
$$|y| = 1$$
$$|x| = 1$$
where the first equation describes the 8 hexagonal faces, and the remaining three equations describe the 6 square faces. My attempt at this is to set up the integral of a function $f(x,y,z)$ as $$\int_{-1}^{1} \int_{-h(z)}^{h(z)} \int_{-g(y,z)}^{g(y,z)} f(x,y,z) dxdydz$$

with $$g(y,z) = \text{min} \{1,\frac{3}{2}-|y|-|z| \}$$
$$h(z) = \text{min} \{1, \frac{3}{2}-|z|\} .$$

Setting the integrand to $f=1$, I should get the volume of the region $V=4$, but this is not the case. Where am I going wrong with the bounds of integration?

答:
In the first octant, we would integrate $$\int_{x=0}^1 \int_{y=0}^{\min(1, 3/2-x)} \int_{z=0}^{\min(1, 3/2-x-y)} f(x,y,z) \, dV. \tag{1}$$  So by symmetry, the full volume is
$$\int_{x=-1}^1 \int_{y=-\min(1, 3/2-|x|)}^{\min(1, 3/2-|x|)} \int_{z=-\min(1, 3/2-|x|-|y|)}^{\min(1, 3/2-|x|-|y|)} f(x,y,z) \, dV. \tag{2}$$  This is the same as what you have, except with a change in order of integration.

I do not recommend evaluating $(2)$ directly, because it is incredibly tedious.  Rather, I recommend evaluating $(1)$ and exploiting symmetry, knowing that the latter arises naturally from the former.  This would give
$$\begin{align}
\frac{V}{8} &= \int_{x=0}^1 \int_{y=0}^{\min(1, 3/2-x)} \min(1, 3/2-x-y) \, dy \, dx \\
&= \int_{x=0}^{1/2} \int_{y=0}^{1/2-x} 1 \, dy \, dx + \int_{x=0}^{1/2} \int_{y=1/2-x}^1 \frac{3}{2}-x-y \, dy \, dx + \int_{x=1/2}^1 \int_{y=0}^{3/2-x} \frac{3}{2} - x - y \, dy \, dx \\
&= \int_{x=0}^{1/2} \frac{1}{2} - x \, dx + \int_{x=0}^{1/2} \frac{3+4x-4x^2}{8} \, dx + \int_{x=1/2}^1 \frac{(3-2x)^2}{8} \, dx \\
&= \frac{1}{8} + \frac{11}{48} + \frac{7}{48} \\
&= \frac{1}{2}.
\end{align}$$


问:
Hartshorne IV.6.4 - no curve of degree 9 and genus 11 in P^3
I'm working on this exercise in Hartshorne: there are no curves of degree 9 and genus 11 in $\mathbb{P}^3$.

The hint says to show that it would have to lie on a quadric surface. This is the part I'm having trouble with. If $X$ is the curve, then there's an exact sequence

$$0 \to H^0(\mathcal{I}_X(2)) \to H^0(\mathcal{O}_{\mathbb{P}^3}(2)) \to H^0(\mathcal{O}_X(2)) \to \ldots$$

and I want $h^0(\mathcal{I}_X(2))$ to be nonzero, which would be true if $h^0(\mathcal{O}_X(2)) < h^0(\mathcal{O}_{\mathbb{P}^3}(2)) = 10$. If $D = \mathcal{O}_X(2)$, then Riemann-Roch says

$$h^0(D) - h^0(K - D) = \deg D + 1 - g = 2 \cdot 9 + 1 - 11 = 8$$

so if $h^0(K - D) = 0$, then all is good. But if $h^0(K - D) \ne 0$, then I can only think of Clifford's theorem, which says

$$h^0(D) - 1 \le \frac{1}{2} \deg D = 9$$

with equality iff $D = 0, K$, or $X$ is hyperelliptic and $D$ is a multiple of the $g^1_2$. If the inequality is strict, then again all is good, so this just leaves the case $X$ hyperelliptic, $D = 9 \cdot g^1_2$. Why is this impossible? Is there a simpler argument altogether?

答:
You just need to exclude the case when $dimH^{0}(X,\mathcal{O}_{X}(2))=10$, $X$ is hyperelliptic and $D=9g_{2}^{1}\in |\mathcal{O}_{X}(2)|$.

If this is the case, by Rieman-Roch theorem, we have

$$H^{0}(X,\mathcal{O}_{X}(D))-H^{0}(X,\mathcal{O}_{X}(K_{X}-D))=8$$

Since $H^{0}(X,\mathcal{O}_{X}(D))=10$, we have $H^{0}(X,\mathcal{O}_{X}(K_{X}-D))=2$. $deg|K_{X}-D|=2g-2-deg(D)=2$. We have $|K_{X}-D|$ is a linear system of degree $2$ dimension $1$. $K_{X}-D=g_{2}^{1}$. We have $$K_{X}=D+g_{2}^{1}$$.

Since $D=2H$ is very ample and $g_{2}^{1}$ is ample, we conclude $K_{X}=D+g_{2}^{1}$ is very ample, which contradicts to our assumption that $X$ is hyperellptic.


问:
What the range of $r$ and $\theta$?
If $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac{y}{x}$ for $0\leq x\leq 1$ and $0\leq y\leq 1$, what the range of $r$ and $\theta$?

I think since $0\leq x\leq 1$ and $0\leq y\leq 1$ then $0\leq r\leq 1$.

since $0\leq x\leq 1$ and $0\leq y\leq 1$, then $\frac{y}{x}>0$ so $0\leq \theta\leq \frac{\pi}{2}$.

Does this answer correct? (Or maybe $r$ depend on $\theta$?)

答:
Because the problem involves Arctan$\left(\frac{y}{x}\right)$, $x=0$ is outlawed.

Therefore, $0 < x \leq 1$ and $0 \leq y \leq 1.$  Therefore,

$~r^2 \in (0,2] \implies r \in (0,\sqrt{2}]$.

$\frac{y}{x} \in [0, +\infty) \implies \theta \in [0, \pi/2).$

Edit

If you allow the interpretation that $x$ may equal $0$, then superficially, the changes are that $r$ may equal $0$, and that $\theta$ may equal $\pi/2$.

The problem with this superficial interpetation is that when $r = 0$, the concept of $\theta$ is meaningless.  Therefore, it could be argued that even when $x=0$ is permitted, $r$ is still not permitted to equal $0$,

Title: What the range of $r$ and $\theta$?


答:
The equation is linear, first order, in $x$.

The solution of the homogeneous equation is easily found to be
$$x=ce^{\dot y/2}.$$

Then by variation of the constant, we find
$$2c'=\left(9y+\frac23\right)e^{-\dot y/2}$$

and

$$x=e^{\dot y/2}\left(\frac12\int\left(9y+\frac23\right)e^{-\dot y/2}dt+c\right).$$

where $y$ is an arbitrary differentiable function of $t$.

Title: Can this differential equation be solved manually?

Body:
$ x\;\ddot{y} - 2\dot{x} + 9y + \frac{2}{3} = 0$

I tried making use of wolfram alpha but it was unable to provide me with a solution.
I also wanted to apply the Laplace transform method, however I am not sure how to apply it given the first term of the equation.


问:
Prove the $\operatorname{length}(\alpha) \ge \|v\|$ so that the line segment from $P$ to $Q$ gives the shortest possible path
Let $P,Q \in \mathbb{R^3} $ and let $\alpha:[a,b] \to \mathbb{R^3}$ be any parameterized curve with $\alpha(a)=P, \alpha(b)=Q.$ Let $v=Q-P$. Prove the length$(\alpha) \ge \|v\|$, so that the line segment from $P$ to $Q$  gives the shortest possible path.

My Work:

From the Cauchy-Schwartz Inequality

$u\cdot v \le \|u\|\|v\|$

If $u=\alpha'(t)$,

$\alpha'(t)\cdot v \le \|\alpha'(t)\|\|v\|$

Then $\int_a^b\alpha'(t)\cdot v dt\le \int_a^b\|\alpha'(t)\|\|v\|dt$***

I believe we can take the $v$ out as a constant and get the following

$v\int_a^b\alpha'(t) dt = v\ (\alpha(b)-\alpha(a))=v(Q-P)=v\cdot v \le \|v\|\int_a^b\|\alpha'(t)\|dt$***

$\mathrm{Arclength} = s(t)=\int_a^b\|\alpha'(t)\|dt$

$v\cdot v \le \|v\|s(t)$

Since $s(t)$ can be any path, the shortest one is the one taken by $v$ where the two inequality becomes a n equality. But for any other path, it will be longer than the segment from $P$ to $Q$.

I think this makes sense, but I'm not sure. If it is correct, I would like to know how I justify the steps with the *** at the end. Say I was a professor and a student came up to me and I had to judge whether this followed from calculus.

Update

$\|v\|=\|Q−P\|=\|\alpha(b)−\alpha(a)\| = \left\|\int_a^b \alpha ' (t)dt\right\| = s(t)$

$\alpha'(t)\cdot v \le \|\alpha'(t)\|\|v\|$

$\alpha'(t)\cdot v \le \|\alpha'(t)\|s(t)$

I believe now we can say that $v$ is the shortest path by the Cauchy Schwartz inequality.

答:
From the equality
$$
\|v\| = \left\|\int_a^b \alpha'(t) \mathrm{d}t\right\|,
$$
and thanks to the triangular inequality for integrals, it follows that
$$
\|v\| \leqslant \int_a^b \left\|\alpha'(t)\right\|\mathrm{d}t = \mathrm{length}(\alpha).
$$


问:
Ambiguity in translating first order logic wfs to english.
So I have some first order logic wfs I want to translate to English[From "Introduction to Mathmatical Logic" by Elliot Mendelson , exercise 2.9] , but I am skeptical whether my translations are correct.Here are my translation:

$(1)$ $(\forall x)(M(x) \land (\forall y) \lnot W(x,y) \to U(x))$

$(t1)$ An unmarried man is unhappy

Here $M(x)$ means "$x$ is man" , $W(x,y)$ means "$x$ is married to $y$", and $U(x)$ means "$x$ is unhappy".

$(2)$ $(\forall x)(V(x) \land P(x) \to A(x,b))$

Here $V(x)$ means $x$ is an even integer, $P(x)$ means "$x$ is a prime number", and $A(x,y)$ means "$x=y$" and $b$ denotes $2$. For me there seems to be two translations, But I am not sure which one is correct or if any of them is even correct.

$(t2.0)$ if $x$ is an even prime integer , then $x$ is equal to $2$.

$(t2.1)$ $2$ is an even prime integer.

$(3)$ $\lnot (\exists y) (I(y) \land (\forall x)(I(x) \to L(x,y)))$

Here $I(y)$ means "$y$ is an integer" and "$L(x,y)$" means "$x \leq y$".The traslation of this one is not mine, I found it in the "solutions to selected exercises". But I am still not sure if it is the correct one (because of the $\leq$ sign having an $=$ portion with it). Someone explain to me why this is precisely the correct one.

$(t3)$ There is no greatest integer.

$(4)$ $\exists x (A^1_1(x)\land\forall y (A^1_1(y)\to A^2_1(x,y)))$

$(5)$ $\forall x(A^1_1(x)\to\forall y(A^1_1(y)\to A^2_1(x,y)))$

$(6)$ $\exists x (A^1_1(x)\land\forall y(A^1_1(y)\to(A^2_1(x,y)\leftrightarrow A^2_1(y,y))))$

Here, $A^1_1(x)$ means "$x$ is a person" and $A^2_1(x,y)$ means "$x$ hates $y$".Here are the translation:

$(t4)$ there is a person who hates everyone.

$(t5)$ everyone hates everyone.

$(t6)$ There is a person who precisely hates those persons who hate themselves.

$(7)$ $(\forall x)(H(x) \to (\exists y)(\exists z)(\lnot A(y,z) \land (\forall u)(P(u,x) \leftrightarrow (A(u,y) \lor A(u,z)))))$

$(t7)$ every person has precisely $2$ distinct parents.

here $H(x)$ means $x$ is a person, $A(u,v)$ means "$u = v$" , and $P(u, x)$ means "$u$ is a parent of $x$".

答:
Long comment

Regarding 2.9(c) [your (3)], a possible symbolization of "There is no greatest integer" will be: $\forall y \ [\text {Int}(y) \to \exists x (\text {Int}(x) \land (y < x))]$, that reads: "for every Integer, there is an Integer that is greater than the first one".

This is equivalent to: $\lnot \exists y \ [\text {Int}(y) \land \lnot \exists x (\text {Int}(x) \land (y < x))]$, that in turn is:

$\lnot \exists y \ [\text {Int}(y) \land \forall x (\text {Int}(x) \to \lnot (y < x))]$.

Now, if we have that "$\text L(x,y)$" means "$x \le y$", "$\lnot (y < x)$" is exactly "$x \le y$", and thus we get Mendelson's version:

$\lnot \exists y \ [\text {Int}(y) \land \forall x (\text {Int}(x) \to \text L (x,y))]$.

Consider what is the meaning of the un-negated formula: "there is an Integer that is greater-or-equal to every Integer". Clearly, this number will be a "greatest integer".

Regarding 2.9(b), it reads: "Every prime even integer is equal to two".


问:
Solving a simple Riccati equation
I am trying to solve a simple Ricatti equation and am not sure i'm correct. Since this is probably the simplest common case I thought I'd post it here:

Write the Riccati equation as
\begin{equation*}
\frac{d E (\tau) }{d \tau} =  e_2 E(\tau)^2 + e_1 E(\tau) + e_0
\end{equation*}
where
we can reduce to a second order linear equation via the substitution $E(\tau) = - \frac{1}{e_2} \frac{w^\prime (\tau)}{w(\tau)}$ which yields
\begin{equation*}
w^{\prime \prime} - e_1 w^\prime + e_2 e_0 w = 0.
\end{equation*}
The general solution to this ODE is $w(\tau) = C_{1} e^{r_{1} \tau} + C_{2} e^{r_{2} \tau}$ where $r_1$ and $r_2$ are the positive and negative roots
\begin{align*}
r_{1,2} = \frac{e_1 \pm \sqrt{e_1^2 - 4 e_2 e_0}}{2} = \frac{e_1 \pm q}{2}
\end{align*}
and $C_{1,2}$ are constants determined by the boundary condition. A general solution for $E(\tau)$ is then
\begin{align*}
E(\tau) &= - \frac{1}{e_2} \frac{C_1 r_{1} e^{r_{1} \tau} + C_2 r_{2} e^{r_{2} \tau}}{C_1 e^{r_{1} \tau} + C_2 e^{r_{2} \tau}}
=  - \frac{1}{e_2} \frac{C r_{1} e^{r_{1} \tau} + r_{2} e^{r_{2} \tau}}{C e^{r_{1} \tau} + e^{r_{2} \tau}}
\end{align*}

Does this look correct? How else can these ODEs be solved?

答:
I suppose that your non linear OE $$\frac{d E (\tau) }{d \tau} =  e_2 E(\tau)^2 + e_1 E(\tau) + e_0
$$ is rewritten as $$\frac{d E }{d x} =  R(x) E^2 + Q(x)E + P(x),~~~(*)$$ just for the simplicity in indexes and the variable $\tau$ is our usual variable $x$. Another method in which you can overcome the OE would be to use a known particular solution of $(*)$, say $y_1$. Then set $y=y_1+u$ wherein $u$ is a solution of $$\frac{du}{dx}-(Q+2y_1R)u=Ru^2$$ Since the latter OE is a Bernoulli equation with $n=2$, it can be turned into a linear OE: $$\frac{dw}{dx}+(Q+2y_1R)w=-R$$ where $w=u^{1-2}=u^{-1}$.

Title: Solving a simple Riccati equation