问:
Basis for Null Space at a Point
I have a plane at a point $\boldsymbol{r}_{0}\in\mathbb{R}^3$ with normal vector $\boldsymbol{n}\in\mathbb{R}^3$.
$$
\boldsymbol{n}^\top (\boldsymbol{r} - \boldsymbol{r}_0) = 0
$$
My aim is to find a basis for this plane. I know that if this plane was through $\boldsymbol{r}_0 = \boldsymbol{0}$ then one way in which I could find a basis for the plane would be via SVD decomposition. The last $2$ columns of $U$ would form a basis for $N((\boldsymbol{n}^\top)^\top) = N(\boldsymbol{n})$, the null space of the "matrix" $\boldsymbol{n}$.

$$
\begin{pmatrix}
    n_1 & n_2 &  n_3
\end{pmatrix}
\begin{pmatrix}
    r_1 \\
    r_2 \\
    r_3
\end{pmatrix} = 0
$$

My question is:

How can I find a basis for the plane at $\boldsymbol{r_0}$ with normal vector $\boldsymbol{n}$ using SVD/QR decompositions? I DO NOT want a different method, I just want to adjust this method to find the correct basis.

答:
Consider the plane at the origin, with basis $\{u_1, u_2\}$ and at some other point $r_0$. These are parallel. So $\{u_1, u_2\}$ serves as a basis for the second one as well.

Well...to be more honest, the second one isn't a linear subspace of $3$-space (it doesn't contain the zero-vector), so it doesn't even have a (linear) basis. But it has an affine basis, or "coordinate system", with origin $r_0$ and the two vectors from the origin-plane as vectors. Or it has an affine "frame", consisting of $r_0, r_0 + u_1, r_0 + u_2$.

Or if you want to consider it as lying in the tangent-space to $\Bbb R^3$ at $r_0$, it is a subspace of that, with basis $(r_0; u_1), (r_0; u_2)$, where I'm using the standard identification of $T\Bbb R^3$ with $\Bbb R^3 \times \Bbb R^3$ here.

Title: Basis for Null Space at a Point


问:
Splitting open sets in perfect spaces
Suppose $V$ is a Hausdorff space which is perfect and $U\subset V$ is a non-empty open set. Can we find two disjoint, non-empty open sets $U_1, U_2\subset U$? Is there any natural class of spaces having this property?</p>

答: Sure: just pick distinct $x,y\in U$, use the fact that $X$ is Hausdorff to choose disjoint open nbhds $V_x$ and $V_y$ of $x$ and $y$, respectively, and let $U_1=V_x\cap U$ and $U_2=V_y\cap U$.</p>

ContentLicense: CC BY-SA 3.0

Title: Splitting open sets in perfect spaces


问:
Dividing $7$ balls into $5$ baskets and finding the probability each basket has at least 1 ball
We are given $7$ balls and $5$ baskets. We want to place each ball into a basket and the probability of each ball assigned to a specific basket is independent of each other. How do we find the probability that there is at least $1$ ball in each basket?

I am aware that we will need to use the inclusion exclusion formula, but how do I find the probabilities?

Let $P(A_i)$ be the probability that the $i^{th}$ basket has an "occupant". We want to find
$$P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5) = 1-P(A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5)^c = 1- \bigcup_{i=1}^5 A_i$$

How do I find the probabilities from here?

EDIT: Balls are distinguishable.

答:
Another way to solve it is using multinomial coefficients,

there are only $2$ cases to consider

$\dfrac{\dbinom {7}{3,1,1,1,1}\dbinom{5}{1,4} + \dbinom{7}{2,2,1,1,1}\dbinom{5}{2,3}}{5^7}= \dfrac{16800}{5^7}$

ContentLicense: CC BY-SA 4.0


问:
Let $V = \{x \in \mathbf{R}^3 \mid x_1+2x_2 -x_3=0 \}.$ Determine $\dim V$.

Let $V = \{x \in \mathbf{R}^3 \mid x_1+2x_2 -x_3=0 \}.$ Determine $\dim V$.

The dimension is the number of vectors in the basis set. Here $x_1,x_2$ and $x_3$ are scalars not vectors so how can the dimension be determined here? Also $x_3 = x_1+2x_2$ so $x_3$ is a linear combination of $x_1$ and $x_2$. This somehow suggest that $\dim V = 2$, but I'm not sure. What can I do?

答:
From a direct isomorphism: the linear map
\begin{align}
\mathbf R^2&\longrightarrow \mathbf R^3,\\
(x_1,x_2)&\longmapsto (x_1, x_2, x_1+2x_2),
\end{align}
is injective and its image is $V$, so it induces an isomorphism from $\mathbf R^2$ onto $V$ by restriction of the codomain.


问:
Solve $ y=({\rm d}y/{\rm d}x)^2 $
It is obvious that we can differentiate both sides of the equation with respect to $x$ and then discuss the result. But can we just make a square root of both sides and integrate?

Sorry the original version is so naive. The following is more detailed.

If we take square root
$$\pm \sqrt y=\frac{{\rm d}y}{{\rm d}x}$$
then
$$\frac{{\rm d}y}{\sqrt y}=\pm {\rm d}x$$
$$2\sqrt y=\pm x+c$$
The solutions are
$$y=(\frac{\pm x+c}{2})^2$$ or $y=0$

If we differentiate both sides with respect to $x$
$$y'=2y'y''$$
$$y'(2y''-1)=0$$
If we choose $y'=0$, then $y=c$. And plug it in to the original equation, we get $c=0$.

If we choose $y''=\frac12$, then $y=\frac14x^2+dx+f$. Plugging in, we get $y=\frac14x^2+dx+d^2$, which is enough.

(How could the non-trivial solutions differ???-this is a solved question now.)

Now, everything is great. From this example, I know that there are something a little bit different with the liner differential equation in a nonlinear one. Sometimes, one needs to plug in the solution with constants under some circumstances to determine some constant. Besides, the form of the solution of a nonlinear differential equation could be more complicated-it could be piecewise-combined. One could choose a set of solution satisfying the equation and match them at every boundary of each segment correspondingly. Thank @Allawonder and @Gae. S. for pointing them out.

答:
Can be factored into two differential equations and so we have two equations of the two parabolas due to the $\pm$ sign.

$$ y'= \pm \sqrt {y} ; \; \dfrac{dy}{\sqrt y}=\pm dx;\;\text{Integrating ,  } 2 \sqrt{y}= \pm x +c $$

which can be separately expressed as

$$ y_1= (x/2-c_1)^2,\quad y_2= (x/2-c_2)^2$$

The parabolas can slide along x-axis so as to have tangential double root

$$ x_{min}= 2c_1,\; x_{min}= 2c_2\;$$

To confirm tangency or double contact we have by differentiation
$$ y= y'^2,\; \to y'= 2y' y''$$

Only when we assume
$$y'\ne 0$$

that is, implying that x-axis is excluded as a solution,  are we allowed a division, and consequently so we have a minima situation:

$$y_{min \;tangency\;point}^{''} = \dfrac12 $$

In the zero arbitrary constant situation both the sets come to the central position of the same parabola of unit focal length.

Title: Solve $ y=({\rm d}y/{\rm d}x)^2 $


问:
Alternative proof of the Fibonacci property $\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}.$
If we consider the standard Fibonacci sequence$$F_0=0;\quad F_1=1;\quad F_n=F_{n-1}+F_{n-2}\quad\forall n\geq 2,$$ it is easy to proof that, given any chunk of $10$ consecutive Fibonacci, its sum is equal to $11$ times the seventh (this statement is also valid for any Fibonacci sequence startinng from any two values). That is $$\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}$$ For example, lets consider the first $10$-numbers chunk of the sequence. Then $$0+1+1+2+3+5+\color{red}{8}+13+21+34=88=11\cdot\color{red}{8}.$$

I proved this by applying the recursive property several times in the following way:

\begin{gather*}\underbrace{F_n+F_{n+1}}_{F_{n+2}}+\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+F_{n+6}+\underbrace{F_{n+7}}_{\underbrace{F_{n+5}}_{F_{n+3}+F_{n+4}}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}}+\underbrace{F_{n+9}}_{\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}+\underbrace{F_{n+8}}_{F_{n+6}+F_{n+7}}}=&\\= F_{n+2}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+3}+F_{n+4}+F_{n+6}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+5}+F_{n+6}+F_{n+6}+F_{n+7}=&\\=\underbrace{F_{n+2}+F_{n+3}}_{F_{n+4}}+\underbrace{2F_{n+4}+2F_{n+5}}_{2F_{n+6}}+7F_{n+6}+\underbrace{F_{n+7}}_{F_{n+5}+F_{n+6}}=&\\=\underbrace{F_{n+4}+F_{n+5}}_{F_{n+6}}+10F_{n+6}=&\\=11F_{n+6}&\end{gather*}

This proof looks a bit naive to me compared to the strength of the statement. That is why I wanted to know if someone knows an alternative way to prove this statement. Thank you.

Body:
After OP changes 10 to 9.

Use $F_j=\frac{a^j-b^j}{\sqrt{5}}, a+b=1, ab=-1, a^2=a+1, b^2=b+1$
$$S_n=\sum_{j=n}^{n+9} F_j= \sum \frac{a^j-b^{j}}{\sqrt{5}}$$
USE sum of finite GP, changing the index as $j-n=k$
$$S_n=\sum_{k=0}^{9} \frac{a^{n+k}-b^{n+k}}{\sqrt{5}}=\frac{1}{\sqrt{5}} \left(a^n \frac{a^{10}-1}{a-1}-b^n\frac{b^{10}-1}{b-1}\right)$$
See $a-1=-b, -1/b=a$, then
$$S_n=\frac{1}{\sqrt{5}} [(a^{n+11}-a^{n+1})-(b^{n+11}-b^n)]=F_{n+11}-F_{n+1}.$$
I hope to come back.

Edit:
Now I prove that $F_{n+11}-F_{n+1}=11F_{n+6}$
Note that $$\left(\frac{1\pm\sqrt{5}}{2}\right)^5=\frac{11\pm5\sqrt{5}}{2}$$
$$F_{n+11}-F_{n+1}-11F_{n+6}=a(a^{10}-11a^5-1)-b(b^{10}-11b^5-1)~~~~(*)$$
As roots of $x^{10}-11x^5-1=0$ are $\frac{11\pm5\sqrt{5}}{2}=a^5,b^5;$
the RHS of (*) is zero.


问:
Order of average of $\mathcal{O}_p(1)$ random variables
Consider a random variable $\mathbf{X} \in R^n$. Let $X_i$ denote the $i$-th element of $\mathbf{X}$. Suppose that $\mathbf{X} = \mathcal{O}_p(1)$ elementwise (i.e. $X_i$ is stochastically bounded or $X_i = \mathcal{O}_p(1)$ for $i = 1, \ldots, n)$ and that $\mathbb{E}[X_i^2] < C$ for $i = 1, \ldots, n$. Based on this, is it possible to show that the average of the elements of $\mathbf{X}$ is also stochastically bounded, i.e.

$$\frac{1}{n} \sum_{i = 1}^n X_i = \mathcal{O}_p(1)?$$

Thank you very much.

答:
If $\{X_i\}$ are i.i.d. with $\mathsf{E}X_1=+\infty$, then
$$
\frac{1}{n}\sum_{i=1}^n X_i\to+\infty \quad\text{a.s.}
$$
(See, e.g., Corollary 3 here.)

If $A\equiv\sup_{i\ge 1}\mathsf{E}|X_i|<\infty$, then for any $C>0$,
$$
\mathsf{P}(|\bar{X}_n|>C)\le \frac{A}{C}.
$$
Now, choose $C$ s.t. $A/C$ is small.


问:
strict order property
I wanted to prove that the theory of ordered abelian group has a strict order property.
I know by the theory of Kikyo and Shellah we have:

a theory is unstable iff it has SOP or NIP

and by the :
https://www.jstor.org/stable/1999281?seq=1

the ordered abelian group does not have IP but is there any way to prove SOP directly?

答:
If you know about imaginary elements, it is not very difficult to see that the strict order property of $T$ is exactly equivalent to the existence of a $T^{\mathrm{eq}}$-definable partial order with an infinite chain (in a saturated model; in an arbitrary model, there may not be an infinite chain, but there are arbitrarily long finite chains).

In particular, if $T$ defines a partial order with an infinite chain, it has SOP, and in particular, if it defines an infinite linear order with an infinite chain, then it has the SOP.

All this follows from considering the formulas $x\leq p$, where $p$ is an element of the infinite chain.

The ordered abelian groups have an infinite linear order built in, so they certainly do have the SOP.

Title: strict order property


问:
The Infinitely Nested Radicals Problem and Ramanujan's wondrous formula
In mathematics, a nested-radical is any expression where a radical (or root sign) is nested inside another radical, e.g. $\sqrt{2 + \sqrt{3}}$.

By extension, an infinitely nested radical (aka, a continued root) is an expression where infinitely many radical expressions are nested within each other. A famous example would be:
$\sqrt{x\sqrt{x\sqrt{x \dots}}} = \sqrt{x x} = x, \forall x \in \mathbb{R}$.

Another common (and easy to solve) variety occurs when the pattern of numbers within the radicals follows a regular repeating pattern.
A simple example of this type is $y = \sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}} = \sqrt{2 + y} = 2$. The key is simply to notice that the expression after one repeated cycle is the same as the original and solve the resulting polynomial equation. (A couple of examples here.

However, while the above examples were simple, nested roots can actually be quite important and satisfying beyond the surface level.

a) The Golden Ratio :
The Golden Ratio, $\phi = \frac{1 + \sqrt{5}}{2}$ is well-known for its regularity and recurring appearance is diverse areas of nature, art and mathematics. The famous continued fraction expression  $\phi = 1 + \cfrac{1}{1+ \cfrac{1}{1 + \cdots}}$ is however, also equivalent to the folowing continued square root $\phi = \sqrt{1 + \sqrt{1+ \sqrt{1 + \dots}}}$

b) Viète's formula: An approximation for $\pi$ that has been called the starting point of mathematical analysis and even "the dawn of modern mathematics", the formula is:
$$\frac{2}{\pi} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2 + \sqrt{2}}}{2} \cdot \frac{\sqrt{2 + \sqrt{2 + \sqrt{2}}}}{2} \cdot \dots$$ which already contains nested radicals.

Interestingly, this can also be expressed as:
$$\pi = \lim_{k\to\infty} 2^k\cdot\sqrt{2-\sqrt{2+\sqrt{2+... (\text{k - 1 times})}}}$$
which not only contains an infinitely nested radical, but it also contains the same elementary example $\sqrt{2+\sqrt{2 + \dots}}$ from above in slightly modified form!

(c) All real numbers in the range $[0,2]$ can be expressed as the infinitely nested radical $\sqrt{2 \pm \sqrt{2 \pm \sqrt{2 \pm \dots}}}$

Examples: $2=\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}}=(+)$, $1=\sqrt{2 - \sqrt{2 - \sqrt{2 - \dots}}}=(-)$, $\phi=\sqrt{2 + \sqrt{2 - \sqrt{2 + \sqrt{2 - \dots}}}}=(+-)$ (The Golden Ratio again!)

(d) Infinitely nested equations are related to polynomial equations and can be sued to find their roots

There can be many more examples of such problems.

However, the difficult problem here is to solve non-repeating continued-roots such as $\sqrt{1 + \sqrt{2 + \sqrt{3 + \dots}}}$. From a cursory look, one might even expect that expression might be divergent! However, it is actually the Nested Radical constant $1.757932 \dots$ (OEIS A072449), for which no closed-form expression is known.

Thus, one can have an arbitrary continued-root, say $\sqrt{a_1 + b_1 \sqrt{a_2 + b_2 \sqrt{a_3 + b_3 \sqrt{a_4 + \dots}}}}$, where the constants might be determined by some complicated set of rules of formulae. And even for the simplest non-repeating cases, one's ability to solve will be entirely dependent on the ability for algebraic manipulation and knowledge of diverse theorems which might sometimes help.

Or so I believed ...

Enter Srinivasa Ramanujan...

In 1911, around 2 years before he first contacted Hardy, Ramanujan published the following problem in the Journal of the Indian Mathematical Society:

$$\sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \dots}}}$$

After waiting 6 months without any solutions, he supplied the following formula from page 105 of his first notebook:

$$x + n + a = \sqrt{ax + (n+a)^2 + x \sqrt{a(x+n) + (n+a)^2 + (x+n) \sqrt{\dots}}}$$

And setting $x=2$, $n=1$ and $a=0$, we get the solution $$\sqrt{1^2 + (2) \sqrt{1^2+ (2+1)\sqrt{1^2 + \dots}}} = 2+1+0 = 3$$.

(Pg 86-87, The Man Who Knew Infinity: A Life of the Genius Ramanujan, Robert Kanigel, 5th ed. 1991)

Right off the bat, we can notice some interesting things here:

While we cannot use this formula for the Nested Radical Constant above, it does allow us to find the values of infinitely many continued-roots. In particular we can immediately solve any fraction where the terms are in an Arithmetic Progression $\sqrt{a + b \sqrt{(a+d) + (a+d) \sqrt{(a+2d) + (a+2d) \sqrt{\dots}}}} = a+d + 1$.

We can generalize this further:
$\sqrt{c.a + b \sqrt{(a+d) + c.(a+d) \sqrt{c.(a+2d) + (a+2d) \sqrt{\dots}}}} = a+d + c$.

For any given natural number $n$, we can use this formula to find a number of unique nested fraction expressions equal to $q(n)$, the number of partitions of that number.

And so on...

My question:

How can we prove this formula? While Ramanujan himself usually derived results with great insight via intuition, I expect that the formula itself should be provable with usual mathematical techniques, especially with the amount of influence Ramanujan had on subsequent mathematics, and the high volume of work on continued fractions which are related to nested-radicals. So, can anyone find any proof for this - either on their own or in existing literature?

If we cannot prove this, is it possible to find some motivation that makes it more intuitively clear. This should be possible since that was the way it was originally obtained.

Any other interesting observations about this formula? Is it related to any other mathematical results? Perhaps it is related to some series, constants or continued fractions?

答:
Method 1:- (More General)

As $$(x+a)^2=x^2+a^2+2ax$$

Replacing $a$ by $n+a$ in above equation where $x,n,a\in R$

$$(x+(n+a))^2=x^2+(a+n)^2+2x(a+n)$$

$\ \implies(x+(n+a))^2=x^2+(a+n)^2+2ax+2nx$

$\ \implies(x+(n+a))^2=ax+(a+n)^2+x^2+2nx+ax$

$\ \implies(x+(n+a))^2=ax+(a+n)^2+x(x+2n+a)$

Taking positive square roots on both sides of above equation

$(x+n+a)=\sqrt{ax+(a+n)^2+x(x+2n+a)}$

Now replacing $x$ by $x+n$

$x+2n+a=\sqrt{a(x+n)+(a+n)^2+(x+n)(x+3n+a)}$

$x+3n+a=\sqrt{a(x+2n)+(a+n)^2+(x+2n)(x+4n+a)}$

$x+4n+a=\sqrt{a(x+3n)+(a+n)^2+(x+3n)(x+5n+a)}$

Inducing these results to $k\in N$ we have,

$x+kn+a=\sqrt{a(x+(k-1)n)+(a+n)^2+(x+(k-1)n)(x+(k+1)n+a)}$

Putting the value for $x+2n+a$ in square root of $x+n+a$

$$(x+n+a)=\sqrt{ax+(a+n)^2+x \sqrt{a(x+n)+(a+n)^2+(x+n)(x+3n+a)}}$$

Now Doing this Process continuously we have our orignal infinite nested radical for $x+n+a$.

On setting $x=2,a=0,n=1$ we get

$$\sqrt{1^2 + (2) \sqrt{1^2+ (2+1)\sqrt{1^2 + \dots}}} = 2+1+0 = 3$$

Method 2:-(slightly less general)

Let $f(n)=\sqrt {1+(n+1)\sqrt{1+(n+2)\sqrt{1+...}}}$

Then $f(n)=\sqrt{1+(n+1)f(n+1)}$

$\implies f(n+1)=\frac{(f(n))^2-1}{n+1}$

Asymptotically we have $(f(n))^2\sim(n+1)f(n+1)$

If we assume $f(n)=an+b$(By equating degrees on both sides) and put in equation for $f(n+1)$
we get,

$n^2(a^2-a)+n(2ab-2a-b)+(b^2-1-a-b)=0$

This gives $a=1$ and $b=2$

so $f(n)=n+2$

For $n=1$ we have

$$\sqrt {1+2\sqrt{1+3\sqrt{1+...}}}=3$$


问:
Canonical vector space structure on stalk of a presheaf of vector spaces
Let $\mathcal{F}$ be a presheaf of complex vector spaces on $X$. The stalk at $x \in X$ is $$ \mathcal{F}_x = \{ (U, s_U) : s_U \in \mathcal{F}(U) \text{ and } x \in U\}/\sim $$ where $(U,s_U) \sim (V,s_V)$ if $s_U \vert_{U \cap V} = s_V \vert_{U \cap V}$.
Supposedly $\mathcal{F}_x$ has a canonical complex vector space structure - I have defined one as setting $[U, s_U] + [V, s_V] = [U, s_U + s_V]$ if $U = V$ and zero otherwise. This is a vector space, but I'm not sure if it's the canonical structure - is it?

答:
You have the right idea but it needs a little modification.

The canonical vector space structure is given by

$$[U, s_U] + r[V, s_V] = [U \cap V, s_U|_{U \cap V} + rs_V|_{U \cap V}]$$

Essentially, you just move to a smaller open set where you can talk about $s_U$ and $s_V$ in the same space. Namely, you are using the vector space structure of $\mathcal{F}(U \cap V)$ to define the thing on the right.


问:
Why does $c^Tx \le d$ on an affine set imply that $c^Tx = d$?
Why does $c^Tx \le d$ on an affine $D = \{x : Ax = b\}$ set imply that $c^Tx = d$?

I tried to proceed with a proof by contradiction assuming $c^Tx \not = d$ i.e. $c^Tx < d$. However, I don't see here any contradiction. Intuitively it is the intersection of a half-space with a hyperplane however, I am struggling to understand the problem.

答: It most certainly does not mean that. Take $c=\vec 0$ and $d=1$ for a counterexample.


问:
Does infinite variance imply infinite expectation?
If a random variable has infinite variance, does it necessarily have infinite expectation or is its expectation not well-defined?</p>

答:
The answer above is excellent but I provide a not so rigorous understanding into why this is the case.

Let us consider the sequence of random variables $X_n \sim $ Uniform($-n,n$)

We see that each $X_n$ is centred around $0$ and thus they all have mean $0$

The variance of a Uniform($a,b$) R.V is $\frac{1}{12}(b-a)^2$ so we see that the variance of our $X_n$ is $\frac{1}{3}n^2$

Hence $\mathbb{E}[X_n] = 0 $ $\forall n$

However Var$(X_n) =\frac{1}{3}n^2 \to \infty$

A similar thing could have be done with $Y_n \sim \mathcal{N}(0,n)$

Loosely speaking the concept of mean is its average value or where the random variable is centred (on mass) and variance talks about how far it deviates from this mean on average. However we can find a random variable that is clearly centred but deviates an infinite amount.

Importantly knowing how far something deviates and what it deviates from are uncorrelated. So the variance tells nothing of the mean.

You may notice that some parameters (like the $a$ and $b$ in our uniform or the $\lambda$ in a poisson) play a part in both the mean and variance so your intuition would (naturally) tell you that the two concepts are linked however they are not.


问:
$\text{SO}(4)$ is homeomorphic to $\text{SO}(3)\times S^3$
Is there a reference for a proof that $\text{SO}(4)$ is homeomorphic to $\text{SO}(3)\times S^3$? Since $\text{SO}(4)$ acts transitively on $S^3$ with stabilizer $\text{SO}(3)$, we have a fiber bundle $\text{SO}(3)\to \text{SO}(4)\to S^3$, but I can't see why this bundle is trivial.

答:
Here is a reference about the topology of SO(4):

http://www.gregegan.net/ORTHOGONAL/03/WavesExtra.html#topology

You can also see this article of wikipedia:

https://en.wikipedia.org/wiki/Rotations_in_4-dimensional_Euclidean_space#Group_structure_of_SO(4)

Which describes precisely the universal cover of SO(4).


答:
The most elementary argument I can think of is the following.

The dimension can be at most $2$ since there are vectors that do not satisfy that relation, e.g. $(1,0,0)$.

On the other hand, $V$ contains at least two linearly independent vectors, e.g. $(1,0,1)$ and $(2,-1,0)$.

Hence the dimension must be $2$.

Body:
Let $V = \{x \in \mathbf{R}^3 \mid x_1+2x_2 -x_3=0 \}.$ Determine $\dim V$.

The dimension is the number of vectors in the basis set. Here $x_1,x_2$ and $x_3$ are scalars not vectors so how can the dimension be determined here? Also $x_3 = x_1+2x_2$ so $x_3$ is a linear combination of $x_1$ and $x_2$. This somehow suggest that $\dim V = 2$, but I'm not sure. What can I do?


问:
How can you rewrite the following equation? $\ln x=-\frac{1}{3}\ln y+C$
So I'm having trouble with rewriting my answers, I simply don't know how to start, it goes as follows :
$$\ln x=-\frac{1}{3}\ln y+C$$
becomes :
$$\sqrt[3]{y}=\frac{C}{x}$$
If someone knows the answer and would like to share it, it would be very much appreciated !

Thanks in advance

答:
Seems more logical to write

$$\ln y=-3\ln x+c.$$

Then

$$e^{\ln y}=y=e^ce^{-3\ln x}=Cx^{-3}.$$


问:
What's the geometrical interpretation of the transitive, faithful action of $S_4$ by conjugation on the set of its four Sylow $3$-subgroups?
A pair of recent posts (here and here) have posed the same question: show that the action of the symmetry group of a cube on pairs of opposite faces defines a surjective homomorphism from $S_4$ to $S_3$. There I guessed that this action is the transitive action of $S_4$ by conjugation on the set of its three Sylow $2$-subgroups. Whether or not this identification is correct, this has made me raise the following, "converse" question: what's the geometrical interpretation of the transitive, faithful action of $S_4$ by conjugation on the set of its four Sylow $3$-subgroups? In particular, what are the four "geometrical elements" acted on by the symmetry group of a cube in this case?

答: In the cube model, each Sylow 3-subgroup corresponds to a pair of opposite vertices of the cube, more specifically it is the group of rotations around the axis connecting those two vertices. $S_4$ acts by rotation on this set of pairs.


问:
Probability example for homework
There's a highway between two towns. To reach the other, people must pay 10 dollars for cars and 25 dollars for bigger vehicles. How big income can we expect if the 60 percent of vehicles are cars and there are 25 incoming vehicles per hour?

What kind of distribution does it follow, and how to calculate with that?

答:
This is a discrete distribution.

$\text{P(Vehicle is car)} = 0.6$ and

$\text{P(Vehicle is a bigger vehicle)} = 0.4$

Expected revenue per vehicle is:

$\text{E(Revenue per vehicle)} = \text{P(Vehicle is car)} (\text{Revenue from car}) +
\text{P(Vehicle is a bigger vehicle)} (\text{Revenue from a bigger vehicles})$

Thus, total expected revenue is:

$\text{Total expected revenue} = (\text{Total number of vehicles}) \text{E(Revenue per vehicle)}$

Title: Probability example for homework


问:
separating disjoint compact set and closed set in regular space

In a regular space $X$, a compact set $A$ and a disjoint closed set $F$ can be separated by disjoint open sets $U,V$.

Since $X$ is regular, for every $x\in A$ there exists disjoint open sets $U_x,V_x\subset X$ such that

$$x\in U_x,\ x\in V_x,\text{ and } U_x\cap V_x = \emptyset.$$

Clearly, $A\subset\bigcup_{x\in A}U_x$, so there exists a finite set $\{x_1,x_2,...,x_k\}\subset A$ such that $A\subset\bigcup_{i=1}^{k}U_{x_i}$. Let $U:=\bigcup_{i=1}^{k}U_{x_i}$ and $V:=\bigcap_{i=1}^{k}V_{x_i}$. Clearly $A\subset U$ and $U\cap V=\emptyset$ but I'm not sure why $F\subset V$.

答:
Apply regularity to $F$ and all $a \in A$:

We have open $U_a,V_a$, such that $a \in U_a, F \subseteq V_a$ and $U_a \cap V_a= \emptyset$. This we do for each $a \in A$.

The $\{U_a\mid a \in A\}$ cover the compact $A$ so finitely many do too:

$$\exists a_1, \ldots,a_n \in A: A \subseteq V:= \bigcup_{i=1}^n U_{a_i}$$

Then define $$U= \bigcap_{i=1}^n V_{a_i}$$

and as for all $a$, $F \subseteq V_a$ we have $F \subseteq U$ and $U$ is open because we have a finite intersection of opens.

Then $U \cap V = \emptyset$: suppose not, then we have some $x \in U \cap V$ and we have that $x \in V$ so for some $a_i$, $x \in U_{a_i}$. But also $x \in V \subseteq V_{a_i}$ and this contradicts the disjointness of all corresponding $U_a$ and $V_a$. So No such $x$ exists and the sets are disjoint.

Your proof was that in a Hausdorff space we can separate a point $x$ from a disjoint compact set, and you needed to intersect the $U_{x_i}$ and union the $V_{x_i}$, not the other way round. Disjointness is proved the same way as above.


问:
Large Deviation, Optimal Transport and Machine Learning Reference
I am looking for references (books/sites/articles) on the following three subjects: Large Deviation, Optimal Transport and Machine Learning References. I would like works which involve any of them individually, or jointly. My motivation for this is the following course description which used to be taught at my school:

Two mathematical frameworks: optimal transport and large deviation
  theory, can be combined to provide tools and methods of analysis for
  some central tasks in data science: classification, clustering and
  regression, dimensionality reduction, conditional probability
  estimation, prediction of rare events. Recurring themes will include
  the power of the composition of simple functions to capture complex
  behavior - as demonstrated by neural networks - and of the
  characterization of both the available observations of the system
  under study and the parameters tuned to model it in terms of
  underlying probability distributions.

If you know of any core books on either Large Deviation or Optimal Trasport that can be grasped at a beginning graduate level, I would much appreciate it!

thank you!

答:
Probably a bit late, but I can supplement the other answers.

For general introduction to the topics you mentioned :

Optimal transport : Villani's "Optimal Transport Old and New" book is great.

Large Deviations : I think Dembo and Zeitouni's book is good. Hollanders book is ok, but contains a bit less content.

Machine Learning : There is a lecture course on youtube which is good https://www.youtube.com/watch?v=_XmGyd4smUs&list=PL8FnQMH2k7jzhtVYbKmvrMyXDYMmgjj_n&index=2 .

There has been a lot of work done w.r.t Large Deviations (Sanov theorem, and Friedlin-Wentzell theory) in determining the cost function of gradient flows  in measure spaces (this topic branches probability theory, thermo-dynamics, differential geometry, optimal transport) . See Peletiers works, in particular https://arxiv.org/abs/1312.7591.


问:
Solving a nonlinear first order ODE
In this text https://www.springer.com/gp/book/9781461454762 on p. 95, it has the following

The text does not explain how to get to this solution. $B, C, D$ are arbitrary constants. With $A>0$, I came to something like this $$Y_0=B\tan\left(\frac{B\bar{x}}{2}+C\right)$$ Is that and the positive case solution the same? Also how did the negative case solution come about?

答:
You can look at this like a Riccati equation and set $Y=-2\frac{u'}{u}$ so that
$$
Y'=-2\frac{u''}{u}+2\frac{u'^2}{u^2}=2\frac{u'^2}{u^2}+A\iff u''+\frac{A}2u=0
$$
This now is a standard linear DE with constant coefficients and you can easily treat the 3 cases $A=2B^2$, $A=-2B^2$ and $A=0$.

\begin{align}
u''+B^2u&=0\implies& u&=\cos(Bx+C),&~~Y&=2B\tan(Bx+C)\\
u''&=0&\implies u&=Dx+C,&~~Y&=-\frac{2D}{Dx+C}\\
u''-B^2u&=0\implies & u&=Ce^{Bx}+De^{-Bx},&~~ Y&= -2B\frac{Ce^{Bx}-De^{-Bx}}{Ce^{Bx}+De^{-Bx}}
\end{align}

Title: Solving a nonlinear first order ODE


问:
Irreducibility of $\operatorname{Spec}$ $\iff$ primality of nilradical
The following is an exercise (Problem 1.19) in Atiyah and Macdonald's Introduction to Commutative Algebra text :

A topological space $X$ is said to be irreducible if $X \neq \emptyset$ and if every pair of non-empty open sets in $X$ intersect, or equivalently if every non-empty open set is dense in $X$. Show that $\operatorname{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal.

I have a specific question about the solution to this problem found here.

Let $X_f$ be the set of prime ideals of $A$ that do not contain $f \in A$, and let $X_g$ be the set of prime ideals of $A$ that do not contain $g \in A$. Suppose $\operatorname{Spec}(A)$ is irreducible. Then the author says that $X_f \cap X_g = \emptyset \Rightarrow X_f$ is empty or $X_g$ is empty. Why does this implication hold ? Is this just a general fact about two prime ideals -- that if their intersection is empty, then one of them must be empty ?

I tried to show this implication by the contrapositive. Suppose that $X_f$ and $X_g$ are both nonempty. Then there is at least one prime ideal of $A$ that does not contain $f$, and there is at least one prime ideal of $A$ that does not contain $g$. However, I don't know how this implies that $X_f \cap X_g \neq \emptyset$ (that is, that there is at least one prime ideal that doesn't contain $f$ nor $g$).

Thanks !

答:
Let $nil(A) \subseteq A$ be the nilradical and let $\pi: A \rightarrow A/nil(A)$ be the canonical map. The map $\pi$ induce a homeomorphism of topological spaces

$$H1.\text{   } Spec(A) \cong Spec(A/nil(A)).$$

Hence $T:=Spec(A)$ is irreduible iff $S:=Spec(B):=Spec(A/nil(A))$ is irreducible.

Assume $B$ is a domain and assume $D(f) \cap D(g)=D(fg)=\emptyset$. It follows $fg$ is nilpotent, and since $B$ is reduced it follows $fg=0$. Since $B$ is a domain it follows $f$ or $g=0$. Hence $D(f)=\emptyset$ or $D(g)=\emptyset$. Hence $S$ is irreducible.

Conversely if $B$ is reduced and $S:=Spec(B)$ is irreducible. Let $f,g \neq 0 \in B$ be elements with $fg=0$. It follows $V(fg)=S$ hence $D(fg)=D(f) \cap D(g)=\emptyset$. Since $S$ is irreducible we must have $D(f)=\emptyset$ hence $f$ is nilpotent. It follows $f\in nil(B)=(0)$ hence $B$ is a domain.

Note: I use the notion "irreducibility" from Hartshorne, Proposition II.3.1 and AM. Ex.1.19:

Definition: A topological $X$ space is "irreducible" iff every pair of nonempty open sets in $X$ intersect.

When is the diagonal a closed immersion with open image?


问:
Irrational inequalities why it is not possible to square without additional condition
Why when I have an irrational inequalities, e.g $\sqrt{f(x)}\geq g(x)$ (for real valued function), instead of an irrational equation I can't square both sides (in case of equation I would do $f(x)=g(x)^2$, not?), but I have to impose that $g(x)\geq 0$ to be able to square? Please can you explain me what is the idea in case of irrational inequalities?

答:
When you are squaring an equation, what you are using is the fact that $f(x)=x^2$ is a well defined function, meaning for each real $x$ there is one and only one real number that is called $x^2$.

In other words, when you start with

$$a=b,$$

you get $a^2=b^2$ because $a$ and $b$ are the same value by assumption, so squaring both will give the same value for $a^2$ and $b^2$.

When you are squaring an inquality, you want to conclude that from

$$a \ge b$$

follows

$$a^2 \ge b^2,$$

which is generally incorrect, as  player3236's comment showed. That conditions as stated is what is called increasing monotonicity of a function $f(x)$:

$$a \ge b \Longrightarrow f(a) \ge f(b),$$

and $f(x)=x^2$ doesn't have the property. Other functions, like $g(x)=2x$ or $h(x)=x^3+2$ do have that property.

Note that $f(x)=x^2$ is a monotone increasing function if you restrict the domain to $x \ge 0$, which is why squaring an inequality "works" if you know or assume that the smaller number is non-negative.


问:
Sufficient statistic for $N(\mu,1).$
So I am answering a problem where
$\underline{X}=(X_1,X_2)'$ be a random variable from $N(\mu,1).$

Here is my approach, the joint density is:

$f(x_1,x_2|\mu) = (2\pi)^{-1} \textrm{exp}(\frac{-1}{2\sigma^2}\sum^{2}_{i=1} x^2_i + \frac{\mu}{\sigma^2} \sum^2_{i=1}x_i-\frac{2\mu^2}{2\sigma^2})$

Since the variance is known as a constant 1 then
$$h(\textrm{x}) = (2\pi)^{-1} \exp(\frac{-1}{2})\sum^2_{i=1} x^2_i$$
and
$$g_u(\bar x,n) = \exp(-\mu^2 + \mu r(x_1.x_2))$$
where $r(x_1,x_2) =\sum^2_{i=1}x_i$

Therefore it can be concluded that sample mean $\bar X_n $ is also a sufficient statistic.

But then after concluding it, the problem asks if $S_2 = S_2(\underline{X})= X_2 - X_1$ is sufficient for $\mu$. Should I still need to prove it ? Or nay nay ?

答:
$$S_2=X_2-X_1\sim N(0;2)$$

Its distribution is is the same $\forall \mu$ thus $S_2$ is ancillary for $\mu$.

In other words, by itself it can tell us nothing about $\mu$

...it is not sufficient

Also note that $S_2=X_2-X_1$ is independent from $S_1=X_1+X_2$.

This because

$S_1$ is Complete and Sufficient for $\mu$

$S_2$ is Ancillary for $\mu$

Thus we can apply Basu's Theorem because all the hypothesis are attained

Title: Sufficient statistic for $N(\mu,1).$


问:
Decomposition of $\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}$

Determine $\alpha,\beta,\gamma,\delta$ in

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

I have come across the partial fraction and trying to decompose it. I have tried the methods used by multiplying out and making one variable to 0 to find the others etc but It does not work here. Any ideas how I could approach this?

答:
Hint:Multiply both sides by the denominator of the l.h.s.:
$$10s+60=(\alpha s+\beta)(s^2+25)+(\gamma s+\delta)\bigl((s+3)^2+25\bigr),\tag{1}$$
and set $s$ to be successively  roots of each factor in the denominator (i.e.  poles of the fraction), $5i$ and $-3+5i$, and identify the real and the imaginary parts of each side,  to  obtain two linear systems in $\alpha,\beta$ and in $\gamma,\delta$ respectively.


问:
Why is this language regular?
I'm at the beginning of Theoretical Informatics and I was given some tasks one of which is following...

Be $k\in \mathbb{N}$, show that following language $L$ with the
alphabet $\Sigma = \{0,1\}$ is regular.

$L= \{w\mid w$ is $2k$ Symbols long and a Palindrome$\}$

In books and on the Internet the consensus is gernerally that Palindromes are not regular (reason is that Finite Machines can't remember much information) but here I am asked to prove it, therefore it should be true.

Why is that ? In my mind I found it impossible to construct a (Non-deterministic) Finite Machine with that properties.

Can someone give me a hint why this is true ?

答: This doesn't contradict the fact that the set of all palindromes is not a regular language, as you are given a fixed length for the palindrome. For each $k$, the corresponding language $L$ is finite and hence regular. If we write $L_k$ rather than $L$ to make the dependency on $k$ explicit, the union $\bigcup_k L_k$ is not a regular language, but each $L_k$ is.

Title: Why is this language regular?


问:
Solving variables for the equilibrium distribution
This question is related to the equilibrium probability distribution of a markov chain. I have: $ \vec {π} * \vec {P} = \vec {π}$

$ [π_0,π_1,π_2,π_3]   $ [ $\begin{matrix} 1/3 & 2/3 & 0 & 0 \\ 1/6 & 1/2 & 1/3 & 0 \\ 0   & 4/9 & 4/9 & 1/9\\ 0 & 0 & 5/6 & 1/6 \end{matrix}$ ] = $ [π_0,π_1,π_2,π_3]  $

Which produces:

$π_0=1/3 π_0 + π_11/6     \\\\\\\\\\\\\\\\π_1=2/3 π_0+π_11/2+π_2 4/9 \ \ \ \ \ \ \ \ \\\\\\π_2=π_11/3+π_24/9+ π_35/6   \\\\π_3 = π_2 1/9 + π_3 1/6
$

I also have the constraint that  $\sum_{i=0}^3 π_i = 1$

How do I solve for each π?, I really only need to solve three of them and the fourth is the difference between the sum of the others and one.

Cheers.

答: Use Gaussian elimination to find all $\pi_i$ are 0.25.</p>


问:
Calculate $\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}$ without using L'Hopital
I can calculate this easily using L'Hopital Rule. Can anyone give me some pointers on how to do this without using L'Hopital?

$$\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}$$

I tried substitution by $n = x+a$ and then $\lim_{n \to 0}$ but that didn't work.

答:
You need to substitute $n=x-a$ instead. Then, note that the required limit is
$$\lim_{n\to 0}\frac{\log n}{\log(e^a\times e^{x-a}-e^a)}$$
Then, you can factor, and proceed with the known identity.

Hope this helps. Ask anything if not clear :)


问:
$\Phi$-extreme weights and the Weyl group orbit of the highest weight
$\newcommand{\g}{\mathfrak{g}}$
Let $P(\g)$ be the weight lattice of $\g$ a semisimple Lie algebra over $\mathbb{C}$, and $P_{++}(\g)$ the set of dominant integral weights.
A subset $\Psi \subset P(\g)$ is $\Phi$-saturated if for all $\lambda \in \Psi$ and $\alpha \in \Phi$, $\lambda - k\alpha \in \Psi$ for all integers $k \in [0, \langle \lambda, \check{\alpha} \rangle]$.
An element $\lambda \in \Psi$ is called $\Phi$-extreme if for all $\alpha \in \Phi$, either $\lambda + \alpha \not\in \Psi$ or $\lambda - \alpha \not\in \Psi$.

I wish to show the following result: for $V$ a finite-dimensional irrep of $\g$ of highest weight $\lambda$, the orbit of $\lambda$ under the Weyl group is the set of $\Phi$-extreme elements of $\chi(V)$, where $\chi(V)$ is the weight lattice of $V$ (the set of all weights of $\g$ on $V$).
I know that $\chi(V)$ is the smallest $\Phi$-saturated subset of $P(\g)$ containing $\lambda$.

I am specifically confused in the following line of the proof of the previous result:

Since $W \cdot \Phi = \Phi$, the set of $\Phi$-extreme elements of $\chi(V)$ is invariant under $W$. Thus it suffices to show that if $\mu \in P_{++}(\g) \cap \chi(V)$ and $\mu$ is $\Phi$-extreme, then $\mu = \lambda$.

I understand how the bolded statement is proven, but I do not see how it implies the result, that $$ W \cdot \lambda = \{\Phi\text{-extreme elements of } \chi(V)\}.$$
I understand that the right-hand side can be described as all weights in $\chi(V)$ for which either adding or subtracting any $\alpha \in \Phi$ is no longer a weight, and I also know that any weight can be transformed into a unique dominant weight by some $w \in W$, however there is no guarantee this would also be an integral weight.

In fact, if $\mu$ is both dominant integral and also a weight of $V$, then $\mu = \lambda$ seems forced by the uniqueness of the highest weight: the assumption that it is also $\Phi$-extreme seems superfluous here.
What am I misunderstanding?

答:
Given the unbolded line in your quote, we know $W$ acts on the set of $\Phi$-extreme elements. What the bolded line shows is then that there is only one $W$-orbit of $\Phi$-extreme elements intersecting $P_{++}(\mathfrak{g}) \cap \chi(V)$—the orbit of $\lambda$:
$$
 \{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V) = \{\lambda\},
$$
so
$$
 W \cdot (\{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V)) = W \cdot \lambda
$$
Then recall that every weight's $W$-orbit intersects $P_{++}(\mathfrak{g})$. So, given that the the 2 of the sets on the LHS are $W$-invariant and the other contains representatives of every $W$-orbit, taking $W$-orbits is equivalent to removing the intersection with $P_{++}(\mathfrak{g})$ on the left-hand side:
$$ W \cdot P_{++}(\mathfrak{g}) = P(\mathfrak{g}) \ ; \  W \cdot \{\text{$\Phi$-extreme}\} = \{\text{$\Phi$-extreme}\} \ ; \ W \cdot \chi(V) = \chi(V)
$$
implies
$$
 W \cdot (\{\text{$\Phi$-extreme}\} \cap P_{++}(\mathfrak{g}) \cap \chi(V)) = \{\text{$\Phi$-extreme}\} \cap \chi(V) = W \cdot \lambda
$$

Edit:
To address OP's last paragraph, it sounds like there's some confusion about the meaning and consequences of integrality for a weight. For me, and it seems like this is the way it's used in the proof, an integral weight $\lambda$ is an element of the vector space spanned by $\Phi$ whose pairing with all roots is an integer. Equivalently, it is an integral combination of fundamental weights. So it's just equivalent to being in $P(\mathfrak{g})$, as opposed to being in the real vector space spanned by $P(\mathfrak{g})$.

So one consequence is that $W$ does preserve integrality, and another is that you typically do have several (integral) weights in the intersection of $\chi(\lambda)$ and $P_{++}(\mathfrak{g})$.

For instance, in the simplest case of $\mathfrak{sl}_2$, we have $\Phi = \pm{\alpha}$ and $P(\mathfrak{sl}_2) = \mathbb{Z}\lambda_{\alpha}$ where $2 \lambda_{\alpha} = \alpha$. Then as long as $V$ is a highest weight representation of a weight $\mu = n \lambda_{\alpha}$ where $n \ge 2$, then $\chi(V) \ni (n - 2)\lambda_{\alpha}$, which is dominant and integral. Almost every highest weight representation $V(n \lambda_{\alpha})$ is included, since $n \lambda_{\alpha}$ is dominant iff $n \ge 0$. However, for all $n$, the only $\Phi$-extreme weights of $V(n \lambda_{\alpha})$ will be $\pm n \lambda_{\alpha}$, and of those only $+n \lambda_{\alpha}$ is dominant.


问:
How did submetrizability come into existence ? What is submetrizability of a topological space used for?
Can anybody tell me how did submetrizability come into existence, and what is its use in topology ? Any examples to make me understand ?</p>

答:
There is a whole subbranch of general topology that came out of finding and proving metrisation theorems, and a lot of properties have been found that are so-called "Generalised Metric Spaces", that have nice properties and are "close to" metrisable in some sense. Monotonically normal spaces are such spaces (I recently talked about them here, in connection with the Michael line),  developable spaces (introduced by Bing), spaces with a $G_\delta$ diagonal, $M$-spaces, $p$-spaces, etc. See Gruenhage's survey paper in the Handbook of Set theoretic Topology.

Many examples of such spaces (non-metrisable ones) do have the property, it was observed, that they have a natural weaker metrisable topology on the same set, e.g. the Sorgenfey line, the Michael line (which have the Euclidean topology as a weaker topology (a subtopology). So it is logical to start studying that feature as a new property in its own right, and call it submetrisable (British spelling). We can then prove interesting equivalent formulations of it (see Gruenhage's article) which make it look like other properties and prove some implications (like: submetrisable implies having a $G_\delta$ diagonal) so that it fits in a network of other properties. I cou;ldn't find who first introduced the property or its name (looks a bit like a property that Arhangel'skij would introduce to me) but it does fit into a wider family of properties.

Added based on comments: different guises under which we can meet this property:

TFAE:

$(X, \tau)$ is submetrisable.

There is a metric $d: X \times X \to \Bbb R$ (so a function obeying the metric space axioms) on $X$ that is continuous on $(X,\tau) \times (X,\tau)$.

There is a continuous injection $f(X,\tau) \to (Y,d)$, where $(Y,d)$ is a metric space (of course having the metric topology $\tau_d)$.

$1 \to 2$: Let $\tau' \subseteq \tau$ be a topology on $X$ that is induced by some metric $d$ on $X$. Then $d$ is continuous as a map on $(X,\tau') \times (X, \tau')$, this is classical, and so the same holds for the stronger topology on $X^2$, $(X,\tau) \times (X,\tau)$.

$2 \to 3$: Let $Y=(X,d)$ and $f(x)=x$. $Y$ is by definition metric and $f$ an injection and $d$-open balls on $X$ are $\tau$-open as $B_d(x,r) = d_x^{-1}[(-\infty,r)]$, where $d_x: (X,\tau) \to \Bbb R, d_x(y)= d(x,y)$ is continuous as $d$ is. It follows that $f$ is continuous, as required.

$3 \to 1$: Given $f:(X, \tau) \to (Y,d)$ continuous and injective, define $\tau'$, a topology on $X$, by $\{f^{-1}[O]\mid O \in \tau_d\}$ and note that $\tau' \subseteq \tau$ and $(X,\tau') \simeq (f[X], d)$ so $X$ is submetrisable.

These are studied for being interesting, they have no "use" whatsoever. What's the use of an $L$-space? It's just expanding our knowledge of what's possible in topology.. I don't think it occurs as a condition in a theorem that is widely applied in analysis or elsewhere e.g. It's just one of the hundreds of properties that topologists have introduced for the sake of exploring.  One interesting theorem (IMO): Every paracompact space with a $G_\delta$ diagonal is submetrisable. (If we replace paracompact by compact we get a metrisable space, then follows)


问:
Necessary condition for orbits to be separated by invariant polynomials.
Suppose we have an algebraic subgroup $G \subset GL(V)$, where $V$ is a finite dimensional vector space over the field of complex numbers $\mathbb{C}$. I'm trying to prove the following:

If $G$-orbits in $V$ can be separated by $G$-invariant polynomials, then $G$ is finite.

I've found out that it's important that the field is $\mathbb{C}$ (I think because $\mathbb{C}$ is closed). Over $\mathbb{R}$ we have a counterexample $SO_1 \subset GL(\mathbb{R}^3)$. But the requirement for subgroup to be algebraic prevented me to construct something of this kind for $\mathbb{C}$. So this condition is also crucial.

I have a feeling it's an easy question, but I didn't succeed in making all those conditions work together. Any hint would be appreciated.

答: We can take homogeneous generators $f_1, \dots, f_n$ of $\mathbb{C}[V]^G$ and consider a map $f : V \to \mathbb{C}^n$, $f(x) = (f_1(x), \dots, f_n(x))$. Since invariant polynomials separate $0$ from other orbits, $f^{-1}(0) = 0$. This condition together with homogeneity implies that $f$ has finite fibers (it is actually finite, When does a homogeneous morphism have only finite fibers?). Hence all orbits are finite. Hence $G$ is finite.


问:
Newton Divided Difference Method of Interpolation in maple
Does anyone know how one can do this in maple?
find Newton divided difference interpolation polynomial for the function cos2x the points {1, 0.6, 0.3, 0.1, 0}</p>

答:
If this is homework then you could show what you've accomplished on your own already.

If you just need the result then you could try the Student:-NumericalAnalysis package. Eg,

```<code>L:=[0, 0.1, 0.3, 0.6, 1]:

p:=Student:-NumericalAnalysis:-PolynomialInterpolation(
               [seq([x,cos(2*x)], x in L)],
               method=newton):

Student:-NumericalAnalysis:-DividedDifferenceTable(p);

  [           1.              0              0             0             0]
  [                                                                       ]
  [ 0.9800665778  -0.1993342220              0             0             0]
  [                                                                       ]
  [ 0.8253356149  -0.7736548145   -1.914401975             0             0]
  [                                                                       ]
  [ 0.3623577545   -1.543259535   -1.539209441  0.6253208900             0]
  [                                                                       ]
  [-0.4161468365   -1.946261478  -0.5757170614   1.070547088  0.4452261980]

interpol:=Student:-NumericalAnalysis:-Interpolant(p,independentvar=x);

      interpol := 1. - 0.1993342220 x - 1.914401975 x (x - 0.1)

         + 0.6253208900 x (x - 0.1) (x - 0.3)

         + 0.4452261980 x (x - 0.1) (x - 0.3) (x - 0.6)

plots:-dualaxisplot(plot([cos(2*x),interpol],x=0..1),
                    plot([cos(2*x)-interpol],x=0..1,legend=`abs. error`));
</code>```

Programming questions on Maple are often better asked on stackoverflow, unless your central question is about the mathematics behind the algorithm.

Body:
If this is homework then you could show what you've accomplished on your own already.

If you just need the result then you could try the Student:-NumericalAnalysis package. Eg,

L:=[0, 0.1, 0.3, 0.6, 1]:

p:=Student:-NumericalAnalysis:-PolynomialInterpolation(
               [seq([x,cos(2*x)], x in L)],
               method=newton):

Student:-NumericalAnalysis:-DividedDifferenceTable(p);

  [           1.              0              0             0             0]
  [                                                                       ]
  [ 0.9800665778  -0.1993342220              0             0             0]
  [                                                                       ]
  [ 0.8253356149  -0.7736548145   -1.914401975             0             0]
  [                                                                       ]
  [ 0.3623577545   -1.543259535   -1.539209441  0.6253208900             0]
  [                                                                       ]
  [-0.4161468365   -1.946261478  -0.5757170614   1.070547088  0.4452261980]

interpol:=Student:-NumericalAnalysis:-Interpolant(p,independentvar=x);

      interpol := 1. - 0.1993342220 x - 1.914401975 x (x - 0.1)

         + 0.6253208900 x (x - 0.1) (x - 0.3)

         + 0.4452261980 x (x - 0.1) (x - 0.3) (x - 0.6)

plots:-dualaxisplot(plot([cos(2*x),interpol],x=0..1),
                    plot([cos(2*x)-interpol],x=0..1,legend=`abs. error`));

Programming questions on Maple are often better asked on stackoverflow, unless your central question is about the mathematics behind the algorithm.


问:
How does the Cooley-Tukey-Algorithm work?
I have tried to understand the Cooley Tukey Algorithm for some time. I read many articles and watched some videos like these:

https://www.youtube.com/watch?v=iTMn0Kt18tg

https://www.youtube.com/watch?v=h7apO7q16V0

https://www.youtube.com/watch?v=E8HeD-MUrjY

https://www.youtube.com/watch?v=mPVtovydY1k

(Source: YouTube)

First my background:
I know how the normal Fourier Transform, the DFT, Laplace and so on work.
I would be able to explain them to other people and have no problems with complex numbers.
I programmed the DFT many times and was able to program a Spectrum-Analyzer where you put in an audio file and it outputs a spectrogram. (Just by using DFT but very limited Frequency range) I use Python. Here is an example DFT in Python:

```<code>import numpy
import matplotlib.pyplot

frequency = 4
samples_power = 6
f_min = 0
f_max = 10

samples = 2**samples_power
signal = numpy.sin(frequency*2*numpy.pi*numpy.linspace(0,1,samples))

f, t = numpy.meshgrid(numpy.linspace(f_min,f_max,samples) ,numpy.linspace(0,1,samples))
n = f*t

dft_matrix = 1/numpy.sqrt(numpy.shape(n)[0])*numpy.power(numpy.e,-2*numpy.pi*1j*n)

dft = numpy.sum(signal*dft_matrix,axis=1)
amplitude = numpy.abs(dft)
phase = numpy.arctan(dft.imag/dft.real)

matplotlib.pyplot.plot(numpy.linspace(f_min,f_max,samples),amplitude)
matplotlib.pyplot.xlabel("Frequency (Hz)")
matplotlib.pyplot.ylabel("Energy / Amplitude")
matplotlib.pyplot.show()
</code>```

But this gets very slow very soon. I want to learn how the FFT works, so that I can use it.
What I know about it, that is has something to do with the "nth root of unity". And even and odd indexes are separated into two separate matrices. There are some redundancies in the DFT, if you analyze just integer frequencies. They are all at 0 in the middle. Even and odd frequencies also cross zero on different fractions. And the Cooley-Tukey-Algorithm/FFT makes use of it.

What I dont understand:

-the connection between all these things (zero points on different fractions, even and odd, nth root of unity ...)

-what seperating even and odd indexes does

-what axis is split into even and odd and what happens with the other because everyone seems only to be talking about one axis.

-what points are not used in the smaller matrices

-if the even and odd matrices are multiplied with the even and odd indexes of the signal

-what operations have to be done with the two resulting vectors to get the output

-what frequencies are used/allowed and if they have to be in a specific order

How does the Cooley Tukey Algorithm Work?

I need it explained in right order with logical connections and as much detail as possible (especially in Math).
I would appreciate it, if someone could do that. Thanks for the answers in advance!

答:
I think the FFT trick becomes obvious once you notice that the DFT of $x(n)1_{2|n}$ (setting to $0$ the odd samples, assuming the signal has even length $N$) is $\frac{X(k)+X(k+N/2)}2$, the DFT of $x(n)1_{2\ \nmid\ n}$ is $\frac{X(k)-X(k+N/2)}2$.

So we can compute the (length $N/2$) DFT of $x(2n)$ and $x(2n+1)$ separately and mix them at the end.

In practice it is a bit more complicated (when $N$ isn't a power of $2$, and there are also some optimizations to compute the $N$th roots of unity) but the idea is there.


问:
Distances are equal in an ellipse.
How would I prove that the distance from the focus F to any point P(x,y) on the ellipse equals the eccentricity times the distance from point P(x,y) to the vertical line x=a/e?</p>

答:
WLOG, the directrix is $x=0$ and the focus $(f,0)$. The given geometric constraint is (considering the squared distances)

$$(x-f)^2+y^2=e^2x^2.$$

By completing the square this is

$$(1-e^2)\left(x-\frac{f}{1-e^2}\right)^2+y^2=\frac{e^2f^2}{1-e^2}.$$

You recognize the equation of an ellipse, and the ratio of the axis is $\sqrt{1-e^2}$ (eccentricity $0$ for a circle).

Title: Distances are equal in an ellipse.


答:
It is always easier to work close to $0$. So, making $x=t+a$
$$\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}=\lim_{t \to 0} \frac{\log(t)}{\log(e^a(e^t-1))}=\lim_{t \to 0} \frac{\log(t)}{a+\log(e^t-1)}$$ Now, using Taylor expansion
$$e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+O\left(t^4\right)$$
$$\log(e^t-1)=\log\big[t+\frac{t^2}{2}+\frac{t^3}{6}+O\left(t^4\right)\big]$$
$$\log(e^t-1)=\log (t)+\frac{t}{2}+\frac{t^2}{24}+O\left(t^3\right)$$
$$\frac{\log(t)}{\log(e^a(e^t-1))}=\frac{\log (t)}{a+\log (t)}-\frac{t \log (t)}{2 (a+\log (t))^2}+O\left(t^2\right)$$ which shows that the limit is $1$ and also how it is approached.

Title: Calculate $\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}$ without using L'Hopital

Body:
I can calculate this easily using L'Hopital Rule. Can anyone give me some pointers on how to do this without using L'Hopital?

$$\lim_{x \to a} \frac{\log(x-a)}{\log(e^x-e^a)}$$

I tried substitution by $n = x+a$ and then $\lim_{n \to 0}$ but that didn't work.


问:
Integral eigenvectors and eigenvalues
I need to find the eigenvalues e eigenvectors of this integral.

a)
$$\int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy$$

b)- Solved thanks
$$\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy$$

答:
Related problem:(I), (II). Here is a start and you have to work out the details. First, you need to find the eigen values $\lambda$ by solving the integral equations

$$ \phi(x) = \lambda\int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy,$$

$$ \phi(x) = \lambda\int_{0}^{1}(x^2y^2-2/45)\phi (y)dy .$$

Now, both of the integral equations have separable kernels which the following method can be used.

Note:

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

$$ \cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y). $$

Added: It seems your book adopting the convention

$$ \lambda \phi(x) = \int_{0}^{2\pi}(\cos^2(x+y)+1/2)\phi (y)dy.$$

Now, using the above hints, we have

$$ \lambda \phi(x) = \int_{0}^{2\pi}\left(\frac{\cos(2x+2y)}{2} + 1/2 \right)\phi(y)dy$$

$$\implies \lambda \phi(x) = \cos(2x)\int_{0}^{2\pi} \frac{\cos(2y)}{2}\phi(y)dy - \sin(2 x) \int_{0}^{2\pi} \frac{\sin(2y)}{2}\phi(y)dy$$

$$ + \int_{0}^{2\pi}\phi(y) dy \longrightarrow (1).  $$

$$ \implies \lambda \phi(x) = \cos(2x) c_1 - \sin(2 x) c_2 + c_3  $$

$$ \implies \phi(x) = \frac{1}{\lambda}\cos(2x) c_1 - \frac{1}{\lambda}\sin(2 x) c_2 + \frac{1}{\lambda}c_3.  $$

Now, subs back in $(1)$ and simplify, you get

$$ \left( c_{{1}}-{\frac {c_{{1}}\pi }{\lambda}} \right) \cos \left( 2\,
x \right) + \left( -c_{{2}}-{\frac {c_{{2}}\pi }{\lambda}} \right)
\sin \left( 2\,x \right) +c_{{3}}-2\,{\frac {c_{{3}}\pi }{\lambda}}=0.$$

Now, by equating the coefficients, we have $\lambda=2\pi,\pi $ and $-\pi$. However, $\lambda=2\pi$ will give the solution $\phi(x)=1$. You can check this solution by substituting in $(1)$.

Title: Integral eigenvectors and eigenvalues


问:
Matrix analysis , inequalities , diagonal entries
Suppose we have an invertible positive definite matrix and we know its diagonal elements when seen as a vector are in increasing order , what can be said about  diagonal vector of its inverse ? , if it's in increasing order or decreasing order etc</p>

答:
No:

for example

$\begin{pmatrix}
3 & 1 & 1 \\
1 & 4 & 3 \\
1 & 3 & 5
\end{pmatrix}$

has an inverse where the diagonal elements are not in order

$\begin{pmatrix}
\;\;\frac{11}{30} & -\frac1{15} & -\frac{1}{30}\\
-\frac1{15} & \;\;\frac{14}{30} & -\frac4{15}  \\
-\frac{1}{30} & -\frac4{15} & \;\;\frac{11}{30}
\end{pmatrix}$


答:
$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

$$10(s+6) =(\alpha s+\beta)(s^2+25)+ (\gamma s+\delta)((s+3)^2+25) $$

$$s=0:\;\; 60= 25\beta + 34 \delta  \tag{1}$$
$$s=1:\;\; 70 =26(\alpha +\beta)+ 41(\gamma +\delta)  \tag{2}$$
$$s=-1:\;\; 50 =26(-\alpha +\beta)+ 29(-\gamma +\delta)  \tag{3}$$
$$s=2:\;\; 80 =29(2\alpha +\beta)+ 50(2\gamma +\delta)  \tag{4}$$

I'm not going to write out the rest of the algebra, but it is simple from there to solve for $\alpha,\beta, \gamma , \delta$, if a bit tedious.

Title: Decomposition of $\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}$

Body:
Determine $\alpha,\beta,\gamma,\delta$ in

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

I have come across the partial fraction and trying to decompose it. I have tried the methods used by multiplying out and making one variable to 0 to find the others etc but It does not work here. Any ideas how I could approach this?


问:
$hf\alpha:I\to S^1$ is not null-homotopic
I was studying this proof, I understood everything, except the part where $hf\alpha:I\to S^1$ is not null-homotopic.

We have, that $f:S^2 \to S^1$ a continous map such that $\space f(-x)=-f(x)$ $ \space \forall \space x \in S^2$,  $h:S^1 \to S^1$ is the continous map $h(z)=z^2$ and $\alpha : I \to S^2 $ is a path with $\alpha(0)=x$ and $\alpha(1)=-x$, ie, has as extremal points two antipodal points.

In fact I have no idea how to show that a map cannot be null-homotopic, the only one I know is the identity map from $S^1$ to $S^1$.

答: You can take any path $\alpha$ in $S^2$ from any $x \in S^2$ to its antipodal point $-x \in S^2$. Then $f\alpha$ is a path in $S^1$ from $y = f(x)$ to $-y = -f(x) = f(-x)$. Let $\gamma$ be the path in $S^1$ running with constant speed counterclockwise from $-y$ to $y$. Then $f\alpha * \gamma $ is a closed path which has a winding number $n \in \mathbb Z$. Hence $h(f\alpha * \gamma) = hf\alpha * h\gamma$ is a closed path which has a winding number $2n$. Note that both parts $hf\alpha$ and $h\gamma$ are closed paths, where $h\gamma$ has degree $1$. But $hf\alpha$ is path homotopic to $h(f\alpha * \gamma) * (h\gamma)^{-1}$ and thus has winding number $2n-1$. Hence it cannot be inessential.


问:
Show that the series of derivatives of an entire function converges
I was asked to show that if $f(z)$ is holomorphic at a neighbourhood of $a$, and $\sum_{n\geq0}f^{(n)}(a)$ converges, then $f$ is entire and $\sum_{n\geq0}f^{(n)}(z)$ converges for all z in the complex plane.It is easy to prove that $f$ is entire since $f^{(n)}(a)$ is bounded and $\frac{|z-a|^{n}}{n!}$ is smaller than a geometric sequence for sufficiently large $n$.
I have trouble with the second assertion. I can only get the result when the convergence at $a$ is absolute. The method was shown Show that the series of derivatives of a certain entire function converges everywhere. But for the non-absolute convergent case, I don't know how to proceed.

Body:
$a=0$
$$\sum_{m=0}^M f^{(m)}(z) = \sum_{m=0}^M\sum_{k\ge 0} f^{(k+m)}(0) \frac{z^k}{k!}
 = \sum_{k\ge 0}\frac{z^k}{k!} \sum_{m=0}^M f^{(k+m)}(0)= \sum_{k\ge 0}\frac{z^k}{k!}(b_k-b_{k+M+1}) $$

where $b_k = \sum_{m=0}^\infty f^{(k+m)}(0)$ converges and $\lim_{k\to \infty} b_k=0$.

Whence $$\sum_{m=0}^\infty f^{(m)}(z)=\sum_{k\ge 0}\frac{z^k}{k!}b_k- \lim_{M\to \infty} \sum_{k\ge 0}\frac{z^k}{k!} b_{k+M+1}$$

$$=\sum_{k\ge 0}\frac{z^k}{k!}b_k- \lim_{M\to \infty} O( \exp(|z|) \sup_{k>M}|b_k|) =\sum_{k\ge 0}\frac{z^k}{k!}b_k$$


问:
What is the probability of a decimal digit in irrational number to repeat itself?
I have this question, by my intuition the probability would be 0, since we do not know what the next irrational number decimal digit will be in base 10.

Does anybody have a more concrete idea, or is there already a solution to this ?

EDIT: Why I think it is not $1/10$ is because we could change the base from base 10 to let's say base 100 and then the probability would be $1/100$. That is also why I think it is zero.

EDIT 2 : What I wanted to ask was, that if we take the decimal digits of an irrational number. Let's say we take those digits in base 10. We do not know what will the next decimal digit be, as they go to infinity. But we could say, that the next digit can be any digit from 0 to 9. But we can change the base of these digits to base 2, where the next digit would have 1/2 probability to be 0 or 1. But that is a contradiction since 1/2 is not the same as 1/10 or if we pick a base 100, it is not the same as 1/100

EDIT 3: So we also have the probability $1/100$ to reconstruct the digits in base 100 back to base 10, since the numbers are always generating at random order since the number is irrational.

答:
First... you seem to be confusing the concept of an irrational number with the concept of a normal number.

It is not true that given a specific irrational number that the digits are equally distributed.  Consider Liouville's Constant written in base 10, $0.11010010001000010000010000001\dots$  This number has no $2$'s, $3$'s, $4$'s, etc... appearing in its decimal expansion anywhere.

Now, you are probably talking about choosing an irrational number at random (which you never clearly specified).  We can in this case refer to the $n$'th digit in base $b$ of such an irrational number and it will be a particular digit with probability $\frac{1}{b}$ and the $n+1$'st digit will be independent of the first and also be that digit also with probability $\frac{1}{b}$.

Let us define random variables.  Let $X$ be our randomly selected irrational number.  Let $X_n^{(b)}$ be the $n$'th digit right of the decimal of $X$ as represented in base $b$.

We do have that $\Pr\left(X_n^{(b)}=X_{n+1}^{(b)}\right)=\frac{1}{b}$ for every $b$.

Your complaint that $\Pr\left(X_n^{(p)}=X_{n+1}^{(p)}\right)=\frac{1}{p}\neq \frac{1}{b}$ is completely irrelevant.  $X_n^{(p)}$ is not the same as $X_n^{(b)}$...  There is no contradiction here.  There is no reason to expect $\Pr\left(X_n^{(b)}=X_{n+1}^{(b)}\right)$ to equal $\Pr\left(X_n^{(p)}=X_{n+1}^{(p)}\right)$.  When you change from one base to another... things change.  Yes, some of the fundamental properties of the number remain the same, however the frequency of the digits does not need to stay the same at all...


问:
Can a number have a prime factor that isn't a part of the number's prime factorization?
Is it possible to have a number, $x$, divisible by some prime, such that that prime does not appear in the unique prime factorization of $x$?</p>

答:
This cannot happen in the natural numbers.  You said your self that factorizations are unique.  If prime $p$ divides $x$, but $p$ doesn't appear in the unique factorization of $x$, then write $x = kp$. Then factor $k$.

Now you have a second factorization of $x$ that does include $p$ violating your uniqueness assumption.


问:
Scaling a compact gives a compact in a locally compact vector space
While wondering about why locally compact vector space are important, I was told that in a locally compact vector space if one scales a compact set by some number, than we still have a compact set, while this might fail in a vector space that is not locally compact. I was trying to see this with the space $\ell^2$ (space of real sequences with finite sum of squares) which is according to this post not locally compact. But then I realized that I don't even know what compact space look like in this vector space.

答: Locally compact topological vector spaces are finite dimensional (see, e.g., Rudin's book) and thus not very interesting for functional analysis.</p>


问:
For real $a$, $b$, $c$ all greater than $1$, show $\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
An inequality question from an olympiad book:

If $a,b,c$ are real numbers satisfying: $a>1$, $b>1$ and $c>1$, then prove that:

$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$

I tried taking $LCM$, but  it doesn't seem to work.

I don't know how to proceed at all. Just a little hint would be very helpful.

答:
Note that $$0=\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}+\ln\dfrac{c^c}{a^a}=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}$$

Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$

We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$

Thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$

Combining these $3$ inequalities gives us $a\ge b$ and $a\ge c$

Observe that (easy to show) $$\ln\dfrac{a^a}{b^b}\ge \ln\dfrac{a}{b}$$

$$\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}\ge \ln\dfrac{a}{b}+\ln\dfrac{b}{c}$$

Then we have

$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)  \prec  \left(\ln\dfrac{a^a}{b^b}, \ln\dfrac{b^b}{c^c}, \ln\dfrac{c^c}{a^a}\right)$$

or (actually only former majorization holds but showing it requires more work, so, including the possibility is better for the sake of shortness)

$$\left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)  \prec  \left(\ln\dfrac{b^b}{c^c}, \ln\dfrac{a^a}{b^b}, \ln\dfrac{c^c}{a^a}\right)$$

Karamata(Majorization) Inequality with using $f(x)=e^x$ ($f''(x)=e^x>0$) gives the desired inequality

$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$


Title: Distances are equal in an ellipse.

Body: How would I prove that the distance from the focus F to any point P(x,y) on the ellipse equals the eccentricity times the distance from point P(x,y) to the vertical line x=a/e?</p>

Body:
From the other direction:

$$\begin{align}
(x-c)^2+y^2&=(x-c)^2+b^2\left(1-\frac{x^2}{a^2}\right)\\
&=x^2\left(1-\frac{b^2}{a^2}\right)-2cx+c^2+b^2\\
&=x^2\frac{c^2}{a^2}-2cx+c^2+b^2\\
&=\frac{c^2}{a^2}\left(x-\frac{a^2}c\right)^2-a^2+c^2+b^2\\
&=\frac{c^2}{a^2}\left(x-\frac{a^2}c\right)^2.
\end{align}$$


问:
$\mathcal{F}_{T}$, when $T$ is constant.
let $T$ be a stopping time, assume it's constant, find $\mathcal{F}_{T}$ :

$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}, A \cap\{T=n\} \in \mathcal{F}_{n}, \forall n \geq 1\right\}$

since $\{T=n\}$ is either the empty set or the universal set and the empty set is always an element of a sigma algebra then :

$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}, A \in \mathcal{F}_{T}\right\}$

so $\mathcal{F}_{T} =\mathcal{F}_{T}$, ?

I'm not sure of the point of the question, is there something I'm missing or is it that simple?

答:
The answer is yes, I think you got lost in notation. Let $j$ be such that $T=j$ in the whole probability space. As you said, $A\cap[T=n]\in \{\emptyset, A\}$. Moreover, to define $\mathcal{F}_T$, we assume that $\{\mathcal{F}_t\}_t$ is a filtration and take $\mathcal{F}_{\infty}= \vee_{t} \mathcal{F}_t$ (the $\sigma$-algebra generated by the union of the finite time filtrations), hence $ \mathcal{F}_\infty \cap \mathcal{F}_n= \mathcal{F}_n$. Therefore,
$$\mathcal{F}_{T}=\left\{A \in \mathcal{F}_{\infty}: A \cap \Omega \in \mathcal{F}_j, A\cap \emptyset \in \mathcal{F}_n, \forall n \neq j  \right\} = \mathcal{F}_j.$$


问:
About the dimension of intersection of null spaces of linear functions
In Page 115 of Axler's book Linear Algebra Done Right (third edition) there is an exercise (numbered 30). See blow:

Suppose $V$ is finite-dimensional and $\phi_1,\dots,\phi_m$ is a linearly independent list in $V'.$ Prove that $$\dim\big((\text{null}\phi_1)\cap\cdots\cap(\text{null}\phi_m)\big)=(\dim V)-m.$$

I have tried for a long time, but failed. In the case of $m=1,$ the result is trivial, just a consequence of the Fundamental Theorem of Linear Maps. But in the case  of $m=2,$ I have tried as follows. Since $\phi_1, \phi_2$ is linearly independent, $$\dim\text{null}(\phi_1)=\dim\text{null}(\phi_2)=\dim V-1.$$ Thus
\begin{align*}
&\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\\
=&\dim(\text{null}(\phi_1))+\dim(\text{null}(\phi_2))-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\
=&\dim V-1+\dim V-1-\dim\big(\text{null}(\phi_1)+\text{null}(\phi_2)\big)\\
\geq &\dim V-1+\dim V-1-\dim V\\
=&\dim -2.
\end{align*}
But how to show the reversed inequality, that is,
$\dim \big(\text{null}(\phi_1)\cap \text{null}(\phi_2)\big)\leq \dim -2?$
Moreover, it seems that the method is hardly valid for $m\geq 3.$
Can anyone help me?

答:
Let $W = \text{null}\:\varphi_{1}\cap\dots\cap \text{nulll}\:\varphi_{m}$. We need to prove that $\text{dim}\:W^0 = m$. It is clear that $\{\varphi_{1},\dots,\varphi_{m}\}$ is a linearly independent subset of $W^{0}$, so $\text{dim}\:W^{0}\geq m$. It can be shown that for any subspaces $U_1$ and $U_2$ of a finite-dimensional vector space $V$ one has $(U_1\cap U_2)^{0} = U_1^{0}+U_2^{0}$. Therefore,
\begin{align}
\text{dim}\:W^0 &= \text{dim}\:(\text{null}\:\varphi_{1}^0+\dots+\text{null}\:\varphi_{m}^{0})\\
&\leq \text{dim}\:\text{null}\:\varphi_{1}^{0}+\text{dim}\:\text{null}\:\varphi_{m}^{0}\\
&= 1+\dots+1\\
&= m.
\end{align}


问:
Question of finding the weakest proposition
could you help me solve this question?

We consider three propositions, $P,Q,R$. We want to find the $R$ that satisfies the following conditions:

a) $P\implies R$ ,

b) $(Q\land R)  \implies P$

c) $¬(R\implies  Q)$

d) $¬(Q\implies  R)$

e) Among all the propositions $R$ that can satisfy the conditions, we want to choose the logically weakest statements such that (a)-(d) hold.

My question is this: A statement $A$ is logically weaker thank $B$ iff $A$ is implied by $B$ but $A$ does not imply $B$.

Given this, we know that  $(Q\implies P)$ could be a good candidate for satisfying a (a) - (d), Since for every $R$ that satisfy the conditions, it holds that $R\implies $ $(Q\implies P)$.

How can we prove that for all $R$ that satisfy the conditions above, $(Q\implies P)$ does not imply any $R$ that implies it, a part from $(Q\implies P)$.

答: $Q \implies P$ is only $\mathbf{false}$ when $Q$ is $\mathbf{true}$ and $P$ is $\mathbf{false}$ . So the only weaker truth function of $P$ and $Q$ is the one that is $\mathbf{true}$ for all values of $P$ and $Q$, which doesn't satisfy conditions (b) and (d).


问:
Integrating over a triangle
Let $\hat T$ be the triangle spanned by $(0,0)$, $(1,0)$ and $(0,1)$. Let
$$I(r,s) = \int_{\hat T} x^r y^s d(x,y)$$
with $r,s\in\mathbb N\cup\{0\}$.
Prove that
$$I(r,s) = \frac{r!s!}{(2+r+s)!}$$

To show this I rewrote the integral to
$$\int_0^1 \int_0^{-x+1} x^r y^s dy dx  = \frac{1}{s+1}\int_0^1 x^r (1-x)^{s+1}dx$$

but I don't know how to continue from this point.
How can I solve this? Or is there another way to prove the proposition?

答:
I think the trick is using integration by parts:

$$I\left[m,n\right] = \int_0^1 x^m (1-x)^{n}dx= \int_0^1 (x^m dx)(1-x)^n$$

Now using the formula$\int{v\ du}= uv-\int{u\ dv}$:

$$I\left[m,n\right]=\frac{x^{m+1}}{m+1}(1-x)^n|_0^1+ \frac{n}{m+1} \int_0^1 x^{m+1}\left(1-x\right)^{n-1}=0+\frac{n}{m+1}I\left[m+1,n-1\right]$$

Now using this formula repeatedly in our case will give us:

$$\frac{1}{s+1}I\left[r,s+1\right]=\frac{1}{r}I\left[r+1,s\right]=\frac{s}{r(r+1)}I\left[r+2,s-1\right]=\frac{s!\ r!}{(r+s+1)!}I\left[r+s+2,0\right]$$

But now,$I[r+s+2,0]$ is simply $\frac{1}{r+s+2}$, ergo:

$$I=\frac{s!\ r!}{(r+s+2)!}$$

Title: Integrating over a triangle


问:
Can we approximate any open set by sub-domains with smooth boundary?
In some books, mainly about PDEs, I read that any open set can be approximated by sub-domain with smooth boundary  (not just piecewise smooth). In 1 dimensional case, this seemly to be quite trivial: for any subdomain, use small open balls to cover its boundary and then mollify the connection parts. But in the higher dimensional case, I think this is not that obvious.

So the first question is:  how can we approximate any open set by sub-domain with smooth boundary?

And the second question is: In what meaning the approximation is? Pointwise, i.e., we can find subdomain $D_n$ with smooth boundary such that $D_n\uparrow A$? uniformly pointwise? Or in the Lebesgue measure sense? etc.

答:
The following proposition can be found in D. Daners, Domain Perturbation for Linear and Semi-Linear Boundary Value Problems, Handbook of differential equations, 2008.

Proposition 8.2.1. Let $\Omega\subset \mathbb{R}^n$ be an open set. Then there exists a sequence of bounded open sets $\Omega_n$ with $C^\infty$-boundary such that $\Omega_n\Subset \Omega_{n+1}\Subset \Omega$ for all $n\in \mathbb{N}$ and $\Omega=\bigcup_{n=1}^\infty \Omega_n$. If $\Omega$ is connected, we can choose $\Omega_n$ to be connected as well.

The proof uses Sard's lemma.


问:
How to determine the existence of certain common eigenvectors in a simpler way?
I met such a linear algebra exercise, and I'm thinking about whether there's any trick to avoid brute-force-computation:

Let $V$ be a finite dimentional vector space over $\mathbb{C}$. Let $W:=V \otimes V \otimes V$, and
$T_{1,2}: W\rightarrow W$ are two linear mappings defined as:
$$T_1(x\otimes y \otimes z)=y\otimes x\otimes z,~~ T_2(x\otimes y \otimes z)=x\otimes z\otimes y.$$

Exercise: determine the pairs $(a_1,a_2)$ of eigenvalues of $T_1$ and $T_2$: $a_j$ is an eigenvalue of $T_j$, such that they share a common eigenvector $w\in W$ nonzero with $T_j(w_j)=a_j w_j$.

So I found that both of them can be diagonalised, with eigenvalues $1$ and $-1$: because we can find a basis for $T_1$ as follows
$$
\left\{\begin{array}{l}
\left(e_{i} \otimes e_{i}\right) \otimes e_{k} \\
\left(e_{i} \otimes e_{j}+e_{j} \otimes e_{i}\right) \otimes e_{k} \\
\left(e_{i} \otimes e_{j}-e_{j} \otimes e_{i}\right) \otimes e_{k}
\end{array}\right., (1\neq i<j\leq d:=dim(V))
$$
and similarly for $T_2$:
$$
\left\{\begin{array}{l}
\left( e_{k} \otimes (e_{i} \otimes e_{i}\right) \\
\left (e_{k} \otimes (e_{i} \otimes e_{j}+e_{j} \otimes e_{i}\right) \\
\left (e_{k} \otimes (e_{i} \otimes e_{j}-e_{j} \otimes e_{i}\right)
\end{array}\right., (1\neq i<j\leq d:=dim(V))
$$
Now the question is: to determine when my choice has a common eigenvector. So obviously $(1,1)$ is correct. But is there any trick to "See" the cases $(1,-1),(-1,1),(-1,-1)$? I'm trying to find a method without too much "brute-force".

Any help/hint is welcome, thanks in advance! (And I guess there's some background behind this exercise. If you have any idea, please give me some clue, thanks!)

答:
Partial answer.

(I) The cases $(1,-1)$ and $(-1,1)$ don't occur.

To see this note that if $w$ is a $\lambda_i$-eigenvector of $T_i$ for $i=1,2$ then $w$ is a $\lambda_1 \lambda_2$-eigenvector of $S=T_1 T_2$. But $S:x\otimes y\otimes z\mapsto z\otimes x\otimes y$ has order $3$ so its eigenvalues lie in $\{1,\omega,\omega^2\}$ (where $\omega$ is a primitive cube root of unity).

(II) Provided $\dim V>2$ the case $(-1,-1)$ does occur.

To see this note that any common eigenvector must be a $1$-eigenvector of $S$. That suggests where we should look for these common eigenvectors.

The $1$-eigenspace of $S$ is spanned by vectors of the forms

(a) $e_i\otimes e_i\otimes e_i$ ;

(b) $e_i\otimes e_i\otimes e_j+ e_i\otimes e_j\otimes e_i+e_j\otimes e_i\otimes e_i$ ;

(c) $v_{(i,j,k)}:=e_i\otimes e_j\otimes e_k+ e_j\otimes e_k\otimes e_i+e_k\otimes e_i\otimes e_j$;

where the indices are distinct.

It can now be checked that $v_{(i,j,k)}-v_{(i,k,j)}$ is a common eigenvector of $T_1$ and $T_2$ both having eigenvalue $-1$.

(III) It would be good to list the triples $(a_1, a_2, w)$ with $T_i w= a_i w$ but this I have not done.

More Complete Answer (For those who know some character theory.)

What we have here is a representation of the group $\langle T_1, S=T_1 T_2\rangle \simeq \langle t, s \mid t^2=s^3=(ts)^2=1\rangle\simeq S_3$ on the space $W=V\otimes V\otimes V$ where $\dim V=n$. Call its character $\psi$.

The character table of $S_3$ is
$$
\begin{array}[l| c c c]
\ & 1 & s & t\\
\iota & 1 & 1 &1\\
\sigma & 1 & 1 & -1\\
\theta & 2& -1 & 0\\
\end{array}.
$$

We can easily calculate the traces $\psi(I)$, $\psi(S)$,  $\psi(T_1)$: as these matrices act by permuting the basis elements $e_i\otimes e_j\otimes e_k$ of $W$ we  need only count how many are fixed. We get at once that $\psi(I)=n^3$, $\psi(S)=n$, and $\psi(T_1)=n^2$. It is then easy to see how $\psi$ decomposes, we  must have
$$
\psi ={n+2 \choose  3}\iota +
{n \choose  3}\sigma +
2{n+1 \choose  3}\theta.
$$

That means that $W$ decomposes into ${n+2 \choose  3}$ $1$ dimensional subspaces on which both $T_1$ and $T_2$ act as multiplication by $+1$; ${n \choose  3}$ $1$ dimensional subspaces on which both $T_1$ and $T_2$ act as as multiplication by $+1$; and $2{n+1 \choose  3}$ $2$ dimensional subspaces where there are no common eigenvectors.


问:
Pulling numbers out of a bag without replacement, probability of pulling a certain sum?
I want to sample without replacement by pulling numbers out of a bag. I want to know the probability of pulling $k$ numbers and having their sum either be equal to, or greater than some value.

This would be directly countable with a small amount of numbers in the bag, but what if there's too many to count?

From this it seems that using Stirling numbers, or something similarly appropriate, would be useful here. But how can I do this without replacement?

In case a direct example is useful:
In a bag of 107 numbers there's:
80 0's,
21 1's,
5 3's,
and one 7.

After pulling 10 numbers without replacement, what is the probability of getting a sum equal to 6? Greater than 6?

答:
Setup for general case.

Let it be that the bag contains $r_i$ numbers $i$ for $i=0,1,\dots,m$ and let $r=\sum_{i=0}^mr_i$.

If you want to calculate the probability on a sum $s$ by pulling $k$ numbers then let $T\subseteq\mathbb N^{m+1}$ be the set of tuples $(n_0,n_1,\dots,n_m)$ that satisfy:

$n_i\leq r_i$ for every $i$.

$\sum_{i=0}^mn_i=k$

$\sum_{i=0}^min_i=s$

Then the probability is:$$\sum_{(n_0,\dots,n_m)\in T}\frac{\binom{r_0}{n_0}\times\cdots\times\binom{r_m}{n_m}}{\binom{r}{k}}$$

This is precise and can be applied to solve the example you mentioned, but will be soon unmanageable in "larger" cases. So actually gives you no answer if "there is too many to count".


问:
The function $w(z)=-(\frac1z+bz),-1<b<1 ,$ maps $|z|<1$ onto ...
I am stuck on the following problem that says:

The function $w(z)=-(\frac1z+bz),-1<b<1 ,$ maps $|z|<1$ onto

  (A) A half plane

  (B) Exterior of the circle

  (C) Exterior of an Ellipse

  (D) Interior of an ellipse

I tried by replacing $z=re^{i \theta}$ but I must admit that I have not gotten very far .Can someone point me in the right direction? Thanks in advance for your time.

答:
Let $z=re^{it}$ with $0 \le r<1$. Then
$$-w(z)=(\frac{1}{r}e^{-it}+bre^{it})=(\frac{1}{r}+br)\cos t+i(-\frac{1}{r}+br)\sin t$$
Let $r=1$, you get the boundary of $-\mathrm{im} \,w$, which is $\big((1+b)\cos t,(b-1)\sin t\big)$. This is an ellipse (circle if $b=0$). Since $-w(z)\to \infty$ as $r\to 0$, $\mathrm{im} \,w$ is the exterior of an ellipse (circle if $b=0$).</p>


问:
Show that for two vector spaces there exist a surjective linear map $\varphi$

If $V = \{(x,y,z) \in \mathbb{R^3} : x+2y-z=0 \}$ and $W=\{(x,y,z) \in \mathbb{R^3} : -x+z=0 \}$ then show that there exists a surjective linear map $\varphi: V\to W$.

How can I show this when $\varphi$ is not given? $\varphi$ is surjective iff $\operatorname{im}(\varphi)=W.$ This is just to show that two sets are the same so i'm thinking of proving this by showing $\operatorname{im}(\varphi) \subset W$ and $W \subset \operatorname{im}(\varphi)$, but i cannot do it. Is this the right way?

答:
Both spaces have dimension 2, since each equation (hyperplane) cuts out one dimension of the ambient space.

Here $\{(1,0,-1),(2,-1,0)\}$ is a basis of $V$ and $\{(1,0,1),(0,1,0)\}$ is a basis of $W$.

Any mapping $V\rightarrow W$ which provides an assignment of the basis vectors is linear, such as $(1,0,-1)\mapsto (1,0,1)$ and $(2,-1,0)\mapsto (0,1,0)$, and in this case also surjective.


问:
$L^2$ version of the Grönwall Lemma
Suppose $I=[0,a]$ is an interval, $u \in C^0[0,a]$ and $B>0$ a constant. If $u$ satisfies
$$
u(t)' \leq  B u(t)
$$
then Grönwall Lemma yields
$$
u(t) \leq u(0) \exp(Bt). \quad (*)
$$
Suppose instead $u$ is a smooth function $u:[0,a] \times [c,d]\to \mathbb{R}$ satisfying
$$
\int_c^d g(s)\partial_t u(t,s) ds \leq B\int_c^d g(s) u(t,s) ds
$$
for all smooth $g:[c,d] \to  \mathbb{R}$. Can one deduce some similar $L^2$-bound on $u$ in this case?

答:
For each fixed $t\in[0,a]$, under your assumptions, it is permissible to set $g(s)=u(t,s)$.
This gives
$$  \int_c^d u(t,s) \partial_t u(t,s) ds \le B \int_c^d u(t,s)^2 ds$$
Note that this can be rewritten as
$$ \partial_t \int_c^d u(t,s)^2 ds \le B \int_c^d u(t,s)^2 ds$$
which is of the previous form (set $v(t)=\int_c^d u(t,s)^2 ds$ to find $v'(t) \le Bv(t)$.)

Title: $L^2$ version of the Grönwall Lemma


问:
Distances are equal in an ellipse.
How would I prove that the distance from the focus F to any point P(x,y) on the ellipse equals the eccentricity times the distance from point P(x,y) to the vertical line x=a/e?</p>

答:
I found here a nice sketch, here it is:

The distance from a focus $S(ae,0)$  to any point $P(x_P,y_P)$ is $\sqrt{(ae-x_P)^2+y_P^2}.$ The eccentricity is $e$ and the distance from point $P(x_P,y_P)$ to the vertical line $x=a/e$ (directrix) is $ae - x_P.$ What you are trying to prove is:

$$\sqrt{(ae-x_P)^2+y_P^2} = e(ae-x_P)$$

Squaring both sides results in

$$y_P^2 = (e^2-1)(ae-x_P)^2$$

Further simplifying would result in the equation of the ellipse, if that's the case, if further simplifying leads to the equation of an ellipse, then the initial claim has to be true! Maybe try using $e = \sqrt{1-b^2/a^2}$?

Title: Distances are equal in an ellipse.


问:
Evaluation of $\lim_{n\to \infty} \frac{1}{n} \text{ln}(1 - a \frac{ n}{(1 - z)^n} )$
I'm trying to find the limit of the function

$$\frac{1}{n} \text{ln}\left(1 - a \frac{ n}{(1 - z)^n} \right)$$

as $n \to \infty$, where a and z are complex and $|z| < 1$.

I plugged this expression into Mathematica and after some experimenting found that it is convergent to some complex number, and that the result is actually independent of a, but I haven't been able to find a rigorous reason why. Can anyone see how you would evaluate this?

答:
Very rough answer: of course it is independent of $a$:

$$\frac{1}{n} \log{\left ( 1-a \frac{n}{(1-z)^n}\right)} \sim \frac{\log{(-a)}}{n} + \frac{\log{n}}{n} - \log{(1-z)} \sim -\log{(1-z)}$$

as $n \to \infty$, for $|1-z| < 1$.  Otherwise, if $|1-z|>1$, the expression goes as

$$-\frac{a}{(1-z)^n} \to 0$$

in that limit.


问:
Show $\cosh([0,\infty))=[1,\infty)$
I have been stuck at this problem for a long time and i cannot figure out a solution.

Prove $\cosh([0,\infty))=[1,\infty)$
Hint: Try to show $\cosh(x)\rightarrow \infty$ for $x\rightarrow\infty$ and the sentence: For a continuously function $f:[a,b]\rightarrow \mathbb{R}$ defined on a closed limited interval $[a,b]\subset \mathbb{R}$ will the codomain contain all numbers between $f(a)$ and $f(b)$.

答:
You could show, that $\cosh$ is strictly increasing (in $[0,\infty)$) by looking at it's derivative.
Furthermore show that it is unbound and use Continuity and the intermediate value theorem.

Title: Show $\cosh([0,\infty))=[1,\infty)$


问:
The difference between validity and entailment in first order logic
I'm a newbie to logic so please forgive me if this is a basic question. I've searched the web and stack exchange but can't seem to find an answer.

The book I'm reading on predicate logic (forallx) says that although the argument '${\rm A}$ therefore ${\rm B}$' may be valid, this doesn't mean that ${\rm A}$ necessarily entails ${\rm B}$.

To me, validity and entailment both seem to have the same definition, namely that ${\rm A}$ entails ${\rm B}$ (or an argument is valid) if it is not possible for the premise to be true and the conclusion false.

Clearly my understanding of the above definition is wrong if entailment and validity are in fact different things, but I can't see the difference between them.

Can anyone enlighten me please? And are the definitions different in propositional logic versus predicate logic?

Many thanks

Max

答:
Consider the following instance:

```<code> (0 != 0) therefore (God exists)
</code>```

Since 0 = 0 the premise is false, hence the implication is valid.  Does this prove God exists?

[note: If you happen to believe in God, change the above conclusion to 'God does not exist'.  The argument still works]

Body:
Consider the following instance:

 (0 != 0) therefore (God exists)

Since 0 = 0 the premise is false, hence the implication is valid.  Does this prove God exists?

[note: If you happen to believe in God, change the above conclusion to 'God does not exist'.  The argument still works]


问:
Decomposition of $\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}$

Determine $\alpha,\beta,\gamma,\delta$ in

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{\alpha s+\beta}{(s+3)^2+25}+\frac{\gamma s+\delta}{s^2+25}$$

I have come across the partial fraction and trying to decompose it. I have tried the methods used by multiplying out and making one variable to 0 to find the others etc but It does not work here. Any ideas how I could approach this?

Body:
Apply your method to proceed as follows

$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}
=\frac{A}{s+5i} + \frac{B}{s-5i} +\frac{C}{s+3+i5}+ \frac{D}{s+3-i5}$$
Then
\begin{align}
&A = \lim_{s\to-5i} \frac{10(s+6)(s+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45+68i}{327}\\
&B = \lim_{s\to5i} \frac{10(s+6)(s-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{-45-68i}{327}\\
&C = \lim_{s\to-3-5i} \frac{10(s+6)(s+3+5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45-41i}{327}\\
&D= \lim_{s\to-3+5i} \frac{10(s+6)(s+3-5i)}{\left[(s+3)^2+25\right](s^2+25)}=\frac{45+41i}{327}\\
\end{align}

and
\begin{align}
& \lambda = A+B = -\frac{90}{327}\\
& \delta =(B -A)\>5i= \frac{680}{327}\\
 & \alpha = C +D = \frac{90}{327}\\
& \beta =(C+D) \>3+(D -C ) \>5i= -\frac{140}{327}\\
\end{align}

Thus
$$\frac{10(s+6)}{\left[(s+3)^2+25\right](s^2+25)}
=\frac{10}{327}\left( \frac{68-9s}{s^2+25} + \frac{9s-14}{(s+3)^3+25}\right)$$


问:
Cube root of $-2+i$
Edit:

My question comes from finding the solutions of this equation using Cardano's method(because our teacher said :D ):

$$x^3-6x+4=0$$

And finally I got:

$$x=(\sqrt[3]{2})\sqrt[3]{-2+\sqrt{-1}}⠀+(\sqrt[3]2)\sqrt[3]{-2-\sqrt{-1}}$$

I want to denest and find the cube roots of this two radicals.

According to this links answers:

First link

Second link

Third link

I tried to find this using norm(actually for dinest):

$$\sqrt[3]{-2+i}$$

So the problem's norm is $5$ and I must find a root with norm = $\sqrt[3]{5}$  .

And I found this :

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)$$

But:

$$(\sqrt{\sqrt[3]{5}⠀-1}⠀+i)^3\neq {-2+i}$$

Anybody can tell me whats my wrong?

And how to correct that?

答:
$-2+i$ has norm (length) equal to $\sqrt{(-2)^2 + 1^2} = \sqrt{5}=5^{\frac12}$.
So the norm of a cube root of it must have norm $(5^{\frac12})^{\frac13} = 5^{\frac16} = \sqrt[6]{5}$.

Now look at the argument of $-2+i$ and take a third of it.

Title: Cube root of $-2+i$


问:
Difference between domain of a function and domain ( a connected open set ) used in complex analysis .
What is the difference between domain of a function and domain ( a connected open set ) used in complex analysis hugely? To be a domain of a function if the set should be always open and connected then my concept of a function is totally contradicted. So there should be a difference between these two domains. Please explain.</p>

答:
Yes, these are slightly different uses of the word (just like how in any language the same word can have multiple meanings based on the context). The domain of a function is just a set. Apriori, no further restrictions are needed. The domain of a function can even be the following set of "symbols" $E=\{\ddot{\smile},\ddot{\frown}, \ddot{\smile}\ddot{\smile}, @, \#\}$. THe domain of a function does not have to be a subset of complex numbers or real numbers or anything.

The use of "domain" to mean a connected open subset of $\Bbb{C}$ is indeed a different meaning to above. I've also heard people use the term "region" to describe a connected open set. For example, people use this terminology when referring to things like "domain of integration/region of integration" etc.

Sometimes, of course, if you're considering a function $f:U\to\Bbb{C}$, where $U\subset \Bbb{C}$ is open and connected, then we can say "the domain of $f$ is a domain", where of course the two uses of the word "domain" in this same sentence have different meanings. Based on the context, you should be able to decipher which meaning is intended.


答: The integral you arrived at is an example of the Beta function. It is easily evaluated for natural numbers $ x, y $ by the method Ali posted.</p>

Title: Integrating over a triangle

Body:
Let $\hat T$ be the triangle spanned by $(0,0)$, $(1,0)$ and $(0,1)$. Let
$$I(r,s) = \int_{\hat T} x^r y^s d(x,y)$$
with $r,s\in\mathbb N\cup\{0\}$.
Prove that
$$I(r,s) = \frac{r!s!}{(2+r+s)!}$$

To show this I rewrote the integral to
$$\int_0^1 \int_0^{-x+1} x^r y^s dy dx  = \frac{1}{s+1}\int_0^1 x^r (1-x)^{s+1}dx$$

but I don't know how to continue from this point.
How can I solve this? Or is there another way to prove the proposition?


问:
Does $A_1\supset A_2 \supset \cdots $ imply $\bigcap A_n \neq \emptyset$?
given a countable collecstion of sets $\{A_n\}$ which satisfies
$$A_1\supset A_2 \supset \cdots \supset A_n\supset \cdots $$

My intuition is
$$
\bigcap A_n  \neq \emptyset
$$
Please show me a counter-example.

答: $A_n = (0, \frac1n) \subseteq \Bbb R$ is a counterexample.


问:
Representation of primes as $2(2^x+n)+1$. Then, what is the generalization of n??
I have written a similar post a while back, but want to highlight this again.
We represent odd numbers as $2n+1$. Therefore, all the composites are in the form of $(2n+1)(2m+1)$ = $2(2mn+m+n)+1$. Let $2mn+m+m$ be a.

Therefore, $2a+1$ is a composite when $a=2mn+m+n$ and $2a+1$ is a prime when a is not expressible as $2mn+m+n$. Now $2mn+m+n$ can be written as $(2mn+m)+n$ and $2mn+m=(2n+1)m$.

The only numbers that CANNOT be represented as $(2n+1)m$ is $2^x$. Now, talking about primes as $2a+1$, $a≠2mn+m+n$ and $2^x≠2mn+m$.

That gives us, $2^x+n≠2mn+m+n$ (inequality is maintained here).
So, if we have a generalization for n, we can represent primes as $2a+1=2(2^x+n)+1$ i.e. $a=2^x+n$.

答:
Comment:

Due to Fermat little theorem, if $(2, x)=1$ then:

$2^x-2\equiv 0\bmod x$

We rewrite a as:

$a=2^x-2 +n+2$

We want a prime, so we must have:

$[x, (n+2)]=1$

Also: $[(n+2),2]=1$

So general form of n can be:

$n=p+2$

Where p is some particular primes; for example:

$x=3, p=3\Rightarrow a=2^3+3+2=13$

$x=3, p=7\Rightarrow a=2^3+7+2=17$

$x=5, p=3\Rightarrow a=2^5+3+2=37$

$x=5, p=7\Rightarrow a=2^5+7+2=41$

$x=7, p=3\Rightarrow a=2^7+7+2=137$

$x=7, p=19\Rightarrow a=2^7+19+2=149$

There can be other forms too. $n=p+2$ is one form.


答: $A_n = \{n, n+1, n+2, \cdots\}$ is a counter-example.

Title: Does $A_1\supset A_2 \supset \cdots $ imply $\bigcap A_n \neq \emptyset$?

Body:
given a countable collecstion of sets $\{A_n\}$ which satisfies
$$A_1\supset A_2 \supset \cdots \supset A_n\supset \cdots $$

My intuition is
$$
\bigcap A_n  \neq \emptyset
$$
Please show me a counter-example.


答:
$\cosh(x)$ is a continuous function.

$\cosh(x)=\dfrac{z+z^{-1}}2$ with $z=e^x>0.$

in the positive $z$, $z+z^{-1}\ge2$ because $\dfrac{(z-1)^2}z\ge0$. The bound is tight because $1+1^{-1}=2.$

$e^x$ is unbounded (take $x\ge n$ and $e^x>e\cdot e\cdots e$), and $2\cosh(x)>e^x.$

So $\cosh(x)$ covers all values from $1$ inclusive to $\infty$.

Title: Show $\cosh([0,\infty))=[1,\infty)$

Body:
I have been stuck at this problem for a long time and i cannot figure out a solution.

Prove $\cosh([0,\infty))=[1,\infty)$
Hint: Try to show $\cosh(x)\rightarrow \infty$ for $x\rightarrow\infty$ and the sentence: For a continuously function $f:[a,b]\rightarrow \mathbb{R}$ defined on a closed limited interval $[a,b]\subset \mathbb{R}$ will the codomain contain all numbers between $f(a)$ and $f(b)$.


问:
Curious Binomial identity that came up in derivative pricing
How can I show that, for a given integer $n$, that the following identity is true?
$$\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]=-(-1)^m$$

This came up in a derivation of arbitrary derivatives of the Black-Scholes function.  It is part of a larger equation for the $n^{th}$ order derivative
$$\sum_{m=0}^{n-2}\frac{H_m(d)}{(\sigma\sqrt{\tau})^m}\left[(-1)^n\binom{n-1}{m}-\sum_{k=m}^{n-2}(-1)^k\binom{n}{k+1}\binom{k}{m}\right]$$
which, in order to make contact with a known result (Carr, Peter. (2000). Deriving Derivatives of Derivative Securities. Journal of Computational Finance. 4. 101 - 128.
https://ieeexplore.ieee.org/document/844609) this identity must hold.  I've also checked that it's true in sagemath.

答:
Have an alternative solution that takes a different approach than the one here. Consider mine more opaque and involved, but worth sharing for the fun of it.

We start with Eq. 1.23 in this reference changing the labeling for convenience
$$\sum_{k=0}^{n'}(-1)^k \binom {n'} k \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m} $$
and apply $\binom n k = \frac{n}{k} \binom {n-1} {k-1}$ as
$$\sum_{k=0}^{n'}(-1)^k \frac{k+1}{n'+1}\binom {n'+1} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n'-m} \binom{n'}{m}. $$
Substitution $n=n'+1$ then yields
$$\sum_{k=0}^{n-1}(-1)^k \frac{k+1}{n}\binom {n} {k+1} \binom {k} m x^k=(-1)^mx^m(1-x)^{n-m-1} \binom{n-1}{m}. $$
Integrate both sides, assume that constants are 0, $x=1$ to get
$$\frac{1}{n}\sum_{k=0}^{n-1}(-1)^k \binom {n} {k+1} \binom {k} m=\frac{(-1)^m}{n}. $$
Integral on the RHS is tricky in general, but evaluates nicely in the present case.

Now we can split off the $n-1$ term from the sum and multiply by $n$ to obtain
$$(-1)^{n-1} \binom {n-1} m + \sum_{k=0}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=(-1)^m. $$
Finally, we multiply by $-1$, note that binomial coefficients with $k < m$ in the sum are 0 and get
$$(-1)^{n} \binom {n-1} m - \sum_{k=m}^{n-2}(-1)^k \binom {n} {k+1} \binom {k} m=-(-1)^m. $$