问:
Best way to calculate number of Young tableaus for a 2 × n grid
My question is:

Let $f(n)$ be the number of Young tableaus for a $2 × n$ grid. As shown below, $f(2) = 2$ and $f(3) = 5$. What is $f(5)$? What can you say about $f(n)$?

Initially, I thought that I'd go and write up the possible combinations for $f(4)$ and $f(5)$ to see what I can say for $f(n)$, but as I'm writing it up, I'm finding it very inefficient to do so. What would be a better way to find $f(4)$ and $f(5)$ other than grinding out all the possible cases?

Any guidance is greatly appreciated.

答:
HINT: These are the Catalan numbers. Suppose that you have a $2\times n$ tableau filled with the integers $1,\ldots,2n$. From it you can construct a string $\pi=p_1p_2\ldots p_{2n}$ of $2n$ parentheses as follows: for $k=1,\ldots,2n$ let $p_k$ be a left parenthesis if $k$ is in the top row, and let $p_k$ be a right parenthesis if $k$ is in the bottom row. Show that the tableau is a Young tableau (i.e., that both rows and each column are increasing) if and only if the string $\pi$ is a valid balanced string of parentheses.

If you prefer to think of the Catalan numbers in terms of paths on the integer grid from $\langle 0,0\rangle$ to $\langle n,n\rangle$ composed of right-steps and up-steps and never rising above the diagonal $y=x$, think think of $p_k$ as a right-step if $k$ is in the top row and an up-step if $k$ is in the bottom row.


问:
Understanding this rational expression without solving
My textbook asks to explain in five words or less, what this rational expression does to each number of x.

$(1/x+1/x^2+1/x^3) \div (1/x^4+1/x^5+1/x^6) $

Im not sure how to go about understanding what the rational expression does to each number of x without performing all the operations.

The answer is "it cubes x"

答:
Hint:$$\frac1{x^4} + \frac1{x^5} + \frac1{x^6} = \frac1{x^3}\left(\frac1x+\frac1{x^2}+\frac1{x^3}\right)$$

Do you get that now?

ContentLicense: CC BY-SA 4.0


问:
Xn binomial limit to infinity
Let $X_n$ be random variables with

$X_n \sim \mathcal{B}(n, 2/n)$

Find $\lim_{n \to\infty }(P(X_n=2))$

I don't know where to start, any help is appreciated.

答:
The hint in the other answer, should be enough. Notice that if $X_{n}\sim \mathcal{B}(n, 2/n)$ then $$P(X_n=2)=\binom{n}{2}\frac{2^2}{n^2}\left(1-\frac{2}{n}\right)^{n-2}$$ So you need to calculate $$\lim_{n\to\infty} \frac{n!}{2!(n-2)!}\frac{4}{n^2} \left(1-\frac{2}{n}\right)^{n-2}=\frac{2}{e^2}$$

This limit actually is not that hard, since $$\frac{n!}{2!(n-2)!}\frac{4}{n^2}=\frac{n-1}{n}\frac{n}{n}\cdot 2\to 2,\ \ \mathrm{as}\ n\to\infty $$

and $$\left(1-\frac{2}{n}\right)^{n-2}=\left(1-\frac{2}{n}\right)^{n}\left(1-\frac{2}{n}\right)^{-2}\to e^{-2}\cdot 1\ \ \mathrm{as}\ \ n\to\infty$$

You can check: https://en.wikipedia.org/wiki/Binomial_distribution and for the limit https://www.researchgate.net/profile/Alvaro_Salas3/publication/265777624_The_exponential_function_as_a_limit/links/55f79c4b08aec948c471542e/The-exponential-function-as-a-limit.pdf

Title: Xn binomial limit to infinity


答:
Hint:

Since $n\cdot \frac 2n = 2$, $X_n$ converges in distribution to the Poisson distribution with $\lambda = 2$.

Hence, $\lim_{n\to\infty}P(X_n=2)= \frac{2^2e^{-\lambda}}{2!} = \frac 2{e^2}$.

Title: Xn binomial limit to infinity

Body:
Let $X_n$ be random variables with

$X_n \sim \mathcal{B}(n, 2/n)$

Find $\lim_{n \to\infty }(P(X_n=2))$

I don't know where to start, any help is appreciated.


问:
Confusion regarding second part of fundamental theorem of calculus
When $A$ is the area function,we define it by $A(x)=\int_{a}^{x}f(t)dt$ and from the first fundamental theorem of calculus we get $A'(x)=f(x)$.Now putting $x=a$,we have $A(a)=0$ since the region has area $0$?But then $A'(a)=0$ which is not equal to $f(a)$ since $f(a)$ doesn't necessarily have to be $0$,isn't this a contradiction?

答:
To see where you're thinking wrongly, apply the same logic to the function $A(x)=x$.  (This would be $f(t)=1$ and $a=0$ if you wish to stay within your area-function scenario, but this doesn't really matter.)  Here, $A(0)=0$.  Is the derivative $A'(0)=0$?  No, as $A'(x)=1$ for all $x$.

You are confusing being $0$ (or being constant) at one point with being $0$ (or constant) everywhere.  We take derivatives of functions at a point, not derivatives of points at a point.  As the derivative $A'(a)$ is defined by the limit
$$
A'(a) = \lim_{h \to 0} \frac{A(a+h)-A(a)}{h}
$$
it necessarily must take into account the values of $A(x)$ in a small interval around $a$ (see the numerator), so what $A(x)$ does at exactly $a$ and nowhere else is not the entire story.


问:
Financial math- Calculating Perpetuity problem

A investor is hesitating between two projects. The first will yield steady
returns of $X$ every $6$ months for the first $10$ years and $X$ every year
after. The second will return $500$ per month for $5$ years, then will
yield $500$ per year in perpetuity. If the effective annual rate is $10$%,
for what $X$ would the investor be indifferent between the two projects?

I am having trouble starting off the problem.

So far I understand that the cash flows for the first project is $X$ for every $6$ months (semi-annual), for the first $10$ years that means $2(10) = 20$ payments for compounding semi-annually, and $X$ every year going on to infinity. This has "semi-annual" interest so $\frac{0.1}{2} = 0.05$

The second project has a cash flow of $500$ per month for $5$ years so it will be $5(12)=60$ months for compoundings and then there is the $500$ per year in perpetuity (goes to infinity). This project has "monthly" interest rate of $\frac{0.1}{12} = 0.833333$

As for timeline(s), I'm guessing you need two? The first project would have a line going from $0$ to $10$ years with 20 intervals of X, and then infinity years afterward. While the 2nd project would have a line from $0$ to $5$ with subintervals of 12 months for sixty months of 1000 each, then after 5 years is 1000 per year to infinity.

答:
$$\require{enclose}$$
Write out the cash flows in terms of the effective monthly rate of interest, which we will call $j$ for now and calculate its value later.  The first flow is
$$PV = X(v^6 + v^{12} + v^{18} + \cdots + v^{120} + v^{132} + v^{144} + v^{156} + \cdots),$$ where $v = 1/(1+j)$ is the monthly present value discount factor.  The second flow is $$PV = 500(v + v^2 + \cdots + v^{60} + v^{72} + v^{84} + v^{96} + \cdots).$$  It is worth noting that both perpetuities contain the payments $$v^{12} + v^{24} + \cdots + v^{60} + v^{72} + \cdots = a_{\enclose{actuarial}{\infty} i}.$$  So we can instead the first flow as
$$PV = X\left((v^6 + v^{18} + \cdots + v^{114}) + a_{\enclose{actuarial}{\infty} i}\right) = X\left((1+j)^6 a_{\enclose{actuarial}{10} i} + a_{\enclose{actuarial}{\infty} i}\right).$$  The second flow can be written as
$$\begin{align}
PV &= 500\left((v + v^2 + \cdots + v^{11}) + v^{12}(v + v^2 + \cdots + v^{11}) + \cdots v^{48}(v + v^2 + \cdots + v^{11}) + a_{\enclose{actuarial}{\infty} i}\right) \\
&= 500\left((1+v^{12} + v^{24} + v^{36} + v^{48})a_{\enclose{actuarial}{11}j} + a_{\enclose{actuarial}{\infty}i} \right) \\
&= 500\left( (1+i)a_{\enclose{actuarial}{5}i}a_{\enclose{actuarial}{11}j} + a_{\enclose{actuarial}{\infty}i} \right).
\end{align}$$
Now we have $$(1+j)^{12} = 1+i,$$ where $i = 0.10$, and employing the usual annuity formulas and equating the present values of these flows, we get $$X \approx 500 \frac{(1.10)(3.79079)(10.4914) + (0.10)^{-1}}{(1.04881)(6.14457) + (0.10)^{-1}} \approx 1634.22.$$

This is not the only possible approach.  We can work in terms of $i$, in which case the cash flows look like this:
$$PV = X(v^{1/2} + v + v^{3/2} + \cdots + v^{10} + v^{11} + v^{12} + \cdots) = X\left((1+i)^{1/2} a_{\enclose{actuarial}{10}i} + a_{\enclose{actuarial}{\infty}i}\right),$$ and $$\begin{align}
PV &= 500(v^{1/12} + v^{2/12} + \cdots + v + v^{13/12} + \cdots + v^{59/12} + v^5 + v^6 + v^7 + \cdots ) \\
&= 500\left( (v^{1/12} + v^{2/12} + \cdots + v^{11/12})(1 + v + v^2 + v^3 + v^4) + a_{\enclose{actuarial}{\infty}i} \right) \\
&= 500\left( (1+i)a_{\enclose{actuarial}{5}i} a_{\enclose{actuarial}{11}j} + a_{\enclose{actuarial}{\infty}i}\right).\end{align}$$  We still get the same result, but the intermediate steps look a little different.  In the first way, $v = 1/(1+j)$ and the exponents on $v$ represent months, whereas in the second, $v = 1/(1+i)$ and the exponents on $v$ represent years.

Yet another way to perform the computation is to compute the present values as the sum of a annuity-immediate with finite term, and a deferred perpetuity-immediate whose deferral period equals the term of the annuity component; e.g.,
$$PV = X\left(a_{\enclose{actuarial}{20}j} + v_i^{10} a_{\enclose{actuarial}{\infty} i}\right),$$ where here $j = (1+i)^{1/2} - 1$ is the effective semiannual or $6$-month rate of interest, and $v_i = 1/(1+i)$ is the effective annual present value discount factor.  Correspondingly, the second flow is $$PV = 500\left(a_{\enclose{actuarial}{60}k} + v_i^5 a_{\enclose{actuarial}{\infty} i}\right)$$ where $k = (1+i)^{1/12} - 1$ is the effective monthly rate of interest.  This approach may be preferable to you since the annuity calculations are simpler, but it uses three rates (monthly, semiannually, and annually) instead of two.


问:
Integrand is sharply peaked
I have a gamma function,
$$n!=\int_0^\infty e^{-x}x^n \mathrm{d}x$$ and if I take the derivative of the logarithm of the integrand, I get $$\dfrac{\mathrm{d}}{\mathrm{dx}}(n\ln x -x)=\dfrac{n}{x}-1.$$  But why I can argument that the integrand is sharply peaked with the contribution important only near $x=n$?

答:
$e^{-x} x^n/\Gamma(n+1)$ for $x > 0$ is the probability density function of the Gamma distribution with scale parameter $1$ and shape parameter $n+1$.  When $n$ is a positive integer, this is the distribution of the sum $X$ of $n+1$ independent random variables
$X_n$, each with exponential distribution with parameter $1$.  The Central Limit Theorem
lets you approximate this with a normal distribution of mean $n+1$ and standard deviation $\sqrt{n+1}$.  This is "strongly peaked" because $\sqrt{n+1}$ is small compared to $n+1$ when $n$ is large.  The Weak Law of Large Numbers says that for any $\epsilon > 0$,
$$ \lim_{n \to \infty} \mathbb P(1-\epsilon < X/(n+1) < 1+\epsilon) = 1$$

Title: Integrand is sharply peaked


问:
Theorem of Catalan - minimal ruled surfaces
I'm trying to prove the Theorem of Catalan of minimal ruled surfaces, which states that the only minimal ruled surfaces are planes and helicoids.

According to Wolfgang Kühnel's Differential Geometry exercise 12, ch. 3, one should first verify that the Gaussian curvature $K$ and the mean curvature $H$ of a ruled surface

$$ f(t,s) = c(t) + sX(t) $$

given by standard parameters (i.e. such that $\lVert X \rVert = \lVert X' \rVert = 1$ and $\langle c', X' \rangle = 0$), are given by

$$ K = \frac{-\lambda}{(\lambda ^2 + s^2)^2} \quad \quad H = -\frac{1}{2 (\lambda^2 + s^2)^\frac{3}{2}} (Js^2 + \lambda' s + \lambda(\lambda J + F))$$

where

$$ F := \langle c', X \rangle $$
$$ \lambda := \langle c' \times X, X' \rangle = \det (c', X, X') $$
$$ J := \langle X'', X \times X' \rangle = \det (X, X', X'') $$

each of which is a function only of $t$.

With this proved, one should be able to determine all minimal ruled surfaces "with ease".

I have proved the expressions for $K$ and $H$ after somewhat lengthy calculations, but I don't know how to use them in order to prove that the plane and the helicoid are the only minimal ruled surfaces. I know that for a surface to be minimal, $H = 0$.

答: $H = 0$ tells you that $Js^2 + \lambda' s + \lambda(\lambda J + F) \equiv 0$. Since this is a quadratic polynomial in $s$ whose coefficients (which depend only on $t$) must vanish, we get $J = 0, \lambda' = 0$, which means $\lambda$ is constant, and $J = \lambda F = 0$. Therefore $X''$ is a linear combination of $X$ and $X'$, but since $\langle  X'', X' \rangle = 0$ (because $||X'|| \equiv 1$) and $\langle X'', X \rangle \equiv - 1$ (because $$||X|| \equiv 1 \implies \langle X, X' \rangle = 0\implies \langle X'', X \rangle = -||X'||^2 = - 1$$), it follows that $X'' = -X$, i.e, $X$ defines a circle. If $\lambda = 0$ then we get $K = H = 0$ and the surface must be a plane. If $\lambda \neq 0$ then we must have $F = 0$, which implies $c' = \lambda X \wedge X'$ and differentiating we get $X'' = 0$, therefore $X$ must define a line perpendicular to the circle defined by $X$. Without loss of generality we can suppose that $c$ and $X$ are given by $c(t) = (0, 0, ct)$ for some $c \neq 0$ and $X(t) = (\cos(t), \sin(t), 0)$. It's easy to see that the correspending parametrization defines a helicoid, which proves Catalan's theorem.

ContentLicense: CC BY-SA 3.0


问:
Representation of $\operatorname{SL}_4$ on $\bigwedge^2 \mathbb{C}^4$
Letting $V = \mathbb{C}^4$ with basis $\{e_1, \dots, e_4\}$, the representation $\rho$ of $G = \operatorname{SL}_4(\mathbb{C})$ on $\bigwedge^4 V$ is given by $g \cdot (v_1 \wedge v_2) = gv_1 \wedge gv_2$.
Set $\Omega = e_1 \wedge e_2 \wedge e_3 \wedge e_4$, and let $B$ be a bilinear form on $\bigwedge^2 V$ defined by $u \wedge v = B(u,v)\Omega$.
I can show that this is a non-degenerate symmetric bilinear form.

I now need to show that $$ \rho: \operatorname{SL}_4(\mathbb{C}) \to \operatorname{SO}\!\big(\!\textstyle\bigwedge^2 V, B\big)$$ is a group homomorphism.
However, I am not even sure that the codomain is correct - why does $\rho(g)$ preserve $B$?
Specifically, I should show that $B(\rho(g) a, \rho(g)b) = B(a,b)$ for all $g \in G$ and $a, b \in \bigwedge^2 V$.

If $g \cdot (a \wedge b ) = ga \wedge gb = a \wedge b$, then I will be done, but I don't think this can be true for all such $a, b \in \bigwedge^2 V$.
Otherwise I have tried decomposing the exterior products, but cannot see where to get equality of scalars.

答:
Setting $\eta=a\wedge b$ and $\nu=c\wedge d$ then

$$ g(\eta\wedge\nu)=(g\eta)\wedge(g\nu)=g(a\wedge b)\wedge g(c\wedge d)=(ga)\wedge(gb)\wedge(gc)\wedge(gd)=g(a\wedge b\wedge c\wedge d). $$

Exercise. Check that $g\xi=(\det g)\xi$ for all $\xi\in\Lambda^4\mathbb{C}^4$ and $g\in \mathrm{GL}_4\mathbb{C}$.

Note all $\xi$s are scalar multiples of the orientation element $\omega=e_1\wedge e_2\wedge e_3\wedge e_4$, so really this is the same as showing $g_1\wedge g_2\wedge g_3\wedge g_4=\det(g_1\,g_2\,g_3\,g_4)\omega$ for column vectors $g_i$.


Title: Understanding this rational expression without solving

Body:
My textbook asks to explain in five words or less, what this rational expression does to each number of x.

$(1/x+1/x^2+1/x^3) \div (1/x^4+1/x^5+1/x^6) $

Im not sure how to go about understanding what the rational expression does to each number of x without performing all the operations.

The answer is "it cubes x"

Body:
Let's se what happens if we simplify the expression.

$(1/x+1/x^2+1/x^3) \div (1/x^4+1/x^5+1/x^6)=$

$\frac {1/x+1/x^2+1/x^3}{1/x^4+1/x^5+1/x^6} =$

$\frac {x^6}{x^6}\frac {1/x+1/x^2+1/x^3}{1/x^4+1/x^5+1/x^6}=$

$ \frac {x^5 + x^4 + x^3}{x^2 + x + 1} =$

$\frac {x^3(x^2 + x + 1)}{x^2 + x+ 1} =$

$\require{cancel}$$\frac {x^3\cancel{(x^2 + x + 1)}}{\cancel{x^2 + x+ 1}}=x^3$.

And that's that

"it cubes $x$" is the appropriate answer.

====

Note though what you actually wrote was dead wrong.

$1/x+1/x^2+1/x^3 \div 1/x^4+1/x^5+1/x^6=$

$1/x+1/x^2+\frac {1/x^3}{1/x^4}+1/x^5+1/x^6=$

$1/x+1/x^2+\frac {x^4}{x^3}+1/x^5+1/x^6=$

$1/x+1/x^2+x+1/x^5+1/x^6=$

$\frac {x^5}{x^6} + \frac {x^4}{x^6} + \frac {x^7}{x^6} + \frac x{x^6} + \frac 1{x^6}= $

$\frac {x^5 +x^4 + x^7 + x + 1}{x^6}$

can't be simplified any further.


问:
Polynomial approximation to formal power series matrix
I noticed that starting with a $2 {\times} 2$ matrix $M$ with a handful of polynomial entries in two variables such that $\det M$ is invertible in $\mathbb C [x] [[y]]$ I can add infinitely many terms of higher orders in $y$ to the entries of $M$ to make $\det M$ a constant.

Further, I noticed that with a bit of tinkering it's actually possible to add finitely many terms to $M$ and still achieve that $\det M$ is constant.

I am now wondering if this is true in general:

Let $M$ be a $2 {\times} 2$ matrix with entries in the ring $\mathbb C [x][[y]]$ and constant determinant. Given $k > 0$, does there exist a matrix $N$ with entries in $\mathbb C [x, y]$ such that $M \,\mathrm{mod} \, (y^{k+1}) = N \,\mathrm{mod} \, (y^{k+1})$ and $\det M = \det N$?

(If it helps, I would be content with assuming that one of the entries is a multiple of $y$ — e.g. if the $(1,2)$ entry is a multiple of $y$, then the determinant being constant implies that the $(1,1)$ and $(2,2)$ entries have a constant term and are both of the form $1 + O (y)$.)

I give a simple example to illustrate:

Example. Let
  $$
M =
\begin{pmatrix}
1 - y & 0 \\
0 & 1 + y + y^2 + \dotsb
\end{pmatrix}
$$
  Then $\det M = 1$ (as the $(1,1)$ entry is invertible in $\mathbb C [[y]]$ with inverse $(2,2)$).

Now let
  $$
N =
\begin{pmatrix}
1-y & -y^{k+1} \\
y^{k+1} & 1 + y + \dotsb + y^{2k+1}
\end{pmatrix}
$$
  Then $\det M = \det N = 1$ and
  $$
M \, \mathrm{mod} (y^{k+1}) = N \, \mathrm{mod} (y^{k+1}) =
\begin{pmatrix}
1 - y & 0 \\
0 & 1 + y + \dotsb + y^k
\end{pmatrix}. \qquad \diamond
$$

This example works equally well if you replace $y$ by a polynomial $p = p (x,y)$ which is a multiple of $y$.

I would also be interested to hear about (textbook/paper) examples computing with matrices of formal power series to see what kind of techniques can be used.

Edit. I have discovered a truly tedious proof of this, which this edit section is too narrow to contain.

It is a "proof by brute force" (solving linear equations for each monomial that may appear in the expression of $\det M \mod (y^{n+1})$), and I think writing up the details should best be left as an Exercise To The Reader.

There ought to be a more elegant proof and although I haven't disclosed the full proof, at least I now believe the result to be true, which might be a reason to look for such a proof.

答:
Notation: For a ring $A$ and ideal $\mathfrak{a}$, let $\operatorname{SL}_{r}(A,\mathfrak{a})$ denote the kernel of the group homomorphism $\operatorname{SL}_{r}(A) \to \operatorname{SL}_{r}(A/\mathfrak{a})$.

Lemma: Let $r \ge 2$, let $A$ be a ring, let $\mathfrak{a}$ be an ideal of $A$. For any $\ell \ge 1$, the map \begin{align} \varphi_{\ell} : \operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell}) \end{align} is surjective.

Proof: We proceed by induction on $\ell$; for $\ell = 1$, we have $\operatorname{SL}_{r}(A/\mathfrak{a},\mathfrak{a}) = \{\mathsf{id}_{r}\}$. Suppose $\ell \ge 2$ and let $\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell},\mathfrak{a}/\mathfrak{a}^{\ell})$; let $\mathsf{W}_{\ell-1} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell-1},\mathfrak{a}/\mathfrak{a}^{\ell-1})$ be the image of $\mathsf{W}_{\ell}$; by induction on $\ell$, there exists $\mathbf{M}_{\ell-1} \in \operatorname{SL}_{r}(A,\mathfrak{a})$ such that $\varphi_{\ell-1}(\mathbf{M}_{\ell-1}) = \mathsf{W}_{\ell-1}$; after replacing $\mathsf{W}_{\ell}$ by $(\varphi_{\ell}(\mathbf{M}_{\ell-1}))^{-1} \cdot \mathsf{W}_{\ell}$, we may assume that $\mathsf{W}_{\ell} \in \operatorname{SL}_{r}(A/\mathfrak{a}^{\ell} , \mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$. Write $\mathsf{W}_{\ell} = \mathsf{id}_{r} + \mathsf{F}$ with $\mathsf{F} = (\mathsf{F}_{i,j}) \in \operatorname{Mat}_{r \times r}(\mathfrak{a}^{\ell-1}/\mathfrak{a}^{\ell})$. Suppose $\mathsf{F}_{1,1} \ne 0$. Choose a lift $\mathbf{f}_{1,1} \in \mathfrak{a}^{\ell-1}$ of $\mathsf{F}_{1,1}$; we may replace $\mathsf{W}_{\ell}$ by $\varphi_{\ell}(\mathbf{E}_{1}) \cdot \mathsf{W}_{\ell}$ where $\mathbf{E}_{1} \in \operatorname{SL}_{r}(A)$ is the matrix which differs from the identity $\mathrm{id}_{r}$ by \begin{align} \begin{bmatrix} 1 - \mathbf{f}_{11} & \mathbf{f}_{11} \\ -\mathbf{f}_{11} & 1+\mathbf{f}_{11} \end{bmatrix} \end{align} in the upperleft $2 \times 2$ submatrix. This ensures that $\mathsf{F}_{1,1} = 0$. If $\mathsf{F}_{1,1} = 0$, then we may add multiples of the first row (i.e. multiply on the left by elementary matrices) to ensure that the first column is equal to the 1st standard basis vector. In this way we may assume that (for $i = 1,\dotsc,r-1$) the $i$th column of $\mathsf{W}_{\ell}$ is equal to the $i$th standard basis vector; for $\mathsf{F}_{r,r}$, the condition $\det \mathsf{W}_{\ell} = 1$ implies that necessarily $\mathsf{F}_{r,r} = 0$ as well; then as before we can make the $r$th column the $r$th standard basis vector using elementary row operations.

Remark: IIRC I got the idea for the above from Cohn's example $\begin{bmatrix} 1+xy & x^{2} \\ -y^{2} & 1-xy \end{bmatrix}$ which is a matrix that cannot be expressed as a product of elementary matrices with coefficients in $\mathbb{Q}[x,y]$.

Corollary: Let $r \ge 2$, let $A$ be a ring, let $\mathfrak{a}$ be an ideal of $A$, let $A^{\wedge}$ be the $\mathfrak{a}$-adic completion of $A$. Then the group homomorphism $\operatorname{SL}_{r}(A,\mathfrak{a}) \to \operatorname{SL}_{r}(A^{\wedge},\mathfrak{a}A^{\wedge})$ is dense for the $\mathfrak{a}$-adic topology.


问:
How do I find the probability of getting an average when selecting from a normal distribution?
If I have that scores follow a normal distribution, I know that the probability of selecting any person at random and them getting a score between 1 and 2 sd's above the mean is 13.6%. If I randomly select 4 people what is the probability that their scores will average to a value between 1 and 2 sd's above the mean? I dont care where any of their individual scores are only that they average to a value between 1 and 2 sds above the mean.</p>

答:
Let's call $\bar X =\frac 14(X_1 + \cdots + X_4)$ with $X_i\sim N(\mu,\sigma^2)$.

Under the given conditions we have

$$\bar X \sim N\left(\mu,\frac{\sigma^2}{4}\right)$$

Hence,

$$P(\sigma \leq \bar X - \mu \leq 2\sigma ) = P\left(2\leq \frac{\bar X - \mu}{\frac{\sigma}{2}}\leq 4\right)=\Phi(4)-\Phi(2)$$

where $\Phi$ is the cumulative distribution function of the standard normal distribution.


问:
A probability density transformation
I run into a problem that is surprising to me.

Suppose three r.v. $Y, U, V$ satisfy $Y = U + V$, $U$ is independent of $V$.

We compute the following conditional density of $V$ given $Y$:

$ f_{V | Y}(v|y) = f_{V|U}(v|y-v) = f_V(v)$.

It's surprising to me because the conditional density $V|Y$ equals the marginal density of $V$.

Is my understanding correct? Thanks!

答:
$$f_{Y,V}(y,v)=f_U(y-v)f_V(v)$$which indicates that the numerators of your first equality are equal but the denominators are not equal since $f_Y(y)\ne f_U(y-v)$ as $f_Y(y)$ does not have $v$. Take for example $U,V\sim B(n,p),Y\sim B(2n,p)$, then $f_Y(y)=\binom{2n}y p^y(1-p)^{n-y}\ne f_U(y-v)$.

Your second equality is true and follows from the independence of $U,V$.

Title: A probability density transformation


问:
How to get the angle between the inner common tangents to two general ellipse
I have been trying to find the angle made by inner tangents to two general ellipse by following the method described by Futurologist to find four homogeneous equations of tangents, this method is presented here and uses the concept of dual conics and degenerate conics. Using that method, I am able to get four tangent equations in homogeneous coordinates. However, I am looking to find the angle between the inner common tangents. For that purpose, currently, I am solving for the intersection of the four tangent lines with the ellipse and then tracing those points back to get the angle made by inner common tangents. This process is a bit computationally expensive and I am looking for a less expensive algorithm.

My question here is that if it is possible to obtain that angle directly instead of going through the hassle of finding the intersection points? I am not interested in getting the equation of tangents. I just need the angle between them.

Graphically in the diagram below, that angle can be represented as $\theta$:

Note: Using the slopes of the lines obtained from above method do not return correct answer because we do not know if that angle is greater one or smaller one  (two angle made by exterior or interior to those line), it depends on the relative position of the two conics. I want a generalize solution which will work for every configuration of two given conics anywhere in the 2D space.

答:
Preliminaries (algorithm for solving a pair of quadratic equations of two variables):

It boils down to finding an algorithm for solving a system of two quadratic equations of two variables, by interpreting it as a projective geometry pencil of conics, then finding the three degenerate conics of the pencil together with a projective transformation that simplifies these three degenerate conics to the point of your being able to read off the solutions very easily in new simplifying coordinate system, and after that transforming them back to the original coordinate system.

I sketched an algorithm in python, I think it seems to work on your example... but I was in a hurry and did not check it properly, so there may be bugs...

```<code>import numpy as np
import math

# technical module, functions, details
def homogenize(x):
    return np.array([x[0], x[1], 1])

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]],
                     [cs_sn[1],  cs_sn[0]]])

# defining the isometry (the rotation plus translation) transformation
# between the coordinate system aligned with the conic and a general (world)
# coordinate system
def isom_inverse(angle, translation):
    '''
    isometry from conic-aligned coordinate system (conic attached)
    to global coordinate system (world system)
    '''
    cos_, sin_ = cos_sin(angle)
    return np.array([[cos_, -sin_, translation[0]],
                     [sin_,  cos_, translation[1]],
                     [   0,     0,             1]])

def isom(angle, translation):
    '''
    isometry from global coordinate system (world system)
    to conic-aligned coordinate system (conic attached)
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]],
                     [ sin_,  cos_, tr[1]],
                     [    0,     0,    1 ]])

# calculating the coinc defined by a pair of axes' lengts,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0],
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))

# calculating the coinc dual to the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def dual_Conic(major, minor, angle, center):
    D_1 = np.array([[major**2,        0,  0],
                    [       0, minor**2,  0],
                    [       0,        0, -1]])
    U_1 =  isom_inverse(angle, center)
    return (U_1).dot(D_1.dot(U_1.T))

# transforming the matrix of a conic into a vector of six coefficients
# of a quadratic equation with two variables
def conic_to_equation(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def equation_to_conic(eq):
    '''
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# given a point (x,y) define the vector (x^2, xy, y^2, x, y, 1)
def argument(x):
    return np.array([x[0]**2, x[0]*x[1], x[1]**2, x[0], x[1], 1])

# given x = (x[0],x[1]) calculate the value of the quadratic equation with
# six coefficients coeff
def quadratic_equation(x, coeff):
    '''
    coeff[0]*x**2 + coeff[1]*x*y + coeff[2]*y**2 + coeff[3]*x + coeff[4]*y + coeff[5] = 0
    '''
    return coeff.dot( argument(x) )

# given a pair of conics, as a pair of symmetric matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the anti-symmetric product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def transform(c1, c2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of the two conics
    '''
    c21 = np.linalg.inv(c2).dot(c1)
    k, U = np.linalg.eig(c21)
    return k, U

# the same as before, but for a pair of equations instead of matrices of conics
def eq_transform(eq1, eq2):
    '''
    eq1 and eq2 = np.array([eq[0], eq[1], eq[2], eq[3], eq[4], eq[5]])
    '''
    C1 = equation_to_conic(eq1)
    C2 = equation_to_conic(eq2)
    return transform(C1, C2)

# realizing the matrix U as a projective transformation
def proj(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_common_points(c1, c2):
    k, U = transform(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = proj(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol

# find the solutions, i.e. the points x=(x[0],x[1]) saisfying the pair
# of quadratic equations
# represented by a pair of vectors eq1 and eq2 of 6 coefficients
def solve_eq(eq1, eq2):
    conic1 = equation_to_conic(eq1)
    conic2 = equation_to_conic(eq2)
    return find_common_points(conic1, conic2)

'''
Esample of finding the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Q1 = Conic(a, b, w, cntr)
dQ1 = dual_Conic(a, b, w, cntr)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Q2 = Conic(a, b, w, cntr)
dQ2 = dual_Conic(a, b, w, cntr)

R = find_common_points(dQ1, dQ2)

print('')
print(R)
print('')
print('checking that the output forms common tangent lines: ')
print('')
print('conic 1: ')
print(np.diagonal(R.dot(dQ1.dot(R.T))) )
print('')
print('conic 2: ')
print(np.diagonal(R.dot(dQ2.dot(R.T))) )
#conic_to_equation(dQ1)
</code>```

Some Explanations: Assume you want to find the intersection points of two conics C1 and C2. Let us assume for simplicity that they are real ellipses that intersect in four different points (to avoid complex numbers)

In the case of finding the common tangents to a pair of conics,
simply convert the two conics two corresponding duals and then find the intersection points
of the duals. These intersection points are the equation coefficients of the tangents of the original conics.

There are possibly several different geometric interpretations of this problem, but let us go with the pencil of conics. The two conics C1 and C2 are represented by 3 by 3 symmetric matrices with non-zero determinants, which I have denoted C1 and C2. The linear combination, called pencil of conics generated by C1 and C2, is the t-parametrized family of conics C1 - t*C2 , where t is just a number. What is crucial is that every conic C1 - tC2 passes through the intersection points of C1 and C2 and these are the only four points they all have in common. You can prove this by observing that if x.T * C1 * x = x.T * C1 * x = 0 then
x.T * (C1 - t*C2) * x = x.T * C1 * x - t * x.T * C2 * x = 0 - t*0 = 0. Furthermore, if you take an intersection point of C1 and C1 - t*C2, then C2 = C1 - t*C2 + s*C2 you can apply the same argument when s = t.

In this family of conics, there are three degenerate conics, that are geometrically three pairs of lines. They occur exactly when t is such that det( C1 - t*C2 ) = 0. This is a polynomial equation of degree 3 with respect to t, so there are three, say different solutions k[0], k[1], k[2], due to the niceness of the C1 and C2. Projectively speaking, each degenerate conic C1 - k[j]*C2 is a pair of lines and they have a common intersection point u[:,j] = [ u[0,j] : u[1,j] : u[2,j] ]. Moreover, rank(C1 - k[j]*C2) = 2, so ker(C1 - k[j]*C2) = 1. This point u[:,j] is characterized as a solution to the equation
(C1 - k[j]*C2) * u[:,j] = 0.
Since C2 is invertible (non-degenerate), multiply both sides of the equation by inverse(C2) and obtain the equivalent equation ( (inverse(C2) * C1) - k[j]*Identity ) * u[:,j] = 0 which is an eigenvalue equation, with k[j] as eigenvalue and u[:,j] as eigenvector. The output of the function transform() is the 1 by 3 array k of eigenvalues and the 3 by 3 matrix U = [ u[:,0], u[:,1], u[:,2] ] of eigenvectors.

Conjugating C1 - k[j]*C2 by U, i.e. (U.T)*(C1 - k[j]*C2)*U, is geometrically equivalent to performing a projective transformation that sends u[:,0] and u[:,1] to infinity and u[:,2] to the origin. Thus, U changes the coordinate system from a general one to a special coordinate system, in which two of the degenerate conics are given by pairs of parallel lines and combined they intersect in a rectangle. The third conic is represeneted by the diagnols of the rectangle.

In this new picture, the intersection problem can be easily solved, just by reading off one solution from the entries of the matrices (U.T)*(C1 - k[j]*C2)*U (the intersection points are the vertices of the rectangle, so if you find one of them, the others are simply mirror symmetric of each other).

Some explanations on how the solutions in special coordinates are obtained.

Since, by construction (by the special choice of eigenvalues k[0] and k[1]) the original conics C1 - k[0]*C2 and C1 - k[1]*C2 are degenerate and thus a pair of lines each, after the projective transformation U, which simplifies the conics and represents them by the matrices L1 = (U.T)*(C1 - k[0]*C2)*U and L1 = (U.T)*(C1 - k[0]*C2)*U, are diagonal and thus geometrically each of these is a pair of lines, parallel one of the coordinate axes, e.g. L1 is a pair of lines parallel to x and L2 is a pair of lines parallel to y. Also for each degenerate conic, the pair of lines are symmetric with respect to the coordinate axes. Hence, the four intersection points form the vertices of a rectangle, mirror symmetric relative to the coordinate axes. Thus, the intersection points are of the projective form

```<code>sol[0, : ] = [ sol[0, 0],  sol[0, 1],  1]
sol[1, : ] = [-sol[0, 0],  sol[0, 1],  1]
sol[2, : ] = [-sol[0, 0], -sol[0, 1],  1]
sol[3, : ] = [ sol[0, 0], -sol[0, 1],  1]
</code>```

so it is enough to find only one of them and then reflect it across the coordinate axes. In the simplifying coordinate system, the conic L1 has the an equation like this

```<code>(a1*y - b1) * (a1*y + b1) = 0
</code>```

where the two lines are a1*y = b1 and a1*y = -b1 so the y coordinate of one of the solutions should be y = abs(b1/a1). If you expand the equation, you get the quadratic equation

```<code>a1^2 * y^2 - b1^2 = 0
</code>```

and in matrix form

```<code>       0     0      0
L1 =   0  L1[1,1]   0
       0     0   L1[2,2]

where   L1[1,1] = a1^2   and   L1[2,2] = b1^2
</code>```

Hence, the y coordinate of one solution is

```<code>y = abs(b1/a1) = sqrt( abs( L1[2, 2] / L1[1, 1] ) )
</code>```

If you do the same for the conic L2, but this time thinking about it being a pair of symmetric lines parallel to the y axis, then you get the x coordinate of a solution

```<code>x = abs(b2/a2) = sqrt( abs( L2[2, 2] / L2[1, 1] ) )
</code>```

That is why, one solution is

```<code>sol[0, :] = [ sqrt(abs( L2[2, 2] / L2[1, 1] )),  sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]
</code>```

Then all solutions are obtained by reflecting in the coordinate axes:

```<code>sol[, :] = [(-1)**i*sqrt(abs( L2[2, 2] / L2[1, 1] )),  (-1)**j*sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]

for i = 0,1 and j=0,1
</code>```

Algorithm for calculating the angle between the internal tangents:

```<code>import numpy as np
import math

'''
####### FUNCTIONS IMPORTANT FOR THE CALCULATION #######
'''

def homogenize(x):
    return np.array([x[0], x[1], 1])

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def conic_from_equation(eq):
    '''
    eq is an np.array vector of the equadratic equation's six coefficients
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    conc is a symmetric matrix of the bilinier form of the quadratic equation
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# obtaining the conic dual to a given conic,
# both represented as symmetric 3x3 matrices
def dual_conic(conic):
    return np.linalg.inv(conic)

# given a pair of conics, as a pair of symmetric 3x3 matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the anti-symmetric product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def trivialization_of(conic1, conic2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of two conics (possibly dual)
    '''
    c21 = np.linalg.inv(conic2).dot(conic1)
    k, U = np.linalg.eig(c21)
    return k, U

# realze the 3x3 matrix U as a projective transformation of projective 2-plane
def projectivity(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_intersection_points(c1, c2):
    k, U = trivialization_of(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = projectivity(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol

# find the three coefficients of all four common tangents to a pair of conics
# make sure that the conics do have four common tangents,
# which is the generic situation
def find_common_tangents(c1, c2):
    dc1 = dual_conic(c1)
    dc2 = dual_conic(c2)
    return find_intersection_points(dc1, dc2)

# if the two conics are ellipses, they do not intersect the line at infinity.
# by polarity, the poles of the line at infinity are
# the (affine and in fact metric) centers of the ellipses and are located
# inside the ellipses each. We use them to detect the inner tangents and
# to orient the vectors normal to these tangents in a way
# so that the angle between these two normal vectors is the vector on the picture
def dual_to_infinity_line(conic):
    p = dual_conic(conic).dot(np.array([0,0,1]))
    return homogenize( p[0:2] / p[2] )

# find the common inner tangents between a pair of conics,
# represented as 3x3 symmetric matrices, with properly oriented normal vectors
def find_common_inner_tangents(c1, c2):
    p1 = dual_to_infinity_line(c1)
    p2 = dual_to_infinity_line(c2)
    tn = find_common_tangents(c1, c2)
    # boolean vector, true if inner tangent false if not
    is_inner_tangent = ( tn.dot(p1)*tn.dot(p2) < 0 )
    tn = tn[ is_inner_tangent, : ]
    # reorient one of the normal vectors so that they give us the proper angle
    if tn[0,:].dot(p1)*tn[1,:].dot(p1) > 0:
        tn[1,:] = - tn[1,:]
    return tn

# calculate the angle between a pair of line, given by their three coefficients
def angle_bn_lines(line1, line2):
    theta = line1[0:2].dot(line2[0:2]) / math.sqrt( line1[0:2].dot(line1[0:2])*line2[0:2].dot(line2[0:2]))
    return math.acos(theta)*180/math.pi

# calculate the proper angle between the inner tangents of two quadratic equations
# that represent a pair of conics, hopefully non-intersecting
def angle_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    return angle_bn_lines(tn[0,:], tn[1,:])

'''
####### END OF FUCTIONS IMPORTANT FOR THE CALCULATION #######
'''

'''
####### auxiliary functions to handle the test scenario:
'''

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]],
                     [cs_sn[1],  cs_sn[0]]])

def isom(angle, translation):
    '''
    isometry from global coordinate system (world system)
    to conic-aligned coordinate system (conic attached)
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]],
                     [ sin_,  cos_, tr[1]],
                     [    0,     0,    1 ]])

# calculating the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0],
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))

# transfromaing the matrix of a conic into a vector of six coeficients
# of a quadratic equation with two variables
def equation_from_conic(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])
'''
####### end of auxiliary functions
'''

'''
Example of finding the angle between the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Eq1 = equation_from_conic(Conic(a, b, w, cntr))
Co1 = conic_from_equation(Eq1)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Eq2 = equation_from_conic(Conic(a, b, w, cntr))
Co2 = conic_from_equation(Eq2)

four_tngs = find_common_tangents(Co1, Co2)

inner_tngs = find_common_inner_tangents(Co1, Co2)

print( angle_bn_inner_tangents_of(Eq1, Eq2) )

</code>```

Calculating the angle bisector line
of the inner tangents.

Here is a routine that, given a pair of lines line1 and line2, represented by vectors with three coefficients, calculates the three coefficients of the angle bisector line that splits the angle in half. It also returns the slope of the tangent line, as the ratio of the y-increase divided by the x-increase.

```<code>def angle_bisector(line1, line2):
    n1 = line1[0:2]
    n2 = line2[0:2]
    n1 = n1 / math.sqrt(n1.dot(n1))
    n2 = n2 / math.sqrt(n2.dot(n2))
    n_bsec = n1 + n2
    p12 = np.cross(line1, line2)
    p12 = p12[0:2] / p12[2]
    c_bsec = n_bsec.dot(p12)
    return np.array([n_bsec[0], n_bsec[1], c_bsec]), np.array([-n_bsec[1], n_bsec[0]])

def angle_and_bsec_vector_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    if abs(np.all(tn.imag)) < 1e-10:
        tn = tn.real
    p1 = dual_to_infinity_line(c1)[0:2]
    p2 = dual_to_infinity_line(c2)[0:2]
    bsec, v_bsec = angle_bisector(tn[0], tn[1])
    angle = angle_bn_C_lines(tn[0,:], tn[1,:])
    if abs(angle.imag) < 1e-10:
        angle = angle.real
    if v_bsec.dot(p2 - p1) < 0:
        v_bsec = - v_bsec
    return angle, v_bsec / np.linalg.norm(v_bsec)
</code>```

Body:
Preliminaries (algorithm for solving a pair of quadratic equations of two variables):

It boils down to finding an algorithm for solving a system of two quadratic equations of two variables, by interpreting it as a projective geometry pencil of conics, then finding the three degenerate conics of the pencil together with a projective transformation that simplifies these three degenerate conics to the point of your being able to read off the solutions very easily in new simplifying coordinate system, and after that transforming them back to the original coordinate system.

I sketched an algorithm in python, I think it seems to work on your example... but I was in a hurry and did not check it properly, so there may be bugs...

import numpy as np
import math

# technical module, functions, details
def homogenize(x):
    return np.array([x[0], x[1], 1])

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]],
                     [cs_sn[1],  cs_sn[0]]])

# defining the isometry (the rotation plus translation) transformation
# between the coordinate system aligned with the conic and a general (world)
# coordinate system
def isom_inverse(angle, translation):
    '''
    isometry from conic-aligned coordinate system (conic attached)
    to global coordinate system (world system)
    '''
    cos_, sin_ = cos_sin(angle)
    return np.array([[cos_, -sin_, translation[0]],
                     [sin_,  cos_, translation[1]],
                     [   0,     0,             1]])

def isom(angle, translation):
    '''
    isometry from global coordinate system (world system)
    to conic-aligned coordinate system (conic attached)
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]],
                     [ sin_,  cos_, tr[1]],
                     [    0,     0,    1 ]])

# calculating the coinc defined by a pair of axes' lengts,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0],
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))

# calculating the coinc dual to the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def dual_Conic(major, minor, angle, center):
    D_1 = np.array([[major**2,        0,  0],
                    [       0, minor**2,  0],
                    [       0,        0, -1]])
    U_1 =  isom_inverse(angle, center)
    return (U_1).dot(D_1.dot(U_1.T))

# transforming the matrix of a conic into a vector of six coefficients
# of a quadratic equation with two variables
def conic_to_equation(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def equation_to_conic(eq):
    '''
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# given a point (x,y) define the vector (x^2, xy, y^2, x, y, 1)
def argument(x):
    return np.array([x[0]**2, x[0]*x[1], x[1]**2, x[0], x[1], 1])

# given x = (x[0],x[1]) calculate the value of the quadratic equation with
# six coefficients coeff
def quadratic_equation(x, coeff):
    '''
    coeff[0]*x**2 + coeff[1]*x*y + coeff[2]*y**2 + coeff[3]*x + coeff[4]*y + coeff[5] = 0
    '''
    return coeff.dot( argument(x) )

# given a pair of conics, as a pair of symmetric matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the anti-symmetric product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def transform(c1, c2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of the two conics
    '''
    c21 = np.linalg.inv(c2).dot(c1)
    k, U = np.linalg.eig(c21)
    return k, U

# the same as before, but for a pair of equations instead of matrices of conics
def eq_transform(eq1, eq2):
    '''
    eq1 and eq2 = np.array([eq[0], eq[1], eq[2], eq[3], eq[4], eq[5]])
    '''
    C1 = equation_to_conic(eq1)
    C2 = equation_to_conic(eq2)
    return transform(C1, C2)

# realizing the matrix U as a projective transformation
def proj(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_common_points(c1, c2):
    k, U = transform(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = proj(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol

# find the solutions, i.e. the points x=(x[0],x[1]) saisfying the pair
# of quadratic equations
# represented by a pair of vectors eq1 and eq2 of 6 coefficients
def solve_eq(eq1, eq2):
    conic1 = equation_to_conic(eq1)
    conic2 = equation_to_conic(eq2)
    return find_common_points(conic1, conic2)

'''
Esample of finding the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Q1 = Conic(a, b, w, cntr)
dQ1 = dual_Conic(a, b, w, cntr)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Q2 = Conic(a, b, w, cntr)
dQ2 = dual_Conic(a, b, w, cntr)

R = find_common_points(dQ1, dQ2)

print('')
print(R)
print('')
print('checking that the output forms common tangent lines: ')
print('')
print('conic 1: ')
print(np.diagonal(R.dot(dQ1.dot(R.T))) )
print('')
print('conic 2: ')
print(np.diagonal(R.dot(dQ2.dot(R.T))) )
#conic_to_equation(dQ1)

Some Explanations: Assume you want to find the intersection points of two conics C1 and C2. Let us assume for simplicity that they are real ellipses that intersect in four different points (to avoid complex numbers)

In the case of finding the common tangents to a pair of conics,
simply convert the two conics two corresponding duals and then find the intersection points
of the duals. These intersection points are the equation coefficients of the tangents of the original conics.

There are possibly several different geometric interpretations of this problem, but let us go with the pencil of conics. The two conics C1 and C2 are represented by 3 by 3 symmetric matrices with non-zero determinants, which I have denoted C1 and C2. The linear combination, called pencil of conics generated by C1 and C2, is the t-parametrized family of conics C1 - t*C2 , where t is just a number. What is crucial is that every conic C1 - tC2 passes through the intersection points of C1 and C2 and these are the only four points they all have in common. You can prove this by observing that if x.T * C1 * x = x.T * C1 * x = 0 then
x.T * (C1 - t*C2) * x = x.T * C1 * x - t * x.T * C2 * x = 0 - t*0 = 0. Furthermore, if you take an intersection point of C1 and C1 - t*C2, then C2 = C1 - t*C2 + s*C2 you can apply the same argument when s = t.

In this family of conics, there are three degenerate conics, that are geometrically three pairs of lines. They occur exactly when t is such that det( C1 - t*C2 ) = 0. This is a polynomial equation of degree 3 with respect to t, so there are three, say different solutions k[0], k[1], k[2], due to the niceness of the C1 and C2. Projectively speaking, each degenerate conic C1 - k[j]*C2 is a pair of lines and they have a common intersection point u[:,j] = [ u[0,j] : u[1,j] : u[2,j] ]. Moreover, rank(C1 - k[j]*C2) = 2, so ker(C1 - k[j]*C2) = 1. This point u[:,j] is characterized as a solution to the equation
(C1 - k[j]*C2) * u[:,j] = 0.
Since C2 is invertible (non-degenerate), multiply both sides of the equation by inverse(C2) and obtain the equivalent equation ( (inverse(C2) * C1) - k[j]*Identity ) * u[:,j] = 0 which is an eigenvalue equation, with k[j] as eigenvalue and u[:,j] as eigenvector. The output of the function transform() is the 1 by 3 array k of eigenvalues and the 3 by 3 matrix U = [ u[:,0], u[:,1], u[:,2] ] of eigenvectors.

Conjugating C1 - k[j]*C2 by U, i.e. (U.T)*(C1 - k[j]*C2)*U, is geometrically equivalent to performing a projective transformation that sends u[:,0] and u[:,1] to infinity and u[:,2] to the origin. Thus, U changes the coordinate system from a general one to a special coordinate system, in which two of the degenerate conics are given by pairs of parallel lines and combined they intersect in a rectangle. The third conic is represeneted by the diagnols of the rectangle.

In this new picture, the intersection problem can be easily solved, just by reading off one solution from the entries of the matrices (U.T)*(C1 - k[j]*C2)*U (the intersection points are the vertices of the rectangle, so if you find one of them, the others are simply mirror symmetric of each other).

Some explanations on how the solutions in special coordinates are obtained.

Since, by construction (by the special choice of eigenvalues k[0] and k[1]) the original conics C1 - k[0]*C2 and C1 - k[1]*C2 are degenerate and thus a pair of lines each, after the projective transformation U, which simplifies the conics and represents them by the matrices L1 = (U.T)*(C1 - k[0]*C2)*U and L1 = (U.T)*(C1 - k[0]*C2)*U, are diagonal and thus geometrically each of these is a pair of lines, parallel one of the coordinate axes, e.g. L1 is a pair of lines parallel to x and L2 is a pair of lines parallel to y. Also for each degenerate conic, the pair of lines are symmetric with respect to the coordinate axes. Hence, the four intersection points form the vertices of a rectangle, mirror symmetric relative to the coordinate axes. Thus, the intersection points are of the projective form

sol[0, : ] = [ sol[0, 0],  sol[0, 1],  1]
sol[1, : ] = [-sol[0, 0],  sol[0, 1],  1]
sol[2, : ] = [-sol[0, 0], -sol[0, 1],  1]
sol[3, : ] = [ sol[0, 0], -sol[0, 1],  1]

so it is enough to find only one of them and then reflect it across the coordinate axes. In the simplifying coordinate system, the conic L1 has the an equation like this

(a1*y - b1) * (a1*y + b1) = 0

where the two lines are a1*y = b1 and a1*y = -b1 so the y coordinate of one of the solutions should be y = abs(b1/a1). If you expand the equation, you get the quadratic equation

a1^2 * y^2 - b1^2 = 0

and in matrix form

       0     0      0
L1 =   0  L1[1,1]   0
       0     0   L1[2,2]

where   L1[1,1] = a1^2   and   L1[2,2] = b1^2

Hence, the y coordinate of one solution is

y = abs(b1/a1) = sqrt( abs( L1[2, 2] / L1[1, 1] ) )

If you do the same for the conic L2, but this time thinking about it being a pair of symmetric lines parallel to the y axis, then you get the x coordinate of a solution

x = abs(b2/a2) = sqrt( abs( L2[2, 2] / L2[1, 1] ) )

That is why, one solution is

sol[0, :] = [ sqrt(abs( L2[2, 2] / L2[1, 1] )),  sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]

Then all solutions are obtained by reflecting in the coordinate axes:

sol[, :] = [(-1)**i*sqrt(abs( L2[2, 2] / L2[1, 1] )),  (-1)**j*sqrt(abs( L1[2, 2] / L1[1, 1] )),  1]

for i = 0,1 and j=0,1

Algorithm for calculating the angle between the internal tangents:

import numpy as np
import math

'''
####### FUNCTIONS IMPORTANT FOR THE CALCULATION #######
'''

def homogenize(x):
    return np.array([x[0], x[1], 1])

# transforming the vector of six coefficients
# of a quadratic equation with two variables into a matrix of
# the corresponding conic
def conic_from_equation(eq):
    '''
    eq is an np.array vector of the equadratic equation's six coefficients
    eq[0]*x**2 + eq[1]*x*y + eq[2]*y**2 + eq[3]*x + eq[4]*y + eq[5] = 0
    conc is a symmetric matrix of the bilinier form of the quadratic equation
    '''
    return np.array([[2*eq[0],   eq[1],   eq[3]],
                     [  eq[1], 2*eq[2],   eq[4]],
                     [  eq[3],   eq[4], 2*eq[5]]]) / 2

# obtaining the conic dual to a given conic,
# both represented as symmetric 3x3 matrices
def dual_conic(conic):
    return np.linalg.inv(conic)

# given a pair of conics, as a pair of symmetric 3x3 matrices,
# calculate the vector k = (k[0], k[1], k[2]) of values for each of which
# the conic c1 - k[i]*c2 from the pencil of conics c1 - t*c2
# is a degenerate conic (the anti-symmetric product of a pair of linear forms)
# and also find the matrix U
# of the projective transformation that simplifies the geometry of
# the pair of conics, the geometry of the pencil c1 - t*c2 in general,
# as well as the geometry of the three degenerate conics in particular
def trivialization_of(conic1, conic2):
    '''
    c1 and c2 are 3 by 3 symmetric matrices of two conics (possibly dual)
    '''
    c21 = np.linalg.inv(conic2).dot(conic1)
    k, U = np.linalg.eig(c21)
    return k, U

# realze the 3x3 matrix U as a projective transformation of projective 2-plane
def projectivity(U, x):
    if len(x) == 2:
        x = homogenize(x)
    y = U.dot(x)
    y = y / y[2]
    return y[0:2]

# find the common points, i.e. points of intersection of a pair of conics
# represented by a pair of symmetric matrices
def find_intersection_points(c1, c2):
    k, U = trivialization_of(c1, c2)
    L1 = (U.T).dot((c1 - k[0]*c2).dot(U))
    L2 = (U.T).dot((c1 - k[1]*c2).dot(U))
    sol = np.empty((4,3), dtype=float)
    for i in range(2):
        for j in range(2):
            sol[i+2*j,0:2] = np.array([math.sqrt(abs(L2[2,2] / L2[0,0]))*(-1)**i, math.sqrt(abs(L1[2,2] / L1[1,1]))*(-1)**j])
            sol[i+2*j,0:2] = projectivity(U, sol[i+2*j,0:2])
    sol[:,2] = np.ones(4)
    return sol

# find the three coefficients of all four common tangents to a pair of conics
# make sure that the conics do have four common tangents,
# which is the generic situation
def find_common_tangents(c1, c2):
    dc1 = dual_conic(c1)
    dc2 = dual_conic(c2)
    return find_intersection_points(dc1, dc2)

# if the two conics are ellipses, they do not intersect the line at infinity.
# by polarity, the poles of the line at infinity are
# the (affine and in fact metric) centers of the ellipses and are located
# inside the ellipses each. We use them to detect the inner tangents and
# to orient the vectors normal to these tangents in a way
# so that the angle between these two normal vectors is the vector on the picture
def dual_to_infinity_line(conic):
    p = dual_conic(conic).dot(np.array([0,0,1]))
    return homogenize( p[0:2] / p[2] )

# find the common inner tangents between a pair of conics,
# represented as 3x3 symmetric matrices, with properly oriented normal vectors
def find_common_inner_tangents(c1, c2):
    p1 = dual_to_infinity_line(c1)
    p2 = dual_to_infinity_line(c2)
    tn = find_common_tangents(c1, c2)
    # boolean vector, true if inner tangent false if not
    is_inner_tangent = ( tn.dot(p1)*tn.dot(p2) < 0 )
    tn = tn[ is_inner_tangent, : ]
    # reorient one of the normal vectors so that they give us the proper angle
    if tn[0,:].dot(p1)*tn[1,:].dot(p1) > 0:
        tn[1,:] = - tn[1,:]
    return tn

# calculate the angle between a pair of line, given by their three coefficients
def angle_bn_lines(line1, line2):
    theta = line1[0:2].dot(line2[0:2]) / math.sqrt( line1[0:2].dot(line1[0:2])*line2[0:2].dot(line2[0:2]))
    return math.acos(theta)*180/math.pi

# calculate the proper angle between the inner tangents of two quadratic equations
# that represent a pair of conics, hopefully non-intersecting
def angle_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    return angle_bn_lines(tn[0,:], tn[1,:])

'''
####### END OF FUCTIONS IMPORTANT FOR THE CALCULATION #######
'''

'''
####### auxiliary functions to handle the test scenario:
'''

def cos_sin(angle_deg):
    return math.cos(angle_deg*math.pi/180), math.sin(angle_deg*math.pi/180)

def rotation(cs_sn):
    return np.array([[cs_sn[0], -cs_sn[1]],
                     [cs_sn[1],  cs_sn[0]]])

def isom(angle, translation):
    '''
    isometry from global coordinate system (world system)
    to conic-aligned coordinate system (conic attached)
    '''
    cos_, sin_ = cos_sin(-angle)
    tr = - rotation((cos_, sin_)).dot(translation)
    return np.array([[ cos_, -sin_, tr[0]],
                     [ sin_,  cos_, tr[1]],
                     [    0,     0,    1 ]])

# calculating the conic defined by a pair of axes' lengths,
# axes rotation angle and center of the conic
def Conic(major, minor, angle, center):
    D = np.array([[minor**2,        0,                 0],
                  [       0, major**2,                 0],
                  [       0,        0, -(minor*major)**2]])
    U = isom(angle, center)
    return (U.T).dot(D.dot(U))

# transfromaing the matrix of a conic into a vector of six coeficients
# of a quadratic equation with two variables
def equation_from_conic(C):
    '''
    c[0]*x**2 + c[1]*x*y + c[2]*y**2 + c[3]*x + c[4]*y + c[5] = 0
    '''
    return np.array([C[0,0], 2*C[0,1], C[1,1], 2*C[0,2], 2*C[1,2], C[2,2]])
'''
####### end of auxiliary functions
'''

'''
Example of finding the angle between the common tangents of a pair of conics:
conic 1: major axis = 2, minor axis = 1, angle = 45, center = (0,0)
conic 2: major axis = 3, minor axis = 1, angle = 120, center = (15,0)
'''

a = 2
b = 1
cntr = np.array([0,0])
w = 45

Eq1 = equation_from_conic(Conic(a, b, w, cntr))
Co1 = conic_from_equation(Eq1)

a = 3
b = 1
cntr = np.array([15,0])
w = 120

Eq2 = equation_from_conic(Conic(a, b, w, cntr))
Co2 = conic_from_equation(Eq2)

four_tngs = find_common_tangents(Co1, Co2)

inner_tngs = find_common_inner_tangents(Co1, Co2)

print( angle_bn_inner_tangents_of(Eq1, Eq2) )

Calculating the angle bisector line
of the inner tangents.

Here is a routine that, given a pair of lines line1 and line2, represented by vectors with three coefficients, calculates the three coefficients of the angle bisector line that splits the angle in half. It also returns the slope of the tangent line, as the ratio of the y-increase divided by the x-increase.

def angle_bisector(line1, line2):
    n1 = line1[0:2]
    n2 = line2[0:2]
    n1 = n1 / math.sqrt(n1.dot(n1))
    n2 = n2 / math.sqrt(n2.dot(n2))
    n_bsec = n1 + n2
    p12 = np.cross(line1, line2)
    p12 = p12[0:2] / p12[2]
    c_bsec = n_bsec.dot(p12)
    return np.array([n_bsec[0], n_bsec[1], c_bsec]), np.array([-n_bsec[1], n_bsec[0]])

def angle_and_bsec_vector_bn_inner_tangents_of(eq_of_conic1, eq_of_conic2):
    c1 = conic_from_equation(eq_of_conic1)
    c2 = conic_from_equation(eq_of_conic2)
    tn = find_common_inner_tangents(c1, c2)
    if abs(np.all(tn.imag)) < 1e-10:
        tn = tn.real
    p1 = dual_to_infinity_line(c1)[0:2]
    p2 = dual_to_infinity_line(c2)[0:2]
    bsec, v_bsec = angle_bisector(tn[0], tn[1])
    angle = angle_bn_C_lines(tn[0,:], tn[1,:])
    if abs(angle.imag) < 1e-10:
        angle = angle.real
    if v_bsec.dot(p2 - p1) < 0:
        v_bsec = - v_bsec
    return angle, v_bsec / np.linalg.norm(v_bsec)


问:
If f is continuous on [a,b] and f(a)=f(b) then show that there exists x,y in (a,b) such that f(x)=f(y)
If f is continuous on $[a,b]$ and $f(a)=f(b)$ then show that there exists $x,y \in (a,b)$ such that $f(x)=f(y)$

It looks obvious if I imagine the graph. But I am not able to prove it.
I am trying to employ intermediate value property, but not able to reach to the conclusion.

答: Hint: Either $f$ is a constant (so you are done), or there is some $t\in(a,b)$ such that $f(t)\neq f(a)$, in which case employ IVT for $f$ on $[a,t]$ and $[t,b]$.


答:
Take $\begin{cases}P(U=1)=&\frac 1 3\\P(U=2)=&\frac 1 3\\P(U=3)=&\frac 1 3\end{cases}$ and $\begin{cases}P(V=-1)=&\frac 1 3\\P(V=0)=&\frac 1 3\\P(V=1)=&\frac 1 3\end{cases}$ and let $Y=U+V$ with U, V being independent.

Then $f_{V|Y=0}(-1)=1$ (and is 0 otherwise). But clearly $f_V\ne f_{V|Y=0}$. Thus $f_{V|Y}\ne f_V$

Title: A probability density transformation

Body:
I run into a problem that is surprising to me.

Suppose three r.v. $Y, U, V$ satisfy $Y = U + V$, $U$ is independent of $V$.

We compute the following conditional density of $V$ given $Y$:

$ f_{V | Y}(v|y) = f_{V|U}(v|y-v) = f_V(v)$.

It's surprising to me because the conditional density $V|Y$ equals the marginal density of $V$.

Is my understanding correct? Thanks!


问:
Stoke's Thm. $\vec F = \langle x^2y,\frac{x^3}{3},xy \rangle $ and $C$ is the boundary of the intersection of $x^2+y^2=1$ and $z = y^2-x^2$
I am having trouble setting up the limit of integration for the following problem.

$\vec F = \langle x^2y,\frac{x^3}{3},xy \rangle $ and $C$ is the boundary of the intersection of $x^2+y^2=1$ and $z = y^2-x^2$.

I found

$$ \operatorname{curl} \vec F = \langle x,-y,0 \rangle $$

and letting $x=\cos \theta, y = \sin \theta , z = z$

the parameterized surface will have a normal vector

$$\vec n =  \langle \cos \theta, \sin \theta, 0\rangle$$.

So, what I need to integrate according to Stoke's Thm. is

$$\iint_S \cos^2 \theta\,dS$$

My question here is, I know that $\theta \in [0, 2\pi]$ but what about $z$?

I want to say that the maximum and minimum values $z$ can take is $1$ and $-1$, respectively, so I want to use $\int_{-1}^1$ but I know that is not right.

I also considered the parameterization $\langle\cos \theta, r\sin \theta, -r^2(\cos 2\theta)\rangle$ but I am not confident going that route.

I believe I have a gap in understanding in this portion of "parameterization" of the curve or surface.

May I have some assistance?

答:
Point 1:

This is a line integral over the boundary of a surface (made by intersection with another surface). So if you want to do line integral for the given vector field then parametrize as

$r(\theta) = (\cos\theta, \sin\theta, - \cos2\theta) \ $, as ($z = y^2 - x^2 = - \cos 2\theta$)

But if you are converting this into a double integral using the curl of the vector field, then you have to parametrize as

$r(\rho, \theta) = (\rho \cos\theta, \rho \sin \theta, - \rho^2 \cos 2\theta)$

Point 2:

you cannot assume normal vector to be $(\cos\theta, \sin\theta, 0)$ as it is intersection of two surfaces. Instead, take derivative with respect to $\rho$ and $\theta$ and do a cross product. That is the standard approach.

$r'_{\rho} = (\cos\theta, \sin\theta, - 2\rho\cos2\theta)$

$r'_{\theta} = (-\rho\sin\theta, \rho\cos\theta, 2\rho^2\sin2\theta)$

$r'_{\rho} \times r'_{\theta} = (2\rho^2 \cos\theta, - 2\rho^2\sin\theta, \rho)$

$curl \vec F = (\rho \cos\theta, - \rho \sin\theta, 0)$

So the double integral is,

$\displaystyle \int_0^{2\pi} \int_0^{1} (\rho \cos\theta, - \rho \sin\theta, 0) \cdot (2\rho^2 \cos\theta, - 2\rho^2\sin\theta, \rho) \ d\rho \ d\theta$

$ = \displaystyle \int_0^{2\pi} \int_0^{1} 2 \rho^3 \ d\rho \ d\theta$


问:
symmetric relation definitions
Definition One: A relation over a set $X$ is symmetric if for all $a,b$ $\in X$, $(a,b)\in R$ if and only if $(b,a)\in R$.

Definition Two: A relation over a set $X$ is symmetric if for all $a,b$ $\in X$, $(a,b)\in R$ if  $(b,a)\in R$.

Am I correct to believe that, if I want to prove a relation over a set $X$ is symmetric, then it does not matter if I use definition one or definition two? I think that those two definitions are the exact same despite the fact that one has a lone "if" and then other has "if and only if." Although this may seem really obvious, I want to double check that my thinking is correct.

答:
Def. $1$ clearly implies Def. $2$.

You can easily show that Def. $2$ implies Def. $1$: we are already given that $(b,a)\in R\rightarrow(a,b)\in R$. Thus $(a,b)\in R\rightarrow (b,a)\in R$, which follows from interchanging $a,b$ in Def. $2$.

Title: symmetric relation definitions


问:
To find an integral that satisfies a substitution property.
Consider the following integral. $$\int \cfrac{1}{9-x^2}dx$$

Substituting $x=3\sin u$ makes the integral $\displaystyle\int\cfrac{1}{3\cos u}du$.

I would like to modify the original integral such that after the substitution $x=3\sin u$, it looks like $\displaystyle\int\cfrac{1}{9\cos^2u}du$.

My only thought so far was that $dx=3\cos u du$ which tells me that I need to have $\cos^3u$ in the denominator of the integrand to get $\cos^2u$ in the denominator after simplification. How I do proceed from here? thanks for your time.

答: Let $u\in[-\pi/2,\pi/2]$. After substitution in the original integral you get$$\int\frac{3\cos u~du}{9\cos^2u}$$so we want to get an additional $3\cos u$ term in the denominator to cancel the $3\cos u$ in the numerator. Note that $x=3\sin u\implies\sqrt{9-x^2}=3|\cos u|=3\cos u$, so modify the original integral to$$\int\frac{dx}{(9-x^2)\sqrt{9-x^2}}$$


问:
The metric space $(l^1 \mathbb{(N)}, d_{l^1})$ is separable.
My goal is to prove the title. The metric space $(l^1 \mathbb{(N)}, d_{l^1})$ is separable. Some definitions:
$$l^1 \mathbb{(N)} = \left\{f : \mathbb{N} \rightarrow \mathbb{R} : \sum^{\infty}_{j = 1} |f(j)| < + \infty\right\}$$
$$d_{l^1}(f,g) = \sum^{\infty}_{j=1}|f(j) - g(j)|$$
Here's what I have so far, but I'm worried that it's incorrect because I never use the metric $d_{l^1}$, and failing to use every aspect of the setup of a problem usually doesn't bode well. I suspect that there may be something incorrect about my assertion that the set $A$ is countable. Am I doing something wrong or is this correct?

$\textit{Proof.}$ Define the set $$A_{m, n} = \left\{f \in l^1 \mathbb{(N)} : \frac{m}{n} \leq \sum^{\infty}_{j=1} |f(j)| \leq \frac{m+1}{n} \text{ where } m \in \mathbb{N}_0, n \in \mathbb{N}\right\}.$$
Then
$$A = \bigcup_{m \in \mathbb{N_0}, n \in \mathbb{N}} A_{m,n}$$
is countable since it is the countable union of countable sets. Next, let $B(f, r)$ be the open ball in $l^1 \mathbb{(N)}$ centered at $f$ with radius $r$. For contradiction, suppose $B(f, r) \cap A = \emptyset$. Then each $f_0 \in B(f, r)$ has the property that $f_0 \notin A_{m,n}$ for all $m \in \mathbb{N_0}, n \in \mathbb{N}$. This means there is no rational number which bounds $\sum^{\infty}_{j=1} |f_0(j)|$ from above, and so $\sum^{\infty}_{j=1} |f_0(j)|$ diverges to $\infty$. This is a contradiction since it implies $B(f,r) \notin l^1 \mathbb{(N)}$.

We have then that $B(f,r) \cap A \neq \emptyset$, and so $A$ is countable and dense in $l^1 \mathbb{(N)}$, proving that the metric space $(l^1 \mathbb{(N)}, d_{l^1})$ is separable.

答:
$A_{m,n}$ is not countable. If you take one $f$ with $\sum |f(j)|=\frac {m+\frac1  2} n$ then $(1+\epsilon)f \in A_{m,n}$ for all $\epsilon$ sufficiently small.

For a correct proof consider sequences $(a_n)$ which are $0$ after some stage with $a_n$ rational for each $n$.


问:
does $\frac{d(p_1,p_2)}{d(p_2,p_3)}=\frac{d(q_1,q_2)}{d(q_2,q_3)}$
the problem is :

if $p_1,p_2,p_3$ are collinier points in the space , and for an arbitrary line $l$ we have that $q_1,q_2,q_3$ are the image of $p_1,p_2,p_3$ on $l$ respectivly. then does the following hold?  $$\frac{d(p_1,p_2)}{d(p_2,p_3)}=\frac{d(q_1,q_2)}{d(q_2,q_3)}$$

i only trid to take a plane $W$ such that $l \in W$ and then put image of $p_1,p_2,p_3$ on $W$.
but i could not move forward.

答: Hint: Let $\hat l$ denote the unit vector along $l$ and $\theta$ be the angle between the line containing $p_i$ and $l$. Then$$d(q_1,q_2)=\left|\left[\vec{p_2}-\vec{p_1}\right]\cdot\hat l\right|=d(p_1,p_2)\left|\cos\theta\right|$$


问:
Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically?
Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically  $$0 \rightarrow \ker \varphi \rightarrow F \xrightarrow{\varphi} P \rightarrow 0? $$ where $F$ is a free $R$-module?

答:
Assume that $P$ is the quotient of a free module.  This means that $P = F/K$ for a free module $F$, and a submodule $K$.  If $i: K \rightarrow F$ is the inclusion map, and $p: F \rightarrow P$ is the canonical quotient map, then the sequence

$$0 \rightarrow K \xrightarrow{i} F \xrightarrow{p} P \rightarrow 0$$

is clearly exact.


问:
Bounding the difference between consecutive terms in the sequence $a_n=(1+1/n)^n$
I'm pretty sure the following estimate holds for $C>1$, but I can't prove it:

$$a_n:=\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1}\leq\frac{C}{n^2}$$

For context, I was looking for some such bound in order to prove that $na_n\to0$.

I tried using the fact that

$$\left(1+\frac{1}{n}\right)^n\leq\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}+\cdots+\color{red}{\frac{1}{n!}}:=S_n$$

because, by some algebra with the binomial theorem,

$$\left(1+\frac{1}{n}\right)^n=\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}\color{blue}{\left(1-\frac{1}{n}\right)}+\cdots+\color{red}{\frac{1}{n!}}\color{blue}{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}$$

where the blue factors are all less than one.

But I also need a good lower bound on the terms $(1+n^{-1})^{n}$, and I can't see how to pick the right lower bound to get, say,

$$a_n\leq S_n-\text{lower bound applied to the term $\left(1+\frac{1}{n-1}\right)^{n-1}$}\leq\frac{C}{n^2}$$

答:
If you use Taylor series
$$b_p=\left(1+\frac{1}{p}\right)^p\implies \log(b_p)=p\log\left(1+\frac{1}{p}\right)$$
$$\log(b_p)=p\Bigg[\frac{1}{p}-\frac{1}{2 p^2}+\frac{1}{3 p^3}-\frac{1}{4
   p^4}+\frac{1}{5 p^3}+O\left(\frac{1}{p^6}\right) \Bigg]$$
$$\log(b_p)=1-\frac{1}{2 p}+\frac{1}{3 p^2}-\frac{1}{4p^3}+\frac{1}{5 p^4}+O\left(\frac{1}{p^5}\right) $$
$$b_p=e^{\log(b_p)}=e-\frac{e}{2 p}+\frac{11 e}{24 p^2}-\frac{7 e}{16
   p^3}+\frac{2447 e}{5760 p^4}+O\left(\frac{1}{p^5}\right) $$

Apply is twice and continue with Taylor series to make
$$a_n=\frac{e}{2 n^2}-\frac{5 e}{12 n^3}+\frac{7 e}{16
   n^4}+O\left(\frac{1}{n^5}\right) $$

So, to make the inequality true for all $n$, we need $C >\frac{1}{2}e$


问:
Partial Integration doubt
I want to integrate an equation but I am unable to understand how do I convert the d(1-c) term to some dx term. Please help me understand such manipulations and share any helpful resources to understand more about them. Equation

$\frac{d(1-c)}{dt}=-1.4\times 10^{-4} \times (1-c)^2(3-\frac{c}{2})$

$\int_1^\frac{1}{2} \frac{d(1-c)}{(1-c)^2(3-\frac{c}{2})}=-1.4\times 10^{-4} \int_0^t dt$

答:
$\frac{d(1-c)}{dc}=-1\implies d(1-c)=-d(c)$.

Alternatively you can substitute $m=1-c$ and your LHS becomes$$\int_0^{1/2}\frac{2~dm}{m^2(5+m)}$$Integrate this with respect to $m$ and substitute $1-c$ back in the final result.

Title: Partial Integration doubt


问:
How do you prove that $\frac{\sin x}{n \sin (\frac{x}{n})}$ converges uniformly?
For $n \in \mathbb N$, define
$$f_n(x) = \frac{\sin x}{n \sin (\frac{x}{n})}.$$
As $n \to \infty$, I have shown that $\{f_n\}$ converges (pointwise) to $\frac{\sin x}{x}$. I am trying to evaluate $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx$, and my attempt is to show the uniform convergence of the integrand on $[0,\pi]$, so that I can interchange the operations of limit and integral to obtain $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx = \int_0^\pi \frac{\sin x}{x} \, dx$.

By plotting $f_n(x)$ with a graphing calculator and letting $n \to \infty$, I am quite convinced that $\{f_n\}$ indeed converges uniformly to $\frac{\sin x}{x}$. Nevertheless, I can't seem to prove this proposition, whether by the definition of uniform convergence or by Cauchy criterion. Perhaps my difficulty stems from the fact that the bounds on $f_n$ depends on the interaction between $\sin x$, $\sin(\frac{x}{n})$, and $n$. That is, finding a bound on one or two expressions can't help in establishing a bound for the whole function. So far I have tried applying the inequalities $0 < \sin x \leq 1$ and $\sin x \leq x$ for $0 < x < \pi$, but without any luck.

Any idea on how to prove this statement? Or is there a theorem on uniform convergence that is applicable for this case?

(Note: This question is related to the Dirichlet kernel in Fourier analysis, which can be expressed as $D_N(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_{n=1}^N \cos nx = \frac{1}{2\pi} \frac{\sin(N+\frac{1}{2})x}{\sin (\frac{x}{2})}$. However, I don't think this is relevant at this point.)

答:
Hint: Choose $m$ such that $n >m$ implies $|1-\frac {\sin (x/n)} {x/n}| <\epsilon$ for all $x \in [0,\pi]$. Then $\frac {\sin x} {(1+\epsilon)x}<f_n(x)<\frac {\sin x} {(1-\epsilon)x}$ for all $x$ if $n >m$. Can you finish?

[Check that $|f_n(x)-\frac {\sin x} x| <\frac {\epsilon} {1-\epsilon} $ for $n >m$ using the fact that $0 \leq \frac {\sin x} x \leq 1$].


问:
Hilbert function for points in projective space
In Lecture 13 of Harris's Algebraic Geometry, we have the Hilbert functions, $h_X$, for the following two cases:

If $X=\{p_1,p_2,p_3\}\subseteq \mathbb{P}^2$ then there are two possible Hilbert functions:

$$h_X(m) = \begin{cases}2 & m=1\\ 3& m\geq 2\end{cases} \quad \text{and} \quad h_X(m)=3 \ \forall m\in \mathbb{N}$$
depending on whether the points are collinear or not, respectively [proof].

If $X=\{p_1,p_2,p_3, p_4\}\subseteq \mathbb{P}^2$ then there are two possible Hilbert functions:

$$h_X(m) = \begin{cases}2 & m=1\\ 3& m= 2\\ 4 & m\geq 3\end{cases} \quad \text{and} \quad h_X(m)=\begin{cases}3 & m=1\\ 4& m \geq 2\end{cases}$$
depending on whether the points are collinear or not, respectively [proof].

Then it leaves the following generalization as an exercise:

Let $X\subseteq \mathbb{P}^n$ be a set of $d$ points. Show that for $m\geq d-1$, the Hilbert function $h_X(m)=d$.

I have no idea about how to approach. How can we generalize the specific arguments given in the proofs (like the case $m=2$)?  To begin with, how to approach the following problem:

If $X=\{p\}\in\mathbb{P}^n$ be a point then $h_X(m)=1$ for all $m\in \mathbb{N}$.

Thank you for your time, please let me know if you have any hints/solution for this problem.

答:
I asked this question when I was an undergraduate student and now I am a graduate student. Though my understanding hasn't improved much, here is an answer from Gathmann's notes (Lemma 6.1.4):

Note that $X$ is a zero-dimensional affine subscheme of $\mathbb{P}_k^n$. That is, $X=\mathrm{Spec}(R)$ for some $k$-algebra $R$.

Claim: $h_X(m)=d$ for large values of $m$, where $d$ is the number of points in $X$ counted with scheme-theoretic multiplicities.

Without loss of generality, we can assume that $X$ is irreducible, i. e. consists of only one point (but may have a non-trivial scheme structure) since in the reducible case $X = X_1 \sqcup \cdots \sqcup X_r$ with
$X_i = \mathrm{Spec}(R_i)$ for $i = 1, \ldots , r$ we have $R = R_1 \times \cdots \times R_r$. Moreover, by a change of coordinates we can assume that this point is the origin in $\mathbb{A}_k^n$. If $X = \mathrm{Spec}( k[x_1,\ldots, x_n]/I)$ we then must have $\langle x_1,\ldots , x_n\rangle = \sqrt{I}$ by the Nullstellensatz. That is, $x_i^m \in I$ for some $m$ and all $i$. Therefore, every monomial of degree at least $D = m\cdot n$
lies in $I$. In other words, $k[x_1 ,\ldots , x_n]/I$ has a $k$-basis of polynomials of degree less than $D$. But the space of such polynomials is finite-dimensional and equals $\dim_k\left(k[x_1,\ldots, x_n]/I\right)=d$ if $I(X)= \langle x_0^d\rangle$ where $I(X)$ be the homogenization of $I$, i.e. $I=I(X)|_{x_0=1}$.

Next, let $S(X)$ be the homogeneous coordinate ring (graded). Hence, for $m\geq m\cdot n$, there is an isomorphism $S(X)^{(m)} \to R$ as vector spaces over $k$ given by

\begin{align*}
\left(k[x_0,\ldots, x_n]/I(X)\right)^{(m)} & \to k[x_1,\ldots, x_n]/I\\
f & \mapsto f|_{x_0=1}
\end{align*}

and the inverse

\begin{align*}
k[x_1,\ldots, x_n]/I &\to \left(k[x_0,\ldots, x_n]/I(X)\right)^{(m)} \\
f &\mapsto f^h \cdot x_0^{m-\deg(f)}
\end{align*}

where $f^h$ denotes the homogenization of a polynomial. Note that the inverse map is well-defined as $k[x_1, \ldots,x_n]/I$ has a basis of polynomials of degree less than $D=m\cdot n$. That is,

$$h_X(m) = \dim_kS(X)^{(m)} = d$$
for large values of $m$.


问:
Binomial identity arising from Catalan recurrence.
The following identity features right at the end of the answer to another question I asked: https://math.stackexchange.com/a/4019598/155881.

$$F(x) = \frac{1}{\sqrt{1-4x}} \left(\frac{1- \sqrt{1-4x}}{2x}\right)^n = \sum\limits_{k=0}^{\infty}{n+2k \choose k}x^k$$

Haven't been able to prove this. We know the expansions of both terms on the left side, making the identity:

$$\left(\sum\limits_{j=0}^{\infty} \left(\binom{n+2j-1}{j} - \binom{n+2j-1}{j-1}\right)x^j\right)
\left(\sum\limits_{l=0}^{\infty}{2l \choose l} x^l\right) = \sum\limits_{k=0}^{\infty}{2n+k \choose k}x^k$$

My attempt:
An obvious thing to do is to try a convolution of the left side and equate coefficients of $x^k$ (also using an alternate expression for the first summation in the product of the left side):

$$\sum\limits_{l=0}^k {n+2l-1\choose l}\frac{l}{n+l} {2k-2l\choose k-l} = {2n+k\choose k}$$

Not sure how to make progress on this. Expanding the binomial terms doesn't lead anywhere.

答:
You can also prove this by showing that both sides are different expressions for the generating function of the same sequence.

Fix a positive integer $n$ from now on. For a nonnegative integer $m$, let $Q_m$ be the number of lattice paths, that touch the main diagonal (starting from the left bottom corner) exactly $n$ times before it goes above it for the first time in an $m\times m + 1$ grid ($m$ horizontal, $m+1$ vertical). Let us call these paths admissible paths. All paths considered here are only allowed to go right or up.

If such an admissible path exists for some $m$ then $m\ge n$ since for every time you touch the diagonal, from the bottom corner up to the last time you touch the diagonal but failed to go up (i.e. n times) you need an horizontal segment that keeps you below it. So, horizontally you need at least $n$ segments, hence $m\ge n$.

The generating function looks then like this: $\displaystyle\sum_{m\ge n} Q_mx^m$.

Define $k = m - n$. We have:

Claim: There is a bijection between the admissible paths and the paths of an $k\times (n + k)$ grid ($k$ horizontal, $n + k$ vertical segments.)

Proof:
Given an admissible path, we remove the following segments from it: the $n$ horizontal segments that follow a failure point and the one vertical segment that follows the first time we cross (i.e the very first segment that is above the diagonal). We concatenate the resulting pieces by gluing their endpoints one to the next. Notice that the first segment we remove always is the first horizontal segment from the left bottom corner.The path that remains has $m - n = k$ horizontal segments and $m + 1 - 1 = m = n + k$ vertical paths. Thus these paths indeed are in the sought grid.

If we have one such path $P$ in order to read the corresponding admissible path we begin by putting an horizontal segment and we follow $P$ until we hit the main diagonal, at which point we put another horizontal segment, and proceed reading $P$. We put horizontal segments until we have touched the diagonal $n$ times (without counting the initial point). At that time we put a vertical segment, and we copy what is left of $P$. Inverting the computations above we see that this is indeed a path in a grid of $m\times m + 1$.

Notice that because our path $P$ goes up $n + k$ times we must always be able to push it horizontally (force it to fail), counting the first segment from the bottom left corner, n times since, in the worst case scenario (k = 0) after pushing the n times we get we have moved $n + k$ in both directions and we still have the extra room to push up once more. This proves that the constructed path out from $P$ is indeed admissible. This concludes proving this correspondence is actually a bijection. $\blacksquare$

Here is a drawing to show this for $n =2, m = 3, k = 1$:

[The yellow edges are those we add to make sure it fails until we want it to succeed]

We have proven then that the generating function is

$$\begin{equation*}
\displaystyle\sum_{m\ge n} Q_mx^m = \displaystyle\sum_{k\ge 0} \binom{n + 2k}{k}x^{n + k}.
\end{equation*}$$

On the other hand, we can forget about the size of the final grid (i.e. about $m$) and construct what happens out of first principles. There are $n + 1$ independent events ocurring: what happens before the first failure, in between the first and second failures, up to what happens in between the $n - 1$ failure and the success moment and then what happens after we crossed the diagonal and went above for the first times.

Each one of the first independent events is some Catalan number $C_{i}$ and contributes $i + 1$ to the horizontal and vertical lengths, since we must not cross the subdiagonal until the right time. Notice this is important, since if we just consider the $i\times i$ square instead we might have intermidiate failures in between endpoints, since catalan paths are allowed to touch the diagonal of the square they live in, and we dont want those extra failures.

Hence the generating function of each of these events is $x\displaystyle\sum_{i = 0}C_ix^i$ (the extra $x$ is the displacement on the contribution to horizontal length!). We have $n$ of these.

Then, after the last Catalan square is done, we have a foced vertical segment and then whatever we want as long as it happens in a $j\times j$ square for some $j$. (This is because at the end, the final grid has to be $m\times m + 1$ and the Catalan part has produced a square grid and the vertical strip takes awaty the $+1$, so we need another square for the dimensions to match. The generating function for this square is
$$\begin{equation*}
 \displaystyle\sum_{j\ge 0} \binom{2j}{j}x^j
\end{equation*}$$
We have thus proved the equality
$$\begin{equation*}
 \left(x\displaystyle\sum_{i = 0}C_ix^i\right)^n\displaystyle\sum_{j\ge 0} \binom{2j}{j}x^j = \displaystyle\sum_{k\ge 0} \binom{n + 2k}{k}x^{n + k}
\end{equation*}$$
Using that we know
$$\displaystyle\sum_{i = 0}C_ix^i = \dfrac{1 - \sqrt{1 - 4x}}{2x}$$
and, from Newton's Binomial theorem, that
$$
\displaystyle\sum_{j\ge 0} \binom{2j}{j}x^j = \dfrac{1}{\sqrt{1 - 4x}}
$$
we conclude
$$
\begin{equation*}
 x^n\left(\dfrac{1 - \sqrt{1 - 4x}}{2x}\right)^n\dfrac{1}{\sqrt{1 - 4x}} = \displaystyle\sum_{k\ge 0} \binom{n + 2k}{k}x^{n + k},
\end{equation*}
$$
and cancelling the factor $x^n$ on both sides we get the desired equality.


问:
Verifying Stokes’ Thm. $\vec F = \left< -y,x,-2\right>$ where $S$ is the region defined by $z^2 = x^2 + y^2, \space z \in [0,4]$
I want to verify Stokes’ Theorem by evaluating both the left and right side of the following eqn.

$$\int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$$

where $\vec F = \left< -y,x,-2\right>$ and $S$ is the region defined by $z^2 = x^2 + y^2, \space z \in [0,4]$.

When I use the parameterization $\vec r = \left< \cos \theta, \sin \theta, z\right>$

I get a normal vector $\vec n = \left< \cos\theta, \sin \theta,0\right>$ and since $curl \vec F = \left<0,0,2 \right>$

$$curlf \vec F \cdot \vec n = 0$$

so the right hand side equals zero.

Here is what I am not sure since it has been a while since I have evaluated line integrals.

I believe that $C$ is the projection of $S$ onto the $xy$-plane, so it will be a circle with radius $16$.

So, the parameterization I choose is $\vec r = \left< 4\cos \theta, 4\sin \theta, 0\right>$

and because $\vec F = \left<-4\sin\theta, 4\cos\theta,-2 \right>$

$$\vec F \cdot \vec r = 0$$

hence the left side is equal to zero.

Can anyone confirm that I did this correctly?

答:
No, both RHS and LHS have mistakes that have led to you getting integrals as zero.

A) For RHS, your parametrization is not correct. The correct parametrization is

$\vec r(\rho,\theta) = (\rho \cos\theta, \rho \sin\theta, \rho)$, as $z = \rho$ from the equation of the cone.

$r'_{\rho} \times r'_{\theta} = (\rho \cos\theta, \rho \sin \theta, - \rho)$

$0 \leq \rho \leq 4, 0 \leq \theta \leq 2\pi$.

Can you now complete the integral?

B) For LHS which is line integral, you should have the $z$ component in your parametrization though it would not matter here.

$\vec r(\theta) = (4\cos\theta, 4\sin\theta, 4)$

Now we are supposed to do $\vec F \cdot \vec {r} \ '$ and not $\vec F \cdot \vec r$. That was the mistake in your working of LHS.

$r'_{\theta} = (- 4 \sin\theta, 4 \cos\theta, 0)$

$0 \leq \theta \leq 2\pi$.

Can you complete the working now?

Last point - the outward normal vector for the given cone surface is pointing downward and that is why we have taken $z$ component negative in the first working. Now align the orientation of the boundary curve to the orientation of the surface in your line integral.


问:
Uniform approximation of continuous function by even polynomials
If f is a continuous function on [a,b], I want to show whether there exists sequence of even polynomials which converges uniformly to f.

I was checking this problem in StackExchange.

The above statement was proved here https://math.stackexchange.com/questions/1693629/uniform-approximation-by-even-polynomial.\

But here Continuous even functions closed and dense  one example of function was produced which can't be approximated uniformly by even polynomials.

Both the solutions seem right to me. I am confused now.

Am I missing something ? Please help me.

答: The reason is that even polynomials are dense on $[0,1]$ or on any other interval contained in the positive semi axis or in the negative semi axis but they cannot be dense on a symmetric interval like $[-1,1]$ since any combination of even functions would be even and, for example an odd function clearly cannot be approximated uniformly.


问:
how can I show that this function is uniformly continuous on $(1,0) $ and $(1,\infty)$
How can I show that $\sqrt x$ is uniformly continuous on the intervals: $(0,1)$ and $(1,\infty)?$

答:
For the first one,
we just need to prove that $\lim_{x \to \ 0^ +} \sqrt(x)$ and  $\lim_{x \to \ 1^ -} \sqrt(x)$ exists.

For the  second one,

$\forall x_0 \in(1,\infty),\forall \epsilon>0 $ we want to find $\delta>0$ s.t.

$\forall x',x'' \in(1,\infty) $when $|x'-x''|<0$ ,

we have$|\sqrt(x')-\sqrt(x'')|<\epsilon$

Let $|\sqrt(x')-\sqrt(x'')|=|\frac{(x')-(x'')}{\sqrt(x')+\sqrt(x'')}|<|\frac{(x')-(x'')}{2}|<\epsilon$

we take $\delta=2\epsilon$


问:
Finding area based on only the sides
I came across this simple looking question:

The perimeter of a triangle is $42$ cm. One side of a triangle is $8$ cm longer than the smallest side and the third side is $1$ cm less than $3$ times the smallest side. Find the area of the triangle.

At first it seemed easy enough. The smallest side is $7$ (which can be derived from:$\frac{42-8+1}{5}$, where subtracting $8$ takes care of the long side and now we have $2$ units of the shortest side, then adding one takes care of the third side now we have $5$ units of the third side and dividing by $5$ we can get the length of the smallest side, $7$) and with that knowledge the length of the first and third side is $15$ and $20$ respectively.

That's when I got stuck. The question gives us no indication of the angles in the triangle. I thought we could use the Pythagorean theorem to see if the triangle is right-angled, but it's not. Is there a rule we can use to find the area? Yes, I'm familiar with trigonometry and $sin()$, $cos()$ and $tan()$ so you can use those in your answer.

If you don't mind can you give me the rule to finding the area, not the answer to the problem? Because I want to solve this question myself. Many thanks to whoever answer's this!

答:
Use Heron's formula.
$$
A=\sqrt{s(s-a)(s-b)(s-c)}
$$
where $s$ is half of the perimeter.

Title: Finding area based on only the sides


问:
Are locally homotopic functions homotopic?
Suppose we have two (smooth) functions $f,g:X\to Y$, where $X,Y$ are smooth (second-countable, Hausdorff) manifolds which are locally homotopic (that is, any point in $X$ has a neighbourhood $U$ such that for any $V$ contained in $U$ with its closure, there is a homotopy which turns $f$ into $g$ in $V$, but keeps them fixed outside $U$).

Are $f,g$ necessarily homotopic? If the answer is yes, how much can we weaken the assumptions?

This seems rather obvious if $X$ is compact, for example, but I can't think of an easy way to show it in general, although it seems to be intuitively true.

答: Take spheres for example. The identity and constant map from $S^n$ to itself satisfy the condition. Because one can always choose $V$ relatively small compared to $U$, so that we can use geodesics to construct the local homotopy.</p>


问:
Using the Cauchy root test
I've got the next series: $$\sum_{n=1}^{\infty} \frac{3^n n!}{n^n} $$ I must to solve it to see if converges or not. So I've got the next process:

Sol (attempt): Let $a_n=\frac{3^n n!}{n^n}$. By the Cauchy root test we've got that:
$$\lim_{n\to\infty}\sqrt[n]{a_n} =\lim_{n\to\infty}\sqrt[n]\frac{3^n n!}{n^n} =\lim_{n\to\infty}\sqrt[n] {\left(\frac{3}{n}\right)^n n!} = \lim_{n\to\infty}\frac{3}{n}\sqrt[n]{n!} = \lim_{n\to\infty}\frac{3}{n}(n!)^\frac{1}{n} = 0$$
If $n$ tends to infinity we've got that $\frac{3}{n}=0$ and $ (n!)^\frac{1}{n}=(n!)^0=1$

Thus, the series converges absolutely.

My question is about if I'm using the test in a good way, and if the limit is correct. If there's any error on my process I'd be grateful to know.

答:
Here is an answer which uses Stirling's approximation, which differs slightly by the other answers. Originally intended to answer a duplicate of this question, I noticed that none of the answers exactly matched mine, and as the question is still unanswered it may be of sight help... I used the formula the straightforward way without any log simplifications as per the other answer....

By Stirling's approximation, we have

$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$

Hence, at the limit to infinity, we can write

$\lim_{n \to \infty} \frac{3^nn!}{n^n}=\lim_{n \to \infty} \frac{3^n}{n^n}\frac{\sqrt{2\pi n}~n^n}{e^n}=\lim_{n \to \infty}\left(\frac{3}{e}\right)^n\sqrt{2\pi n}$

As 3>e, the limit at infinity clearly diverges, hence the entire sum diverges......

Title: Using the Cauchy root test


问:
What is the simplest logical translation for "even if"?
I am looking for the propositional logic translation of "even if", not the epistemic logic meaning.

Specifically, how do I translate

"If I will die, then I will die even if I am careful"

"If I will die even if I am careful, then being careful is pointless"

答:
D= your dying

C= being careful

I have uncovered an aporia within Logic by that the proper way to translate "I will die even if I am careful." is D & ~[(C v ~C)→ D], as you have effectively stated the contradiction, D & ~D.

When you say "I will die even if I am careful", you are effectively stating that you will die and that it doesn't matter as to whether or not you are careful.  You're both making the claim that you will die and that you're being careful has no bearing upon your dying.  The negation of the conditional, within most systems of Logic, however, seems to result in the exact opposite of what you are attempting to express.  This, I believe, should result in the creation of an entirely other field of Logic, or a radical reconception of it as it stands, but, if you are asked this question on a test, just write "D", as that is what every Logic textbook will tell you to do.


问:
Serre FAC book by McLennan
I am planning to read Serre's FAC paper. I came to know that there is a book/notes by Andy McLennan, containing the commutative algebra background as well as an english translation of the FAC (Ravi Vakil's blog mentions it: https://math216.wordpress.com/2014/12/30/2014-version/)

But the link to the book seems to be broken. I could not find it elsewhere. Does someone know of a working link?

答:
The Algebra of Coherent Algebraic Sheaves with an Explicated Translation of Serre's "Faisceaux Algébriques Cohérents"  is now available at my personal website

https://andymclennan.droppages.com/Books/

or more directly from

https://andymclennan.droppages.com/fac_trans.pdf  (thanks to Martin for pointing this out)

Thanks for your interest!

Title: Serre FAC book by McLennan


问:
What is the number of degree $1$ representations of a finite group?
Let $G$ be a finite group and $G'$ be the derived subgroup of $G.$ Then I know that every degree $1$ representation of $G$ factors through $G/G'.$ How does it imply that the number of degree $1$ representations of $G$ is same as $\left \lvert G/G' \right \rvert\ $?

Would anybody please help me in this regard? Thanks in advance.

答:
A general fact for finite groups is that they have a finite number of irreduicible representations and if the dimensions of these are $d_1,..., d_n$ then we have
$$d^2_1 + ... + d^2_n = |G|.$$
This is a consequence of Schur orthogonality relations among characters.

If $G$ is abelian, then all the irreducible reps are one dimensional and so the previous equation tells you $|G| = n$, the number of irreducible reps.

Now, in your context, $G/G'$ is abelian and as you say every 1 dimensional rep of $G$ corresponds, bijectively since you can always compose with the quotient, with those 1 -dimensional reps of $G/G'$, but his being abelian has $|G/G'|$ of those, by the argument above.