问:
Calculating probability of getting 3 pairs in 13 cards(similar and related to another question I asked before)
Assuming there are 52 cards and dealt with 4 different players. Each of them will receive 13 cards. What are the possibilities of a player receive exactly 3 pairs?

*four-of-a-kind is considered 2 pairs.

This means we have  3 different pairs and 2 same pairs and one different pair.

For 3 different pairs,

${{13}\choose{2}}$ is for two ranks showing on the two pairs.

$ {{10}\choose{7}}$ is with 3 ranks being chosen and 10 ranks left, we choose from there.

$ {{4}\choose{2}}$ is the combination for the pair in the rank.

$ {{4}\choose{1}}$ is the suit for the non-matching cards.

For 2 same pairs and one different pair,

${{13}\choose{2}}$ is for two ranks showing on the two pairs.

$ {{10}\choose{7}}$ is with 2 ranks being chosen and 11 ranks left, we choose from there.

$ {{4}\choose{4}}$ is the combination for the 2 same pairs in the rank.

$ {{4}\choose{2}}$ is the combination for the pair in the rank.

$ {{4}\choose{1}}$ is the suit for the non-matching cards.

$$
P=\frac{\binom{13}{2}\cdot\binom{10}{7}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot\binom{4}{2}\cdot {{4}\choose{1}}^7}{\binom{52}{13}}+\frac{\binom{13}{2}\cdot\binom{11}{7}\cdot\binom{4}{4}\cdot\binom{4}{2}\cdot\binom{4}{1}^7}{\binom{52}{13}}=1.3707 \times{10}^{-5}
$$

Does this way of calculating probability include both situations or it doesn't?

答:
I agree with your count with the following remarks.

For 3 different pairs instead of “${{13}\choose{2}}$ is for two ranks showing on the two pairs”
should be “${{13}\choose{3}}$ is for three ranks showing on the three pairs”.

For 2 same pairs and one different pair instead of “${{13}\choose{2}}$ is for two ranks showing on the two pairs” should be “$13$ is for the rank showing on the doubled pair and $12$ is for the rank showing on the remaining pair” and instead of “$ {{10}\choose{7}}$” should be $ {{11}\choose{7}}$.


问:
Why if a polynomial is divided by x-1 and gives the remainder 3 then p (1) = 3?
Why if a polynomial is divided by $x-1$ and gives the $remainder=3$ then $p(1) = 3$? Does it have to do with ruffini? Could you describe the steps that lead me to affirm this?

ContentLicense: CC BY-SA 4.0

Body:
$$P(x)=(x-a)\times Q(x)+R$$

$$P(a)=(a-a)\times Q(a)+R=R$$

here $a=1, R=3$


问:
concave function is positive on convex set
I have seen the following sentence somewhere, but I can not prove it.

Let $f$ be a nonnegative concave function  on a convex domain $D$. Assume that $f$ achieves a positive value at some point $x\in D$. Then $f(y)>0$ for all $y\in D^o$.

答:
I presume we are working in $\mathbb R^n$.

As the statement is already valid for the point $y$ itself, choose another point $x\in D^{o}, x\ne y$. The open ball $B(x,\epsilon)$ is contained in $D$ for some $\epsilon > 0$.

Now, look at the line going through $x$ and $y$: $l(t)=x+t(y-x)$. For small negative values of $t$ (say $-\delta<t\le 0$) we will still have $l(t)\in B(x,\epsilon)$ because $|l(t)-x|=|t||y-x|$ and by making $|t|\lt\delta=\frac{\epsilon}{|y-x|}$ this will be true.

Find one such point $z=l(t)\in B(x,\epsilon)\subseteq D$ for a small negative $t$ so we have $z, x, y$ colinear, with $x$ between $y$ and $z$, and as both $y$ and $z$ belong to $D$, the whole segment between $y$ and $z$ is contained in $D$ (convexity). With a suitable re-parameterisation of the line $xyz$ (new parameter: $\tau$) we can have $x=y\tau+z(1-\tau)$ with $0\lt\tau\lt 1$. We know that $f(z)\ge 0$ and $f(y)>0$, so $f(x)=f(y\tau+z(1-\tau))\ge\text{ (concavity) }\tau f(y)+(1-\tau)f(z)>0$ - since at least the first term $\tau f(y)$ is $>0$ (and the second is $\ge 0$).


问:
Simple Partial Derivative Chain Rule Question
Say $f: \mathbb{R}^n \rightarrow \mathbb{R}$ by some $f(x_1,x_2,\cdots,x_n)$. Further suppose that each $x_i: \mathbb{R}^m \rightarrow \mathbb{R}$ by some $x_i=x_i(\eta_1,\eta_2,\cdots,\eta_m)$. Does it then follow that
$$\frac{\partial f}{\partial \eta_i} = \sum_{k=1}^{n} \frac{\partial f}{\partial x_k}\frac{\partial x_k}{\partial \eta_i},$$
assuming all of $\frac{\partial f}{\partial x_k}$ and $\frac{\partial x_k}{\partial \eta_i}$ exist on some subset of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively?

I don't have much formal education in way of partial derivatives and aside from the fact of treating variables that we aren't taking the derivative with respect to to be constants.

Really, my goal is to be able to successfully take some PDE like:
$$u_{xx}+u_{yy}=0$$
with $u=u(x,y)$ and doing some coordinate change $\xi = x+y$ and $\eta=x-y$ and doing the change of variable to the homogenous equation. What do people call this? Doing a change of variable to the PDE or equation?

If this is correct I will be able to move forward in at least being able to do the chain rule I need to for my class.

答:
Yes; since you're taking only one $\eta_i$ to vary everything just reduces to the case where you just had $x_k(\eta)$ in fact.

One thing that is worth stressing is that the $\partial/\partial x_k$ derivatives are holding the other $x_m$ constant, rather than the other $\eta_j$ variables held constant elsewhere. This is probably implicit, but important to realize.

And I would just call it a change of variables!

ContentLicense: CC BY-SA 3.0


问:
Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.

$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$

Is there some other more effective way for this limit?

答: You can simply use L'Hospital's Rule $$\lim_{x\to 0}\frac{1-(\cos x)^\frac13}{1-\cos (x)^\frac13}\space=\space\lim_{x\to 0}\frac{\frac{(\cos 0)^\frac13-(\cos x)^\frac13}{0-x}}{\frac{\cos (0)^\frac13-\cos(x)^\frac13}{0-x}}\space=\space\frac{\lim_{x\to 0}\frac{(\cos x)^\frac13-(\cos 0)^\frac13}{x-0}}{\lim_{x\to 0}\frac{\cos(x)^\frac13-\cos(0)^\frac13}{x-0}}$$ If you set $(\cos x)^\frac13$ and $\cos(x)^\frac13$ equal to $f(x)$ and $g(x)$ respectively, you will get $$\lim_{x\to 0}\frac{f'(x)}{g'(x)}\space=\space\lim_{x\to 0}\frac{\frac{(\cos x)^\frac{-2}{3}(-\sin x)}{3}}{\sin(x)^\frac{1}{3}x^\frac{-2}{3}}\space=\space\lim_{x\to 0}\frac{\sin x (x^\frac23)}{3\sin(x)^\frac13 (\cos x)^\frac23}$$ One more time, divide both denominator and nominator by $x$ $$\lim_{x\to 0}\frac{\frac{\sin x (x)^\frac23}{x}}{\frac{3\sin(x)^\frac13 (\cos x)^\frac23}{(x^\frac13)(x^\frac23)}}\space=\space\lim_{x\to 0}\frac{x^\frac43}{3(\cos x)^\frac23}\space=\space 0$$


问:
Derivative of $F$ with respect to $W$
Let $g$ be the derivative of a function $F$ with respect to $Z$ where $Z = WX$

(i.e. $\frac{\partial}{\partial Z}F = g$ where $Z = WX$ )

If I want to find the derivative of $F$ with respect to $W$. How do I find that?

(i.e. What is the value for $\frac{\partial}{\partial W}F$ ?)

Note: I know chain rule is involved somehow but I don't know how exactly is it involved

答:
Logarithmic differentiation and Chain rule

$$ Z=WX, \; \log Z = \log W+\log X$$

$$ \dfrac{\frac{\partial Z}{\partial W}}{Z} = \dfrac{1}{W} + \dfrac{\frac{\partial X}{\partial W}}{X} $$

Again plug in $$Z=WX.$$

Title: Derivative of $F$ with respect to $W$


问:
Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational?
$$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$

Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.

I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$

Sorry if this is a stupid question.

答:
Note that
$$
\sqrt[3]{2+\sqrt{5}}=\frac{1}{2}\sqrt[3]{16+8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3+3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2+(\sqrt{5})^3}=\frac{1+\sqrt{5}}{2}.
$$
Similarly,
$$
\sqrt[3]{2-\sqrt{5}}=\frac{1}{2}\sqrt[3]{16-8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3-3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2-(\sqrt{5})^3}=\frac{1-\sqrt{5}}{2}.
$$
Thus,
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1,
$$
as desired.

As for the second question, we can use the @player3236's approach. Let
$$
a=\sqrt[3]{x+\sqrt{y}},~b=\sqrt[3]{x-\sqrt{y}}.
$$
Then, for $c=a+b$ we have
$$
c^3=2x+3c\sqrt[3]{x^2-y},
$$
so $c$ is a root of the polynomial $t^3-3\sqrt[3]{x^2-y}\cdot t-2x$. If $c=a+b=1$, then
$$
1-3\sqrt[3]{x^2-y}-2x=0,
$$
or
$$
\sqrt[3]{y-x^2}=\frac{2x-1}{3},
$$
or
$$
y=\frac{1}{27}(27x^2+(2x-1)^3).
$$
We've obtained some necessary condition. However, I'm not sure whether this condition is sufficient or not (I guess that proof should be similar to case $x=2$, $y=5$, which was discussed above).


问:
Double Summation Simplification
I want to simplify the double summation
$$\sum_{k=1}^n (\sum_{i=1}^k 3i)$$

I started by doing

$$ 3 \sum_{k=1}^n(\sum_{i=1}^k i)$$

which would give me the following, I think

$$ 3 \sum_{k=1}^n(\sum_{i=1}^k \dfrac{n(n+1)}{2})$$

I don't know how to continue from here, I just know the final answer is supposed to be

$$\dfrac{n(n+1)(n+2)}{2}$$

Can someone give me some tips on how to continue after the second step?

答:
We are using the hockey-stick identity for binomials:

$\sum_{i=1}^k i = \binom{k+1}{2}$

$\sum_{k=1}^n \binom{k+1}{2} = \sum_{k=2}^{n+1} \binom{k}{2} = \binom{n+2}{3}$

$\binom{n+2}{3} = \frac{(n+2)!}{3! (n-1)!} =\frac{n (n+1)(n+2)}{6}$

Finally multiply by $3$ and you will get your result.

Title: Double Summation Simplification


问:
Finding integrating factor (IF) when integrating factor IF will be a function of both $x$ and $y$
I'm not finding any resource or description or systemic methodology to find integrating factors when the integrating factor will be a function of both x and y.

I'm on this problem,

$$
( y - xy^2 ) dx + (x + x^2y^2) dy = 0
$$

In its present form its not exact, but multiplying through by an integrating factor $ \frac{1}{x^2y^2} $ does make it exact.

Now I want to know how you find that, when the integrating factor is both a function of x and y.

The technique on this page for finding an integrating factor $u(x)$ fails.

Which means there's no function of purely x that will make this equation exact.

So I'm trying something like this:

1. Multiply the original equation by a function u(x,y):

$$
u(x,y)( y - xy^2 )dx + u(x,y)(x + x^2y^2)dy = 0
$$

2. Attempt to set $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $

$$
\frac{\partial}{\partial y} u(x,y)( y - xy^2 ) = \frac{\partial}{\partial x} u(x,y)(x + x^2y^2)
$$

(The two sides must be equal for u(x,y) to make the original equation exact, since $ \frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x} $ for an exact equation)

3. Now you have:

$$
u_y(x,y)(y-xy^2) + u(x,y)(1-2xy) = u_x(x,y)(x+x^2y^2) + u(x,y)(1+2xy^2)
$$

I'm kind of stuck now. What is the next step, or is this an incorrect start?

答:
The equation is of the form $M(x,y)dx + N(x,y)dy = 0$. Set $M(x,y) = 0$ to obtain $y_0(x) = \frac{1}{x}$. Set $N(x,y) = 0$ to obtain $x_0(y) = \frac{-1}{y^2}$.

Let $g(y) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M}$, given $x = x_0(y)$. Similarly, let $f(x) = \frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{-N}$, given $y = y_0(x)$. We obtain $g(y) = \frac{-2}{y}$ and $f(x) = \frac{-2}{x}$

Note that $f(x)$ and $g(y)$ satisfy $$\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = M(x,y)g(y) - N(x,y)f(x)$$. Hence an integrating factor, $\mu$, can be found by $e^{\int f(x)dx + \int g(y)dy + c}$. Doing this gives us $\mu = \frac{k}{x^2y^2}$, for arbitrary k.

The method used here can be found at https://www.ijesm.co.in/current_issue.php


问:
prove that f is constant if $f(2^x)=f(3^x)$
I am trying to prove that when $f$ is continuous for all non-negative real number and
$$f(2^x)=f(3^x),$$ then $f$ is constant.

If $f(2^x)=f(3^x)$ then from here I can say that $f(2^x/2^n)=f(3^x/3^n)$ and then I can use limit when $n\to\infty$, but I am not sure how to make connection with $f(x)$ function itself to prove.

Could you give me any hints?

Thanks in advance.

答:
Your question is problematic as stated. I suspect that you mean to ask the following:

Suppose that $f:[0,\infty)\to \Bbb R$ is continuous and that for all $x \in \Bbb R$, we have
$$
f(2^x) = f(3^x).
$$
Prove that $f$ is constant.

Here is a hint:

With the substitution $y = 3^x$, rewrite this condition as $f(y) = f(g(y))$ for all $y > 0$, where $g(y) = y^{\alpha}$, and $\alpha = \log(2)/\log(3)$.

Note that for all $0<y<1$, we have $y < g(y) < 1$. Similarly, for all $y > 1$, we have $1 < g(y) < y$.


问:
Why equivalence $\ln(1+u)_{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$ holds for $\color{red}{u\to 0}$?
Taylor series for $\ln(1+u)$ is:

$$\ln(u+1)=u-\cfrac12 u^2+\cfrac13 u^3-\cdots+\cfrac{(-1)^k(u-1)^{k+1}}{k+1}+\cdots$$
and this holds for $u\in(-1,1)$ (because $\ln(u+1)=\int\frac{1}{u+1}du=\int\sum_{n=0}^{\infty}(-u)^ndu$ and the geometry series holds for $|-u|<1$ or $u\in(-1,1)$).

But when we want to use it as equivalence for example to evaluate limits we say $\ln(1+u)_\color{red}{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$

My question is why this  holds for $\color{red}{u\to 0}$ ? Is there any reason why we choose $u$ approaches to zero?

答:
Because when approximating, we usually ignore the terms in the infinite series from a certain power. So that leftover will stay as an error between your estimation and  $\ln(1+u)$. And we can only be sure of the convergence to 0 of that error when $u \rightarrow 0$

Remark:Taylor's formula also provides  a formula to control that error.


答:
(1) : $$\sum_{i=1}^k{ i } = \frac{{(k+1)k}}{2}$$
Plugging (1) into the double sum , we get :

$${3}\sum_{k=1}^n \frac{{k^2+k}}{2} = {\frac{{3}}{2}}(\sum_{k=1}^nk^2 + \sum_{k=1}^nk )$$

$$\sum_{k=1}^n  k^2 = {\frac{{1}}{6}}({{n}*(n+1) *(2n+1)} )$$

All what you have to do is  to replace the 2 sums by their expressions and then you will get it

Title: Double Summation Simplification

Body:
I want to simplify the double summation
$$\sum_{k=1}^n (\sum_{i=1}^k 3i)$$

I started by doing

$$ 3 \sum_{k=1}^n(\sum_{i=1}^k i)$$

which would give me the following, I think

$$ 3 \sum_{k=1}^n(\sum_{i=1}^k \dfrac{n(n+1)}{2})$$

I don't know how to continue from here, I just know the final answer is supposed to be

$$\dfrac{n(n+1)(n+2)}{2}$$

Can someone give me some tips on how to continue after the second step?


问:
Average of rolling $N$ $K$-sided dice and only taking the value of the lowest die roll
This question is related, but mine is about a much more general case.

I want to roll $N$ dice $D_1, D_2, ..., D_{N}$. Each die $D_i$ has $K_i$ sides (so different dice possibly have different number of sides). Once I've rolled them, I want to keep the lowest value.

The sides of die $D_i$ are labeled $0, 1, ..., K_i-1$.

Example:

There are 4 dice (with, respectively, 4 sides, 7 sides, 2 sides, 9 sides)

I roll them, and get the values 2, 3, 1, 0

0 is the lowest values, so that's the final result of the experiment.

What is the expected value of the result? On average, what is my final result?

答:
Let $X$ be the minimum of the $N$ rolls. It takes values in $0,1,\ldots, \min_i(K_i) - 1$.

$$E[X] = \sum_{m=1}^{\min_i(K_i) - 1} P(X \ge m) = P(X \ge 1) + P(X \ge 2) + \cdots + P(X \ge \min_i(K_i) - 1)$$

If you can compute each probability $P(X \ge m)$ you can get the expectation. Note that $$P(X \ge m) = P(\text{every dice roll is $\ge m$}) = P(\text{1st roll is $\ge m$}) P(\text{2nd roll is $\ge m$}) \cdots P(\text{$N$th roll is $\ge m$}).$$

The probability that the $i$th roll is $\ge m$ is $\frac{\max(K_i - m, 0)}{K_i}$.

In the end you get a pretty gnarly expression due to the generality of the $K_i$ possibly being all different.


问:
The failure of product lemma for Sacks forcing
I am reading Jörg Brendle's Bogota note, and the author claimed that Sacks forcing with countable support product does not satisfy a product lemma. Especially, he mentioned that if $\mathbb{S}_I$, $|I|$-many countable support product of Sacks forcing $\mathbb{S}$, adds $|I|$ Sacks reals $\langle s_i\mid i\in I\rangle$ and $i\neq j$, then $s_j$ is not $\mathbb{S}$-generic over $V[s_i]$.

Let me take $I=2$ for simplicity. One subtle point of the claim is that $\mathbb{S}$ need not be absolute between $V$ and $V[s_0]$. In fact, the product lemma for general forcing shows that $s_1$ would be $\mathbb{S}^V$-generic over $V[s_0]$. Hence I think the author intends to claim that $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$.

However, $\mathbb{S}^V\neq\mathbb{S}^{V[s_0]}$ does not automatically mean $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$. My question is:

How to prove $s_1$ is not $\mathbb{S}^{V[s_0]}$-generic over $V[s_0]$?

I am new at Sacks forcing, so I have no idea how to start. I would appreciate your help!

答:
The point is that in the two-step iteration $s_1$ codes $s_0$ over the ground model in the sense that there is a Borel function $f \colon 2^\omega \to 2^\omega$ in $V$ so that $f(s_1) = s_0$. This can't happen in the product (for example $s_1$ is not a minimal real over $V$).

I don't know if there is a simple way to see this but I will sketch how I would prove it (if you want me to add details I can do that).

Lemma 1. Sacks forcing has the continuous reading of names, i.e. whenever $\dot y$ is an $\mathbb{S}$-name for an element of an arbitrary Polish space $X$ and $p \in \mathbb{S}$, then there is $q \leq p$ and a continuous function $g \colon [q] \to X$ so that $q \Vdash g(\dot s_0) = \dot y$.

This is usually formulated with $X = \omega^\omega$ but it doesn't make a difference since every Polish space is the continuous image of $\omega^\omega$. Also $[q]$ is the set of branches through $q$ and $\dot s_0$ is a name for the Sacks real that is added by $\mathbb{S}$.

Now a condition in the two-step iteration is a pair $(p,\dot q)$ where $p$ is a perfect tree and $\dot q$ is a $\mathbb{S}$-name for a perfect tree. Thus $\dot q$ is a name for an element of the Polish space $X$ of perfect trees and we find $p' \leq p$, $g \colon [p'] \to X$ interpreting $\dot q$, in the sense that $p' \Vdash g(\dot s_0) = \dot q$.

Now consider the following lemma.

Lemma 2. There is a continuous function $g' \colon [p'] \to X$ so that for every $x \in [p']$, $$g'(x) \subseteq g(x)$$ and for every $x\neq y \in [p']$, $$[g'(x)] \cap [g'(y)] = \emptyset.$$

Now we can define a name $\dot q'$ so that $p' \Vdash \dot q' = g'(\dot s_0)$. The point is that now we can read off $s_0$ from $s_1$. Define $f(x) = y$ if $y \in [p']$ and $x \in [g'(y)]$, and else $f(x) = 0$. Then $f$ is a Borel function in $V$ and $f(s_1) = s_0$.


问:
Every image of an immersion is an image of an injective immersion?
Let $M,N$ be smooth connected* manifolds, with or without boundary.
Let $f:M \to N$ be a smooth immersion. Can we realize $f(M)$ as an image of some injective immersion into $N$?

That is, does there exist a manifold $\tilde M$ and an injective smooth immersion $j:\tilde M \to N$, such that $j(\tilde M)=f(M)$?

I am particularly interested in the case where $\dim M=\dim N$, and $\partial M \neq \emptyset$. (If $\partial M = \emptyset$, the image is open, hence an embedded submanifold of $N$).

*As commented below, if we assume $M$ is not connected, there probably are examples where the image cannot be realized as an image of an injective immersion.

答:
About the immersion $f \sqcup g : \mathbb{R} \sqcup \mathbb{R} \to \mathbb{T}^2$ as two dense wrapped lines, it turns out that it can be made injective! This is done by progressively restricting the domain of $f$ and $g$ to avoid self-intersections.

By changing the speed of the curves, we may suppose that $f(\mathbb{Z}) \cap g(\mathbb{R}) = f(\mathbb{R}) \cap g(\mathbb{Z}) = \varnothing$ and define $A_n = [-n,-n+2] \cup [n-2,n]$ for every $n \in \mathbb{Z}^+$. Start by removing $A_2 \cap g^{-1}(f([-1,1]))$ from the domain of $g$. Next, erase from the domain of $f$ the set $A_3 \cap f^{-1}(g([-2,2]))$. Then erase from the domain of $g$ the set $A_4 \cap g^{-1}(f([-3,3]))$, and so on. The final domains of $f$ and $g$ will be open, because the removed set is a locally finite union of closed sets, hence closed, so $f \sqcup g$ will still be a smooth immersion of a manifold. To see injectivity, suppose that $f(x) = g(y)$, with $x, y \in \mathbb{R}\setminus\mathbb{Z}$ in the final domains of $f$ and $g$, respectively. Let $n$ be odd such that $x \in A_n$ and $m$ even such that $y \in A_m$. If $n < m$ then $A_m \cap g^{-1}(f([-m+1,m-1]))$ was erased from the domain of $g$, and this contains $y$ because $x \in A_n \subseteq [-n,n] \subseteq [-m+1,m-1]$. The other case is handled equally.


问:
Factorial simplification help; how did wolfram get this?
in this problem n is assumed to be odd:
Here's what I am trying to simplify:
$$ \frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}
$$

Wolfram is telling me that this should simplify to $$\frac {1}{2} $$ but I am very confused about how it gets there. I understand that the two factorial terms combine, but then I have $$ (n-2) (\frac {n-1} {2}) $$ as my final simplification and I really have no idea how to move from that to $$\frac{1}{2}$$ Any help would be appreciated.

答:
Here is a variation based upon the functional equation: $n!=n(n-1)!$.

We obtain
\begin{align*}
\color{blue}{\frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}}
&=\frac {(n-2)!}{(n-1)(n-2)!}\cdot \frac {\frac{n-1}{2}\left(\frac{n-3}{2}\right)!}{\left(\frac{n-3}{2}\right)!}\\
&=\frac{1}{n-1}\cdot\frac{n-1}{2}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}
and the claim follows.


答:
This involves the direct application of the remainder theorem. The remainder theorem is a generalisation of the factor theorem. Recall that the factor theorem states

Let $P(x)$ be a polynomial function. If $(x-a)$ is a factor of $P(x)$, then $P(a)=0 \, .$

It is easy to see why this theorem is true. Let $Q(x)$ be the  the quotient of $P(x)$ and $(x-a)$. Then we can write
$$
P(x) = Q(x)(x-a)
$$
This means that $P(a)=Q(a)(a-a)=0 \, .$ The remainder theorem is similar. Let's say we divide $P(x)$ by $(x-a)$, but this time we are left with the remainder $r$, meaning that
$$
P(x)=Q(x)(x-a)+r
$$
We can see that $P(a)=r$. Thus we can state the remainder theorem:

Let $P(x)$ be a polynomial function. If dividing $P(x)$ by $(x-a)$ leaves a remainder of $r$, then $P(a)=r$

To apply this to the problem at hand, set $a=1$ and $r=3$.

Title: Why if a polynomial is divided by x-1 and gives the remainder 3 then p (1) = 3?

Body: Why if a polynomial is divided by $x-1$ and gives the $remainder=3$ then $p(1) = 3$? Does it have to do with ruffini? Could you describe the steps that lead me to affirm this?


问:
Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$
Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.

This can be proved by showing

$\hspace{130pt}f(a)<\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}<f(b)$

and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.

I'm not sure how to show this though. Any suggestions?

答:
I'll leave here an alternative proof, It might be helpful to someone:

Lemma 1:
If $\{x_1,...,x_n\} \subset [a,b]$ then: $\exists x \in [a,b]:$

$$x = \frac{x_1 + ... + x_n}{n}$$

Proof:

Let $g: [a,b] \to \Bbb R$ such that:

$$g(x) = \sum_{k = 1}^n x_k - nx $$

note that $g$ is continuous. Now, for each $i \in \{1,...,n\}$, we know that $a \leq x_i \leq b$. So if we sum all $x_i$ we must get the following:

$$na \leq \sum_{k = 1}^n x_k \leq nb$$

So now we have:

$g(a) = \sum_{k = 1}^n x_k - na \geq na - na = 0$

$g(b) = \sum_{k = 1}^n x_k - nb \leq nb - nb = 0$.

So, as $g(a) \geq 0 \wedge g(b) \leq 0$ we conclude that: $\exists x \in [a,b]: g(x) = 0$.

$$\begin{align}
g(x ) &= 0
\\
\sum_{k = 1}^n x_k - nx &= 0
\\
x &= \frac{1}{n}  \sum_{k = 1}^n x_k
\end{align}$$

So we conclude that $\exists x \in [a,b]:$

$$x = \frac{x_1 + ... + x_n}{n}$$

Now we can easily prove the result.

Let's denote $ m:= \min\{f(x_1),...,f(x_n)\}$ and $M:= \max\{f(x_1),...,f(x_n)\}$.

And let $i,j \in \{1,...,n\}$ such that $f(x_i) = m$ and $f(x_j) = M$.

Now, we know that $\{f(x_1),...,f(x_n)\} \subset [m,M]$. So according to lemma 1: $\exists c \in [m,M]: $

$$c = \frac{f(x_1) + ... + f(x_n)}{n}$$

We have $c \in [m,M]$ which is the same as saying that $m \leq c \leq M$ and this means that $f(x_i) \leq c \leq f(x_j)$.

Because $f$ is continuous, using the intermediate value theorem, we know that $\exists z \in [a,b]: f(z) = c$, so we have:

$$f(z)= \frac{f(x_1) + ... + f(x_n)}{n}$$

q.e.d.


问:
LCM Inequality proof
i have a question that i cannot solve

let a,b,c be positive integers such that b divides c
prove that

```<code>LCM(a,b) <= LCM(a,c)
</code>```

so far i tried to go with the definition of gcd and division but it does not help
because i cannot prove that gcd(a,b) > gcd(a,c)
so if anyone can give me a lead or a tip that would be great!

Thank you for your help.

答: $\mathrm{lcm}(a,c)$ is by definition a multiple of $a$ and $c$. Because $b|c$, it is also a multiple of $a$ and $b$. By definition of $\mathrm{lcm}(a,b)$ as the least common multiple of $a$ and $b$, you have directly $$\mathrm{lcm}(a,b) \leq \mathrm{lcm}(a,c)$$

Title: LCM Inequality proof


问:
Infinite sum of square root
I tried demonstrate that $\displaystyle \sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}$ is convergent with $a_n$ and $b_n$ are the fourier coefficients of continuos $f(x)$ with $f(x)\in L_2(-\pi, \pi)$ and $f(\pi)=f(-\pi)$. My attempt to show it is to proof that the sum $$\sum_{k=0}^\infty a_n^2+b_n^2$$ is convergent (using Bessel inequation). But I'm not sure if $$\sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}\leq \sum_{k=0}^\infty a_n^2+b_n^2$$

or is it not necessary?

答:
NOT TRUE.

The series
$$
f(x)=\sum_{n=1}^\infty \frac{\sin nx}{n}
$$
converges conditionally for every $x\in\mathbb R$ and defines a $2\pi-$periodic function $f$, which belongs to $L^2[0,2\pi]$, since
$$
\sum_{n=1}^\infty (a_n^2+b_n^2)=\sum_{n=1}^\infty \frac{1}{n^2}<\infty.
$$

But
$$
\sum_{n=1}^\infty \sqrt{a_n^2+b_n^2}=\sum_{n=1}^\infty \frac{1}{n}=\infty.
$$

Title: Infinite sum of square root


问:
Showing that $x= y$
Suppose that $x,y,a,b \in \mathbb{R}^n$ with the Euclidean norm such that $||x-a||=||y-a||$ and $||x-b||=||y-b||$. Is it true that $x= y$? Intuitively this is true to me, but I'm not too sure how to show that. Is there any suggestion or hint?

答: Hint: If $n=2$, your question is whether or not the intersection of two circles always consists of a single point.

Title: Showing that $x= y$


问:
Manifolds with boundary and definition

Can you help for understanding this definitions in a good way.What is the my problem is that I can not image this definition and proposition in my mind and also ı dont understand the reason that ı write gren writing.Also what is actually $i_{x}w$.These are basi thinks but can someone give me light these definition and propostiton,I will so happy thanks

答:
The orientation form is defined as an every where non-zero n form.

This means that its value on any n linearlly independent vectors is always non-zero.

The boundary of the manifold has a unit inward point normal vector. At each point of the boundary the unit normal can be extended to n linearly independent vectors by adding any n-1 linearly independent vectors that are tangent to the boundary. The orientation form on the manifold is never zero on these n frames and thus defines by contraction with the unit normal an orientation form on the boundary

Title: Manifolds with boundary and definition


答: This is false even in two dimensions: set $a=(0,1),b=(0,-2),x=(1,0),y=(-1,0)$.

Title: Showing that $x= y$

Body: Suppose that $x,y,a,b \in \mathbb{R}^n$ with the Euclidean norm such that $||x-a||=||y-a||$ and $||x-b||=||y-b||$. Is it true that $x= y$? Intuitively this is true to me, but I'm not too sure how to show that. Is there any suggestion or hint?


问:
write vector as linear combination of null and column space vectors
I understand that the nullspace N(A) and row space of a matrix are orthogonal to each other and am trying to write a vector x as a linear combination of the two.

My matrix A:

$
A = \begin{bmatrix}
1 & 1 & 1 & 1 \\
1 & -1 & -1 & 1 \\
0 & 1 & 0 & 1
\end{bmatrix}
$

The rank of A = 3 and thus the Null space dimension = 1.  I have found the null space basis Z =

$
\begin{bmatrix}
-1 \\ -1 \\ 1 \\ 1
\end{bmatrix}
$

If given a point $x_1$ = $(1 \ 3 \ 1 \ 2)^{\top}$, how do I write $x = p_i + q_i $ where $p \in N(A) \ \And \ q \in R(A^{\top})$?

答:
You want $a\pmatrix{1\\1\\1\\1}+b\pmatrix{1\\-1\\-1\\1}+c\pmatrix{0\\1\\0\\1}+d\pmatrix{-1\\-1\\1\\1}=\pmatrix{1\\3\\1\\2};$ i.e.,

$$a+b-d=1\tag1$$

$$a-b+c-d=3\tag2$$

$$a-b+d=1\tag3$$ and

$$a+b+c+d=2.\tag4$$

You could solve these equations by adding $(1)$ and $(3)$ to solve for $a$ and $b-d$, subtracting $(2)$ from $(4)$ to solve for $b+d$ (hence solving for $b$ and $d$), and then using $(2)$ or $(4)$ to solve for $c$.


问:
Combinatorics , choose 5 numbers out of positive and negative numbers
Following numbers are given: $-8,-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6$

In how many ways can you choose $5$ numbers of the numbers above, if the product of these are suppost to be positive.

a) Repetition is not allowed

b) Repetition is allowed

For a) and b) order does not matter.

c) What is the answer in a) if it is important in wich order the numbers are choosen?

My attempt:
a) We need to have $0,2 $ or $4$ negative numbers if the product is suppost to be positive. I have $8$ negative numbers and $6$ positive. Order does not matter here so it should be:
$C(8,0)\times C(6,5) + C(8,2) \times C(6,3) + C(8,4) \times C(6,1)$

b) Repetition is allowed and order does not matter.

Case 1: There are 0 negative numbers and 5 positive. If $x_i$ =The amount of a postive number, $i$ is between 1 and 6, and $x_i$ is equal or larger then zero. Then I get $x_1+x_2+x_3+...+ x_6 = 5$ , wich will give me $C(10,5)$ ways

Case 2: There are 2 negative numbers and 3 postive. If $Z_i = $ The amout of $A$ negative number, i is between 1 and 8. Then I get $z_1+z_2+...+ z_8 = 2$ and $x_1+x_2+...+x_6= 3$. This gives me $C(9,2) \times C(8,3)$

Case 3: There are 4 negative numbers and 1 positive. Then I get $z1+z2+...+z8 = 4$ and $x1+x2+...+x6 =1$. This gives me $C(11,4) \times C(6,1)$

The sum= $4248$

c) Order is important and repitition is not allowed.

Case 1: 5 postive, 0 negative
$P(6,5) = 720$

Case 2: 2 negative, 3 positive
$C(5,2) \times P(8,2) \times P(6,3) = 67200$

Case 3: 4 negative, 1 positive
$C(5,4) \times P(8,4) \times P(6,1) = 50400$

Total =$118320. $

Am I correct on a), b) and c)?

答:
Yes your answers are correct. For part (c), you may want to just use your answer from part $a$ -

$\displaystyle {8 \choose 0} {6 \choose 5} + {8 \choose 2} {6 \choose 3} + {8 \choose 4} {6 \choose 1} = 986$

Now once you have the $5$ numbers you can arrange them in $5!$ ways as order matters for part (c).

So, total number of ways for part (c) $ = 986 \times 5! = 118320$ which is same as your answer.


问:
Uniqueness of adjoint functors up to isomorphism
Suppose we are given functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G,G':\mathcal{D}\rightarrow\mathcal{C}$ such that $G$ and $G'$ are both right adjoint to $F$. To show that $G$ and $G'$ are isomorphic, we wish to come up with a natural isomorphism between the two functors. This is how Awodey does it:

For any $D\in\mathcal{D}$ and any $C\in\mathcal{C}$, we have that $Hom(C,GD)\cong Hom(FC,D)\cong Hom(C,G'D)$ simply using the hom-set definition of adjunctions. By the Yoneda principle, this means that $GD\cong G'D$ for all $D\in\mathcal{D}$.

Awodey then states that

But this isomorphism is natural in $D$,  again by adjointness.

I dont' quite understand how the naturality follows from adjointness. Any help is appreciated.

答:
The bijection $\hom(C,GD) \cong \hom(C,G'D)$ is natural in $C$, thus it is induced by an isomorphism $GD \cong G'D$ (Yoneda Lemma). But it is also natural in $D$, and since the Yoneda embedding is faithful, this means that $GD \cong G'D$ is natural in $D$. More details:

If $D \to E$ is a morphism, then

$\begin{array}{ccc} GD & \rightarrow & G'D \\ \downarrow & & \downarrow \\ GE & \rightarrow & G'E \end{array}$

commutes iff for every $C$ the diagram

$\begin{array}{ccc} \hom(C,GD) & \rightarrow & \hom(C,G'D) \\ \downarrow & & \downarrow \\ \hom(C,GE) & \rightarrow & \hom(C,G'E) \end{array}$

commutes.


问:
How to relax an Integer Linear Programming model and describe it as a minimal cost flow problem?
I have the following Integer Linear Programming model:

$\text{min} Z = 6x_{1,3} + 5x_{1,4} + 2x_{1,2} + 3x_{2,3} + 5x_{2,4} + 2x_{3,5} + 4x_{4,5} + 100(y_{1,3} + y_{1,4} + y_{1,2} + y_{2,3} + y_{2,4} + y_{3,5} + y_{4,5})$

The constraints are :

$y_{i,j} = \begin{cases}1 \text{ if  } x_{i,j} > 0 \\ 0 \text{ if } x_{i,j} = 0 \end{cases}$

$x_{1,3} + x_{1,4} + x_{1,2} \leq 20$

$x_{1,3} \leq 10$

$x_{1,4} \leq 20$

$x_{1,2} \leq 20$

$x_{2,3} + x_{2,4} \leq 30$

$x_{2,3} \leq 25$

$x_{2,4} \leq 30$

$x_{3,5} \leq 30$

$x_{4,5} \leq 30$

Now I'm tasked with relaxing the above model and to describe it as a minimal cost flow problem.

What I'm trying to do is to first come up with a minimal cost flow problem without taking into account the binary variables of the previous model.

This is what I get:

$\text{min} Z = 6x_{1,3} + 5x_{1,4} + 2x_{1,2} + 3x_{2,3} + 5x_{2,4} + 2x_{3,5} + 4x_{4,5} $

With the following constraints :

$ x_{1,3} + x_{1,4} + x_{1,2} = 20 $

$ x_{2,3} + x_{2,4} - x_{1,3} = 10 $

$ x_{3,5} - (x_{1,3} + x_{2,3}) = 0$

$ x_{3,5} - (x_{1,4} + x_{2,4}) = 0$

$ -(x_{3,5} + x_{4,5}) = -30$

How can I take into account the binary variables of the first model in this minimal cost flow problem?

答:
To link the $x$ and $y$ variables, let $M_{i,j}$ be a constant upper bound on $x_{i,j}$ and introduce linear big-M constraints
$$x_{i,j} \le M_{i,j} y_{i,j}$$
that enforce $x_{i,j} > 0 \implies y_{i,j} = 1$.
For example $M_{1,3} = 10$.


问:
$f: F[x] \to F[x]$ is a isomorphism if $F$ is a domain?
Let $F$ be a field and $f: F[x] \to F[x]$ with $f\left(p\right) := p\left(ax\right)$. In fact, $f$ is a isomorphism and this is a elementary proof, with the "one-to-one" part of it just by showing that $f(g) = f(h) \Rightarrow g=h$ and the properties of a homomorphism did by using the identities of sum and product of polynomials (the last one was found here: https://math.stackexchange.com/q/1937571).

Ok, now comes the question: If $F$ is, now, a domain, the result still true? Because in my proof I can't see any necessity of it been an field. If someone have some clue, I'll aprecciate!

答:
You can prove the following more general result:

Lemma Let $R$ be a commutative ring with unity. Then $f: R[x] \to R[x]$ defined by $f(p)=P(aX)$ is an isomomorphism if and only if $a$ is invertible in $R$.

For the direct implication you can show that $g(P)=P(a^{-1}X)$ is the inverse.

For the inverse implication, use the fact that there exists some $P \in R[X]$ such that
$$f(P)=X$$


问:
Question about offspring and making a function regarding the population of offspring
A homologous gene has two alleles $e$ and $E$, with $e$ being recessive and harmful. The proportion of individuals of genotype $Ee$ in the healthy population is $\lambda$. If two parents are both healthy & they have one healthy child. If they then have a second child, what is the probability it will also be healthy as a function of $\lambda$?

So I know that if both parents are healthy their genotypes will either be $Ee$ or $EE$ but I don't really know how to continue on from this fact.

答:
What do we know before the first child is born?  At that point, we have no information at all regarding the genetic type of the parents so we must assume that they follow the average population statistics.  That is to say, they are both $Ee$ with probability $\lambda^2$, one is $Ee$ and the other is $EE$ with probability $2\lambda(1-\lambda)$ and they are both $EE$ with probability $(1-\lambda)^2$.

If they are both $Ee$ then the probability that the first child is healthy is $\frac 34$.  In all other cases, the probability is $1$.

Thus our a priori probability that the first child is healthy is $$\Psi=\frac 34\lambda^2+2\lambda(1-\lambda)+(1-\lambda)^2$$

The portion of that which is explained by both parents having type $Ee$ is, of course, $\frac 34\lambda^2$.

Thus, after seeing the first healthy child, we re-estimate the probability that the parents are both $Ee$ as $$P'=\frac 34\lambda^2\Big /\Psi$$

After some calculating we see that this is $$P'=\frac 3{\lambda+2}-\frac 3{\lambda-2}-3$$

Exercise:  confirm that $P'$ is in fact a possible probability, i.e. that it is between $0$ and $1$.  Show also that it is $≤\lambda^2$ with equality only at $\lambda=0,1$. (that makes sense since, for sensible $\lambda$, the first healthy child is evidence, albeit fairly weak, that the parents are not both $Ee$.)

Of course, now that we have the revised estimate for the probability that they are both $Ee$ we see that the probability that the second child is unhealthy is $\frac 14P'$, so the probability that the second child is healthy is $$\boxed {1-\frac 14P'}$$


问:
Showing that a sequence is decreasing
I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $

is decreasing. The fact that $2 < a_n < 3 $ has been proven already.

I now just need to show that $a_n$ is decreasing.

I have tried the usual ways of showing $a_{n+1} - a_n \le 0 $ and $\frac{a_{n+1}}{a_n} \le 1 $

but could not get anywhere with these.

答:
Hint

Define $$e_n=a_n-2$$which leads to
$$e_{n+1}=a_{n+1}-2={a_n^2-4\over 5}={e_n}{e_n+4\over 5}$$and prove $e_n$ is decreasing to $0$.

Remark

The reason I subtracted $2$ from $a_n$ is that the sequence seems to tend to $2$, hence this is a good starting point.

Title: Showing that a sequence is decreasing


问:
Invert probabilities (lowest value having highest probability in a set)
I have a set of values $(v_1,v_2,..,v_n)$ positive integers. I want to assign probabilities depending on the value, with the lowest having the highest probability. The sum of probabilities should be $1$ for the whole set.

I can do it the other way around if I find $t = \sum v_i$ then $p(v_i) = \frac{v_i}{t}$ but the lowest value has the least probability.

答:
A natural assignement is
$$p(v_i)=\frac{1/v_i}{\sum_i1/v_i}$$
Since all $v_i$ are positive, we can safely reciprocate them, which turns high values into low values and vice versa. Then we apply the "other way round" assignment.

This has the advantage that same values have same probabilities.


Title: Showing that a sequence is decreasing

Body:
I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $

is decreasing. The fact that $2 < a_n < 3 $ has been proven already.

I now just need to show that $a_n$ is decreasing.

I have tried the usual ways of showing $a_{n+1} - a_n \le 0 $ and $\frac{a_{n+1}}{a_n} \le 1 $

but could not get anywhere with these.

Body: If $2<a_n<3$, then $(a_n-2)(a_n-3)<0$, so $a_n^2-5a_n+6<0$, so $a_{n+1}=\frac15(a_n^2+6)<a_n.$


问:
Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.

$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$

Is there some other more effective way for this limit?

Body:
Writing $x=u^3$ for convenience, we have

$$\begin{align}
{1-\sqrt[3]{\cos x}\over1-\cos\sqrt[3]x}
&={1-\sqrt[3]{\cos u^3}\over1-\cos u}\\
&={1-\cos u^3\over1-\cos u}\cdot{1\over1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3}}\\
&={1-\cos^2u^3\over1-\cos^2u}\cdot{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&=u^4\left(\sin u^3\over u^3\right)^2\left(u\over\sin u\right)^2{1+\cos u\over(1+\cos u^3)(1+\sqrt[3]{\cos u^3}+\sqrt[3]{\cos^2u^3})}\\
&\to0\cdot1\cdot1\cdot{1+1\over(1+1)(1+1+1)}\\
&=0
\end{align}$$


问:
Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$
Is there a closed form for the following infinite product?
$$\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}$$</p>

答:
Since
$$
\Gamma\left(n+\frac12\right)=\frac{(2n)!}{n!4^n}\sqrt\pi\quad\text{and}\quad\Gamma(n)=(n-1)!
$$
We have
$$
\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n)}=\color{#C00000}{\binom{2n}{n}}\frac{\color{#0000FF}{n}}{\color{#00A000}{4^n}}\sqrt\pi
$$
Letting $n=2^k$, and taking logs, we get the log of the partial product to be
$$
\begin{align}
&\sum_{k=1}^n\left(\color{#C00000}{\frac1{2^k}\log\left(2^{k+1}!\right)-\frac1{2^{k-1}}\log\left(2^k!\right)}\color{#00A000}{-\log(4)}\right)+\sum_{k=1}^\infty\left(\color{#0000FF}{\frac{k}{2^k}\log(2)}+\frac1{2^{k+1}}\log(\pi)\right)\\
&=\color{#C00000}{\frac1{2^n}\log\left(2^{n+1}!\right)-\log(2)}\color{#00A000}{-n\log(4)}\color{#0000FF}{+2\log(2)}+\frac12\log(\pi)\\
&\stackrel{\text{Stirling}}\sim\frac1{2^n}\left(2^{n+1}(n+1)\log(2)-2^{n+1}+\frac12\log(2^{n+2}\pi)\right)+(1-2n)\log(2)+\frac12\log(\pi)\\
&=(2n+2)\log(2)-2+\frac1{2^{n+1}}\log(2^{n+2}\pi)+(1-2n)\log(2)+\frac12\log(\pi)\\
&=\log(8)-2+\frac12\log(\pi)+\frac1{2^{n+1}}\log(2^{n+2}\pi)\\
&\stackrel{n\to\infty}\to\log(8)-2+\frac12\log(\pi)
\end{align}
$$
Thus, the limit of the product is
$$
\frac{8\sqrt\pi}{e^2}
$$</p>


问:
Show sequential characterisation of closure from some definition of the closure of a subset
Let $E$ be a metric space and $A \subset E$.

The closure of $A$ is defined as: $\bar{A} := \underset{F \textrm{ closed subset} \\ A  \subset F}{\bigcap} F$

Starting from this definition, how can I show that $x \in \bar{A}$ iff there exists a sequence $(x_k)$ with values in $A$ which converges towards x?

I don't see how I can use the definition to make a link with any sequence. Thanks for the help!

EDIT:
For me, a closed set is the complementary of an open set.
An open set is a set which is a vicinity of each of its elements.

答:
SKETCH: Suppose first that $x\in E\setminus\operatorname{cl}A$; then there is a closed set $F$ such that $A\subseteq F$ and $x\notin F$. Use this to show that there is an $\epsilon>0$ such that $A\cap B(x,\epsilon)=\varnothing$ and conclude that no sequence in $A$ can converge to $x$. This will show that if $x$ is the limit of a sequence in $A$, then $x\in\operatorname{cl}A$.

It remains to show that if $x\in\operatorname{cl}A$, then there is a sequence in $A$ that converges to $x$. Let $\epsilon>0$; $E\setminus B(x,\epsilon)$ is a closed set. If $A\subseteq E\setminus B(x,\epsilon)$, we would have $x\in\operatorname{cl}A\subseteq E\setminus B(x,\epsilon)$, which is absurd. Thus, $A\nsubseteq E\setminus B(x,\epsilon)$, and therefore $A\cap B(x,\epsilon)\ne\varnothing$. This is true for every $\epsilon>0$, so for each $n\in\Bbb Z^+$ there is an $x_n\in A\cap B\left(x,\frac1n\right)$; now consider the sequence $\langle x_n:n\in\Bbb Z^+\rangle$.


问:
The set $D = \{ p \in (0,\infty) \mid \int_X |f|^p ~d\mu < \infty\}$
Let $\mu$ be a (positive real) measure on a space $X$ and $f$ a measurable function. Put $F: (0,\infty) \rightarrow \overline{\mathbb R}_+, p \mapsto \int_X |f|^p ~d\mu$. Write $D = \{ p \in (0,\infty) \mid F(p) < \infty\}$.

It is easy to see that $F(p) \le F(r) + F(s)$ if $r<p<s$, whence the fact that $D$ is connected alias an interval. Moreover, $\log F$ is convex by Hölder's inequality.

Is there $\mu$ and $f$ such that $D$ has exactly one element?

The example of $\mu$ Lebesgue-measure on $(0,1)$ resp. $(1,\infty)$ and $f(x) = x^\alpha$ shows that $D$ can be an open interval. Can it be closed? Left-open, right-closed? Left-closed, right-open?

In my intuition, $D$ should be open, at least left-open in general. What do you think?

This is an (as I guess, standard) exercise, taken from the Big Rudin. I apologize if this is a duplicate, thank you for indicating the answering article then.

答:
It's a fun exercise. It is mainly based on two measures $\mu_1,\mu_2$ defined on $\mathbb{R}$ as follows :
$$ \mu_1= \sum_{n=2}^{\infty} \frac{1}{n^2ln(n)^2}\delta_{n}$$
and
$$ \mu_2= \sum_{n=2}^{\infty} \frac{1}{ln(n)^2} \delta_{\frac{1}{n}}$$
With $f(x)=x$ and $\mu:= \mu_1 +\mu_2$. We see clearly that :

$$ D^{\mu}_f = \{1\}$$
(which is also a closed set)
$$ D^{\mu_1}_f = (0,1]$$
$$ D^{\mu_2}_f= [1,\infty)$$


答:
The inequality holds in the opposite direction with minor changes. Since $\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n=0$ for $f(x)\in L_2(-\pi,\pi)$, we have some $M$ such that $$n\ge M\implies \sqrt{a_n^2+b_n^2}<1$$hence
$$
\sum_{n=M}^\infty {a_n^2+b_n^2}<\sum_{n=M}^\infty \sqrt{a_n^2+b_n^2}
$$

Title: Infinite sum of square root

Body:
I tried demonstrate that $\displaystyle \sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}$ is convergent with $a_n$ and $b_n$ are the fourier coefficients of continuos $f(x)$ with $f(x)\in L_2(-\pi, \pi)$ and $f(\pi)=f(-\pi)$. My attempt to show it is to proof that the sum $$\sum_{k=0}^\infty a_n^2+b_n^2$$ is convergent (using Bessel inequation). But I'm not sure if $$\sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}\leq \sum_{k=0}^\infty a_n^2+b_n^2$$

or is it not necessary?


问:
Conditional Probabilities defective items
A company sends 30% of its product to Client A and 70% to Client B. Client A reports that 5% of the products it received are defective, whereas Client B reports that 4% of products received are defective. The defective products are returned back to the company.

What is the probability of a returned defective product coming from client B?

probability of an item being sent to customer A  = $P(C_a) =  0.30$ & $P(D|C_a) = 0.05$

probability of an item being sent to customer B  = $P(C_b) =  0.70$ & $P(D|C_b) = 0.04$

$P$(Returned) = $P$(Defective)= $P(D)$ = $0.3 \times 0.05 + 0.7\times 0.04 = 0.043$

My thinking is that I need to find $P$( $C_b$ | Defective ) $= P( C_b | D )$

$P( C_b | D ) = \frac{P(D \,\cap \,C_b)}{P(D)} = \frac{P(D \,\cap \,C_b)}{0.043}\;$ so I need to find $P(D \,\cap \,C_b)$

$P(D|C_b) = \frac{P(D \,\cap \,C_b)}{P(C_b)}\;$ so $0.04 = \frac{P(D \,\cap \,C_b)}{0.70}\;$ which implies $P(D \,\cap \,C_b) = 0.028$

$P( C_b | D ) = \frac{0.028}{0.043} = \frac{28}{43}$

however I feel my answer may be wrong. Am i on the right track?

Body: That's right. You could just say (.04)(.70) / [(.04)(.70) + (.05)(.30)] = .028 / .043 = .651


问:
find triangle in a complete graph having minimum edge weight sum
Given a complete undirected graph $G(V, E)$, with $n$ vertices and every edge having a positive weight $E_{ij} \gt 0$.

How can one efficiently find a triangle/three vertices $v_1, v_2, v_3$ such that the sum of edges between them is minimum?

$\arg\min_{i, j, k} E_{ij}+E_{ik} + E_{jk}, s.t. i\neq j \neq k$

答:
One approach is to use integer linear programming as follows.  Let binary decision variable $x_i$ indicate whether vertex $i$ is selected, and for $(i,j)\in E$, let $y_{i,j}$ represent $x_i x_j$.  The problem is to minimize $$\sum_{(i,j)\in E} E_{i,j} y_{i,j}$$ subject to linear constraints
\begin{align}
\sum_i x_i &= 3 \tag1 \\
x_i + x_j - 1 &\le y_{i,j} &&\text{for $(i,j)\in E$} \tag2 \\
\end{align}
Constraint $(1)$ selects three vertices.
Constraint $(2)$ enforces $(x_i \land x_j) \implies y_{i,j}$.


问:
Why if a polynomial is divided by x-1 and gives the remainder 3 then p (1) = 3?
Why if a polynomial is divided by $x-1$ and gives the $remainder=3$ then $p(1) = 3$? Does it have to do with ruffini? Could you describe the steps that lead me to affirm this?

答:
What does $P(x)$ having remainder $3$ after being divided by $3$ mean?

It means that there is a polynomial $\frac {P(x)}{x-1} = Q(x) + \frac 3{x-1}$ or in other words that $P(x) = Q(x)(x-1) + 3$.

So $P(1) = Q(1)(1-1) + 3 = Q(1)\cdot 0 + 3 = 0 + 3 =3$.

That's all.

======

If you really want details.  Suppose $P(x) = a_nx^n + a_{n-1}x^{n-1} + ..... + a_1x + a_0$ is an $n$th degree polynomial and $a_k$ are all rational numbers.

Now if $T(x) = b_{n-1}x^{n-1} + b_{n-2}x^{n-1} + .... + b_1x + b_0$ is some $n-1$ degree polynomial where $b_k$ are all rational numbers.

Then $T(x)(x-1) = b_{n-1}x^n + (b_{n-2}-b_{n-1})x^{n-1} + ......+ (b_1-b_2)x^2 + (b_0-b_1)x -b_0$ is an $n$ degree polynomial.

Now that means if we set $b_{n-1} = a_n; b_{n-2} =a_{n-1}+b_{n-1};....b_{k-1}= a_k+b_k$ down to $b_0=a_1 +b_1$ and then if we let $R=a_0 + b_0$, we get a unique $n-1$ degree polynomial, $Q(x)$ and a unique constant $R$, where$P(x) = Q(x)(x-1) + R$.

It is pretty easy to see this is the only such polynomial.  We must have $b_{n-1} = a_n$ and that means $b_{n-2}$ must equal $a_{n-1}+b_{n-1}$ and so on.

Note: It will always be the case that $P(1) = Q(1)(1-1) + R = R$.

So if we are told $R=3$, that's it.  $P(1) = 3$.


问:
Why is it necessary to use sin or cos to determine heading? (dead reckoning)
Here's the problem:

(see pic for problem): https://fbcdn-sphotos-c-a.akamaihd.net/hphotos-ak-ash3/21281_10152793202590262_1804321932_n.jpg

You have a robot that is moving forward at a variable rate delta distance and you need to work out dead reckoning i.e. how far you're moving to the left or the right (slide).

What I don't understand is what makes sin appropriate for y and not x and vica versa. I wanted to ask for a simple example or explanation to explain what is going on here.

答:
It just depends on how you're defining your axes and your heading.

If you go with the standard definitions, i.e., the $x$-axis is horizontal, the $y$-axis is vertical, and heading is measured as "the counter-clockwise angle from the positive $x$-axis", then $\Delta D \cdot \cos{\theta} = \Delta x$ and $\Delta D \cdot \sin{\theta} = \Delta y$. That relies on the unit circle definition of sin and cos, which use the same conventions.

Using other conventions will throw in opposite signs, or switch sin and cos. For example, defining heading as "clockwise angle from the positive $y$-axis (like North)" means you switch sin and cos.

EDIT: Diagram! In each of these, I used SOH-CAH-TOA to find $x$ and $y$. In the third, I had to convert from an obtuse angle to an acute one, and to do so, I used its supplementary angle ($180^\circ - \theta$), and some trig identities. Notice that the first and second ones have the same form, even though all the definitions are different.

I would advise sticking to the first form, because it is standard in most of geometry and (in theory) programming. The third might be useful for screen coordinates, and the fourth might be good for systems which use deviation from north as their heading. You could also just convert those $x$s, $y$s, and $\theta$s into a form suitable for the first system though.


问:
Prove the system of ODE is asymptotically stable using Lyapunov function.
Given the system of ODE
$$x_1'=x_2$$
$$x_2'=-a \sin x_1- b x_2$$
Where $a,b >0$ real numbers.
Using the Lyapunov function $$v(x_1,x_2)=\int_0^{x_1} a \sin{s}\,ds+\frac{1}{2} X^TPX$$
where $X=(x_1,x_2)$ and P some matrix, prove the system is asymptotically stable.

I need to find an appropriate matrix $P$ so that $v$ satisfies $v(0,0)=0$, $v(x_1,x_2)>0$ and $\frac{dv(0,0)}{dt}=0$, $\frac{dv}{dt}<0$,

I tried many values for $P$ but couldn't couldn't come up with anything.
Can you help me prove this?

Body:
Let
$$ v=Ax_1^2+2Bx_1x_2+Cx_2^2+a\int_0^{x_1}\sin sds. $$
Then
\begin{eqnarray}
\frac{dv}{dt}&=&2Ax_1x_1'+2B(x_1'x_2+x_1x_2')+2Cx_2x_2'+a\sin x_1\cdot x_1'\\
&=&2Ax_1x_2+2B(x_2^2-ax_1\sin x_1-bx_1x_2)+2C(-ax_2\sin x_1-bx_2^2)+ax_2\sin x_1\\
&=&2(A-bB)x_1x_2+a(1-2C)x_2\sin x_1-2bBx_1\sin x_1+2(B-bC)x_2^2.
\end{eqnarray}
Let
$$ A-2bB=0, 1-2C=0$$
which implies $A=2bB,C=\frac12$. The positive definiteness of $Ax_1^2+2Bx_1x_2+Cx_2^2$ requires
$$ B^2-AC<0 $$
or $$B^2-bB<0 \text{ or }B<b.$$
Choose $B=\frac14b, A=\frac12b^2$ and then
$$ \frac{dv}{dt}=-\frac12b^2x_1\sin x_1-\frac14bx_2^2\le0. $$


问:
Pair of functions with same values and same derivatives at distinct points
Let $f,g:[0,1]\to \mathbb{R}$ be two functions of class $C^1$ such that $f'(x)>0$ and $g'(x)>0$ for all $x\in [0,1]$. Assume that
$$
f(0) = g(0) \qquad \text{and} \qquad f(1) = g(1).
$$
Show that there exist $x,y\in[0,1]$ such that
$$
f(x) = g(y) \qquad \text{and} \qquad f'(x) = g'(y).
$$

I made the following attempts (non of them worked): First, define a function $h:[0,1]\times [0,1]\to \mathbb{R}$ given by
$$
h(x,y) = [f(x)-g(y)]^2 + [f'(x)-g'(y)]^2
$$
and try to show that $h(x,y) = 0$ for some $(x,y)$. Then I defined $k:[0,1]\times [0,1] \to \mathbb{R}$ by
$$
k(x,y) = \frac{1}{2}[f(x)-f(y)]^2
$$
and tried to show that at some point $(x,y)$ the divergence of $k$ is zero, because in this case we have that
$$
0 = \frac{\partial k}{\partial x}(x,y) + \frac{\partial k}{\partial y}(x,y) = (f(x)-g(y))f'(x)-(f(x)-f(y))g'(y) = (f(x)-f(y))(f'(x)-g'(y))
$$
but it didn't work either. May be I'm not following the right way to solve this exercise.

答:
Define the paths $\gamma_f : [0,1] \rightarrow \mathbb{R}^2$ and $\gamma_g : [0,1] \rightarrow \mathbb{R}^2$ by
$$\forall x \in [0,1], \quad \gamma_f(x)=\left(f(x), \frac{1}{f'(x)}\right) \quad \quad \text{and} \quad \quad \gamma_g(x)=\left(g(x),\frac{1}{g'(x)}\right)$$

Because $f$ and $g$ are strictly increasing, then these two paths can be seen as the graph of two functions $F$ and $G$, defined on the interval $[a,b]$, where $a=f(0)=g(0)$ and $b=f(1)=g(1)$.

Moreover, using substitutions $t=f(u)$ and $y=g(u)$, one has $$\int_a^b F(t)dt = \int_0^1 \frac{f'(u)}{f'(u)} \mathrm{du} = 1 = \int_0^1 \frac{g'(u)}{g'(u)} \mathrm{du} = \int_a^b G(y)\mathrm{dy}$$

So the graphs of $F$ and $G$ intersect, which means that the paths $\gamma_f$ and $\gamma_g$ intersect, and you are done.


问:
Volume of a box inside an ellipsoid: is it correct?
I gotta find the maximum volume of a box which edges are parallels to the coordinate axes and that is inside of the ellipsoid
$$\frac{x^{2}}{4}+\frac{y^{2}}{81}+\frac{z^{2}}{196}=1$$

Since the volume is
$$V=xyz$$
I found that $\lambda=42\sqrt{3}$ using Lagrange's method. So
$$x=\frac{2\sqrt{3}}{3},\quad y=3\sqrt{3},\quad z=\frac{14\sqrt{3}}{3}$$
and then
$$V=28\sqrt{3}$$
I want to know if what I've done is right.

答: Transform the ellipsoid into the unit sphere – the scale factors are $2,9,14$ in the $xyz$ directions respectively. The cube inscribed in the unit sphere has edge length $\frac2{\sqrt3}$ and volume $\frac8{3\sqrt3}$. Multiplying this by $2,9,14$ gives the volume of the largest axis-aligned box in the ellipsoid as $224\sqrt3$ – eight times your answer, since you only calculated one octant.


问:
Where in my work is my math falling apart when solving for the partial derivative of Residual sum of squares (Linear Algebra)
I am given the following equation:
$$RSS(B, \alpha) = \sum_{i=1}^{N} (y_{i} - B^{T}x_{i} - \alpha)^{2} $$

My steps are as follows. I provided the images to my work at the bottom and this is just a description of my thought process.

I define $\lambda_{i} =  y_{i} - B^{T}x_{i} - \alpha$

I then notice that $\lambda^{T}\lambda = \lambda_{1}^{2} + \lambda_{2}^{2} + \ldots + \lambda_{n}^{2}$

Since I defined earlier that $\lambda_{i} =  y_{i} - B^{T}x_{i} - \alpha$, I can substitute that back in and I am now able to represent the original summation.

I then look at the vector $\lambda$ and I notice that each individual element is:
$$\begin{bmatrix}y_{1} - B^{T}x_{2} - \alpha \\ y_{2} - B^{T}x_{2} - \alpha \\ \vdots \\ y_{n} - B^{T}x_{n} - \alpha\end{bmatrix}$$
and that can be broken down into 3 different column vectors $Y, B^{T}X,\overline{\alpha}$

So $\lambda = Y - B^{T}X - \overline{\alpha}$ and I get that:
$$RSS(B, \alpha) = (Y - B^{T}X - \overline{\alpha})^{T}(Y - B^{T}X - \overline{\alpha})$$

After distributing the transpose and multiplying everything out, once I take the partial derivative with respect to $B$ and $\alpha$ I get the following two equations.

$$\frac{\partial}{\partial B} = X^{T}Y - X^{T}XB - X^{T}\overline{\alpha}$$
$$\frac{\partial}{\partial \alpha} = Y - B^{T}X - \overline{\alpha}$$

but after solving for $B$ and $\alpha$ I get that $0 = 0$ and I am lost on how to continue from there. I know there is something wrong with my math but I am having trouble identifying it. I never learned how to take derivatives of matrices in school so I am basing all my knowledge on http://www.gatsby.ucl.ac.uk/teaching/courses/sntn/sntn-2017/resources/Matrix_derivatives_cribsheet.pdf

work part 1

work part 2

答:
$\def\p{\partial}$
Let's use a naming convention where an uppercase Latin letter denotes a matrix, lowercase Latin a column vector, and lowercase Greek a scalar.

Then define the variables
$$\eqalign{
J &= {\tt11}^T
 &\quad\big({\rm All\,Ones\,Matrix}\big)\\
C &= I-\tfrac 1nJ
 &\quad\big({\rm Centering\,Matrix}\big) \\
M &= X^+C \\
X &= \big[\,x_1\;x_2\;\ldots\;x_n\big]^T \\
b &= B \\
w &= Xb + \alpha{\tt1} - y
 &\quad\big({\rm Residual\,Vector}\big) \\
}$$
where $X^+$ is the Moore-Penrose inverse of $X$.

Write the RSS function in terms of these new variables
and calculate its differential.
$$\eqalign{
 \rho &= w^Tw \\
d\rho &= 2w^Tdw \\
 &= 2w^T(X\,db + {\tt1}\,d\alpha) \\
}$$
Holding $b$ constant (so that $db=0$)
yields the gradient with respect to $\,\alpha$.
$$\eqalign{
d\rho &= 2w^T{\tt1}\,d\alpha \\
\frac{\p \rho}{\p \alpha}
 &= 2(w^T{\tt1}) = 2({\tt1}^Tw) \\
 &= 2\left({\tt1}^TXb +n\alpha -{\tt1}^Ty\right) \\
}$$
Set this gradient to zero and solve for the optimal $\alpha$.
$$\eqalign{
\alpha &= \tfrac 1n\,{\tt1}^T\big(y-Xb\big) \\
\alpha{\tt1} &= \tfrac 1n\,J\big(y-Xb\big) \;=\; (I-C)\,(y-Xb) \\
y-\alpha{\tt1} &= Cy + (I-C)Xb \\
}$$
Similarly, holding $\alpha$ constant yields
the gradient with respect to $b$.
$$\eqalign{
d\rho &= 2w^TX\,db = 2(X^Tw)^Tdb \\
\frac{\p \rho}{\p b}
 &= 2X^Tw = 2\big(X^TXb +\alpha X^T{\tt1} -X^Ty\big) \\
}$$
Set the gradient to zero and solve for the optimal $b$.
$$\eqalign{
&X^TXb = X^T(y-\alpha{\tt1}) \\
&b = X^+(y-\alpha{\tt1}) \\
&b = X^+\Big(Cy - (C-I)Xb\Big) \\
&\Big(I+X^+(C-I)X\Big)b = X^+Cy \\
&\Big(X^+CX\Big)b = X^+Cy \\
&b = \big(X^+CX\big)^+X^+Cy \\
}$$
The following parameter combinations will be very useful.
$$\eqalign{
Xb &= \big(X^+C\big)^+\big(X^+C\big)y &\doteq\; M^+My \\
\alpha{\tt1} &= (I-C)\,(y-Xb)  &\doteq (I-C)(I-M^+M)y \\
}$$
Substituting the optimal parameter values yields
$$\eqalign{
w &= Xb +\color{red}{\alpha{\tt1}} -y \\
 &= M^+My +\color{red}{(I-C)y +(C-I)M^+My} -y \\
 &= C(M^+M-I)\,y \\
\rho &= w^Tw \\
 &= y^T(M^+M-I)^TC^TC(M^+M-I)\,y \\
 &= y^T\left(I-M^+M\right)C\left(I-M^+M\right)y \\
\\
}$$

So that's how you would solve the problem if you treat $\alpha$ as a separate variable. But what most people do instead is use augmented variables by prepending ${\tt1}$ to each $x_k$ vector and prepending $\alpha$ to the $b$ vector.

Then the algebra becomes much simpler, i.e.

$$\eqalign{
X &= \big[\,\hat x_1\;\hat x_2\;\ldots\;\hat x_n\big]^T,\qquad
\hat x_k = \pmatrix{{\tt1}\\x_k},\qquad b = \pmatrix{\alpha\\B} \\
w &= Xb - y \\
\rho &= w^Tw \\
d\rho &= 2(X^Tw)^Tdb \\
\frac{\p \rho}{\p b} &= 2X^T(Xb-y) = 0 \\
b &= X^+y \\
w &= (XX^+-I)y \\
\rho &= y^T(I-XX^+)^T(I-XX^+)y \\
  &= y^T(I-XX^+)y \\
}$$


问:
Definition of Cell Decomposition?
In Chapter 5 of Lee's Intro to Topological Manifolds (page 130), he defines a cell decomposition as follows:

I've been struggling to properly unpack this characterization. I have the two following confusions.

a) When he says the map $\phi:D\to X$ such that $\phi|_{\mathrm{Int}D}$ is a homeomorphism between $\mathrm{Int} D$ and $e$, I'm not sure what exactly is the interior of $D$, as $D$ is simply some topological space that is homeomorphic to a closed unit ball in $\mathbb{R}^n$. Thus, in a sense, wouldn't $\mathrm{Int}D= D$ because $D$ is open in $D$ itself? But of course this doesn't make much sense.

b) When Lee says $\phi$ maps $\partial D$ into the union of all cells of $\mathcal{E}$ of dimensions strictly less than $n$, which of the following does he mean?
$$\Phi(\partial D) \subseteq \bigcup_{e\in \mathcal{E}'}e\text{ such that }\mathcal{E}' = \{e\in \mathcal{E}:\text{dimension of }e\text{ is}<n\}$$
$$\Phi(\partial D)  = \bigcup_{e\in \mathcal{E}''}e\text{ such that }\mathcal{E}'' \subseteq \mathcal{E}'\text{ where }\mathcal{E}'\text{ is as defined above}$$
$$\Phi(\partial D) = \bigcup_{e\in \mathcal{E}'}e\text{ where }\mathcal{E}'\text{ is as defined above}$$
Any guidance is appreciated, thanks!

答:
(a) Note that I explained these uses of the notations $\partial D$ and $\operatorname{Int} D$ on page 129, in the paragraph just before the subsection heading. It says there that if $D$ is a closed $n$-cell, then $\partial D$ and $\operatorname{Int} D$ denote the images of $\mathbb S^{n-1}$ and ${\mathbb B}^n$, respectively, under some homeomorphism $F\colon \overline {\mathbb B}{}^n \to D$.

(b) Your first interpretation is correct (as is your third; I don't see any difference between them). The union of all cells of dimension less than $n$ is a subset of $X$, and the requirement is just that $\Phi(\partial D)$ be contained in that subset. It doesn't say anything about the image being equal to the entire subset or being equal to a union of cells.

Title: Definition of Cell Decomposition?


问:
How to solve a convection-diffusion equation analytically?
For a 1-d convection equation $u_t + c u_x = 0$ on $\mathbb{R}_+ \times \mathbb{R}$ with a constant $c$, we know the solution is simply $u(x,t)=u_0(x-ct)$ for an initial condition $u_0 := u(x,0)$. However, if we add some diffusion to the RHS of the equation, with a constant $b$, to have
$$
u_t + c u_x = b u_{xx}
$$
then how can we solve this type of convection-diffusion equation analytically?

In my point of view, we may solve the heat equation $u_t = b u_{xx}$ first, by separation of variables, and then add the "transport" features to our solution. It seems requiring some knowledge from Fourier transform, which I am not familiar with, to get the solution of the heat equation, and the form is not fundamental but in a form of convolution. Anyway, I guess we may expect to have a final solution to the convection-diffusion equation in the form like:
$$
u(x,t) = \frac{1}{\sqrt{4 \pi b t}} \int_{-\infty}^{\infty} e^{-\frac{(x-ct-s)^2}{4 b t}} u(s,0) ds
$$
based on Prof. Gilbert Strang's lecture note.

However, I am questioning about the proof in details that the form above is exactly the solution to the convection-diffusion equation given.

答: If you consider $v(t,x)=u(t,x+ct)$, then $v_t=u_t+cu_x$ and $v_{xx}=u_{xx}$ so that the equation in $u$ is equivalent to the usual heat equation in $v$. Translating the standard solution back should result in the given solution formula.


答:
Using the definition of a decreasing sequence, we starter by computing $a_{n+1}-a_n.$

$$\begin{align}
a_{n+1}-a_n & \implies \frac{1}{5}(a_n^2+6)-a_n\\
& \implies \frac{a_n^2-5a_n+6}{5}\\
& \implies \frac{(a_n - \frac{5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(\frac{2a_n-5}{2})^2-\frac{1}{4}}{5}\\
& \implies \frac{(2a_n^2-5)-1}{20}
\end{align}$$

Next, since you have proved that $2 < a_n < 3,$ observe that

$$\begin{align}
2 < a_n < 3 & \implies 4 < 2a_n < 6\\
& \implies -1 < 2a_n -5 < 1\\
& \implies (2a_n-5)^2<1\\
& \implies (2a_n-5)^2-1<0\\
& \implies \frac{(2a_n-5)^2-1}{20}<0
\end{align}$$

Therefore you have that

$$a_{n+1}-a_n < 0$$

and by definition you conclude that $a_n$ is a decreasing sequence. $\square$

Title: Showing that a sequence is decreasing


问:
Model theory question with finiteness
It should be pretty elementary, but I can't really see it.. Cheers to anyone who can help me

Let $\Psi$ be a transitive set that is a model of $ZF$. Then, $a$ is finite if and only if $\Psi \models (\underline{a}\ \text{is finite})$.

答:
[Edit: I address first the original version of the question, where the requirement that $\Psi$ is transitive was not stated. The last paragraph addresses this case.]

As stated, this is false, and it requires too many wrinkles to fix appropriately:

(Assume that) $\mathsf{ZF}$ is consistent, and that $\Psi$ is a model of $\mathsf{ZF}+\lnot\mathrm{Con}(\mathsf{ZF})$. Define $a$ to be the set of natural numbers that code proofs in $\mathsf{ZF}$ of the inconsistency of $\mathsf{ZF}$. It is then the case that (in the true universe of sets) $a$ is the empty set, so it is finite. However, in $\Psi$, the object $\underline{a}$ corresponding to $a$ not only is non-empty, it is in fact infinite, because by assumption $\Psi$ sees a proof of the inconsistency of $\mathsf{ZF}$, and any such proof can be "padded" by adding as many irrelevant deductions as wanted before its conclusion.

On the other hand, any (standard) hereditarily finite set $a$ has a (parameter free) definition that, provably in $\mathsf{ZF}$, is satisfied by a unique set, and that set is hereditarily finite, and we can pick these definitions so they decode correctly in all models $\Psi$ of $\mathsf{ZF}$, in the following sense: Letting $\underline{t}$ denote the unique element of $\Psi$ that satisfies the definition of $t$, we can recursively prove that for any such $a$, say $a=\{b_1,\dots,b_n\}$, where the $b_i$ are pairwise distinct, we have $\Psi\models\underline{a}=\{\underline{b_1},\dots\underline{b_n}\}$, and the $\underline{b_i}$ are distinct in $\Psi$, and $\underline{n}$ is in $\Psi$ the $n$-th natural number and, in $\Psi$, $\underline{a}$ is finite and has size $\underline{n}$.

(The previous paragraph can be further generalized as well. And there are appropriate versions for other theories. For example, any model $\Lambda$ of $\mathsf{PA}$ has an initial segment isomorphic to the true $\mathbb N$, and any $n\in\mathbb N$ has a canonical $\underline{n}$ corresponding to it in $\Lambda$, and all true identities $n+m=k$, $n\times m=p$, etc, hold in $\Lambda$ of these interpretations.)

This being said, in practice there is a very usual case where we can prove the result you mentioned, without needing to assume that $a$ is in any way definable: If $\Psi$ is transitive, and the membership relation of $\Psi$ is the restriction to $\Psi$ of the usual membership relation, then finiteness is absolute between the universe $V$ and $\Psi$. You can indeed prove by induction on $n$ that if $a\in\Psi$, then $|a|=n$ iff $\Psi\models|a|=n$. Note that for this to even make sense one first needs to verify that, among other things, $\omega\subset\Psi$ and that each $n\in\omega$ is in $\Psi$ the $n$-th natural number.

Title: Model theory question with finiteness


问:
Prove that some subsets contain two positive integers $ a $ and $ b $ such that $ a|b$
Let $ n $ be a positive integer $(n\ge 2)$ and $ A $ be the set
$$A=\{1,2,3,...,2n\}$$
Prove that

$$(\forall B\subset A)\;$$
$$ \Bigl(\# B=n+1\implies \exists (a,b)\in B^2 \;:\;a\ne b \wedge a|b\Bigr)$$

It is easy to prove it when $ 1\in B $ and when $ 2\in B $ but i have no idea for the other cases.

Any help will be appreciated.
Thanks in advance.

答:
Let $B=\{x_1,x_2...x_{n+1}\}\subseteq A$.

Let $x_i=2^{a_i}p_i$ where $2\not|~p_i$ and $a_i\in\Bbb Z_{\ge0}$. Then $p_1,p_2,...,p_{n+1}$ are $n+1$ odd numbers that belong to $S=\{1,3,5,...,2n-1\}$. But $|S|=n$. By the pigeonhole principle, $p_k=p_m$ for some $k\ne m$. This gives $x_k|x_m$ or $x_m|x_k$.


问:
How to calculate the intersection of two parametric curves?
Let $f_1:\Bbb R\to\Bbb R^3$ and $f_2:\Bbb R\to \Bbb R^3$.
The curves are given by:
$f_1(t)=(t,t^2,t^3)$ and $f_2(t)=(e^t,\cos(t),\sqrt{1+t^2})$

So far, I represented each of the parametric variables as different entities (I named t=s for f2(t)), but now I am stuck regarding the system of equations. I am struggling to transform the system of equations into a matrix, because it contains an exponential, cosine and a squareroot.

I'd be glad if someone could help me out with this problem!(:

答:
The given parametrizations produce the following equations:

$$t=e^s$$
$$t^2=e^{2s}=\cos s$$
$$t^3=e^{3s}=\sqrt{1+s^2}$$

For $s>0$ we have $2s>0$, and so $e^{2s}>1\ge\cos s$ i.e. $e^{2s}\ne\cos s$ - a contradiction.

For $s<0$ we have $3s<0$ and so $e^{3s}<1\le\sqrt{1+s^2}$ i.e. $e^{3s}\ne\sqrt{1+s^2}$ - a contradiction.

The only remaining value is $s=0$, which implies $t=1$, and gives you an intersection point $(1,1,1)$.


问:
Showing that a sequence is decreasing
I need to show that the sequence defined by $a_1 = \frac{5}{2} $ and $a_{n+1} = \frac{1}{5}(a_n^2 + 6) $

is decreasing. The fact that $2 < a_n < 3 $ has been proven already.

I now just need to show that $a_n$ is decreasing.

I have tried the usual ways of showing $a_{n+1} - a_n \le 0 $ and $\frac{a_{n+1}}{a_n} \le 1 $

but could not get anywhere with these.

Title: Showing that a sequence is decreasing

Body:
If $x\in (2,3)$, then
$$
x^2-5x+6=(x-2)(x-3)<0 \quad\Longrightarrow\quad \frac{1}{5}(x^2+6)<x
$$
So, if $a_n\in(2,3)$, then
$$
a_n > \frac{1}{5}(a_n^2+6)=a_{n+1}
$$


问:
Calculating The Galois Group of the Splitting Field of $f=x^3-3$
If we let $f=x^3-3$ the let L be the splitting field for this polynomial I am trying to find $\Gamma(L:\mathbb{Q})$ and all intermediate field extensions.

Now as this is a splitting field and finite it is a finite normal extension and so we can use the fundemental theorem of Galois Theory so we know that:

$|\Gamma(L:\mathbb{Q})|=[L:\mathbb{Q}]$

Now we can see that if we define $\zeta=\exp{\frac{2\pi i}{3}}$ then we have that $L=\mathbb{Q}(\zeta,(3)^{\frac{1}{3}})$ and the minimal polynomial of $(3)^{\frac{1}{3}}$ over $\mathbb{Q}$ has degree 3 and the minimal polynomial of $\zeta$ has degree 2 over $\mathbb{Q}((3)^{\frac{1}{3}})$ so from the tower law we can see that:

$[L:\mathbb{Q}]=6$

Now we can see that a basis for $L$ is given by $\displaystyle{\{1,\zeta,(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},(3)^{\frac{1}{3}}\zeta,(3)^{\frac{2}{3}}\zeta\}}$

So if we now consider $\mathbb{Q}$ automorphisms of $L$ then we need only consider it's action on the basis elements, so we can see that if $f$ is a $\mathbb{Q}$ automorphism then we must have that:

$f(\zeta)=\zeta,(3)^{\frac{1}{3}}\zeta$ and

$f((3)^{\frac{1}{3}})=(3)^{\frac{1}{3}},(3)^{\frac{2}{3}},-\zeta$

which will give us the 6 elements of $\Gamma(L:\mathbb{Q})$

Is what I have done so far correct? and if so how do I now find the intermediate fields? I know that they are related to the subgroups but I am unsure how?

Thanks for any help

答:
I'll add a few things to this. You know that the Galois group has order 6, so there are only two choices: $S_3$ and $Z_3\times Z_2$. The latter group is abelian, so this means that every subgroup is normal. Thus, let $L/\mathbb{Q}$ be the splitting field and denote the Galois group by $G$.

The fundamental theorem also says that there's a bijective correspondence between subextensions $L/K/\mathbb{Q}$ such that $K/\mathbb{Q}$ is Galois and normal subgroups of $G$. Take the subextension $\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}$. This extension is not normal, so the subgroup $\textrm{Aut}(L/\mathbb{Q}(\sqrt[3]{3}))\leq G$ is not a normal subgroup. This implies that $G$ can't be abelian, so it must be $S_3$.

Now $S_3$ has the following subgroups:

A unique subgroup of order $3$.

$3$ subgroups of order $2$ that are all conjugate. These are the ones generated by $(1\, 2)$, $(1\, 3)$ and $(2\, 3)$.

The unique order $3$ subgroup is normal, so it corresponds to a unique quadratic Galois extension $K/\mathbb{Q}$. This must be the field extension $\mathbb{Q}(\zeta_3)/\mathbb{Q}$, which we know is Galois.

Next, we find the three non-Galois cubic subextensions of $\mathbb{Q}$. These are the field extensions corresponding to each root of $X^3-3$, so we get the extensions:

$$\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3\sqrt[3]{3})/\mathbb{Q},\;\;\mathbb{Q}(\zeta_3^2\sqrt[3]{3})/\mathbb{Q}$$

To show that these extensions are all distinct, note that if e.g. $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\zeta_3\sqrt[3]{3})$, then $\zeta_3 = \zeta_3\sqrt[3]{3}/\sqrt[3]{3}\in \mathbb{Q}(\sqrt[3]{3})$. This is impossible, since $L\neq \mathbb{Q}(\sqrt[3]{3})$. The same argument applies to the other two pairs.


问:
If $f$ is a continuous, either $f$ is constant or $\int_{0}^{1} f(t) \, dt < \max_{x\in[0,1]}f$.
Our situation:

We have a a function $f$ that is continuous on [0,1] and
$M=\max\{f(x) : x \in [0,1]\}$.

We WTS that either f is constant ($\forall x \in [0,1] \space f(x)=M$) or that $\int_{0}^{1} f(t) \, dt < M.$

Intuitively I know this is true but it becomes a bit harder to prove. This is what I have so far:

Firstly f can be the function $g(x)=M$ because g is continuous and non-neg on [0,1] and M is naturally the maximum of $g$ on $[0,1]$.

The harder part is if $\exists x \in [0,1]$ such that $f(x) < M$ (f is not constant):

Since f is continuous it is Riemann integrable which means if $U(f)$ is the upper integral of $f$ on $[0,1]$ ($U(f) = inf\{U(f,P) : P \in X\}$ where $X$ is a collection of all partitions of $[0,1]$ and $U(f,P)$ is a upper sum) then $U(f) = \int_{0}^{1} f(t) \, dt.$ (This is probably the part I am least confident about)

Let P be a partition of [0,1] then $U(f,P)= \sum_{i=0}^{k-1} M_i(a_{i+1} - a_{i})$ where $M_i$ is the supremum of $f$ on that interval. Then $U(f,P)<  M\sum_{i=0}^{k-1}(a_{i+1} - a_{i})=U'(f,P).$ Note that we have a strict less than because f is not constant.

Then we have $\int_{0}^{1} f(t) \, dt = U(f) \leq U(f,P) < U'(f,P)=M$ which means $\int_{0}^{1} f(t) \, dt < M$.

Does this proof look ok?

答: HINT: Try to prove that if $0\le F(x)$ is continuous and strictly positive elsewhere in $[0,1]$, then $\int_0^1F(x)dx>0$. And apply it to $M-f(x)$.


问:
How do I see that $F$ is a vector field defined on all of $\mathbb{R}^3$?
$$\vec{F}(x,y,z)= y^3z^3\mathbf{i} + 2xyz^3\mathbf{j} + 3xy^2z^2\mathbf{k}$$

How do I see that $F$ is a vector field defined on all of $\mathbb{R}^3$? And then is there an easy way to check if it has continuous partial derivatives?

I am looking at a theorem and it states, If F is a vector field defined on all of R^3 whose component functions have continuous partial derivatives and curl F = 0, then F is a conservative vector field.

I don't know how to check those conditions. Could someone show with that problem given?

答: The vector field $F$ is defined on all of $\mathbb{R}^3$ because all of its component functions are (there are no pints where the functions are undefined, i.e., they make sense when plugging in any point of $\mathbb{R}^3$ into them). If say one of the component functions was $\frac{1}{x-y}$ then the vector field wouldn't be defined along the plane $x=y$. Moreover the component functions are continuous, as every polynomial function is continuous. To calculate the curl of the vector field just use the definition of curl, which involves just computing partial derivatives.</p>


问:
Why is it actually important for $p$ to be prime in the ring of $p$-adic numbers?
In the proof that the set $Z_{[p]}$ of all $p$-adic numbers is a ring, it is not necessary for $p$ to be prime. Is it actually important for $p$ to be prime for the ring of $p$-adic numbers?

答:
Let $R$ be any ring and $\mathfrak a\subset R$ be any ideal.
Then for all $n\in \Bbb N$, $\mathfrak a^n$ is also an ideal, hence we have quotient rings $R/\mathfrak a^n$, and of course we have natural  homomorphisms $R/\mathfrak a^n\to\ R/\mathfrak a^m$ whenever $m>n$. From this, we arrive at the projective limit
$$ \lim_\leftarrow R/\mathfrak a^n.$$
This does not require $\mathfrak a$ to be prime.


问:
$\lim\limits_{n\rightarrow\infty} {\frac{24n-8}{n^2+n+1}} \ne 0.5$
Got stuck in disproving limit, hope you can help/hint me. Disprove $$\lim\limits_{n\rightarrow\infty} {\frac{24n-8}{n^2+n+1}} \ne 0.5$$

My effort :
So I did common factor and got to $\left|{\frac {-n^2+47n-17}{2n^2+2n+2}}\right|$

Now I don't know how to proceed.
I tried to do something and got to  $\left|{\frac {n^2-47n+17}{2n^2+2n+2}}\right|$ but it doesn't help.

Body:
$\dfrac{24n-8}{n^2+n+1}=\dfrac{n(24-\frac{8}{n})}{n(n+1+\frac{1}{n})}=\dfrac{24-\frac{8}{n}}{n+1+\frac{1}{n}}$

So $\lim\limits_{n\to\infty}\dfrac{24n-8}{n^2+n+1}=\lim\limits_{n\to\infty}\dfrac{24-\frac{8}{n}}{n+1+\frac{1}{n}}=0$,

since $24-\dfrac{8}{n}\xrightarrow{n\to\infty}24$ and $n+1+\dfrac{1}{n}\xrightarrow{n\to\infty}+\infty$


问:
Find the sum of two angles created by equilateral triangles on a line.
Point A, C, E are on line G, in that order. Point B and D are on the same side of line G so that ABC and CDE are equilateral triangles. Find the sum the of <BAD and <BED.

I tried to solve this problem by making <DAE = x. However, this gets me nowhere because I have to introduce a second variable, y, no matter what I do. Even then, I can not find all the angles in terms of x and y.

答:
It helps to draw a diagram.

Observe that $\triangle ACD \cong \triangle BCE$ (by SAS) and $AB //CD$.

This shows that $\angle BEC = \angle ADC = \angle BAD$, which is sufficient to solve your problem.


答:
If $f$ is not constant, then there exists an $x_0\in (a,b)$, such that
$$
f(x_0)<\max_{x\in [a,b]} f(x)=M
$$
If we set $2\varepsilon=M-f(x_0)$, then there exists a $\delta>0$, such that
$$
|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-f(x_0)|<
\varepsilon=\frac{1}{2}\big(M-f(x_0)\big)
\quad\Longrightarrow\quad
|f(x)-f(x_0)|<\frac{1}{2}M-\frac{1}{2}f(x_0)
\quad\Longrightarrow\quad
f(x)<\frac{1}{2}M+\frac{1}{2}f(x_0)
=M-\varepsilon
$$
Hence
$$
\int_a^b f=\int_a^{x_0-\delta} f+\int_{x_0-\delta}^{x_0+\delta} f+\int_{x_0+\delta}^b f \le (x_0-\delta-a)M+2\delta (M-\varepsilon)+(b-x_0-\delta)M \\=(b-a)M-2\delta\varepsilon<M.
$$

Title: If $f$ is a continuous, either $f$ is constant or $\int_{0}^{1} f(t) \, dt < \max_{x\in[0,1]}f$.

Body:
Our situation:

We have a a function $f$ that is continuous on [0,1] and
$M=\max\{f(x) : x \in [0,1]\}$.

We WTS that either f is constant ($\forall x \in [0,1] \space f(x)=M$) or that $\int_{0}^{1} f(t) \, dt < M.$

Intuitively I know this is true but it becomes a bit harder to prove. This is what I have so far:

Firstly f can be the function $g(x)=M$ because g is continuous and non-neg on [0,1] and M is naturally the maximum of $g$ on $[0,1]$.

The harder part is if $\exists x \in [0,1]$ such that $f(x) < M$ (f is not constant):

Since f is continuous it is Riemann integrable which means if $U(f)$ is the upper integral of $f$ on $[0,1]$ ($U(f) = inf\{U(f,P) : P \in X\}$ where $X$ is a collection of all partitions of $[0,1]$ and $U(f,P)$ is a upper sum) then $U(f) = \int_{0}^{1} f(t) \, dt.$ (This is probably the part I am least confident about)

Let P be a partition of [0,1] then $U(f,P)= \sum_{i=0}^{k-1} M_i(a_{i+1} - a_{i})$ where $M_i$ is the supremum of $f$ on that interval. Then $U(f,P)<  M\sum_{i=0}^{k-1}(a_{i+1} - a_{i})=U'(f,P).$ Note that we have a strict less than because f is not constant.

Then we have $\int_{0}^{1} f(t) \, dt = U(f) \leq U(f,P) < U'(f,P)=M$ which means $\int_{0}^{1} f(t) \, dt < M$.

Does this proof look ok?


问:
Show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$
Define the Fibonacci sequence via $f_1 = f_2 = 1$, and if $n > 2$, then $f_n =
f_{n−1} + f_{n−2}$ and show that for every positive integer $ f_n=\frac{\left ( \frac{1+\sqrt5}{2} \right )^n-\left ( \frac{1-\sqrt5}{2} \right )^n}{\sqrt5}$

I have been solving this problem using strong induction. This is my attemp.

For $n=1$.

$f_1=1$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )-\left ( \frac{1-\sqrt5}{2} \right )}{\sqrt5}=1$

For $n=2$

$f_1=2$ and,

$\displaystyle \frac{\left ( \frac{1+\sqrt5}{2} \right )^2-\left ( \frac{1-\sqrt5}{2} \right )^2}{\sqrt5}=1$ as well.

the statement is also true for $n = 3$

Now, assume that the statement is true for some $n-1$ and $n-2$. We get

$\displaystyle f_n =f_{n-1}+f_{n-2}=\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-1}-\left ( \frac{1-\sqrt5}{2} \right )^{n-1}}{\sqrt5}+\frac{\left ( \frac{1+\sqrt5}{2} \right )^{n-2}-\left ( \frac{1-\sqrt5}{2} \right )^{n-2}}{\sqrt5}$

From here, my intention is to show that the statement for $f_n$ is true. But I couldn't move on.

答:
For homogeneous second order linear difference equations there is a shortcut. I wonder why it is not well known and almost everybody is using the "standard way" of solving a linear equation system. BTW it just so happened somebody else applied it earlier today in this post: https://math.stackexchange.com/a/3898819/843178

This technique can be used in higher order equations (along with a treatment of duplicate roots) but it gets messy. So for higher order LDEs it's best to use the "standard solution".

The characteristic equation $^2−−1=0$ has two roots $\varphi=\frac{1+\sqrt 5}{2}, \psi=\frac{1-\sqrt 5}{2}$.
Vieta’s formulas give $\varphi+\psi=1, \varphi \psi = -1$.

Therefore $$f_{+2}−(\varphi+\psi)f_{n+1}+\varphi \psi f_=0.$$

Rearranging the terms, we get
$$f_{n+2}−\psi f_{+1}=\varphi(f_{+1}−\psi f_) $$

$$f_{n+2}−\varphi f_{+1}=\psi (f_{+1}−\varphi f_) $$

Both are geometric sequences, so
$$f_{+1}−\psi f_=\varphi^ (f_1−\psi f_0 )=\varphi^ \tag 1$$
$$f_{+1}−\varphi f_=\psi^ (f_1− \varphi f_0 )=\psi^ \tag 2$$

(1)-(2), $$(−) f_=\varphi^−\psi^⇒ f_=(\varphi^−\psi^)/(\varphi−\psi).\blacksquare$$


问:
Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$?
I was doing some problems in Big-O notation, and I got intrigued by this question as it was never really addressed. If we plot $x^2$, it's obviously bigger than $x\log x$ for all $x > 0$. However, I started plotting graphs for smaller and smaller values of the exponent, and saw that the crossing point between $x^k$ and $x\log x$ got further and further away. I wondered if maybe there was some particular constant which was the smallest value for $k$ such that the two graphs would cross. But then I found out that even $x^{1.1} >x\log x$ when $x> 3.43 \times 10^{+15}$. This has lead me to believe that for any $k>1$, $x^k$ will eventually overtake $x\log x$, but I think it would be beyond my capabilities to prove it, and I don't know where to start. Does anyone know the answer to this, and is the proof understandable for an undergrad math student?

Edit: Also the functions cross at two points, and I noticed that the first crossing point converges to $e$ as $k$ goes to $1$. What's the explanation for this?

答:
If $k > 1$, then $k = 1 + \varepsilon$ for some $\varepsilon > 0$. So $x^k = x^{1 + \varepsilon} = xx^\varepsilon$.

Thus, to show $x^k = xx^\varepsilon > x\log(x)$, eventually, amounts to showing $x^\varepsilon > \log(x)$, eventually. And, this latter fact is true because:
\begin{align*}
\lim_{x \to \infty}\frac{\log(x)}{x^\varepsilon} &= \lim_{x \to \infty} \frac{1/x}{\varepsilon x^{\varepsilon - 1}}\quad \text{ (L'Hôpital's rule)} \\
&= \lim_{x \to \infty} \frac{1}{\varepsilon x^\varepsilon} = 0\quad \text{ ($x^\varepsilon \to \infty$ as $\varepsilon > 0$)}
\end{align*}

Answer to Edit:

The reason you have a crossing point converging to $e$ as $k \to 1$ is quite simple. Just let $k = 1$ and let us calculate the crossing points. Crossing points occur when: $x^k = x = x \log(x)$. Or rearranging:
$$
x - x\log(x) = x (1 - \log(x)) = 0
$$ If you ignore $x = 0$, then the above is true when $1 - \log(x) = 0$ or $\log(x) = 1$ or $x = e$.


问:
Classifying the representations of $G=\langle(123), (456), (23)(56) \rangle \subset S_6$.
This is exercise 3 of chapter 2 of the book "Representations of finite and lie groups" By B. Thomas. My questions are in bold. I am essentially confused by two steps but I think I understand the entire question apart from that.

Let G be a subgroup of order $18$ in the symmetric group $S_6$ given by
$$ G = \langle(123), (456), (23)(56)\rangle.$$
Show that G has a normal subgroup of order $9$ and four normal subgroups
of order $3$. By considering quotients show that G has two representations of degree one and four inequivalent irreducible representations of degree $2$, none of which is injective/faithful.

My approach is as follows.

Observe  $|G|=18=2 \cdot 3^2$.  Denote $n_3$ for the number of Sylow $3$-subgroups. We see that by Sylow's theorem $n_3 \equiv 1 \bmod 3$, so candidates are $n_3=1, 4, 7$  etc. However we also have that $n_3|2$, so $n_3=1.$  This means there is one sylow subgroup of order $9$, and another consequence of the Sylow theorems is that this group is then normal. We can construct such a group by considering the two generators of order $3$:
$$ H= \langle(123), (456) \rangle  \cong C_3 \times C_3.$$
Now this group has index $|G:H|= \frac{|G|}{|H|}=18/9=2$ and therefore it must indeed be normal in $G$. We can also use the generators (and one of its inverses) of order $3$ to construct $4$ cyclic subgroups of order $3$.

$$ N_1=\langle (1 2 3)\rangle=\{e, (123), (132)\} \cong C_3$$
$$ N_2=\langle (4 5 6)\rangle=\{e, (456), (465)\}  \cong C_3$$
$$ N_3=\langle (1 2 3)(4 5 6)\rangle=\{e, (1 2 3)(4 5 6), (1 3 2)(4 6 5)\} \cong C_3$$
$$ N_4=\langle (1 2 3)(4 6 5)\rangle=\{e, (1 2 3)(4 6 5), (1 3 2)(4 5 6 )\}  \cong C_3$$
We have thus found four subgroups of order $3$, but we still have to prove they are normal in $G$.

(how do I know these exhaust all the subgroups of $G$ one can have of order $3$? I do not know how to formulate this/the exact reason). I also thought earlier I knew why these were normal, but I was mistaken.

We can now use the fact that we may lift representations of quotients to the original group using the canonical quotient homomorphism.

$|G/H|=2$ and it is an abelian group (specifcally $C_2$) therefore by a corollary/consequence of Weddderburn's theorem we must have $2$ irreducible complex representations of $G/H$, which are both $1$ dimensional (I prefer to use Dummit and Foote corollary 11(1), page 861 for this). We lift these to $G$.

Now we can consider the quotients $$G/N_1, G/N_2, G/N_3, G/N_4 $$

According to the hints and solutions section in the back, these each give a $2$-dimensionial representation, why?

Once we accept this we can also lift these $4$ representations to $G$ to end up with $4$ (distinct/inequivalent) representations of $G$. The dimensions of the representations of $G$ we have so far are thus $1, 1, 2, 2, 2, 2$. Then observe we must also have by theorem $10.(4)$ of page 861 of Dummit and Foote that:
$$ \sum_{i=1}^r n_i ^2 = |G $$
However if we compare this to what we already have, we find:
$$ 2 \cdot 1^2 + 4 \cdot 2^2 =18 =|G|$$
So we indeed find that these are all representations by exhaustion. A lift of a quotient $G/N$ to $G$ always has $N$ contained in its kernel. We therefore know that lifted representations never have a trivial kernel and therefore they are not faithful (since they are not injective homomorphisms).

答:
To answer my own questions in bold:

The list of elements in $N_1$, $N_2$, $N_3$ and $N_4$ exhaust all elements of order $4$ in $G$ and therefore these are the only four groups we can form of order $4$.

We can compute normality directly by looking what conjugation by the generators does to each of these subgroups. One can compute directly for $g\in \{(123),(456), (23)(56)\}$ that:
$$ g N_i g^{-1}\subseteq N_i$$
For instance:
$$(456)N_1(456)^{-1}= (456)\{e, (123), (132)\}(465) =$$ $$\{(456)(465),(456)(123)(465), (456)(132)(465)  \}= \{e, (123), (132)\}=N_1 $$

And then for a product of two  generators $a, b$ we have:
$$(ab)N_i (ab)^{-1}= a(bN_ib^{-1})a^{-1}\subseteq aN_i a^{-1} \subseteq N_i. $$
So if we know that the subgroups are left invariant under conjugation by the generators, the conjugation action of the entire group on the subgroups $N_i$ leaves these subgroups invariant, therefore they are normal subgroups by the subgroup test.

If we look at the quotients $G/N_i $ we notice they are of order $18/3=6$. There are two groups (up to isomorphism) of order $6$, that is $S_3$ (non-abelian) and $C_6$(abelian, cyclic). In $G$ we get that:
$$(23)(56) (456)= (23) (4 6).$$
But also:
$$(456) (23)(56)=(456)(56)(23)= (45)(23). $$
In the quotient $G/N_1$ these elements do not get mapped to $0$ or to the same elements. We therefore know that the quotient is not abelian. One can also do a similar check for the other quotients or investigate the order of the elements and realise that no element has order $6$. Upon inspection all the quotients have the structure of $S_3$, which has three irreps of dimension $1, 1, 2$ as was one of the examples in the lecture notes/textbook.


问:
Solving equations involving fractional exponents?
Let's say we have the following equation (with $a,b \in \mathbb{Z}$): $$x^\frac{a}{b} = 2$$

If the exponent is an even integer, we get two solutions. If it is an odd integer, we get only one solution. How about the other cases and how do I write the result in closed form?

答: Then do $$\left (x^{a/b}\right )^b=2^b,$$ gives you the equation $x^a=2^b.$ This one you know how to solve.


问:
Linear independence of $\sin(x)$ and $\cos(x)$
In the vector space of $f:\mathbb R \to \mathbb R$, how do I prove that functions $\sin(x)$ and $\cos(x)$ are linearly independent. By def., two elements of a vector space are linearly independent if $0 = a\cos(x) + b\sin(x)$ implies that $a=b=0$, but how can I formalize that? Giving $x$ different values? Thanks in advance.</p>

答:
Does this work?

Assume that $\sin(x)$ and $\cos(x)$ are linearly dependent. Then one must be a scalar multiple of the other, that is

$$\sin(x) = a * \cos(x)$$

But for $x=\pi/2$, we have

$$1 = a * 0$$

which is impossible. Thus $\sin(x)$ and $\cos(x)$ are linearly independent.

I have a feeling that there is some error in my reasoning here, please point it out if you see one :)


问:
Why equivalence $\ln(1+u)_{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$ holds for $\color{red}{u\to 0}$?
Taylor series for $\ln(1+u)$ is:

$$\ln(u+1)=u-\cfrac12 u^2+\cfrac13 u^3-\cdots+\cfrac{(-1)^k(u-1)^{k+1}}{k+1}+\cdots$$
and this holds for $u\in(-1,1)$ (because $\ln(u+1)=\int\frac{1}{u+1}du=\int\sum_{n=0}^{\infty}(-u)^ndu$ and the geometry series holds for $|-u|<1$ or $u\in(-1,1)$).

But when we want to use it as equivalence for example to evaluate limits we say $\ln(1+u)_\color{red}{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$

My question is why this  holds for $\color{red}{u\to 0}$ ? Is there any reason why we choose $u$ approaches to zero?

答:
Based on my questions in the comments to the original question, I don't think I understand the original question. However, I'll provide an answer anyway.

$x \in (-1,1) \implies \ln(x+1) \color{red}{=} x-\cfrac12 x^2+\cfrac13 x^3-\cdots \quad (*)$

This says a few things.

$\quad \ln\left(\frac{18}{11}\right) = \ln\left(\frac{7}{11} + 1\right) \color{red}{=} \frac{7}{11} - \frac{1}{2}\left(\frac{7}{11}\right)^2 + \frac{1}{3}\left(\frac{7}{11}\right)^3 -\cdots$.  Check your calculator if you don't believe me. And this works for every number $x \in (-1,1)$. This is, in fact, exactly what $(*)$ says.

The function $f(x) = \ln(x+1)$ in a neighbourhood $(-\frac{4}{5}-\delta, \ -\frac{4}{5}+\delta)$ with $\delta \leq \frac{1}{5}$ is the same as $g(x) \ (\text{which}= f(x) \ ) = x-\cfrac12 x^2+\cfrac13 x^3-\cdots $ in the same neighbourhood: the two functions are equal.

$|x| < 1 \implies x^3, x^4 $ etc are small. Therefore we sometimes write:

$\ln(1+x) \sim x-\frac12 x^2+\frac13 x^3 \quad (1) \ , $  or

$\ln(1+x) \sim x-\frac12 x^2 \quad (2)$,

the latter being appropriate only if $x$ is very small.

So now I think I am getting to the crux of your question, which I believe is the following: I just wrote: $|x| < 1 \implies x^3, x^4 $ etc are "small". However, this is not true, especially for $|x|$ close to, but just less than $1$. So then, how do we justify $(1)$ and $(2)$ for $|x|$ close to, but just less than $1$? And I'm not sure. Like Paresseux Nguyen said, Taylor's Theorem gives a formula to see what the error is between e.g. $x-\frac12 x^2+\frac13 x^3$ and $\ln(x+1)$.


问:
Is every Borel set a countable union of intervals?
Since the Borel $\sigma$ algebra is generated by all open subsets of $\mathbb{R}$ and all open sets are the countable union of disjoint open intervals, I figured that any Borel set is the countable union of intervals. I also used the fact that if we only use '$\sigma $ algebra operations' on intervals then we get an interval or countable union of intervals.
But I ask if this is actually true?

答:
The Cantor set $C$ is an uncountable Borel set and it does not contain any non-trivial interval (i.e. a singleton).

Hence $C$ can not be written an as countable union of intervals.

However, its complement is indeed a countable union of disjoint open intervals!


答:
$$AC=BC$$

$$\angle ACD=\angle BCE=120$$

$$CD=CB$$

by S.A.S congruence theorem

$$\triangle ACD\cong \triangle BCE$$

$$\angle DAC=60-\angle BAD=\angle CBE=\angle BED$$

$$\angle BED+\angle BAD=60-\angle BAD+\angle BAD=60$$

Title: Find the sum of two angles created by equilateral triangles on a line.

Body:
Point A, C, E are on line G, in that order. Point B and D are on the same side of line G so that ABC and CDE are equilateral triangles. Find the sum the of <BAD and <BED.

I tried to solve this problem by making <DAE = x. However, this gets me nowhere because I have to introduce a second variable, y, no matter what I do. Even then, I can not find all the angles in terms of x and y.


问:
Set of fixed points is convex.
Problem: Suppose that $f:\mathbb{R}^n \to \mathbb{R}^n$ has Lipschitz constant $1$. Is it true that the set of fixed points of $f$ is convex?

My attempt: I managed to show that if $x,y$ are two fixed points of $f$ and $z  = (1-\lambda)x+\lambda y$ for $\lambda \in (0,1)$, then $||f(z)-x||=||z-x||$ and $||f(z)-y||=||z-y||$. But I'm not quite sure what to do after this. Is there any hint or suggestion?

答: So $f(z)$ is on the sphere around $x$ passing through $z$, and is also on the sphere around $y$ passing through $z$. These two spheres "kiss" at $z$ so that $f(z)$ must be $z$.

Title: Set of fixed points is convex.


问:
$\gcd(2k_1,2k_2−1)=\gcd(k_1,2k_2−1)$
Prove $\gcd(2k_1,2k_2−1)=\gcd(k_1,2k_2−1)$

Both $k_1$ and $k_2$ are positive integers.

I tried to start from the fact that: $\gcd(2,2k-1) = 1$ because they are relatively prime.
But I am not sure how to expand my idea

答: If $p$ is a  divisor of $2k_1$ and $2k_2-1$, $p$ is odd since $2k_2-1$ is odd, this implies $p$ divides $k_1$

Title: $\gcd(2k_1,2k_2−1)=\gcd(k_1,2k_2−1)$


问:
Limit $\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$

What is the limit of:
  $$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$

So what I tried is:

$$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos x}-1)\over x\cdot x^2}$$

From here, using the rule $\lim_\limits{x\to0}{\sin{x}\over{x}}=1$ it remains to evaluate
$$\lim_\limits{x\to0}{{1\over\cos{x}}-1\over x^2}.$$
I tried changing it a lot of ways but it always gets messier so I'm not sure what to apply here to complete the question.

答:
$\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac12$

$\lim_{x\to 0}\frac{\sin x}{x}=1$

$$\lim_{x\to 0}\frac{\tan x - \sin x}{x^3}\space=\space\lim_{x\to 0}\frac{\sin x}{x^3 \cos x}\space-\space\lim_{x\to 0}\frac{\sin x \cos x}{x^3 \cos x}\space=\space\ \Biggl(\lim_{x\to 0}\frac{1}{\cos x}\Biggl)\Biggl(\lim_{x\to 0}\frac{\sin x }{x}\Biggl)\Biggl(\lim_{x\to 0}\frac{1}{x^2}\Biggl)-\Biggl(\lim_{x\to 0}\frac{1}{\cos x}\Biggl)\Biggl(\lim_{x\to 0}\frac{\sin x}{x}\Biggl)\Biggl(\lim_{x\to 0}\frac{\cos x}{x^2}\Biggl)\space=\space\Biggl(\lim_{x\to 0}\frac{1}{\cos x}\Biggl)\Biggl(\lim_{x\to 0}\frac{\sin x}{x}\Biggl)\Biggl[\lim_{x\to 0}\frac{1}{x^2}\space-\space\lim_{x\to 0}\frac{\cos x}{x^2}\Biggl]\space=\space\lim_{x\to 0}\frac{1-\cos x}{x^2}\space=\space \frac12$$


答:
For
$|x| < 1$,
$\ln(1+x)
=\sum_{n=1}^{\infty} \dfrac{(-1)^{n-1}x^n}{n}
$.
Therefore
the error after $m$ terms is
$e_m(x)
=\ln(1+x)-\sum_{n=1}^{m} \dfrac{(-1)^{n-1}x^n}{n}
=\sum_{n=m+1}^{\infty} \dfrac{(-1)^{n-1}x^n}{n}
$.

If $x > 0$
this is an
alternating sum
of decreasing terms
so the value is less than
the first term,
so
$|e_m(x)|
\lt \dfrac{|x|^{m+1}}{m+1}
$.

If
$x < 0 $ then

$\begin{array}\\
|e_m(x)|
&=\left|\sum_{n=m+1}^{\infty} \dfrac{(-1)^{n-1}x^n}{n}\right|\\
&\le\sum_{n=m+1}^{\infty} \dfrac{|x|^n}{n}\\
&\le\sum_{n=m+1}^{\infty} \dfrac{|x|^n}{m+1}\\
&=\dfrac1{m+1}\sum_{n=m+1}^{\infty} |x|^n\\
&=\dfrac{|x|^{m+1}}{(m+1)(1-|x|)}\\
\end{array}
$

In both cases,
the error after $m$ terms
is of order $|x|^{m+1}
$.

Title: Why equivalence $\ln(1+u)_{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$ holds for $\color{red}{u\to 0}$?

Body:
Taylor series for $\ln(1+u)$ is:

$$\ln(u+1)=u-\cfrac12 u^2+\cfrac13 u^3-\cdots+\cfrac{(-1)^k(u-1)^{k+1}}{k+1}+\cdots$$
and this holds for $u\in(-1,1)$ (because $\ln(u+1)=\int\frac{1}{u+1}du=\int\sum_{n=0}^{\infty}(-u)^ndu$ and the geometry series holds for $|-u|<1$ or $u\in(-1,1)$).

But when we want to use it as equivalence for example to evaluate limits we say $\ln(1+u)_\color{red}{u\to 0} \sim u-\frac12 u^2+\frac13 u^3-\cdots$

My question is why this  holds for $\color{red}{u\to 0}$ ? Is there any reason why we choose $u$ approaches to zero?


问:
Likelihood test ratio for discrete random variables
Just would like to know whether the following statement is true or not:

For a discrete random variable, the numerator of the likelihood ratio test (which is a supremum on a set) is equal to the maximum likelihood of the sample for the null hypothesis.

Please provide a counter example if false. I, unsuccessfully, tried proving that the supremum is the maximum by proving that the parameter set will be compact for a discrete random variable.

Thanks for the help!

答: The answer to your question is no. To see why, let's look at a simple example. Suppose the proportion of people with red hair among a population is $\theta\in [0,1]\cap \mathbb{Q}$. Let $\Theta_0=\Big\{\frac{1}{n}\Big\}_{n\in\mathbb{N}}.$ We perform the hypothesis test $H_0:\theta \in \Theta_0$ versus $H_a:\theta \notin \Theta_0$ by taking a SRS of size $n=5$ from our population. Suppose that, among our sample of five individuals, we observe nobody with red hair. Our likelihood function becomes $\mathcal{L}(\theta)=(1-\theta)^5$. Notice $\max_{\theta \in \Theta_0}\mathcal{L}(\theta)$ does not exist whereas $\sup_{\theta \in \Theta_0}\mathcal{L}(\theta)=1$