问:
How do I create a truth table in Mathematica out of the logical statement "Exactly two out of three people are truth-tellers"?
There are three people. One of them states 'Exactly two of us are truth-tellers." Every person is either a truth-teller or a liar. I was hoping someone could help me figure out how to create a truth table in Mathematica for this statement. I found that this logical statement is equivalent to the statement "Exactly two people are truth-tellers" from this link: https://mathematica.stackexchange.com/questions/14051/how-to-solve-the-liar-problem ∃x∃y((x≠y∧P(x)∧P(y))∧∀z(P(z)→(z=x∨z=y))).
I used Boolean Table but couldn't generate the right input for Boolean Table to get a truth table. I am trying to figure out the following puzzle using Mathematica's computational abilites:
There are three people.
Person A states, "Exactly two people are truth-tellers." Person B states,"I and Person C are truth-tellers." Person C states "Person A or B is a liar."
How can I use the solving features in Mathematica to input logical statements and figure out who is telling the truth and who is lying?</p>

答:
Truth-Tables are for propositional logic only. They don;t work for full first-order logic statements with quantifiers.

If you are ok doing this in propositional logic, you can use $A,B,C$ for the three people being truth-tellers respectively, and use the statement:

$$(A \land B \land \neg C) \lor (A \land \neg B \land C) \lor (\neg A \land B \land C)$$

So, with $A$ being the person making the making that 'exactly two of us are truth-tellers, you can encode that bit of information as:

$$A \leftrightarrow [(A \land B \land \neg C) \lor (A \land \neg B \land C) \lor (\neg A \land B \land C)]$$

$B$ saying that "I and C are truth-tellers" would be:

$$B \leftrightarrow (B \land C)$$

$C$ saying that "Person A or B is a liar" would be:

$$C \leftrightarrow (\neg A \lor \neg B)$$


问:
If we don't know that two sets are independent but there intersection is equal to their multiplication, can we say they are independent?
Lets say $P(A) = 0.2$ and $P(B) = 0.3$, and we know that $$ P(A \cap B ) = 0.06 $$
Does this mean the events are independent?

答:
In general, by definition, two events $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$.

But $0.2\times 0.3\ne 0.6$ in your example. [Notes: the question was edited later.] If you have instead
$$
P(A)=0.2,\quad P(B)=0.3,\quad P(A\cap B)=0.06
$$
then you can conclude that $A$ and $B$ are independent.

ContentLicense: CC BY-SA 4.0


问:
Find the constant $c$ if you have a given limit

If $\displaystyle\lim_{x\to11}\frac{|x^2-6x|-|x^2+cx|}{x-11}.=22$, find the value of the constant $c$.

I have a problem getting the correct result which is $-16$. I tried absolutely everything and the closest I got was $-6$. My idea was to just check limits from both sides (although one side would be enough because I was sure an answer is a final number, thus I chose $11$ from the positive side so the absolute values cancel out easier). When I simplify as much as I can, I really don't get anywhere. I don't know what to do with the $x-11$ in the denominator.

答:
First, as $x$ is close to $11$ you have that $x^2-6x>0$, so your expression is
$$
\frac{x^2-6x-|x^2+cx|}{x-11}.
$$
As $x-11\to0$, the numerator needs to be or approach a multiple of $x-11$. In particular, it goes to zero as $x\to11$. So
$$
0=\lim_{x\to 11} x^2-6x-|x^2+cx|=121-66-|121+11c|=55-|121+11c|
$$
This requires
$$
|121+11c|=55.
$$
So either
$
-121-11c=55
$, giving $c=-16$, or
$121+11c=55$, giving $-6$. The second choice makes the numerator identically zero, so the limit would be zero. That leaves $c=-16$.


问:
Limit to infinity with absolute value
I am learning math all by myself, no school, no teacher, and I come across a problem that I do not understand:

$\lim _{x \to -\infty } \frac{  | x^{2} - x  | +2 x^{2} }{1 - 3x^{2}}$

From my understanding, $|x²-x|$ should be translated as $-(x² - x)$, as we are approaching negative infinity. That will result in $-\frac{1}{3}$.
The book solutions is  $-1$, and I can confirm the solution by graphing the function. If I translate as $(x²-x)$ I get the $-1$.

So, why is it not $-(x²-x)$?

Thanks.

答:
We do have to consider whether $x^2 - x \ge 0$ and  therefore $|x^2 - x|$ should be translated as $x^2 -x$.  Or if $x^2 -x < 0$ and $|x^2 -x|$ should be translated as $-(x^2 -x) =x-x^2$.

Which is is it?

If $x < 0$ then $x^2 > 0$. And $-x > 0$.  And so $x^2 - x = x^2 + (-x) > 0 + 0 = 0$.

So $|x^2 -x| > 0$.  And so $|x^2-x|$ should be translated as $(x^2 -x)$.

If that seems counter intuitive try a value.  Let $x = - 10$.  Then $x^2 - x = (-10)^2 -(-10) = 100 + 10 = 110> 0$.

Title: Limit to infinity with absolute value


答:
Let $A$ and $B$ two events, so the events are independent if and only if any of the three conditions is met :

$\mathbb{P}[A|B]=\mathbb{P}[A]$.

$ \mathbb{P}[B|A]=\mathbb{P}[B]$.

$\mathbb{P}[A\cap B]=\mathbb{P}[A]\cdot \mathbb{P}[B]$.

Therefore, you need only verify in this case the condition number $3$.

Title: If we don't know that two sets are independent but there intersection is equal to their multiplication, can we say they are independent?

Body:
Lets say $P(A) = 0.2$ and $P(B) = 0.3$, and we know that $$ P(A \cap B ) = 0.06 $$
Does this mean the events are independent?


问:
every function in $C^k_0(U)$ is $C^k_0(\mathbb{R}^d)$
Often in many proofs we treat functions in $C^k_0(U)$ (the set of $C^k$ functions with compact support where $U$ is an open set in $\mathbb{R}^d$) as functions in $C^k_0(\mathbb{R}^d)$.They extend it by zero without saying why this new function if of compact support.
$$
\DeclareMathOperator{\supp}{\operatorname{supp}}
\begin{align}
C^k_0 &=\{f\in C^k(U):\: \supp(f) \text{ is compact }\} \\
\supp(f) & =\overline{\{x\in U,f(x)\neq 0\}}
\end{align}
$$
Here are my questions that might help me understand. I

Is the closure in the definition of the support taken relatively to $\mathbb{R}^d$ topology, or to the induced topology over $U$?

Is it true that $\supp(f)$ need not to be closed in $\mathbb{R}^d$  if it is taken relatively to the subspace topology, because
$$
\supp(f)=\overline{\{x\in U,f(x)\neq 0\}}^{\,U}=\overline{\{x\in U,f(x)\neq 0\}}^{\,\mathbb{R}^d}\cap U
$$
and if we take the colsure in $\mathbb{R}^d$ then $\supp(f)$ might be going outside $U$?

Thank you.

答: Actually, it's impossible for the support to go outside $U$ in either case, if $U$ is open and the support is assumed compact. If a function $f$ has compact support $K = \text{supp}(f)$ in $U$ then the distance from $K$ to $\partial U$ is strictly positive. This is because, viewed as subsets of $\mathbb{R}^d$, $K$ compact, $\partial U$ closed implies the distance between the two sets is attained by some $p \in K, q \in \partial U$. If the distance were zero, then $p = q$ and $K \cap \partial U \neq \emptyset$. But as $K \subset U$, where $U$ is open, this says that $U \cap \partial U \neq \emptyset$, a contradiction (open sets don't intersect their boundaries). Therefore the distance from $K$ to $\partial U$ is strictly positive. It follows from this that compactness of the support in $U$ under the subspace topology is equivalent to compactness of the support in the topology of $\mathbb{R}^d$ (I think this also follows from general topological arguments). As you point out, this is not true for closedness of the support. This is one indication that compactness is stronger/more important in some senses than closure.


问:
Proving that if $a^n \equiv 1 \pmod m$ then $d\mid n$
I'm having some trouble proving the following:

Let $d$ be the smallest positive integer such that $a^d \equiv 1 \pmod m$, for $a \in \mathbb Z$ and $m \in \mathbb N$ and with $\gcd(a,m) = 1$. Prove that, if $a^n \equiv 1 \pmod m$ then $d\mid n$.

The first thing that came to my mind was Euler's theorem but I couldn't conclude anything because I'm not very skilled when it comes to using Euler's totient function. Can someone give me any tips or show me how to solve this?

答:
Let us try a constructive proof (also):

Firstly, an edge case: If $d=1$, we know that $a\equiv 1$ and $a^n \equiv 1 \bmod m$ for any $n$, and also that $1\mid n$ always, so the assertion is true in this case.

So otherwise we know that $a^0 \equiv 1, a^1\equiv a \bmod m$ and that $d$ is defined such that not only $a^d\equiv 1 \bmod m$ but also that $a^k \not\equiv 1 \bmod m$ for $1\leq k<d$.

Now consider the value of $a^n \bmod m$, where $n>d$, since the assertion is true by definition for smaller $n$. Find the closest multiple of $d$ no greater than $a$ and the residue so that $n=bd+c$, with $c<d$. Now $a^n \equiv a^{bd+c} \equiv a^{bd}\cdot a^c \equiv  (a^d)^b\cdot a^c \equiv 1^b\cdot a^c \equiv a^c\bmod m$. Now since $c<d$, the only value for which $a^c\equiv 1 \bmod m$ is $c=0$, in which case $n=bd$; that is, $a^n\equiv 1 \bmod m \implies d\mid n$ as required.


问:
Standard Normal Distribution Help
The question is:

Let X be normally distributed with a mean of 10 and a standard deviation of 6.
Calculate the value x such that P(x<X<14) = 0.4329.

The correct answer in the memo is given as:

I understand up to the point of P(Z < Zx) = 0.3146, but I don't understand the jump to the next step Zx = -0.4828. My original answer was (x-10)/6 = 0.3146 -> x=11.8876, but it seems that it is wrong.

答:
Define $\Phi$ as
$\Phi(z) := P(Z \leq z),$
and note, that
$$\Phi(z_x) = 0.3146...$$
and thus we get that
$$z_x = \Phi^{-1}(0.3146) = -0.4828...$$
The function $\Phi^{-1}$ is refered to as the quantile function or as the inverse CDF for the standard normal distribution and it is implemented in most statistical software.

Title: Standard Normal Distribution Help


答:
It easy to to see the if $a^d\equiv 1 \pmod m$ then for any multiple of $kd$ of $d$ we have $a^{kd}\equiv (a^d)^k\equiv 1^k \equiv 1 \pmod m$.

So we have to prove that if $n$ is not a multiple of $d$ and $a^n\equiv 1$ is impossible.

If $d\not \mid n$ then $kd < n < (k+1)d$ for some $k$.

So we need a way to either prove directly that $a^n\not \equiv 1 \pmod n$ or athat if $a^n\equiv 1\pmod m$ we can contradict that $d$ is the least such positive integer.

Hint:

And if we consider that if $kd < n < (k+1)d$ then $n = kd + r$ for some $r:  0 < r < d$

And with that hint:

 And $a^n \equiv a^{kd+r} \equiv a^{kd}a^r\equiv a^r\pmod m$.

 And $0< r < d$ so $a^n \equiv a^r \not \equiv 1$.

And that's that.

Title: Proving that if $a^n \equiv 1 \pmod m$ then $d\mid n$

Body:
I'm having some trouble proving the following:

Let $d$ be the smallest positive integer such that $a^d \equiv 1 \pmod m$, for $a \in \mathbb Z$ and $m \in \mathbb N$ and with $\gcd(a,m) = 1$. Prove that, if $a^n \equiv 1 \pmod m$ then $d\mid n$.

The first thing that came to my mind was Euler's theorem but I couldn't conclude anything because I'm not very skilled when it comes to using Euler's totient function. Can someone give me any tips or show me how to solve this?


问:
how do I prove that $1 > 0$ in an ordered field?
I've started studying calculus.
As part of studies I've encountered a question.
How does one prove that $1 > 0$?

I tried proving it by contradiction by saying that $1 < 0$,
but I can't seem to contradict this hypothesis.

Any help will be welcomed.

答:
Using that
$$ x \neq 0 \Rightarrow x^2 > 0 $$
leads to
$$ 1^2 > 0 $$
and thereby
$$ 1 > 0 $$

ContentLicense: CC BY-SA 3.0


问:
$\{r_n: n = 1,2,...\}$ enumeration of the rational numbers in $[0, 1]$. Show $\sum_{n = 1}^\infty \frac{1}{2^n}f(|x - r_n|)$
$f$ is a non-negative Lebesgue integrable function on $[0, 1]$, and $\{r_n: n = 1,2,...\}$ is an enumeration of the rational numbers in $[0, 1]$. I want to show:
\begin{equation}
       \sum_{n = 1}^\infty \frac{1}{2^n}f(|x - r_n|)
   \end{equation}

converges for a.e. $x \in [0, 1]$.

I know that since $f$ is Lebesgue integrable, it follows that $\int_0^1 |f|dx < \infty$. However, I'm not sure how I would use this to solve the problem.

答:
Rather than solving the problem entirely I will give a hint:

You will use the idea that since f is integrable, the measure of the set of values x which take f(x) bigger than an upper bound M will go to zero in measure as the upper bound goes to infinity. In the case where the sum doesn't converge the values that f takes on will have to exceed the bounds too fast because of that one over 2 to the n in the front of the sum, this will force the measure of such places to approach 0.


问:
Understanding the fundamental lemma of labelled counting
From page-79 of generating functionology,

And the following proof,

The first paragraph is understandable for me, but I don't get how from it he concluded that summation. Here are the parts I understand:

The weight of hands in $h$ and the number of hands in $h'$ must add upto $n$

The number of cards in $h$ and the number of cards in $h'$ must add upto $k$

The part I don't get is how the binomial coefficent came and multiplication of two terms came.

答:
$\newcommand{\h}{\mathcal{H}}$
$\newcommand{\f}{\mathcal F}$We start with the expression $$h(n,k) = \sum_{n' \leq n,k'\ leq k} \binom{n}{n'}h'(n',k')h''(n-n',k-k') $$

and how it comes from what's been explained previously. I will then explain the rest. It is possible that you already know some of what I write, but to be complete I will still write it.

So $h(n,k)$ counts the number of hands with weight $n$, comprised of $k$ cards coming from $\f$, the exponential family associated to $\h$. Now, the set of cards in $\f$,  is by definition the (disjoint) union of the set of cards of $\f'$  and the set of cards of $\f''$.

Therefore, a hand which would be counted under $h(n,k)$, would be formed uniquely as follows :

Pick an $n' \leq n$ and $k' \leq k$. (Therefore the summation)

Pick , a hand of weight $n'$ and size $k'$ from $\h'$.  (Therefore, $h'(n',k')$ and $h''(n-n',k-k')$).

Now, contrary to expectation, we aren't done yet. This is because a hand counted under $h(n,k)$ needs to be a collection of cards whose labels are a disjoint decomposition of $[n]$.

Now, a hand ,say $H'$, counted in $h'(n',k')$ has total weight $n'$, and all label sets of this hand put together make exactly $[n']$. On the other hand (double meaning intended!) $H''$ counted in $h''(n-n',k-k')$, all the label sets put together give $[n-n']$. So now, we must recombine the $[n']$ and $[n-n']$, to give us $[n]$. How is this done?

The answer is simple : instead of $[n']$, we choose any subset of $[n]$ of size $n'$,which we use to change all the labels of hand $H'$. Then, the complement of that subset is a subset of size $n-n'$, which we use to label $H''$. The choice of that subset is independent of the choices of $H'$ and $H''$, because by relabelling the number of hands is not going to change if I change which set the labels have to be a superset of, as long as the size of the superset is retained.

I must give an example, to keep the sanity of my clientele. Let's say we are creating a hand of weight $n=7$, with $k=4$ cards.

I pick hand $H'$ , which has total weight $n'=3$ and $k'=2$ cards, from $F'$. Then I pick $H''$ which has total weight $4 = n-n'$ and $k-k' = 2$ cards.

Now, let's take an example : $H'$ has cards with labels $\{1,3\}$ and $\{2\}$ (so they form a decomposition of $[n'] = [3]$), while $H''$ has cards with labels $\{1,4\}$ and $\{2,3\}$, say, so they form a decomposition of $[n-n'] = [4]$. (The pictures don't matter in our situation, we keep them out of context).

Now, putting them together to get a decomposition of $[n] = [7]$, how do I do that? I decide on a subset of $[n] = [7]$ of size $n' = 3$, which I will use to relabel the first part. The remaining I use to relabel the second part. Let's use the subset given by $\{2,5,7\}$.

Now, I create a mapping like so : $\{1,2,3\} \leftrightarrow \{2,5,7\}$. Then, the cards in $H'$ get relabelled like $\{1,3\} \to\{2,5\}$ and $\{2\} \to \{7\}$.

Then for $H''$ I get the mapping $\{1,2,3,4\} \leftrightarrow \{1,3,4,6\}$ (the complement of $\{2,5,7\}$). This gives the relabelled cards of $\{1,4\} \to \{1,6\}$ and $\{2,3\} \to \{3,4\}$.

Now, we can put the relabelled cards together, and that gives us a hand.

It is clear that for each subset of $[n]$ of size $n'$ the relabelling can be done, and is independent of hand choice. It follows that the number of subsets of such type i.e. $\binom{n}{n'}$ is to be multiplied with the earlier expression, leading to the expression for $h(n,k)$ stated at the starting of the answer.

The last line

Let us write down $\h'(x,y)\h''(x,y)$ , multiply term-by-term, and then do a little rewrite using $\frac{1}{s!t!} = \frac{\binom {s+t}s}{(s+t)!}$ :
$$
\h'(x,y)\h''(x,y) = \left(\sum_{n',k' \geq 0} h'(n',k')\frac{x^{n'}}{n'!}y^{k'} \right) \left(\sum_{n'',k'' \geq 0} h''(n'',k'')\frac{x^{n''}}{n''!}y^{k''}  \right) \\
= \sum_{n',n'',k',k'' \geq 0} h'(n',k')h''(n'',k'') \frac{x^{n'+n''}}{n'!n''!} y^{k'+k''} \\
= \sum_{n',n'',k',k'' \geq 0} h'(n',k')h''(n'',k'') \binom{n'+n''}{n'}\frac{x^{n'+n''}}{(n' + n'')!} y^{k'+k''}
$$

Now, what is the coefficient of $\frac{x^n}{n!}y^k$? Well, note that all we require $n' + n'' = n$ and $k'+k'' = k$ i.e. $n'' = n- n'$ and $k'' = k' - k$. Therefore, we can choose any $n'\leq n, k' \leq k$ and take $n'',k''$ as above. Furthermore, it is clear that this covers all terms containing $\frac{x^n}{n!}y^k$. Combining all such terms the answer is:
$$
\sum_{n' \leq n, k' \leq k} h'(n',k')h''(n-n',k-k') \binom{n}{n'} \color{green}{ = h(n,k)}
$$

Therefore, $\h'(x,y)\h''(x,y) = \sum_{n,k} h(n,k)\frac{x^n}{n!}y^k = \h(x,y)$, as desired.


问:
Dimension of a subspace of $K^S$
Let $S$ be a set and $V$ a subspace of the $K$-vector space $K^S$. Suppose there exist a finite family $(s_i)_{1\leq i\leq p}$ of elements of $S$ and a finite family $(f_i)_{1\leq i\leq p}$ of elements of $V$ such that $f_i(s_j)=\delta_{ij}$ for all $i,j\in[1,p]$. I want to show that this implies $p\leq\text{dim}(V)$.

Let $(e_s)_{s\in S}$ be the canonical basis of $K^{(S)}$ and $(e_s^*)_{s\in S}$ the associated coordinate forms. There exists an isomorphism $\varphi:(K^{(S)})^*\rightarrow K^S$ sending $\varphi(e^*_s)$ to $(\langle e_{s'},e^*_s\rangle)_{s'\in S}$. It suffices to show that there exists a family $(x_i)_{1\leq i\leq p}$ of $K^{(S)}$ such that
$$\langle x_i,\varphi^{-1}(f_j)\rangle=\delta_{ij}.$$
The book proposes to take $x_i:=e_{s_i}$. I don't see how $\langle e_{s_i},\varphi^{-1}(f_j)\rangle=\delta_{ij}$. (The book asserts that $f(s)=\langle e_s,\varphi^{-1}(f)\rangle$ for all $s\in S$ and $f\in K^S$.) Why does it hold?

Edit:

Silly mistake. Ofc $f=\varphi(\varphi^{-1}(f))=(\varphi^{-1}(f)(e_s))_{s\in S}$. Solved.

答:
There is no need to go to the dual, we can show that the family $(f_i)_{1\le i\le p}$ is linearly independent directly. Assume we have a linear combination
$$
0 = \sum_{i=1}^p \lambda_i\, f_i.
$$
Let $k\in\{1,\dots,p\}$ and evaluate both sides at $s_k$ to obtain
$$
0 = \left(\sum_{i=1}^p \lambda_i\, f_i\right)(s_k) = \sum_{i=1}^p \lambda_i\, f_i(s_k) = \sum_{i=1}^p \lambda_i\, \delta_{ik} = \lambda_k.
$$
Hence $\lambda_k=0$ for $k=1,\dots,p$ and the $f_i$ are linearly independent.

Having found $p$ linearly independent vectors in $V$ we conclude that $p \le \dim(V)$.

Title: Dimension of a subspace of $K^S$


问:
Torus removed two points is homotopy equivalent to wedge sum of three circles
What is an easy way to see that torus with two points removed is homotopy equivalent to wedge of three circles?

I am trying to see it by viewing the torus as a square with two sides identified, but the intuition isn't clear to me, unlike the case with one point removed.

I am trying to see it as each triangle being retracted to its boundary, but I have no idea how that results in wedge of three circles.

Thanks!

答: You can actually see the three circles in this picture! Look at the three edges incident to the bottom-left corner in your picture. These all become circles after we do the identifications, and they meet at a single point (all four vertices of the rectangle become the same point after identifications). And, after identifications, these three edges are the entire space!</p>


问:
Proof that $z+z^2$ is algebraic over a field
I need to show / or disprove, that if $z\in L$ for a Field extension $L\supseteq K$ is algebraic over $K$, then $z+z^2$ is also algebraic over $K$.

I thought of writing the expression $\sum_{i \in I} k_i(z+z^2)^i$ and forming it around to get a solution I can work with.. but I kinda doubt that this works after trying for a while..

can someone give a hint?

kind regards

答: Sum and product of algebraic numbers are algebraic, to see this, if $a,b$ are algebraic, it is equivalent to $[K(a):K]$ is finite dimensional, $K(a,b):K(a)]$ is finite dimensional and $[K(a,b):K]\leq [K(a,b):K(a)][K(a):K]$, you deduce that $K(a,b)$ is algebraic.


问:
A problem on indefinite integral: $\int(\cos x)^m\sin(nx) \mathrm dx$

If
$$I(m,n)=\int(\cos x)^m\sin(nx) ~\mathrm dx,$$
How do I get $7I(4,3)-4I(3,2)$?

答:
A general solution

$$\begin{align}
I(m,n)
&=\int(\cos x)^m \sin(nx) ~\mathrm{d}x \\
&=\int(\cos x)^m\sin((n-1)x+x) ~\mathrm{d}x \\
&=\int(\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x+\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&\stackrel{\color{red}{\text{IBP}}}{=}-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m+1}{n-1}-1\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&=-\dfrac{\cos((n-1)x)}{n-1}(\cos x)^{m+1}-\left(\dfrac{m-n+2}{n-1}\right)\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x
\end{align}$$

Now

$$\begin{align}
&\int(\cos x)^m\cos((n-1)x)(\sin x)~\mathrm{d}x \\
&\stackrel{\color{red}{\text{IBP}}}{=} -\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right)\int (\cos x)^{m+1}\sin((n-1)x) ~\mathrm{d}x \\
&=-\dfrac{(\cos x)^{m+1}\cos((n-1)x)}{m+1}+\left(\dfrac{n-1}{m+1}\right) I(m+1,n-1)
\end{align}$$

This is tedious but now everything can be evaluated by successively putting values into it. It would be great if someone shows a nice way of evaluating the given value by some manipulation.


问:
Mathematical formulation of an optimization problem
I have the following optimization problem at hand.

There are N integer numbers, $a_{i},a_{i+1},\dots,a_{n}$ where $a_{i} > 0$.

We need to partition these numbers into d sets and partitions should be contiguous. Please assume valid combination of $N$ and $d$. For example, a valid partition would be like (assume $N = 10$ and  $d = 3$):

$a_{1}, a_{2}, a_{3} \quad |\quad a_{4}, a_{5}, a_{6} \quad |\quad a_{7}, a_{8}, a_{9},a_{10}$

Now, in each partition, there will be a maximum value, call it as $a^{k}$ for the $k$th partition where $1 \le k \le d$. We need to minimize the $\sum_{k = 1}^{d} a^{k}$.

So, my approach was to break down this optimization problem into subproblems:

for example, If $f(i,k)$ = Optimal sum when we partition $1$ to $i$ numbers into $k$ partitions.

Then, $f(i,k+1) = \text{min}\left[f(j,k) + \text{max}\left(a_{j+1},a_{j+2},\dots,a_{i} \right) \right] \quad \text{where} \quad j < i$.

With this formulation, we can calculate smaller subproblems first and then construct the bigger subproblems.

But, I don't consider discussing the above approach here, because of its algorithmic nature. I want to know if there is another way of handling this optimization problem in a more mathematical way instead of algorithmic treatment.

Sorry for the term "mathematical way", because I am not aware of such techniques if exist.

Thanks!

答:
You can solve the problem via integer linear programming as follows.  For each pair $(i,j)$ with $1 \le i \le j \le n$, let binary decision variable $x_{i,j}$ indicate whether part $\{a_i,\dots,a_j\}$ appears in the partition.  The problem is to minimize $$\sum_{i,j} \max(a_i,\dots,a_j) x_{i,j}$$ subject to
\begin{align}
\sum_{i,j} x_{i,j} &= d \tag1 \\
\sum_{i,j: i \le k \le j} x_{i,j} &= 1 &&\text{for $k\in \{1,\dots,n\}$} \tag2\\
\end{align}
Constraint $(1)$ selects exactly $d$ parts.
Constraint $(2)$ forces each element to appear in exactly one part.


问:
Groups of $3$ out of $23$ s.t. all students cooperate exactly once

A class consists of $23$ students. We want to create groups of $3,$
in a way  that all students cooperate (two students cooperate iff they coexist in some group) exactly once. How many groups of $3$ as above are there in total?

Of course $\binom{23}{3}$ is the total number of ways to create groups of $3.$ But how do we exclude all the repetitions?

Thanks in advance.

答: I may be missing something (or not understanding the problem), but it doesn't seem possible. Each of the $23$ individual students must participate in $11$ different groups of $3$ (so that they can work with each of the other $22$ students exactly once).  This suggests $23\cdot 11$ total groups, BUT that counted each group three times (once for each member).  Thus the actual number of groups would be $\frac{23\cdot 11}{3}$. That's not a whole number!


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答:
You can do this: Write $x = 2x-x$. And then you have $$x^{2} + 2x - x-2 = x(x+2) -1 (x+2) = (x+2)(x-1)$$

Suppose  you have a quadratic like $x^{2} + (a+b)x + ab$, then you actually want to see how $ab$  can be split up into some numbers such that their product is $ab$ and their sum is $a+b$. Now you have $-2$ which can be written as $2 \times -1$ and $2 - 1=-1$ which the co-efficient of $x$. That's why you write it as $x^{2} + 2x -x -2$. Another example,  would be consider

$x^{2} + 8x+12$. You want to factorize this. $12$ can be split up as $4 \times 3$ or $2 \times 6$. But $4+3=7$ which is not our $x$ coefficient. So we have to go with $2$ and $6$ and see that $x^{2} + 6x + 2x +12 = (x+6)\cdot (x+2)$.

Title: How to factor quadratic $ax^2+bx+c$?


问:
Show that $e^{\pi}>\pi^e$.

Show that $e^{\pi}>\pi^e$.

Note that you don't know $e=2.7182...$ nor $3.1415...$. How can we show this?

Thank you.

p.s. This is just a question came upin my mind sudden;y, it may not have a nice solution.

答: Hint: Consider the function $f(x)=x^{1/x}$. It has a maximum value.</p>

Title: Show that $e^{\pi}>\pi^e$.


答:
Overall there are $\frac{23*22}{2}=253$ pairs of students in the class. And a group of $3$ students will contain $3$ pairs of students. But $253$ is not divisible by $3$, so the desired assignments are impossible without some repetition.

See also: Steiner Triple Systems.

Title: Groups of $3$ out of $23$ s.t. all students cooperate exactly once

Body:
A class consists of $23$ students. We want to create groups of $3,$
in a way  that all students cooperate (two students cooperate iff they coexist in some group) exactly once. How many groups of $3$ as above are there in total?

Of course $\binom{23}{3}$ is the total number of ways to create groups of $3.$ But how do we exclude all the repetitions?

Thanks in advance.


问:
How can I tell if a set is compact and connected?
For expample if I consider the subspace $A_n$ of $\mathbb{R}^2$ defined by
$$A_n=\{(x,y)\in\mathbb{R}^2:y=|x|^{2n+1}\}$$
and
$$X=\bigcup_{n\ge0} A_n$$
Using Heine-Borel I just have to prove it's closed and it's limited, for the first consideration, but how should I do that?

答:
The set is not bounded so I don't think that it will be possible to show that it is bounded.

If you draw a picture of the set you will see that there are infinitely many distinct unbounded curves originating at the origin.

It may also be possible to imagine a small ball at the origin and very thin open sets surrounding each distinct curve emitting from the origin, these open sets being thin enough to not overlap the other curves coming out of the origin (as long as you are away from the origin to begin with which that original ball around the origin lets you do).  These thin open sets along with that ball will provide an open cover of the set which is infinite and has no subcover at all, since leaving out one of the thin open sets will leave out the curve that goes along with it.


答: Hint: $z$ is algebraic over $K$ means the $K\!$-algebra $K[z]$ is a finite dimensional $K$-vector space. Observe that $K[z+z^2]$ is a subalgebra of $K[z]$.

Title: Proof that $z+z^2$ is algebraic over a field

Body:
I need to show / or disprove, that if $z\in L$ for a Field extension $L\supseteq K$ is algebraic over $K$, then $z+z^2$ is also algebraic over $K$.

I thought of writing the expression $\sum_{i \in I} k_i(z+z^2)^i$ and forming it around to get a solution I can work with.. but I kinda doubt that this works after trying for a while..

can someone give a hint?

kind regards


问:
Why is there a second basis step for k = 2 in this problem?
I'm new to forum-posting, and apologize in advance for poor formatting. I'm a bit confused on a proof involving Strong Induction and Fibonacci numbers, which displays as an image here (apologies again for formatting).

My impression is that because $k = 2$ generates $F_3 \leq 2F_2$, it should be covered by the assumption that  $F_i \leq 2F_{i-1}$ for $1 \leq i \leq k$, where $k \geq 2$. However, we still take the extra step to confirm that $k = 2$ generates a valid inequality. Why?

It seems to me that the only reason for this would be that the definition of Fibonacci numbers inherently relies on two prior numbers. However, I encountered other examples of "Strong" induction, albeit from YouTube examples, that failed to establish two base cases, which has left me confused.

Any explanations would be greatly appreciated, and once again, sorry for the formatting.

答:
You are right. The claim is

$$\forall\,n\ge2,F_n\le2F_{n-1}.$$

Hence the base case seems to be

$$F_2\le2F_1,$$ which is true, and the inductive step

$$F_k\le2F_{k-1}\implies F_{k+1}\le2F_k.$$

If we limit ourselves to simple induction, about all we can write is
$$F_{k+1}=F_k+F_{k-1}\le3F_{k-1}$$ and this does not establish the claim.

Hence the correct inductive step is

$$F_{k}\le2F_{k-1}\land F_{k+1}\le2F_{k}\implies F_{k+2}=F_{k+1}+F_k\le 2F_{k}+2F_{k-1}=2F_{k+1}.$$

And for the same reason, the base case has to be

$$F_2\le2F_1\land F_3\le 2F_2.$$


问:
Find constant k if given a continuous piecewise function

Seems like an easy question but I'm stumped. The answer is $5$ and I can't seem to get to it. Of course, I know I must take the limit of the top function and that it should be equal to k but I can't evaluate the limit.

答:
Separate the fraction into two parts, and use the fact that $\lim_{t\to 0} \frac{\sin t}{t} =1 $:

$$\frac{x^2\sin(5/x)}{\sin kx} +\frac{25x}{\sin kx}\\ \to x^2 \cdot \frac{\sin(5/x)}{kx} +\frac{25x}{kx} \\ =\frac xk \sin (5/x)+\frac{25}{k} \\ \to0+\frac{25}{k}$$

Then for continuity, $$\frac{25}{k} = k \implies k=5$$


问:
Understanding the modal necessitation rule
I understand that the modal necessitation rule is:

$ \vdash p  \Rightarrow  \vdash \Box p $

That is, if p is a theorem then necessarily p is a theorem.

But I don't quite understand what this means, or how I could use it in a modal proof.  Obviously if I derive $x$ on a line I cannot then move to necessarily $x$.  (Just because it's actually raining does not imply that it's raining in all possible worlds.)

Does the rule mean that only if $p$ is a tautology then I can move to necessarily $p$?  Just trying to figure out exactly what is meant by "theorem" and how i can use the necessitation rule in a proof.  Thanks.

答:
The rule is that if $\varphi$ is provable from no assumptions [other than logical axioms], i.e. if $\varphi$ is a theorem, then $\Box\varphi$ is also a theorem. That's a plausible rule to have in the modal logic of necessity: it formally echoes the idea that if something is demonstrable by logical reflection alone it is necessarily true.

Thus, the following is a proof in standard modal systems:

$(p \lor \neg p)$

$\Box(p \lor \neg p)$

$\Box\Box(p \lor \neg p)$

where the first line is a tautology derivable from no assumptions [other than axioms], so using necessitation we can infer (2). But note that (2) is still derived from no assumptions (logical reflection alone is required to see it is true) so we can use necessitation again to infer (3) in turn.

The input to an application of the necessitation rule has to be a theorem but needn't be a truth-functional tautology.

[There are some wonderful books on elementary modal logic which will explain this very clearly -- there are some suggestions in §2.2 of the Teach Yourself Logic Guide you can download here.]


问:
How to compute the gradient of a dot product?
Let  $a\in \Bbb{R}^n$ be a vector consisting of constant values, $b ∈ \Bbb{R}^n$ and  $C\in \Bbb{R}^{n\times n}$ .

How can the gradient be calculated for

$∇_ba \boldsymbol{\cdot} b$ ( dot-product of a and b)
and/or $∇_bb \boldsymbol{\cdot} b$

I understand that this would be taking the derivate of the respective functions. However what process would I go through to solve such a problem and what should I expect to arrive at?

I know it can be started like...
$∇_ba·b =(d/db_1 a·b, d/db_2 a·b, d/db_3 a·b...) $

$d/db_1 a·b  = d/db_1 (a_1b_1+a_2b_2...)$

How do i get to the scalar and how can i understand the process going forward?

答: Let's use @Deane's strategy. Summing over repeated indices,$$\frac{\partial(a_ib_i)}{\partial b_j}=\frac{\partial a_i}{\partial b_j}b_i+a_j$$.A matrix $M_{ij}:=\frac{\partial a_i}{\partial b_j}$ satisfies $\nabla_b(a\cdot b)=M^Tb+a$. Meanwhile,$$\frac{\partial(a_ia_i)}{\partial b_j}=2M_{ij}a_i=2(M^Ta)_j.$$


问:
How to rotate curve around a point
Given the equation of a curve in x y, say a parabola, how to I re-write the equation if the curve is rotated around some point by some number of degrees ?</p>

答:
Let the point to rotate around be $(x_{0}, y_{0})$.

First, translate everything so $(x_{0},y_{0})$ is at the origin. This entails using the shifted coordinate system $(x',y') = (x-x_{0}, y-y_{0})$. Now, we use the formula for rotation about the origin through an angle $\theta$:

$$(x',y')\to(x'\cos\theta - y'\sin\theta, x'\sin\theta + y'\cos\theta)$$

Finally, we must shift back to the original coordinate system. Thus, to rotate an angle $\theta$ about $(x_{0}, y_{0})$, you must substitute:

$$\boxed{x\to(x- x_{0})\cos\theta - (y - y_{0})\sin\theta + x_{0},\ y\to(x- x_{0})\sin\theta + (y - y_{0})\cos\theta + y_{0}}$$

A faster way to accomplish this would be to use transformation matrices:

$${x\brack y}\to {\cos\theta\ \ -\sin\theta\brack\sin\theta\ \ \ \ \ \cos\theta}{x - x_{0}\brack y-y_{0}} + {x_{0}\brack y_{0}}$$

Simplifying the RHS will get you the formula we derived earlier.

Title: How to rotate curve around a point


答:
Start by writing the equation in a new coordinate system that is centered on $(x_0,y_0)$, your rotation point. For example $$y=x^2$$ will become $$(y-y_0)+y_0=[(x-x_0)+x_0]^2$$
We have now the coordinates in this new reference frame: $x'=x-x_0$ and $y'=y-y_0$.
If we rotate this by an angle $\alpha$ to a new coordinate frame, we get $$x''=x'\cos\alpha+y'\sin\alpha\\y''=y'\cos\alpha-x'\sin\alpha$$
Therefore, by replacing $y'$ and $x'$ by $y''$ and $x''$:
$$(y-y_0)\cos\alpha-(x-x_0)\sin\alpha+y_0=[(x-x_0)\cos\alpha+(y-y_0)\sin\alpha+x_0]^2$$
All you need to do is to simplify the equation and group the terms together.

Title: How to rotate curve around a point

Body: Given the equation of a curve in x y, say a parabola, how to I re-write the equation if the curve is rotated around some point by some number of degrees ?</p>


问:
Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$
How to get a closed form for the following sum:

$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$

Where $F_k$ is the $k$th Fibonacci number.

I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$

Body:
$F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1,ab=-1, a^2=a+1, b^2=b+1$
We will make multiple use of there relations below:

$$S=\frac{1}{5}\sum_{k=0}^{n} (-1)^k[a^{2k}+b^{2k}-2(ab)^k]~~~~(1)$$ $$S=\frac{1}{5}\sum_{k=0}^{n}(-1)^k\left[a^{2k}+b^{2k}-2(n+1)]\right]~~~~(2)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2-1}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2-1}-2(n+1)\right]~~~~~(3)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)\right]~~~~~(4)$$
$$S=\frac{1}{5(a-b)}\left[\frac{(-1)^{n+1}a^{2(n+1)(a-b)}}{-a^2+ab}+\frac{(-1)^{n+1}b^{2(n+1)}-1}{-b^2+ab}-2(n+1)(a-b)\right]~~~~(5)$$
$$=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-1/a-1/b-2(n+1)(a-b)\right]$$
$$S=\frac{1}{5(a-b)}\left[(-1)^{n}[a^{2n+1}-b^{2n+1}]-(2n+1)(a-b)\right]~~~~(6)$$
$$\implies S=\frac{F_{2n+1}-(2n+1)}{5}~~~~~(7)$$


问:
Why is the sqrt. of negative one relation important for imaginary numbers?
Why is the square root of negative one an important relation for imaginary numbers and quaternions, etc.?

I mean, if we are going to have imaginary numbers, couldn't they be imaginary without needing to be defined in terms of the square root of negative one?

I.e. why couldn't we have imaginary numbers without them having any definition in terms of a relation to the real numbers?

Seems to me that you could say imaginary numbers are based on the square root of x, where x is some number that's not on the real number line (but not necessarily square root of negative one—maybe instead, 1/0).

答: Because historically, the question that needed to be addressed was not "What definition can we give to a second continuum of numbers?" with the answer "Let's say they're multiples of $\sqrt{-1}$."  Rather, it was "What can we do when multiples of $\sqrt{-1}$ emerge from algebraic computations?" with the answer "Let's treat them as a second continuum of numbers."


问:
Relationship between Gram determinant and cross product
I am trying to prove that the determinant of the Gram matrix is equal to the norm of the cross product squared for $v_1, v_2 \in \mathbb{R}^3$:

$$\|v_1 \times v_2\|^2 = \det\Bigl(\begin{smallmatrix}
\langle v_1, v_1 \rangle & \langle v_1, v_2 \rangle\\
\langle v_2, v_1 \rangle & \langle v_2, v_2 \rangle
\end{smallmatrix}\Bigr)$$

But the only idea I have seems very technical and unpleasant, any suggestions?

I have already seen this thread but I don't find it helpful.

答:
$\def\e{\varepsilon}\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}$
The Levi-Civita symbol allows one to write the cross product as
$$\eqalign{
c &= a\times b \\&= \e:ab^T \\
c_i &= \e_{ijk}\;a_jb_k\quad\big({\rm index\,notation}\big) \\
}$$
where the colon denotes the double-dot product.
The symbol has another interesting property
$$\eqalign{
\e_{ijn}\,\e_{nk\ell}
  &= \delta_{ik}\,\delta_{j\ell} - \delta_{i\ell}\,\delta_{jk}
\quad\big({\rm Kronecker\,delta\,symbols}\big) \\
  &= \alpha_{ijk\ell}-\beta_{ijk\ell} \\
\e\cdot\e &= \alpha-\beta \\
}$$
The fourth-order tensors $(\alpha,\beta)$ have some useful and easily verified properties.

Given an arbitrary matrix $(M)$
$$\eqalign{
\alpha:M &= M:\alpha &= M\qquad&\big({\rm identity\,operation}\big) \\
\beta:M &= M:\beta &= M^T\qquad&\big({\rm transpose\,operation}\big) \\
}$$
Putting all of this together
$$\eqalign{
(a\times b)\cdot(f\times g)
 &= (ab^T:\e)\cdot(\e:fg^T) \\
 &= ab^T:(\e\cdot\e):fg^T \\
 &= ab^T:(\alpha-\beta):fg^T \\
 &= (ab^T:\alpha:fg^T) - (ab^T:\beta:fg^T) \\
 &= (ab^T:fg^T) - (ab^T:gf^T) \\
 &= (a\cdot f)(b\cdot g) - (a\cdot g)(b\cdot f) \\
 &= \det\left(\m{a\cdot f&a\cdot g\\b\cdot f&b\cdot g}\right)\\
}$$
Now set $\;(a=f=v_1)\;$ and $\;(b=g=v_2)\;$ to obtain the desired result.


问:
Combinatorial/ Binomial Identity
I am finding difficulty in showing the following identity:

$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1} = \frac{2}{n+1}.$$

Mainly because of the inverse part.

Title: Combinatorial/ Binomial Identity

Body:
We have by binomial coefficient algebra that

$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1}
= \sum_{k=1}^n \frac{(n-1)!/(n-k)!/(k-1)!}
{(2n-1)!/(2n-1-k)!/k!}
\\ = {2n-1\choose n}^{-1}
\sum_{k=1}^n k \frac{(2n-1-k)!}{n! (n-k)!}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
\sum_{k=1}^n k {2n-1-k\choose n-k}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
[z^n] (1+z)^{2n-1} \sum_{k=1}^n k \frac{z^k}{(1+z)^k}.$$

Now the coefficient extractor enforces the range and we get
for the extractor only

$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
= [z^n] (1+z)^{2n-1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
\\ = [z^n] (1+z)^{2n-1}
\frac{z(1+z)}{(1+z-z)^2}
= [z^{n-1}] (1+z)^{2n}
= {2n\choose n-1}.$$

Restoring the scalar in front now yields

$$\frac{1}{n} {2n-1\choose n}^{-1} {2n\choose n-1}
= \frac{1}{n} 2n \frac{1}{n+1} = \frac{2}{n+1}.$$

This is the claim.


问:
This book CONFLATES true and conditional distribution?
Consider the following excerpt from the book Mathematical Theory of Bayesian Statistics by Watanabe. Right on the second page, the following is stated:

What does the second sentence really mean?

It seems to me that the author wants to transmit an informal idea (but I don't know which one), since if we consider the statement literally, it doesn't make sense: One cannot condition on an a distribution, only on an event.

Why, then, is this "representation" used (and moreover, a misleading term "conditional probability distribution used") if the first sentence already describes in a perfectly clear manner what is going on?

答:
The second sentence means that given the distribution is q, here is the probability of getting the set of data points.  It is a very careful way to remind the reader that there is typically an assumed distribution underlying the process in question.

The "event" here is that the process $q$ is present, and not say some process $r$, $s$, or $t$.

At a more general level, one can always add an explicit condition, but we generally do not do so just to clarify notation and simplify a discussion.  We could include the condition that the measuring system is the same (or not!).

Explanation:  Every statistical measurement, such as concerned here, is based on some assumptions, or elements that can be expressed as "given that...".  For instance, when we say that the probability of a coin coming up heads is $0.5$, we often write it as $P(H) = 0.5$.  But what we really mean is that the probability is $0.5$ GIVEN THAT the coin is fair.  So we might write $P(H|{\rm fair}) = 0.5$.  Is that enough?  No!  We assume that the flipping procedure is fair too... that the flick of the thumb does not favor heads or tails.  Thus we could write $P(H|{\rm fair} \& {\rm fair~flip}) = 0.5$.  Is that enough?  No!  We assume that the person reporting the flip is honest, and has good eyesight.  Thus we could write $P(H|{\rm fair} \& {\rm fair~flip} \& {\rm accurate~reporting}) = 0.5$.  Are we done yet?  No!...  Well you get the idea.

In statistics we take certain assumptions as given, needing no explicit expression in our equations.

That's all this overall question refers to:  Watanabe is making it explicit that the model is $q$.  He doesn't have to do this, but it cannot hurt.  It is like writing $P(H|{\rm fair}) = 0.5$, above.  The "fair coin" would be one possible "model," $q$.

Frankly, OP, I think you're spending far too much time on this issue, and your discussion of "product measures," $\sigma$ algebras, and such is an irrelevant diversion.


问:
Norm of a reproducing kernel Hilbert space in terms of supremum over unit ball
Let $H$ be a reproducing-kernel Hilbert space (RKHS) of bounded continuous functions $g:X \to \mathbb{R}$, where $X$ is some topological space.

The norm of $g \in H$ is given by

$$
\lvert\lvert g \rvert\rvert_H = \sqrt{\langle g, g \rangle_H}.
$$

This paper (in the final equality of the second proof on p. 727) effectively states the following property

$$
\sup_{\lvert\lvert f \rvert\rvert \le 1} \langle g, f \rangle_H = \lvert\lvert g \rvert\rvert_{H}. \quad\quad\quad (\star)
$$

How does $(\star)$ follow?

One thought is that the sup-norm of $g$ is given by

$$
\lvert\lvert g \rvert\rvert_{\infty}
= \sup_{x \in X}|g(x)|
= \sup_{x\in X}\langle g, \phi_x \rangle_H,$$

where $\phi_x \in H$ is the projection of $x$ by RKHS.  If $\phi: X \to H$ is onto (i.e.,, $\{\phi_x \mid x \in X\} \equiv H$) then $$\sup_{x \in X} \langle g, \phi_x \rangle_H = \sup_{f \in H}\langle g, f \rangle_H.$$

However, it is not clear to me that the sup-norm is relevant, that $\phi$ is onto, or how the unit ball in $(\star)$ relates to the norm.

答: By Cauchy-Schwarz, we have that $|\langle g,f\rangle_H|\leq\|g\|_H\|f\|_H\leq \|g\|_H$ if $\|f\|_H\leq 1$. On the other hand, every Hilbert spaces are reflexive. By Kakutani's Theorem, the closed unit ball is weakly compact, hence ($\star$) says that the function $f\mapsto\langle g,f\rangle_H$ attains its maximum over all $f$ with norm $\leq 1$. There is no square as in the question.


问:
Prove that $g(x)=(x^2−1)^x$ is increasing on $(1,+\infty)$
Let $A = \mathbb{R}\setminus[−1,1]$. Let $g : A\to\mathbb{R}$ be defined by
$g(x)=(x^2-1)^x$
for all $x\in A$.

Prove that $g(x)=(x^2-1)^x$ is increasing on $(1,\infty)$

I have currently attempted to prove this by showing $g'(x)\geq 0 $ for all $ x\in A $ which gives $g'(x)= (x^2-1)^{x-1}(2x^2+(x^2-1)ln(x^2-1))$ which should be greater than or equal to $0$ however I am unsure how to show this or whether I have gone about this the right way.

Any Help will be grateful.

答: Follow what you left off, $2x^2 + (x^2-1)\ln(x^2-1)= 2(x^2-1)+(x^2-1)\ln(x^2-1)+2= 2t+t\ln t + 2=h(t), t = x^2-1>0$. Computing $h'(t) = 3+\ln t= 0 \iff t=e^{-3}$. Observe that $h''(t) = \dfrac{1}{t} > 0, t > 0$. Thus $h_{\text{min}} = h(e^{-3})= 2e^{-3}+2+-3e^{-3}= 2-e^{-3} > 0$. By calculus's $2$nd second derivative test, it shows that $h(t) > 0$, and thus your $g'(x) > 0$ and the function is increasing over $(1,\infty)$.


问:
For which p>0 does $S_t=W_t+t^p$ admit an equivalent martingale measure?
Let $W$ be a brownian motion and $p>0$.

For which $p$ does $S_t=W_t+t^p$ admit an equivalent martingale measure?

I recently saw at my lectures that:

NFLVR cond:
There does not exist a sequence $\{H_n\}_{n \geq 1}$ of predictable processes, integrable wrt  S, such that there exists $t_0 ,  b , \epsilon >0$   $$ \int_0^{t_0} H_n(s) S_s > -1/n$$
and $$ P(\int_0^{t_0} H_n(s) S_s >b)> \epsilon$$

Is equivalent to having such a measure and I think that is pretty much my only tool, so I guess it has to be that. On the other hand I can't get any ideas on how to use the condition.

答:
Let me just give a short answer since probably nobody cares:

It can bee seen as a corollary to (at least the proof i have seen) of Girsanov that if we are looking for a equivalent martingale measure for a continuous semimartingale $S_t = M_t + A_t$ where $M_t$ is the local martingale part and $A_t$ is the bounded variation part, then $A_t$ must satisfy that there exists predictable $h_t$ such that $A_t=\int _0^t h_s d<M_s>$ and $\int_0^t h_s^2 d<M_s> < \infty $. ($<M>$ is quadratic variation). In our case this rules out $p\leq 1/2$ since $A_t=\int_0^t p\cdot s^{p-1} ds $ which is only square integrable if $p > 1/2$.

Assuming from now on $p>1/2$. We will see an Martingale measure does indeed exist. This is simply and application of Girsanov  - e.g. by applying Novikovs condition which is here trivial - on $h_s = - p s^{p-1}$. This directly yields that $W_t - \int h_s d<W>_s=W_t + t^p$ is a brownian motion under $\mathbb{Q}$ (in particular a local martingale) and $\mathbb{P}$ and $\mathbb{Q}$ are equivalent.


问:
Matrix, Null space, Col space
Regarding:

Prove or provide a counterexample: let $A \in \mathbb{M}_{n \times n}$. Then $\mathbb{R}^n=ColA + NullA$.

--

So I think this is false, because I can take a matrix:

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

And we have $\dim ColA=1$, $\dim NullA=0$, and $\dim \mathbb{R}^3=3$,

and we know that $\mathbb{R}^n=ColA + NullA$ if and only if $\dim (ColA+NullA)= \dim \mathbb{R}^n$.

The only problem that I have here, is that I am not 100% sure that $\dim NullA=0$ and that $\dim ColA + \dim NullA = \dim (ColA+NullA)$

I would love to here some explanation regarding those kind of spaces :)

Thanks a lot!

答:
You are correct that the statement is false, but the counterexample you give does not work. Instead, consider the matrix
$$
A = \pmatrix{0&1\\0&0}.
$$

Title: Matrix, Null space, Col space


答:
The modern approach to complex numbers doesn't use square roots at all. The modern approach is to say $i^2=-1$ rather than $i=\sqrt{-1}$. It is more indirect, but actually makes a lot more sense once you get used to working with complex numbers.

How do you work with this? Essentially, complex numbers are polynomials in the variable $i$. They behave like any other polynomials we know and love. With one exception: whenever you see $i^2$, you're allowed to substitute it with $-1$ (this includes cases like $i^3=i^2\cdot i$ and $i^{1000}=(i^2)^{500}$).

That's it. That's what the complex numbers are.

Note that $i^2=-1$ is essential to make complex numbers something other than pure polynomials. So powers of $i$ must be related to real numbers. (Many other quadratic relations can be used instead if you wish, like $i^2=i-1$; basically, any relation that doesn't have a real solution if you interpret it as an equation will give you the complex numbers as a result, but they won't look exactly like the standard complex numbers.)

My gut says it's a better way to teach complex numbers as well. But I haven't tried that enough to actually say for certain.

Title: Why is the sqrt. of negative one relation important for imaginary numbers?

Body:
Why is the square root of negative one an important relation for imaginary numbers and quaternions, etc.?

I mean, if we are going to have imaginary numbers, couldn't they be imaginary without needing to be defined in terms of the square root of negative one?

I.e. why couldn't we have imaginary numbers without them having any definition in terms of a relation to the real numbers?

Seems to me that you could say imaginary numbers are based on the square root of x, where x is some number that's not on the real number line (but not necessarily square root of negative one—maybe instead, 1/0).


问:
Covariant derivative do Carmo help clarification concept
Let $M$ be a Riemannian manifold and let $\nabla$ be a connection compatible with the metric. Then for any vector fields $V,W$ along the smooth curve $c:I\rightarrow M$ , we have

$\frac{d}{dt}\langle V, W\rangle =\langle \frac{DV}{dt},W\rangle+ \langle V, \frac{DW}{dt}\rangle$

However my question is: Why does $\frac{d}{dt} \langle V, W \rangle$ make sense?

$V$ and $W$ are vector fields along a curve, so I suppose $\langle V, W\rangle$ is viewed as a map $I\rightarrow \mathbb{R}$ , where $\langle V,W \rangle(t)=\langle V_t, W_t \rangle$ . But how do we show that it is smooth?  Also compatibility of the metric is defined on smooth vector fields.

答:
If $M$ is a smooth manifold then $TM$ also has a natural smooth structure, hence, does $E=TM\times TM$. The projection $\pi: TM\to M$ defines a smooth map
$$
\pi\times \pi: E\to M^2.
$$
I will leave it to you to verify that the preimage of the diagonal under $\pi\times \pi$ is a smooth submanifold $A$ in $E$.  Explicitly:
$$
A=\{(u,v)\in TM\times TM: \pi(u)=\pi(v)\},
$$
i.e. $A$ consists of pairs of vectors in the same tangent space $T_pM$, $p\in M$.

A Riemannian metric $g$ on $M$ is a smooth function $g: A\to {\mathbb R}$ (satisfying further linearity conditions which are irrelevant for us). Given a curve $c: I\to M$, a vector field $X$ along $c$ is a smooth map $X: I\to TM$ such that $\pi\circ X=c$. A pair of such vector fields $X, Y$ defines a smooth map $P: I\to A$. Hence, the composition of two smooth maps $g\circ P$ is again smooth. This composition is nothing but the function
$$
t\mapsto g(X,Y)(t), t\in I.
$$


问:
Gauss Theorem used in the proof of Liouville's Theorem
I am using the 2nd Edition of PDEs by L.C.Evans.

on Page 29 the following theorem is stated:

Theorem 7 (Estimates on derivatives).

Assume $u$ is harmonic in $U$. Then

$$|\nabla u(x_0)|\leq \frac{C}{r^{n}}\|u\|_{L^1(B(x_0,r))}$$
for each ball in $B(x_0,r) \subset U$

note: this is a special case of the original version, which is a more general kth order of gradient.

And the Proof:

Proof

$u_{x_i}(x_0)$ is harmonic if $u$ is harmonic

apply the mean value formula for $u_{x_i}(x_0)$
$$\begin{split}
|u_{x_i}(x_0)| & = |\frac{2^n}{\alpha(n)r^n}\int_{B(x_0,r/2)}u_{x_i}dx| \\
 & = |\frac{2^n}{\alpha(n)r^n}\int_{\partial B(x_0,r/2)}u\nu_idS|
\end{split}$$
...

I stop because my question is here.

I know here Gauss Theorem is applied but I think Gauss Theorem is for divergence and not partial derivative on a single dimension.

答:
Inspired by RRL, I mananaged to understand the steps

let $v$ be a n-dimensional constant vector field with the $i$-th element being $1$ and others $0$. Then
$$u_{x_i} = \nabla u\cdot v$$
And for the vector field $uv$ we have the following (note that $u$ is $\mathbb{R}^n \to \mathbb{R})$)

$$\nabla \cdot (uv) = \nabla u \cdot v + u\textbf{div}(v) = \nabla u\cdot v$$

so
$$u_{x_i} = \nabla \cdot (uv)$$
$\nabla \cdot (uv)$ fits the divergence theorem and and we can use it to achieve the 2nd "$=$"


问:
Direct summands of successive extensions of line bundles
On a projective variety any direct summand of a direct sum of line bundles is going to be direct sum of a subset of initial line bundles. This follows from the Krull-Schmidt property of vector bundles on projective varieties. Now my question is a little bit similar to this. Let's assume the vector bundle is constructed by taking successive extensions of a set of line bundles in the form $\mathcal{O}(n)$. Is any direct summand of this vector bundle in the similar form?

答:
Of course not. Let $X$ be a smooth curve and take any line bundle $\mathcal{L}$. Then for $n \gg 0$ the bundle $(\mathcal{L} \oplus \mathcal{L}^{-1}) \otimes \mathcal{O}(n)$ is globally generated. Therefore, a general morphism
$$
\mathcal{O}(-n) \to \mathcal{L} \oplus \mathcal{L}^{-1}
$$
is a fiberwise monomorphism, hence its cokernel is a line bundle, which by computing the determinant can be identified with $\mathcal{O}(n)$. Thus, we have an exact sequence
$$
0 \to \mathcal{O}(-n) \to \mathcal{L} \oplus \mathcal{L}^{-1} \to \mathcal{O}(n) \to 0
$$
which shows that any line bundle on $X$ is a direct summand of an extension of $\mathcal{O}(n)$.


问:
Prove that $f(f(x))=x$ has no roots .... $f$ having a general form
This problem gave me some headache, especially because $f$ have its own general form :

let $f(x) = ax^2 + bx + c$. Suppose that $f(x) = x$ has no real roots.

Show that equation $f(f(x))=x$ has also no real roots.

答:
Consider $g(x) = f(x)-x$. If $f$ is continuous, then $g$ will be continuous as well. A continuous function without zeroes is either strictly positive or strictly negative. This means that either $f(x)>x$ for all $x$ or $f(x)<x$ for all $x$.

Hence either $f(f(x))<f(x)<x$ for all $x$ or $f(f(x))>f(x)>x$ for all $x$. Either way, $f(f(x))\neq x$ for all $x$.


问:
What are some of the Websites/web-tools which must be a part of every mathematician's arsenal?
What are some of the web-tools which must be a part of every mathematician's arsenal? Tools which help mathematicians research, solve problems or even collaborate together!

Here is a list I gathered. Many among you may already know these, But still, in case.

MathQuill--
This is a great website for easy typing in LaTeX.

Integral calculator and the Derivative calculator-- Especially useful when one wants to crack a tough indefinite question. Steps included for those who need them!

OEIS--A list of all mathematically interesting sequences that contains over 3 million of such. Many things are discovered thanks to OEIS.

Desmos--One of the best online graphing tool with loads of computing power.

WolframAlpha and Mathematica-- Who doesn't know them! A unique engine for computing answers and providing knowledge.

MATLAB--Math. Graphics. Programming.
Whether you’re analyzing data, developing algorithms, or creating models, MATLAB is designed for way you think and the work you do.

Software such as Maple, MATLAB, Mathcad, Pari/GP, R, and the WolframAlpha website, etc are famous, and this question is not asking for similar software alternatives.

I haven't added math stack exchange to this list  for obvious reasons :~)

Thanks in advance to all for the over-enthusiastic help :-)

You might want to check out
Online tools for doing symbolic mathematics

PS-No intention of advertisement or promotion of any website or software.

答:
The Math-aware search engine: Approach Zero is great for online investigations of mathematical expressions.

Example: Looking for $[x^n]\frac{1}{\sqrt{1-x}}$ provides this result.


问:
dual of an ovoid in $PG(3,q)$
In $PG(3,q)$, a $(q^2+1)$-cap is an ovoid. A $(q^2+1)$-cap is a set of $q^2+1$ points, no three of which are collinear. What is the dual of an ovoid?

I know how to get the dual of a statement involving points, lines and incidence between them by interchanging the words "point" and "line". So a $(q^2+1)$-cap would correspond to a set of $q^2+1$ lines no three of which intersect at a given point. Is this the dual of an ovoid?

答:
In 3 dimensions the duals of points are the planes, and the duals of lines are the lines.
So in this case the dual of an ovoid is a set of $q^2+1$ planes, no three of which have a common line.
(As the ovoid is a generalization of quadrics; and in classical geometry the dual of a quadric is the set of the tangent planes of a quadric.)

Title: dual of an ovoid in $PG(3,q)$


问:
What are the main differences between finite and compact group representation?
I know the fact that the number of equivalence classes of a compact group may be countably infinite and for finite groups we have a finite number. But I am not sure how to see that there are infinitely many for the compact case. The equivalence relation is that two representations of a group $G$ are equivalent iff there is a intertwining isomorphism between the resp. vector spaces.
edit: not by an example but from theory itself

答:
It is actually quite nontrivial to see that any infinite compact (Hausdorff) group $G$ has infinitely many (inequivalent) representations; a priori it's not clear that an arbitrary such group has any (finite-dimensional) representations.

But this is true and follows from the Peter-Weyl theorem, which implies that the matrix coefficients of irreducible finite-dimensional representations of $G$ must separate points on $G$. That is, if $g \neq h$, then there exists some irreducible finite-dimensional representation $\rho : G \to GL_n$ such that $\rho(g) \neq \rho(h)$. This is impossible if $G$ is infinite but there are only finitely many irreducibles.

An alternative way to phrase mostly the same argument: the Peter-Weyl theorem also implies that every irreducible (is finite-dimensional and) occurs in $L^2(G)$ with multiplicity equal to its dimension. So this again gives that if there are finitely many irreducibles then $L^2(G)$ is finite-dimensional, so $G$ is finite.


问:
How to find mutually absolutely continuous measures?
The problem is as follows :
Let μ be a σ-finite measure on (X,M). Then there exists a finite measure ν on (X,M) such that μ and ν are mutually absolutely continuous ; that is μ<<ν and ν<<μ

I am not sure how to even start the problem. Any help is appreciated

答: Hint: If you let $\nu = D \: d\mu$ where $D: X \rightarrow \left (0,\infty \right]$ is an element of $L^1(\mu)$ then this will be true. So if you can prove that such a function exists for $\mu \; \sigma$-finite then you have solved the problem.


答:
Let $\mbox{Id}_\mathbb{R}:\mathbb{R}\to \mathbb{R}$ the function $\mbox{Id}_\mathbb{R}(x)=x$. Note that :

$\mbox{Im}(\mbox{id}_{\mathbb{R}})\supset \mbox{Im}(f)$,

$\mbox{Im}(f)=\mbox{Im}(f|_{\mbox{id}_{\mathbb{R}}})\supset \mbox{Im}(f|_{\mbox{Im}(f)})=\mbox{Im}(f\circ f)$

$\mbox{Im}(f\circ f-\mbox{id}_\mathbb{R})\supset\mbox{Im}(f|_{\mbox{id}_\mathbb{R}} -\mbox{id}_\mathbb{R})$.

Then if $f|_{\mbox{id}_\mathbb{R}} -\mbox{id}_\mathbb{R}$ no have real roots then $f\circ f-\mbox{id}_\mathbb{R}$ too no have real roots.

Title: Prove that $f(f(x))=x$ has no roots .... $f$ having a general form

Body:
This problem gave me some headache, especially because $f$ have its own general form :

let $f(x) = ax^2 + bx + c$. Suppose that $f(x) = x$ has no real roots.

Show that equation $f(f(x))=x$ has also no real roots.


问:
The value of $r$ for which $\binom {20}{r} \binom{20}{0} +\binom{20}{r-1}\binom{20}{1}...\binom{20}{0} \binom{20}{r}$ is maximum is?
Most solutions write the given expression as
$$(1+x)^{20}(1+x)^{20}=(1+x)^{40}$$

But I can’t wrap my head around has to how it is even correct to do so.

When you expand the factors, what about all the other residual terms such $(\binom {20}{0})^2$ that aren’t included in the question expression?

I know this question already exists on MSE, no it does not clear the specific problem I have

答:
You can choose $r$ elements from $40$ by choosing $r$ from the first $20$; or $r-1$ from the first $20$ and $1$ from the remaining $20$; or $r-2$ from the first $20$ ...

This is also reflected in the coefficients of the two different expressions $(1+x)^{20}(1+x)^{20}=(1+x)^{40}$ which collect and combine.

$\left(\binom {20}0\right)^2$ is what you get from choosing $0$ from $40$ and gives the coefficient of $1$ (ie the coefficient of $x^0$ with $r=0$)


答:
So here's the story without using pullback bundles: A vector field $V$ along the curve $c: I \rightarrow M$ is defined to be a smooth map $V: I \rightarrow T_*M$, where $V(t) \in T_{c(t)}M$. In local coordinates, the covariant derivative of $V = V^k\partial_k$ along $c$ is defined to be
$$
\frac{DV}{dt} = \left(\frac{dV^i}{dt} + \Gamma^i_{jk}\dot{c}^j V^k\right)\partial_i, $$
where $\dot{c}$ is the derivative of $c$ and $\Gamma^i_{jk}$ are the Christoffel symbols. You can check that this definition is co-ordinate-independent and satisfies all the expected properties.

If you want the story using pullback bundles, see, for example, https://www.math.nyu.edu/~yangd/papers/pullback.pdf

Title: Covariant derivative do Carmo help clarification concept

Body:
Let $M$ be a Riemannian manifold and let $\nabla$ be a connection compatible with the metric. Then for any vector fields $V,W$ along the smooth curve $c:I\rightarrow M$ , we have

$\frac{d}{dt}\langle V, W\rangle =\langle \frac{DV}{dt},W\rangle+ \langle V, \frac{DW}{dt}\rangle$

However my question is: Why does $\frac{d}{dt} \langle V, W \rangle$ make sense?

$V$ and $W$ are vector fields along a curve, so I suppose $\langle V, W\rangle$ is viewed as a map $I\rightarrow \mathbb{R}$ , where $\langle V,W \rangle(t)=\langle V_t, W_t \rangle$ . But how do we show that it is smooth?  Also compatibility of the metric is defined on smooth vector fields.


问:
Convergence of signed measures implies inequality between variations
I'm trying to tackle Junghenn's Principles of Real Analysis' exercise 5.11:

Let $\mu, \mu_1, \mu_2, \dots$ be signed measures on $(X, \mathfrak{F})$ with $\mu_n \to \mu$. Show that $\mu^{\pm}(E) \leq \liminf_{n} \mu_n^{\pm}(E)$ and $|\mu|(E) \leq \liminf_{n} |\mu_n|(E)$.

I started off trying to prove that $\mu = \liminf_{n} \mu_n^+ - \liminf_{n} \mu_n^-$ to conclude the first inequality, but this didn't seem to pay off, since I couldn't prove that $\liminf_{n} \mu_n^+$ and $\liminf_{n} \mu_n^-$ were measures. I've also tried to use the fact that
$$\mu^+(E) = \sup\{\mu(A) : A \in \mathfrak{F} \cap E\}$$
along with
$$\liminf_{n} \mu_n^+ = \sup\bigg\{ \inf\Big\{ \sup\{ \mu_k(A) : A \in \mathfrak{F} \cap E \} : k \geq n \Big\} : n \geq 1 \bigg\},$$
but I kind of got tangled up between all the supremums and infimums, and I couldn't make much progress. Any help/hints would be greatly appreciated.

Edit: I was finally able to do it. I'll post my solution in case someone might want to take a look at it.

答:
We will first show that $\mu^{\pm}(E) \leq \liminf_{n} \mu_{n}^{\pm}(E)$. Let $E \in \mathfrak{F}$. We know that for each $n \in \mathbb{N}$,
$$
\mu_n^{+}(E) = \sup\{ \mu_n(A) : A \in \mathfrak{F} \cap E \}
$$
$$
\mu_n^{-}(E) = - \inf\{ \mu_n(A) : A \in \mathfrak{F} \cap E \}.
$$
Thus, it is clear that for each $A \in \mathfrak{F} \cap E$,
$$
\mu_n(A) \leq \mu_n^{+}(E) \quad \forall n \in \mathbb{N}.
$$
$$
-\mu_n(A) \leq \mu_n^{-}(E) \quad \forall n \in \mathbb{N}.
$$
Since $\mu_n \to \mu$, this implies that
$$
\mu(A) \leq \liminf_{n} \mu_n^{+}(E) \quad \forall A \in \mathfrak{F} \cap E
$$
$$
-\mu(A) \leq \liminf_{n} \mu_n^{-}(E) \quad \forall A \in \mathfrak{F} \cap E.
$$
That is, $\liminf_{n} \mu_n^{+}(E)$ is an upper bound for $\{ \mu(A) : A \in \mathfrak{F} \cap E \}$ and $\liminf_{n} \mu_n^{-}(E)$ is an upper bound for $\{ -\mu(A) : A \in \mathfrak{F} \cap E \}$. Then,
$$
\sup\{ \mu(A) : A \in \mathfrak{F} \cap E \} \leq \liminf_{n} \mu_n^{+}(E)
$$
$$
\sup\{ -\mu(A) : A \in \mathfrak{F} \cap E \} \leq \liminf_{n} \mu_n^{-}(E).
$$
Using $\sup\{ -\mu(A) : A \in \mathfrak{F} \cap E \} = -\inf\{ \mu(A) : A \in \mathfrak{F} \cap E \}$, this means that
$$
\mu^{+}(E) \leq \liminf_{n} \mu_n^{+}(E)
$$
$$
\mu^{-}(E) \leq \liminf_{n} \mu_n^{-}(E),
$$
and so $\mu^{\pm}(E) \leq \liminf_{n} \mu_{n}^{\pm}(E)$.

To show $|\mu|(E) \leq \liminf_{n} |\mu_{n}|(E)$ let $E \in \mathfrak{F}$. Hence, for each $n \in \mathbb{N}$,
$$
|\mu_n|(E) = \sup\bigg\{ \sum_{j=1}^{k} |\mu_n(E_j)| : E_1, \dots, E_k \text{ is a measurable partition of } E \bigg\}.
$$
Thus, let $E_1, \dots, E_k$ be a measurable partition of $E$. It is then clear that
$$
\sum_{j=1}^{k} |\mu_n(E_j)| \leq |\mu_n|(E) \quad \forall n \in \mathbb{N}.
$$
Since $\mu_n \to \mu$, we get $\lim_{n\to\infty} \sum_{j=1}^{k} |\mu_n(E_j)| = \sum_{j=1}^{k} |\mu(E_j)|$, and so
$$
\sum_{j=1}^{k} |\mu(E_j)| \leq \liminf_{n} |\mu_n|(E).
$$
Given that $E_1, \dots, E_k$ was an arbitrary measurable partition of $E$, it follows that $\liminf_{n} |\mu_n|(E)$ is an upper bound for $\bigg\{ \sum_{j=1}^{m} |\mu(E_j)| : E_1, \dots, E_m \text{ is a measurable partition of } E \bigg\}$, and so
$$
|\mu|(E) = \sup\bigg\{ \sum_{j=1}^{m} |\mu(E_j)| : E_1, \dots, E_m \text{ is a measurable partition of } E \bigg\} \leq \liminf_{n} |\mu_n|(E).
$$


问:
Finding $\lim_{x\to \infty} e^{\bigl( x+\frac{1}{x} \bigr)}- e^x$ by L'Hôpital's Rule
I am unable to evaluate this limit:
$$\lim_{x\to \infty} e^{\bigl( x+\frac{1}{x} \bigr)}- e^x$$
I tried to solve by L'Hôpital's rule, but I am ending nowhere. The  steps are obvious, so I am not re-writing them here. Graphically, this should go to infinity. I tried a few online calculators too, but those were without steps.  Please help out.

EDIT: My steps:

$\frac{e^{1/x}-1}{e^-x}$

$\frac{e^(x +1/x)}{x^2}$

$e^{\bigl( x+\frac{1}{x} \bigr)} \frac{x^2-1}{2x^3}$

I am unable to get rid of the term: $e^{x + 1/x}$

答:
HINT (for the use of Hôpital's rule)

$$e^{\bigl( x+\frac{1}{x} \bigr)}- e^x=e^x(e^{1\over x}-1)=\frac{e^{1\over x}-1}{e^{-x}}$$

EDIT

Compute $$\begin{aligned}\lim\limits_{x\to \infty}e^{\bigl( x+\frac{1}{x} \bigr)}- e^x&=\lim\limits_{x\to \infty}\frac{e^{1\over x}-1}{e^{-x}}=\\\\&=\lim\limits_{x\to \infty}\frac{e^x}{x^2}\cdot e^{1\over x}\end{aligned}$$
Now, use

$\lim\limits_{x\to \infty}e^{1\over x}=e^0=1$

$\lim\limits_{x\to \infty}\frac{e^x}{x^2}=\infty$ (repeated use of l'Hôpital's rule)


问:
Is there a nice definition of a homotopy type?
We often talk about specific mathematical objects, even though the way we formalise them in set theory actually means that we're talking about multiple objects. For example "the" complete graph on $n$-nodes is actually many graphs, one for each set of cardinality $n$. We can solve this problem by taking the proper class of all such graphs, but I find proper classes ontologically unconscionable. In general, in this vein, we could talk about the isomorphism classes in any category. We could also take a representative of each class, but this is "cheating", and probably difficult, if not generally impossible, at least in $\textrm{ZFC}$. I suspect that while keeping a set-theoretic foundation, then we have to make due with this multiplicity. However, graphs and most other objects we talk about are structured sets, and usually they do have the nice property that if they are isomorphic, then only the underlying sets differ, in a sense I make explicit in the appendix. On the other hand, homotopy types are not structured sets. A point and a line are radically different objects but they have the same homotopy type. And, due to my objections, we can't define a homotopy type as the equivalence class of all its spaces, since that's a proper class, and we can't just take a representative. So, is there a nicer way to define homotopy types?

Appendix: We take an underlying set $U$, and build another set out of it. A set can be thought of as a rooted tree with no infinite branches (and an asymmetry condition corresponding to extensionality), so we can think of building a new set as taking such a rooted tree $T$ (without the asymmetry condition), and attaching members of the underlying set to it. Explicitly, we take a surjective (partial) function $f:L\to U$ where $L$ is the set of leaves of $T$. We can then think of the set we've built as that obtained by taking copies of the trees representing the members of $U$ and attaching their roots to the corresponding members of $L$. Of course, this tree needs to satisfy the asymmetry condition to represent a set, although it doesn't really matter. An isomorphism between $f:L\to U$ with $L$ the leaves of $T$ and $f':L'\to U'$ with $L'$ the leaves of $T'$ is a rooted tree isomorphism $i:T\to T'$ and a bijection $b:U\to U'$ such that $bf=f'i$. In all examples I can think of, an isomorphism in a category of structured sets (e.g. graphs, topological spaces, groups) is an isomorphism in this sense, or at least there is a way of representing the objects of the category so that it is one.

答:
On the other hand, homotopy types are not structured sets. A point and a line are radically different objects but they have the same homotopy type.

There is a subtle point here that is worth addressing. It's true that the usual definition of a homotopy type does not produce a definition in terms of structured sets. However, that doesn't preclude the possibility that there is some other definition which does produce a structured set.

But this is known to be impossible in the following precise sense: Freyd showed in Homotopy is Not Concrete that there is no faithful functor from the homotopy category (in fact from any variant of the homotopy category which contains finite-dimensional CW complexes) to $\text{Set}$. In other words, there is no way to describe homotopy types as structured sets in such a way that homotopy classes of maps are exactly the structure-preserving functions.


问:
Complex Analysis - Derivation of Bode Integral
I am trying to work through the steps deriving the Bode Integral in the following paper:

https://www.sciencedirect.com/science/article/pii/S2405896315025045

I am stuck on page 261, left column, on the section for the Gamma_pk contour integral, specifically the following unjustified statement:

$$ S(s)=S'(s)(p_{k}-s)$$

S(s) is the sensitivity transfer function.  Pk is a pole of the open-loop transfer function, and therefore a zero of S(s), which the contour was chosen to go around.  Please see figure 3 in the paper for a visual depiction.

Anyhow, I think it's just the complex plane throwing me off here, but I am unable to arrive at, or justify, that statement.  I come close, but can't quite get all the way there.

Thanks for any insights!

答:
It's helpful to consider one of the examples provided in the paper, namely that of an open-loop unstable system with sensitivity function $S(s)=\dfrac{s^2-1}{s^2+s+9}$. Then $$\frac{S'(s)}{S(s)}=\frac{d}{ds}\ln S(s)=\frac{2s}{s^2-1}-\frac{2s+1}{s^2+s+9}=\frac{s^2+20s+1}{(s^2-1)(s^2+s+9)}$$

or $$S(s)=S'(s)\cdot \frac{(s^2-1)(s^2+s+9)}{s^2+20s+1}.$$

It should be evident that this doesn't agree with their stated formula. That said, the reason $S(s)$ is open-loop unstable is due to the zero at $s=1$, and with this in mind we write $S(s)=S'(s)(s-1)\cdot f(s)$ where $$f(s)=\frac{(s+1)(s^2+s+9)}{s^2+20s+1}$$ is analytic in the right half-plane. This form seems entirely sufficient for the purposes of their argument.

Moreover, if we look to Lewis's notes on control theory (which your paper cites!) then the above is a case of their first equation on page 366: Given a zero of $S_L(s)$ at $s=p_j$ with positive real part, one may write

$$\ln\frac{S_L(s)}{S_L(\infty)}=\ln(s-p_j)+\ln f_j(s)$$
with $f_j(s)$ being analytic on and within the contour which runs around $p_j$. So I would discount the paper's arguments for the Bode integral formula in favor of those in Lewis's notes.


问:
Prove that $\sqrt{1+\pi^2}, \hspace{2pt}\pi-\sqrt\pi,\hspace{2pt} \pi^2+\pi+\sqrt{1+2\pi}$ are not algebraic
Considering the fact that $\pi$ is not an algebraic number, I need to prove these expressions are not algebraic :

$$\sqrt{1+\pi^2},\hspace{10pt} \pi-\sqrt\pi,\hspace{10pt} \pi^2+\pi+\sqrt{1+2\pi}$$

Thanks

Some explanations will be appreciated ! Thank you all for your time .

答:
Set $c=\pi^2+\pi+\sqrt{1+2\pi}$. Then

$$
\sqrt{1+2\pi}=c-\pi^2-\pi
$$
and
$$
1+2\pi=(c-\pi^2-\pi)^2
$$
and, rearranging terms, you get a polynomial with coefficients in $\mathbb{Q}[c]$ that is satisfied by $\pi$. This means that $\pi$ is algebraic over $\mathbb{Q}[c]$. But you know that $\pi$ is transcendental, so also $c$ must be transcendental, otherwise you'd get that
$$
[\mathbb{Q}[\pi]:\mathbb{Q}]=
[\mathbb{Q}[\pi]:\mathbb{Q}[c]][\mathbb{Q}[c]:\mathbb{Q}]
$$
would be finite.


答: A somewhat indirect approach: since$$\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\lim_{y\to0^+}\frac{e^y-1}{y}=\lim_{y\to0^+}e^y=1$$by L'Hôpital's rule (or even just the definition of $\left.(e^z)^\prime\right|_{z=0}$),$$\lim_{x\to\infty}(e^{x+1/x}-e^x)=\lim_{x\to\infty}e^x=\infty.$$

Title: Finding $\lim_{x\to \infty} e^{\bigl( x+\frac{1}{x} \bigr)}- e^x$ by L'Hôpital's Rule

Body:
I am unable to evaluate this limit:
$$\lim_{x\to \infty} e^{\bigl( x+\frac{1}{x} \bigr)}- e^x$$
I tried to solve by L'Hôpital's rule, but I am ending nowhere. The  steps are obvious, so I am not re-writing them here. Graphically, this should go to infinity. I tried a few online calculators too, but those were without steps.  Please help out.

EDIT: My steps:

$\frac{e^{1/x}-1}{e^-x}$

$\frac{e^(x +1/x)}{x^2}$

$e^{\bigl( x+\frac{1}{x} \bigr)} \frac{x^2-1}{2x^3}$

I am unable to get rid of the term: $e^{x + 1/x}$


问:
Doubt on proof of theorem 14.22 of Rudin's Real and Complex Analysis
I don't understand what I've marked in yellow in the following theorem of Rudin's Real and Complex Analysis

I think I know why $\{f(z_n)\}$ has no limit point in $A_2$, because if there exists a subsequence $\{f(z_{n_k})\}$ convergent to some point $w \in A_2$, then as $f^{-1}$ is continuous, we have that
$$z_{n_k}=f^{-1}(f(z_{n_k})) \longrightarrow f^{-1}(w) \in A_1$$
but $|z_{n_k}|$ converges to $1$, so $|f^{-1}(w)|=1$ having a contradiction with $f^{-1}(w)$ being in $A_1$.

What I don't get is why $|f(z_n)|$ converges to $1$ and what does the assumption that $1< |z| < 1+ \varepsilon$ and the note that $f(z_n) \in V$ have to do with this convergence. Any ideas?

答:
I think what is confusing you is the line:

"In the latter case replace f by R_2/f"

You see only the former case is discussed, the latter case is left to the reader to use the same reasoning after a transformation. So everything that follows assumes the former case.

The former case doesn't let V anywhere near the outer boundary, that is what V subset of A(1,r) means. r is strictly smaller than that outer boundary.

f(z_n) has no limit points in A_2 implies that |f(z_n)| has no limit points in (1,R_2), but also |f(z_n)| is bounded within [1,r] and so must have a limit point by compactness. [1,r] take away (1,R_2) is the singleton {1}. |f(z_n)| is bounded in the real numbers has one and only one limit point which is 1 so it converges to 1. The proof of that may take a lemma if you haven't seen it before.


问:
Construct a "natural" bijection between homomorphisms with domain $G$ and homomorphisms with a factor group $G/H$ as domain.
Assume $G$ is a group with a normal subgroup $H$. Let $K$ be another group.
Consider the sets $$X = \{ \varphi: G/H \rightarrow K \mid \varphi~\text{ is a homomorphism.} \} $$
and
$$Y = \{ \varphi: G \rightarrow K \mid \varphi~\text{ is a homomorphism and takes value}~e_K~\text{on}~H.  \}.$$
We want to show that there is a "natural" bijection between the two sets.

To construct a mapping from $X$ to $Y$, we can use the canonical projection $\pi : G \rightarrow G/H$
with $g \mapsto gH$.
This yields $X \ni \varphi \mapsto \varphi \circ \pi \in Y$.

How do I construct the inverse of this map?

Context: This is a claim made during a lecture after discussing isomorphism theorems. Constructing the inverse should require something similar to what one does in proving the first isomorphism theorem.

答: The first isomorphism theorem tells you that for any group homomorphism $\varphi : G \to K$ with $H \subseteq \ker \varphi$ there exists a unique group homomorphism $\tilde{\varphi} : G/H \to K$ such that $\tilde{\varphi} \circ \pi = \varphi$, where $\pi : G \to G/H$ is the canonical projection, namely, $\tilde{\varphi}(gH) = \varphi(g)$ for every $g \in G$ (check that this is well-defined!). Thus, the map $$X \to Y; \quad\phi \mapsto \phi \circ \pi$$ is a bijection (since for every element of $Y$ there is a unique element of $X$ which is in its preimage).


问:
Can any continuous function be represented as an infinite polynomial?
Can any continuous function be represented as an infinite polynomial?

Motivation: the antiderivative

$
\int^\ e^{-x^2}dx\
$

can be expressed as an infinite polynomial(write Taylor series for integrand function and integrate) but this antiderivative has no closed/elementary form expression according to Liouville's theorem but is clearly continuous. So are the rest of the non-elementary functions expressible as infinite polynomials? Fascinating.Any insights on how to proceed????

答:
Very late reply but might be somehow helpful for future understanding. As pointed out in the comments by @Tyler, the answer of your question would be it depends on your settings.

In the theory of functional analysis, the set of polynomials are dense in the vector space of continuous functions with the infinity norm. This follows directly from the Stone Weierstrass theorem. Thus, if you consider any compact interval (assumed it to be path connected) you can conclude there exists such an approximation. However, in my knowledge, the method to compute this might not be unique.

This idea can be extended to general functions taking functions in compact sets.
Check wikipedia as pointed by the user in the comments


问:
Inequality for determinant of sum of two orthogonal matrices
My goal is to show that :

$$\forall A,B\in O_n(\mathbb{R}), |{\rm det}(A+B)| \le 2^n.$$

How can I access the determinant of the sum of two matrices ? I don’t know many ways to establish inequalities in Algebra, except with Bessel’s inequality and Cauchy Schwarz inequality, but they don’t seem to be very helpful here.

答:
another proof, no eigenvalues required:

since this is over $\mathbb R$, instead of working with absolute values, work with squares

i.e. prove  ${\rm det}(A+B)^2\leq 2^{2n}$

special case: consider if both matrices are the identity, then

$\big \vert{\rm det}(I+I)|^2 = \det\Big(\big (I+I)^T\big (I+I)\Big) = \det\big(I+2I+I\big)=\prod_{k=1}^n (1+1+1+1) = 2^{2n}$

in general:

$\det\big((A+B)^2\big) $

$= \det\Big(\big (A+B)^T\big (A+B)\Big) $

$= \det\big(I+A^TB + B^TA+I\big)$

$\leq \prod_{k=1}^n (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)$

$=\prod_{k=1}^n \big \vert (1+(A^TB)_{k,k}+(B^TA)_{k,k}+1)\big \vert$

$\leq \prod_{k=1}^n \Big(1+\big \vert (A^TB)_{k,k}\big \vert+\big \vert(B^T A)_{k,k}\big \vert+1\Big)$

$\leq \prod_{k=1}^n \Big(1+1+1+1\Big)$

$= 2^{2n}$

justification:

1.) Hadarmard Determinant Inequality

2.) any matrix of the form $C^T C$ is PSD so it has real non-negative diagonals

3.) Triangle inequality (and multiplying over a point-wise bound)

4.) $(A^TB)$ and $(B^TA)$ are real orthogonal matrices so each component has modulus $\leq 1$


答:
let g(x)=f(x)-x, because f(x)=x has no real roots, then g(x)>0 or g(x)<0 for all x belongs to R.

suppose g(x)>0 for all x. then g(f(x))>0 for all x.

f(f(x))-x=f(f(x))-f(x)+f(x)-x=g(f(x))+g(x)>0 for all x

then f(f(x))-x>0 for all x, so f(f(x))=x also has no real roots.

the same proof for g(x)<0 for all x.


问:
Determinant of the matrix of an orthogonal operator.

Let $V$ be a real vector space with inner product and $\dim(V)=n$. Let $T$ be an orthogonal operator on $V$. If $A$ is the matrix of $T$ relative to the basis $\beta=\{ \beta_1,\ldots, \beta_n\}$ of $V$, prove that $\det(A)=\pm 1$

I think that if we prove that $A$ is an orthogonal matrix then the answer is clearly but $\beta$ is not an orthogonal basis of V so I don't know how can I prove that $\det(A)$ is $\pm 1$, REMEMBER THAT $\beta$ is any basis of $V$ and is not orthonormal.

答: Since $T$ is an orthogonal operator, the determinant of its matrix with respect to an orthonormal basis is $\pm1$. But the determinant is independent of the choice of the basis, and so…


答:
The point is that if you have some $\varphi: G \rightarrow K$ with kernel containing $H$ then there exists some unique $\varphi' : G/H\rightarrow K$ satisfying
$$\varphi'(g\cdot H)=\varphi(g)$$
where $g\cdot H$ is a coset in $G/H$. You can check this in the usual way one checks such definitions - first that it is well defined, since if $g\cdot H=g'\cdot H$ then $\varphi(g)=\varphi(g')$, and then that this is a homomorphism (which is not too hard once we know it's well-defined).

This association between functions $\varphi:G\rightarrow K$ with kernel containing $H$ and maps $\varphi':G/H\rightarrow K$ is the map $Y$ to $X$ that you're looking for.  You can check that it is inverse to the map you already have.

Title: Construct a "natural" bijection between homomorphisms with domain $G$ and homomorphisms with a factor group $G/H$ as domain.

Body:
Assume $G$ is a group with a normal subgroup $H$. Let $K$ be another group.
Consider the sets $$X = \{ \varphi: G/H \rightarrow K \mid \varphi~\text{ is a homomorphism.} \} $$
and
$$Y = \{ \varphi: G \rightarrow K \mid \varphi~\text{ is a homomorphism and takes value}~e_K~\text{on}~H.  \}.$$
We want to show that there is a "natural" bijection between the two sets.

To construct a mapping from $X$ to $Y$, we can use the canonical projection $\pi : G \rightarrow G/H$
with $g \mapsto gH$.
This yields $X \ni \varphi \mapsto \varphi \circ \pi \in Y$.

How do I construct the inverse of this map?

Context: This is a claim made during a lecture after discussing isomorphism theorems. Constructing the inverse should require something similar to what one does in proving the first isomorphism theorem.


问:
$\rho(A \circ A) \ge \rho(A)^2.$
Suppose we have a nonnegative matrix $A$. Can we claim the following:
$$\rho(A \circ A) \ge \rho(A)^2.$$ Here $A \circ A$ is the Hadamard product.

Context: For a nonnegative matrix $A$ I know the spectral radius. Can we use that to give a bound for the spectral radius of $A \circ A$.

I computationally verified for a few examples. It seems to be true.

答: The reversed inequality is true. Note that $A\circ A$ is a principal submatrix of $A\otimes A$. If we embed $A\circ A$ inside a zero matrix of the same size as $A\otimes A$, we can view $A\circ A$ as a nonnegative matrix $B$ that is entrywise bounded above by $A\otimes A$. It follows that $B^k\le(A\otimes A)^k$ and hence $\|B^k\|^{1/k}\le\|(A\otimes A)^k\|^{1/k}$ using Frobenius norm or the induced maximum norm. Therefore, by Gelfand's formula, we always have $\rho(A\circ A)=\rho(B)\color{red}{\le}\rho(A\otimes A)=\rho(A)^2$.

Title: $\rho(A \circ A) \ge \rho(A)^2.$


问:
Relationship between Gram determinant and cross product
I am trying to prove that the determinant of the Gram matrix is equal to the norm of the cross product squared for $v_1, v_2 \in \mathbb{R}^3$:

$$\|v_1 \times v_2\|^2 = \det\Bigl(\begin{smallmatrix}
\langle v_1, v_1 \rangle & \langle v_1, v_2 \rangle\\
\langle v_2, v_1 \rangle & \langle v_2, v_2 \rangle
\end{smallmatrix}\Bigr)$$

But the only idea I have seems very technical and unpleasant, any suggestions?

I have already seen this thread but I don't find it helpful.

答:
Here's a straightforward argument without horrendous algebra. Assume $v_1,v_2$ are linearly independent; I leave the degenerate case to you. Let $n=\dfrac{v_1\times v_2}{\|v_1\times v_2\|}$. Consider the matrix
$$A = \left[\begin{array}{ccc} | & | & | \\ v_1 & v_2 & n \\ | & | & |\end{array}\right].$$
Then your Gram determinant is the determinant of $A^\top A$. On the other hand, $\det (A^\top A) = \det (A^\top)\det(A) = \det(A)^2$. But
$$\det A = n\cdot (v_1\times v_2) = \|v_1\times v_2\|.$$


问:
A certain passage in Carathéodory's book concerning analytic continuation
The fragment below is from the book 'Theory of functions of a complex variable', by Carathéodory.

The situation is this. We have an analytic function $f(z)$ in some region $G_z\subset\mathbb{C}$. Then $G_w=f(G_z)$ is a region in the $w$-plane. We assume that $G_w^*$ is a simply-connected subregion of $G_w$ that does not contain points $w=f(z)$ such that $f'(z)=0$. We know,
by the inverse function theorem, that $f$ has a local inverse around such points. But then
Carathéodory claims (highlighted below) that we can start at any such point $w_0$ and continue analytically such a local inverse along any "polygonal train" -- (i.e any polygonal path) starting from $w_0=f(z_0)$ (as long as we stay inside $G_w^*$). He then deduces -- by the monodromy theorem -- that $f$ has a single-valued analytic inverse defined on the whole of $G_w^*$.

Carathédory's argument says this:

If $f$ is an analytic function defined on a region $\Omega$ such that $f'(z)\neq 0$ for every $z\in\Omega$, then $f$ has a single-valued analytic inverse defined in any simply connected subregion of $f(\Omega)$.

As far as I know, there are examples of functions defined in simply connected regions that cannot be analytically continued.

So my question is this:

-- Is the statement above true? How does Carathédory know that we can extend a local inverse analytically along every polygonal path that lies in $G_w^*$?

Added
In order to clarify the question, here is the final comment Carathéodory
makes:

答:
This is not true without additional hypotesis. If it were true,
it would imply Caratheodory's claim about the existence of a single valued inverse, which is false in general:

Let $f(z)=z^2(z-1)$ on $\mathbb{C}$. This map is surjective, and it has only two singular points: $0$ and $\frac23$. However, if we remove this values from the domain, we get a surjective map from $\mathbb{C}-\{0,\frac23\}$ to $\mathbb{C}$ which is non injective. Using the
fact that $\mathbb{C}-\{0,\frac23\}$ has $\mathbb{D}$ as its universal cover, we get a map
$\varphi:\mathbb{D}\twoheadrightarrow  \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and not injective: thus $\varphi$ does not admit a single valued inverse on $G_w^*=Gw=\mathbb{C}$ as such a function should be entire and bounded and thus by Liouville's theorem, constant.

One can also construct $\varphi:\mathbb{C}\twoheadrightarrow \mathbb{C}$ such that $\varphi'\neq 0$, $\varphi$ is surjective and non linear (and thus non injective, as the only injective entire functions are linear): see this mathoverflow post for such an example (the error function is an explicit example).

$\varphi$ does not admit a single valued inverse on $G_w^*=G'w=\mathbb{C}$, as such a function should be entire and it should avoid more than one point (since for almost every $w, \varphi^{-1}(w)$ is infinite)
and thus by Picard's little theorem it should be constant.

If, however, we let $f$ satisfy a few more assumptions, the result follows. For example, if $f:G_z\to G_w^*$ is proper, it is a well known result that $f$ is actually a covering map. As $G_w^*$ is simply connected, it's its own universal cover, and thus the holomorphic lifting lemma implies the existence of an inverse and thus in particular the possibility of analytic continuation along polygonal paths.


问:
Probability to be infected in series of meetings
I have difficulty interpreting a simple probabilistic calculation and would be happy to help. We are in the midst of the Covid-19 epidemic and I am trying to assess someone's probability to be infected in a time window. For simplicity, I guess the probability of getting infected during a single session depends on its duration alone.

In math terms:

The probability to be infected in a given meeting is $P(t) = t\cdot\Theta$, where $t$ is meeting duration and $\Theta$ is only a calibration parameter of the virus (and keep the probability under 1).

Hence, the probability to be infected in a series of meetings is the complementary event to be not-infected in any meeting:
$P(t_1...t_N) =  1-\prod_{k=1}^N(1-t_k\cdot\Theta)\tag{*}$

Does this inference sound reasonable? It seems to me that this is the material of a basic course in probability.

Surprisingly and incomprehensibly to me, I saw in an article a calculation that actually reaches the following result for the same question (and the same assumptions):

$P(t_1...t_N) = 1- \exp(-\Theta\sum_{k=1}^{N}t_{k})\tag{**}$

I do not understand what the relationship is between $(*)$ and $(**)$ and how in the calculation of a complementary event they arrived at the equation $(**)$?

Thanks a lot!

答:
It is not actually true that the probability of getting infected in a meeting is a linear function of the length $t$ of the meeting; this is an approximation only valid when the product of the time and the infection probability is small. The simplest update to this linear model is the following: think about the problem in small time steps of length $\frac{1}{n}$, and suppose that in a small time step you have a small probability $\frac{\Theta}{n}$ of being infected, and that these infections are independent. Then the probability that you are infected after the entire meeting of length $t$ is $1$ minus the probability that you are not infected, which is

$$1 - \left( 1 - \frac{\Theta}{n} \right)^{nt}$$

and for large $n$ this approaches

$$1 - e^{- \Theta t}.$$

This says that the time to infection follows an exponential distribution. The same argument applies to a sequence of meetings (after all, a sequence of meetings is just one long meeting) and gives $1 - e^{\Theta \sum t_i}$. Really this should be $1 - e^{\sum \Theta_i t_i}$ as the risk-per-unit-time parameter $\Theta_i$ could differ dramatically between meetings also. I have not checked if this is what the article you link is doing.

This is only the next simplest model after the linear model (which it reduces to if $\Theta t$ is small), and very importantly it does not account for the fact that the risk-per-unit-time increases over time if people are indoors and expelling droplets into the air and the room is poorly ventilated. I think the risk-per-unit-time after an hour or two may actually be much higher than the risk-per-unit-time at the beginning of a meeting, but don't quote me on that. To account for this we should be computing something more like

$$1 - \exp \left( \int_0^{\sum t_i} \Theta(t) \, dt \right)$$

where the risk $\Theta(t)$ now varies over time; above we considered the special case that $\Theta(t)$ is constant or piecewise constant. But we would need more modeling assumptions to figure out a good choice of shape for $\Theta(t)$, then more data to estimate it.

For practical risk assessments I recommend the microCOVID project, which some of my friends were involved with, and which has been updated to take B117 into account.


问:
Doubt in proof while showing that injective continous map sends open sets to open sets.
I was trying to prove the following statement

If $U \subset \mathbb{R}^n $ and suppose $f : U \to \mathbb{R}^n$ be a continous map which is injective. Then $f(U)$ is open in $\mathbb{R}^n$.

I considered an closed $\epsilon$-ball around a point $u \in U$, that is $$D_n=\{x \in \mathbb{R}^n:||x-u|| \leq \epsilon \} \subset U$$. Then it is clear that boundary of $D_n$, say $B=\{x \in \mathbb{R}^n:||x-u|| = \epsilon \}$ is homeomorphic to $S^{n-1} \subset \mathbb{R}^n$.

What I cannot understand is the fact that  $f(B)$ too is isomorphic to $S^{n-1}$. Remember that $f$ is given to be injective and continous, but how does that help?

答: There is a homeomorphism $\varphi:S^{n-1}\to B$, and a bijective continuous map $f:B\to f(B)$. So the composition $f\circ\varphi:S^{n-1}\to f(B)$ is continuous and bijective. Since $S^{n-1}$ is compact and $f(B)$ is Hausdorff it follows that this composition is a homeomorphism.


问:
Graded commutator defines graded derivation?
${}$

I'm having difficulties understanding what to do on a problem for my exercise class. The problem is as follows:

${}$

Consider a $\mathbb{Z}_2$ graded algebra $\mathcal{A} = \mathcal{A}_0 \oplus \mathcal{A}_1$ together with its graded derivations
$$ \delta \in \text{Der}_{\pm} (\mathcal{A}) : \; \mathcal{A}_i \longrightarrow \mathcal{A}_{i+\deg(\delta)} \quad ,$$
$$\delta(ab) \enspace = \enspace \delta(a)b + (-1)^{\deg(a)\deg(\delta)}a\delta(b) \quad , \qquad  (\ast)$$
where
$$\deg(a) \enspace = \enspace \begin{cases} 0 \; , \quad a \in \mathcal{A}_0 \\ 1 \; , \quad a \in \mathcal{A}_1 \end{cases}$$
and $\deg(\delta) = 0$ for derivations and $\deg(\delta) = 1$ for antiderivations.

${}$

Show that the graded commutator
$$\big[ \delta_1, \delta_2 \big]_{\pm} \enspace = \enspace \delta_1 \delta_2 - (-1)^{\deg(\delta_1)\deg(\delta_2)} \delta_1 \delta_2$$
defines again a graded derivation (with $\deg\big( [\delta_1, \delta_2]\big) = \deg(\delta_1) + \deg(\delta_2) \; \text{mod} \; 2$).

${}$

I don't quite get the following: $\;$ My commutator $f(x,y) \equiv [x,y]$ needs two arguments. To show that a map is a graded derivation, I need to show that my map fullfills equation $(\ast)$. However, this equations is defined for a map having only one argument. What do I misunderstand? What do I need to do?

${}$

答: You are required to prove that the commutator of two derivations is again a derivation, no that the commutator (seen itself as a antisymmetric bilinear map) is a derivation.


问:
Intersection of intervals
Let $a\in\mathbb{N}_{\geq3}$.

How can one prove that $$\bigcap_{i = 1}^{a} \bigcup_{j = 0}^{i-1} \left[\frac{1+aj}{i},\frac{a(j+1)-1}{i}\right] = \varnothing,$$
where $\varnothing$ is the empty set?

Example for $a=5$:

Red is the interval, Yellow are the gaps between the intervals that cause the intersection to be a null set, white gaps do not effect the intersection.

答:
First, a lemma. Let $a\ge 1$ be an integer and let $x\in[0,1]$. Then there exist integers $m$ and $n$ such that $1\le n\le a$ and $|n x - m| < \frac{1}{a}$. To prove this, let $z = \exp(2\pi i x)$ and consider the numbers $1, z,z^2,\ldots,z^a$ in the complex plane. Let $\phi_k = \arg(z^k)$ with $0\le \phi_k<2\pi$. Reorder $\phi_0,\ldots,\phi_a$ so that $\phi_{\pi(0)}\le \phi_{\pi(1)}\le\ldots\le\phi_{\pi(a)}$ for an appropriate permutation $\pi$ of $0,\ldots,a$. Observe

$\ \ \sum_{k=1}^a (\phi_{\pi(k)} - \phi_{\pi(k-1)}) < 2\pi$.

Since there are $a$ non-negative terms on the left hand side, at least one of these terms is bounded above by $\frac{2\pi}{a}$. Thus,

(*) $ \ \ 0 \le \arg(z^{\pi(k)}) - \arg(z^{\pi(k-1)}) = \arg(z^{{\pi(k)}-{\pi(k-1)}}) < \frac{2\pi}{a}$

for some $k\in\{1,\ldots,a\}$. Let $n = |\pi(k)-{\pi(k-1)}|$. Note $1\le n\le a$. By definition of $z$ and (*) above, it follows $|n x - m| < \frac{1}{a}$ for some integer $m$.

Now the main theorem is easy. Let

$ S = \cap_{i=1}^a \cup_{j=0}^{i-1} \left[\frac{1+aj}{i},\frac{a(j+1) -1}{i}\right].$

Suppose $\beta \in S$. With $i=1,j=0$, we see $1\le \beta \le a-1$. Let $x=\beta/a$ and apply the lemma above. Then there are integers $m,n$ with $1\le n \le a$ such that

(**) $ \left| n \frac{\beta}{a} - m \right| < \frac{1}{a}$.

Thus, $n\frac{\beta}{a}$ is within $\frac{1}{a}$ of some integer.

Now focus on the union term with $i=n$ in the definition of $S$:

$T = \cup_{j=0}^{n-1} \left[\frac{1+aj}{n},\frac{a(j+1) -1}{n}\right]$.

Observe that
$T \subset \{y\in\mathbb{R}: \left|n \frac{y}{a}\right| \ge \frac{1}{a}\}$.
In other words, for every member $y$ of $T$, $n \frac{y}{a}$ is at least $\frac{1}{a}$ from the closest integer. Recalling (**), we see
$\beta \notin T$. By definition of $S$ and $T$, $S\subset T$, so $\beta\notin S$.

Title: Intersection of intervals


问:
Integral of a $1$-form on $S^1$ with respect to a parametrization (Guillemin and Pollack, 4.4.6)
I have a question on problem 4.4.6 in Guillemin and Pollack's Differential Topology. Alright, so we've been given a parameterization of $S^1$ $$h: \mathbb{R} \to S^1,\ t\mapsto (\cos t,\sin t)$$
and we want to show that for any $1$-form $\omega$, we have $$\int_{S^1}\omega = \int_0^{2\pi}h^*\omega$$
Discussing with my peers, we have the following two strategies:

Let $U_N$ and $U_S$ be the charts on $S^1$ obtained by excluding the north and south pole, let $h$ induces a diffeomorphism of $(0,2\pi)$ on $U_N$ and $(-\pi,\pi)$ on $U_S$. So
$$\int_{U_N}\omega = \int_0^{2\pi}h^*\omega,\qquad \int_{U_S}\omega = \int_{-\pi}^{\pi}h^*\omega = \int_0^{2\pi}h^*\omega$$
the latter inequality coming from the relevant substitution of the integral over $\mathbb{R}$. Is this enough to deduce the answer, if it is, I don't see why.

Since $h$ is a subjective quotient map and periodic, so it factors through a diffeomorphism of $S^1$, say $\tilde{h}$ i.e. $h = \tilde{h}\circ \pi$, where $\pi:\mathbb{R} \to \mathbb{R}/2\pi\mathbb{Z}$ is the quotient map. So we have $$\int_{S^1}\omega = \int_{q(0)}^{q(2\pi)}\tilde{h}^*\omega$$
Can we say
$$\int_{q(0)}^{q(2\pi)}\tilde{h}^*\omega = \int_{0}^{2\pi}q^*\tilde{h}^*\omega$$

I suppose a third strategy would be verify in local coordinates but we're a bit confused on that as well. Any remarks will be appreciated!

答:
I think the following Proposition (16.8) from John Lee's Introduction to Smooth Manifolds will prove helpful:

Let $M$ be an oriented smooth $n$-manifold w/ or w/o boundary, and let $\omega$ be a compactly supported $n$-form on $M$. Suppose $D_1,\ldots,D_k$ are open domains of integration in $\mathbb{R}^n$ (bounded subset with boundary measure zero), and for $i=1,\ldots,k$ we are given smooth maps $F_i:\overline{D_i}\rightarrow M$ satisfying

$F_i$ restricts to an orientation-preserving diffeomorphism from $D_i$ onto an open subset $W_i\subset M$

$W_i\cap W_j=\emptyset$ when $i\neq j$

$\operatorname{supp}\omega\subset\overline{W_1}\cup\dots\cup\overline{W_k}$

Then
\begin{equation}
\int_M\omega=\sum_{i=1}^k\int_{D_i}F_i^*\omega.
\end{equation}

At first glance, assuming the orientations are the standard ones, the $F$'s and $D'$s we pick should be quite simple.