答:
$y = 2 + \frac{x^2}{2} \,$ (from $(0,2)$ to $(2,4))$

Parametrize it using $x = t$, $y = \frac{t^2}{2} + 2$

So your points on the curve are given by $(t, \frac{t^2}{2} + 2)$. Starting point is at $t = 0$ and the end point is at $t = 2$.

Now vector field is $(y, x^2)$ or $ (\frac{t^2}{2} + 2, t^2)$.

$r'(t) = (1, t)$.

Now you can do a dot product for your line integral from $t = 0$ to $t = 2$

Title: How do I calculate an integral of $F$ where $F = \langle y,x^2\rangle$?

Body:
I have to calculate the above integral and I'm not sure how to do it. First I find the curl of $\langle y,x^2\rangle$ and I get:

$$\langle0,0,2x-1\rangle$$

Therefore it's not conservative because it's not all $0$s and I can't solve the integral.

Is that right? I feel like that isn't right.


问:
Can you determine that ANY matrix with no free-variables MUST have a leading varible in each row?
The question is:
Let $A \in M_{n×n}(\mathbb{R})$ and let $d \in \mathbb{R}^n$. Prove that if the equation $Ax = d$ has a unique solution, then for
every $b ∈ \mathbb{R}^n$, the equation $Ax = b$ has a unique solution

So, one way I thought of for proving this, is determining that A has no free variables (relying on the fact that $Ax = d$ has a unique solution) and from there claim that every row in the matrix has a leading variable. Then, according to the theorem:

(1) $Ax = b$ has a solution for all b in $\mathbb{R}^m$.

(2) The span of the columns of $A$ is all of $\mathbb{R}^m$.

(3) $A$ has a leading variable in each row.

Then I can conclude that the original statement ( If the equation $Ax = d$ has a unique solution, then for
every $b ∈ \mathbb{R}^n$
, the equation $Ax = b$ has a solution) is TRUE.

My problem is I'm not sure how to prove that if a matrix has no free-variable it MUST have a leading variable in each row.
Is it even possible to prove that? If so how?

答:
The answer is: $Ax = d$ has a unique solution if and only if $A$ is invertible. Since $A$ is invertible, for every $b, x:=A^{-1}b$ is the unique solution.

If you are curious about the iff statement...

...the easiest way to see why is to cycle through the properties of the RREF of $A$.

Let $A_{\text{inv}}$ be the product of the series of elementary row operations required to transform $A$ to $\text{RREF}(A)$.

$Ax=d$ has a unique solution if and only if $\text{RREF}(A)$ is the identity matrix. And you can see that by noting
$Ax=d$ is equivalent as $\text{RREF}(A)x=A_{\text{inv}}d$. And for square matrices whose RREF is the identity, they are invertible because, by definition, $I=\text{RREF}(A)=A_{\text{inv}}\cdot A$.

If you are not familiar with elementary row operations or RREF, here is the essence.

First, you should read from the link or elsewhere what elementary row operations on a matrix look like. Next..

.. an elementary row operation is achieved by left-multifying with the matrix produced from the same elementary operation on identity.

For example, starting with

$$
\begin{equation}\tag{1}
\begin{bmatrix}
-1 & 1\\
1 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
0\\
1
\end{bmatrix}
\end{equation}
$$

suppose you want to make row 2 add row 1. Perform that to the identity matrix, you get:

$$
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}
\longrightarrow
\begin{bmatrix}
1 & 0\\
1 & 1
\end{bmatrix}
$$

The row operation on the original equation is thus

$$
\begin{bmatrix}
1 & 0\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 1\\
1 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
1 & 0\\
1 & 1
\end{bmatrix}
\begin{bmatrix}
0\\
1
\end{bmatrix}
$$

which simplifies to

$$
\begin{equation}\tag{2}
\begin{bmatrix}
-1 & 1\\
0 & 3
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}
=\begin{bmatrix}
0\\
1
\end{bmatrix}
\end{equation}
$$

(1) is true if and only if (2) is true. Thus, you can solve (2) and get the answer for (1). You do that because the answer for (2) is (more) obvious.

Every matrix has a RREF because there is an algorithm that uses elementary row operations to produce RREF for every matrix.

ContentLicense: CC BY-SA 4.0


问:
Prove that structure $(Q/{0}; =;*)$ has infinite number of automorphisms
Prove that structure $(Q/{0}; =;*)$ has infinite number of automorphisms
I think it's about we don't have $<$ or $>$, so we don't have strict order in structure, so we can do whatever we want? but i dont know how to say it more formal/

答: Hint: Take any permutation (i.e. bijection to itself) of the set of the prime numbers $P$: $\phi:P\to P$. Now map every rational number $q=\pm p_1^{\lambda_1}\cdots p_n^{\lambda_n}$, where $\lambda_1,\ldots\lambda_n\in\mathbb Z$ into $\pm \phi(p_1)^{\lambda_1}\cdots \phi(p_n)^{\lambda_n}$. Prove that this map is an automorphism.


问:
if $\sum_{i=1}^n a_i$ converges but $\sum_{i=1}^n (a_i)^2$ diverges, does that mean that $\sum_{i=1}^n a_i$ is conditionally
I don't think this is true, but cannot find an example to disproof it.
Either that or i need to prove that $\sum_{i=1}^n |a_i|$ diverges, which I am unclear how to approach the proof.

答: Yes, it is true. That's so because, if $\sum_{n=1}^\infty a_n$ was absolutely convergent, then we would have $\lim_{n\to\infty}a_n=0$ and therefore $a_n^{\,2}\leqslant |a_n|$ if $n$ is large enough. So, by the comparison test (this is where absolute convergence is used), $\sum_{n=1}^\infty a_n^{\,2}$ would converge too.


问:
Proof a function is convex if and only if it meets this condition?
How to proof that a function f:R->R is convex if and only if dom(f) is convex and for each a, b, c in it's domain that are $a<b<c$, we have:

Determinant of matrix:
$$
\begin{vmatrix}
 1 &     1  &     1\\
  a &     b  &    c\\
 f(a) & f(b)  &f(c)
\end{vmatrix}\ge 0.
$$

Determinant is:

$$    bf(c)-cf(b)+cf(a)-af(c)+af(b)‐bf(a) >= 0$$

Then:

$$   f(a)(c-b) + f(b)(a-c) + f(c)(b-a) >=0$$

Then according to the a<b<c, we can say:

$$    f(a)(c-b) + f(c)(b-a) >= f(b)(c-a)$$ [edited]

So I went until here but I don't know how to connect this to the Jensen's inequality to prove that f is convex.

答:
You need almost no more steps to go beyond there. The definition of convexity of a function includes the following:
$$
f(\theta x+(1-\theta)y)\le \theta f(x)+(1-\theta)f(y)
$$
Now, try to rearrange the last inequality you obtained with $x=a$, $y=c$ and a proper choice of $\theta$.


问:
Inductive proof $f(n) = n$ with function composition.
I need help with a inductive proof. Currently got no idea how to start with this assignment.

Assignment:

Let $f : \mathbb{N}_0 \to \mathbb{N}_0$ be a function, so that the inequality:

$$f \circ f(n) < f(n+1)$$ exists. $\forall n \in \mathbb{N}_0$

Show that $f(n) = n$. $\forall n \in \mathbb{N}_0$

I am currently struggling to understand where to start with this one. Thanks for any help in advance.

答:
First, note that it is enough to show that any such $f$ is strictly increasing, for then $f(f(n))<f(n+1)$ implies $f(n)<n+1$ and an easy induction shows that $f$ is the identity.

Since the image of $f$ is a non-empty set of natural numbers, it has a a smallest element, $m_0$.  Suppose that $f(n)=m_0$ for some $n>0$.  Then $n-1\in \mathbb{N}_0$, and we have $f(f(n-1))<f(n)=m_0$ which contradicts the definition of $m_0$.  However, $f(n)=m_0$ for some natural number $n$, so $f$ attains its minimum at, and only at, $n=0$.

For $k=1,2,\dots$, let $\mathbb{N}_k=\{n\in \mathbb{N}_0\mid n\geq k\}$, and let $f_k$ be the restriction of $f$ to $\mathbb{N}_k$.   I claim that the minimum value of $f_k$ is attained at, and only at $n=k$.  We have already proved the basis.  Suppose that $k>0$ and that the claim is true for $n=0,1,\dots,k-1$.  $f_k$ attains its minimum $m_k$ at some natural number $n\geq k$.  Suppose, by way of contradiction, that $n>k$.  Then $n-1\geq k$ and we have $f(f(n-1))<f(n)=m_k$.  To show that this is a contradiction, we have only to show that $f(n-1)\geq k$.  The induction hypothesis implies that $f(0), f(1), \dots, f(k-1)$ is an increasing sequence of natural numbers, so that $f(k-1)\geq k-1$.  By the induction hypothesis, $f_k$ attains its minimum only at $k-1$, so that $n-1\geq k$ implies $f(n-1)>f(k-1)\geq k-1$, which completes the proof.

I'd like to say a few words about your statement that you couldn't figure out where to start.  That gave me trouble, too.  I tried to start by proving $f(0)=0$, but all I could prove was that $f(n)\neq0$ for $n>0$.  For a short while, I tried to construct a counterexample, but I quickly became convinced that the statement is true.  So then I looked for a weaker condition that would imply the theorem.


问:
Number of numbers $ 0 \leq n \leq 10,000,000$ such that the sum of the digits is at most $34$

Calculate how many natural numbers $n$ where $ 0\leq n \leq 10,000,000$ such that the sum of their digits is AT MOST $34$

And I said the following:

We have 8 digits in numbers between $0$ and $10,000,000$. For the first digit there can be only $0$ or $1$. For all rest digits we have all possibilties $[0,...,9]$. So let us do a generating function: $F(x) = (1+x)(1+x+x^2+...+x^9)^7$. We are searching for the coefficients of $1$ and $x$ and $...$ and $x^{34}$. But that is like searching for the coefficient of $x^{34}$ in the generating function $G(x) = (1+x+x^2+...)(1+x)(1+x+x^2+...+x^9)^7$

But how do I continue from there? How can I know the coefficient of $x^{34}$ in $G(x)$?

答:
In comments you have correctly reduced the question to determining the coefficient of $x^{34}$ in $(1-x^{10})^7(1-x)^{-8}$. We can reduce modulo $x^{35}$, in which case the first factor becomes $1-\binom71x^{10}+\binom72x^{20}-\binom73x^{30}=1-7x^{10}+21x^{20}-35x^{30}$. Then it suffices to determine in the factor $(1-x)^{-8}$ the terms of degree $34-10k$ for $k=0,1,2,3$, which can be seen to be repectively $\binom{-8}{34}(-x)^{34}=\binom{41}7x^{34}=22481940x^{34}$, $\binom{-8}{24}(-x)^{24}=\binom{31}7x^{24}=2629575x^{24}$, $\binom{21}7x^{14}=116280x^{14}$, and $\binom{11}7x^4=330x^4$. Utlimately I find the coefficient
$$
  22{,}481{,}940x-7\times2{,}629{,}575+21*116{,}280-35*330=6{,}505{,}245.
$$
Just a sanity check: that's more than half of all numbers $n<10^7$; indeed the average sum of $7$ digits is $7\times4.5=31.5$, so it is more likely the sum is at most $34$ than the contrary. The final outcome of course is $6{,}505{,}245+1=6{,}505{,}246$.</p>

ContentLicense: CC BY-SA 3.0


问:
How to taylor expand $\Pi_{i=1}^d\frac{\sin x_j}{x_j} = 1-\frac{1}{6}|x|^2 + O(|x|^4)$ as $|x| \to 0$?
How do we integrate $\int_C (1-e^{i \langle w,x \rangle})dw$ where $C = [-1,1]^d$?

I am given that the answer is $2^d (1- \Pi_{i=1}^d\frac{\sin x_j}{x_j})$ but I cannot see why.

Also, why do we have that $\Pi_{i=1}^d\frac{\sin x_j}{x_j} = 1-\frac{1}{6}|x|^2 + O(|x|^4)$ as $|x| \to 0$?

I cannot find any explanation on these results and I would greatly appreciate if anyone could show how to derive these results.

Answer to first question.

The first integral can be calculated by splitting $1$ from the integrand, and looking at $\int_{-1}^1 e^{iwx}dw = \frac{1}{ix}(e^{ix}-e^{-ix}) = \frac{1}{ix} 2i \sin x = \frac{\sin x}{x}$.

But I am stil struggling with the taylor expansion.

Body:
We know that
$$
{\sin u\over u}=1-{u^2\over 6}+O(u^4)
$$
hence
$$
\prod_{i=1}^d{\sin x_i\over x_i}{=\prod_{i=1}^d[1-{x_i^2\over 6}+O(x_i^4)]
\\=1-\sum_{i=1}^d{x_i^2\over 6}+\sum_{i=1}^dO(x_i^4)+O(||x||^6)
}
$$
hence the result.


问:
Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$
The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?

Body:
Hint:

If $\sin x+\cos x=y,$

$ y^2=?,$

$-\sqrt2\le y\le\sqrt2\ \ \ \ (1)$

$$\dfrac{y^2-1}2=y\iff y^2-2y-1=0\iff(y-1)^2=2$$

$\implies y=1\pm\sqrt2$

By $(1), y\ne1+\sqrt2$


问:
Solving linear congruences by hand: modular fractions and inverses
When I am faced with a simple linear congruence such as
$$9x \equiv 7 \pmod{13}$$
and I am working without any calculating aid handy, I tend to do something like the following:

"Notice" that adding $13$ on the right and subtracting $13x$ on the left gives:
$$-4x \equiv 20 \pmod{13}$$

so that $$x \equiv -5 \equiv 8 \pmod{13}.$$

Clearly this process works and is easy to justify (apart from not having an algorithm for "noticing"), but my question is this: I have a vague recollection of reading somewhere this sort of process was the preferred method of C. F. Gauss, but I cannot find any evidence for this now, so does anyone know anything about this, or could provide a reference? (Or have I just imagined it all?)

I would also be interested to hear if anyone else does anything similar.

答:
When presented with

$\tag 1 ax \equiv b \pmod{n}$

if $a \mid b$ the solution is right in front of you.

But there is also a 'plug in' solution if $a \mid n-1$ or  $a \mid n+1$:

If $a \mid n-1$ then $x = \Large(\frac{n-1}{a}) \normalsize (-b)$ solves $\text{(1)}$.

If $a \mid n+1$ then $x = \Large(\frac{n+1}{a}) \normalsize (b)$ solves $\text{(1)}$.

Can we 'make hay' with the OP's linear congruence?

$\quad 9x \equiv 7 \pmod{13} \; \text{ iff } \; -4x \equiv 7 \pmod{13} \; \text{ iff }$

$ \quad  4x \equiv -7 \pmod{13} \; \text{ iff } \; 4x \equiv 6 \pmod{13}$

We are in business now with $4x \equiv 6 \pmod{13}$ since $4 \mid 12$; a solution is

$\quad x = \Large(\frac{n-1}{a}) \normalsize (-b) = (3)(-6) = -18 \equiv 8 \pmod{13}$

Here is an example where the $n + 1$ manipulation can be used:

$\quad 5x \equiv 1 \pmod{17} \; \text{ iff } \; -12x \equiv 1 \pmod{17} \; \text{ iff }$

$ \quad  12x \equiv -1 \pmod{17} \; \text{ iff } \; 12x \equiv 16 \pmod{17} \; \text{ iff } \; 6x \equiv 8 \pmod{17}$

We are in business now with $6x \equiv 8 \pmod{17}$ since $6 \mid 18$; a solution is

$\quad x = \Large(\frac{n+1}{a}) \normalsize (b) = (3)(8) = 24 \equiv 7 \pmod{17}$


问:
How to find the first term given the common ratio and the sum

The ratio of a GP is $r$ = $2$ and the sum to eight terms is $1785$. Find the first term.

I have no real idea on how to approach this problem, so far I tried:

Since the $r$ is given I tried dividing $1785$ by $2$ and then divided that answer so on and so forth by $2$ 8 times, but is there an easier way to do this?

Body:
$$S_n=a_1\times\frac{1-r^n}{1-r}$$

where $r\not=1$

$$S_8=a_1\times\frac{1-2^8}{1-2}=1785$$

$$255a_1=1785$$

$$a_1=7$$


问:
Is this the best notation?
I am working with two types of conjugate numbers:

$p + q$ in which $p \in \mathbb{Q}$ and $q \in \mathbb{R}$

$p + q$ in which $p \in \mathbb{Q}$ and $q \in i\mathbb{R}$

I would like to define a set which contains both types of numbers. But I would like to stress their conjugate forms. Is the following enough to describe it? Also, is this the best notation available?

$R = \{ p \pm q | p \in \mathbb{Q} \wedge ( q \in \mathbb{R}_{\ge 0} \vee q \in i \mathbb{R}_{\ge 0}  )  \}$

Thanks

答:
Your set is
$
(\mathbb Q + \mathbb R) \cup (\mathbb Q + i\mathbb R) =
\mathbb R \cup (\mathbb Q + i\mathbb R) =
\{ z \in \mathbb C : \operatorname{Im}(z)=0 \text{ or } \operatorname{Re}(z) \in \mathbb Q \}
$.

Choose the expression that makes your intention clear to the reader. I guess it is the first one.

Title: Is this the best notation?


问:
Is the process of determining whether a singular is regular or not same for the homogenous and non-homogenous ODEs?
I have this second-order non-homogeneous ODE:
$$0.5x^5A''(x)+\left(2x^4+x^3\right)A'(x)+\left(x^3+x^2+x-1\right)A(x)=-1$$

I have determined that $x=0$ is a singular point and divided both sides by $0.5x^5$. What I am now trying to do is determine whether it is regular or not.

I am aware of the process with homogenous ODEs, about coefficients being divisible by $(x-x_0)$. But I am wondering, does the same property retain with non-homogeneous ODEs? I have sought the answer on the internet but all examples are of homogeneous DEs.

答:
$0.5x^5A''(x)+(2x^4+x^3)A'(x)+(x^3+x^2+x-1)A(x)=-1$

$A''(x)+\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)A'(x)+\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)A(x)=-\dfrac{2}{x^5}$

$\lim\limits_{x\to 0}\left(x\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)\right)=\lim\limits_{x\to 0}\left(4+\dfrac{2}{x}\right)=\infty$

$\lim\limits_{x\to 0}\left(x^2\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)\right)=\lim\limits_{x\to 0}\left(2+\dfrac{2}{x}+\dfrac{2}{x^2}-\dfrac{2}{x^3}\right)=\infty$

$\therefore$ the finite singular point $x=0$ is irregular.

That's not the end. We should also check the singularities at infinity. Because these also act as one of the key points of distinguishing the ODE type.

Let $t=\dfrac{1}{x}$ ,

Then $\dfrac{dA}{dx}=\dfrac{dA}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{dA}{dt}=-t^2\dfrac{dA}{dt}$

$\dfrac{d^2A}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{dA}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{dA}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)(-t^2)=t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}$

$\therefore t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}-(4t+2t^2)t^2\dfrac{dA}{dt}+(2t^2+2t^3+2t^4-2t^5)A=-2t^5$

$t^4\dfrac{d^2A}{dt^2}-(2t^2+2t)t^2\dfrac{dA}{dt}-(2t^5-2t^4-2t^3-2t^2)A=-2t^5$

$\dfrac{d^2A}{dt^2}-\left(2+\dfrac{2}{t}\right)\dfrac{dA}{dt}-\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)A=-2t$

$\lim\limits_{t\to 0}\left(-t\left(2+\dfrac{2}{t}\right)\right)=\lim\limits_{t\to 0}(-2t-2)=-2$

$\lim\limits_{t\to 0}\left(-t^2\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)\right)=\lim\limits_{t\to 0}(-2t^3+2t^2+2t+2)=2$

$\therefore$ the singularities at $x=\pm\infty$ are regular.


问:
Evaluating $\lim_{(x,y)\to(0,0)} \frac {3xy^2}{2x^2+9y^4}$
I have:
$$\lim_{(x,y)\to(0,0)} \dfrac {3xy^2}{2x^2+9y^4}=
I$$ tried replacing $y$ with $x\cdot m$ and I got $\lim\limits_{x\to 0}\frac{3xm^2}{2+9x^2m^4}$ and I thought this limit is $0$, but my teacher told me that this limit is not $0$. What can I do?

答:
Hint

The limit doesn't exist actually. If $f$ is your function, try to see how behave $f(0,t)$ and $f(t^2,t)$ when $t\to 0$.


问:
Is it possible to have a walk between every two vertices of any arbitrary length in a graph?
I have a very basic question. I was going through the topic Walks  in Graphs. I was just wondering: Is it possible to have a walk between every two vertices $x$ and $y$ of any arbitrary length $l$ in $G$? Is it really possible for every connected graph? Here $G$ is connected graph. Thanks in advance. Please correct me if I am wrong.

答:
If the graph is connected and non-bipartite (i.e., contains an odd length cycle), then for all pairs of vertices $x$ and $y$, there exists some $L$ such that for all $\ell \geq L$ there is a walk from $x$ to $y$ of length $L$.

We can achieve $\mathrm{dist}(x,y)+2k$ for all $k \geq 0$ by walking from $x$ to $y$, then walking back and forth along an edge $yz$ for some neighbour $z$ of $y$.

We can achieve $L:=\mathrm{dist}(x,y)+2k+t$ for all $k \geq 0$ for some odd number $t$ by walking from $x$ to the odd cycle, around the odd cycle and back to $x$, then continuing as per the first bullet point.

By definition, it's not possible to walk from $x$ to $y$ in less than $\mathrm{dist}(x,y)$ steps.

If the graph is bipartite, then any walk on the graph alternates between colour classes; so either the number of odd-length walks or the number of even-length walks from $x$ to $y$ is $0$ (depending on whether or not $x$ and $y$ are in the same colour class).


答:
Hint:

If the first term was $1$, the sum would be $255$. (Why ? How is this helpful ?)

Title: How to find the first term given the common ratio and the sum

Body:
The ratio of a GP is $r$ = $2$ and the sum to eight terms is $1785$. Find the first term.

I have no real idea on how to approach this problem, so far I tried:

Since the $r$ is given I tried dividing $1785$ by $2$ and then divided that answer so on and so forth by $2$ 8 times, but is there an easier way to do this?


Title: Evaluating $\lim_{(x,y)\to(0,0)} \frac {3xy^2}{2x^2+9y^4}$

Body:
I have:
$$\lim_{(x,y)\to(0,0)} \dfrac {3xy^2}{2x^2+9y^4}=
I$$ tried replacing $y$ with $x\cdot m$ and I got $\lim\limits_{x\to 0}\frac{3xm^2}{2+9x^2m^4}$ and I thought this limit is $0$, but my teacher told me that this limit is not $0$. What can I do?

Body:
$$\lim_{(x,y)\to (0,0) } \dfrac {3xy^2}{2x^2+9y^4}$$
For $y=0$:
$$\lim_{(x,y)\to (0,0) }  \dfrac {3xy^2}{2x^2+9y^4}=\dfrac {0}{2x^2} = 0$$
For $x=y^2$ we have that:

$$\lim_{(x,y)\to (0,0)} \dfrac {3xy^2}{2x^2+9y^4}=\lim_{(x,y)\to (0,0) }\dfrac {3y^4}{11y^4}=\dfrac {3}{11} \ne 0$$

The limit does not exist.


问:
Find intersection of two semi-circles
I know how to calculate the intersection points of two circles. But I wasn't able to adjust the formula in my Python program so it would also apply for semi-circles.

How can I find the intersection points of two semi-circles?

For example, with which formula can I calculate the intersection points of semi-cirle $a$ and semi-circle $b$? Notice that the intersection points on the red line should not be calculated. You can actually call it arc $a$ and arc $b$.

答: Compute the intersection points of the full circles to find candidates. Then throw away the candidates that don’t lie on the semicircles.</p>

Title: Find intersection of two semi-circles


答:
To see why the limit doesn't exist, it can be useful define

$\sqrt 2\,x = u$

$3y^2=v$

to obtain

$$\lim_{(x,y)\to(0,0)} \dfrac {3xy^2}{2x^2+9y^4}=\frac{1}{\sqrt 2}\,\lim_{(u,v)\to(0,0)} \dfrac {uv}{u^2+v^2}$$

which is simpler to handle.


问:
Finding diagonalizable matrix's basis
If $T: R^3 \rightarrow R^3$ is diagonalizable and has distinct eigenvalues, find $0 \neq v \in R^3$ such that

{$v,T(v), T^2(v)$} is a basis for $R^3$.

I think that statement: {$v,T(v), T^2(v)$} is a basis for $R^3$, is obvious by theorem, but don't know what the question asks to find. It means that I really have to find exact value of $v$, $v=(1,0,0)$ something like that?

答:
Since you have distinct eigenvalues, you have a complete eigenbasis, which I will denote as $\{ v_1, v_2, v_3 \}$

Claim: $ v = v_1 + v_2 + v_3 $ works.

Complete this. Show that $av + bTv + cT^2 v = 0 \Leftrightarrow a, b, c = 0 $.

Hint: You have a quadratic equation that is satisfied by 3 distinct values.

Title: Finding diagonalizable matrix's basis


问:
How to write a multivariable polynomial as a matrix expression
Say I have a polynomial $2x^2 + 3xy + 9y^2$, I was wondering how I would go about converting this into a matrix expression of the form $(x,y)S$, where $S$ is a $1 \times 2 $ matrix.

Any help would be appreciated and I apologise for being unable to figure out how to put a matrix in

答: $$2x^2 + 3xy + 9y^2 = (x,y)\begin{pmatrix}2&\frac32\\\frac32&9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$


答:
Actually, since you have a quadratic polynomial, you need to have an expression where $(x,y)$ occurs twice, not once. Something like this:
$$
\begin{bmatrix} x & y \end{bmatrix}
\begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} x \\ y \end{bmatrix}
$$
Can you multiply this out and compare the coefficients to your expression?

Title: How to write a multivariable polynomial as a matrix expression

Body:
Say I have a polynomial $2x^2 + 3xy + 9y^2$, I was wondering how I would go about converting this into a matrix expression of the form $(x,y)S$, where $S$ is a $1 \times 2 $ matrix.

Any help would be appreciated and I apologise for being unable to figure out how to put a matrix in


问:
Hypothesis testing on sampling from exponential distribution
Problem:
Investigating the lifetime of some product, we're interested in testing the hypotheses
\begin{align}
 H_0: \mu=2\\
 H_1: \mu<2
\end{align}

Where $\mu$ is the expectation value of the lifetime in some arbitrary unit. We assume that the lifetimes follow an exponential distribution with parameter $\lambda$, i.e.
\begin{equation*}
 f_X(x)=
 \begin{cases}
 \lambda e^{-\lambda x} &\text{ for }x\geq 0\\
 0      &\text{ ellers }
 \end{cases}
\end{equation*}
and $E[X]=\mu=\frac{1}{\lambda}$.

Now assume we have sampled the lifetime of 10 sample products and found a sample mean $\bar{X}$. Let $c>0$ be some number such that, if $W\geq c$ we accept the null hypothesis, otherwise discard it.

Determine the number $c$ such that our test has signifiance level of 95%.

Solution (attempt)

Our book mostly covers how to find the sigifiance intervals for Normal distributions, or cases of known mean with n large, which is hardly applicable for $n=10$, and the assignment asks for an exact test of the nullhypothesis.

I've only noted the following, that if $X_1,X_2,....X_n$ are indenpendent Exponential($\lambda$) stochastic variables, then their sum follows a Gamma distribution. I.e.
\begin{equation*}
 \bar{X}=\frac{X_1,X_2,...X_n}{n}=\frac{1}{n}\sum_{i=1}^{n}X_i\sim \frac{1}{n}Gamma(n,\lambda) \sim Gamma(n,n\lambda)
\end{equation*}

That is, it follows some distribution
\begin{equation}
 f_Y(y) = \frac{(n\lambda)^n y^{n-1}e^{-n\lambda y}}{\Gamma(n)}, y>0
\end{equation}

But how does this help me determine $c$?

P.S.
The assignment problem also states that we might use the property:
For $X\sim Gamma(\alpha_1,\lambda), Y\sim Gamma(\alpha_2,\lambda) \implies X+Y\sim Gamma(\alpha_1+\alpha_2,\lambda)$.

However, I'm not sure how to make use of this, as I don't see what is to be summed, any way.

答:
Your hypothesis system can be rewritten as follows:

$$\begin{cases}
\mathcal{H}_0: &\lambda=\frac{1}{2} \\
\mathcal{H}_1: &\lambda>\frac{1}{2}
\end{cases}$$

Now let's fix  $\lambda'<\lambda''$ and derive the likelihood ratio:

$$\frac{L(\lambda'|\mathbf{x})}{L(\lambda''|\mathbf{x})}=\frac{(\lambda')^n e^{-\lambda'\Sigma x}}{(\lambda'')^n e^{-\lambda''\Sigma x}}=\Bigg(\frac{\lambda'}{\lambda''}\Bigg)^nExp(\lambda''-\lambda')\Sigma x$$

It is evident that this ratio is a monotone increasing function in $T=\Sigma x$

Now it is enough to apply a well known theorem on UMP test for families having a monotone likelihood ratio to derive the critical region, that in this case is

$$\mathbb{P}\Bigg[T<k|\lambda=\frac{1}{2}\Bigg]=\alpha$$

In your case, you know that

$$T=\Sigma x\sim Gamma(10;\frac{1}{2})=\chi_{(20)}^2$$

Thus you can easy solve your exercise using the chi-square table

It results to me that we reject the Hypotheses that $\mu=2$ against the alternative that $\mu<2$ iff $\overline{X}_{10}<1.085$

The theorem I refer to is the following taken from Mood Graybill Boes, Chapter IX, Tests of Hypotheses


问:
Smooth functions on a nonopen set
I am reading the Loring Tu, Introduction to manifolds where you can find this excersise:

Using a partition of unity, show that a function $f:S\rightarrow \mathbb{R}^m$ is $C^{\infty}$ on $S\subset \mathbb{R}^n$ if and only if there exist an open set $U$ in $\mathbb{R}^n$ containing $S$ and a $C^\infty$ function $F:U\rightarrow \mathbb{R}^m$ such that $f=F|_S$

I was trying to prove the $\Rightarrow$ part of this using the $C^\infty$ extension of a function (Proposition 13.2) and taking a bump function with support in U and that is identically 1 in S. My question is: why shouldn't I take this approach and instead use a partition of unity?

答:
There seems to be missing some asumptions in here. I will then add my own assumptions and proove this result: I will work in the case $S\subset \mathbb{R}^n$ is a compact embedded submanifold of $\mathbb{R}^n$ without boundary.

Let $NS = \{ (x,v) \in S\times \mathbb{R}^n,~ v\in T_xS^{\perp}\}$ the normal bundle of $S$. Let $g : NS \to \mathbb{R}^n$ be the function $g(x,v) = x +v$. Its tangent bundle $T(NS)$ is canonically isomorphic to $TS \oplus TS^{\perp} = T\mathbb{R}^n$, and we can consider $\mathrm{d}_{x,0}g$ as the identity map of $\mathbb{R}^n$. Thus, $g$ is a local diffeomorphism at $(x,0)$ and thus is a local diffeomorphism on the zero section $S\times\{0\} \subset NS$. Because of the compactness of $S$, there exists a positive number $\varepsilon >0$ such that
\begin{align}
g : \{(x,v) \in NS ~|~ \|v\| < \varepsilon\} & \longrightarrow \mathbb{R}^n \\
(x,v)&\longmapsto x+v
\end{align}
is an embedding.
We denote by $U$ the image of $g$ of this embedding. This is a tubular neighbourhood of $S$ which can be parametrized by $S\times B\left(0,\varepsilon \right)$ where $B(0,\varepsilon)$ is the ball of radius $\varepsilon$ in $\mathbb{R}^{n-\dim S}$. If $p\in U$, we write $g^{-1}(p) = (x_p,v_p)$, and $ p \mapsto x_p$ and $p \mapsto v_p$ are smooth functions.

Let $\varphi : \mathbb{R} \to [0,1]$ be a smooth function such that:

$\varphi(0)=1$

$|t| \geqslant \varepsilon^2 \Rightarrow \varphi(t) = 0$

Let $f : S \to \mathbb{R}^m$ be a smooth function. Then
\begin{align}
F : \mathbb{R}^n &\longrightarrow \mathbb{R}^m \\
 p & \longmapsto \left\{\begin{array}{lcl}
0 & \text{if} & p\notin U \\
\varphi\left(\|v_p\|^2\right)f\left(x_p\right) & \text{if} & p \in U
\end{array}\right.
\end{align}
is a smooth function such that $F|_S=f$.

In fact, the key is that if $S$ is a submanifold of $\mathbb{R}^n$, then there exists a tubular neighbourhood of $S$ in $\mathbb{R}^n$, and there exists a smooth function with compact support in this tubular neighbourhood which is constant (and equal to $1$) on $S$.

You can try to adapt this result (using partition of unity) in the case $S$ is an embedded non compact submanifold without boundary.

Title: Smooth functions on a nonopen set


问:
Sufficient conditions for linear mapping $T:X \to Y$ being continuous
I can neither prove nor disprove the following statement.
I would be grateful if anyone could give me some advice.

Let $X$ and $Y$ be complex Banach spaces,
and let $T:X \to Y$ be a linear mapping.
Let $D \subset X$ be a dense subset of $X$ ($D$ is NOT necessarily a subspace of $X$),
and $\|Tx\|_Y \leq C \|x\|_X$ for all $x \in D$.
Then, $\|Tx\|_Y \leq C\|x\|_X$ for all $x \in X$.

答:
It's false if you assume $D$ to be a proper dense subspace of infinite codimension, let alone more general dense sets. As an example, if we take $X = C[0, 1]$ (with the $\infty$-norm), and $D$ to be the subspace of polynomials on $[0, 1]$, then Stone-Weierstrass implies $D$ is a dense subspace of $X$. As $D$ has countable dimension, it must have (uncountably) infinite codimension, as infinite-dimensional Banach spaces cannot have uncountably infinite dimension (by Baire Category Theorem).

If we suppose $D$ is a dense, proper subspace of Banach space $X$ with infinite codimension and $Y$ is non-trivial, you'd be asserting that every extension of a bounded linear map $S : D \to Y$ to $X$ must also be bounded, with the same norm. This is patently untrue.

Suppose we have such a linear map $S$. Start with a hamel basis $B$ of $D$, and let $B' \subseteq X$ such that $B \cup B'$ is a basis of $X$. Without loss of generality, assume every vector in $B \cup B'$ has length $1$. Since $D$ has infinite codimension, $B'$ is infinite.

Select a sequence $x_n \in B'$, and some $y \in Y \setminus \{0\}$. Define $\hat{S}$, an extension of $S$ to $X$, on the Hamel basis $B \cup B'$, by letting
$$\hat{S}(x) = \begin{cases} S(x) & \text{if }x \in B \\ ny & \text{if } x = x_n \\ 0 & \text{if }x \in B' \setminus \{x_1, x_2, \ldots\}.\end{cases}$$
The above uniquely defines a linear map $\hat{S} : X \to Y$, which cannot be bounded, as $\|x_n\| = 1$, but $\|\hat{S}(x_n)\| = n\|y\| \to \infty$ as $n \to \infty$.

So, in specific reference to your question, let $T = \hat{S}$, and note that $\|Tx\| \le \|S\|\|x\|$ for $x \in D$, but $T$ is unbounded on $X$.


问:
Distinct roots of $ \int\limits_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt} = 2x -1$
The total number of distinct $x\in[0,1]$ for which $ \int\limits_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}  = 2x -1$ is ________.

My approach is as follow

$y = \int\limits_0^x {\frac{{{t^2}}}{{1 + {t^4}}}dt}  - 2x + 1$

$y' = \frac{{{x^2}}}{{1 + {x^4}}} - 2$

$y' = \frac{1}{{{x^2} + \frac{1}{{{x^2}}}}} - 2$

${x^2} + \frac{1}{{{x^2}}} \ge 2\left( {AM \ge GM} \right)$

Hence $y'<0$ decreasing

$y\left( 0 \right) = \int\limits_0^0 {\frac{{{t^2}}}{{1 + {t^4}}}dt}  - 2 \times 0 + 1 = 1$

The official answer is $1$ but how we will find whether $y(1)<0$

答:
Your $AM\ge GM$ approach in effect already shows it, just apply it to

$${t^2\over1+t^4}={1\over t^2+{1\over t^2}}\le{1\over2}$$

so

$$y(1)=\int_0^1{t^2\over1+t^4}\,dt-2\cdot1+1\le{1\over2}-2+1=-{1\over2}$$


问:
If $V \hookrightarrow H$ are Hilbert spaces and $f \in H'$ then $f \in V'$?
Let $H=(H, (\cdot, \cdot)_H)$ and $V=(V, (\cdot, \cdot)_V)$ be Hilbert spaces such that $V \hookrightarrow H$ that is $V$ is  continuously embedding in $H$. By Riesz representation theorem we can identify $H=H'$. Thus we have
$$V \hookrightarrow H=H' \hookrightarrow  V'.$$

Since $H' \hookrightarrow  V'$ then, in particular, $H' \subset V'$.

Question. If $f \in H'$ then we can to say that $f \in V'$?

I think so, in virtue of $H' \subset V'$. But more formally, due to $f \in H'$ then $f$ is linear. Now, I want to prove that $f$ is bounded (in $V'$). So I can to write, for every $v \in V$,
$$\langle f , v \rangle _{V',V}= \langle f , v \rangle _{H',H} \leq \|f\|_{H'}\cdot \|v\|_{H} \leq \|f\|_{H'}\cdot c\cdot \|v\|_{V}?$$
Here, $c>0$ is the constant of the embedding $V \hookrightarrow H$.

答:
The proof you have written works. Denoting the inclusion by $i:V\to H$ you have for $f\in H$ and $v\in V$ that:

$$|f(i(v))| ≤\|f\|_{H'} \cdot \|i(v)\|_H ≤ \|f\|_{H'}\cdot \|i\|_{L(V,H)}\cdot \|v\|_{V}$$

where you applied boundedness of $f$ and $i$ as linear maps to get that $v\mapsto f(i(v))$ is bounded as a linear map (with bound $\|f\|_{H'} \cdot \|i\|_{L(V,H)}$).

Depending on how precise you want to be it is not entirely correct to say that $f$ is an element of $V'$. Rather you should say that any $f\in H'$ induces an element of $V'$ by the prescription $f\mapsto f\circ i$.

Understanding the problem in that way also gives you a more immediate understanding why you get an element of $V'$, for $f\circ i:V\mapsto \Bbb K$ is continuous and linear as a composition of continuous linear maps.


问:
How to prove that this condition implies that the function is constant
I am trying assignments of my complex analysis course which will not be discussed and I got struck on this question:

Let D be the open unit disc centred at origin and $f : D \to \mathbb{C}$ be an analytic function. Let f=u+iv . If f(z)=$\sum_{n=0}^{\infty} a_n z^n $ is the power series of f and for any closed curve p in D , $\int_{p} \frac{ f(z) dz}{(z-a)^2} =0 $ for all a in D with $|a|\geq 1/2$ then f must be constant.

Cauchy theorem implies that $a_{1}=0$ for any defined power series with a as centre such that $|a|\geq 1/2$. But I don't have any reason to believe that f is constant and so I am unable to understand which result I should use.

Kindly help me to prove the deduction.

It is my first course in complex analysis.

答:
Observe that Cauchy's formula gives for all $|z| \geq 1/2$

$$f'(z) = \frac{1}{2\pi i} \int_{p} \frac{f(\zeta)}{(\zeta-z)^2} d\zeta=0,$$

which shows that $f$ is constant in the annulus $1/2 < |z| < 1$.

Now, you can use the maximum modulus principle on $f'$ for the disc $|z| < 1/2$. Since $f'$ is itself analytic, it achieves its maximum at the boundry of this disc, which we have shown to be 0. Thus, $f'(z)=0$ for all $z \in D$, which shows that $f$ is constant.


答:
Apply the half-angle substitution $t= \cot \frac x2$ to $1-\frac{1}{\sin x}-\frac {1}{\cos x}=0$
$$t^4+ 4t -1=0$$
which factorizes as

$$(t^2+2a t+\frac{2a^3-1}a)(t^2-2a t+\frac{2a^3+1}a)=0$$
with $a^6+\frac14a^2-\frac14=0$, which yields $a^2=\frac12$. Plug it into above equation to obtain a pair of real roots
$$t=\frac{-1\pm\sqrt{2\sqrt2-1}}{\sqrt2}$$
Thus, the solutions for $x$ are
$$x=2n\pi+2\cot^{-1} \frac{-1\pm\sqrt{2\sqrt2-1}}{\sqrt2} $$

Title: Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$

Body: The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?


问:
Prove that there are exists uncomputable partial function $U(n,x)$ such that $ \forall n \in N: U(n,x)$ and $U(x,n)$ are computable.
Prove that there are exists uncomputable partial function $U(n,x)$ such that $ \forall  n \in N:  U(n,x)$  and  $U(x,n)$ are computable.

It's new material for me and i don't have any ideas how to solve tasks like this..

答:
The intuitive idea is this: take a non-computable function and "spread it out" across $\mathbb{N}^2$ so that the horizontal and vertical slices each encode only finitely much non-trivial information, and hence are computable.

Here's one solution: Let $H\subseteq \mathbb{N}$ be a non-computable set. Define $$U(x,y) = \begin{cases} 1 & \text{if $x=y$ and $x\in H$}\\ 0 & \text{otherwise}.\end{cases}$$

Now $U$ is not computable. Indeed, if it were computable, then the function $x\mapsto U(x,x)$ would be computable, but this is the characteristic function of $H$.

On the other hand, for any fixed $n$, the functions $x\mapsto U(n,x)$ and $x\mapsto U(x,n)$ are computable. Indeed, if $n\in H$, we have $$U(n,x) = U(x,n) = \begin{cases} 1 & \text{if $x=n$}\\ 0 & \text{otherwise}\end{cases}$$
which is computable: it is the characteristic function of the set $\{n\}$. And if $n\notin H$, then $U(n,x) = U(x,n) = 0$ for all $x$.


问:
Prove $\int e^{x^{2}} = \sqrt{\pi}$
I have to prove $\int e^{x^{2}} = \sqrt{\pi}$.

I aready know that $\int e^{x^{2}} = \frac{\sqrt{\pi} * erfi(x)}{2}$. What i cannot understand is how $\frac{erfi(x)}{2}$ get cancelled.

Thank you.

答:
$erf$ is a name we sometimes give to the primitive (anti-derivative) of the function $x \mapsto e^{-x^2}$, which cannot be expressed in the usual way. The reason we give it a name is because we know a few interesting properties about it. But in any case, since we don't know its expression, it's not gonna help us compute the integral of $e^{-x^2}$. You have to proceed otherwise...

Here's a hint: think about the expression $\left(\displaystyle\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x\right)^2$, and try to turn it into a simple integral, instead of a product of two integrals...

Title: Prove $\int e^{x^{2}} = \sqrt{\pi}$


答:
I'm assuming you meant ${\int e^{-x^2}dx}$ (correct me if I'm wrong). When you talk about the expression
$$
\int e^{-x^2}dx
$$
This is not going to just be ${\sqrt{\pi}}$. This is an anti-derivative - that is,
$$
\frac{d}{dx} \int e^{-x^2}dx = e^{-x^2}
$$
But clearly
$$
\frac{d}{dx}\sqrt{\pi}=0
$$
What your professor is probably meaning to ask you is to show
$$
\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}
$$
which is true. There is surprisingly more than one way to calculate this integral - polar coordinates, feynmans trick.... so it's difficult to give you the right guidance towards an answer that complements the material you have covered. Please edit the question and add more context

Title: Prove $\int e^{x^{2}} = \sqrt{\pi}$

Body:
I have to prove $\int e^{x^{2}} = \sqrt{\pi}$.

I aready know that $\int e^{x^{2}} = \frac{\sqrt{\pi} * erfi(x)}{2}$. What i cannot understand is how $\frac{erfi(x)}{2}$ get cancelled.

Thank you.


答:
you have a simple way of showing this with double integrals. Let :
$$I=\int_{-\infty}^\infty e^{-x^2}dx $$ Now we get:
$$I^2=\int_{-\infty}^\infty\int_{\infty}^\infty e^{-x^2}e^{-y^2}dxdy $$
Now you can do a polar change of variable, with $r^2=x^2+y^2 $, and $dxdy=rdrd\theta$

$$I^2=\int_{r=0}^{\infty}\int_{\theta=0}^{2\pi} re^{-r^2}drd\theta=\pi\int_0^\infty 2re^{-r^2}dr\\
=\pi\left[-e^{-r^2} \right]_0^\infty=\pi$$

Hence the result :
$$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi} $$

Title: Prove $\int e^{x^{2}} = \sqrt{\pi}$


答:
Since $T$ is diagonalizable, let $T=diag (t_1, t_2, t_3)$. Suppose $v=(v_1, v_2, v_3)\in R^3$ is a desired vector, then
$$
\det
\pmatrix{
v_1& v_2& v_3\\
t_1v_1& t_2v_2& t_3v_3\\
t_1^2v_1& t_2^2v_2& t_3^2v_3
}=v_1v_2v_3\text{Vandermonde matrix}
$$
so $v_1v_2v_3\ne 0$ and $t_1, t_2, t_3$ are distinct.</p>

Title: Finding diagonalizable matrix's basis

Body:
If $T: R^3 \rightarrow R^3$ is diagonalizable and has distinct eigenvalues, find $0 \neq v \in R^3$ such that

{$v,T(v), T^2(v)$} is a basis for $R^3$.

I think that statement: {$v,T(v), T^2(v)$} is a basis for $R^3$, is obvious by theorem, but don't know what the question asks to find. It means that I really have to find exact value of $v$, $v=(1,0,0)$ something like that?


问:
Stabiliser subgroup of an element in the set of all positive real numbers $\mathbb{R}$?
Consider the group of positive real numbers $(\mathbb{R}>0)$ acting on the real numbers by multiplication. What is the stabiliser subgroup of $x=0$?

What I've tried so far

So as I know, the stabiliser subgroup of an element $x$ is the set of all elements that leave $x$ invariant. The element $x=0$ can be multiplied by any real number to give $0$ (itself). So would this mean that the stabiliser subgroup for $x=0$ in the Group in my case is the set of all real numbers $\mathbb{R}$?

答:
Your reasoning is definitely correct until the last statement. Stabilizer of $0$ is not $\mathbb{R}$ but it should be $\mathbb{R}^+$. Note that if a group $G$ acts on a set $A$, then stabilizer of $a \in A$ in $G$ is
$$G_a = \{g\in G\ |\ g \cdot a = a\}$$
So, the elements that fix $a$ are all in the group $G$. Also note that $G_a \le G, \forall a \in A$.


问:
Assuming that A rolls first, find the probability that the final roll is made by A
I've a question about this probability's problem:

A and B alternate rolling a pair of dice, stopping either when A rolls the sum 9 or when B rolls the sum 6. Assuming that A rolls first, find the probability that the final roll is made by A.

Now, I know how to solve this problem by considering every possible turn for the A's final roll. Summing all the possibilities will lead to an infinite geometric series, hence, the answer.

However, there's an easier way to solve this problem:

Let $P_A$ be the probability that A wins when A rolls first, and let $P_B$ be the probability that B wins when B rolls first. Using that the sum of the dice is 9 with probability $\frac{1}{9}$, we obtain upon conditioning on whether A rolls a 9 that

$P_A = \frac{1}{9}+\frac{8}{9}(1-P_B)$

Similarly,

$P_B=\frac{5}{36}+\frac{31}{36}(1-P_A)$

Solving these equations gives that $P_A=\frac{9}{19}$ (and that $P_B=\frac{45}{76}$.)

This solution is by the author Sheldon M. Ross of the A First Course in Probability textbook.

How he got this solution? Especially, the part $(1-P_B)$. A detailed explanation would be appreciated.

Thank you!

答:
That solution considers only the first toss.  You might find it easier to consider the first two tosses.  In that case, three things are possible:  $A$ wins, $B$ wins, the game restarts.  Thus $$P_A=\frac 19\times 1+\frac 89\times \frac 5{36}\times 0 +\frac 89\times \frac {31}{36}\times P_A$$

from which you can easily compute $P_A$.

To do it the book's way, consider only the first toss.  Either $A$ wins  (prob. $\frac 19$) or not (prob. $\frac 89$).  In the latter case we are in the game in which $B$ is starting.  By definition, the probability that $B$ wins in that game is $P_B$ so the probability that $A$ wins in that game is $1-P_B$.  But of course the probability that $A$ wins is the $\frac 19$ times the probability that $A$ wins on the first toss plus $\frac 89$ times the probability that $A$ wins given that $A$ eventually wins given that it is now $B's$ turn.   That's what the given formula says.

Worth working it out in both ways, just for practice.


问:
To find a specfic equation $z-\frac{1}{s\ln(s)}-n\ln(s)==0$
Here I want to find the solution to the following equation
$$z-\frac{1}{s\ln(s)}-n\ln(s)=0,$$
where $z$ and $n$ are constants.
Here the aim is to find the solution $s=?$.

Any suggestion is welcome.

答:
You are looking for the zero's of function
$$f(s)=z-\frac{1}{s\log(s)}-n\log(s)$$
Let $s=e^x$ and consider the function
$$g(x)=n x^2- zx+e^{-x}$$ Its first derivative
$$g'(x)=2 n x-e^{-x}-z$$
cancels at
$$x_*=\frac{z}{2 n}+W\left(\frac{1}{2 n}e^{-\frac{z}{2 n}}\right)$$ where $W(.)$ is Lambert function.
We also have
$$g''(x)=e^{-x}+2n$$ which is always positive if $n \geq 0$.

If $g(x_*)<0$, make a series expansion of $g(x)$ around $x_*$
$$g(x)=g(x_*)+\frac{1}{2} (x-x_*)^2 g''(x_*)+O\left((x-x_*)^3\right)$$ and solve it to get, as an approximation,
$$x_\pm=x_*\pm \sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$

Trying for $n=3$ and $z=7$ this gives
$$x_*=\frac{7}{6}
+W\left(\frac{1}{6 e^{7/6}}\right)\approx 1.21607 \implies g(x_*) \approx-3.77962$$ This gives
$$x_-\approx 0.120362 \qquad \text{and}\qquad x_+\approx 2.31177$$ while the "exact" solutions, obtained by Newton method, are

$$x_-\approx 0.132651 \qquad \text{and}\qquad x_+\approx 2.31920$$

OwnerDisplayName: Blueka


问:
how to solve this probability exercise -selecting perfect parts from 2 boxes?
there are two boxes of parts, there are 50 parts in box A and 10 are perfect, 30 parts in box B and 18 are perfect. A box is selected randomly and two parts are taken out from it at a time without replacement.

1- find the probability that the first part is perfect.
2- suppose that the first part is found to be perfect. Find the probability that the second part is still perfect

I have resolved the first question and I got 0.4, but I still trying to resolve the second one, I would appreciate it If I get help or at least some hints

答:
I have an alternative way of thinking about this. The first part you have got correct, well done,  so I will provide only discussion for the second. This solution requires Bayes' theorem and The law of total probability

Once we are told that we found a perfect first piece we gain some information. What exactly have we gained? We are more likely to be in box B. Why? As it more fruitfully produces good parts. Let us ask ourself another question. Given that we have found a perfect piece, how likely are we to have initially selected box B?

Let A and B denote the events we selected box A or B respectively and let P denote the event we found a perfect piece.

$P(B|P)$ (the probability we selected box B given we found a perfect piece) = $P(P|B) \cdot \frac{P(B)}{P(P)} $ This is from Bayes Theorem. We know that $P(B) =\frac{1}{2} $ as the boxes are selected uniformly randomly. We also have $P(P) = \frac{2}{5}$ from your initial answer. Now onto $P(P|B)$. This is the probability we selected a perfect piece given we had box $B$ and is clearly $\frac{18}{30} = \frac{3}{5}$

Putting this all together yields: $P(B|P) = \frac{3}{5} \cdot \frac{1}{2} \cdot \frac{5}{2} = \frac{3}{4}$ Hence if we find a perfect piece in a box we are 75% sure we selected box $B$ and thus 25% sure we selected box $ A$. You could also for extra safety do the same calculation for $P(A|P) = P(P|A) \cdot \frac{P(A)}{P(P)} = \frac{1}{5} \cdot \frac{1}{2} \cdot \frac{5}{2} = \frac{1}{4} $

From the law of total probability, the probability we drawn another perfect piece is:

$P$(Another perfect piece | We are in box B ) $\cdot P($We are in box B) +   $P$(Another perfect piece | We are in box A ) $\cdot P($We are in box A) = $\frac{17}{29} \cdot \frac{3}{4} + \frac{9}{50} \cdot \frac{1}{4}  = \frac{2811}{5800} \approx 0.485$

Hope this helped.


问:
$\mathbb{Q}(\xi_n)\cap\mathbb{Q}(\xi_m) \subset \mathbb{Q}(\xi_{\gcd(n,m)})$

$\mathbb{Q}(\xi_n)\cap\mathbb{Q}(\xi_m) \subset \mathbb{Q}(\xi_{\gcd(n,m)})$ where $\xi_i$ is a primitive $i$th root of unity.

How can I show this? Could you give any hint?

答:
For $E/F$ Galois and $L/F$ then $[EL:L]=[E:E\cap L]$.

For simplicity assume that $E=F(a)$, let $g\in L[x]$ be the $L$-minimal polynomial of $a$, since all the roots of $g$ are in $E$ we get that $g\in E\cap L[x]$ so it is in fact the $E\cap L$-minimal polynomial of $a$ thus of degree $[E:E\cap L]$.

Then use it to find $$[\Bbb{Q}(\xi_n):\Bbb{Q}(\xi_n)\cap \Bbb{Q}(\xi_m)]$$ and compare with $$\frac{\phi(n)}{\phi(\gcd(n,m))} = [\Bbb{Q}(\xi_n):\Bbb{Q}(\xi_{\gcd(n,m)}]$$


问:
Find the basis of the intersection of two vector subspaces
I know there is lots of topics about intersection of two vector subspaces and basis but i still dont fully understand how we should handle these question.
So this is my homework:

Suppose U and W are subspaces of $R^3:\\$
$U=[(1,0,-1),(0,1,1)]\\W=[(2,4,0),(0,0,\sqrt{3})]\\$

Find a basis of $U\cap W \\$

So i know $U \cap W$ -> $a_1*(1,0,-1)+a_2*(0,1,1)-b_1*(2,4,0)-b_2*(0,0,\sqrt{3})=0\\$
a possible combination of coefficients $a_1=1,a_2=2,b_1=\frac12,b_2=-\sqrt{3}\\$

then i put the $1*(1,0,-1)+2*(0,1,1)$ in a matrix $$
        \begin{bmatrix}
        1 & 0 \\
        0 & 2 \\
        -1 & 2  \\
        \end{bmatrix}
\\$$

then i bring it to row-achelon form

$$
        \begin{bmatrix}
        1 & 0 \\
        0 & 2 \\
        0 & 0  \\
        \end{bmatrix}
\\$$

so Basis=$\{ (1,0,-1),(0,1,1) \}$ (i m not sure if the last vector should be $(0,1,1)$ or $(0,2,2)) $?

So is the solution correct ?

答:
Let me try a comprehensive answer. The Zassenhaus algorithm does this job, but it is somehow difficult to explain. Another approach is to remember that elementary row operations preserve linear dependence between columns of a matrix. Suppose that
$$U=[u_1=(1,0,0,-1),u_2=(0,1,0,-1),u_3=(0,0,1,1)]$$ and $$W=[w_1=(1,0,-1,0),w_2=(0,1,-1,0),w_3=(0,0,0,1)].$$
Note that the generators of $U$ and $W$ are linearly independent, otherwise we could exclude the dependent vectors. Now, form a matrix whose columns are the vectors of U followed by the vectors of W:
$$M=\left[\begin{matrix}
1 & 0 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & -1 & -1 & 0 \\
-1 & -1 & 1 & 0 & 0 & 1 \\
\end{matrix}\right].
$$
Computing the reduced row echelon form of $M$ we obtain that
$$\text{rref}(M)=\left[\begin{matrix}
1 & 0 & 0 & 0 & -1 & -\frac{1}{2} \\
0 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & 1 & 1 & \frac{1}{2} \\
\end{matrix}\right].
$$
Therefore we can conclude from the aforementioned fact, that
$$\begin{aligned}
w_2&=-u_1+u_2+w_1 \\
w_3&=-\frac{1}{2}u_1+\frac{1}{2}u_3+\frac{1}{2}w_1
\end{aligned}$$
and that $u_1$, $u_2$, $u_3$ and $w_1$ form a basis of $U+W$.
We cannot say that $w_2$ and $w_3$ form a basis of $U\cap W$, because they are not linear combinations of vectors of $U$, but defining
$$\begin{aligned}
w_2'&:=w_2-w_1 = -u_1+u_2 \\
w_3'&:=w_3-\frac{1}{2}w_1=-\frac{1}{2}u_1+\frac{1}{2}u_3
\end{aligned}$$
we see that $w_2'\in U\cap W$ and $w_3'\in U\cap W$. It is easy to see from general considerations that $w_2'$ and $w_3'$ are linearly independent and from
$$\text{dim}(U+W)+\text{dim}(U\cap W)=\text{dim}(U)+\text{dim}(W),$$
it turns out that they form a basis of $U\cap W$. This procedure can clearly be generalized and even transformed into a proof of the theorem about the dimension of the sum of subspaces.


问:
Modelling a horizontal mass damper using differential equations
I am trying to solve this second-order differential equation:

$y'' + y'+ y + (y')^2 =0$

I was able to solve the equation $y'' + y'+ y $, by substituting $y$ as $Ae^{kt}$.
But now I have this new term $(y')^2$.

Note: This equation represents the simplified equation of forces on a horizontal mass damper.

答:
Let $u=\dfrac{dy}{dt}$ ,

Then $\dfrac{d^2y}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dy}\dfrac{dy}{dt}=u\dfrac{du}{dy}$

$\therefore u\dfrac{du}{dy}+u^2+u+y=0$

$(y+u^2+u)\dfrac{dy}{du}=-u$

This belongs to an Abel equation of the second kind.

Let $v=y+u^2+u$ ,

Then $y=v-u^2-u$

$\dfrac{dy}{du}=\dfrac{dv}{du}-2u-1$

$\therefore v\left(\dfrac{dv}{du}-2u-1\right)=-u$

$v\dfrac{dv}{du}-(2u+1)v=-u$

$v\dfrac{dv}{du}=(2u+1)v-u$

Let $r=u+\dfrac{1}{2}$ ,

Then $v\dfrac{dv}{dr}=2rv-r+\dfrac{1}{2}$

Let $s=r^2$ ,

Then $\dfrac{dv}{dr}=\dfrac{dv}{ds}\dfrac{ds}{dr}=2r\dfrac{dv}{ds}$

$\therefore2rv\dfrac{dv}{ds}=2rv-r+\dfrac{1}{2}$

$v\dfrac{dv}{ds}=v-\dfrac{1}{2}+\dfrac{1}{4r}$

$v\dfrac{dv}{ds}=v-\dfrac{1}{2}\pm\dfrac{1}{4\sqrt s}$

This belongs to an Abel equation of the second kind in the canonical form.


问:
How do I compute the area of a cross which contains several disks which may overlap?

I have found this in a Facebook group :)
The only thing we are given is the marked distance and we need to find the shaded area.
I have tried several options but I am always missing some data.
Then I tried the following: I considered two different conditions:

The two side circles do not overlap; they are tangent to each other. In this case, the given distance corresponds to 3 diameters, so each is 3.33 units and the requested area is 100 sq. units.

The two side circles overlap by 100%. Again the area is 100 sq. units.

It is therefore obvious that the area is independent from the amount of overlapping, i.e. from the angle of the specific line segment to the horizontal line.

Can you provide a geometrical solution? Many thanks!!

答:
Let's call $r$ the radius of a circle, and $2x$ the overlap region between two circles. Then the area is $$A=(2r)^2+4(2r)(4r-2x)=4r^2+32r^2-16rx=36r^2-16rx$$
Now let's take a look at one of the circles on the sides. Draw the perpendicular from the intersection point to the axis connecting circles. This has a length $h$. Using Pythagoras, $$h^2=r^2-(r-x)^2=2rx-x^2$$
Now looking at the given line, just take half, and we apply Pythagoras again:
$$h^2+(3r-x)^2=5^2$$
Using $h^2$ from above, $$5^2=2rx-x^2+9r^2-6rx+x^2=9r^2-4rx$$
Then area is exactly four times this, so $A=4\cdot5^2=100$

In the above picture $DE=x$, $CD=h$, $AC=r$. Then $AD=r-x$


问:
What is the notation $D^2$'s meaning in Evans' Partial Differential Equation?
In his book, he did explain that $Du=D_xu=(u_{x_1},\cdots,\,u_{x_n})$. But what it means by writing $D^2u$?

答: It's explained in appendix A, on p. 617.</p>


问:
Relation between continuity and connection between topologies
If $X$ is a set and $\tau_1,\tau_2$ two topologies on $X$. What does it mean to put the continuity of the identic map on $X$ (i.e $id_X(x)=x\forall x\in X$) in a relation to the comparability fo two topologies (in our case $\tau_1,\tau_2$)</p>

答: The identity map being continuous means that every open set in $\tau_2$ is also open in $\tau_1$. This can also be stated as "Topology $\tau_2$ is weaker than topology $\tau_1$". See http://en.wikipedia.org/wiki/Weaker_topology .  </p>


问:
How I prove absolute convergence of this infinite product?
Why does the 1/n term needs to be 0 in order to have absolute convergence for the product? $$P=\prod_{n=1}^{\infty }u_n=\prod \frac{(n-a_1)(n-a_2)...(n-a_n)}{(n-b_1)(n-b_2)...(n-b_n)}$$
$$u_n=1-\frac{a_1+a_2+...a_n-b_1-b_2-...b_n}{n}+O(n^{-2})$$

I also tried to solve it but I get $P=\coprod_{i=1}^{k}\frac{\Gamma (1-b_i)}{\Gamma (1-a_i)}e^\frac{b_i}{a_i} $
Could you show me the calculations in order to get the right answer?
Whitch is $P=\coprod_{i=1}^{k}\frac{\Gamma (1-b_i)}{\Gamma (1-a_i)}$ and help calculate with this product: $pi /2=\prod_{n=1}^{\infty }\frac{4n^2}{4n^2-1}$

答:
Tha infinite product $\prod (1+\alpha_n)$ converges absolutely (definition) iff
the series $\sum \alpha_n$ converges absolutely.

In this case, the rational function $u_n$ satisfies
$$
u_n = 1 + \frac{A}{n} + O\left(\frac{1}{n^2}\right)\qquad\text{as }n\to\infty
$$
for some constant $A$.
Now $u_n = 1+\alpha_n$.  If $A \ne 0$, then $|\alpha_n| \sim \frac{|A|}{n}$,
so $\sum |\alpha_n|$ diverges by comparison with the harmonic series $\sum \frac{1}{n}$.  So $\sum \alpha_n$ does not converge absolutely.
So $\prod u_n$ does not converge absolutely.  So it remains only
to compute $A$ in this case:
$$
A = a_1 + a_2 + \dots+ a_k - b_1 - b_2 -\dots - b_k .
$$
If this is not zero, then the product does not converge absolutely.


问:
prooving the harmonic inequality
My real analysis textbook shows this result but gives no proof for it, it would be great if someone could show me why the following result holds.

Show

$$\sum_{n=1}^{\infty} \frac{H_n}{n(n+1)} = \sum_{n=1}^{\infty}
 \frac{1}{n^{2}}$$

where $H_n = \sum_{k=1}^{n} 1/k$.

Title: prooving the harmonic inequality

Body: Using $\sum_{n\ge1}\sum_{k=1}^na_{nk}=\sum_{k=1}^n\sum_{n\ge k}a_{nk}$, the sum is$$\sum_{n\ge1}\left(\frac1n-\frac{1}{n+1}\right)\sum_{k=1}^n\frac1k=\sum_{k\ge1}\frac1k\sum_{n\ge k}\left(\frac1n-\frac{1}{n+1}\right)=\sum_{k\ge1}\frac1k\frac1k.$$


问:
Integral of meromorphic function.
Let $f(z)$ be meromorphic function in $\Omega\subset\mathbb{C}$ with a pole $z=z_0$. Then, is it true that there is a number $N$ such that
$$\sum_{n=1}^{\infty} \int_C f(z)(z-z_0)^{n-1}dz=\sum_{n=1}^{N}\int_C f(z)(z-z_0)^{n-1}dz$$
where $C$ is closed contour containing $z_0$. This integral and summation are principal part of coefficient of Laurent series. I guess, for any meromorphic function $f$, $$f(z)=(z-z_0)^N g(z)$$
for some $N\in \mathbb{N}$ and holomorphic function $g(z)$ in $\Omega$. But, I am failing it.

答:
No try with $z_0=0$, $f(z)= \frac1{z(z-1)}$ and $C$ the circle of radius $2$. Then for all $n\ge 2$, $$ \int_C f(z)z^{n-1}dz= 2i\pi $$
If you meant that $0$ is the only pole of $f$ and $\Omega$ is simply connected then yes sure, for $n$ large enough $f(z)z^{n-1}$ is analytic on $\Omega$ and its integral on a closed-contour contained in $\Omega$  is $0$.

Title: Integral of meromorphic function.


问:
Expected position of 1D random walk
This is problem taken from Chap. 03 from Larry Wasserman's All of Statistics:

A particle starts at the origin of the real line and moves along the line in jumps of one unit. For each jump the probability is $p$ that the particle will jump one unit to the left and the probability is $1-p$ that the particle will jump one unit to the right. Let $X_n$ be the position of the particle after $n$ units. Find $\mathbb{E}[X_n]$.

So simply applying the definition of expectation, after 1 jump we have:

$\mathbb{E}[X_1] = -1 \cdot (1-p) + 1 \cdot p = 2p-1$

At this point my intuition tells me that the expected position after $n$ jumps will be $\mathbb{E}[X_n] = n(2p-1)$, however if I take the expectation after 2 jumps I have:

$\mathbb{E}[X_2] = -2 \cdot (1-p)^2 -1 \cdot (1-p) + 1 \cdot p + 2 \cdot p^2 = -2(1 -2p + p^2) -1 + p + p + 2p^2 = -2 + 4p - 2p^2 -1 + 2p + 2p^2 = 6p - 3 = 3(2p - 1)$

which would mean $\mathbb{E}[X_n] = (n+1)(2p-1)$. So I'm not sure which is correct? I can't spot something problematic in my algebra so is it a case of naive intuition or a case naive extrapolation to further jumps?

答: Revisit your second equation (In particular the second and third terms -1 (1-p) +1 p ) and you can figure it out.</p>

Title: Expected position of 1D random walk


问:
Why aren't these two matrix multiplications equal?
I'm not sure why this isn't working as I expected. If $W$ is a diagonal matrix

$$
B = WA \\
C = W^{\frac{1}{2}}AW^{\frac{1}{2}}
$$

If W is diagonal, I cannot see a reason why $B \ne C$. I did it on paper and it came out to be equal, and then I tried it in numpy to test it and it came out not equal. I must be missing something simple. Does $B = C$?

答:
In general, no. Multiplying $A$ on the left by a diagonal matrix rescales the rows of $A$, whereas multiplying $A$ on the right by a diagonal matrix rescales the columns of $A$.

However, $B$ and $C$ are equal if $W$ is a multiple of the identity matrix.


问:
Find the probability that $m^2-n^2$ is divisible by $4$
Two integers are $m$ , $n$ chosen at random with replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$ .

my solution : Two integers can be chosen (with replacement ) in $9\times 9=81$ ways .

$m^2-n^2 $ is divisible iff either both $m,n$ are  odd or both are even .
There are $4\times 4=16 $ ways way of obtaining both even integers and $5\times 5=25 $ ways of obtaining both odd integer.

So total number of favorable outcome $=16+25=41$ .

Hence required probability = $\frac{41}{81}$ .

Now i would like to slightly change the question :

Two integers are $m$ , $n$ chosen at random without replacement from the set of integers $1,2,3..9$ . Find the probability that $m^2-n^2$ is divisible by $4$

my solution when order consideration is relevant  :

Total number of obtaining two integers $= 2\times 9\times 8=2\times 72$

Numbers ways of obtaining two even numbers $=2\times 4\times 3 $ and number of ways obtaining odd numbers $=2\times 5\times 4$ .

So total number of favorable cases= $2(4\times 3+5 \times 4)=2\times 32 $ .
Hence required probability $=\frac{2 \times 32 }{2 \times 72 }=\frac{32}{72}$

my solution when order is considered irrelevant

Total numbers of cases $=9\times 8=72$

Numbers of ways obtaining both even and both odd integers is $4\times3=12$ and $5\times4=20$ respectably .

So number of favorable cases$=12+20=32$ .

Hence probability =$\frac{32}{72}$

Is my understanding in above problems correct ?
Please point out if i calculated anything wrong?

Thank you

答:
We will use the following three things:

1)
$(x^2-y^2) = (x-y)(x+y)$

2) Odd + Odd = Even

Odd - Odd = Even

Even + Even = Even

Even - Even = Even

Odd + Even = Even + Odd = Odd

Odd - Even = Even - Odd = Odd

3)
Even $\cdot$ Even = Multiple of $4$

Odd $\cdot$ Odd = Odd $\not = $ Multiple of $4$

Hence if $x,y $ both Odd or both  Even we have $(x^2-y^2) = (x-y)(x+y) =$ Even $\cdot$ Even = Multiple of $4$

And if one is Even and the other is Odd we have $(x^2-y^2) = (x-y)(x+y) =$ Odd $\cdot$ Odd $\not =$ Multiple of $4$

Hence for $(x^2-y^2) $ to  be divisible by 4 we need them to be "same" , with respect to divisibility by two. Think of this as there being $9$ cards. $4$ Red and $5$ Black. What is the probability that if, with replacement, we pick the same colour twice? Clearly this is $\frac{5}{9}\cdot \frac{5}{9} + \frac{4}{9}\cdot\frac{4}{9} = \frac{41}{81} \approx 0.5 $


问:
Normal Line to a surface
What is the equation of the normal line to the surface r(u,v)=(cosu, sinv+sinu ,cosv) at the point (1,1,0)?

What I have done so far:

I have found the partial derivatives of the surface separately with respect to u and respect to v. I know I need to multiply the two to get the i,j,k coordinates. Should I sub in the point given before doing this or after? should I change to polar coordinates? how do I sub these values in to cos and sin so that it is defined?

答:
Some quick corrections and answers:

I have found the partial derivatives of the surface separately with respect to u and respect to v. I know I need to multiply the two to get the i,j,k coordinates.

There is no operation that is referred to as "multiplying vectors". Presumably you mean that you compute the cross-product $\mathbf r_u \times \mathbf r_v$.

Should I sub in the point given before doing this or after?

It is easiest to substitute in the correct values for $u,v$ before taking the cross-product

Should I change to polar coordinates? How do I sub these values in to cos and sin so that it is defined?

I have no clue what you mean by either of these questions.

Here's a complete answer. We are given $\mathbf r(u,v) = (\cos u, \sin v + \sin u, \cos v)$. Presumably, we are also given a domain, i.e. the values of $u$ and $v$ that we consider. I suspect that this was given to you as $u,v \in [0,2\pi)$.

Begin by computing the partial derivatives
$$
\mathbf r_u(u,v) = (-\sin u, \cos u, 0), \quad \mathbf r_v(u,v) = (0,\cos v, -\sin v).
$$
Now, we want to find these vectors at the point $(1,1,0)$ by plugging in the values of $u$ and $v$ associated with this point. This means that we need to find these values. That is, we need to find values of $u,v$ between $0$ and $2\pi$ for which
$$
(\cos u, \sin v + \sin u, \cos v) = (0,1,1).
$$
Verify that this only occurs when $u = \pi/2$ and $v = 0$. We compute
$$
\mathbf r_u(\pi/2,0) = (-1,0,0), \quad \mathbf r_v(\pi/2,0) = (0,0,1).
$$
To find a normal vector to the surface, we take the cross product of these vectors to get
$$
(-1,0,0) \times (0,0,1) = (0,1,0).
$$
To answer the question, give an equation for the line through $(0,1,1)$ in the direction of $(0,1,0)$. For instance, we can write
$$
\ell(t) = (0,1,1) + (0,t,0).
$$

Title: Normal Line to a surface


问:
How to solve the recurrence relation $a_{1}=2, a_{n}=\frac{a_{n-1}+2}{2 a_{n-1}+1}(n \geq 2)$ with generating functions?
There's already a way to solve it, called "fixed point method", that is, from the relation we define its characteristic equation as $x=\dfrac{x+2}{2x+1}$，then we have $x_1=1,x_2=-1$. So the following relation established:
$$
\frac{a_{n}-1}{a_{n}+1}=\frac{\frac{a_{n-1}+2}{2 a_{n-1}+1}-1}{\frac{a_{n-1}+2}{2 a_{n-1}+1}+1}=-\frac{1}{3} \cdot \frac{a_{n-1}-1}{a_{n-1}+1}
$$
It is obvious that $\displaystyle \frac{a_{n}-1}{a_{n}+1}=\frac{1}{3} \cdot\left(-\frac{1}{3}\right)^{n-1}$, and then we have $a_{n}=\dfrac{3^{n}-(-1)^{n}}{3^{n}+(-1)^{n}}$.

My question is, how to solve this kind of recurrence relations with generating functions? Also, "fixed points" can be applied to solving recurrences like $a_{n+1}=\dfrac{a_{n}^{2}+b}{2 a_{n}+d}$, which seems impossible to solve using generating functions.

答:
Let $a_0:=0$, and let
$$f(x):=\sum_{n=0}^\infty a_nx^n$$
be the generating function of the sequence $(a_n)_{n\geq0}$. Since $\lim_{n\to\infty} a_n=1$ this $f$ is even a bona fide analytic function in the unit disc. We have
$$a_n\bigl(1+(-1/3)^n\bigr)=(1-(-1/3)^n\bigr)\qquad(n\geq0)$$
and therefore
$$a_n\bigl(x^n+(-x/3)^n\bigr)=\bigl(x^n-(-x/3)^n\bigr)\qquad(n\geq0)\ .$$
This implies
$$f(x)+f\left(-{x\over3}\right)={1\over1-x}-{1\over1+{x\over3}}={4x\over(1-x)(3+x)}\qquad\bigl(|x|<1\bigr)$$
and shows that your generating function fulfills a certain functional equation.

I don't know what to make with this.


答:
If the identity $(X,\tau_1)\to(X,\tau_2)$ is continuous, then the preimage of each open set in the codomain, that is a set in $\tau_2$, is an open set in the domain, i.e. an element of $\tau_1$. So $\tau_2$ is coarser than $\tau_1$.

If you know how to characterize continuity locally, then this means that for every $x\in X$ and every neighborhood of $x$ w.r.t. $\tau_2$ there is a neighborhood of $x$ w.r.t. $\tau_1$ contained in the given neighborhood, which means that the local basis is finer.

Title: Relation between continuity and connection between topologies

Body: If $X$ is a set and $\tau_1,\tau_2$ two topologies on $X$. What does it mean to put the continuity of the identic map on $X$ (i.e $id_X(x)=x\forall x\in X$) in a relation to the comparability fo two topologies (in our case $\tau_1,\tau_2$)</p>


问:
How many configurations are possible for this unique circle-arrow arrangement?
8 unique points - A,B,C,D,E,F,G,H are arranged symmetrically on a circle. You are given 8 arrows. You have to connect these 8 points with these arrows in such a way that:

You can connect a point only to its adjacent points (A->B connection is possible, but A->C connection is not possible, unless A and C are adjacent points).

There should be at least AND at most 1 arrow directed outwards from all points (Directed outwards means that tail of the arrow has to be on the point and tip of the arrow on its adjacent point).

How many such configuration are possible? (You have to use all 8 arrows in a configuration)

Here is one such configuration:

enter image description here

答:
Count the number of ways to position the eight points on the circle.

For each point, there are two choices for the arrow directing outward from that point.

Account for over-counting. If you rotate a configuration, does it count as a different configuration? If you flip a configuration so that the arrangement of the points is in reverse order, but the arrows still start and end at the same points, does it count as a different configuration? [This will depend on the question-asker, it is not particularly clear to me from the question statement as-is.]


问:
Relatively Simple But I Don't Know What To Do.
Can anyone help me solve
$$
 2\begin{pmatrix}
   \frac{1}{h}I^{n\times n}  & 0^{n\times n}
    \\
    0^{n\times n}   & I^{n\times n}
    \end{pmatrix} \ddot{\sigma}(t)=\begin{pmatrix}
    0^{n\times n} & \frac{1}{h}I^{n\times n}
    \\
    -\frac{1}{h}I^{n\times n}   & 0^{n\times n}
    \end{pmatrix}\dot{\sigma}(t)
$$

with initial conditions $\sigma(0)=\begin{pmatrix} q \\ p \end{pmatrix},$ $\sigma(1)=\begin{pmatrix} q' \\ p' \end{pmatrix}$. Here $h>0$, I really don't know how to go about it. With $p,q,q',p'\in\mathbb{R}^n$.

答:
Let
$$
 A:=2\begin{pmatrix}
   \frac{1}{h}I^{n\times n}  & 0^{n\times n}
    \\
    0^{n\times n}   & I^{n\times n}
    \end{pmatrix},\quad B:=\begin{pmatrix}
    0^{n\times n} & \frac{1}{h}I^{n\times n}
    \\
    -\frac{1}{h}I^{n\times n}   & 0^{n\times n}
    \end{pmatrix}
$$

Integrate both sides over $t$, you get

$$
A\dot{\sigma}(t)=B\sigma(t)+C
$$

for some $C\in\mathbb{R}^{2n}$.

Thus

$$
\dot{\sigma}(t)=A^{-1}B\sigma(t)+C'
$$

for some other $C'\in\mathbb{R}^{2n}$.

Now the usual solution for linear systems kicks in.


问:
Understanding equality sign in context of big-O-notation
I am not quite sure if I have understood correctly the equality sign when using big-O-notation. Let's consider the following example of a sequence:

$\left(a_n\right)_{n\in\mathbb{N}}:= 5+f(n)$, where $f(n)$ is some function of $n$.

Now, we assume that we can prove that $f(n)$ has the same asymptotic behavior like $\frac{1}{n}$ and hence $f(n)=O\left(\frac{1}{n}\right)$ (or some would say $f(n)\in O\left(\frac{1}{n}\right)$). In many proofs this makes the authors write:
$$
 \lim\limits_{n\to\infty}5+f(n)=\lim\limits_{n\to\infty}5+O\left(\frac{1}{n}\right)=5.
$$
However, our tutor said that if we write this and then argue that the limit is $5$ he will deduct points. (note that I am 2nd year undergraduate student) So I wonder if the following reasoning is what justifies this "equality":

As $f(n)\in O\left(\frac{1}{n}\right)$ we know that there exists a constant $C$ and an index $n_0$ such that for all $n>n_0$: $$|f(n)|\leq C\Big|\frac{1}{n}\Big|\implies 5-C\frac{1}{n}\leq f(n)\leq 5+C\frac{1}{n}~\underset{\text{lemma}}{\underset{\text{squeeze}}{\implies}}~ \lim\limits_{n\to\infty}5+f(n)=5.$$

So whenever we make use of big-O-notation we should bear in mind that the $O(\cdot)$ sign represents a function which has certain properties. Exploiting those properties allows us to derive the limit (e.g. $f\in O\left(\frac{1}{n}\right)$) or sometimes these properties doesn't help us at all when deriving a limit (e.g. $f\in O(1)$).

Is this correct? Or is my understanding wrong?

答: I can't read minds, but the tutor's concern may be that $O(1/n)$ is a set of functions rather than an arbitrary function in that set (unless we achieve the latter with an abuse of notation), so $\lim_{n\to\infty}(5+O(1/n))$ is undefined even though $\lim_{n\to\infty}(5+f)$ has the same value for all $f\in O(1/n)$.


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
These numberphile videos explain the irregularity quite well:

Problem with french numbers

In english 1-12 have individual names and after that we have the -teens (4 and 10, 5 and 10, etc.) similar break point happens at the 17th number in the French number system. The major problem which occurs is at 70s. *The French have no word for 70, they call is 60 and 10 - soixante-dix.*So on, 71 is 60 and 11 till 79.

Similarly, 80 is also not invented, it is 4 20s, quatre-vingt. 81 is 4 20s 1, quatre-vingt-un. 90 is, 4 20s 10, quatre-vingt-dix. 91 is 4 20s 11, quatre-vignt-onze.

Finally, they have a word for 100, ceut.

It looks similar to roman numbers Roman number system, where you repeat the letters to combine the values. X is 10 XX is 20 and so on.

Another peculiar thing in french grading system, 19 out of 20

Title: Why do the French count so strangely?


问:
What is a condition for two real functions $f,g$ to "commute", so $f(g(x))=g(f(x))$?
Say I'm given two functions $f,g$. Can I tell if they "commute" without actually trying them in the formula
$f(g(x))=g(f(x))$?

And given a function $f$, is there a way to find all functions $g$ such that
$f(g(x))=g(f(x))$?

I've tried using derivatives and the chain rule but haven't got anything interesting yet.

答:
If $f(x) = g^{-1}(x)$, then $f(g(x)) = g(f(x))$ but the converse is not true.

If $f(x) = g(x)$, then $f(g(x)) = g(f(x))$.


问:
Linear transformation that finds symmetry of a line segment?
Im trying to find 2x2 matrices that when you multiply the line segment by the matrix you get a symmetry of the initial line segment. I know that the identity matrix will give you the identity symmetry, but I know there are 3 other symmetries in R2 so I am not sure what else those are.</p>

答:
I know there are 3 other symmetries in R2 so I am not sure what else those are

Reflection by bisecting perpendicular line. （aka perpendicular bisector?)

Reflection by the line

Rotation of 180 degrees about the mid-point.

Btw, per explanation of the symmetry-group tag, what you are seeking is not standard definition of symmetry groups. Though we can easily see the answer without formal definitions, if you want more details, you should define what you mean by symmetry groups.


问:
Reason of notation for quotient group
Let's consider a group $G$ and a subgroup $H$.

We note the quotient group $G/H$

My question is: why this notation and the notion of "quotient" ?

Is the motivation only because from Lagrange theorem for finite groups, where we have $|G/H|=|G|/|H|$ or there are deeper reasons ?

Maybe I am lacking some intuition to see further motivations.

Note: I am by far not an expert of group theory, I only know the basics so I expect an answer following the same logic =)

答:
The notation working out nicely for Lagrange's theorem is one reason. There's a couple others:

$\mathbb{Z}/n\mathbb{Z}$ is our prototypical case for quotient groups, and here it really is very closely related to division; the equivalence class of $x$ mod $n$ is the (equivalence class of the) remainder of $x$ divided by $n$.

The notation also works out very nicely for the third isomorphism theorem (which, if you haven't learned it yet, you will shortly). Namely, if $G$ is a group, and $K < H < G$, with everything normal in everything else, then

$$\frac{G}{H} \simeq \frac{G/K}{H/K}$$

That is, we can "cancel" subgroups in a quotient group just like we can cancel factors in an actual fraction.

Title: Reason of notation for quotient group


问:
$R^2$ is not isometric to $R^3$
Is there a direct proof for showing that $R^2$ is not isometric to $R^3$ (with the usual metrics)? I know that they are not homeomorohic but I think there should be some direct and easy proof for showing that they are not isometric.</p>

答: Suppose there exists an isometry $T:\mathbb R^3 \to\mathbb R^2$. Then $$T(\mathbf x)=\binom{\mathbf u}{\mathbf v}\mathbf x+\mathbf b\,,\text{ where } \,\mathbf{u,v,x}\in\mathbb R^3,\mathbf u\cdot\mathbf v=0,\text{ and }\, \mathbf b\in \mathbb R^2.$$Now notice that $|T(\mathbf u \times \mathbf v)-T(\mathbf 0)|=0\neq|(\mathbf u\times \mathbf v)-\mathbf 0|$, so $T$ doesn't preserve the distance between $(\bf u\,\times\,\bf v) $ and the origin. This a contradiction.

Title: $R^2$ is not isometric to $R^3$


答:
To supplement the other answer: you will end up with a general solution of the form
$$
\sigma(t) = e^{At}D + C
$$
for undetermined vectors $C,D \in \Bbb R^{2n}$. Plugging in initial conditions yields
$$
\sigma(0) = e^{A\cdot 0}D + C = D + C, \\
\sigma(1) = e^{A \cdot 1}D + 1 \cdot C  = e^A D + C.
$$
Solve this to get $D = (e^A - I)^{-1}(\sigma(1) - \sigma(0))$ and $C = \sigma(0) - D$.

Note that $e^A - I$ is invertible in most cases.

Consider the case where $e^A - I$ fails to be invertible. This means that $A$ has an eigenvalue $\lambda$ such that $e^\lambda = 1$. Since the eigenvalues of $A$ are $\pm \frac 1{2 \sqrt{h}} i$, this occurs if and only if there is an integer $k$ for which
$$
\frac 1{2 \sqrt{h}} = 2 \pi k \implies h = \frac 1{(4\pi k)^2}.
$$
In this case, we find that
$$
e^{tA} = \cos(2 \pi k t) I + 2 \sqrt{h}\sin(2 \pi k t)A.
$$
This means that any solution $\sigma(t) = e^{At}D + C$ is periodic with period 1.  It follows that the initial conditions of $\sigma(0)$ and $\sigma(1)$ are not sufficient to find $\sigma(t)$, and there can only be such a solution if $\sigma(0) = \sigma(1)$.

Title: Relatively Simple But I Don't Know What To Do.

Body:
Can anyone help me solve
$$
 2\begin{pmatrix}
   \frac{1}{h}I^{n\times n}  & 0^{n\times n}
    \\
    0^{n\times n}   & I^{n\times n}
    \end{pmatrix} \ddot{\sigma}(t)=\begin{pmatrix}
    0^{n\times n} & \frac{1}{h}I^{n\times n}
    \\
    -\frac{1}{h}I^{n\times n}   & 0^{n\times n}
    \end{pmatrix}\dot{\sigma}(t)
$$

with initial conditions $\sigma(0)=\begin{pmatrix} q \\ p \end{pmatrix},$ $\sigma(1)=\begin{pmatrix} q' \\ p' \end{pmatrix}$. Here $h>0$, I really don't know how to go about it. With $p,q,q',p'\in\mathbb{R}^n$.


问:
Showing that compatible germs are the image of a section.
I'm currently self-studying Ravi Vakil's Rising Sea. I have been stuck on exercise 2.4.C, which ask one to prove that any compatible germs is the image of a section. The following definition etc are all taken from the notes. If I have understood it correctly we have the map $$I : \mathscr{F}(U) \rightarrow \prod_{p \in U} \mathscr{F}_p$$ given by $ s \mapsto (\bar{s})_p$, i.e mapping a section to its germ in the respective points. Now germs $(s_p)$ are defined to be compatible if there exists a covering $\{U_i\}$ of $U$ and sections $f_i \in U_i$ such that the germ of $f_i$ for all $p \in U_i$ is $s_p$. I think that one is supposed to glue together the sections given in the definition of compatible germs, though I haven't been able to show that the given sections' restrictions are equal on the overlaps/intersections. Many thanks for any help or hint.

答:
Let $(s_p)_{p \in U} \in \prod_{p \in U} \mathscr F_p$ be a compatible germ. By definition, there exists a cover $\{U_i\}_{i \in I}$ for $U$, and elements $f_i \in \mathscr F(U_i)$ such that for all $q \in U_i$ we have $s_q = [f_i, U_i] \in \mathscr F_q$.

If $q \in U_i \cap U_j$, then $[f_i, U_i]=s_q=[f_j, U_j]$ in $\mathscr F_q$, hence $$[f_i|_{U_i \cap U_j}, U_i \cap U_j] = [f_j|_{U_i \cap U_j}, U_i \cap U_j] \in \mathscr F_q.$$

Since $q$ was arbitrary, this holds on all of $U_i \cap U_j$. Since $\mathscr F$ is a sheaf, the map $\mathscr F(U_i \cap U_j) \to \prod_{p \in U_i \cap U_j} \mathscr F_p$ is injective and thus we have $f_i|_{U_i \cap U_j}=f_j|_{U_i \cap U_j}$.

Since $\mathscr F$ is a sheaf, there exists a unique $f \in \mathscr F(U)$ such that $f|_{U_i} = f_i$. Therefore, the map $\mathscr F(U) \to \prod_{p \in U} \mathscr F_p$ takes $$f \mapsto ([f, U])_{p \in U} = ([f_i, U_i])_{p \in U} = (s_p)_{p \in U}.$$


问:
Bayes rules with more than two variables
In my lectures I saw this equation.

$$
p(j=1, x|\theta) = p(x|j=1, \theta)p(j=1)
$$

I tried to apply Bayes rule on the left hand side, but I couldn't get the right hand side of the equation.

答:
The right equation is
$$
p(j=1, x|\theta) = p(x|j=1, \theta){\color{red}{p(j=1|\theta)}}
$$
since $\theta$ is considered a global information in this equation and must hold anywhere. A simple proof follows by using the definition of conditional probability with the events $x|\theta$ and $j=1|\theta$.


问:
Combinatorics with dices
Let's say that we play dice 30 times.

How many ways can I play the dice 30 times and the sum of all throws is 105.
I started by subtracting 30 to 105, since at least the result in every throw is '1'  and then I use the Bars and Stars Method to distribute the remaining 75 points in 30 boxes and I get (104,75). Now I need to subtract the cases whenever those boxes get more points than 5.

I am supposed to use the Principle Of Inclusion and Exclusion somewhere, but I don't know how to use it. I am in my 1st year at college, so I would really appreciate if you could tell me how to think when faced with this kind of problems.

答:
The way to face this problems is as follows. You count all possibilities and then subtract the ones you do not like. As you have suggested. The problem is how you subtract them. So the way to think about this is assume the $i-$th dice  gives you more than $6$ so you can create that property. First abstract your problem to be the following
$$A=\{(d_1,\cdots ,d_{30}):\sum _{i=1}^{30}d_i=105, \, 1\leq d_i\leq 6\},$$
so you apply your very smart observation by doing $c_i=d_i-1$ to get a set
$$B=\{(c_1,\cdots ,c_{30}):\sum _{i=1}^{30}c_i=105-30=75, \, 0\leq c_i\leq 5\}.$$
that assignment is a bijection, so $|A|=|B|.$

Now consider $B_j=\{(c_1,\cdots , c_{30}):\sum _{i=1}^{30}c_i=75,\, c_j>5\},$
because this is exactly the properties that you do not want, you want to count all possibilities (without any single restriction) and then take out the sequences that lie inside any of the $B_j.$ So you want
$$\left |\text{All}\setminus \bigcup _{i=1}^{30}B_i\right |=\left |\text{All}\right |- \left | \bigcup _{i=1}^{30}B_i\right |.$$
By using stars and bars you can compute All.

Now you focus in what happens with the $B_j'$s Notice that using your idea, you can set $e_j = c_j-6$ because you know that in $B_j$ $c_j>5$ so $e_j\geq 0,$ so again, using stars and bars you can get your result.

Then you try to do it for two sets say $n\neq m$ you want to compute
$$\left | B_n\bigcap B_m \right |,$$ notice that it is analogous to the last one but here you are doing $e_n = c_n-6$ and $e_m=c_m-6.$ If you notice, this will not depend on $n$ or $m,$ but just in the fact that you choosed two. This is called homogeneous, so whenever you are doing the PIE, you will just have to consider $\binom{30}{2}|B_1\cap B_2|=\sum _{1\leq i<j\leq 30}|B_i\cap B_j|.$

Can you get then the formula?

Title: Combinatorics with dices


问:
System of equations involving 4 variables

If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$

I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.

答:
Define $A_i$ as follows
\begin{eqnarray*}
a+b=A_0 \\
ax+by=A_1 \\
ax^2+by^2=A_2 \\
ax^3+by^3=A_3 \\
ax^4+by^4=A_4 .
\end{eqnarray*}
Multiply the first & third equations & subtract the square of the second equation
\begin{eqnarray*}
A_0 A_2 -A_1^2 = ab(x-y)^2.
\end{eqnarray*}
Multiply the first & fourth equations & subtract the product of the second & third equations
\begin{eqnarray*}
A_0 A_3 -A_1 A_2 = ab(x+y)(x-y)^2 \\
(x+y) = \frac{A_0 A_3 -A_1 A_2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now multiply the first & fifth equations & subtract the square of the third equation
\begin{eqnarray*}
A_0 A_4 - A_2^2 = ab(x^2-y^2)^2 = ab(x+y)^2(x-y)^2 = \frac{(A_0 A_3 -A_1 A_2)^2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now plug in your values of $A_0,A_1,A_2,A_3$ and solve for $A_4$.


问:
Conjecture: $a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$
Hi it's a conjecture inspired by a work of Vasile Cirtoaje :

Claim :

Let $0.5\geq a \geq b \geq c\geq 0$ such that $a+b+c=1$ and $n\geq 1$ a natural number then we have :

$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$$

To prove it I have tried Bernoulli's inequality as we have :

$$a^{(2(1-a))^{\frac{1}{n}}}\leq 1-(1-a)(2(1-a))^{\frac{1}{n}}$$

But it's not enough and wrong.

I have tested for $n=1,2,3,4,5$.Moreover the equality case is $a=b=0.5$.

Case $n=1$ :

As in my previous posts we have the inequality $x\in[0,0.5]$ :

$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$ applying this for each variable we want to show :

$$2^{2a+1}a^2(1-a)+2^{2b+1}b^2(1-b)+2^{2(1-(a+b))+1}(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Now by Bernoulli's inequality we have:

$$2^{2x+1}\leq 2(1+2x)$$

Remains to show :

$$2(1+2a)a^2(1-a)+2(1+2b)b^2(1-b)+2(1+2(1-(a+b)))(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Wich is wrong because we have an equality case at $b=c=0.25$ and $a=0.5$!

My question :

Have you a counter-example or later (if it's true) a proof ?

Thanks in advance !

Body:
I have found a counterexample:
$$a=\frac{2}{5},\ b=\frac{3}{10},\ c=\frac{3}{10},\ n=1.$$
This gives:
$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n} = \left(\frac{2}{5}\right)^{\frac{6}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\frac{3}{10}\approx 1.0037 >1.$$


问:
Example of filtration
Could someone please provide an example of such filtration:

答:
The sigma-algebras $\mathfrak I_t$ make up the filtration in that link.  To get an example, you should take an example of a one-to-one ergodic measure-preserving transformation $\mathbb U$, and a "starting" sigma-algebra $\mathfrak I_0$, and then plug them into the formula.

We say that an event $\Lambda$ is "known" at a future time $t$ iff $\mathbb U^{-t} (\Lambda)$ is known at time $0$.

An example.

It seems we are working with two-sided sequences.  Let $\Omega = \{0,1\}^{\mathbb Z}$ be the space of two-sided sequences of $0,1$.  $\mathfrak G$ is the product sigma-algebra (equivalently, the Borel sets for the product topology).  Let $\pi_t$ be the projection onto the $t$ coordinate.  That is, if $\omega \in \Omega$ is
$$
\omega = (\dots,\omega_{-2},\omega_{-1},\omega_0;\omega_1,\omega_2,\cdots),
\tag1$$
then $\pi_t(\omega) = \omega_t$.  I put a semicolon in $(1)$ to mark where the $0$ coordinate is.  We think of $\omega_0$ as what is happening now, $\omega_{-1}$ is what happend yesterday, $\omega_{-2}$ what happened the day before yesterday, and so on.  And $\omega_1$ us what will happen tomorrow, $\omega_2$ the day after tomorrow, and so on.  The events $\Lambda \in \mathfrak G$ that we know now make up the sigma-algebra
$$
\mathfrak I_0 = \sigma\{\pi_t\;:\;t \le 0\}
\tag2$$
So saying that $\Lambda \in \mathfrak I_0$ means that it depends
only on what happens today, what happened yesterday, and so on backward.  These are the events we know now.  (Assuming we have kept perfect records of the past, but have no crystal ball to fortell the future.)

Let $\mathbb P : \mathfrak G \to [0,1]$ be the "product" measure on $\Omega$, where each factor has the $(1/2,1/2)$ measure.  In probabilistic language, the random variables $\pi_t$ are i.i.d Bernoulli trials.
The transformation $\mathbb U: \Omega \to \Omega$ moves everything one step into the past.
$$
\mathbb U\big(
(\dots,\omega_{-2},\omega_{-1},\omega_0;\omega_1,\omega_2,\cdots)
\big) =
(\dots,\omega_{-1},\omega_0,\omega_1;\omega_2,\omega_3,\cdots).
$$
Equivalently, $\pi_{t}(\mathbb U(\omega)) = \pi_{t-1}(\omega)$.
(If there is an event that the Tomorrow People think will happen three days in their future, we think it will happen four days in our future.)
I leave it to you to check that it is an ergodic measure-preserving transformation.  So
$$
\mathfrak I_1 = \{\Lambda \in \mathfrak G \;:\;
\mathbb U^{-1}\Lambda \in \mathfrak I_0\} =
\sigma\{\pi_t\;:\;t \le 1\}
\tag3$$
That is, saying $\Lambda \in \mathfrak I_1$ means that we will know tomorrow whether $\Lambda$ occurs or not.  Similarly,
$$
\mathfrak I_s = \sigma\{\pi_t\;:\;t \le s\}
$$
and saying $\Lambda \in \mathfrak I_s$ means that we know on day $s$ whether $\Lambda$ occurs or not.  (Here $s \in \mathbb Z$. It could be negative.)

Title: Example of filtration


问:
When will there be more people alive than dead? (Solving an exponential function)
According to this source, there are 108 billion people who have ever been born. So, way more dead (101 billion) than alive (7 billion). In how many years will that no longer be true?

For the sake of simplicity (and this is drastically simplifying it), let's assume the world started with 1 person in "year 0" and keep the birth rate fixed throughout all time at 2%. So, we can represent the current population $P$ in year $n$ as $r^n$ (where r=1.02, for 2% growth), and the total population of all prior years as $\sum_{i=0}^{n-1} r^i$. We're looking for $n>0$ such that:

$$r^n \geq \sum_{i=0}^{n-1} r^i$$
$$\implies r^n \geq \frac{r^n - 1}{r-1} \text{(using the geometric series formula)}$$
$$\implies r^{n+1} - 2r^n + 1 \geq 0$$

How do we solve this equation for $n$?

答:
Exponential growth is scale-independent. It doesn't matter where you are on the curve.

If $n$ people are alive at any given moment and the population exactly doubles every generation, then there are the same number dead as alive; this is the convergent series

$n/2 + n/4 + n/8 +...$

If growth is slower then there are more dead, and if growth is faster then there are more alive.

In the real world it is more complicated because you have to take account of changes in both growth rates and  the proportion of people who die childless. Various guesstimates have been made over the years, and they differ wildly. Forty years ago people thought there were more alive than dead, now we think it far the other way. In truth, nobody has enough historical data to make accurate statements about the past.

Technology has brought us one further principle: the past is no longer any guide to the future.


问:
If $T$ is a maximal $L$ theory such that every finite subset of $T$ is consistent, then either $\sigma \in T$ or $\lnot \sigma \in T$.
Prove without using compactness that if $T$ is a maximal $L$ theory such that every finite subset of $T$ is consistent, then either $\sigma \in T$ or $\lnot \sigma \in T$. Here, by maximal, we mean maximal with respect to this property.

Attempt:

Let $T$ be an $L$-theory with the given property and fix an $L$ sentence $\sigma$. Suppose both $\sigma, \lnot \sigma \in T$. Then $S = \{\sigma, \lnot \sigma \} \subseteq T$ is a finite subset of $T$, so $S$ is consistent. Thus there exists a model of $S$, say $\mathcal{S}$ such that $\mathcal{S}$ satisfies both $\sigma$ and $\lnot \sigma$. This is impossible, so we conclude that for any $L$ sentence $\sigma$, at most one of $\sigma, \lnot \sigma \in T$.

Suppose $\sigma, \lnot \sigma$ are both not in $T$. Then consider $T_1 = T \cup \{\sigma\}$ and $T_2 = T \cup  \{\lnot \sigma\}$. $T$ is contained in $T_1$ and $T_2$ and it is maximal, so it must be that $T_1, T_2$ each contain finite subsets which are not consistent. These subsets must contain $\sigma, \lnot \sigma$ respectively because otherwise they would be finite subsets of $T$, which are consistent by assumption. Thus we can write them as $T' \cup \{\sigma\}, T'' \cup \{\lnot \sigma\}$ resp.

I am not sure how to proceed from here. Any hints would be great!

答:
For the second part, we will prove that $T' \cup T''$ is a finite inconsistent subset of $T$. This contradicts $T$ having only finite consistent subsets, so we will be done.

$T′∪T′′∪\{σ\}$ is inconsistent because it is a superset of $T′∪\{σ\}$ and $T′∪T′′∪\{¬σ\}$ is inconsistent because it is a superset of $T′′∪\{¬σ\}$. Thus $T′∪T′′$ must be inconsistent, otherwise if a model existed, then it would satisfy either $σ$ or $¬σ$, so one of $T′∪T′′∪\{σ\}, T′∪T′′∪\{¬σ\}$ would be consistent.

Note : Supersets of inconsistent theories are inconsistent. If a model exists for $T$, a theory which contains $Q$, then this model satisfies all the $L$ sentences in $T$, so it also satisfies all the $L$ sentences in $Q$. Thus it must be a valid model for $Q$ as well, so $Q$ must be consistent too.


问:
The solution to the equation $ - \Delta u = (\lambda - \log u)u$
$\Omega$ is a bounded domain in $\mathbb R^n$ with smooth boundary. Consider the Dirichlet problem:$$-\Delta u = (\lambda  - \log u)u ~~~{\rm on}~~~  \Omega~~~ {\rm and}~~~u=0 ~~~{\rm on}~~\partial {\Omega}$$ where the solution $u\in {C^\infty }(\bar \Omega )$ and $u>0$ on $\Omega$ and $\lambda$ is a positive constant.

My question is: what conditions do we need for the existence and uniqueness of the solution $u$
?

答:
Define $$
f(t) = \left\{ \begin{array}{rl}
 -t\log{t} &\mbox{ if $t\in[0,1]$} \\
  0 &\mbox{ otherwise}
       \end{array} \right.
$$

Note that $f$ is continuous. COnsider the problem

$$\tag{1}
 \left\{ \begin{array}{rl}
 -\Delta u-\lambda u=f(u) &\mbox{ in $\Omega$} \\
  u=0 &\mbox{ in $\partial\Omega$}
       \end{array} \right.
$$

1 - Existence

I - There is some methods avaliable to find a solution to $(1)$. For example, we can use the fact that $f$ is continuous with compact support, to prove that the energy functional $I:H_0^1\to\mathbb{R}$ associated with $(1)$ is (under some additional restriction) weakly lower semi continuous and coercive. Remember that if $F(t)=\int_0^t f(s)ds$, then $$Iu=\frac{1}{2}\int_\Omega |\nabla u|^2-\frac{\lambda}{2}\int_\Omega u^2-\int_\Omega F(u)$$

The additional restriction necessary here is $\lambda<\lambda_1$ (to guarantee coercivity).

II - (This is just a guess - I dont have verified the calculations): Also, you can prove existence, by noting that if $\lambda$ is not a eigenvalue of $-\Delta$, then the operator $-\Delta-\lambda I$ is invertible, and you can use arguments similar to this one.

Remark 1: The original function $t\log{t}$ is not monotone, so I dont know if this problem has uniqueness.

Remark 2: The function $f$ being continuous, we can conclude that $u\in C^1(\overline{\Omega})$.

Now, let's find conditions such that the solution $u$ of $(1)$ satisfies $00$, you can conclude by the maximum principle that $u>0$. On the other hand, by regularity theory, you have that the solution $u$ of $(1)$ (See Breziz chapter 9) satisfies $(1)$ almost everywhere, i.e. $u$ is a Strong solution. If we suppose that the set where $u>1$ is non0empty, then we would have $-\Delta u-\lambda u=0$ almost everywhere in this set. If $\lambda$ is not a eingevalue, then this is a absurd, which implies that $u\geq 1$.

From the last paragraph, we can conclude that in fact $u$ is a solution to the original problem and satisfies $0<u\leq 1$. Hence, by using the first method proposed, we can conclude that if $\lambda<\lambda_1$, then you problem has a solution $u\in C^{2,\alpha}(\overline{\Omega})$ satisfying $0<u\leq 1$. I dont know if we can get more regularity, because the functions $-t\log{t}$ is only Holder continuous. A argument of bootstrap can be applied in the interior of $\Omega$, but in the boudary I dont know, because $-t\log{t}$ is not even Lipschitz for $t=0$.

Body:
Define $$
f(t) = \left\{ \begin{array}{rl}
 -t\log{t} &\mbox{ if $t\in[0,1]$} \\
  0 &\mbox{ otherwise}
       \end{array} \right.
$$

Note that $f$ is continuous. COnsider the problem

$$\tag{1}
 \left\{ \begin{array}{rl}
 -\Delta u-\lambda u=f(u) &\mbox{ in $\Omega$} \\
  u=0 &\mbox{ in $\partial\Omega$}
       \end{array} \right.
$$

1 - Existence

I - There is some methods avaliable to find a solution to $(1)$. For example, we can use the fact that $f$ is continuous with compact support, to prove that the energy functional $I:H_0^1\to\mathbb{R}$ associated with $(1)$ is (under some additional restriction) weakly lower semi continuous and coercive. Remember that if $F(t)=\int_0^t f(s)ds$, then $$Iu=\frac{1}{2}\int_\Omega |\nabla u|^2-\frac{\lambda}{2}\int_\Omega u^2-\int_\Omega F(u)$$

The additional restriction necessary here is $\lambda<\lambda_1$ (to guarantee coercivity).

II - (This is just a guess - I dont have verified the calculations): Also, you can prove existence, by noting that if $\lambda$ is not a eigenvalue of $-\Delta$, then the operator $-\Delta-\lambda I$ is invertible, and you can use arguments similar to this one.

Remark 1: The original function $t\log{t}$ is not monotone, so I dont know if this problem has uniqueness.

Remark 2: The function $f$ being continuous, we can conclude that $u\in C^1(\overline{\Omega})$.

Now, let's find conditions such that the solution $u$ of $(1)$ satisfies $0<u\leq 1$. Because $f\geq 0$ and there is a open set where $f>0$, you can conclude by the maximum principle that $u>0$. On the other hand, by regularity theory, you have that the solution $u$ of $(1)$ (See Breziz chapter 9) satisfies $(1)$ almost everywhere, i.e. $u$ is a Strong solution. If we suppose that the set where $u>1$ is non0empty, then we would have $-\Delta u-\lambda u=0$ almost everywhere in this set. If $\lambda$ is not a eingevalue, then this is a absurd, which implies that $u\geq 1$.

From the last paragraph, we can conclude that in fact $u$ is a solution to the original problem and satisfies $0<u\leq 1$. Hence, by using the first method proposed, we can conclude that if $\lambda<\lambda_1$, then you problem has a solution $u\in C^{2,\alpha}(\overline{\Omega})$ satisfying $0<u\leq 1$. I dont know if we can get more regularity, because the functions $-t\log{t}$ is only Holder continuous. A argument of bootstrap can be applied in the interior of $\Omega$, but in the boudary I dont know, because $-t\log{t}$ is not even Lipschitz for $t=0$.