问:
Expected number of tosses to get a head from a coin using integration formulae?
I recently started learning Expectation Probability , first of all Any Good resources to study it will be appreciated if any one can share

What I have learnt so far the expected  value of some Unknown Variable say $x$ be $ E(x)$

which boils down this equation  $ E(x) = \int_{-\infty}^\infty x*P(x) dx $ where P(x) is the Probability of some specific $x$ .

So I wanted to find the Expected Number of tosses to Get a  Heads from an Unbiased Coin So I used this equation to solve it

$E(Getting Heads) =>  \int_{0}^\infty\dfrac{x}{2^x}  dx$ , **IT IS A PRETTY STANDARD RESULT THAT E(GETTING  HEADS) = 2, But this integral is giving me another answer , which is $\dfrac{1}{\ln^2\left(2\right)}$
 Can Anyone tell me where am I going wrong with understanding stuff.

Body:
$\sum_{x=1}^{\infty} x*2^{-x} = \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{128}+\frac{8}{256}+\frac{9}{512}+\frac{10}{1024}+\frac{11}{2048}+\frac{12}{4096}\dots$

$= \frac{1}{2}+ \frac{1}{2}+\frac{6}{16}+\frac{2}{8}+\frac{16}{64}+\frac{22}{256}+\frac{28}{1024}+\frac{34}{4096}+\dots$

$= \frac{2}{2}+\frac{2}{8}+\frac{2}{16}+\frac{2}{4}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}+\frac{2}{256}+\frac{2}{512}+\frac{2}{4096}+\dots$

Here another and maybe a more easy way to see it this way that the picture and the adjusted line of fractions is correct:

$\sum_{x=1}^{\infty} x*2^{-x} = \frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\frac{7}{128}+\frac{8}{256}+\frac{9}{512}+\frac{10}{1024}+\frac{11}{2048}+\frac{12}{4096}+\frac{13}{8192}+\frac{14}{16384}+\frac{15}{32768}+\frac{16}{65536}+\dots$

$=\frac{16}{32}+\frac{16}{32}+\frac{12}{32}+\frac{8}{32}+\frac{5}{32}+\frac{48}{512}+\frac{28}{512}+\frac{16}{512}+\frac{9}{512}+\frac{80}{8192}+\frac{44}{8192}+\frac{24}{8192}+\frac{13}{8192}+\frac{56}{65536}+\frac{30}{65536}+\frac{16}{65536}+\dots$

$=\frac{256}{512}+\frac{256}{512}+\frac{192}{512}+\frac{128}{512}+\frac{80}{512}+\frac{48}{512}+\frac{448}{8192}+\frac{256}{8192}+\frac{144}{8192}+\frac{80}{8192}+\frac{352}{65536}+\frac{192}{65536}+\frac{104}{65536}+\frac{56}{65536}+\frac{30}{65536}+\frac{16}{65536}+\dots$

$=\frac{960}{512}+\frac{928}{8192}+\frac{750}{65536}+\dots$

$=\frac{122880}{65536}+\frac{7424}{65536}+\frac{750}{65536}+\dots$

$=\frac{131054}{65536}+\dots\approx 1,999725341796875$

$=\frac{65536}{65536}+\frac{65518}{65536}+\dots$

$=\frac{2}{2}+\frac{65518}{65536}+\dots$

$=\frac{2}{2}+\frac{32768}{65536}+\frac{32750}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{16384}{65536}+\frac{16366}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{8192}{65536}+\frac{8174}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{4096}{65536}+\frac{4078}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2048}{65536}+\frac{2030}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{1024}{65536}+\frac{1006}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}+\frac{512}{65536}+\frac{494}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}+\frac{2}{256}+\frac{256}{65536}+\frac{238}{65536}+\dots$

$=\frac{2}{2}+\frac{2}{4}+\frac{2}{8}+\frac{2}{16}+\frac{2}{32}+\frac{2}{64}+\frac{2}{128}+\frac{2}{256}+\frac{2}{512}+\frac{128}{65536}+\frac{110}{65536}+\dots$

The integral formula will result in some small deviations due to irrationalities of it.


问:
What guarantees the existence of maximal atlas of a smooth manifold?
Given a smooth manifold $M$ of dimension greater than 1, if $\mathfrak{U}$ is an atlas of $M$, then there exists a unique maximal atlas $\mathfrak{W}$ that contains $\mathfrak{U}$. What I am unsure is the existence of $\mathfrak{W}$. Of course if I assume the existence of such $\mathfrak{W}$, then uniqueness is clear. My question is, what guarantees the existence?

Clearly, a maximal atlas of $M$ must be uncountable, so I cannot inductively append chart by chart on $\mathfrak{U}$ to get $\mathfrak{W}$. Any help would be immensely appreciated.

答:
Declare that a pair $(U, \phi)$ where $U$ is an open in $M$ and $\phi: U \to V$ is an identification of $U$ with some open $V$ in $\mathbb{R}^n$, is in $\mathfrak{M}$ if and only if it is compatible with every such pair in $\mathfrak{U}$.

Here, compatibility means that if $(U', \phi')$ is some such pair in the atlas $\mathfrak{U}$ that you started with and $U'$ meets $U$, then $\phi'^{-1} \circ \phi$ is smooth (where the composition makes sense).

If this is an atlas, it's clearly maximal (anything compatible with all the charts in it is in particular compatible with all the charts in $\mathfrak{U}$, so in the $\mathfrak{M}$), but we have to check that any two charts we just added are compatible with each other.

This is true (omitting some details here) because the opens in $\mathfrak{U}$ cover $M$, so you can always compare (suitable refinements of) two charts in $\mathfrak{M}$ with a common chart in $\mathfrak{U}$, and compositions of smooth maps are smooth.

ContentLicense: CC BY-SA 4.0


问:
graph of $f$ dividing rectangle area.

Find all the monotone increasing functions $f(x),x\in(0,\infty)$ that satisfy the following:

for every $x>0$ the rectangle $(0,f(x)),(0,0),(x,0),(x,f(x))$ is divided by the graph of the function to two parts where one of them has double the area of the other

I've set the following equation $\int_{0}^{x}f(s)ds=2\int_{0}^{x}f(x)-f(s)ds\to 3y-3c=2xy'$ and then solving this equation and the second one is the in the same way if the upper area is the larger.

Am I right?

答:
You're on track except for the $c$, which isn't necessary since we're working with definite integrals.

To make things clearer, instead of jumping to notation involving $y$, start by defining $F(x)=\int_0^xf(s)ds$. Then your integral equation is equivalent to $F(x)=2xF'(x)-2F(x)$. Note that you also have the initial condition $F(0)=\int_0^0f(s)ds=0$.

Title: graph of $f$ dividing rectangle area.


问:
Group action definition
The Wikipedia for group action and several posts here have what seems to me to be two, slightly different definitions for a group action.$\ $   Let G be a group acting on space X.$\ $  These two definitions are:

A group homomorphism $\varphi$ from the group to the automorphism group of the space.
$$\varphi : G\ \longrightarrow \text{Aut}(X)$$

A map that takes an element of the group and an element of the space, and returns an element of the space.
$$\alpha : G\, \times X\ \longrightarrow X$$

Which of these is the the group action?$\ $   Is it a function of one argument or two?$\ $  What is the return type?  Does it return a function (automorphism) or an element of the space?$\ $ The first definition is in line with what I think of as a representation, ex. from a symmetry group C$_\text{2v} \longrightarrow $ GL(V) that returns a matrix, rather than a vector.

I (think) I understand the basic idea of what is going on, but I'd like to know what the actual mathematical definition is.

Thanks in advance.

答:
There's a more general setting for this equivalence. Let $G$ be a group and $X$ a set; consider:

D1) There is a map: $\cdot:G\times X\to X$, $(g,x)\mapsto g\cdot x$, such that:

$e\cdot x=x$, for every $x\in X$;

$g\cdot(h\cdot x)=(gh)\cdot x$, for every $g,h\in G, x\in X$.

D2) There is a group homomorphism $\varphi\colon G\to\operatorname{Sym}(X)$.

Claim. D1) $\iff$ D2).

Proof.

D1) $\Longrightarrow$ D2). Let's define $\varphi(g)(x):=g\cdot x$; then (injectivity):

$$\varphi(g)(x)=\varphi(g)(y)\Rightarrow g\cdot x=g\cdot y\Rightarrow g^{-1}\cdot(g\cdot x)=g^{-1}(g\cdot y)\stackrel{\text{(D1)}}{\Longrightarrow}x=y$$

and (surjectivity), for every $y\in X$, $y\stackrel{\text{(D1)}}{=}\varphi(g)(g^{-1}\cdot y)$. Therefore, indeed $\varphi(g)\in\operatorname{Sym}(X)$ for every $g\in G$. Now, for every $g,h\in G,x\in X$:

\begin{alignat}{1}
\varphi(gh)(x) &= (gh)\cdot x \\
&\stackrel{\text{(D1)}}{=} g\cdot(h\cdot x) \\
&= \varphi(g)(\varphi(h)(x)) \\
&= (\varphi(g)\varphi(h))(x) \\
\end{alignat}

whence $\varphi(gh)=\varphi(g)\varphi(h)$.

D2) $\Longrightarrow$ D1). Let's define $g\cdot x:=\varphi(g)(x)$; then, for every $x\in X$:

$$e\cdot x=\varphi(e)(x)\stackrel{\text{(D2)}}{=}Id_X(x)=x$$

Moreover, for every $g,h\in G,x\in X$:

$$(gh)\cdot x =\varphi(gh)(x)\stackrel{\text{(D2)}}{=}(\varphi(g)\varphi(h))(x)=\varphi(g)(\varphi(h)(x))=g\cdot(h\cdot x)$$

$\Box$

Therefore, feel free to choose any between D1 and D2 as definition of group action.

Title: Group action definition


问:
Showing an elliptic curve has infinitely many points over $\mathbb{Z}_p$
I stumbled upon this question, and I can't think of how to do it, or what kind of results to use. The question is as follows:

Let $$y^2=x^3+ax+b$$ be an elliptic curve ($a,b$ integers), and let $p \geq 5$ be a prime with good reduction. Then I have to show that this curve has infinitely many solutions in $\mathbb{Z}_p$, but I cant think of what results to use to try and prove this, I'm guessing something like Hensel's Lemma might work, but I can't see how. So I was wondering if I could get some hints on this.

Thank you

答:
Here is a solution that works for $p\geq 11$. You will need to modify the proof to make it work for $p=5$ and $p=7$.

Since $p\geq 11$, and $p$ is a prime of good reduction, the Hasse bounds tell us that
$$|E(\mathbb{F}_p)|\geq p+1-2\sqrt{p}\geq 5.$$
Since there are at most $3$ solutions modulo $p$ with $y\equiv 0 \bmod p$, it follows that there is at least one point in $E(\mathbb{F}_p)$ with non-zero $y$ coordinate.

Let $(\alpha,\beta)$ be a point on $E(\mathbb{F}_p)$ with $\beta\not\equiv 0 \bmod p$, and let $\gamma\in \mathbb{Z}_p$ be any $p$-adic integer such that $\gamma\equiv \alpha \bmod p$. Put $f(x)=x^3+ax+b$, and consider the polynomial $q(y)=y^2-f(\gamma)$. Then any $\delta_0\equiv \beta \bmod p$ satisfies $q(\delta_0)=\delta_0^2-f(\gamma)\equiv 0 \bmod p$, and $q'(\delta_0)=2\delta_0\equiv 2\beta\not\equiv 0 \bmod p$. Hence, by Hensel's Lemma $q(y)$ has a root $\delta$ in $\mathbb{Z}_p$, and so $(\gamma,\delta)\in E(\mathbb{Z}_p)$. Since the choice of $\gamma$ was arbitrary among all elements of the form $\alpha+p\mathbb{Z}_p$, it follows that $E(\mathbb{Z}_p)$ is infinite.

PS. One can also do this using formal groups. For instance, Proposition 6.3 of Chapter VII of Silverman's "The Arithmetic of Elliptic Curves" implies that $E(\mathbb{Z}_p)$ is infinite.

ContentLicense: CC BY-SA 3.0


问:
Brussel Sprout Random Variable, Mean, SD, and Probability
A local farmer's market sells 1 pound bags of Brussels Sprouts. The farmer believes that the mean number of Brussels Sprouts in each bag is 18 with a standard deviation of 3.4. You buy 2 pounds of Brussels Sprouts.

a. You want to explore the range of sizes between brussels sprouts. You find the difference between the number of sprouts in your two bags. Define your random variable for this situation, and find its mean and standard deviation.

I'm a bit confused with this question. Would the random variable be the # (number) of brussels sprouts in each bag? Wouldn't the mean be 18 and the standard deviation be 3.4?

b. IF the number of Brussels Sprouts per bag is normally distributed, what is the probability that the difference in number of Brussels Sprouts between two bags is more than 4?

I have no idea how to answer b. Would we make a normal probability distribution and use normalcdf?

答:
If $X_i$ is the number of sprouts in bag $i\in\{1,2\}$, then the random variable representing the difference in the number of sprouts in the two bags is $X_1-X_2$. We are given $X_i\sim N(18,3.4^2)$, so$$E[X_1-X_2]=E[X_1]-E[X_2]=0$$and assuming independence$$V[X_1-X_2]=V[X_1]+V[X_2]=3.4^2+3.4^2$$giving $\sigma_{X_1-X_2}=(3.4)\sqrt2$.

For solving $(b)$, use the fact that the difference of two independent normal random variables $X_{1,2}$ is a normal variable $X$ with expectation $\mu_X=\mu_{X_1}-\mu_{X_2}$ and variance $\sigma_X^2=\sigma_{X_1}^2+\sigma_{X_2}^2$.


问:
When does l'Hospital's rule work for series?
This question stems from a response a colleague of mine received as a (flawed) solution to a problem on his calculus exam. The student was to determine convergence of a series of the form $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$. The student instead considered the series  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ and thought that this series and the original must behave the same way. This, of course, is false in general.

What conditions on $f$ and $g$ are required so that the series  $\displaystyle\sum_{n=1}^\infty \frac{f(n)}{g(n)}$ converges iff  $\displaystyle\sum_{n=1}^\infty \frac{f'(n)}{g'(n)}$ converges?

答:
I'm the guy who had the confused student.

In my original posing of the question, I ask for $f'$ and $g'$ to be monotonic, not just $f$ and $g$.  Are you still able to create a counterexample?

Someone mentioned usefulness.  If something like this worked, I'd have students apply it to series like n/(n^3-1).  The Comparison Test is awkward to use correctly.  The Limit Comparison Test is great for this one, but it's nice to have options.

Maybe the list of conditions to make this conjecture true becomes too lengthy to bother with.  But I have yet to see a counterexample to these conditions:

$f'$ and $g'$ are monotonic.

$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=0$

$f$ and $g$ have "L'Hopital conditions": $\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}g(x)=0$ or $\pm\infty$.


问:
Frobenius coin problem
Suppose that you only have coins worth, say 3 and 5 euros. According to Sylvester result we can find the Frobenius nr $g(3,5)=15-3-5=7$ so 7 is the largest integer that cannot be written as $a_{1}k_{1}+a_{2}k_{2}$ for $k_{1},k_{2}\in\mathbb{N}$ and $a_{1},a_{2}$ are the values of these coins.

a) how do you pay 8€,9€ and 10€ with these coins?

b)use a) to show that it is possible to pay all amounts that are greater than 10€ with the coins 3€ and 5€.

c) show that it is impossible to pay the amount of 7€ with these coins.

I am afraid I do not understand 100% the whole idea behind the Frobenius numbers.

a) can we just take 3€+5€=8€ and 3€+3€+3€=9€ and 5€+5€=10€ this seems suspicious of how easy it is....

b)do I have to use both coins? or just 3€ or 5€?
11€=3€+5€+3€

12€=3€+3€+3€+3€

13€=3€+5€+5€

14€=5€+3€+3€+3€

.

.

c) if we could pay 7€ with these coins we could have written

$7€=k_{1}5€+k_{2}3€$ but this is impossible as $k_{1},k_{2}\in\mathbb{N}$

can someone please explain to me what should be done in this exercise and how?

答:
For part c, consider the three cases:

No €5 coin is used

Exactly one €5 coin is used

At least two €5 coins are used

In each  case, you can easily show that the total can't be €7.

Title: Frobenius coin problem


答: Divide  numerator and denominator by $3^{n}$ and take the limit. The $\pm $ signs won't matter now and you get the limit as $\frac {0+1} {0+3}= \frac 1 {3}$.

Title: Find limit with alternating sign

Body:
Please, help proceed further

$\lim_{n \to +\infty}\frac{(-2)^n+3^n}{(-2)^{n+1}+3^{n+1}}=\lim_{n \to
+\infty}\frac{(-2)^n}{(-2)^{n+1}+3^{n+1}}+\lim_{n \to
+\infty}\frac{3^n}{(-2)^{n+1}+3^{n+1}}=\lim_{n \to
+\infty}\frac{1}{-2+3*(-1.5)^{n}}+\lim_{n \to +\infty}\frac{1}{(-2)*{(\frac{-2}{3})}^{n}+3}$


答:
What you are seeing here is a special case of the very general concept of currying: For three sets $A, B, C$ there is a natural bijection between maps $A\times B\to C$ and maps $A\to \operatorname{Maps}(B,C)$, where $\operatorname{Maps}(B,C)$ denotes the set of maps $B\to C$.

The bijection is rather trivial and works as follows: Given a map $f\colon A\times B\to C$ we can define for each $a\in A$ the map $f_a\colon B\to C$ given by $f_a(b)=f(a,b)$. This defines a map $A\to\operatorname{Maps}(B,C)$ by $a\mapsto f_a$. Conversely, from a map $g\colon A\to\operatorname{Maps}(B,C)$ you can always define a map $A\times B\to C$ by $(a,b)\mapsto \big(g(a)\big)(b)$.

These two constructions are inverse to each other and hence establish a bijection between maps $A\times B\to C$ and maps $A\to\operatorname{Maps}(B,C)$.

In your case $A=G$ and $B=C=X$ so that maps $G\times X\to X$ correspond to maps $G\to\operatorname{Maps}(X,X)$. However, you are only interested in the subset of those maps that actually map into $\operatorname{Sym}(X)$ (which is the subset of $\operatorname{Maps}(X,X)$ consisting of only the bijections) and have the property of being a group homomorphism. In the setting of maps $G\times X\to X, (g,x)\mapsto g\cdot x$, these criteria translate to the group action axioms $1\cdot x=x$ and $(gh)\cdot x=g\cdot(h\cdot x)$ for all $g,h\in G$ and $x\in X$.

Title: Group action definition

Body:
The Wikipedia for group action and several posts here have what seems to me to be two, slightly different definitions for a group action.$\ $   Let G be a group acting on space X.$\ $  These two definitions are:

A group homomorphism $\varphi$ from the group to the automorphism group of the space.
$$\varphi : G\ \longrightarrow \text{Aut}(X)$$

A map that takes an element of the group and an element of the space, and returns an element of the space.
$$\alpha : G\, \times X\ \longrightarrow X$$

Which of these is the the group action?$\ $   Is it a function of one argument or two?$\ $  What is the return type?  Does it return a function (automorphism) or an element of the space?$\ $ The first definition is in line with what I think of as a representation, ex. from a symmetry group C$_\text{2v} \longrightarrow $ GL(V) that returns a matrix, rather than a vector.

I (think) I understand the basic idea of what is going on, but I'd like to know what the actual mathematical definition is.

Thanks in advance.


问:
Differentiate $\frac{e^{-2x}}{\sqrt x}$

Differentiate, with respect to $x$, $\frac{e^{-2x}}{\sqrt x}$.

I  a having difficulties with differentiation. The answer is

$$-\frac{e^{-2x}(4x+1)}{2x\sqrt{x}}$$

But I am looking at my working out and can't seem to solve the question. This is what I did:

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\times \frac{d}{dx}(e^{-2x})-e^{-2x}\times \frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2} = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}\frac12(\sqrt{x})^{-1/2}}{x}$$
$$ = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}}{2x\sqrt{x}}$$

image of my work

If someone could please help me solve this question or give advise please help me! Thank You!

答:
We have $f(x)=\dfrac{e^{-2x}}{\sqrt x}$.
Use Quotient rule as you suggested in your answer too:
$$f'(x)=\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}$$
Now multiply the fraction by $\frac{2\sqrt x}{2\sqrt x}$ to get rid of $\frac{-1}{2\sqrt x}$ in numerator:

$$\frac{-2e^{-2x}\sqrt x -\frac{1}{2\sqrt x}\times e^{-2x}}{x}\times\frac{2\sqrt x}{2\sqrt x}=\frac{-4e^{-2x}\times x-e^{-2x} }{2x\sqrt x}=-\frac{e^{-2x}(4x+1)}{2x\sqrt x}$$

Title: Differentiate $\frac{e^{-2x}}{\sqrt x}$


问:
How can I solve this equation involving a distribution?
I'm trying to find all distributions $u \in \mathcal{D'}(\mathbb{R})$ such that $$(x^2+1)u = \delta$$ and I'm having some trouble. I can see that since $x \mapsto \frac{1}{x^2+1}$ is smooth, we have a particular solution given by $u = \frac{1}{x^2+1} \delta$. So then I look for solutions to the homogeneous equation $(x^2+1)u =0$ to add to this, and this is what I can't do. I know that if $x^2 u = 0$ then $x = c_0 \delta + c_1 \partial \delta$ for some constants $c_0$ and $c_1$ but I don't know how to apply this in this case. Any help appreciated!

答:
Since $f\delta=f(0)\delta$, the solution $u = \frac{1}{x^2+1} \delta$ reduces to $u=\delta$.

There are no other solutions. The homogeneous equation $(x^2+1)u=0$ is equivalent with $u=0$ since $x^2+1\neq 0$ for all $x$.


问:
Find the radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

Radius of convergence of $\sum_{n=1}^\infty (\frac{5n^3+2n^2}{3n^3+2n})z^n$

I know Hadamards lemma says that $\frac{1}{R}=\lim\sup \vert a_n\vert^{1/n}$

But I'm not sure how to compute $\lim\sup \vert a_n\vert^{1/n}$

I can see the $(a_n)\to 5/3$, so I believe $\lim\sup \vert a_n\vert$ should just be the limit of the sequence? But I'm not sure how to actually use the lemma to find $R$.

答:
Since $a_n \to 5/3$, there is $N \in \mathbb N$ such that

$$1 \le a_n \le 2$$

for $n>N.$ Hence

$$1 \le a_n^{1/n} \le 2^{1/n}$$

for $n>N.$  This gives

$$ a_n^{1/n} \to 1.$$


问:
Is it possible that each coordinate is independent for von Mises Fisher distribution?
Let $(X_1,...,X_n)$ be an $n$-dimensional random vector whose distribution is a von Mises Fisher distribution on $S^{n+1}$

Is it possible that $X_1,...,X_{n-1}$ are independent?

答:
Presumably you meant to write

For $n > 1$, let $(X_1,...,X_n)$ be an $n$-dimensional random vector whose distribution is a von Mises Fisher distribution on $S^{n-1}$

Is it possible that $X_1,...,X_{n}$ are independent?

To see that $X_1,...,X_{n}$ are not independent, note that
$P\Bigl(x_2 > \frac{3}{4}\Bigr) > 0$,
but
$P\Bigl(x_2 > \frac{3}{4}{\,{\Large{\mid\,}}}x_1 > \frac{3}{4}\Bigr)=0$.


问:
Probability of conditional complement
I know that $$P(A^{C}|B) = 1 - P(A|B)$$ so I was wondering if can we also say that $P(B^{C}|A) = 1 - P(B|A)$? Can we just swap the terms like that?

Thanks

答: If you know the first equality for all $A$ and $B$ you can surely interchange $A$ and $B$. Here is the actual proof: $P(A)=P(A\cap B)+P(A\setminus B)$. Dividing by $P(A)$ this gives $1=P(B|A) +P(B^{c}|A)$. Hence $P(B^{c}|A)=1-P(B|A)$. Of course, you have to assume that $P(A)>0$.

Title: Probability of conditional complement


问:
There are infinitely many choices of $(\alpha_1,\dots,\alpha_n)$ such that $f(\alpha_1,\dots,\alpha_n)\neq 0$
I'm trying to solve this exercise in the page 10 of this book

Maybe I'm forgetting something, but I couldn't solve this exercise, I need a hint or something to begin to solve this question.

Thanks in advance

答: Hint: the case $\,n=1\,$ is clear, by a polynomial $\neq 0\,$ over a field has no more roots than its degree. For $\,n>1,\,$ write $f = c_k\, x_n^k + \cdots + c_0,\ c_i \in K[x_1,\ldots,x_{n-1}],\ c_k\neq 0.\,$ By induction we can specialize $\,(x_1,\ldots,x_{n-1}) = (a_1,\ldots,a_{n-1})\in (K \setminus \Lambda)^{n-1}$ so  the leading coeff $\,c_k\,$ stays $\ne 0,\,$ yielding $\, 0\ne g(x_n) = f(a_1,\ldots,a_{n-1},x_n) \in K[x_n].\,$ By induction there are infinitely many $\, a_n \in K \setminus \Lambda\,$ such that $\,g(a_n) = f(a_1,\ldots,a_n) \ne 0.\ \ $ QED


问:
$u(x,t)\cdot(\nabla\rho(x,t))$ versus $(u(x,t)\cdot\nabla)\rho(x,t)).$
I am new in the Navier Stokes Equation

In the Eulerian description of the fluid, if I have $g(t)=\rho(x(t),t)$, then:
$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x}(x(t),t)x'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t)\cdot(\nabla\rho(x,t)).$$

$u(x,t)=x'(t)$ is the velocity field of the flow of the fluid, $\rho:\Omega×[0,T] \rightarrow \mathbb{R}$ is the mass density, and $\nabla$ is the nabla operator.

MY QUESTION IS how did
$u(x,t)\cdot (\nabla\rho(x,t))$ become $(u(x,t)\cdot\nabla)\rho(x,t))?$

I searched on internet and they said that:

Generally the convective derivative of the field u·∇y, the one that contains the covariant derivative of the field, can be interpreted both as involving the streamline tensor derivative of the field u·(∇y), or as involving the streamline directional derivative of the field (u·∇) y, leading to the same result.

BUT I don't know why and how did this lead to the same result!!

答:
This is just because $\rho$ is a scalar function (i.e. takes values in $\mathbb R$). If you check
\begin{align}
(\nabla \rho )_i &= \partial_i \rho,
\\
u\cdot\nabla &=\sum_i u_i \partial_i,
\end{align}
so
$$u\cdot (\nabla \rho) = \sum_i u_i \partial_i \rho = \sum_i(u_i \partial_i)\rho = (u\cdot\nabla) \rho.$$

The situation is slightly different if you were computing $g'$ for $g(t) = v(x(t),t)\in\mathbb R^n$. Then, you might wonder, what exactly is the meaning of
$$ u \cdot (\nabla v)?$$
In this case, $u$ is a vector, but $\nabla v$ is a matrix. As a matrix product, we have
$$ (u\cdot\nabla)v = (\nabla v)u.$$
The computation: writing components out in full (and using chain rule)
\begin{align} \frac d{dt} g_i'(t)
&= \partial_t v_i(x,t) + \sum_j\partial_jv_i(x,t)\partial_t x_j
\\&= \partial_t v_i(x,t) + \sum_j\partial_jv_i(x,t)u_j(x,t)
\\&= \partial_t v_i(x,t) + \sum_ju_j(x,t)\partial_jv_i(x,t) \end{align}
In the last line you see exactly $\sum_ju_j(x,t)\partial_j =: u\cdot\nabla $ appear, and in the second last line, you see $\nabla v=( \partial_j v_i)_{i,j}$ multiplied on the right by the vector $u$. The two lines are equivalent because we wrote all the components out and at the level of scalars, everything commutes.

IMO $u\cdot(\nabla v)$ is dangerous notation in this case, it does not match the matrix product $u^T \nabla v$. Somehow it looks like you should be able to relate this with doing a "contraction" with the "left index" i.e. $a\cdot M := a_i M_{ij}$, but this is wrong because the Jacobian $(\nabla v)_{ij} = \delta_{\color{red} j} v_{\color{red} i}$ is the other way.


问:
Where to start when learning math (again)?
I have a few questions I hope you can help me answer.

First, I'll introduce myself. I'm a finance undergraduate student in Australia, but I'm originally from Norway. Throughout school I always loved math, but I ended up studying finance. The last year or so I have started to realise that I should have done Computer Science or Engineering in stead, as I would like to see myself in a quant role when I finish studying. Even so, I've decided to finish this finance degree.

The last month or so I've started programming in C++, and refreshing up on my maths knowledge (Khan Academy, algebra, precalculus). My problem is I don't know where to go from here, and what order I should be learning the different branches of math.

I've picked up a book on linear algebra today; "Linear algebra and its applications" by Gilbert Strang. Do you think this is an alright place to start? Or do I start with basic calculus? And where can I go next? Differential equations?

I hope some of you can help me in the right direction. Thanks.

答:
I answered a similar question recently:

Which topics of mathematics should I study?

However, this might be more on the theoretical side and I'm not sure if that is what you'd like.

Hope this helps.


Title: Differentiate $\frac{e^{-2x}}{\sqrt x}$

Body:
Differentiate, with respect to $x$, $\frac{e^{-2x}}{\sqrt x}$.

I  a having difficulties with differentiation. The answer is

$$-\frac{e^{-2x}(4x+1)}{2x\sqrt{x}}$$

But I am looking at my working out and can't seem to solve the question. This is what I did:

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\times \frac{d}{dx}(e^{-2x})-e^{-2x}\times \frac{d}{dx}(\sqrt{x})}{(\sqrt{x})^2} = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}\frac12(\sqrt{x})^{-1/2}}{x}$$
$$ = \frac{\sqrt{x}(-2e^{2x})-e^{-2x}}{2x\sqrt{x}}$$

image of my work

If someone could please help me solve this question or give advise please help me! Thank You!

Body:
Note that $\frac{1}{\sqrt x} = x^{-1/2}$. Now use the product rule:

$$\frac{d}{dx} \left(e^{-2x} \cdot x^{-1/2}  \right)= e^{-2x} \cdot -\frac{1}{2}x^{-3/2} + (e^{-2x} \cdot -2) \cdot x^{-1/2}$$

which is already correct, but can be written in a nicer form:
$$=e^{-2x} \left(-\frac{1}{2}x^{-3/2} - 2x^{-1/2} \right)=e^{-2x} x^{-3/2} \left(-\frac{1}{2} - 2x \right)= -\frac{1/2 + 2x}{e^{2x} x^{3/2}} = -\frac{1+4x}{2e^{2x} x^{3/2}}$$


问:
Supremum norm for the set of continuous functions

Let $D \ne \emptyset$ and $(X,d)$ be a metric space. Let $Z =(D,X)$, that is the set of all $\textbf{bounded}$ continuous functions $f:D \to X$. Now $Z$ isn't a vector space, but we can  define the sup metric  $e$ for $Z$ by setting $$e(f,g) = \sup_{x\in D} d(f(x),d(g(x)).$$

This was a definition on a book I'm reading without a proof and I'm trying to figure out if this satisfies the conditions for a metric space. If it does it should satisfy $$d(x,y) \geqslant 0 \text{ with equality iff $x=y$} \\ d(x,y) = d(y,x) \\ d(x,y) \leqslant d(x,z) + d(z,y)$$

I'm slightly confused about how to show these. For the first one I would want to show that $\sup_{x\in D} d(f(x),d(g(x)) \geqslant 0$, which seems intuitive that the distance between two functions wouldn't be negative, but I'm not sure how to formalize this. Also the second one seems self explanatory that the distance between $f$ and $g$ is the same as the distance between $g$ and $f$. If anyone happens to know where this would be proven I would be appreciate a link.

答:
Your arguments are superficial. You say for example that symetry is plausible because we are talking about a distance. Don't forget: you are trying to prove that $e$ is a metric, don't use it as an assumption!

For instance, the first one goes as follows: for every $x\in D$, we have $d(f(x),f(g))\geq 0$, because $d$ is a metric on $X$. The supremum of a set of non-negative numbers is non-negative (or infinity...). Moreover, it is zero iff all numbers in the set whose supremum is considered are 0. So indeed, $e(f,g)\geq 0$ (including the possibility that it is infinite). Equality holds iff $d(f(x),f(y))=0$ for all $x\in D$, that is, iff $f(x)=g(x)$ for all $x\in D$, that is, iff $f=g$.

As @Kavi Rama Murthy  already pointed out, the statement itself is false in general, because of the possibility of infinite distances. This is usually fixed by restricting the attention to bounded (continuous) functions. For example, if $D$ is compact, then continuous functions are bounded, so the finiteness of the supremum metric is guaranteed.


问:
Finding optimal strategies of matrix game
So I'm trying to wrap my head around the basics of game theory, and would like to know if I argue correctly:

Say I have the game

$$
A = \begin{pmatrix}
0 & 0 & 0 & 0  \\
0 & -1 & 2 & 1  \\
0 & 1 & -2 & -1  \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
and that I would like to find optimal strategies for this game. Suppose that player $1$ plays the mixed strategy $(p_1,p_2,p_3,p_4)$. If $p_2$ and $p_3$ are greater than $0$, player $1$ is not guaranteed to receive a non-negative value. However, if they are zero, then player $1$ is guaranteed to receive $0$. This means that optimal strategies for player $1$ are of the form $(u, 0, 0, 1-u)$, where $u \in [0,1]$. In the same way one would find that the only optimal strategy for player 2 is $(1,0,0,0)$.

Is this correct?

答:
Your solution is correct, since the elements $ a_ {11} = 0 $ and $ a_ {41} = 0 $ are both saddle points of the matrix, so your strategies are optimal solutions.

However, I am not entirely sure of your reasoning. A priori it is not enough that the row player is guaranteed a non-negative payout. He would have to get the biggest payout possible to accept his strategy. Certainly in this case that highest possible payout is zero, since that is the value of the game according to the saddle points. Maybe that's what you had in mind, and in that case everything would be fine. But you should probably justify that this is the highest payout that the row player can guarantee (saddle points argument again).


问:
Is $f(n)=\frac{(-1)^n (2n-1)+1}{4}$ surjective?
We have $f:\Bbb{N}\to\Bbb{Z}$. By definition this map is surjective, if $\forall z\in\Bbb{Z}\exists n\in\Bbb{N}$ such that $f(n)=z$. The direct proof means I need to show the equation $$f(n)=\frac{(-1)^n (2n-1)+1}{4}=z$$

has a solution for each $z$, which I don't think is possible. Assuming the contradiction didn't help either? Ideas? I strongly believe this map is surjective.

答: When $z \geq 1$ we have $f(2z)=z$. Also $f(2z+1)=-z$ for $ z  \geq 1$.  And $f(1)=0$ so $f$ is surjective.


问:
dot product in a general coordinate system
NOTE: Einstein summation is assumed. Also, the component expanded vector in the standard basis will be expressed with square brackets and in the curvilinear basis with round brackets. E.g, in the case of plane polar coordinates,
$$\begin{bmatrix}
1\\
\sqrt{3}
\end{bmatrix} =\begin{pmatrix}
2\\
\pi /3
\end{pmatrix}$$

Links to similar questions: 1 and 2. The reason I'm asking this question is because the answers given in these posts were pretty unsatisfactory - I don't want to assume the coordinates are orthogonal and I'm ok with the result being somewhat circular in nature. All I'm looking for is a nice succinct formula that gives correct results.

I'm currently studying differential geometry. I'm sure we are all aware of the formula for the inner product of two vectors in the standard basis
$$\langle\underline{u},\underline{v}\rangle_{\text{std.}}=\delta_{ij}u^iv^j$$
However, this is only true if the vectors are expressed in the standard basis. For instance we know that in plane polar coordinates, the dot product is obviously not $r r'+\theta\theta'$. Wikipedia cites the formula
$$\langle\underline{u},\underline{v}\rangle_{\mathscr{C}}=g_{ij}u^iv^j$$
As the inner product for a general coordinate system $\mathscr{C}$.
Now obviously, it shouldn't matter what coordinate basis we take the inner product with respect to. The only reason for the above notation is to emphasize that the form of the inner product will change from one coordinate system to another, but indeed for a particular pair of vectors its value will not.

However when this formula is applied to plane polar coordinates, this gives the strange formula
$$\langle\underline{u},\underline{v}\rangle_{\text{polar}}=u_rv_r+r^2u_\theta v_\theta$$
Which doesn't really make sense. In this context, what even is $r$? In fact, the above formula fails for obvious reasons - the metric tensor varies from point to point, e.g for the vector $[~1~~~1~]^{\mathrm{T}}=(~\sqrt{2}~~~\pi/4~)^{\mathrm{T}}$ the metric tensor is
$$\mathbf{g}(\sqrt{2},\pi/4)=\begin{bmatrix}
\begin{bmatrix}
1 & 0
\end{bmatrix} & \begin{bmatrix}
0 & 2
\end{bmatrix}
\end{bmatrix}$$
Whereas for the vector $[~5~~~12~]^{\mathrm{T}}=(~13~~~\arctan(12/5)~)^{\mathrm{T}}$ it is
$$\mathbf{g}(13,\arctan(12/5))=\begin{bmatrix}
\begin{bmatrix}
1 & 0
\end{bmatrix} & \begin{bmatrix}
0 & 169
\end{bmatrix}
\end{bmatrix}$$
So what are Wiki referring to by $g_{ij}$ in the referenced formula?

But, even if the two vectors have the same metric tensor, i.e, same $r$ component, the formula still fails, as we presumably get
$$\langle(r,\theta_1),(r,\theta_2)\rangle_{\text{polar}}=r^2(1+\theta_1\theta_2)$$
Which is clearly not correct.

So here is my question: Let's suppose we have two vectors in $\mathbb{R}^n$ which in the standard basis are
$$\underline{v}=\begin{bmatrix}
x^{1}\\
\vdots \\
x^{n}
\end{bmatrix} \ ;\ \underline{v'} =\begin{bmatrix}
{x'} ^{1}\\
\vdots \\
{x'} ^{n}
\end{bmatrix}$$
And in some curvilinear basis are
$$\underline{v}=\begin{pmatrix}
q^{1}\\
\vdots \\
q^{n}
\end{pmatrix}~;~\underline{v'}=\begin{pmatrix}
{q'}^{1}\\
\vdots \\
{q'}^{n}
\end{pmatrix}$$
Let's suppose we know the coordinate transformation functions $q^i=q^i(x^1,...,x^n)$ and also the reverse coordinate transformation functions $x^i=x^i(q^1,...,q^n)$ and we are given the explicit form of the metric tensor $\mathbf{g}$ for the coordinate transformation $[~x^1~~~...~~~x^n~]^{\mathrm{T}}\to(~q^1~~~...~~~q^n~)^{\mathrm{T}}$. Given the two vectors in their curvilinear components, is there any way to work out $\langle\underline{v},\underline{v'}\rangle$ without resorting to converting back to the standard basis? That is, is there any magic formula
$$\langle\underline{v},\underline{v'}\rangle_{\mathscr{C}}=f\left((q^1,...,q^n),({q'}^1,...,{q'}^n)\right)$$

答:
Summary:
The magic formula you seek really is that $g(v,w)=g_{ij}v^iw^j$, provided you understand the definition of every symbol appearing in this formula.

Seeing as I wrote the answer in the first link, it seems fitting I should say a few words (which unfortunately aren't as few as I'd hoped). In general if you have an $n$-dimensional (pseudo) Riemannian manifold $(M,g)$, then for each $p\in M$, by definition, $g_p$ is an (pseudo)inner product on the tangent space $T_pM$. So, if you have two vectors $v,w\in T_pM$, the quantity $g_p(v,w)$ is an element of $\Bbb{R}$. Now, if you have a chart $(U,x)$ on the manifold (i.e a coordinate system), then for each $p\in U$, we can consider the chart-induced basis $\left\{\frac{\partial}{\partial x^i}(p)\right\}_{i=1}^n$ of $T_pM$. So, if $v,w\in T_pM$ we can write these vectors as linear combinations of the basis vectors as
\begin{align}
v&= v^i\frac{\partial}{\partial x^i}(p) \quad \text{and} \quad w= w^j\frac{\partial}{\partial x^j}(p)
\end{align}
for some $v^1,\dots v^n,w^1,\dots, w^n\in\Bbb{R}$
(in fact $v^i= dx^i(p)[v]$; i.e the evaluation of the covector $dx^i(p)\in T_p^*M$ on the vector $v\in T_pM$). So, now if we want to calculate their inner product, we have by bilinearity,
\begin{align}
g_p(v,w)&=g_p\left(v^i\frac{\partial}{\partial x^i}(p), w^j\frac{\partial}{\partial x^j}(p)\right)\\
&= g_p\left(\frac{\partial}{\partial x^i}(p),\frac{\partial}{\partial x^j}(p)\right)\cdot v^i w^j
\end{align}
This is clearly alot to write, so we write the last line as $g_{(x)\, ij}(p)$ to indicate that it's the $i,j$ component of $g$ at the point $p$, taken relative to the chart $(U,x)$. It is common to omit the $(x)$, and write this equation as $g_p(v,w)=g_{ij}(p)v^iw^j$, with the understanding that $v=v^i\frac{\partial}{\partial x^i}(p)$ and similarly for $w$.

Also, we usually understand that the point $p$ is always present hence we write this succinctly as $g(v,w)=g_{ij}v^iv^j$; but seeing as your confusion is essentially arising from not carefully distinguishing points of tbe manifold from vectors in the tangent space at that point, I explicitly wrote it all out.

By the way, even if we take $M=\Bbb{R}^n$ to be a trivial manifold you should still be careful to distinguish between points of $M=\Bbb{R}^n$ and vectors in $T_pM=T_p\Bbb{R}^n$. You're right that there is indeed a nice isomorphism $T_p\Bbb{R}^n\cong \Bbb{R}^n$ (where now we think of $\Bbb{R}^n$ as a vector space, not a manifold), but we only get this isomorphism after a choice of chart: the obvious one is the global chart $(\Bbb{R}^n,\text{id}_{\Bbb{R}^n})$. But of course, if you want to use a different chart, then you'll get a different isomorphism. So, blurring the distinction between points of the manifold and vectors in the tangent space (I'd say especially in simple cases like $\Bbb{R}^n$, when you're first learning these things) is never a good idea unless you already know what you're doing (in which case you can of course do whatever you want), or unless you're using the identity chart (i.e cartesian coordinates), because in this case everything will work out nicely though you may not fully appreciate why. So, if you plan on using polar/parabolic/hyperbolic/elliptic/spherical or any other common coordinate systems, then you must be especially careful regarding the distinction (since the $g_{ij}$'s are no longer constant functions).

Let's work through an example to see how the calculations work. Let $M=\Bbb{R}^2$ which we think of as a manifold. Your very first equality $\begin{bmatrix}
1\\
\sqrt{3}
\end{bmatrix} =\begin{pmatrix}
2\\
\pi /3
\end{pmatrix}$ is already riddled with confusion, because you write

Also, the component expanded vector in the standard basis will be expressed with square brackets and in the curvilinear basis with round brackets.

From this it seems that you think this is an equality of vectors in some tangent space of $\Bbb{R}^2$, when in fact it is not. What you mean is you're considering the manifold $M=\Bbb{R}^2$, on which you have the polar coordinate chart $(r,\theta)$ defined on the open set $U:= \Bbb{R}^2\setminus [0,\infty)\times\{0\}$. Note that $r,\theta$ are functions on $U$, i.e $r:U\to \Bbb{R}$ is defined as $r(a,b):= \sqrt{a^2+b^2}$, and $\theta$ is of course given by a piecewise formula which if restricted to the positive quadrant is just $\theta(a,b)= \arctan\left(\frac{b}{a}\right)$.

Here, when I write $(a,b)$, don't think of this as "a vector in the standard/cartesian basis" or whatever. What I mean is $U$ is a (open) subset of $\Bbb{R}^2$, and $\Bbb{R}^2$ is by definition a set whose elements are tuples of real numbers. That's it. So $(a,b)$ is just a particular element (point) of the set $U$, and $r(a,b)$ is the value of the function $r$ on the element $(a,b)\in U=\text{domain}(r)$. So, your very first equality actually means that if you let $p=(1,\sqrt{3})\in U$, then $r(p)=2$ and $\theta(p)=\frac{\pi}{3}$. In particular it DOES NOT represent the relationship between components of a vector in the tangent space.

So, now, let's fix the point $p=(1,\sqrt{3})$ and consider the vectors $v,w\in T_pM=T_p\Bbb{R}^2$ given by
\begin{align}
v&= \frac{\partial}{\partial r}(p)+ \frac{\partial}{\partial \theta}(p) \quad \text{and} \quad w=\frac{\partial}{\partial \theta}(p)
\end{align}
(i.e in the language of what I wrote above, these are the basis vectors of $T_pM$ induced by the polar coordinate chart). i.e if you write $v=v^r\frac{\partial}{\partial r}(p)+v^{\theta}\frac{\partial}{\partial \theta}(p)$, then it should be clear that I'm saying $v^r=v^{\theta}=1$, and likewise that $w^r=0,w^{\theta}=1$. In polar coordinates, the components of the metric tensor are $g_{\text{polar},11}(p)=1$, $g_{\text{polar},22}(p)=[r(p)]^2$ and $g_{\text{polar},12}(p)=g_{\text{polar}21}(p)=0$ i.e we just write $[g]_{\text{polar}}=\text{diag}(1,r^2)$ more succinctly. Then, using the above formula, we have
\begin{align}
g_p(v,w)&= g_{\text{polar}11}(p)v^rw^r + g_{\text{polar}22}(p)v^{\theta}w^{\theta}\\
&= 1\cdot 1\cdot 0 + [r(p)]^2\cdot 1\cdot 1\\
&= [r(p)]^2\\
&= 4.
\end{align}

Just for fun, let's see how things work out if we express $v,w$ in terms of $\frac{\partial}{\partial x}(p)$ and $\frac{\partial}{\partial y}(p)$. So, we have the following "change of basis formula":
\begin{align}
\begin{cases}
\frac{\partial}{\partial r}(p)&= \frac{\partial x}{\partial r}(p)\cdot \frac{\partial}{\partial x}(p) + \frac{\partial y}{\partial r}(p)\cdot \frac{\partial}{\partial y}(p)\\
&= \cos(\theta(p))\frac{\partial}{\partial x}(p)+ \sin(\theta(p))\frac{\partial}{\partial y}(p)\\
&= \frac{1}{2}\frac{\partial}{\partial x}(p) + \frac{\sqrt{3}}{2}\frac{\partial}{\partial y}(p)\\\\
\frac{\partial}{\partial \theta}(p)&= \frac{\partial x}{\partial \theta}(p)\cdot \frac{\partial}{\partial x}(p) + \frac{\partial y}{\partial \theta}(p)\cdot \frac{\partial}{\partial y}(p)\\
&= -r(p)\sin(\theta(p))\frac{\partial}{\partial x}(p) + r(p)\cos(\theta(p))\frac{\partial}{\partial y}(p)\\
&= -\sqrt{3}\frac{\partial}{\partial x}(p) + \frac{\partial}{\partial y}(p)
\end{cases}
\end{align}
So, if you do the substitutions, you should find
\begin{align}
v&= \left(\frac{1}{2}-\sqrt{3}\right)\frac{\partial}{\partial x}(p) +\left(\frac{\sqrt{3}}{2}+1\right)\frac{\partial}{\partial y}(p)
\end{align}
and
\begin{align}
w&= -\sqrt{3}\frac{\partial}{\partial x}(p) + \frac{\partial}{\partial y}(p)
\end{align}
In other words, we have $v^x=\frac{1}{2}-\sqrt{3}, v^y=\frac{\sqrt{3}}{2}+1, w^x=-\sqrt{3}, w^y=1$. Now, since in cartesian coordinates, the components of the metric are just $[g]_{\text{cart}}=\text{diag}(1,1)$ (almost by definition), we have
\begin{align}
g_p(v,w)&= v^xw^x + v^yw^y = 4,
\end{align}
which is precisely what we found above.

I hope this shows you the importance of being careful about what the point $p$ of the manifold is and what the vectors $v,w$ in the tangnet space are, and why it is important not to conflate the two (and also how to do the change of basis within the tangent space).


问:
Ancient Egyptian Method for getting an Area of Quadrilateral
I am currently learning the history of calculus, and I am really curious about the Ancient Egyptian way to get an Area of Quadrilateral. So, I hope I could get some enlightenment.

Here is the Egyptian Way:

Consider Quadrilateral with 4 different lengths a,b,c,d. (a,c are facing each other, and b,d are facing each other.) (Each set of sides may not be parallel.)

They calculated the area as follows,
$$A=\frac{a+c}{2}\times \frac{b+d}{2}$$

However, we know this is wrong, but I want to show if this method is overestimated or underestimated, and why.

P.S. I think they tried this way because getting an Area of rectangular is the easiest way, and they decided to make rectangular with lengths of averages of two opposite sides. And this new area somehow looks similar (close) to the original area.

答:
Let's see on a rhombus how much the area can be off.

Let call $c$ the side of the rhombus. Its area is half the product of the diagonals, so, with Pythagoras' theorem, if the half-diagonals are $a$ and $b$,

$$c^2=a^2+b^2$$

And the area is

$$A=2ab=2a\sqrt{c^2-a^2}$$

While the approximate area is

$$A'=c^2$$

Hence

$$\frac{A}{A'}=2\frac{a}{c}\sqrt{1-\frac{a^2}{c^2}}=2\sqrt{\frac{a^2}{c^2}\left(1-\frac{a^2}{c^2}\right)}=2\sqrt{x(1-x)}$$

with $x=\frac{a^2}{c^2}$.

The last expression is maximum for $x=\frac12$, and equals then $1$.

This ratio tends to $0$ as $\frac{a}{c}$ tends to $0$ or $1$, that is, when the rhombus is nearly flat. One can expect this: the approximate area is constant, while the exact area varies from $0$ to a maximum of $c^2$ obtained for $a=\frac{c}{\sqrt2}$. For the rhombus, the approximate area is then always larger or equal to the true one, and is correct in only one case, i.e. when it's a square.


问:
Show that $S$ is a subspace iff $u+r\cdot v \in S$ for all $u,v \in V$ and $r \in \Bbb F$.
Let $S$ be a subset of a vector space $V$ over a field $\Bbb F$. Show that $S$ is a subspace iff $u+r\cdot v \in S$ for all $u,v \in V$ and $r \in \Bbb F$.

For the right direction i.e. "if" conditions:
Since $S$ is a subspace of $V$ over a field $\Bbb F$, then the vector addition and scalar multiplication in $V$ are holds in $S$. On the otherhand, the ten axioms of vector spaces are holds in $S$. Hence, $u+v \in S$ and $r \cdot v \in S$ for all $u,v \in S$ and for all $r \in \Bbb F$. Thus, $u+r \cdot v \in S$.

For the left direction i.e. "only if" conditions: Let $u,v \in S$ and $r \in \Bbb F$ with $u+r\cdot v \in S$. Since $u$ and $u+r\cdot v$ both are in $S$, then $r \cdot v$ also in $S$. Since $S$ is a subset of $V$, it suffices to show that the additive identity and additive invers exists in $S$. Since $S \ne \emptyset$, then $0 \cdot v = o \in S$, where $o$ is the zero vector in $V$. Next, $(-1) \cdot v = -v \in S$. Hence, $S$ is a subspace.

Am I true?

答:
$S$ is a subspace of $V$ if $S\neq \emptyset$ and $qu+rv\in S$ for each $u, v\in S$, $q, r\in \Bbb F$

With this in mind, $(\implies)$ is trivial, it is the case when $(q, r) =(1, r)$.

For $(\Leftarrow)$, set $v=u$, the condition $u, v\in S \implies u+rv\in S$ therefore gives $(1+r)u\in S$ for each $r\in \Bbb F$. In other words $qu\in S$ for each $q\in \Bbb F$. Then, applying the condition again we see $u, v\in S \implies qu+rv\in S$ is met.


答:
You need to use $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{d}{dx}u-u\frac{d}{dx}v}{v^2}.$$

When this formula is compared to your question, you have $u = e^{-2x}$ and $v=\sqrt{x}.$ Just substitute and simplify.

That is $$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}\frac{de^{-2x}}{dx}-e^{-2x}\frac{d\sqrt{x}}{dx}}{(\sqrt{x})^2}.$$

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{\sqrt{x}e^{-2x}(-2)-e^{-2x}\frac{1}{2\sqrt{x}}}{(\sqrt{x})^2}.$$

After simplification, one should get

$$\frac{d}{dx}\left(\frac{e^{-2x}}{\sqrt{x}}\right) = \frac{-4xe^{-2x}-e^{-2x}}{2x\sqrt{x}} = -\frac{e^{-2x}(4x+1)}{2x\sqrt{x}} $$

Title: Differentiate $\frac{e^{-2x}}{\sqrt x}$


问:
If $\ker f=\ker g$ and $\mbox{im} f=\mbox{im}g$, does it yield $g=\lambda f$ for some $\lambda$?
The question is, really, contained in the title : let $f$ and $g$ be 2 linear functions on $\mathbb R^n$, if we suppose that $\ker f=\ker g$ and $\mbox{im} f=\mbox{im}g$, does it yield $g=\lambda f$ for some $\lambda$? Is it at least true when $n=2$ ?

答:
What about

$$A=\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix} \text{ and } B=\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{pmatrix}?$$

Or $$C=\begin{pmatrix}
1 & 0\\
 0 & 1
\end{pmatrix} \text{ and }
D=\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}$$ in dimension $2$?


答: No. Let $n=2$ and $f$ the diagonl matrix $f=diag(1,1)$ and $g$ the diagonal matrix $g=diag(1,2)$

Title: If $\ker f=\ker g$ and $\mbox{im} f=\mbox{im}g$, does it yield $g=\lambda f$ for some $\lambda$?

Body: The question is, really, contained in the title : let $f$ and $g$ be 2 linear functions on $\mathbb R^n$, if we suppose that $\ker f=\ker g$ and $\mbox{im} f=\mbox{im}g$, does it yield $g=\lambda f$ for some $\lambda$? Is it at least true when $n=2$ ?


答:
No.

If $f$ and $g$ are bijective, then $f$ and $g$ satisfy the above conditions.


问:
Bases with Negative Exponents in Exponential Functions
Is it correct to state that:
$$a^x=\frac{1}{a^{-x}}=\left(\frac{1}{a}\right)^{-x}$$
and:
$$a^{-x}=\frac{1}{a^{x}}=\left(\frac{1}{a}\right)^{x}$$

Even though
$$\left(\frac{1}{a}\right)^{-x}=\frac{1^{-x}}{a^{-x}}$$
and
$$\left(\frac{1}{a}\right)^{x}=\frac{1^x}{a^x}$$
?

Thanks!

答:
Yes, this is true as for any finite value of $m$, we have $1^m=1$, so the latter reduces to
$$\left(\frac1a\right)^{-x}=\frac{1}{a^{-x}}=a^x$$
And by our original identity,
$$\left(\frac1a\right)^{-x}=\frac{1}{\left(\frac1a\right)^x}=\frac{1}{\frac{1}{a^x}}=a^x$$
So, there is no contradiction.

Hope this helps. Ask anything if not clear :)


问:
What does a probability of $1$ mean?
From a textbook on probability on the Law of Large Numbers:

Theorem 3-19 (Law of Large Numbers): Let $X_1,X_2, \ldots , X_n$ be mutually independent random variables (discrete or continuous), each having finite mean  and variance. Then
  if $S_n = X_1 + X_2 +\dots+ X_n$,

$$ \lim_{n \to\infty} P\left(\left|\frac{S_n}{n} - \mu\right| \geq \varepsilon\right) = 0 $$

Since $S_n$ is the arithmetic mean of $X_1,X_2, \ldots , X_n$, this theorem states that the probability of the arithmetic mean $\frac{S_n}{n}$ differing from its expected value $\mu$ by more than $\varepsilon$ approaches zero as $n \to \infty$. A stronger result, which we might expect to be true, is that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $  but this is actually false. However, we can prove that $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $
with probability one.

The only difference between the last sentence and the one before that is the phrase 'with probability one'. What does probability one mean here ? The usual definition is that the event occurs with 100% certainty. If that is the case, why is the original assertion $ \lim_{n \to\infty} \frac{S_n}{n} = \mu $ false ?

答: $S_n$ is itself a random variable. Lets say $\mu$ is 0.5. $S_n = 0$ is a possible event iff $X_i=0$ is possible for every i. That means that $lim_{n\rightarrow ∞}\frac{S_n}{n}$ is also a random variable! It is $\mu$ with probability one, but there are events for which it is not $\mu$!</p>

Title: What does a probability of $1$ mean?


问:
Does laplacian, curl and gradient commute in spherical coordinate?
I just cannot get my head around this seeming contraction:

First, we expect gradient, curl and lapalcian etc. to commute in $R^n$ as $R^n$ is flat space, even when we use spherical coordinate.
However, by inspecting the formula for laplacian and gradient etc. in spherical corrdiantes, it seems that they do not commute. For example, take a function $f(r)$ which only depends on $r$, then
\begin{align}
\Delta \nabla f &= \frac{1}{r^2}\partial_r r^2 \partial_r \partial_r f \boldsymbol{e}_r\\
&=\frac{1}{r^2} (r^2 \partial_r^3 + 2r \partial_r^2) f \boldsymbol{e}_r
\end{align}
while
\begin{align}
 \nabla \Delta f &= \partial_r (\frac{1}{r^2}  \partial_r r^2 \partial_r  f )\boldsymbol{e}_r\\
&=\partial_r (\partial_r^2 + \frac{2}{r}\partial_r) f \boldsymbol{e}_r \\
&= (f\partial_r^3 +\frac{2}{r}\partial_r^2-\frac{2}{r^2}\partial_r)f \boldsymbol{e}_r
\end{align}
and the two differ by a term of $-\frac{2}{r^2}\partial_r$! Similar things happen to laplacian and curl.

Any suggests or comments? Thanks!

答:
The gradient of a radial scalar function $f=f(r)$ in spherical coordinates is
$$ \nabla f(r,\theta)= f'(r)\hat {\mathbf r}(\theta)$$
Note that this is a vector. The vector Laplacian, in general is a slight mouthful- from trusty wikipedia if $\mathbf  F = A_r \hat{\mathbf r} + A_\theta  \hat{\boldsymbol \theta} + A_\varphi  \hat{\boldsymbol \varphi}  $,
$$\Delta \mathbf F = \left(\Delta A_{r}-\frac{2 A_{r}}{r^{2}}-\frac{2}{r^{2} \sin \theta} \frac{\partial\left(A_{\theta} \sin \theta\right)}{\partial \theta}-\frac{2}{r^{2} \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\right) \hat{\mathbf{r}}
\\+\left(\Delta A_{\theta}-\frac{A_{\theta}}{r^{2} \sin ^{2} \theta}+\frac{2}{r^{2}} \frac{\partial A_{r}}{\partial \theta}-\frac{2 \cos \theta}{r^{2} \sin ^{2} \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\right) \hat{\boldsymbol{\theta}}\\+\left(\Delta A_{\varphi}-\frac{A_{\varphi}}{r^{2} \sin ^{2} \theta}+\frac{2}{r^{2} \sin \theta} \frac{\partial A_{r}}{\partial \varphi}+\frac{2 \cos \theta}{r^{2} \sin ^{2} \theta} \frac{\partial A_{\theta}}{\partial \varphi}\right) \hat{\boldsymbol{\varphi}}$$
In our case $A_\theta = A_\varphi = 0$, and $A_r$ is a function of $r$, leading to the simplification
$$ \Delta \nabla f = \left(\Delta f'(r)-\frac{2 f'(r)}{r^{2}}\right) \hat{\mathbf{r}}(\theta) = \left( f'''(r)+\frac{2 f''(r)}r -\frac{2 f'(r)}{r^{2}}\right) \hat{\mathbf{r}}(\theta)
$$
where we used the scalar Laplacian $\Delta g(r)=\frac1{r^2}(r^2 g'(r))' = g''(r) + \frac{2}{r}g'(r)$.

Lets start from the other side: the scalar Laplacian of $f=f(r)$ is as before, the radial scalar function
$$ \Delta f(r)=\frac1{r^2}(r^2 f'(r))' = f''(r) + \frac{2}{r}f'(r)$$
And now take the gradient:
$$ \nabla \Delta f (r)=\left(f'''(r)+\left(\frac2r f'(r)\right)' \right)\hat {\mathbf r}(\theta)$$
Upon expanding $\left(\frac2r f'(r)\right)'=\frac2rf''(r) - \frac{2}{r^2} f'(r)$, we again obtain the same result. Maths is consistent for another day...!


答:
Wikipedia article Quadrilateral, section Area says that the area of a convex figure is always less than or equal to the Egyptian estimate: $$A\le\tfrac 14(a+c)(b+d)$$
Of course, a concave one has even less area.

Title: Ancient Egyptian Method for getting an Area of Quadrilateral

Body:
I am currently learning the history of calculus, and I am really curious about the Ancient Egyptian way to get an Area of Quadrilateral. So, I hope I could get some enlightenment.

Here is the Egyptian Way:

Consider Quadrilateral with 4 different lengths a,b,c,d. (a,c are facing each other, and b,d are facing each other.) (Each set of sides may not be parallel.)

They calculated the area as follows,
$$A=\frac{a+c}{2}\times \frac{b+d}{2}$$

However, we know this is wrong, but I want to show if this method is overestimated or underestimated, and why.

P.S. I think they tried this way because getting an Area of rectangular is the easiest way, and they decided to make rectangular with lengths of averages of two opposite sides. And this new area somehow looks similar (close) to the original area.


答:
You just have the apply the fact that $d$ is a metric, plus the definition of sup.
You don't need continuity of the functions, just their boundedness.

As $f$ and $g$ are bounded, this means that there are $M,N \in \Bbb R$ such that $f[D]$ is contained in some ball $B(x_1,M)$ and $g[D]$ is contained in some ball $B(x_2,N)$, say.

But then for all $x \in D$ we have $$d(f(x), g(x)) \le d(f(x), x_1) + d(x_1, x_2) + d(x_2, g(x)) < M+ d(x_1,x_2) + N$$

so that the set $\{d(f(x), g(x)\mid x \in D\}$ is bounded above in $\Bbb R$ (the right hand side is fixed and does not depend on $x$).

So $e(f,g)$ is well-defined, as in $\Bbb R$ every set that is bounded above has a supremum in $\Bbb R$. For any fixed $x\in D$ we have $0 \le d(f(x), g(x)) \le e(f,g)$ (as a sup is an upperbound for all values and $d$ is a metric) so $e(f,g) \ge 0$ follows immediately.

Suppose that $f \neq g$. Then for some $x \in D$ we have that $f(x) \neq g(x)$; this is what equality of functions means. But then for that $x$, $0 < d(f(x), g(x)) \le e(f,g)$ as $d$ is a metric and so $e(f,g)>0$ for distinct functions, which is what we had to prove (in contrapositive form).

Now if $f,g \in Z$ for all $x$ we have $d(f(x), g(x)) = d(g(x), f(x))$ as $d$ is a metric. So the sets $\{d(f(x), g(x)) \mid x \in X\}$ and $\{d(g(x), f(x)) \mid x \in X\}$ are exactly the same set, so their sups, resp $e(f,g)$ and $e(g,f)$ are the same too. This shows symmetry.

Finally, for the triangle inequality, let $f,g,h \in Z$.

For any fixed $x \in D$ we have

$$d(f(x), h(x)) \le d(f(x), g(x)) + d(g(x), h(x)) \le e(f,g) + e(g,h)$$

using that $d$ obeys the triangle inequality and the sups are upperbounds. So the real number $e(f,g)+e(g,h)$ is an upperbound for the set $\{d(f(x), h(x)) \mid x \in X\}$ and $e(f,h)$ is the smallest upperbound of that same set so $e(f,h) \le e(f,g)+e(g,h)$ follows and we're done.

Title: Supremum norm for the set of continuous functions

Body:
Let $D \ne \emptyset$ and $(X,d)$ be a metric space. Let $Z =(D,X)$, that is the set of all $\textbf{bounded}$ continuous functions $f:D \to X$. Now $Z$ isn't a vector space, but we can  define the sup metric  $e$ for $Z$ by setting $$e(f,g) = \sup_{x\in D} d(f(x),d(g(x)).$$

This was a definition on a book I'm reading without a proof and I'm trying to figure out if this satisfies the conditions for a metric space. If it does it should satisfy $$d(x,y) \geqslant 0 \text{ with equality iff $x=y$} \\ d(x,y) = d(y,x) \\ d(x,y) \leqslant d(x,z) + d(z,y)$$

I'm slightly confused about how to show these. For the first one I would want to show that $\sup_{x\in D} d(f(x),d(g(x)) \geqslant 0$, which seems intuitive that the distance between two functions wouldn't be negative, but I'm not sure how to formalize this. Also the second one seems self explanatory that the distance between $f$ and $g$ is the same as the distance between $g$ and $f$. If anyone happens to know where this would be proven I would be appreciate a link.


问:
Why is taking the derivative inside of this improper integral justified? In proving variance of gaussian random variable.
In deriving the variance of a gaussian random variable $x$, following series of argument is used:

(1):  $var(x) = \int_{-\infty}^{\infty}x^2\frac{1}{\sigma\sqrt{2\pi}}exp(-\frac{x^2}{2\sigma^2})dx$.

(2): The integral on RHS of (1) can be evaluated by differentiating w.r.t. $\sigma$ on the identity

$$
\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}exp(-\frac{x^2}{2\sigma^2})dx = \sigma.
$$

(3): Then, the LHS and RHS become (after differentiating w.r.t. $\sigma$)

$$
 \int_{-\infty}^{\infty}x^2\frac{1}{\sigma^3\sqrt{2\pi}}exp(-\frac{x^2}{2\sigma^2})dx = 1.
$$

(4): Multiplying by $\sigma^2$ on both sides, I can recover $var(x) = \sigma^2$.

The step I am having trouble validating is the LHS from (2) to (3). I think what's going on is that the differentiation is done inside of the integral as follows:
$$
\int_{-\infty}^{\infty}\frac{\partial}{\partial\sigma}\left((\frac{1}{\sqrt{2\pi}}exp(-\frac{x^2}{2\sigma^2})\right)dx = 1.
$$
But I do not know how this is justified. Leibniz's rule does not seem to fit as the integral is improper.

Why and when is this okay?

Need help with multi-variable calculus is related but the reasoning in this answer is not understandable for me. Can I get a better answer?

答: The $\pm\infty$ limits are no problem since you can choose functions $g,\,h$ with$$\int_{-\infty}^\infty f(x)dx=\int_0^1f(g(y))h(y)dy,$$e.g. $g(y):=\ln\frac{y}{1-y},\,h(y):=\frac{1}{y(1-y)}$, so$$\begin{align}\partial_\sigma\int_{-\infty}^\infty f(\sigma;\,x)dx&=\partial_\sigma\int_0^1 f(\sigma;\,g(y))h(y)dy\\&=\int_0^1 \partial_\sigma f(\sigma;\,g(y))h(y)dy\\&=\int_{-\infty}^\infty\partial_\sigma f(\sigma;\,x)dx.\end{align}$$


问:
How do I compute this integral without Γ?
Let $M = \{(x, y) \in \mathbb{R}^2 \mid 0 \leq y \leq \sqrt2, y^2 \leq x \leq 2\}$.
Compute $\iint_M y^3 e^{x^3} \,dx \,dy$

答:
Observe that the integral is

$$\int_0^2e^{x^3}\Bigg[\int_0^{\sqrt{x}}y^3 dy\Bigg]dx=\frac{1}{4}\int_0^2 x^2 e^{x^3}dx$$

which looks like an immediate integral


问:
3D sphere with a hole volume
I am currently studying Calculus 3.
I need to calculate the volume of 3D sphere with radius $r$ and with a hole.
The hole's diameter is $r$ from the sphere and it's locating between sphere center and sphere edge.

Is there any formula for this question?

Any help would be greatly appreciated.

答:
I don't really understand what you mean, but lets say what you want is this (if it's not, ignore this answer)

This PDF is great for answering your question. Click here to access it. http://www.sfu.ca/~adebened/funstuff/sphere_cyl.pdf

Basically if you want to skip all the Math Jargon in the PDF, the volume is calculated by:

$$V = \frac{4}{3}\pi h^3$$

Hopefully this answers your question.

Title: 3D sphere with a hole volume


问:
denumerable set: If A, B are denumerable sets, AXB is a denumberable set. Prove ZXN, ZXZ and QXQ are also denumerable sets.
So I prove $\mathbb{Z}\times \mathbb{N}\sim\mathbb{N}$ part:
$f:\mathbb{Z}\times \mathbb{N} \to \mathbb{N}\times\mathbb{N}$
such that

$$f:(z,n)\mapsto\begin{cases} (2z,n), & z>0 \\ (2(-z)+1,n), & z\leq 0\end{cases} $$

$f: \text{bijective},
\implies (z_1, n_1) \neq (z_2, n_2)
\implies f(z_1, n_1) \neq f(z_2, n_2)
\implies f: \text{injective}$.

$f(\mathbb{Z}\times \mathbb{N}) = \mathbb{N}\times \mathbb{N}
\implies \mathbb{Z}\times \mathbb{N}\sim \mathbb{N}\times \mathbb{N} \wedge \mathbb{N} \times \mathbb{N} \sim \mathbb{N}$

i am not sure how to prove that $\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Q}\times \mathbb{Q}$ are denumerable sets.

Thank you so much for your help.

答:
Let $A,B$ be denumerable sets, and assume $\mathbb{N}\times\mathbb{N}\sim\mathbb{N}$. Then choose bijections $k:\mathbb{N}\times\mathbb{N}\longrightarrow\mathbb{N}$, $h:A\longrightarrow\mathbb{N}$ and $g:B\longrightarrow\mathbb{N}$ be bijections. Then: $$(h,g)\circ k:A\times B\longrightarrow\mathbb{N}$$
Is a bijection.


答:
Connect the vertices $A$ and $C$ to observe the following:
$$ab+cd \geq ab\sin(x)+cd\sin(y)=2\bigtriangleup ABC+2\bigtriangleup ACD=2\cdot Area(ABCD)$$
Likewise: $$bc+ad \geq2\cdot Area(ABCD)$$
So, this method always overestimates. Also, note that the equality occurs only when quadrilateral $ABCD$ is a rectangle.


问:
How to find generators in simplicial cohomology? For example, for $S^1$ and $S^2\dots$
Is there a way to find generators in cohomology groups? Any algorithm or even philosophical remark would be highly appreciated.

For example, I have $S^1\cong K$ represented by $1$-dimensional simplicial complex $K$ in the form of a triangle. Let its vertices be $1,2,3$. I want to find a generator of $H^1(K,\mathbb{Z})\cong \mathbb{Z}$. After some time thinking I decided to check whether cochain dual to edge $\{1,2\}$ (i.e. cochain $\{1,2\}$ taking value $1$ on the edge $\{1,2\}$ and value $0$ on other two edges) is a generator of $H^1(K,\mathbb{Z})$ and was able to prove it directly, that is, I can prove that any element $a\{12\}^*+b\{23\}^*+c\{31\}^*\in H^1(K,\mathbb{Z})$ can be written as $x\{3,1\}^*+d(y)$ by explicitly finding $y\in C^0(K,\mathbb{Z})$ and $x\in \mathbb{Z}$. However, at first I only guessed that $\{1,2\}^*$ is a generator. Is there a way how I could understand it without guessing?

Also, if I have, for example, $S^2\cong L$, where $L$ is a scary looking simplicial complex (imagine boundary of $3$-dimensional polyhedron with triangle faces scary enough to not have a name), then I have no idea what a candidate for generator of $H^2(L,\mathbb{Z})$ looks like. How could I see it in this case? I mean, as dimension increases, difficulty of guessing and direct proof must increase too.

Thank you.

答:
Ok! Well, this was more long-winded than I intended, haha. There's a LOT to cover here, so don't feel bad if it takes a while to sink in! Everything will come with time. For now, though, let's get started!

How do we compute simplicial cohomology? Moreover, when we compute it, what are the generators physically?

The first question won't be too hard to answer -- one of the big draws of cohomology (and part of its great success) is its effectiveness. We really can compute the hell out of these things, and there are computer programs (such as sage) that will happily compute the cohomology of your favorite complex. Indeed, in sage, the following works:

```<code># Let's use a premade complex for simplicity
# you could build your own if you wanted to, though!
T = simplicial_complexes.Torus()

T.cohomology()

# the above outputs: {0: 0, 1: Z x Z, 2: Z}
# this is the reduced cohomology!

# If you want the unreduced cohomology, then run
T.cohomology(reduced=False)

# which outputs {0: Z, 1: Z x Z, 2: Z}
# (as expected)
</code>```

A lot of ink has been spilled (see here and here for instance) about what cohomology is intuitively. I think there is a good answer for manifolds (De Rham Cohomology) but things are less clear for simplicial cohomology.

Here's my take:

Let's look at a simple graph -- say

then 0-cochains are (basically) functions on the vertices. Similarly, 1-cochains are (basically) functions on the edges.

Now a 1-cochain is a coboundary exactly when we can "integrate" along the cochain. That is, exactly when we can find a function on the vertices which is compatible with the edges. Let's see two examples in action:

If we want to "integrate" this, we want to assign integers to $v_0$, $v_1$, and $v_2$ so that the difference along each edge agrees with our function. Of course, it's obvious how we can do this:

$v_0$ can get whatever value it likes ($C$, say)

which forces $v_1$ and $v_2$ to get $C+5$

Luckily, then the difference along $e_{12}$ is $0$. Which is what we needed!

Let's see another example where we're less lucky.

In this case we can try the same procedure. You can convince yourself that no function on the vertices can possibly be compatible with this edge-function.

This element, then, survives to the cohomology. It is a cocycle (since it lives in dimension 1 and there are no simplices of dimension 2 for it to map to) but it is not a coboundary (since there is no vertex function which induces it).

A generator of the cohomology group is some edge-function (or family of edge functions, more generally) with the property that every non-integrable edge function is a linear combination of the generators.

You can see that "locally" we can always solve this problem. Along any given edge it's easy to find a function on the vertices which works with that one edge.

The issue comes from trying to glue these "local" solutions together. It's possible that there is no way to choose the choices in a way that's globally compatible.
As we saw in the above example, we could solve the problem for $e_{02}$ and for $e_{01}$, but this forced us into a corner where $e_{12}$ was unsatisfiable. Notice this issue occurred because of the cycle. It is in this (slightly roundabout) way that cohomology "detects holes".

This is what cohomology does. It measures the ways that "gluing" together locally easy decisions might fail. So then generators for our cohomology group encapsulate all the ways (up to a natural notion of "equivalence", which you should work out!)
an edge-function might fail to be solvable. That is, the generators are "basic" obstructions from which all other unsolvable edge-functions spawn.

Now I'll wave my hands and say the higher dimensional case is analogous. An element of $H^2$ is a function defined on triangles which can't be "integrated" to a function on the edges bounding those triangles, etc.

It's at this point that things begin to get hard to visualize, thankfully the computation remains easy. Let's see why!

For concreteness we'll work with real valued functions. Then we have a vector space of such functions and everything is nice. You can imagine functions valued in an arbitrary ring will work too, at the cost of doing module theory instead of linear algebra (though for nice rings like $\mathbb{Z}$, things are still extremely tractable).

Now let $K$ be a simplicial complex.

Build the chain complex
$$
\cdots \to C_{n+1} \to C_n \to C_{n-1} \to \cdots
$$
where each $C_n$ is free abelian, generated by the $n$-simplices in $K$.

Notice our boundary map $\partial : C_{n+1} \to C_n$ is given by a matrix.
We know what $\partial \sigma$ is for each individual $n+1$-simplex $\sigma$, but since the $\sigma$ form a basis for the space, that's all we need. For the triangle example we've been working with, can you compute $\partial$?
It should be a $3 \times 3$ matrix mapping
$\mathbb{Z}(e_{01}, e_{12}, e_{02}) \to \mathbb{Z}(v_0, v_1, v_2)$.

Dualize, by looking at functions from each of these groups to $\mathbb{R}$.
This gives us a new complex
$$
\cdots \leftarrow
\text{Hom}(C_{n+1}, \mathbb{R}) \leftarrow
\text{Hom}(C_n, \mathbb{R}) \leftarrow
\text{Hom}(C_{n-1}, \mathbb{R}) \leftarrow \cdots
$$

Note, by free-abelian-ness, each $C^n = \text{Hom}(C_n, \mathbb{R})$ is basically the same as looking at functions from individual $n$-simplices to $\mathbb{R}$ (like we were doing in the previous section). You should convince yourself that the coboundary operator for this complex is exactly $\partial^{T}$ (the transpose of the matrix from step 1).

Compute homology of this sequence. That is, look at
$$
\frac{\text{ker}(\partial^T : C^n \to C^{n+1})}{\text{im}(\partial^T : C^{n-1} \to C^{n})}
$$

Of course, computing the kernel of a matrix is easy. Then after we have our two kernels, computing the quotient of two vector spaces is also easy (though regrettably much less frequently taught... See here for a quick description).

But now we're done! The linear algebra that we just did endowed us with a basis for the cohomology group. If you like, you can simply look at the size of the basis and carry on with your life. But as a (fun?) exercise, do you see how to answer your own question from here? The basis elements actually correspond to functions on $n$-simplices! Do you see how?

I hope this helps ^_^

Body:
Ok! Well, this was more long-winded than I intended, haha. There's a LOT to cover here, so don't feel bad if it takes a while to sink in! Everything will come with time. For now, though, let's get started!

How do we compute simplicial cohomology? Moreover, when we compute it, what are the generators physically?

The first question won't be too hard to answer -- one of the big draws of cohomology (and part of its great success) is its effectiveness. We really can compute the hell out of these things, and there are computer programs (such as sage) that will happily compute the cohomology of your favorite complex. Indeed, in sage, the following works:

# Let's use a premade complex for simplicity
# you could build your own if you wanted to, though!
T = simplicial_complexes.Torus()

T.cohomology()

# the above outputs: {0: 0, 1: Z x Z, 2: Z}
# this is the reduced cohomology!

# If you want the unreduced cohomology, then run
T.cohomology(reduced=False)

# which outputs {0: Z, 1: Z x Z, 2: Z}
# (as expected)

A lot of ink has been spilled (see here and here for instance) about what cohomology is intuitively. I think there is a good answer for manifolds (De Rham Cohomology) but things are less clear for simplicial cohomology.

Here's my take:

Let's look at a simple graph -- say

then 0-cochains are (basically) functions on the vertices. Similarly, 1-cochains are (basically) functions on the edges.

Now a 1-cochain is a coboundary exactly when we can "integrate" along the cochain. That is, exactly when we can find a function on the vertices which is compatible with the edges. Let's see two examples in action:

If we want to "integrate" this, we want to assign integers to $v_0$, $v_1$, and $v_2$ so that the difference along each edge agrees with our function. Of course, it's obvious how we can do this:

$v_0$ can get whatever value it likes ($C$, say)

which forces $v_1$ and $v_2$ to get $C+5$

Luckily, then the difference along $e_{12}$ is $0$. Which is what we needed!

Let's see another example where we're less lucky.

In this case we can try the same procedure. You can convince yourself that no function on the vertices can possibly be compatible with this edge-function.

This element, then, survives to the cohomology. It is a cocycle (since it lives in dimension 1 and there are no simplices of dimension 2 for it to map to) but it is not a coboundary (since there is no vertex function which induces it).

A generator of the cohomology group is some edge-function (or family of edge functions, more generally) with the property that every non-integrable edge function is a linear combination of the generators.

You can see that "locally" we can always solve this problem. Along any given edge it's easy to find a function on the vertices which works with that one edge.

The issue comes from trying to glue these "local" solutions together. It's possible that there is no way to choose the choices in a way that's globally compatible.
As we saw in the above example, we could solve the problem for $e_{02}$ and for $e_{01}$, but this forced us into a corner where $e_{12}$ was unsatisfiable. Notice this issue occurred because of the cycle. It is in this (slightly roundabout) way that cohomology "detects holes".

This is what cohomology does. It measures the ways that "gluing" together locally easy decisions might fail. So then generators for our cohomology group encapsulate all the ways (up to a natural notion of "equivalence", which you should work out!)
an edge-function might fail to be solvable. That is, the generators are "basic" obstructions from which all other unsolvable edge-functions spawn.

Now I'll wave my hands and say the higher dimensional case is analogous. An element of $H^2$ is a function defined on triangles which can't be "integrated" to a function on the edges bounding those triangles, etc.

It's at this point that things begin to get hard to visualize, thankfully the computation remains easy. Let's see why!

For concreteness we'll work with real valued functions. Then we have a vector space of such functions and everything is nice. You can imagine functions valued in an arbitrary ring will work too, at the cost of doing module theory instead of linear algebra (though for nice rings like $\mathbb{Z}$, things are still extremely tractable).

Now let $K$ be a simplicial complex.

Build the chain complex
$$
\cdots \to C_{n+1} \to C_n \to C_{n-1} \to \cdots
$$
where each $C_n$ is free abelian, generated by the $n$-simplices in $K$.

Notice our boundary map $\partial : C_{n+1} \to C_n$ is given by a matrix.
We know what $\partial \sigma$ is for each individual $n+1$-simplex $\sigma$, but since the $\sigma$ form a basis for the space, that's all we need. For the triangle example we've been working with, can you compute $\partial$?
It should be a $3 \times 3$ matrix mapping
$\mathbb{Z}(e_{01}, e_{12}, e_{02}) \to \mathbb{Z}(v_0, v_1, v_2)$.

Dualize, by looking at functions from each of these groups to $\mathbb{R}$.
This gives us a new complex
$$
\cdots \leftarrow
\text{Hom}(C_{n+1}, \mathbb{R}) \leftarrow
\text{Hom}(C_n, \mathbb{R}) \leftarrow
\text{Hom}(C_{n-1}, \mathbb{R}) \leftarrow \cdots
$$

Note, by free-abelian-ness, each $C^n = \text{Hom}(C_n, \mathbb{R})$ is basically the same as looking at functions from individual $n$-simplices to $\mathbb{R}$ (like we were doing in the previous section). You should convince yourself that the coboundary operator for this complex is exactly $\partial^{T}$ (the transpose of the matrix from step 1).

Compute homology of this sequence. That is, look at
$$
\frac{\text{ker}(\partial^T : C^n \to C^{n+1})}{\text{im}(\partial^T : C^{n-1} \to C^{n})}
$$

Of course, computing the kernel of a matrix is easy. Then after we have our two kernels, computing the quotient of two vector spaces is also easy (though regrettably much less frequently taught... See here for a quick description).

But now we're done! The linear algebra that we just did endowed us with a basis for the cohomology group. If you like, you can simply look at the size of the basis and carry on with your life. But as a (fun?) exercise, do you see how to answer your own question from here? The basis elements actually correspond to functions on $n$-simplices! Do you see how?

I hope this helps ^_^


问:
Monte-Carlo integration
Let a function $f$ to be $x\in \left[a,b\right],\:0\le f\left(x\right)\le c$.

We want to calculate the approximation of the definite integral of the function in the range $[a,b]$, we can suppose that the exact integral is very difficult to calculate in this range, but we can for all $x$ calculate $f(x)$ easily.

We can sample a lot of points randomly $\left\{X_i,Y_i\right\}\:_{i=1}^N$ from the rectangle in the range: $x\in[a,b], y\in[0,c]$.

First of all, we need to find a way to calculate approximately the integral in the range $[a,b]$

My way:

I succeeded to calculate and this will be $s'$. ($s$ - the original integral, $s'$ - the approximate integral)

$s = \int _a^b\:f\left(x\right)dx$

$s'=c\left(b-a\right)\cdot \frac{1}{n}\sum _{j=1}^n\:I_j$

We need to calculate with $a,b,c,s,\epsilon,\delta$ how many points we need to sample for: $p(|s'-s|>\epsilon)<\delta$ and we need also to be helped by Chebyshev's inequality, but I have no idea how to go on with it.

答:
Since $f$ is bounded, you can bound the variance of your Monte-Carlo estimate, $ I_N = (1/N) \sum_i f(x_i) $:

$ \sigma^2 = Ε(I_N - Ef)^2 \le c^2(b-a)^2/N $

To see this, use the i.i.d. zero mean variable $y_i = f(x_i) - Ef, |y_i| \le c$:

$Ε(I_N - Ef)^2 = (1/N^2)E\Big[ \Big( \sum_i y_i  \Big)^2 \Big] = (1/N^2)E\Big[\sum_i y_i^2 + 2 \sum_{i, j > i} y_iy_j \Big] = (1/N)E[y_1^2]$

You then use this in the Chebyshev approximation:

$ P(|I_N - Ef| > \epsilon) < \sigma^2/\epsilon^2 \le \frac{c^2(b-a)^2}{\epsilon^2N} = \delta$

Title: Monte-Carlo integration


问:
prove that $a_1 + \cdots +a_n \le n^2. $
Let $n$ be a positive integer, and consider a sequence $a_1 , a_2 , \cdots , a_n $ of positive integers. Extend it periodically to an infinite sequence $a_1 , a_2 , \cdots $ by defining $a_{n+i} = a_i $ for all $i \ge 1$. If $a_1 \le a_2 \le \cdots \le a_n \le a_1 +n $  and $a_{a_i } \le n+i-1 \quad\text{for}\quad i=1,2,\cdots, n, $ prove that $a_1 + \cdots +a_n \le n^2. $

Any help? I tried and noticed that for all $i \le a_1$, we have $a_i \le n$. Then for all $i \in \{2, 3, \cdots, a_1\},$ we have $a_{a_i} \le n+i-1$.

答:
Notice that in the range $a_1$ to $a_1+n$ there are $(a_1+n)-(a_1)+1=n+1$ positive integers
Hence from the inequality $a_1\leq a_2 \leq a_3 \leq ... \leq a_n \leq a_1+n$ we can directly conclude that $a_i=a_1+i-1$ for all $1<i\leq n$ and $a_1$ can be any positive integer. Hence we get
$$a_1+a_2+...+a_n=a_1+(a_1+1)+(a_1+2)+...+(a_1+n-1)=na_1+\frac{n(n-1)}{2}$$
Now we are left to show that $a_1+\frac{n-1}{2}\leq n$. Which is true as
$$a_1+\frac{n-1}{2}\leq n$$
$$\frac{n-1}{2}\leq n-a_1\leq n-1$$
Which is definitely true. Hence proved!


问:
What is the name of the theorem of tridiagonal reduction of symmetric matrices?
I have over the last year been implementing a code to solve the many-body Schrödinger equation for applications in Nuclear physics, as part of my PhD-studies. The most important step in that code is the use of the Lanczos algorithm to find the lowest eigenvalues of the Hamiltonian matrix. Like most numerical algorithms for diagonalization of symmetric matrices Lanczos reduces the matrix to a tri-diagonal matrix with the same eigenvalues which then can be diagonalized easily. In a summer course I took back in 2017 a lecturer cited a specific theorem that states that any symmetric (diagonalizable) matrix can be reduced to a tri-diagonal matrix. Unfortunately I did not write that down. Now that I'm writing my dissertation I would like to cite that theorem when I explain the Lanczos algorithm. I have been looking for it in every source I can think of but can't find any mentioning of such a theorem. Every text on numerical symmetric matrix diagonalization seem to take that property of symmetric matrices as obvious and don't cite any theorem. I wonder if anyone here knows the name of that theorem, or a source where it is shown?</p>

答:
No theorem needs to be invoked in order to perform the similarity transformation in the Lanczos algorithm.  It is correct by construction.

In the comments, you cited this link as a reference.  In section 2.1, $α_i$, $β_i$, and $v_i$ are defined by Eq. (1). $T$ is defined as the tridiagonal matrix using $α_i$ and $β_i$. $V$ is defined using $v_i$. When you have defined $T$ and $V$ in this way, if you re-write Eq. (1) as a matrix equation using these matrices, you get Eq. (3). And Eq. (3) is an orthogonal similarity transform.

This is why the algorithm is so brilliant.  If you pick the construction correctly, you automatically get a similarity transformation.

If you still have questions about the algorithm, then the misunderstanding involves one of the steps above, probably the step where you re-write Eq. (1) as a matrix equation.

As an aside, I want to point out that there are important things left to be proved; it's just that they come after Eq. (3). There are at least three things that pop out at me:

Typically we halt the iteration before getting to $n$ and argue that $T$'s extreme eigenvalues and eigenvectors are similar to $A$'s. This needs proof, and I have never read a full proof of this, but I have heard that it is easy for Hermitian matrices and hard for non-Hermitian matrices (Arnoldi iteration).

How do we know that none of the $v_i$ are zero vectors?  (If any of them were zero, then $V$ would not be an orthogonal matrix.  We do, however, know that they are all orthogonal by construction.)

How do we know that the algorithm will be numerically stable in practice? (The naive approach isn’t.)


问:
Random numbers when parameter changes in time
Suppose we want to simulate a poisson stochastic process which happens in time with a fixed rate $\lambda$. We can simulate this process generating the delays between discrete events randomly, sampling from an exponential distribution with fixed parameter $1/\lambda$. Then calculate the time of a given event summing over all the previous delays.

As an example, this is readily done in python with:

```<code>import numpy as np

def make_exp_inf_sequence(lambda= 70):
    x = 0
    while True:
        x = x + np.random.exponential(1/background)
        yield x
</code>```

The following code snippet will sample from a poisson distribution with rate parameter $\lambda$:

```<code>stream = make_exp_inf_sequence()
j = 0
for i in stream:
    j += 1
    if i > 1:
        print(j)
        break
</code>```

Is it possible to extend this technique when the rate parameter change in time $\lambda = \lambda(t)$? What is the correct way to simulate stochastic processes with continuously changing parameters?

Thank you

答:
In general, one can use an acceptance-rejection method. Suppose $\Lambda=(\lambda(t))_{t\geq0}$ is bounded by $\kappa < \infty$.

Initialize with $t=0$.

Set $t \leftarrow t - \frac{1}{\kappa} \log U_1$, where $U_1 \sim \text{Unif}(0,1)$.

If $\lambda(t)/\kappa \geq U_2$, where $U_2 \sim \text{Unif}(0,1)$ independently of $U_1$, then deliver $t$.

Goto step 2.

Classic reference: https://apps.dtic.mil/dtic/tr/fulltext/u2/a059904.pdf.


问:
phase portrait of linear system and normal vector

given the following system
$\left\{\begin{array}
>x_1'=cx_2-bx_3\\x_2'=-cx_1+ax_3\\x_3'=bx_1-ax_2\end{array}\right.$
and we know that $\sqrt{a^2+b^2+c^2}=1$ prove that all the phase portrait of the system are close curves which belongs to parallels planes in $\mathbb{R}^3$and find the normal vector to these plane

I've found that the eigenvalues are $0,\pm i$ but I never delt with 3d phase portrait.

any hint please?

答:
Your system can be written using the cross product as
$$
\dot {\vec v}=\vec v\times \vec n,~~\vec v=\pmatrix{x\\y\\z},~\vec n=\pmatrix{a\\b\\c}.
$$
It immediately follows that the scalar product with the factors has to give zero on the right and thus also on the left, so that
$$
r^2=\vec v·\vec v~\text{ and }~ d=\vec n·\vec v
$$
are constants. The solutions thus lie in the plane given by the second equation, and on the sphere given by the first equation. The intersection of both are circles.


问:
Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$
Let a,b,c be three non-zero real number such that the equation, $\sqrt 3 a\cos x + 2b\sin x = c$, $x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, has two distict roots $\alpha$ and $\beta $ where $\alpha +\beta=\frac{\pi}{6}$. Then find the value of $\frac{b}{a}$.

My approach is as follow

$\sqrt 3 a\cos x + 2b\sin x = c$

$3{a^2}{\cos ^2}x = {c^2} + 4{b^2}{\sin ^2}x - 4bc\sin x$

$ {\sin ^2}x\left( {4{b^2} + 3{a^2}} \right) - 4bc\sin x + {c^2} - 3{a^2} = 0$

Roots are real when $16{b^2}{c^2} + 4\left( {4{b^2} + 3{a^2}} \right)\left( {{c^2} - 3{a^2}} \right) > 0$

$\sin \alpha  + \sin \beta  = \frac{{4bc}}{{4{b^2} + 3{a^2}}};\sin \alpha \sin \beta  = \frac{{{c^2} - 3{a^2}}}{{4{b^2} + 3{a^2}}}$, how I will proceed from here

答:
let $\alpha =A,\beta =B$

as $$c-c=0$$ $$\sqrt{3}a \cos A+2b\sin A-\sqrt{3}a\cos B-2b\sin B=0$$

$$\sqrt{3}a \cos A-\sqrt{3}a\cos B+ 2b\sin A-2b\sin B=0$$

$$-2\sqrt{3}a \sin\left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)+2b\sin \left(\frac{A-B}{2}\right)\cos \left(\frac{A+B}{2}\right)=0$$  Can you end it now? :cancel $\sin (\frac{A-B}{2})$


答:
You can see the equation as defining the intersections of the right half of the unit circle and a straight line. The bisector of the two points makes the angle $\dfrac{\alpha+\beta}2$ with the horizontal axis, and the line is perpendicular to it.

Hence

$$\dfrac{2b}{\sqrt3 a}=\tan\dfrac\pi{12}.$$

Analytical solution:

$$\cos\left(x-\arctan\dfrac{2b}{\sqrt3a}\right)=\frac c{\sqrt{3a^2+4b^2}}$$

and

$$x=\arctan\dfrac{2b}{\sqrt3a}\pm\arccos\frac c{\sqrt{3a^2+4b^2}}.$$

Obviously,

$$\alpha+\beta=2\arctan\dfrac{2b}{\sqrt3a}.$$

Title: Quadratic equation in trigonometric form $\sqrt 3 a\cos x + 2b\sin x = c$

Body:
Let a,b,c be three non-zero real number such that the equation, $\sqrt 3 a\cos x + 2b\sin x = c$, $x \in \left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$, has two distict roots $\alpha$ and $\beta $ where $\alpha +\beta=\frac{\pi}{6}$. Then find the value of $\frac{b}{a}$.

My approach is as follow

$\sqrt 3 a\cos x + 2b\sin x = c$

$3{a^2}{\cos ^2}x = {c^2} + 4{b^2}{\sin ^2}x - 4bc\sin x$

$ {\sin ^2}x\left( {4{b^2} + 3{a^2}} \right) - 4bc\sin x + {c^2} - 3{a^2} = 0$

Roots are real when $16{b^2}{c^2} + 4\left( {4{b^2} + 3{a^2}} \right)\left( {{c^2} - 3{a^2}} \right) > 0$

$\sin \alpha  + \sin \beta  = \frac{{4bc}}{{4{b^2} + 3{a^2}}};\sin \alpha \sin \beta  = \frac{{{c^2} - 3{a^2}}}{{4{b^2} + 3{a^2}}}$, how I will proceed from here


问:
Does gradient descent for linear regression select the minimal norm solution?
I was told that Gradient Descent finds the weights of smallest norm.

This is what I understood in the linear regression setting:

$f_w(x)=w^\top x$ are the linear functions $ \mathbb{R}^n \rightarrow \mathbb{R} $.

$ \mathcal{D}=\{(x_i,y_i) \in \mathbb{R}^d\times\mathbb{R} \mid i=1,...,n\} $ is a dataset

Then I choose some loss function $L$ to get a cost function
$ C_0(w)=\frac{1}{n} \sum_i L(f_w(x_i),y_i) $

Finally, I define a regularized cost $ C_\lambda(w)=C_0(w)+\frac{\lambda}{2n}||w||^2 $

Then there must be some sort of equivalence between (and it looks it should be some classical result):

minimizing $C_\lambda(w)$ for an opportune $ \lambda \ge 0 $

minimizing $C_0(w)$ with Gradient Descent

My questions:

Can somebody clarify a bit this fact? By providing some hypothesis (e.g. on $L$), some insight, some reference...

Does this apply in a similar way to the linear classification setting ($y_i\in\{-1,+1\}$)? In that case, I would expect some equivalence bewteen:

maximizing $M(w)= \min_{i=1...n} y_i \langle w, x_i \rangle $ over $||w||=1$

minimizing $||w||$ over $ y_i \langle w, x_i \rangle \ge 1 $

maybe the previous 1. and 2. points above (those regarding regression)

答:
Gradient descent finds the minimum of any differentiable function, therefore if you use it to minimize $C_0(w)$ (assuming $L$ is differentiable), you will get the weights of the linear regression.

The weight minimization comes about when you add the term $(\lambda/2n)||w||^2$. As $\lambda$ increases, you're trying to minimize the error in the regression by keeping the value of the weights lower than before.

If $||w|| = ||w||_2$ (differentiable), then you can do the minimization with gradient descent.

But you could have $||w|| = ||w||_1$ (which is non-differentiable) and you can't use gradient descent then. Also, in that case the penalty that is used is $\lambda'||w||_1$ (no squaring of the norm).


问:
writing $M : \Gamma_{n,0} \backslash \Gamma_n$
Let $\Gamma_n = \operatorname{Sp}_n(\Bbb Z)$ and $\Gamma_{n,0} \subset \Gamma_n$ be a subgroup. We write
$$ M = \begin{pmatrix}A & B\\ C & D \end{pmatrix} \in \operatorname{Sp}_n(\mathbb{Z})$$
where $M$ is a $(2n)\times(2n)$ matrix and $A,B,C,D$ are $n\times n$ matrices.

I am reading some function theory script and I encountered the writing (definition of the Eisensteinseries for Siegel modular forms)

$$ E_k(Z) := \sum_{M : \Gamma_{n,0} \backslash \Gamma_n} \det(CZ + D)^{-k} . $$

What does $M : \Gamma_{n,0} \backslash \Gamma_n$ mean? What is $M$?

I know "$\backslash$" as a set-substraction but that doesn't seem to make sense here.

It is probably meant as some form of Quotient space but I'm not sure what exactly. Also, I only know "$/$" for the notation of Quotient spaces.

Also, if $\Gamma_{n,0} \backslash \Gamma_n$ means something (a set), why does the author write "$M:$" and not "$M \in$"? (The same author uses "$\in$" everywhere else.)

答:
In the theory of automorphic forms, quotients or coset spaces on both sides arise, and $X\backslash Y$ is written for example when $X$ is a subgroup of $Y$ and we want cosets of the form $Xy$ for $y\in Y$.

The group $Sp_n(\mathbb Z)$ is the group of $2n$-by-$2n$ integer matrices $g$ such that $g^t J_n g=J_n$ where $J_n=\pmatrix{0_n&-1_n\cr 1_n&0_n}$, where $1_n$ and $0_n$ are $n$-by-$n$ identity and $0$ matrices. Here, $\Gamma_n$ is another notation for $Sp_n(\mathbb Z)$, and $\Gamma_{n,0}$ is elements $g=\pmatrix{*&*\cr 0&*}$ therein.

The use of "$M:$" is indeed a bit inconsistent, but not surprising as a descriptor of summation. Here, $M$ is meant to range over representatives for that quotient, with lower $n$-by-$n$ block entries $C,D$.

That quotient has representatives uniquely determined by the lower halves $\pmatrix{C&D}$, and the pair $(C,D)$ is left-modulo $GL_n(\mathbb Z)$.


问:
How is a change of variables performed in Lotka-Volterra normalizations?
How is the change of variables really performed in Lotka-Volterra?

I'm trying to understand what happens in the following document:

https://www.maths.dur.ac.uk/~ktch24/term1Notes(10).pdf, p. 17

I tried plugging the variables in, but I don't see how it produces the result.

答:
Start with the first equation

\begin{align}
\frac{{\rm d}x}{\rm {d}t} &= ax - bxy \\
\frac{{\rm d}(dx/c)}{{\rm d}t} &= a(dx/c) - b(dx/c)y & \text{multiply by } (d/c) \\
\frac{{\rm d}\hat{x}}{{\rm d}t} &= a\hat{x} - b\hat{x}y & \text{define } \hat{x}= dx/c \\
\frac{{\rm d}\hat{x}}{{\rm d}(at)} &= a\hat{x}/a - \hat{x}(by/a) & \text{divide by } a \\
\frac{{\rm d}\hat{x}}{{\rm d}\hat{t}} &= \hat{x} - \hat{x}\hat{y} & \text{define } \hat{t}=at \text{ and } \hat{y}=by/a\\
\end{align}

you can use the same definitions of $\hat{x}$, $\hat{y}$ and $\hat{t}$ for the second equation, will leave that one to you


问:
What formulas are required to calculate a 3d transformation?
Considering the point of view of square A and B, what math tranformations must be applied (either to the 3d camera or world) to transition from A to B?

I can tell that for the B viewpoint I had to move right and up, but I lack the math background to know what formulas can give me accurate values. Thanks

答:
A way to do it is to rotate the cube around the $z$ axis, then around the $x$ axis, with a $\pi/4$ rotation angle.

It means that if we define classically the rotation matrices around these resp. axes as:

$$R_z=\begin{pmatrix}\cos(a)& -\sin(a)& 0\\
   \sin(a)& \cos(a)& 0\\
   0& 0& 1\end{pmatrix}
R_x=\begin{pmatrix}1& 0& 0\\
   0& \cos(b)& -\sin(b)\\
   0& \sin(b)& \cos(b)\end{pmatrix},$$

with $a=b=\pi/4$, the rotation to be applied to the cube is the following product:

$$R=R_x R_z$$

(Recall: matrices must be applied from right to left), giving:

$$R \approx \begin{pmatrix}0.7071  & -0.7071  &     0\\
    0.5000  &  0.5000 &  -0.7071\\
    0.5000  &  0.5000 &   0.7071 \end{pmatrix}$$

But, if we keep $a=\pi/4$ and take different values of $b$, it will make your cube do a kind of curtsy in front of you, helping you to select the most convenient $b$.

A particular case: $b=\operatorname{acos}(1/\sqrt{3})$, gives a position with 3 identical lozenges:

$$R \approx \begin{pmatrix}0.7071  & -0.7071  &       0\\
    0.4082  &  0.4082 &  -0.8165\\
    0.5774  &  0.5774 &    \ \ 0.5774\end{pmatrix}$$


问:
Derivative of the Normal CDF w.r.t. its mean.
I want to know how to take the derivative of a Normal CDF w.r.t its mean using a more general method than using the normalization function.

A Normal CDF is definied by $\Phi(\frac{x-\mu}{\sigma})$. Using chain rule is straight forward to calculated its derivative w.r.t. its mean:
\begin{equation}
\frac{\partial}{\partial \mu} F(x)= \frac{\partial}{\partial \mu} \Phi(\frac{x-\mu}{\sigma})  = - \frac{1}{\sigma} \phi(\frac{x-\mu}{\sigma})
\end{equation}
I would like to get the same result using the proper CDF Function.
\begin{equation}
\frac{\partial}{\partial \mu} F(x)=\frac{\partial}{\partial \mu} \int_{-\infty}^x \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} dx
\end{equation}
MY TRIAL (verified by online calculators):

The bounds of the integral are not a function of $\mu$ so (by Leibiniz Rule) the derivative can go inside:
\begin{equation}
\frac{\partial}{\partial \mu} \int_{-\infty}^x \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} dx=\int_{-\infty}^x  \frac{\partial}{\partial \mu} \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} dx = \int_{-\infty}^x  \frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} · (- \frac{-2x+2\mu}{2\sigma^2} )dx
\end{equation}
Simplifying terms and rearrengin conviniently the last term becomes
\begin{equation}
\frac{1}{\sigma\sqrt{2 \pi}}\int_{-\infty}^x  e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} ·  \frac{x-\mu}{\sigma} \frac{1}{\sigma}  dx
\end{equation}
Using Change of Variable:
\begin{equation}
z=\frac{x-\mu}{\sigma} \Rightarrow \frac{dz}{dx}= \frac{1}{\sigma} \Rightarrow dx= \sigma dz
\end{equation}
\begin{equation}
\frac{1}{\sigma\sqrt{2 \pi}} \int_{-\infty}^x  e^{-\frac{1}{2}z^2} ·  z \frac{1}{\sigma} \sigma dz = \frac{1}{\sigma\sqrt{2 \pi}}\left[-e^{-\frac{1}{2}z^2} \right]^x_{-\infty} =\frac{1}{\sigma\sqrt{2 \pi}}\left[-e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} \right]^x_{-\infty} = -\frac{1}{\sigma\sqrt{2 \pi}}e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2} = -\phi(\frac{x-\mu}{\sigma})
\end{equation}
If we compare it with the inital statement, there is a $1/\sigma$ missing. Introducing the derivative in an online calculator and the result into a calculuator of integrals gives exactly the same result. So I'm going crazy with this.

Thanks in advance!

答:
The last equality in your answer is false.
$$\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2}=\phi(x)$$ so that: $$-\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}=-\frac1{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)$$


问:
In how many ways can two particular players be in the first 5 of the selection list?
The starting 5 in a basketball team is to be picked, in order, from the 10 players in the squad. Determine how many ways can this be done if Jamahl and Anfernee must be in the first 5 players (starting 5)

Textbook's answer: $\mathbb336$

My approach: As it has stated "starting 5", let $P_1$ and $P_2$ be the positions of both of them. However, they have not stated to select the exact position, so there are 2 ways to arrange them in the 2 position.

$P_1$$P_2$

$P_2$$P_1$

The remaining three players can be arranged in $^8P_3$ ways.

The answer:
2 $\cdot$ $^8P_3$ = 336 $\cdot$ 2 = 672

My answer differs as I have included  the additional 2 arrangements of Jamahl and Anfernee, but the question did not stated whether their positions are fixed respectively. I know I am wrong, and it does not mean the textbook is wrong. But, if you could provide me a justification, then it would be appreciated.

答:
My assuption is that: either the book did not consider the positions of Jamahl and Anfernee can be switched and be the same, OR the book assumed that people would not think of the additional 2 arrangements. So the first option is the book is wrong and the latter is the one where you are wrong.

I advise you to seek advice from your teacher and see what they've got to say about that. :)


问:
Check convergence of $\sum_{k=2}^\infty \frac{1}{k^2-1}$
How can I check the convergence of

$$\sum_{k=2}^\infty \frac{1}{k^2-1}$$

? As it's almost the same as

$$\sum_{k=2}^\infty \frac{1}{k^2}$$

which converges, it should converge, but I can't use a comparison between the two here, as

$$\sum_{k=2}^\infty \frac{1}{k^2-1} \geq \sum_{k=2}^\infty \frac{1}{k^2}$$

so it wouldn't help. So how should I go about it?

Body: $\frac 1 {k^{2}-1} \leq \frac 2 {k^{2}}$ for all $k \geq 2$.


问:
Pontryagin dual of the multiplicative group of a local field.
Let $K$ be a local non-Archimedian field.  Let $K^{\times}$ be the group of invertible elements of $K$.

Is there an explicit description of the Pontryagin dual of $K^{\times}$?

答:
Assume first $k$ is a $p$-adic field. Then as topological groups $k^\times$ is isomorphic to $\mathbb{Z} \oplus \mathbb{Z} / (q-1) \oplus \mathbb{Z} / p^a \mathbb{Z} \oplus \mathbb{Z}_p^d$ for $a \geq 0$ and $d = [k : \mathbb{Q}_p]$. On the other hand, if $k$ has characteristic $p$, then $k^\times$ splits as $\mathbb{Z} \oplus \mathbb{Z} / (q-1) \oplus \mathbb{Z}_p^N$ for some $N$. This is Proposition II.5.7 of Neukirch's Algebraic Number Theory. It should thus suffice to consider each of these components. The only difficult part here is $\mathbb{Z}_p$, whose dual is the Prüfer group $\mathbb{Z}[
p^{-1}]/\mathbb{Z}\cong\mathbb{Q}_p/\mathbb{Z}_p$.


Title: Check convergence of $\sum_{k=2}^\infty \frac{1}{k^2-1}$

Body:
How can I check the convergence of

$$\sum_{k=2}^\infty \frac{1}{k^2-1}$$

? As it's almost the same as

$$\sum_{k=2}^\infty \frac{1}{k^2}$$

which converges, it should converge, but I can't use a comparison between the two here, as

$$\sum_{k=2}^\infty \frac{1}{k^2-1} \geq \sum_{k=2}^\infty \frac{1}{k^2}$$

so it wouldn't help. So how should I go about it?

Body:
Since $1\leq (\frac{k}{2})^{2}$ for k bigger than 1:

$$\sum_{k=2}^\infty \frac{1}{k^2-(\frac{k}{2})^{2}}\geq\sum_{k=2}^\infty \frac{1}{k^2-1} \geq \sum_{k=2}^\infty \frac{1}{k^2}$$