问:
Order statistics and their CDF
Here's a problem that seems rather peculiar to me. This time I have no initial idea about how to solve it.

Let $X_1$, ..., $X_n$ be independent, real valued random variables with density $f$ and CDF $F$. Let $F_i$ denote the CDF of $X_{(i)}$.

a) What is the distribution of $F(X_{(i)})$ and $F_i(X_{(i)})$?

b) What is the variance of $F(X_{(i)})$?

[edit: I have realised by now that my approach was flawed and furthermore that my question needs clarification. So here is my second try.]

As I understand the question, we have real valued, i.i.d. random variables $X_1$, ..., $X_n$ with density $f$ and CDF $F$.

$X_{(1)}< ... < X_{(n)}$ are the corresponding order statistics with CDF $F_i:=F_{X_{(i)}}$.

I know the general formula for the CDF of order statistics. It is given by $$F_i(t)=\sum_{k=i}^n \binom{n}{k}F(t)^k(1-F(t))^{n-k}.$$

Now, the CDF of a random variable is a measurable function. Thus $F(X_{(i)})$ and $F_i(X_{(i)})$ are real valued random variables again. And a) asks for their distribution.

Per definition, we have $$F(t)=\mathbb{P}(X_i\leq t).$$ Hence $$F(X_{(i)})=\mathbb{P}(X_i\leq X_{(i)}).$$ This is the probability of the event that any $X_i$ is less or equal $X_{(i)}$. By definition of order statistics, we have $$F(X_{(n)})=\mathbb{P}(X_i\leq X_{(n)})=1,$$ as $X_{(n)}$ is the maximum of the $X_i$. But how do I derive the distribution of $F(X_{(i)})$ for $i \in \{1, ..., n-1\}$?

I think, if they were uniformly distributed, the answer would simply be $$F(X_{(i)})=i/n.$$ But their distribution ist unknown, so I'm stuck and have no idea how to proceed from here.

答:
First recall the most important result in order statistics:

For every random variable $Z$ with continuous CDF $H$, $H(Z)$ is uniform on $(0,1)$.

Thus, if $(X_i)_{1\leqslant i\leqslant n}$ has continuous CDF and $(U_i)_{1\leqslant i\leqslant n}$ is an i.i.d. sample uniform on $(0,1)$, then, for every $1\leqslant i\leqslant n$ and every $x$ in $(0,1)$,
$$
P(F_i(X_{(i)})\leqslant x)=P(U_1\leqslant x)=x.
$$
And $(F(X_i))_{1\leqslant i\leqslant n}$ is distributed as $(U_i)_{1\leqslant i\leqslant n}$, thus, the CDF $G_i$ of $F(X_{(i)})$ is such that, for every $x$ in $(0,1)$,
$$
G_i(x)=P(F(X_{(i)})\leqslant x)=P(U_{(i)}\leqslant x)=\sum_{k=i}^n{n\choose k}t^k(1-t)^{n-k}.
$$
Recall a most useful result to compute expectations:

For every $(0,1)$-valued random variable $Z$ with CDF $H$,
  $E[Z]=\displaystyle\int_0^1(1-H)=1-\int_0^1H$.

Hence,
$$
E[U_{(i)}]=1-\sum_{k=i}^n{n\choose k}\int_0^1t^k(1-t)^{n-k}\mathrm dt.
$$
Recall now that:

For every $k\leqslant n$,
  $\displaystyle\int_0^1t^k(1-t)^{n-k}\mathrm dt=\frac1{n+1}{n\choose k}^{-1}$.

Hence,
$$
E[U_{(i)}]=1-\sum_{k=i}^n\frac1{n+1}=\frac{i}{n+1}.
$$
Recall finally the analogue for second moments of our result for expectations:

For every $(0,1)$-valued random variable $Z$ with CDF $H$,
  $\displaystyle E[Z^2]=\int_0^12x(1-H(x))\mathrm dx$, that is, $\displaystyle E[Z^2]=1-2\int_0^1xH(x)\mathrm dx.$

Hence,
$$
E[U_{(i)}^2]=1-2\sum_{k=i}^n{n\choose k}\int_0^1t^{k+1}(1-t)^{n-k}\mathrm dt=1-2\sum_{k=i}^n{n\choose k}\frac1{n+2}{n+1\choose k+1}^{-1},
$$
that is,
$$
E[U_{(i)}^2]=1-\frac2{(n+2)(n+1)}\sum_{k=i}^n(k+1)=\frac{i(i+1)}{(n+1)(n+2)},
$$
from which you can probably guess an expression of $E[U_{(i)}^k]$ valid for every nonnegative integer $k$, and from which, independently, one deduces that
$$
\mathrm{var}(F(X_{(i)}))=\mathrm{var}(U_{(i)})=\frac{i(i+1)}{(n+1)(n+2)}-\left(\frac{i}{n+1}\right)^2=\frac{i(n+1-i)}{(n+1)^2(n+2)}.
$$

Title: Order statistics and their CDF


问:
Factorising a spacial polynomial
I have to factorise $2(x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$. It seems that there a very simple expression to which simply this polynomial.

The only thing I found so far is $2 x^2 (1 + x + x^2 + x^3 + x^4 + x^5)$. Can you explain to me in  details how to factorise this even more? If so, is there a general rule for such similar polynomial?

ContentLicense: CC BY-SA 4.0

Title: Factorising a spacial polynomial

Body:
$$2(x^2+x^3+x^4+x^5+x^6+x^7)$$

$$2x^2(1+x+x^2+x^3+x^4+x^5)$$

$$2x^2\frac{x^6-1}{x-1}$$


问:
Prove $2^n > n^2 + n + 1$
I did this but I'm pretty sure it's wrong

$P(5) $ is true
Assume $P(k)$ is true
for some $k \ge 5$, ie. $2^k > k^2 + k + 1$

$$2^{k+1} = (k + 1)^2 + (k + 1) + 1$$
$$2^{k+1} = k^2 + 3k + 3$$

Then:

$$2^{k+1} > (k^2 + k + 1) \cdot 2 > k^2 + 3k + 3$$

$$2^{k+1} > k^2 > k + 1$$

From the onset we know that:

$$2^{k+1} > 2^k > k^2 + k + 1$$

Therefore:

$$2^{k+1} > 2^k$$ (which is true)

Therefore, by the PMI, $P(k)$ is true for all $k \ge 5$

答: You proof is correct to the point where you need to prove that $2(k^2+k+1)>{(k+1)}^2+k+1+1$, or that $k^2-k-1>0$. For this last part you either redo the induction or use $k^2>2k$, which is true for k>2.

Title: Prove $2^n > n^2 + n + 1$


问:
Computing The Expected Value and The Variance
I've done the half of this though question but I couldn't figure out the other half. Any help would be much appreciated!</p>
```<code>Mike is a salesperson for a large retailer and he is covering long distances with his car every day.
The number of highway tolls Mike will need to go through on any day can be modelled as a random
variable X and it is assumed that it has the PMF function shown below, (I computed the mean and the
variance, mean E(X) = 4.91 and variance = 0.9419). At each toll, Mike will be ticketed a $7 bill and, at
the end of the day, he will get a single $3 discount, thanks to his frequent-user discount card for
highway tolls.
</code>```
```<code>k         6      5       4         3        2

Px(k)    0.3    0.4     0.25      0.01     0.04
</code>```
```<code>Let the total highway costs incurred by Mike on a single day be modelled by another discrete random
variable T. Compute the expected value and variance of T.
</code>```

答:
You said that $\mathbb{E}(X) =4.91$ and $Var(X)=0.9419$ (I didn't check that they were correct).

Since each toll costs 7, $X$ is the number of tolls, and he has a single discount of 3 at the end of the day, we can say that the total cost incurred by Mike in a day will be
$$ T = 7X - 3 .$$

Then, we can compute the mean and variance of $T$ using the properties of mean and variance:

$$ \mathbb{E}(T) = \mathbb{E}(7X-3) = 7 \mathbb{E}(X) -3 = 31.37 $$
$$ Var(T) = Var(7X-3) = 7^2 Var(X) = 46.1531 $$

EDIT:
I used the following properties:

$ \mathbb{E}(X+Y) = \mathbb{E}(X) + \mathbb{E}(Y) $,
for any couple of random variable $X$ and $Y$.

$ \mathbb{E}(a X) = a \mathbb{E}(X)$, for any random variable $X$, for any real number $a \in \mathbb{R}$.

$ Var(X+Y) = Var(X) + Var(Y) $, for random variable $X$ and $Y$ which are independent.

$ Var(aX) = a²Var(X)$, for any random variable $X$, for any real number $a \in \mathbb{R}$.

$ Var(a) = 0$, for any constant random variable $a \in \mathbb{R}$.


答:
You started well but the end of your proof is not right and it is unclear when you write 2 inequalities.

We want to prove $2^{k+1}>(k+1)^2+(k+1)+1$ knowing already that $2^k>k^2+k+1$.

$2^{k+1}=2\times2^k>2\times(k^2+k+1)=2k^2+2k+2$ but we wanted to prove $2^{k+1}>(k+1)^2+(k+1)+1=k^2+3k+3$ so it is enough to prove that $2k^2+2k+2>k^2+3k+3$.

Which is the same as $k^2>k+1\iff k(k-1)-1>0$ which is always true for any $k\geq5$.

Then $2^{k+1}>2\times(k^2+k+1)>(k+1)^2+(k+1)+1$ and so the proof is done.

Title: Prove $2^n > n^2 + n + 1$

Body:
I did this but I'm pretty sure it's wrong

$P(5) $ is true
Assume $P(k)$ is true
for some $k \ge 5$, ie. $2^k > k^2 + k + 1$

$$2^{k+1} = (k + 1)^2 + (k + 1) + 1$$
$$2^{k+1} = k^2 + 3k + 3$$

Then:

$$2^{k+1} > (k^2 + k + 1) \cdot 2 > k^2 + 3k + 3$$

$$2^{k+1} > k^2 > k + 1$$

From the onset we know that:

$$2^{k+1} > 2^k > k^2 + k + 1$$

Therefore:

$$2^{k+1} > 2^k$$ (which is true)

Therefore, by the PMI, $P(k)$ is true for all $k \ge 5$


问:
Zero set of ideal of an algebraic set is the algebraic set itself
An algebraic set defined by $S \subseteq k[x_1,...,x_n]$ is given by

$$X = \{a \in \mathbb{A}^n(k) | f(a) = 0, \forall f\in S\}$$

So this would the set of points that make polynomials in $S$ vanish.

The ideal of a set $X \subset \mathbb{A}^n(k)$ is given by:

$$I(X) = \{ f \in k[x_1,...,x_n] | f(a) = 0, \forall a \in X \}$$

So this is the set of polynomials that vanish on points in $X$.

Lastly, the zero set of $I(X)$ is given by:

$$V(I(X)) = \{a \in \mathbb{A}^n(k) | f(a) = 0, \forall f \in I(X)\}$$

which is the set of points that make polynomials in $I(X)$ vanish.

Now I am trying to prove that if $X$ is an algebraic set then $V(I(X)) = X$, which is an exercise that I have taken from a book I'm reading ("Algbraic Geometry: A Problem Solving Approach"). My intuition tells me that this would be the case if I could prove that $S = I(X)$. I can see that $S \subseteq I(X)$ since if a polynomial $f$ is in $S$ then of course we have that $f(a) = 0, \forall a \in X$. Then it would naturally follow that $X = V(S) = V(I(X))$, however I am struggling to see why $I(X) \subseteq S$ since $I(X)$ might contain a polynomial $g$ s.t. $g(a) = 0$ but $g \notin S$.

答:
Observe that $V$ and $I$ are inclusion reversing: (a) if $X\subseteq X'\subseteq \mathbb{A}^n(k)$, then $I(X')\subseteq I(X)$, and (b) if $Y\subseteq Y'\subseteq k[x_1,\dots,x_n]$, then $V(Y')\subseteq V(Y)$.

Next, observe that (c) for $X\subseteq \mathbb{A}^n(k)$ and $Y\subseteq k[x_1,\dots,x_n]$, we have $X\subseteq V(Y)$ if and only if $Y\subseteq I(X)$, since both hold if and only if every  polynomial in $Y$ vanishes on every point in $X$.

This gives the structure of a Galois connection, and your desired result follows purely formally.

Since $I(X)\subseteq I(X)$, by (c), $X\subseteq V(I(X))$. For the other inclusion: since $X$ is an algebraic set, $X = V(Y)$ for some $Y\subseteq k[x_1,\dots,x_n]$. Now $X\subseteq X = V(Y)$, so by (c), $Y\subseteq I(X)$, and by (b), $V(I(X)) \subseteq V(Y) = X$.

The same argument (reversing the roles of $X$ and $Y$ and using (a) instead of (b) in the last step) shows that for $Y\subseteq k[x_1,\dots,x_n]$, if $Y = V(X)$ for some $X\subseteq \mathbb{A}^n(k)$, then $I(V(Y))= Y$. If $k$ is algebraically closed, then the ideals which are "closed" in this sense (i.e. which arise as $V(X)$ for some $X\subseteq \mathbb{A}^n(k)$) are exactly the radical ideals, by Hilbert's Nullstellensatz.


Title: Factorising a spacial polynomial

Body:
I have to factorise $2(x^2 + x^3 + x^4 + x^5 + x^6 + x^7)$. It seems that there a very simple expression to which simply this polynomial.

The only thing I found so far is $2 x^2 (1 + x + x^2 + x^3 + x^4 + x^5)$. Can you explain to me in  details how to factorise this even more? If so, is there a general rule for such similar polynomial?

Body:
Hint:

$x^2+x^3=x^2(1+x)$

$x^4+x^5=x^4(1+x)$

You can continue...


问:
Representing $n$-linear functions as some sort of array?
Let $T$ be a linear transformation between vector spaces $$T: V \rightarrow W$$

Once we select a basis for $V$ and $W$, we can represent $T$ by a matrix.

Given a bilinear functional $$B:V \times W \rightarrow \mathbb{R}$$
Once we select a basis for $V$ and $W$ we can represent $B$ by a matrix.

But what if we had a third vector space equipped with a basis, say $Z$, and we had a bilinear function $$M: V \times W \rightarrow Z$$

I believe we could no longer represent this function by a matrix. But what if we had an array in 3 dimensional space? Then could we represent $M$ by a matrix?

What if we had an $n$-linear function $$V_1 \times ... \times V_n \rightarrow W$$ or perhaps a $n+1$-linear functional $$V_1 \times ... \times V_n \times V_{n+1} \rightarrow \mathbb{R}$$

Could you represent each of these as an $n+1$-dimensional array?

Thank you

答:
Yes. To keep it simple, let's assume all the $V_i$ are the same, OK? Pick a basis $v_1, \ldots, v_k$ for $V$. Take your $n$-linear functional $f$, and compute
$$
f(v_1, v_1, \ldots, v_1) = c_{1,1,1, \ldots, 1} \\
f(v_1, v_1, \ldots, v_2) = c_{1,1,1, \ldots, 2} \\
\vdots\\
f(v_k, v_k, \ldots, v_k) = c_{k,k,k, \ldots, k}
$$
i.e., compute all $k^n$ coefficients by plugging in all possible combinations of subscripts. This is an $n$-index array of coefficients. For $n = 1$, this is a list; for $n = 2$ (the bilinear case) it's a matrix; for larger $n$ it's a cube or hypercube or whatever of coefficients.

And these completely determine your function, because of $n$-linearity.

In the event that your function has some degree of symmetry (e.g., for a 2-arg function, if you have $f(v_1, v_2) = f(v_2, v_1)$), then there are some relations among the coefficients. (In the 2-arg case, it means that the coefficient matrix is symmetric; arguably the anti-symmetric case is more interesting, at least for geometry, where my interests lie.)


问:
Find probability mass function of random variable

Given two lottery tickets, out of which one can be bought. The first lottery \$100 can be won with probability 0.1, and the price of ticket is \$10. In the second lottery, \$50 can be won with probability 0.1 and \$500 with probability 0.01. The price of ticket is \$20. To decide which ticket to buy a fair coin can be tossed. In case of head, first ticket is chosen else the second ticket. Let $X$ be random variable that denotes the net payout (taking into account price of a ticket). Find probability mass function of $X$?

I tried doing the above by using law of total probability by using the conditional probability $P(X|H)$ and $P(X|T)$ and the respective probabilities $P(H)=\frac{1}{2}$ and $P(T)=\frac{1}{2}$. So here $P(X)=P(X|H)*P(H)+P(X|T)*P(T)$. Here I assumed $P(X|H)=p$ and $P(X|T)=q$. So here $\Sigma p=1$ and $\Sigma q=1$ for the first and the second lottery ticket.

Would the expected value of $X$ be dependent on different payout or probabilities? In the above case, it is coming as average of the expected values of net payouts for each of two lotteries.

答:
What are the possible discrete values that random variable $X$ (net payout) can take on and what are their probabilities? For example, one can have net payout of $30$ with probability $\frac{1}{2} \times 0.1 = 0.05$. The same is true for net payout of $90$.

$p(x)$ = $\left
\{\begin{array}
{l}0.05 \ \ \ \ \ \ \ \ x \in \{30, 90\} \\
... \\
...
\end{array}
\right.$

Can you complete this by finding other discrete values of $X$ and associated probabilities?


问:
Explanation for $k{n \choose k - 1} = (k-1){n \choose k - 1} + {n \choose k - 1}$
Does anyone have a good explanation for why this makes sense? I am trying to remember this fact in a proof but simply can't seem to make a good explanation for it.

NOTE: I have already shown that  $k{n \choose k - 1} =  (k-1){n \choose k - 1} + {n \choose k - 1}$ through mathematically reducing the RHS to the LHS. But I would like a meaning/explanation in words for why it makes since that the LHS would still equal the RHS after manipulation. My professor mentioned something about if we take away a k, then we have to add it back or something.

Any attempt to explain is great! Thanks!

答: $k=(k-1)+1$, now multiply both sides by $\binom{n}{k-1}$


问:
contruced a $*$-isomorphism between two von neumann algebras
Let $M$ be a von Neumann algebra and $p$ is a projection in $M$. Can we contruct a $*$-isomorphism between $M$ and $pMp$?

答:
Try with $M=\mathbb C\oplus\mathbb C$ and $p=(1,0)$.

The question is more relevant in the case of II$_1$-factors, and it was been studied already by Murray and von Neumann 80 years ago (the Fundamental Group). From the uniqueness of the hyperfinite II$_1$-factor $R$ it follows that $R\simeq pRp$ for all nonzero projections $p\in R$. At the other end of the spectrum, Popa has constructed a II$_1$-factor $M$ such that $M\not\simeq pMp$ for all nontrivial projection $p\in M$.


问:
How to expand this tensorial Taylor expansion to the $n$th term?
Wikipedia describes the use of the Hessian matrix in a Taylor series expension. I've noted that the first term is written in terms of the original function, the second term uses the gradient of the function, and the third term uses the Hessian of the function. While each term is a scalar, I see a pattern of using higher rank tensor operations as the order of approximation increases.

$$y = f(\vec{x} + \Delta \vec{x}) \approx f(\vec{x}) + \nabla f(\vec{x}) \Delta \vec{x} + \frac{1}{2} \Delta \vec{x}^T H(\vec{x})\Delta \vec{x}$$

How can we expand this approximation to an arbitrary $n$th term? I am expecting, or hoping, there is a way to do this in terms of tensors.

答:
You could write
$$
f(x+Δx)=f(x)+f'(x)Δx+\tfrac12f''(x)[Δx,Δx]+\tfrac16f'''(x)[Δx,Δx,Δx]+...
$$
where the derivatives are tensors, specifically vector valued symmetric multi-linear forms. The non-linear arguments are separated from the linear arguments of the tensor.

For instance for the exact solution of $x'=f(x)$ you get by recursively inserting the differential equation into derivatives of itself
$$\begin{align}
x(t+h)&=x(t)+hf(x)+\tfrac12h^2f'(x)f(x)+\tfrac16h^3\Bigl(f''(x)[f(x),f(x)]+f'(x)^2f(x)\Bigr)\\
&~~~+\tfrac1{24}h^4\Bigl(f'''(x)[f(x),f(x),f(x)]+3f''(x)[f'(x)f(x),f(x)]+f'(x)f''(x)[f(x),f(x)]+f'(x)^3f(x)\Bigr)+..
\end{align}$$
and latest at this point the motivation for B-trees and B-series begins to make sense.


问:
Metric entropy: Question to partition if $N\to\infty$
I have a question concerning this Wikipedia on metric entropy.

There, the following object is defined:
$$
\bigvee_{n=0}^NT^{-n}Q=\{Q_{i_0}\cap T^{-1}Q_{i_1}\cap\ldots\cap T^{-N}Q_{i_n}: \textrm{where }i_l=1,\ldots,k, l=0,\ldots,N\}
$$
and each element has positive measure.

I am wondering if this is also defined if $N\to\infty$ and how it looks like: Does all elements still have positive measure for $N\to\infty$? In other words, what is
$$
\bigvee_{n=0}^{\infty}T^{-n}Q?
$$

EDIT: I think $\bigvee_{n=0}^{\infty}T^{-n}Q$ is usually defined to be the smallest $\sigma$-algebra which contains $\bigvee_{n=0}^N T^{-n}Q$ for each $N\in\mathbb{N}$ (see, for example, here, Definition 5).

答:
Indeed, $\bigvee_{n=0}^{\infty}T^{-n}Q$ denotes the $\sigma$-algebra generated by the elements of $\bigvee_{n=0}^NT^{-n}Q$ for all $N\in\mathbb N$.
It may happen for example that all elements of $\bigvee_{n=0}^{\infty}T^{-n}Q$ besides the empty set have positive measure or that only the empty set has zero measure. That depends on each measure and on each map.

For example, take $T(x)=2x\bmod1$ on $[0,1]$ with the Lebesgue measure. The Borel $\sigma$-algebra has the cardinality of $\mathbb R$ and so for example for $Q$ composed of $[0,\frac12]$ and $[\frac12,1]$ it is impossible that the sets in $\bigvee_{n=0}^NT^{-n}Q$ for all $N\in\mathbb N$ contain all the elements of the $\sigma$-algebra.

Two corrections on your post: you need to erase $l=0,\ldots,N$ in the definition of  $\bigvee_{n=0}^NT^{-n}Q$ and it is false in general that all the sets in $\bigvee_{n=0}^NT^{-n}Q$ have positive measure (note that in fact some may even be empty).


问:
Bounds on the zeros of a polynomial
I'm stuck on the following problem:

Let $p(z) = z^n + a_{n-1}z^{n-1}+...+a_0$ be a polynomial of degree $n$, and let $A = \max\{|a_i|: 0\leq i \leq n-1\}$. Show that $p(z)$ has $n$ roots (counted with multiplicity) in the disk $|z|<A+1$.

I have attempted (but failed) to show that $|p(z)|>0$ on $|z|\geq A+1$. Then I would be done by the fundamental theorem of Algebra.

I wonder whether this is just a slick application of Rouche's theorem.

There is a proof on Wikipedia which uses Holder's inequality, which I haven't seen yet.

Could anyone provide some hints for a proof which does not use this inequality? Am I even on the right track?

答:
Only the triangle inequality is needed, not even Rouché's theorem. First note that we can assume that $A>0$, because the statement surely holds for the polynomial $p(z) = z^n$.

Now assume that $p(z) = 0$.

Case 1: $|z| \le 1$. Then $|z| \le 1 < 1+A$.

Case 2: Then
$$
 |z^n| \le |a_{n-1}| |z|^{n-1} + \cdots + |a_1||z|+ |a_0| + \le A (|z|^{n-1} + \cdots + |z|+1) = A \frac{|z|^n-1}{|z|-1}
$$
and it follows that
$$
 |z|-1 \le A \frac{|z|^n-1}{|z|^n} < A \, .
$$

This shows that $|z| < A+1$ for all roots of $p$.

Title: Bounds on the zeros of a polynomial


问:
What is the difference between an unbounded solution and infinite optimal solutions?
I am currently studying Linear Programming, and one of the few things that I haven't found a solution for (pun obviously intended) is the difference between a solution being either unbounded or having infinite optimal solutions.

My thinking is, if something is unbounded, is that not the same as it being infinite? I must not have a proper understanding of both, as I cannot see why my thinking would correct.

If possible, could someone illustrate the differences with a graph?

答:
Unbounded Solution refers to one solution being unbounded, in the sense that the $y = x$ is unbounded.

Infinite Optimal Solutions refers to an infinite family of functions that are all optimal solutions, in the sense that $x = \pi + 2n\pi$ is an infinity family of solutions to the problem $\min\cos(x)$.


问:
normally hyperbolic invariat manifold
I am studying a flow in discrete time over a cylinder $ X= \mathbb{R} \times \mathbb{S}^1$ of the form:
$$\begin{cases}y_{n+1}= \frac{1}{2}y_n
\\ \theta_{n+1}= \theta_n + by_n + \frac{1}{2} \end{cases} $$
where the last identity is intended modulo 1, i.e. identifying $$\mathbb{S}^1= \mathbb{R}/\mathbb{Z}.$$
This system is very easy because I have that y contracts and I have unique invariant limit cycle in $y=0$ which is asymptotically stable and is an normally hyperbolic invariant manifold, since I can split the tangent space to the curve $y=0$ in $X$ as the direct sum of the tangent space in such curve "plus" the contractive direction.
Chosen smooth function $ f=f(\theta)\, \colon\,  \mathbb{S}^1 \to \mathbb{R}$  and taken $\sigma $ small I want to perturbate the $y$ cooordinate such that I consider the system:
$$\begin{cases}y_{n+1}= \frac{1}{2}y_n  + \sigma f(\theta_n)
\\ \theta_{n+1}= \theta_n + by_n + \frac{1}{2} \end{cases} $$

It is well known that a normally hyperbolic invariant manifold $M \subset X$ is preserved under small perturbation, in the sense that for $\sigma$ small enough there exists $M_{\sigma}$ sucht that it is normally hyperbolic and invariant for the perturbed system and is diffeomorphic to $M$. If such exists,
it is easy -I think- to see that trajectories  converge exponentially to it, because:
$T_{M_{\sigma}}X= TM_{\sigma} \oplus M_{\sigma}^{s}$, where the latter is the perturbed stable direction and a trajectory starting  in $(y,\theta) \in X$ has no other choice that getting to $M_{\sigma}$ by following the contractive direction (since I have no unstable manifold).

First question is : Is this argument true? Because I feel like I am missing something;

Second question:
The proof of persistence of invariant manifold relies on the fact that stable and unstable manifolds persist and I take the intersection. In such case I have no unstable manifold in the non perturbed  system, hence I do not understand how this manifold "persists", in the sense explained above.
Basically the problem is, I perturb a system which has exponential converge to a set and I want to prove there still has exponential convergence to some set which is homeomorphic to the first set using persistence of normally hyperbolic invariant manifolds, but I am not understanding exactly everything. I would appreciate any sort of help.

答:
Note that the map, even before considering $f$, need not be continuous, let alone differentiable.

But for the perturbation to also have a normally hyperbolic manifold you need the map to be differentiable and to also perturb in that class. You are right that you have a single limit cycle but for the rest you need differentiability (or at least to repeat the argument and see whether they can be generalized, which usually is not possible).

On your second question, there is no problem if there is only expansion or only contraction. In that case simply the perturbation has also only expansion or only perturbation, respectively.

Title: normally hyperbolic invariat manifold


问:
Center of gravity $\frac{1}{\lambda_d(E)}\int_E xdx \space$
We are looking at $\mathbb{R}^d$ with the Lebesgue measure. Let $E \subset \mathbb{R}^d$ be bounded and measurable with $\lambda_d(E) > 0$. The center of gravity is component-wise defined as $$\frac{1}{\lambda_d(E)}\int_E xdx \space.$$

How do I calculate the center of gravity of the cube $[0,T]$, $T>0$?

答:
The cube $[0,T]^{3}\subset \mathbb{R}^{3}$ has centre of mass at some point $y=(y_{1},y_{2},y_{3})$. We can compute $y_{i}$ by the integral

$$ y_{i} = \frac{\displaystyle\int_{[0,T]^{3}}p_{i}(x) dx}{\displaystyle\int_{[0,T]^{3}}dx} = \frac{\displaystyle\int_{[0,T]^{3}}p_{i}(x) dx}{\lambda_{3}([0,T]^{3})}$$

where $p_{i}$ is the projection of $x$ into the $i$th (canonical) basis vector of $\mathbb{R}^{3}$ i.e. for $x=(x_{1},x_{2},x_{3})$, $p_{i}(x)=x_{i}$.

I assume you are familiar with triple integrals and thus this simply becomes an exercise in computing them.


问:
Why can inverse functions be determined by manipulating equations?
A common method for determining an inverse function $f^{-1}$ of a function $f$ is to write $f$ as an equation and solve for x. For example, if $f:=\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$, $x\mapsto x^2$, then $g:=\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$, $x\mapsto \sqrt{x}$ is an inverse function of $f$ because

$y=x^2$ iff $x=\sqrt{y}$.

(for another example see this post: https://math.stackexchange.com/a/2890915/493672)

Why does this method always work? The references I consulted seem to only state that this is how to do it, but not why it works. It's simply not evident to me. One problem I have is that functions are not equations, but in this case the authors write some function $f$ as an equation, treat it as an equation and get correct results anyway. It would be nice to have some justification for why  inverse functions can be determined like this.

Edit: Fixed the range.

答:
Let me answer on two levels.

First, if $y = f(x)$ for an invertible function $f$, then $x = f^{-1}(y)$. When you write out the equation $$y = f(x)$$
where $f(x)$ is some expression in $x$, then you solve this equation for $x$, the result will be in the form $$g(y) = x$$
or $x = g(y)$, for some other expression $g(y)$. But that expression defines a function $g$. And the very meaning "solve for $x$" is that the resultant equation $x = g(y)$ is equivalent to the original equation $y = f(x)$. But if we substitute for $x$ and $y$ in these equations from the other we get:
$$x = g(f(x))\\y=f(g(y))$$
which is exactly the condition that $f$ and $g$ are inverse functions.

The second level is this. When you "solve $y = f(x)$ for $x$", what do you do? You make a series of changes to the equation to get an equivalent equation. The key rule is that if you do something to one side of the equation, you must do the same thing to the other side of the equation. But those "things" that you do can be thought of as applying some function to both sides:

Add something? $$g(x,y) = f(x,y) \longrightarrow g(x,y) + a = f(x,y) + a$$
is another way of writing $$g(x,y) = f(x,y) \longrightarrow h(g(x,y))= h(f(x,y))$$
where $h(t) = t + a$. Multiplying is the same, where $h(t) = at$. Taking the square is $h(t) = t^2$, etc.

So you start with $y = f(x)$, and step-by-step apply a sequence of functions:
$$y = f(x)\\h_1(y) = h_1(f(x)) = h_1\circ f(x)\\h_2\circ h_1(y) = h_2\circ h_1\circ f(x)\\\vdots\\h_n\circ \dots \circ h_2\circ h_1(y) = h_n\circ \dots \circ h_2\circ h_1\circ f(x)$$

But there is a point to this. If you do it right, in the end $h_n\circ \dots \circ h_2\circ h_1\circ f(x) = x$, so you end up with
$$h_n\circ \dots \circ h_2\circ h_1(y) = x$$
Letting $g = h_n\circ \dots \circ h_2\circ h_1$, this is $$x = g(y)$$ As before, this means $g$ is the inverse of $f$.


问:
Method for estimating number of fish in my Aquarium
Context: I have an aquarium (50 Liters) with lots of fishes from the same specie (Guppy) and i have noticed that recently their number as grown exponentially due to their reproductive cycle. I would like to estimate their number.

Problem: It's not possible for me to count them individually, and i cannot dry the aquarium due to the fact that the newborns are translucid and have sizes bellow 4 milimeters, it would be an impossible task to distinguish them from the grains of smalls rocks in my aquarium.

Pre-Condition: Due to the fact that newborns stay relatively close to the ground and as they grow they will occupy (swim) in places increasingly closer to the surface level, we can consider that the fishes are uniformly distributed in the aquarium. Fish Tank

Possible method: Take a cup of water (for example 150 milliliters) consistently from the same place of the aquarium, and count the number of fishes from inside the cup. By repeating this method multiple time i could in theory estimate the total number of fishes in my aquarium.

Question: Is this a good method? Are there any big flaws in my lines of thinking? Could you think of a better, more pratical method?

Thanks in advance for your help :)

答: Why not take a photograph or video and count them using that?</p>


问:
Is being 'economical' a good strategy for epsilon-delta proofs?
Today I was trying to compute the following limit:
$$
\DeclareMathOperator{\epsilon}{\varepsilon}
\lim_{x \to 1}\frac{1-\sqrt{x}}{1-x}
$$
A single application of L'Hôpital's rule tells us that the limit is equal to $1/2$, but I wanted to practise proving that it is equal to $1/2$ directly using the epsilon-delta definition of a limit. The strategy that I used seems somewhat unorthodox, since I have never seen it in a calculus textbook. My idea was to find the maximum value of $\delta$ for which it is still true that if $x\in(1-\delta,1+\delta)$ then $\frac{1-\sqrt{x}}{1-x}\in(1/2-\epsilon,1+\epsilon)$, provided that $x\neq1$. Consider the graph of $f(x)=\frac{1-\sqrt{x}}{1-x}$:

If $x\in[1-\delta,1+\delta]$, then the value of $x$ for which $f(x)$ is furthest away from $1/2$ is $1-\delta$ (this is because the gradient of $f$ is strictly increasing). Hence, if $f(1-\delta)-1/2=\epsilon$, then it must be the case that if $x\in(1-\delta,1+\delta)$, then $\frac{1-\sqrt{x}}{1-x}\in(1/2-\epsilon,1+\epsilon)$. All we have to do is solve $f(1-\delta)-1/2=\epsilon$:
$$
\frac{1-\sqrt{1-\delta}}{\delta}-\frac{1}{2} =\epsilon \, .
$$
After some algebraic manipulation, we find that
$$
\delta = \frac{8\epsilon}{(2\epsilon+1)^2} \, ,
$$
which completes the proof. My questions are:

Is this a valid technique for epsilon-delta proofs?

Are there simpler ways of approaching this problem?

答:
Simpler way is:

$$\frac{1-\sqrt x}{1-x}=\frac{1-\sqrt x}{(1-\sqrt x)(1+\sqrt x)}=\frac 1{1+\sqrt x}$$

$\Rightarrow Lim_{x\rightarrow 1}\frac{1-\sqrt x}{1-x}=\frac 12 $


问:
Is $(Q,.)$ an abelian and cyclic group?
My teacher said that an example of an abelian group that is not cyclic is $(Q/\{0\},.)$.

So I was curious why did he use $Q/\{0\}$ instead of just $Q$. Does using $Q$ make it cyclic?

答:
No.

Every group has exactly one idempotent, namely its identity. For suppose $x^2=x$ is an idempotent. Then $xx=x^2=x=ex$, so multiplying on the right by $x^{-1}$ gives $x=e$.

In $(\Bbb Q,\cdot)$ both $0\cdot 0=0$ and $1\cdot 1=1$ but $0\neq 1$. Hence $(\Bbb Q,\cdot)$ is not a group.

An example of an abelian group that is not cyclic is the Klein four group.

Title: Is $(Q,.)$ an abelian and cyclic group?


问:
For a non-vanishing field $X$ and $a\in C^1(\mathbb R^n;\mathbb C)$, how to reduce the study of $X\cdot \nabla u +au= f$ to the study of $Xv=g$?
I'm doing exercises in the first chapter of Serge Alinhac's "Hyperbolic Partial Differential Equations", and while most of the other exercises for this chapter are straightforward, I'm stuck on Q3:

Let $X$ be a non-vanishing field in $\mathbb R^n$ and $a\in C^1(\mathbb R^n)$ a complex function. Explain how the study of the equation $Xu+au = f$ can be (locally) reduced to the study of the equation $Xv=g$.

Here, Serge is using the notation $Xu \triangleq X\cdot\nabla u$; $u$ is a scalar function (presumably, due to $a$, taking values in $\mathbb C$). I would assume $f,g$ can be taken to be smooth.

Things I've tried:

The "(locally)" that appears in the question suggests to me a change of variables. But if I say let tildes mean composition with some nice $\phi$, e.g. $\tilde u=u\circ \phi$ and so on, all I end up with is $(\tilde X \cdot ((\nabla \phi)^{-1}\circ \phi)\nabla \tilde u) +  \tilde a \tilde u = \tilde f$ which doesn't seem helpful.

I tried to consider the real and imaginary parts. (Earlier in the book, it is remarked that $\partial_t + i\partial_x$ is the Cauchy-Riemann operator; the real and imaginary parts of  $\partial_tu + i\partial_xu =0$ form precisely the standard system called the Cauchy-Riemann equations) But I don't see any way to continue from the resulting system. If say $u=v+iw$, $f=g+ih$, and $a=b+ic$, then the two equations you get for $(v,w)$  are
$$ Xv+bv - cw = g, \\ Xw + cv + bw = h.$$

I tried to use what I know from basic ODE theory, namely integrating factors. If I could solve $Xz = az$, with $z\neq 0$ everywhere, then by Leibniz rule,
$$ X  (uz) = zXu + uX z = z Xu + au z $$
Then $$Xu + au = f\iff zXu + auz = fz \iff X(uz) = fz$$
Of course, one takes $uz=v$ and $fz=g$, to match the question's notation, so this feels like the most promising approach. But I don't get what the complex part of $a$ is supposed to do for me, and the book did not consider solving things like $Xz = az$ yet, nor am I 100% sure its always possible to solve this equation.

Are any of these a step in the right direction, and does anyone have some pointers?

(edit after sleeping) So in the case where $X=\partial_1$ and $a$ takes purely imaginary values, it works out like this. You start the solution of $Xz=az$ at whatever $z_0$ say $z_0=1$. The solution is $z(x) = z_0 \exp(\int_0^{x_1} a(s,x_2,\dots,x_n) ds)$. My stronger assumption on $a$ means that
$$\partial_1\left( \frac{|z|^2}2 \right) = \Re(\partial_1 z\overline z)= \Re ( a |z|^2) = 0$$
so that $|z|^2$ is a constant, and in particular $z\neq 0 $ everywhere.

My lack of formal training in differential geometry is showing, but I suppose from try #1 above, the nonvanishing of $X$ allows me to reduce to this case? I guess I need to solve $\partial_1 = \tilde X \cdot ((\nabla \phi)^{-1}\circ \phi)$?

And also, my hunch now is that the result is false if $a$ can take  more general complex values...

答:
I suppose that Alignac provides details for showing that $Xv=g$ admits smooth local solutions in a neighborhood of a point where $X$ is non-zero? (e.g. by integrating along the flow of X).

After that you pose and solve locally, as suggested in comments, $XA=a$ and then $XB=e^{A}f$ (there is no need to separated into real and imaginary parts). After that $u=Be^{-A}$ provides a local solution to the problem since
$$ X u + a u = (X + a) (B e^{-A}) = (XB) e^{-A}-B a  e^{-A} +a B e^{-A} = f $$

The non-vanishing of $X$ ensures that the problems are equivalent. Otherwise you run into  counter examples like:
$$  x u'(x) + u(x) = 1 $$
which does have a smooth solution ($u\equiv 1$) but for which $x A'(x)=1$ does not have a solution in a neighborhood of zero.


问:
Can we always approach probability problems treating each event serially?
In many probability problems where there are a lot of parallel events, I have seen that the solution uses the approach/trick to consider events one at a time instead. E.g. as when throwing 6 dice, consider instead trowing one dice at a time.

Another example is the following problem:

How many people should we have in a room so that at least two of the
people in the room have the same birthday with probability greater
than $50\%$?

The solution given also considers the events serially

For the first individual in the room, he can have birthday on any day
and still the statement: "all individuals in the room have different
birthdays" would be true.

Hence the probability that "all individuals
in the room have different birthdays" with just one individual is $1$.

Now we take the second individual: the probability that he has a
birthday on a different day than the first individual is exactly
$\frac{364}{365}$ as there are exactly $364$ non-conflict days
available (we are assuming here all birthdays are equally likely).
With three individuals the probability that "all individuals in the
room have different birthdays" is:

$1\cdot\frac{364}{365}\cdot\frac{363}{365}$ and we can continue this
process with the fourth, fifth etc

I think the solution is basically the same idea as in problems which state throwing of multiple dices, where we approach of what is going to happen instead of making the throw of N dice, but N sequential throws of $1$ dice.

So what are the per-requisites to be able to correctly use a thinking/solution approach from concurrent events to serial events? I.e. is there some twist of this problem that such an approach would fail?

答:
I think what you're getting at here is that the method you quote, is essentially assigning an arbitrary order to your random events, as a tool for thinking about them at one at a time.

The key scenario in which assigning an arbitrary order to events will fail, is when the order is important in affecting the outcome.  This happens when event $x_n$ is not independent of event $x_{n-1}$ in some sequence.

For example, if you want to know the probability that a man had solved a Rubik's cube at least once before turn $m$ of the cube, you could no longer assign an arbitrary order to your random events.  This is because once he has solved it on the $p^{th}$ turn, the probability is one for all $m\geq p$.

Indeed you can say more.  Suppose he has a method of turning which ensures he never returns the cube to a previous state, the probability of solving the cube is a strictly monotonic increasing function of $n$.


问:
Solve a Linear Homogeneous Equation Initiate Value Problem (complex)
I have this exercise on Linear Homogeneous Equation Initiate Value, and i'm kind of stuck. I don't know if what i'm about to ask exists already here as an answer, but i'm kind of newbie and i don't use the same symbols yet so i can't understand neither the question nor the answer. If you think it already exists, just point me that way. So here is the problem:

$y_{x+3} - y_{x+2} + 4y_{x+1} -4y_x=0, y_0=2, y_1=3, y_2=-3$

The corresponding characteristic equation is:

$l^3 - l^2 +4l -4 =0$, with roots  $l=1$ or $l=2i$ or $l=-2i$

Now, i'm supposed to apply these:

$a=r\cos u$, $b=r\sin u$ where $l=a+bi$ $=>r=\sqrt(a^2+b^2)$ and $tgu=b/a$ ,  $r>0$, $u \in [0,2π).$

Pretty simple so far.

I said that $n=π/2 $, $r=2$ and this is my general solution:

$y_x=c_1*1^x + c_2*2^x*\cos(π*x/2) + c_3*2^x\sin(π*x/2)$,  where $c_2*2^x*\cos(π*x/2) = 0$

Then, i apply the first conditions and i have:

$c_1 + c_3  = 2 = y_0$

$c_1 + 2c_3 = 3 = y_1$

$c_1 + 4c_3 =-3 = y_2$

So i cannot solve the system and i can't find out what i've done wrong. It is supposed to be possible to solve of course.

Thanks in advance

答:
Thanks to everyone who took time to help me. I finally figured it out by myself,pretty dumm mistake. So here is the general solution:

$y_x=c_1*1^x + c_2*2^x*\cos(π*x/2) + c_3*2^x\sin(π*x/2)$,  where $c_2*2^x*\cos(π*x/2) = 0$

I thought that:

$c_2*2^x*\cos(π*x/2) = 0$.

Well, that's wrong because:

e.g. $x=2\\c_2*2^2*\cos(π*2/2) = -4c_2$

ContentLicense: CC BY-SA 3.0


问:
A multiple integral problem in $6$D: $I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$
Let $B_r$ to be a ball centered at origin with radius $r$ in dimension $3$. My problem is the following multiple integral:
$$I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$$
The result says that $I$ is exactly equivalent to $r^5$. The lower bound for $I$ is trivial but I have trouble in the upper bound. Moreover, can this result be extended to dimension $n$?

答:
By setting $(ra,rb )=(x ,y)$ we see that $|x-y|^{-1}dxdy = r^{-1}|a-b| (r^3da)(r^3db) = r^5 |a-b|^{-1}dadb$ and the integration domain becomes the product of unit balls. So if the integral is finite when $r=1$, then for general $r$ it scales like $r^5$. In higher dimensions the answer is of course $r^{2n-1}. $

As for this unit radius case we can change variables to $u=x-y$ and $v=x+y$. Being a linear orthogonal change of coordinates it introduces a constant scale factor which I ignore. Then the integral is like

$$\int_{B-B}\int_{B+B}\frac{ dudv }{|u|}$$
Which is finite because $1/|u|$ is locally integrable in dimension 3. Written in this form it could be computed explicitly but I’ll leave that to you.


问:
How to show the well-definedness of a lift of a continuous map via a covering map?
Given a topological space $X$ is path-connected and locally path-connected. Say $p:Z \rightarrow Y$ is a covering map such that $p(z_0)=y_0$. A map $f:X \rightarrow Y$ is continuous and $x_0 \in X$ be such that $f(x_0)=y_0$. Also given that $f_*(\pi_1(X,x_0))\subset p_*(\pi_1(Z,z_0))$.

I am trying to show that $f$ can be lifted to a $Z$ via $p$ and the lift maps $x_0$ to $z_0$.

I have proceeded like this:

As $X$ is path connected so given $x \in X$, we have a path $\alpha$ from $x_0$ to $x$. Then I have a unique lifting of $f \circ \alpha$ to a path $\widetilde{f \circ \alpha}$ in $Z$ starting from $z_0$. I defined $\tilde{f}$ from $X$ to $Z$ as $\tilde{f}(x)= \widetilde{f \circ \alpha}(1)$.

I believe that $\tilde{f}$ will be the required lifting of $f$. And to show only the existence of such a lift local path-connectedness of $X$ is redundant. All it remains to finish the proof is that $\tilde{f}$ is well-defined i.e. independent of the path $\alpha$.

Am I proceeding correctly ? How do I show the well-definedness of $\tilde{f}$?

答:
I have found the answer to this problem. This is proved as a theorem in Allen Hatcher's Algebraic Topology (Proposition 1.33)

Anyway, thanks.


问:
Proving DeMorgan's laws, basic case with 2 sets (From Casella and Berger) (Typo in solution?)
Required to prove the following:

$$(A \cup B)^c = A^c \cap B^c$$

The textbook's solution:

Casella and Berger Ex 1.3c

Isn't the first iff statement incorrect?

$$ x \in (A \cup B)^c \iff x \notin A \text{ or } x \notin B $$

I am under the impression that the second statement implies x can be in A or B (as x is not in A or x is not in B)

If this is indeed a typo, what would a proof look like? I'm struggling with this.

答:
the only typo is that or it should be an and istead :
$$
x \in (A \cup B)^c \iff x \notin A \text{ and } x \notin B
$$


问:
Action of a 1-form on vector : changing of coordinates - tensor calculus
I try currently to revise my classic. Could you help me please to demonstrate :

$$\frac{\partial T^{\prime\mu}}{\partial x^{\prime \alpha}}=\frac{\partial}{\partial x^{\prime \alpha}}\left(\frac{\partial x^{\prime \mu}}{\partial x^{v}} T^{v}\right)$$

i.e by using a changing of coordinates between $x^{\mu}$ and $x^{\prime\mu}$. $T^{\mu}$ represents the contravariant components of a "vector" (I mean a tensor of rank 1).

This demostration would be the same than proving :

$$T'^{\mu}\text{d}x^{\nu}=T^{\nu}\text{d}x^{\prime\mu}$$

How to justify this equality if I apply a 1-form $\text{d}x^{\mu}$ or another 1-form $\text{d}x^{\prime\mu}$ on a vector $T$ ?

答:
This appears to be a little over-complicated. You have a basis change with coordinate transformation $(x')^μ=C^μ{}_νx^ν$. Accordingly the contravariant tensor $T$ transforms as $(T')^μ=C^μ{}_νT^ν$. Now compute partial derivatives using the chain rule and $x=C^{-1}x'$
$$
\frac{∂(T')^μ}{∂(x')^α}=\frac{∂(C^μ{}_νT^ν)}{∂x^κ}\frac{∂x^κ}{∂(x')^α}=C^μ{}_ν\,(C^{-1})^κ{}_α\frac{∂T^ν}{∂x^κ}
$$

Let's try in a larger context. ${\bf T}=T^ν{\bf e}_ν=T'^μ{\bf e}'_μ$ is a vector field in tangent space. The basis vectors are the coordinate derivatives ${\bf e}_ν=\frac{∂}{∂x^ν}$. In that sense one gets by the chain rule
$$
{\bf e}_ν=\frac{∂x'^μ}{∂x^ν}{\bf e}'_μ
$$
which then gives
$$
T'^μ=T^ν\frac{∂x'^μ}{∂x^ν}
$$
Now put parentheses around it and the derivative operator for $x'^α$ before it to get the claimed formula.

If the coordinate change is linear as in the first try, then $\frac{∂x'^μ}{∂x^ν}=C^μ{}_ν$ and
$$
C^μ{}_κ\frac{∂x^κ}{∂x'^α}=\frac{∂(C^μ{}_κx^κ)}{∂x'^α}
=\frac{∂x'^μ}{∂x'^α}=\delta_α^μ
\implies
\frac{∂}{∂x'^α} = \frac{∂x^κ}{∂x'^α}\frac{∂}{∂x^κ}
=(C^{-1})^κ{}_α\frac{∂}{∂x^κ}
$$


问:
Intersection of ideals,why $\bigcap _{\alpha \in A}\:I_{\alpha }\:=\sum _{\alpha \in A}\:I_{\alpha }\:=\left\{a_{\alpha _1}+...+a_{\alpha _n}\right\}$
We've been given the definition of intersection of ideals, that it is equal to:

$$\bigcap _{\alpha \in A}\:I_{\alpha }\:=\sum _{\alpha \in A}\:I_{\alpha }\:=\left\{a_{\alpha _1}+...+a_{\alpha _n}\::\:n\in \mathbb{N},\:a_{\alpha _i}\in I_{\alpha _i}\right\}$$

I see that those $a_{\alpha _i}$ are arbitrary elements of Ideals $I_{\alpha }$, but let's say that for every element $a_{\alpha _i}$ except $a_{\alpha _1}$ we have $a_{\alpha _i}$ = 0. Then $a_{\alpha _1}$ would be an element of $\bigcap _{\alpha \in A}\:I_{\alpha }$, but it doesn't have to be true that $\forall _{i\:}a_{\alpha _1}\in \:I_{\alpha _i}$.

So what's the deal here?

答: Your doubts are completely justified. Take a very simple example: $\mathbb{Z}$ as the  ring, $A=\{1,2\}$ and $I_1=(2),I_2=(3)$ the ideals generated by the primes $2$ and $3$ respectively. Clearly $I_1\cap I_2=(6)$, but $I_1+I_2=\{2x+3y:x,y\in\mathbb{Z}\}$. Obviously $I_1\cap I_2\subsetneq I_1+I_2$, since e.g. $2\in I_1+I_2$. So there must be something wrong with the "definition" that you got.


问:
Why is the formula for the area of a cardioid $ \int_a^b \frac{1}{2} r^2 d \theta$
I've seen this expression in many places :$\int_a^b \frac{1}{2} r^2 d \theta$ and was wondering if someone can explain where this came from? I've noticed that it's sometimes explained in conjunction with double integrals but I'm not very familiar with double integrals (as in what it means using a picture). </p>

答:
Answer based on my answer to the question Explain $\iint \mathrm dx\,\mathrm dy = \iint r \,\mathrm \,d\alpha\,\mathrm dr$.

1. Area in Cartesian and in polar coordinates. Let $T$ be a bounded closed region of the $xy$-plane. If we decompose $T$ into $n$ rectangular cells the area $A$ of $R$ is the limit

$$ A=\lim_{n\to\infty}\; \sum_{i=1}^{n }\Delta A_{i}$$

which by definition of a double integral is equal to

$$A=\iint_T \mathrm{d}x\;\mathrm{d}y,\tag{1}$$

where $\Delta A_{i}$ is the area of a generic rectangular cell. The limits of integration are to be found in terms of $x,y$.

Figure 1: Generic $i^{th}$-cell in polar coordinates with the shape of a circle sector

Let's compute the same area using polar coordinates $r,\theta$. If we decompose $T$ into cells with a shape of sectors of a circle defined by two radii whose difference is $\Delta r_{i}$ for the generic $i^{th}$ cell and two rays making an angle $\Delta \theta _{i}$ with each other, the area of the  cell (see figure 1 above), using the formula of a circle sector, is

$$\frac{1}{2}\left[ \left( r_{i}+\frac{1}{2}\Delta r_{i}\right) ^{2}-\left(
r_{i}-\frac{1}{2}\Delta r_{i}\right) ^{2}\right] \Delta \theta
_{i}=r_{i}\Delta r_{i}\Delta \theta _{i},$$

where $r_{i}$ is the radius of the middle point of the cell. The same area $A$ can be expressed as the limit

$$A=\lim_{n\to\infty}\;\sum_{i=1}^{n }r_{i}\Delta r_{i}\Delta \theta _{i},$$

which by definition of a double integral is equal to

$$A=\iint_T r\;\mathrm{d}r\;\mathrm{d}\theta. \tag{2}$$

The limits of integration of $T$ are expressed in terms of $r$ and $\theta$. The relation between Cartesian and polar coordinates is

$$x=r\cos\theta,\qquad y=r\sin\theta.$$

In 2. and 3. we evaluate $(2)$ when $T$ is a circle and a cardioid.

2. Example. Area of a circle with radius $R$.  Observing that $r\in [0,R]$ and $\theta\in[0,2\pi]$ the integral $(2)$ reduces to

$$\begin{eqnarray*}A_{\text{circle}}=\int_{0}^{2\pi }\left( \int_{0}^{R}r\ \mathrm{d}r\right) \mathrm{d}\theta
&=&\int_{0}^{2\pi }\frac{R^{2}}{2}\ \mathrm{d}\theta\tag{$\ast$}\\ &=&\pi R^{2}.\end{eqnarray*}$$

Equation $(\ast)$ corresponds to the formula in the question. In this case $R$ is a constant.

3. Area of a Cardioid. To compute the area of a cardioid with equation $r=4a\cos ^{2}\frac{\theta }{2}$ (see figure 2 below) we need to find the limits of integration. From the figure we see that $r\in [0,4a\cos ^{2}\frac{\theta }{2}]$ and $\theta\in[0,2\pi]$. The integral $(2)$ reduces to

$$\begin{eqnarray*}
A_{\text{cardioid}}=\int_{0}^{2\pi }\left( \int_{0}^{4a\cos ^{2}\frac{\theta }{2}}r\ \mathrm{d}
r\right)\ \mathrm{d}\theta  &=&\int_{0}^{2\pi }\left( \left. \frac{r^{2}}{2}\right\vert _{0}^{4a\cos ^{2}\frac{\theta }{2}}\right)\ \mathrm{d}\theta\tag{$\ast\ast$} \\
&=&\int_{0}^{2\pi }\left( 8a^{2}\cos ^{4}\frac{\theta}{2} \right)\ \mathrm{d}\theta  \\
&=&6\pi a^{2}.
\end{eqnarray*}$$

Equation $(\ast\ast)$ corresponds to the formula in the question. Now $r$ is not constant, rather a function of $\theta$.

$$\text{Figure 2: cardioid }r=4a\cos ^{2}\frac{\theta }{2}$$

--

Added: The factor $r$ in $(2)$ is the Jacobian of the transformation. Here is an  evaluation:
$$\begin{eqnarray*}
\frac{\partial \left( x,y\right) }{\partial \left( r,\theta \right) }
&=&\det
\begin{pmatrix}
\partial x/\partial r & \partial x/\partial \theta  \\
\partial y/\partial r & \partial y/\partial \theta
\end{pmatrix}
\\
&=&\det
\begin{pmatrix}
\cos \theta  & -r\sin \theta  \\
\sin \theta  & r\cos \theta
\end{pmatrix}
\\
&=&r\cos ^{2}\theta +r\sin ^{2}\theta  \\
&=&r.
\end{eqnarray*}$$


问:
Find a convex function $f$ which satisfies $\lim\limits_{x\to+\infty}f'(x)=a$ and $\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$
Question 1.

Find a function  which satisfies the following conditions:

$f:\mathbb{R}\to \mathbb{R} ,\quad f\  \cup$

$\lim\limits_{x\to+\infty}f'(x)=a$

$\lim\limits_{x\to+\infty}\Big[x\Big(f'(x)-a\Big)\Big]\neq0$

It relates to this  problem.
I'm unable to find a such function.

Thanks in advance.

Edit: One final question (Perhaps what I should have asked).

Question 2.

Find a function $g$ such that:

$g:\mathbb{R}\to \mathbb{R}, \  g \  \cup $

$\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}\quad$

$\lim\limits_{x\to+\infty}\Big[x\Big(g'(x)-a\Big)\Big]\neq0$

Let's note that if $\lim\limits_{x\to+\infty}\Big[g(x)-ax \Big]=\lambda\in\mathbb{R}$ $\ $then $\lim\limits_{x\to+\infty}g'(x)=a$

Also it can be shown that $\exists\lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big) $

Second Edit ((Question 2. is impossible)

set $\lambda = k+c, c\neq 0$
then

$\lim\limits_{x\to+\infty}\Big[g(x)-ax-k \Big]=c\neq 0\quad$

So,  $c=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k\Big]=\lim\limits_{x\to+\infty}\Big[{x(g(x)-ax-k)\over x} \Big]\overset{{\infty}\over \infty}{=}\lim\limits_{x\to+\infty}\Big[{\Big(x(g(x)-ax-k)\Big) '\over \Big(x\Big) '} \Big]=\lim\limits_{x\to+\infty}\Big[g(x)-ax-k+x(g'(x)-a) \Big]=c + \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)\\ \Longrightarrow \lim\limits_{x\to+\infty}x\Big(g'(x)-a \Big)=0$

答:
I'm not the most au fait with convex analysis but here is a function which I believe satisfies your conditions.

Consider the continuously differentiable function

$$f(x) = \begin{cases} x-\log(x+1) & x >0 \\ 0 &x\leq 0\end{cases}$$

whose derivative

$$f'(x) = \begin{cases} \frac{x}{x+1} & x >0 \\ 0 &x\leq 0\end{cases} $$

is monotone increasing on $\mathbb{R}$ and hence $f(x)$ is convex.

Moreover, $f'(x)\to 1$ as $x\to +\infty$ and

$$x(f'(x)-1) = -\frac{x}{x+1} \to -1$$

as $x\to+\infty$.


问:
Rank inequality

Let $ A, B \in M_n (\mathbb{C})$ such that $(A-B)^2 = A -B$. Then $\mathrm{rank}(A^2 - B^2) \geq \mathrm{rank}( AB -BA)$.

I tried to apply the basic inequalities without results. How to start? Thank you.

答:
$\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
$\newcommand{\M}{\begin{pmatrix} I_{(r)} & 0 \\ 0 & 0 \end{pmatrix}}$
The key to this problem is to use the property of idempotent matrix $A - B$. As @Berci suggested, denote $A - B = P$, then $P^2 = P$, which implies there exists an order $n$ invertible matrix $T$, such that $P = T\diag(I_{(r)}, 0)T^{-1}$, where $r
= \rank(P)$ (this is a standard result for idempotent matrix, if you need a proof for that, I can add it as appendix).

Now $A = B + P$, substituting it to $A^2 - B^2$ and $AB - BA$ respectively, we have $A^2 - B^2 = P + BP + PB, AB - BA = PB - BP$. Equivalently, we have
\begin{align*}
& T^{-1}(A^2 - B^2)T = \M + T^{-1}BT\M + \M T^{-1}BT, \tag{1} \\
& T^{-1}(AB - BA)T = \M T^{-1}BT - T^{-1}BT\M. \tag{2}
\end{align*}

Partition $T^{-1}BT$ as $\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}$, where $A_{11}$ is of order $r$, then $(1)$ and $(2)$ become
\begin{align*}
& T^{-1}(A^2 - B^2)T = \begin{pmatrix} I_{(r)} + 2A_{11} & A_{12} \\ A_{21} & 0
\end{pmatrix}, \\
& T^{-1}(AB - BA)T = \begin{pmatrix} 0 & A_{12} \\ -A_{21} & 0 \end{pmatrix}.
\end{align*}

In view of this, to show that $\rank(A^2 - B^2) \geq \rank(AB - BA)$, it suffices to show
\begin{align*}
\rank\begin{pmatrix} B_1 & B_2 \\ B_3 & 0\end{pmatrix}
\geq \rank\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}
\end{align*}
for any order $r$ matrix $B_1$, $r \times (n - r)$ matrix $B_2$, and $(n - r) \times r$ matrix $B_3$. To this end, let $\begin{pmatrix} 0 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} 0 \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ be the maximal linear independent group of $\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}$, where $e_1, \ldots, e_i$ are columns of $B_3$, and $e_{i + 1}, \ldots, e_{i + j}$ are columns of $B_2$, $0 \leq i \leq r$, $0 \leq j \leq n - r$.
It can then be shown that columns $\begin{pmatrix} f_1 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} f_i \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ are linearly independent, where $f_1, \ldots, f_i$ are corresponding columns of matrix $B_1$. To wit, suppose
\begin{align*}
a_1\begin{pmatrix} f_1 \\ e_1 \end{pmatrix} +
\cdots + a_i\begin{pmatrix} f_i \\ e_i \end{pmatrix} + a_{i + 1}\begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix} + \cdots + a_{i + j}\begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix} = 0,
\end{align*}
then
\begin{align*}
& a_1f_1 + \cdots + a_if_i + a_{i + 1}e_{i + 1} + a_{i + j}e_{i + j} = 0 \tag{3} \\
& a_1e_1 + \cdots + a_ie_i = 0. \tag{4}
\end{align*}
By linear independence of $e_1, \ldots, e_i$, from $(4)$ we obtain $a_1 = \cdots = a_i = 0$. Substituting these back to $(3)$, we have $a_{i + 1}e_{i + 1} + \cdots +
a_{i + j}e_{i + j} = 0$, which implies $a_{i + 1} = \cdots = a_{i + j} = 0$ by linear independence of $e_{i + 1}, \ldots, e_{i + j}$. This shows $\begin{pmatrix} f_1 \\ e_1 \end{pmatrix},
\ldots, \begin{pmatrix} f_i \\ e_i \end{pmatrix}, \begin{pmatrix} e_{i + 1} \\ 0 \end{pmatrix}, \ldots, \begin{pmatrix} e_{i + j} \\ 0 \end{pmatrix}$ are linearly independent, whence
\begin{align*}
\rank\begin{pmatrix} B_1 & B_2 \\ B_3 & 0\end{pmatrix} \geq i + j =
\rank\begin{pmatrix} 0 & B_2 \\ -B_3 & 0\end{pmatrix}.
\end{align*}

This completes the proof.

Title: Rank inequality


问:
How can I prove that $\lim_{n \to \infty} a_n = 21$ using the $\epsilon - N$ definition?
The exercise goes like this...

Let $(a_n) \subset \mathbb{R}$ given by $a_1=\sqrt{21}$ and $a_{n+1}=\sqrt{21a_n}$

a) Find the value of $L=\lim_{n \to \infty} a_n$

b) Prove that $L=\lim_{n \to \infty} a_n$

I managed to prove that the sequence is monotonic increasing and bounded above by $21$ by induction so i can assume that $(a_n)$ converges and show that $L=21$ using a little bit of algebra. But for b) I'm asked to prove the fact using the $\epsilon - N$ definition of limit

I'm used to struggle a lot proving limits for non-recursive sequences using the formal definition of limit, so in this case I don't really know how to proceed.

Any suggestions?

答: Try to show that $a_{n}=21^{1-\frac{1}{2^{n}}}$. What is the limit thereof?


问:
finding all possible routes. Is it a travelling salesman problem?
I'm not very clever on mathematics and I'm not being able to start building a solution to this. I'm trying to develop an algorithm with which I can solve the following problem.
Let's say I have a set of number $\{1, 2, 3, 4\}$, I want to go from $x$ ($x$ being in the subset $\{1,2,3\}$) to $4$ using all possible routes given that the possible routes are:

starting from $1$
$$
\{1,4\}, \{(1,2),(2,4)\},\{(1,3),(3,4)\},\{(1,2),(2,3),(3,4)\}
$$
starting from $2$
$$
\{2,4\}, \{(2,3),(3,4)\}
$$
starting from $3$
$$
\{3,4\}
$$

As I said I'm not being able to develop a proper solution without doing it by hand, I would like to build an algorithm in which the input would be the boundaries of the set (in the case I presented $1$ and $4$) and the output would be the subsets with all possible combinations. Where can I start? Is this traveling salesman problem? What literature could I use to find out about this problem?

答:
Unless I misunderstand your question, I think you missed e.g. $1\to3\to2\to4$.

You are using here permutation groups of the set $\{1,2,3,4\}$ and that would be a good thing to search for in the literature.  These are subgroups of the symmetric group.

Note that this is often not always a commutative or abelian group because it matters in which orders you permute elements.

The transpositions are the elements of the permutation group that exchange two elements and leave all the others fixed.  E.g. $(1,2)$ is the transposition that exchanges $1$ for $2$.  You are constructing a chain of transpositions that exchanges $4$ with any element in $\{1,2,3\}$

A simple algorithm for your purposes begins with the observation that there will never be more than three steps (assuming you do not want to retrace your steps.
If you did, there would be infinitely many solutions).  Knowing this you can construct an exhaustive list of every chain of transpositions of length three.  Then you can check each in turn to determine whether it satisfies your required conditions.

All your examples are increasing, but you don't give this as a rule.  If you require your sequences to be increasing then you can just use this rule to filter out appropriate solutions.


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答:
Even though you seem to be happy with the given answers, I can't resist pointing out the following construction which I already mentioned in this thread. If this example is already contained in one of the links provided in the other answer, I apologize for the duplication:

Choose a space-filling curve $c: \mathbb{R} \to \mathbb{R}^2$ and compose it with the projection $p$ to one of the coordinate axes. This gives you an example of a continuous and surjective function $f = p \circ c: \mathbb{R} \to \mathbb{R}$ all of whose pre-images are uncountable.


问:
Transfer Function from input to node.
From the IIR filter flow graph below i don't understand how the transfer function is calculated in every node:

The circles contains 'X' inside are multiplications.

The circles contains 'Σ' inside are additions.

The triangles are delayers.

For example for node1, my textbook states that the transfer function is:

For node2:

For node3:

For node4:

I really don't understand how we have calculate this.

答:
OK, I've had a try. You need to write down equations for what happens in the circuit. There is some feedback loop in the circuit which will help you solve the circuit in the end.

Let's call the state at the nodes $n_i$. Then, the state at $n_4$ is clearly the sum of $bx$ and what comes in from node 3 times some delay $z^{-1}$, so

$$n_4=z^{-1} n_3 + bx$$

Similarly, at node 2 we have

$$n_2 = bc_1 x -a_1 n_4$$

and at node 1 we have

$$n_1 = bc_2 x - a_2 n_4 \; .$$

Finally node 3 gives

$$n_3 = n_2 + z^{-1} n_1$$

This gives 4 linear equations in 4 unknowns. Solving them, I got the following for $n_1$

$$n_1=bx\frac{(c_2-a_2)+(c_2a_1-c_1a_2)z^{-1}}{1+a_1z^{-1}+a_2z^{-2}} \; .$$

Title: Transfer Function from input to node.


问:
The action of the Modular Group $PSL(2, \mathbb{Z})$ on $\mathbb{Q} \cup \infty$
How to prove that the action of the Modular Group $PSL(2, \mathbb{Z})$ on $\mathbb{Q} \cup \infty$ is transitive, i.e., for every $q_1,q_2 \in \mathbb{Q} \cup \infty$, there is $g\in PSL(2, \mathbb{Z})$ such that $g(q_1)=q_2$.

答:
To recall, the action of the modular group $\text{PSL}(2,\mathbb Z)$ on $\mathbb Q \cup \{\infty\}$ is defined for $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ (with determinant $ad-bc=1$) and for $q \in \mathbb Q \cup \{\infty\}$ by the formula
$$g(q) = \frac{aq+b}{cq+d}
$$

Step 1: First consider $q_1 = 0$ and any $q_2 = \frac{m}{n}$, with $\frac{m}{n}$ in lowest terms, of course. Use the fact that $\gcd(m,n)=1$ to find $a,b \in \mathbb Z$ such that $am+bn=1$.

Then let $g = \begin{pmatrix} b & m \\ -a & n \end{pmatrix} \in \text{SL}(2,\mathbb Z)$ (check that the determinant is, actually, $+1$, by using the equation $am+bn=1$).

Now compute:
$$g(0) = \frac{b \cdot 0 + m}{-a \cdot 0 + n} = \frac{m}{n}
$$

This works except in the very special case that $\frac{m}{n} = \frac{1}{0}$, but in that case just use $g = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$

Step 2: Next, consider arbitrary $q_1,q_2$.

Use Step 1 to choose $g_1 \in \text{SL}(2,\mathbb Z)$ so that $g_1(0)=q_1$

Use Step 1 again to choose matrix $g_2 \in \text{SL}(2,\mathbb Z)$ so that $g_2(0)=q_2$.

It follows that $g_2 \circ g^{-1}_1(q_1) = g_2(0)=q_2$.

Step 3: Since the group operations on $\text{PSL}(2,\mathbb Z)$ are just matrix operations, you can then compute the matrix $g_2 \circ g^{-1}$ using ordinary matrix inversion and matrix multiplication.


答:
You proved that the limit $L$ exists, now only observe that the limit must satisfy
$$L=\sqrt{21 L}$$
because of course $\lim_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty}a_{n+1} = \sqrt{21\lim_{n\rightarrow\infty}a_n}$ by the continuity of the square root and multiplication.

Title: How can I prove that $\lim_{n \to \infty} a_n = 21$ using the $\epsilon - N$ definition?

Body:
The exercise goes like this...

Let $(a_n) \subset \mathbb{R}$ given by $a_1=\sqrt{21}$ and $a_{n+1}=\sqrt{21a_n}$

a) Find the value of $L=\lim_{n \to \infty} a_n$

b) Prove that $L=\lim_{n \to \infty} a_n$

I managed to prove that the sequence is monotonic increasing and bounded above by $21$ by induction so i can assume that $(a_n)$ converges and show that $L=21$ using a little bit of algebra. But for b) I'm asked to prove the fact using the $\epsilon - N$ definition of limit

I'm used to struggle a lot proving limits for non-recursive sequences using the formal definition of limit, so in this case I don't really know how to proceed.

Any suggestions?


问:
Are there any famous/notable mathematicians who have their own YouTube channel?
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.

答:
Here's one that I have found named Eddie Woo.

His most popular videos define the case of $0$ in mathematics and some others and I suggest you watch them.


问:
Lemma regarding density function
Lemma:  Let Z be an n-dimensional continuous random variable. If, for every $B \subset \mathbb{R}^n$,
\begin{equation}
P(\textbf{Z} \in B)=\int_B h(\textbf{x}) d\textbf{x}
\end{equation}
then $h$ is the density of Z.

This lemma is in my textbook and it was stated without proof and I wonder if anyone can refer me to a proof of it. Perhaps it is trivially easy to prove but I haven't figured it out yet.

My difficulty in proving it lies in that we are dealing with $\textbf Z \in B$ where $B$ is an arbitrary subset of $\mathbb{R}^n$ and not $[a_1,b_1]\times...\times[a_n,b_n]$.

答:
Let $F(x_1,...,x_n) = P[Z_1 \leq x_1, \ldots, Z_n \leq x_n]$ be the cumulative distribution function. Since $h$ satisfies the relationship you specified, this means that for
$B = (-\infty,x_1]\times \cdots \times (-\infty,x_n]$,
$$
  F(x_1,\ldots,x_n)
= \int_B h(t) dt.
$$
This is the multidimensional analogue of what you learned for one dimensional densities.</p>

Title: Lemma regarding density function


问:
Explanation about completeness and incompleteness theorems in logic
There must be some fundamental notion that I don't understand, because I have these doubts about Godel's completeness and incopleteness theorems in first order logic. The completeness theorem states that in a first order theory everything that is true is provable; the incompleteness theorem states that in the Robinson arithmetic (for example) there is a formula $A$ such that both $A$ and $\lnot A$ are not provable. Since one of $A$ and $\lnot A$ must be true, why it isn't true anymore that everything true is provable? Thanks in advance for your patience

答:
"The completeness theorem states that in a first order theory everything that is true is provable." No it doesn't.

To say standard first order logic is complete is to say that if some premisses $\Gamma$ semantically entail a conclusion $\varphi$, then there is a formal proof from the premisses $\Gamma$ to the conclusion $\varphi$.

To say that a first order theory with axioms $\Gamma$ is (negation) incomplete is to say that there is some $\varphi$ such that $\Gamma$ entails neither $\varphi$ nor $\neg\varphi$.

It is obvious you can have an incomplete theory with a complete first-order logic (just omit some needed axioms from the theory).

What is not obvious, of course, but which we get from Gödel, is that some first order theories are in a good sense incompletable.

Any introductory text should explain this. Try for example my Gödel Without (Too Many) Tears which you can freely download from https://www.logicmatters.net/igt -- see page 6.


问:
Why can inverse functions be determined by manipulating equations?
A common method for determining an inverse function $f^{-1}$ of a function $f$ is to write $f$ as an equation and solve for x. For example, if $f:=\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$, $x\mapsto x^2$, then $g:=\mathbb{R}_0^+\rightarrow\mathbb{R}_0^+$, $x\mapsto \sqrt{x}$ is an inverse function of $f$ because

$y=x^2$ iff $x=\sqrt{y}$.

(for another example see this post: https://math.stackexchange.com/a/2890915/493672)

Why does this method always work? The references I consulted seem to only state that this is how to do it, but not why it works. It's simply not evident to me. One problem I have is that functions are not equations, but in this case the authors write some function $f$ as an equation, treat it as an equation and get correct results anyway. It would be nice to have some justification for why  inverse functions can be determined like this.

Edit: Fixed the range.

答:
@PaulSinclair's answer is great for the specific case you're looking at, but I think it's worthwhile to address the "why" more generally.

Mathematics is a language, in the sense that it builds up information through both syntax and semantics; and distinguishes true from false through propositions (relative to a collection of axioms). Syntax is the rules of what makes "a correct sentence". Semantics is the informational meaning of the sentence. Propositions are truth statements: sentences that can be evaluated as either true or false (or undecidable, in rare cases).

For example, in vernacular language:

"Ate dog boy" is syntactically invalid.

"Green ideas sleep furiously." (Chomsky) is syntatically valid, but semantically invalid. Do note that syntactic validity in no way implies semantic validity.

"Am I hungry ?" is both syntactically and semantically valid, but not a proposition.

"Santa Claus exists." is syntactically, semantically, and propositionally valid, but propositionally false (at least in this realm of existence; and with Santa Claus defined as the mythical flying man, rather than the mall version).

"Some birds can fly" is syntactically, semantically, and propositionally valid, and propositionally true.

As for mathematical language, similar thing can be said:

"$x \; 2 \to =$" is syntactically invalid

"$x \in \mathbb{R}, \frac{x}{0}$" is syntactically valid, but semantically invalid

"Is the theorem $2 + 2 = 5$ true ?" is syntactically valid, semantically valid, but propositionally invalid.

"$2 + 2 = 5$" is syntactically, semantically, propositionally valid, but propositionally false.

"$2 + 2 = 4$" is syntactically, semantically, and propositionally valid and propositionally true.

So how is this relevant to your problem specifically ? Consider the following:

"The boy is petting the dog."

"The dog is being pet by the boy."

and

$A = \pi r^2$, the area of a circle is the radius squared times the constant $3.14 \dots$

$r = \sqrt{\frac{A}{\pi}}$, the radius of a circle is the square root of its area-divided-by-$3.14 \dots$

We have:

$$\text{"The boy is petting the dog."} \Leftrightarrow \text{"The dog is being pet by the boy."}$$

and

$$A = \pi r^2 \Leftrightarrow r = \sqrt{\frac{A}{\pi}}$$

Both pairs of statements, while syntactically distinct (active form vs passive form), are semantically identical (equivalent). This is what logical equivalence (your "iff", and more generally, isomorphisms) does for you in mathematics: it allows you to know when distinct mathematical representations "mean" (or "refer to") the same thing.

Finding the reciprocal of a bijective function is precisely this sort of passage "from the active form to the passive form". That's "why" it works.


Body:
$$a_n=\sqrt{21\sqrt{21\sqrt{21\sqrt{21\sqrt{21...}}}}}$$

$$\implies \dfrac{ (a_n)^2}{21}=a_n$$

$$\implies a_n=21$$

Or more straightforward,

$$\sqrt{21\sqrt{21\sqrt{21\sqrt{21\sqrt{21...}}}}}=21^\frac{1}{2}\times 21^\frac{1}{4}\times 21^\frac{1}{8} \cdots=21^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots}=21^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}=21.$$


答: Insights into Mathematics: by N J Wildberger

Body:
Are there any famous/notable mathematicians who have their own YouTube channel?

I found this amazing  video regarding the book.
The YouTube channel name is The Math Sorcerer,
but I don't know what the background of The Math Sorcerer is. From the analysis of the video I think he is a famous/notable mathematician.

I want to know more about YouTube channels of notable mathematicians.


问:
About the constants in the Language of Set theory (with ZF)
Accordingly, the language of Set Theory is made up only with the relational symbol "$\in$". But then how to deal with constants? Let's consider for example the case when defining "$\emptyset$". From the axiom of extensionality and the axiom of existence we get to the expression $\exists! X\forall Y(Y\notin X)$. In the first, place what is the meaning of saying  that we define this "$X$" to be $\emptyset$? Also, what is $\emptyset$ if not a constant? What happen when adding more axioms to build a new theory, like say group theory in which we need to use constants?  </p>

答: I think you are confusing between constants in the language (of sets) and constants in the meta-language. The language of $ZF$ does not have constants. That means that you are not obliged to interpret any constant as a particular set. There are simply no constants to interpret. You only need to interpret the symbol $\in$. Now, it turns out that once that is done it is a theorem that $\exists !X\forall Y \neg (X\in Y)$. That means that in the interpretation there is a unique set $X$ with some property. Now we turn to the meta theory (where we have plenty of constants) and assign this unique set $X$ to the constant $\emptyset$ of the meta language. We then (confusingly perhaps) freely use $\emptyset$ when we argue formally in $ZF$. This is fine as long as you remember that when you see $\emptyset$ used in a $ZF$ formula, you need to replace it (appropriately) by the sentence of $ZF$ that defines what $\emptyset $ is. Once you do that you'll get an honest $ZF$ formula (i.e., with no constants). Of course, this is very tedious and so is avoided. We use definition to avoid writing the same thing again and again and again and again.


问:
Rational systole of a manifold
Given a Riemannian manifold $(M^n,g)$ and an integer $1 \leq k \leq n-1$, the $k$-systole of $M$ is given by

$$\operatorname{sys}_k(M,g) = \inf \{ \mathcal{H}^k(\Sigma) : \Sigma^k \in \mathcal{S} \text{ and } [\Sigma] \neq 0 \in H_k(M; \mathbb{Z}) \},$$

where $\mathcal{S}$ denotes the set of all closed $k$-dimensional submanifolds of $M$ and $\mathcal{H}^k$ denotes the $k$-Hausdorff measure.

Is it possible to make sense of a "rational $k$-systole" by changing the homology coefficients to be $\mathbb{Q}$? I worry because if $\Sigma \in \mathcal{S}$, then $H_k(\Sigma; \mathbb{Q}) \cong \mathbb{Q}$ is not a finitely generated abelian group, and we cannot pushforward a "fundamental class" by the inclusion $i : \Sigma \to M$ to impose $[\Sigma] \neq 0 \in H_k(M; \mathbb{Q})$.

答:
Ok, let's try to make sense of rational systoles. First, one needs to define mass on rational chains. Let $r\otimes \sigma$ be a smooth rational singular $k$-simplex, where $r\in {\mathbb Q}$. Then only meaningful definition of the mass would be $|r| \cdot |\sigma|$ where $|\sigma|$ is the $k$-volume of the simplex $\sigma(\Delta^k)$ (one can be more creative here, but the end-result will be the same). Then for a $k$-chain
$$
c= \sum_i a_i \sigma_i
$$
we get $|c|= \sum_i |a_i|\cdot |\sigma_i|$. This definition will have the property that for an integer chain we get exactly Gromov's definition, but also
$$
|r \cdot c|= |r|\cdot |c|
$$
for every $r\in {\mathbb Q}$. Now, you can define
$$
sys_{k, {\mathbb Q}}(M)= \inf_{c} |c|,
$$
where the infimum is taken over all cycles $c\in Z_k(M,  {\mathbb Q})$ which are not null-homologous. But then, whenever
there exists one such cycle $c$, you get
$$
\lim_{m\to\infty} |\frac{1}{m} \cdot c|= \lim_{m\to\infty} \frac{1}{m} |c|  =0.
$$
Hence, $sys_{k, {\mathbb Q}}(M)=0$.

One can modify this definition as follows: Use integer cycles only, but impose a different nonvanishing condition: Require that $[c]\ne 0$ in $H_k(M, {\mathbb Q})$. In other words, you want $c$'s which aare non-torsion classes:
$$
sys_{k, {\mathbb Q}}(M)= \inf_{c} |c|,
$$
where the infimum is taken over all $c\in Z_k(M,  {\mathbb Z})$ whose images in $Z_k(M,  {\mathbb Q})$ represent nonzero elements of $H_k(M,  {\mathbb Q})$.

Such systolic quantity will be, in general, nonzero. Maybe this is what you want.

Title: Rational systole of a manifold


问:
A multiple integral problem in $6$D: $I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$
Let $B_r$ to be a ball centered at origin with radius $r$ in dimension $3$. My problem is the following multiple integral:
$$I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$$
The result says that $I$ is exactly equivalent to $r^5$. The lower bound for $I$ is trivial but I have trouble in the upper bound. Moreover, can this result be extended to dimension $n$?

答:
Solution for 3-dimension space

$I(A,B)=\int\limits_{V_{x}}\int\limits_{V_y} d^3\textbf x\,d^3\textbf y \frac{1}{|\textbf x-\textbf y|}=\int\limits_{V_{y}}(\int\limits_{V_x} d^3\textbf x\, \frac{1}{|\textbf x-\textbf y|})d^3\textbf y=$$=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1\left(\int\limits_{0}^B\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2d\phi_2\frac{1}{\sqrt{R^2+r^2-2rR\cos\theta_2}}\right)$
$$I(A,B)=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1F(B,r)$$
$A$ and $B$ are the radii of balls; for the convenience $B\geqslant{A}$ and  $z$ axis of the polar system of coordinates ($X$ space) is oriented along the vector $\textbf y$.

Integrating over $\theta_2$ and $\phi_2$

$$F(B,r)=2\pi\int\limits_{0}^B\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2\frac{1}{\sqrt{R^2+r^2-2rR\cos\theta_2}}=$$$$\frac{2\pi}{\sqrt{2r}}\int\limits_{0}^B\int\limits_{-1}^1\frac{R^2}{\sqrt{R}\sqrt{\frac{r^2+R^2}{2Rr}-x}}dxdR=$$
$$=\frac{2\pi}{r}\int\limits_{0}^BR\Bigl(R+r-|R-r|\Bigr)dR=\frac{2\pi}{r}\Bigl(rB^2-\frac{r^3}{3}\Bigr)$$

$$I(A,B)=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\frac{2\pi}{r}\Bigl(rB^2-\frac{r^3}{3}\Bigr)r^2dr\sin\theta_1d\theta_1d\phi_1=\frac{8\pi^2}{3}A^3\Bigl(B^2-\frac{1}{5}A^2)\quad (B>A)$$
For $B=A$ $$I=\frac{32\pi^2}{15}A^5$$


问:
I is ideal,if $\left[r\right]=r+I$, why $\left[a\right]+\left[b\right]=\left[a+b\right]$ & $\left[a\right]\cdot \left[b\right]=\left[a\cdot b\right]$?
I can't find a way to prove it for myself and it seems kind of counter intuitive

I can think of why $\left[a\right]+\left[b\right]=\left[a+b\right]$ would be true (because of $2I\:=\:I$ maybe) but what about $\left[a\right]\cdot \left[b\right]=\left[a\cdot b\right]$?

答:
Hint

If $i,j \in I$

$$(a+i)(b+j)= ab +aj+ib+ij$$ which belongs to $[a][b]$ if $I$ is a two-sided ideal by definition of a (two-sided) ideal as $aj, ib,ij \in I$.

Same for the sum.


问:
Weather Temperature Expectation
In the below question, I found the answers of the sub-questions as follows,

```<code>a) (-3*(1/8))+(6*(1/2))+(9*(3/8)) = 6

b) 6^2 = 36

c) (3*6+1)^2 = 19^2 = 361
</code>```

But they turned out to be wrong, I appreciate the help in advance!

```<code>Tomorrow's weather temperature is modelled as a discrete random variable X and the forecasts reflect
that three values can be considered with certain probabilties, as given in the PMF table below. Let
be a random variable with the probability distribution below.
</code>```
```<code>m      -3    6     9
Px(m) 1/8   1/2   3/8
</code>```
```<code>a) The expected temperature for tomorrow will be E(X)?

b) You want to model another random event as a function of X such that Y = X^2. Then, E(Y)?

c) You through that your approach in part b was wrong, and Y should be (3X+1)^2 instead of just X^2.
Then, E(Y)?
</code>```

答:
b)

$$E(X^2)=(-3)^2\times\frac{1}{8}+6^2\times\frac{1}{2}+9^2\times\frac{3}{8}$$

You calculated $E^2(X)$ instead

c) similarly

Title: Weather Temperature Expectation


问:
Proving Jensen's inequality for the general case starting from the finite case
For a finite convex combination, Jensen’s inequality is given by: $$f\left(\sum_i^n a_i x_i\right) \leq \sum_i^n a_i f(x_i)$$ for a convex $f$. Proving this is not so bad starting from the definition of a convex function (iteratively break the convex combination into a convex combination between just two points and apply the definition of a convex function).

For the case of general probability measures: $$f(\mathbb E X) \leq \mathbb E f(X)$$ where $X$ is a real valued random variable, the approach I know is pretty different (show that a tight affine lower bound of $f$ exists then apply linearity of expectation) but I feel that the method for the finite case could be made to work here.

Here is my attempt, drawing inspiration from the construction of the Lebesgue integral:

Given a real valued random variable $X$, let $X^+,X^-$ be such that $X = X^+ - X^-$ so that it suffices to show the claim for non-negative random variables (apply the definition of convexity to the convex combination $\mathbb P [X>0] X^+ + \left(1-\mathbb P[X>0]\right) X^-$). Let $X_j$ be a sequence of simple random variables (defined as a finite sum of indicator functions on the sample space) dominated by $X$ such that $$\mathbb E X_j \rightarrow \sup \left\{\mathbb E Y: Y \leq X,\, Y\text{ is simple}\right\}$$

Now we have:
$$f\left(\mathbb E X\right) = f\left(\mathbb E X_j + o(1)\right) = f\left(\mathbb E X_n\right) + o(1) \leq \mathbb E f\left( X_j\right) + o(1)$$

Where the first equality is by definition of the sequence $X_j$, the second equality is from the fact that $f$ is convex and so continuous on the interior of its domain, the inequality is from the finite case of Jensen's inequality.

Next, I would like to claim that $X_j \rightarrow ^p X$ and apply the continuous mapping theorem to complete the argument, but I'm not sure how to justify that claim.

I could use feedback on my reasoning as well as some help justifying the last claim.

EDIT: there is a simpler argument. Instead of defining $X_j$ as I have, I could instead define it as a sequence of simple random variables increasing towards $X$ pointwise (for every event in the sample space). I had not realize that we could define such a sequence in general, I am not sure how to show that this is true. In that case we have:

$$\mathbb E f\left( X_j\right) = \mathbb E f\left( X\right) + o(1)$$

by the continuous mapping theorem.

Any feedback on this followup approach would be very appreciated.

答:
In the usual approximation of $X\ge 0$ by simple $X_j\ge 0$, gotten by successively finer partitions of the range of $X$ into dyadic intervals, one has $X_j$ increasing pointwise to $X$. You then have $\lim_j \Bbb EX_j = \Bbb E X$, and $\lim_j f(X_j)=f(X)$ by the continuity of $f$. But to claim that $\lim_j\Bbb Ef(X_j) =\Bbb Ef(X)$ would seem to require something like dominated convergence (in the absence of any monotonicity assumption about $f$).

The convexity of $f$ does imply that if the right-hand derivative $f'_+(0)$ is larger than $-\infty$ then there are constants $a,b$ such that $g(x):=f(x)+a+bx\ge 0$ for all $x$, and such that $g$ is increasing. You can now use the monotone convergence theorem (applied to $g$) to conclude  that $\Bbb Ef(X)=\lim_j\Bbb Ef(X_j)$.

But what about, for example, $f(x) = -\sqrt{x}$, $x\ge 0$?


问:
Two teams A and B play series of five games...
Two teams A and B play series of five games. A tie is impossible for each game. When one of the teams wins three games the series of games are finished and no more games are played. What is the expected number of games played in this series if on average, team A wins 3 times more games than team B?

This problem appeared in my textbook and I'm confused about how to solve it.

Edit: It's 3/4 and 1/4. Sorry, I just woke up.

答:
If A is "three times more likely to win a game than B then A must have probability of winning 3/4 and B 1/4

Writing "A" for a win by A an "B" for a win by B, the possible series are

AAA which has probability (3/4)^3= 27/64.

BBB which has probability (1/4)^3= 1/64.

BAAA, ABAA, AABA each of which has probability (3/4)^3(1/4)= 27/256.

ABBB, BABB, BBAB each of which has probability (3/4)(1/4)^3= 3/256.

There are 6 different ways B can win two games before A wins three
BBAAA, BABAA, BAABA, ABBAA, ABABA, AABBA each of which has probability (3/4)^3(1/4)^2= 27/256.

There are 6 different ways A can win two games before B wins three
AABBBL ABABB,, BAABA, BAABB, BABAB, BGAAB each of which has probability (3/4)^2(1/4)^3= 3/256.


答:
Effectively we use a lot more symbols in the language of set theory then what we have in the language.

For example, $\varnothing,\omega,\omega_1,\aleph_\omega,\subseteq,\mathcal P,\bigcup$ and so on. The reason we can do that is that all of these things are definable elements, relations, functions, whatever.

This means that if we add those symbols to the language and add axioms stating that these symbol satisfy a particular formula, then we really didn't change anything substantial in our systems, at least in terms of provability.

So these symbols are actually shorthand for formulas. Much like often we agree that the logic only has $\lnot,\land,\exists$ and the rest of the connectives and $\forall$ are just shorthand for more complicated formulas.

But to your second question, when we want to build the axioms of group theory in $\sf ZF$, we don't add axioms to $\sf ZF$. No, instead we formalize the notion of a language, and logic, and all that and we define a set of symbols with which we can write the axioms of group theory. In that set we can decide to have a symbol for the neutral element (although that too is a definable constant from the axioms of group theory).

Title: About the constants in the Language of Set theory (with ZF)

Body: Accordingly, the language of Set Theory is made up only with the relational symbol "$\in$". But then how to deal with constants? Let's consider for example the case when defining "$\emptyset$". From the axiom of extensionality and the axiom of existence we get to the expression $\exists! X\forall Y(Y\notin X)$. In the first, place what is the meaning of saying  that we define this "$X$" to be $\emptyset$? Also, what is $\emptyset$ if not a constant? What happen when adding more axioms to build a new theory, like say group theory in which we need to use constants?  </p>


答:
Hints:

What is the expected number of games played in this series if on average, team A wins 3 times more games than team B?

First of all, I interpret this to mean that if $p = 0.75, q = 0.25$, then in each game, A's chance of winning is $p$ and B's chance of winning is $q$.

First, you need the general formula for Bernoulli trials:

Given $n$ trials of an independent event, with probability of success or failure $= p$ and $q$, respectively, then the chance of exactly $k$ successes in the $n$ trials is

$$\binom{n}{k}p^kq^{(n-k)}.$$

Next, you need to know that there are exactly 6 possibilities:
either team A wins in either 3,4, or 5, games, or team B wins in either 3,4, or 5 games.

Next, you need to know that (for example) for team A to win in exactly $k$ games, two things must happen:

team A won exactly 2 games in the first $(k-1)$ games.

team A won the $k$-th game.

The over all computation will be

$\sum_{k=3}^5 k \times f(k) +
\sum_{k=3}^5 k \times g(k).$

Here, $f(k)$ is the probability of team A winning in exactly $k$ games and $g(k)$ is the probability of team B winning in exactly $k$ games.

At this point, for the new student, it is simply a matter of forgoing elegance, and getting down in the mud.  Later, with experience, you will see shortcuts that needn't concern you now.

Title: Two teams A and B play series of five games...

Body:
Two teams A and B play series of five games. A tie is impossible for each game. When one of the teams wins three games the series of games are finished and no more games are played. What is the expected number of games played in this series if on average, team A wins 3 times more games than team B?

This problem appeared in my textbook and I'm confused about how to solve it.

Edit: It's 3/4 and 1/4. Sorry, I just woke up.


答:
Here is a description of the incompleteness theorems for a layman:

Four words: provable, there exists a proof that an proposition is true; refutable, there exists a proof that an proposition is false; decidable, is said of a proposition that is either provable or refutable; undecidable, is said of a proposition that is neither provable nor refutable.

Incompleteness theorem 1: by formulating the following theorem "G: the theorem called G is not provable within the theory T", Gödel shows that G is necessarily undecidable, because G being provable leads to a contradiction and G being refutable leads to a contradiction.

Incompleteness theorem 2: a proof for the coherence of theory T is a proof that no proposition in T is both True and False. However, the existence of theorem G inside T makes this exercise of "knowing if it is not both True and False at the same time" impossible for G. Thus, there does not exist a proof of coherence for T; otherwise through this proof of coherence, one could know the truth value of G.

Title: Explanation about completeness and incompleteness theorems in logic

Body: There must be some fundamental notion that I don't understand, because I have these doubts about Godel's completeness and incopleteness theorems in first order logic. The completeness theorem states that in a first order theory everything that is true is provable; the incompleteness theorem states that in the Robinson arithmetic (for example) there is a formula $A$ such that both $A$ and $\lnot A$ are not provable. Since one of $A$ and $\lnot A$ must be true, why it isn't true anymore that everything true is provable? Thanks in advance for your patience


问:
Why is the gradient of this determinant, $\det(I - AA^\dagger)$, off by factor of $2$?
I have banged my head on this for a couple of hours and I can't find what's wrong:

$$
\begin{align}
\frac{\partial}{\partial A_{pq}}\det(\mathbb{1} - AA^\dagger) &= \det(\mathbb{1} - AA^\dagger)\mathrm{tr}\left[(\mathbb{1} - AA^\dagger)^{-1}\frac{\partial(\mathbb{1} - AA^\dagger)}{\partial A_{pq}}\right]\\
&=\det(\mathbb{1} - AA^\dagger)\sum_{mnk}(\mathbb{1} - AA^\dagger)^{-1}_{mn}\frac{\partial(\mathbb{1}_{nm} - A_{nk}A^\dagger_{km})}{\partial A_{pq}}\\
&= \det(\mathbb{1} - AA^\dagger)\sum_{mnk}(\mathbb{1} - AA^\dagger)^{-1}_{mn}\delta_{np}\delta_{kq}(-A^\dagger_{km})\\
&=\det(\mathbb{1} - AA^\dagger)\sum_{m}(\mathbb{1} - AA^\dagger)^{-1}_{mp}(-A^\dagger_{qm})\\
&=-\det(\mathbb{1} - AA^\dagger)(\mathbb{1} - AA^\dagger)^{-T}A^*
\end{align}
$$

When I try it out numerically though, it's off by a factor 2:

```<code>def det(A):
    return np.linalg.det(np.identity(A.shape[0]) - A @ np.conj(A.T))

def ddet(A):
    return - det(A) * np.linalg.inv(np.identity(A.shape[0]) - A@np.conj(A.T)).T @ np.conj(A)

A1 = np.array([[0.1, 0.2],[0.2, 0.3]])
A2 = np.array([[0.1, 0.2],[0.2, 0.3+delta]])
delta = 1e-6

np.isclose((det(A2) - det(A1))/delta, 2*ddet(A1)[1,1])
>>> True
</code>```

When the entries of $A$ are complex it's even more wrong, but let's solve the 2x issue first.

答:
For typing convenience define the variables
$$\eqalign{
P &= \Big(I-AA^H\Big) = P^H \quad\implies\quad P^T=P^*\\
\phi &= \det(P)\\
}$$
and denote the trace as a colon product
$$A:B = {\rm Tr}(A^TB)$$
Then calculate the Wirtinger differential via Jacobi's formula
$$\eqalign{
d\phi &= \phi\;P^{-T}:dP \\
  &= -\phi\;P^{-T}:\big(dA\,A^H+A\,dA^H\big) \\
  &= -\phi\;P^{-T}A^*:dA -\phi\;A^TP^{-T}:dA^H \\
  &= -\phi\;\big(P^{-1}A\big)^*:dA -\phi\;\big(P^{-1}A\big):dA^* \\\\
}$$
When $A$ is real, $\,A^*=A,\,$ $P^*=P\,$ and
$$\eqalign{
d\phi &= -2\phi\;\big(P^{-1}A\big):dA \\
\frac{\partial \phi}{\partial A} &= -2\phi\;P^{-1}A \\
}$$
In the general case
$$\eqalign{
\frac{\partial \phi}{\partial A} &= -\phi\;\big(P^{-1}A\big)^* \\
\frac{\partial \phi}{\partial A^*} &= -\phi\;P^{-1}A \\\\
}$$
NB:   In order to test the complex case, your Python code should estimate the total differential instead of the gradient, i.e. by validating expressions such as the following
$$\eqalign{
A &= A_1 \quad\qquad&\implies\quad dA = A_2-A_1 \\
P &= I - AA^H \quad\qquad&\implies\quad dP = -dA\,A^H-A\,dA^H \\
}$$
$$\eqalign{
\frac{-d\phi}{\phi}
=
\frac{\det(P)-\det(P+dP)}{\det(P)}
  \quad \overset{?}{=} \quad
{\rm Tr}\Big(\big(P^{-1}A\big)^HdA
 +
\big(P^{-1}A\big)^T\,dA^*\Big) \\
}$$
If you insist on using the gradient, then you must use multivariate Wirtinger derivatives in order to code the correct expression.


问:
How do you divide a complex number with an exponent term?
Ok, so basically I have this:

$$ \frac{3+4i}{5e^{-3i}} $$

So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.

I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).

So my real question is, the exponent, does it have to be in radian or angle?

I ended up with $e^{3.927i}$.

答:
Put the numerator also in e-power form. r = 5 and the argument follows from arctan(4/3)
Now you have an e-power in the Numerator and an e-power in the Denominator. Divide the r-values, subtract the arguments. Don't expect a nice looking answer; arctan(4/3) does not come out good</p>


问:
How to find this integral over an annulus
i have $v=\sum_{g\in G_k} g v_r\in W^{1,N}_{0,G_k}(\Omega_r)\setminus\{0\}$.

$$\Omega_r=\{x\in \mathbb{R}^N, r<|x|<r+1\}, r>0, N\geq 2, N\neq 3$$
$$O_k=\{g\in O(2): g(x)=\left(x_1 \cos\frac{2\pi l}{k}+x_2\sin\frac{2\pi l}{k},-x_1\sin\frac{2\pi l}{k}+x_2\cos\frac{2\pi l}{k} \right)$$

Where  $O(2)$ Is the group of $2\times 2$ orthogonal matrices, $x=(x_1,x_2)\in\mathbb{R}^2$  and $l\in \{0,\ldots, k-1\}$

and $G_k=O_k\times O(N-2), 1\leq k <\infty$.

$W^{1,N}_{0,G_k}(\Omega_r)=\{u\in W^{1,N}_0(\Omega_r): u(x)=u(g^{-1}x),\, \text{for all} \, g\in G_k\}$

$v_r\in W^{1,N}_0(B_{\delta,r})\setminus\{0\}$ where $B_{\delta,r}=B_\delta(((2r+1)/2,0,\ldots,0))\subset \Omega_r$ satisfies
$$
g^i B_{\delta,r}\cap g^j B_{\delta,r}=\emptyset,\, for all \, g^i\in G_k, i\neq j, i,j=0,1,...,k-1
$$

My question is why $\int_{\Omega_r}|\nabla v|^N dx=\int_{\Omega_r} |\nabla v_r|^N dx $

答:
Suppose that $f$ and $g$ have disjoint support. There are $3$ cases:

$f(x)\ne0$ and $g(x)=0$

$f(x)=0$ and $g(x)\ne0$

$f(x)=0$ and $g(x)=0$

In each case, $|f(x)+g(x)|=|f(x)|+|g(x)|$ and $|f(x)||g(x)|=0$.

Apply the Binomial Theorem to get
$$
\begin{align}
|f(x)+g(x)|^n
&=(|f(x)|+|g(x)|)^n\\
&=\sum_{k=0}^n\binom{n}{k}|f(x)|^k|g(x)|^{n-k}\\
&=|f(x)|^n+|g(x)|^n
\end{align}
$$
since all terms except $k=0$ and $k=n$ contain a product of $|f(x)||g(x)|$, so those terms are $0$.

Note that the change of variables is due to the rotations, but the determinant of a rotation matrix is $1$, so they don't affect the value of the integral.

Therefore,
$$
\begin{align}
\int_{\Omega_r}|\nabla v(x)|^N\,\mathrm{d}x
&=\int_{\Omega_r}\left|\sum_{g\in G_k}\nabla v_r(g(x))\right|^N\,\mathrm{d}x\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(g(x))|^N\,\mathrm{d}x\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(x)|^N\,\mathrm{d}g^{-1}(x)\\
&=\int_{\Omega_r}\sum_{g\in G_k}|\nabla v_r(x)|^N\det(g)^{-1}\,\mathrm{d}x\\
&=k\int_{\Omega_r}|\nabla v_r(x)|^N\,\mathrm{d}x
\end{align}
$$


答:
The problem is with the use of the word "true".

The completeness theorem says that $T$ proves $\varphi$ if and only if $\varphi$ is true in all the models of $T$.

The incompleteness theorem says that there is $\varphi$ that is true in a specific model, usually taken to be $\Bbb N$, which is not provable from Robinson arithmetic.

Truth is always relative to a structure, but in the case of arithmetic, when we say "true" without qualifying it, we mean in the standard model: the natural numbers. But there are other models, and in those $\varphi$ will be false. Exactly because it is not provable from Robinson arithmetic.


问:
Uniformly continuous on two sets(Union)
Well... I got a idea from this question If $f$ is uniformly continuous on two sets, show that $f$ is also uniformly continuous on the union of two given sets. So Let me think more the general case. I would suggest you the statement that I made and regard it as true.

$f$ is a uniformly continuous on both $D_1$ and $D_2$ (Here the $D_1 \cap D_2  \not= \phi$)
Then, $f$ is a uniformly continuos on $D_1 \cup D_2$

You might look above statement is trivial though, still I can't prove it. :(
Does anyone have a idea or hint to prove that?

p.s.) If the statement is false, What is the counterexamples?

答:
Let $D_1 = [-1, 0] \cup (1, 2]$ and $D_2 = [0, 1]$

Define $f(x) = 0$ on $[-1, 0]$; $f (x) = x $ on $[0, 1]$; $f(x) = 0 $ on $(1, 2]$

$f$ is separately UC on $D_1, D_2$ but discontinuous on $[-1, 2]$

The key to constructing the counterexample is that $D_1, D_2$ must share some part of their boundary in order to have at least one point in common, but then uniform continuity will be preserved at the intersection. So, it's set up so that $D_1, D_2$ can meet without intersecting somewhere else.  One could construct counterexamples in the plane where $D_1, D_2$ are each connected sets and intersect on part of their boundary but only meet on another part. For example let the line $(0, -1)$ to $ (0, +1)$ be the boundary between $D_1, D_2$ with $D_1$ to the left and $D_2$ to the right. Let both include the line $(0, -1) $ to $(0, 0)$ so the intersection is non-empty, but only one include $(0, 0 )$ to $(0, 1)$ with the other being open up to this line.

Title: Uniformly continuous on two sets(Union)


问:
Optimality — Hamilton-Jacobi-Bellman (HJB) versus Riccati
Most of the literature on optimal control discuss Hamilton-Jacobi-Bellman (HJB) equations for optimality. In dynamics however, Riccati equations are used instead. Jacobi Bellman equations are also used in Reinforcement learning.

Are there any comparisons or parallels between the two class of equations as far as optimality is concerned? How do they differ?

答:
My simple answer would be that they are quite the same thing.

Explanation

Riccati Equations can be derived from Hamilton-Jacobi-Bellman equations in the particular case of LQR problem, an optimal control problem where the dynamics is linear and the cost is quadratic.

Consider the finite horizon LQR problem. In this particular case, the Hamilton-Jacobi-Bellman equation has the expression (for semplicity I put $N=0$)

$$ \partial_t V(x,t) + \min_u \left\{ \partial_x V(x,t) \cdot (Ax+Bu) + x^T Q x + u^T Ru \right\} = 0 $$
with the terminal condition
$$ V(x,T) = x^T Q_f x . $$

Now we look for solutions of the form $V(x,t) = x^T P(t) x$, where $P(t)$ is a symmetric matrix for each $t \in [0,T]$. If we substitute this expression in the (HJB) equation, we get

$$ x^T P'(t) x + \min_u \left\{ 2 P(t)x \cdot (Ax+Bu) + x^T Q x + u^T Ru \right\} = 0 . $$

We can explicitly find the minimum of the expression inside the curly brackets. For a given couple $(t,x)$, let us define $\Phi$ as
$$ \Phi(u) = 2 P(t)x \cdot (Ax+Bu) + x^T Q x + u^T Ru .$$

The minimum is obtained when $∇\Phi(u) = 0$, that is when
$$ 2B^T P(t) x + 2 Ru = 0,$$
so the optimal control is
$$ u^*(t,x) = -R^{-1} B^T P(t)x ,$$
with
$$ \Phi(u^*(t,x)) = 2 x^T P(t)^T \left(Ax-BR^{-1} B^T P(t)x \right) + x^T Q x + x^T P(t)^T B R^{-1} B^T P(t)x  $$

So we can rewrite again the (HJB) equation without the minimization term:
$$ x^T P'(t) x + 2 x^T P(t)^T \left(Ax-BR^{-1} B^T P(t)x \right) + x^T Q x + x^T P(t)^T B R^{-1} B^T P(t)x = 0 , $$
and by grouping the $x^T$ and the $x$ term and doing some simple algebraic steps ($P(t)$ is symmetric) we get
$$ x^T \left( P'(t) + 2 P(t) A - P(t) BR^{-1} B^T P(t) + Q  \right) x = 0 , $$

Since the above equation must hold for each $x$, it is equivalent to the matrix differential equation

$$ P'(t) + 2 P(t) A - P(t) BR^{-1} B^T P(t) + Q = 0 .$$

Finally, in order to satisfy the final condition, it must be
$$ P(T) = Q_f .$$

Comments:

This is not a rigorous proof that the two equations are equivalent, but it shows that they are quite the same thing.

I couldn't get the term $A^TP(t) + P(t)A$ of the Riccati equations, instead I found $2P(t) A$.

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