问:
Produce infinite distinct limit sets from an original set on the real line
Limit set Suppose that $P\subseteq \mathbb{R}$, The limit set of $P$
$$P':=\{\mathrm{all\ the\ limit\ points\ of\ }P\}.$$
Now we ask for a subset $A\subseteq\mathbb{R}$ such that
$$A^{(1)}=A',A^{(2)}=(A^{(1)})',A^{(3)}=(A^{(2)})',\cdots,A^{(n+1)}=(A^{(n)})',\cdots$$
are distinct$(n=1,2,\cdots)$.

The idea occured to me was that the limit set of $\{\frac{1}{m}+\frac{1}{n}\}(m,n\in \mathbb{N}^+)$ is $\{\frac{1}{n}\}(n\in\mathbb{N}^+)$, so I tried $A=\{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}|a_i\in\mathbb{N}^+,n\in\mathbb{N}^+\}$, only to find that $A'=[0,+\infty)$, whose limit set is the same as itself.

This is just so tricky to find such a set which is able to carry out the limit-set operation infinitely without repetitiveness. I wonder if anyone can help me on it. Thank you in advance! Any idea would be highly appreciated!

答:
Henno Brandsma has suggested one approach. Here is another; I’ll leave it to you to verify the details.

Let $A_0=\{0\}\cup\left\{2^{-k}:k\in\Bbb Z^+\right\}$; clearly $A_0^{(1)}=\{0\}$, and $A_0^{(2)}=\varnothing$.

If $X\subseteq\Bbb R$ and $a,b\in\Bbb R$, let $a+bX=\{a+bx:x\in X\}$.

Let $A_1=\{1\}\cup\bigcup_{k\in\Bbb Z^+}\left(1+2^{-k}+2^{-k}A_0\right)$. If you draw a sketch, you’ll see that $A_1$ squeezes a copy of $A_0$ into each of the intervals

$$\left[1+2^{-k},1+2^{-k}+2^{-(k+1)}\right]$$

for $k\in\Bbb Z^+$. And you can show further that $A_1^{(1)}=\{1\}\cup\left\{1+2^{-k}:k\in\Bbb Z^+\right\}$, $A_1^{(2)}=\{1\}$, and $A_1^{(3)}=\varnothing$.

Continue in this fashion, recursively constructing sets $A_n$ for $n\in\Bbb N$ by letting

$$A_{n+1}=\{n+1\}\cup\bigcup_{k\in\Bbb Z^+}\left(n+1+2^{-k}+2^{-k}A_n\right)$$

for each $n\in\Bbb N$; $A_{n+1}$ squeezes a copy of $A_n$ into each of the intervals

$$\left[n+1+2^{-k},n+1+2^{-k}+2^{-(k+1)}\right]$$

for $k\in\Bbb Z^+$. The construction ensures that $A_n^{(n+1)}=\{n\}$ and $A_n^{(n+2)}=\varnothing$ for each $n\in\Bbb N$. Thus, the set $A=\bigcup_{n\in\Bbb N}A_n$ has the desired property.


问:
Clique cover problem with general clique weights
I consider a graph $G(V,E)$ and each clique has a general weight. The problem is to find a clique cover that maximizes the sum of the weights of the cliques. That is, I want to select a set of cliques such that all nodes are covered in exactly one clique while the sum of the clique weights is maximized. This can be solved by an integer linear program.

My question is how to show that this problem is NP-hard (or not). I suspect that this problem is NP-hard since the minimum clique cover problem is NP-hard.
https://en.wikipedia.org/wiki/Clique_cover

Note: I have limited knowledge of graph theory and computational complexity.

答:
It is NP-hard. Lets reduce independent set with bounded degree to this problem.

We are given graph $G$ with $n$ vertices and want to know if it has an independent set of size $k$.

For each edge $u-v$ add new vertex $x_{uv}$ and replace this edge with two: $u-x_{uv}$ and $v - x_{uv}$. For each vertex $u$ and it's neighbors $v$ and $v'$ add edge $x_{uv} - x_{uv'}$. This will be our new graph.

Inclusion-maximal cliques in our new graph are of form $\{u\} \cup \{x_{uv} | v \in N(u)\}$. As $G$ has bounded degree, number of cliques will be polynomial (really linear) in $n$.

Lets say that weight of such inclusion-maximal clique is $1$, and weight of any other clique is $0$. Note that for any two vertices $u$, $v$ s.t. there was edge $u-v$ in $G$, in any clique-cover at most one of $u$, $v$ is covered with inclusion-maximal clique. At the other hand, if we have some independent set, we can cover vertices from it with inclusion-maximal cliques (and cover rest however we want, with singletons, for example).

Thus independent sets of size $m$ corresponds to clique-covers with weight $m$, so $G$ has independent size of size $k$ iff our new graph has clique-cover of weight at least $k$.

ContentLicense: CC BY-SA 4.0


问:
Is $\{0\}$ a *prime* ideal in $(\mathbb{Z}_2,\oplus,\odot)$?
In my textbook (Abstract Algebra by Saracino), it gives $(\mathbb{Z}_2,\oplus,\odot)$ as an example of a commutative ring that has a maximal ideal that is not prime. It claims that the trivial ideal $\{0\}$ (which is maximal because it's the only ideal) is not prime. But the only way to express $0$ as the product of elements in $\mathbb{Z}_2$ is $1 \cdot 0 = 0 \cdot 1 = 0 \cdot 0$. In each case, $0$ appears as a factor, and $0 \in \{0\}$. So wouldn't this imply that $\{0\}$ is a prime ideal of $(\mathbb{Z}_2,\oplus,\odot)$?  I just want to make sure I'm not crazy here...

This is the textbook's definition of prime:

Let $R$ be a ring, $I$ an ideal in $R$. Then $I$ is prime if whenever $a,b \in R$ and $ab \in I$, then at least one of $a$ or $b$ is in $I$.

答:
Usually, we consider $\{0\}$ to be a prime ideal in any (commutative, unital) ring without zero divisors. The alternative is just cumbersome.

A ring is an integral domain iff $\{0\}$ is prime. $I\subseteq R$ is prime iff $R/I$ is an integral domain. Formulating these and similar statements while trying to take into account that $\{0\}$ can't be prime just isn't worth the hassle.

This also makes every maximal ideal into a prime ideal.


问:
Probability of Sum being $4$ with unfair dice
You roll two six-sided dice.  Die $1$ is fair.  Die $2$ is unfair such that the probability of rolling an odd number is $2/3$ and the probability of rolling an even number is $1/3$, though the probability rolling of each odd number is the same, and the probability of rolling each even number is the same.  What is the probability of the sum of both dice adding up to $4$?

My answer:

I wrote out the combinations to sum $4$ between both dies:
(F1, U3), (F2, U2), (F3, U1) [F = Fair die, U = Unfair die].

My thinking is we need to find the OR probability of these pairs:
P((F1 and U3) or (F2 and U2) or (F3 and U1)):
$$(\frac16 \times \frac23) + (\frac16 \times \frac12) + (\frac16 \times \frac23) = 0.30555555555.$$

Is this correct?

答:
The computations of @user2661923 and @mathlover are confirmed (+1 each) by a simulation in R. With a million rolls of the two dice, one can expect two place accuracy.

```<code>set.seed(313)
d1 = sample(1:6, 10^6, rep=T)
d2 = sample(1:6, 10^6, rep=T, p = c(2,1,2,1,2,1))
s = d1+d2
mean(s == 4)
[1] 0.092494     # aprx P(S = 4)
5/54
[1] 0.09259259   # exact

cutp = (1:12)+.5
hist(s, prob=T, br=cutp, col="skyblue2",
     main="Two Dice: One Fair")
</code>```

Body:
The computations of @user2661923 and @mathlover are confirmed (+1 each) by a simulation in R. With a million rolls of the two dice, one can expect two place accuracy.

set.seed(313)
d1 = sample(1:6, 10^6, rep=T)
d2 = sample(1:6, 10^6, rep=T, p = c(2,1,2,1,2,1))
s = d1+d2
mean(s == 4)
[1] 0.092494     # aprx P(S = 4)
5/54
[1] 0.09259259   # exact

cutp = (1:12)+.5
hist(s, prob=T, br=cutp, col="skyblue2",
     main="Two Dice: One Fair")


问:
Difference Quotient with $x-a$

I'm having trouble understanding how to do the step highlighted in red if anyone could help

答: OK, that one is: $\dfrac{-2(x-a) - 1(x-a)(x+a)}{x-a}= \dfrac{-2(x-a)}{x-a} - \dfrac{1(x-a)(x+a)}{x-a} = -2 -1(x+a)= -2 -(x +a)=-2-x-a$ where we used the property fraction subtraction with the common denominator and the multiplicative identity property of the reals, namely:  $\dfrac{x-y}{z} = \dfrac{x}{z}-\dfrac{y}{z}$, and $1\cdot x = x$.

Title: Difference Quotient with $x-a$


问:
Karnaugh tables ; number of groups
Hey guys i was just wondering why is the number of all groups in a Karnaugh table equal to $3^n$ ; for example :

In a $2×2$ table with $2$ variables we have $4$ groups of one and $4$ groups of two and $1$ group of four which is equal to $9$.

Or in a $2×4$ table with $3$ variables we have $8$ groups of one and $12$ groups of two and $6$ groups of four and $1$ group of eight which is equal to $27$.

答:
For readers not familiar with these Karnaugh tables, see here.

First of all, congratulations for this non-evident observation.

There is simple proof based on a certain coding. I understand what you mean by a group (which is the advantage of thinking geometrically with Karnaugh tables), but I will switch to the more tractable equivalent boolean algebra expressions.

Let us understand the situation in the case of 3 variables (corresponding to your $2 \times 4$ table).

In this case, the monomials are :

$$A,B,C,\bar{A},\bar{B},\bar{C},A\bar{B},\bar{A}B,..., ABC,A\bar{B}C,....$$

(please note that variables are listed in the alphabetic order: $A$, then $B$ then $C$)

Let us consider any boolean monomial in the above list.

We are going to code it by a list of three numbers $-1,0,1$ in the following way.

When a variable:

is present in it (without a bar on it), code it +1,

is present in it as negated (i.e., with a bar), code it -1,

is absent, code it $0$.

In this way, to any monomial having any number of variables, we can associate a specific code with three numbers.

Examples: $A\bar{B}C$ is coded $(1,-1,1)$, $A\bar{C}$ is coded $(1,0,-1)$, $C$ is coded $(0,0,1)$ etc...

resulting in a $3^3$ count, and more generaly a $3^n$ count if there are $n$ variables.

Remarks:

In fact, we implicitly take into account the "void" monomial. Otherwise the count would be $3^n-1$...

One could have given a (more complicated) proof by distinguishing the different cases (as you did) and using relationship $\displaystyle \sum_{i=0}^n {n\choose i} 2^i = (1+2)^n=3^n$.

I discovered almost the same question here.

Title: Karnaugh tables ; number of groups


答:
This is the distributive property xy+xz=x(y+z)

was what I was looking for, thank you.

Title: Difference Quotient with $x-a$

Body: I'm having trouble understanding how to do the step highlighted in red if anyone could help


问:
How many squarings are needed to exceed 2?
Given the base 2 representation of $x\in\mathbb Q$ ($1<x<2$ and has a finite number of digits in base 2), find a number $k\in\mathbb N$ such that $2<x^{2^k}<4$.

The final answer/algorithm is not allowed to use powers, roots and logarithms.

I managed to find a range of possible values for $k$.

Looking at the digits after he decimal point, if the $n$th digit is the first non-zero digit, then squaring this number will necessarily turn at least one of the two preceding digits into a 1 but none of the digits before it. I figured this out by looking at the edge cases: if all the digits after the $n$th digit are zero then the $n-1$ digit becomes 1 when the number is squared, and if all these digits are 1 it is equivalent to a number where only the $n-1$ digit of the fractional part is 1, and squaring this number makes the $n-2$ digit be 1.

So the algorithm for finding the upper and lower limits of $k$ is:

Looking at the digits after the decimal point, count the leading zeros (let's call the result of the count $z$). Then $\lfloor\frac{z+2}{2}\rfloor ≤ k ≤ z+1$.

答:
Better proof of the same conclusion as my previous answer, this time without using approximation.

Again, there is no known solution for all numbers, but I found a quick solution for a significant amount of cases, which is useful for optimizing computer programs (for example: try this answer's method, if it fails do something else, like squaring repeatedly until you can determine the solution using this answer's method).

$$C_{z,n} \stackrel{\text{def}}{=} \frac{2^n-1}{2^{n+z}}+1 = 1.\underbrace{0000000000}_{z\text{ zeros}}\underbrace{111111111}_{n\text{ ones}}\text{ (base 2)}$$

These numbers are useful edge cases because if we know that $k$ (the amount of squarings needed to exceed 2) is the same for $C_{z,n}$ and $C_{z,n+1}$ then it is also the same for all the numbers in between, and that lets us know the answer for all these numbers by counting the first consecutive zeros and ones.

For every number of the form $C_{z,n}$ we know that $k$ is either $z$ or $z+1$. Solving the inequality $C_{z,n} > \sqrt[2^z]{2}$ will tell us the if squaring $z$ times is enough or not. Scaling the inequality around the value 1 (subtract 1, multiply by the scale factor, then add 1) by a factor of $2^z$ results in $C_{0,n} > 2^z (\sqrt[2^z]{2}-1)+1$.

$k=\begin{cases}
z + 1 & C_{0,n} < 2^z (\sqrt[2^z]{2}-1)+1\\
z & C_{0,n} > 2^z (\sqrt[2^z]{2}-1)+1
\end{cases}$

The following chart shows the solution to the inequality for $n=1,2,3$:

This gives us the following solution:

(Same as in my previous answer, just written differently.)

$$k =
\begin{cases}
z+1=1 & & z = 0\\
z+1 & n = 1\\
z+1 & n = 2 & z ≤ 2\\
z & n = 2 & z > 2\\
z & n > 2 & z > 0
\end{cases}$$

Now that we know the value of $k$ for all the edge cases, we can fill in (most of) the gaps. If $k$ is the same for $C_{z,n}$ and $C_{z,n+1}$ then it is also the same for all the numbers in between:

$$C(z,n) \stackrel{\text{def}}{=} \{x|C_{z,n} ≤ x < C_{z,n+1}\}$$

Every number in $C(z,n)$ starts with the digits of $C_{z,n}$ followed by a zero followed by anything.

The following table shows the solutions for all the gaps we can fill:


问:
$\text{Nil}(R)=\text{rad}(0), I$ - prime ideal $\iff R/I$ integral domain, I - max ideal $\iff R/I$ field, and other questions on Ideals and Rings
So I've analysed my last lecture in Commutative Algebra and those are the following questions I'd have:$\DeclareMathOperator{\Max}{Max}$

Like e.g. $I+J=\left\{a+b\::\:a\:\in \:I,\:b\:\in \:J\right\}$, is there a way to precisely define the elements in $\bigcap _{\alpha \in A}\:I_{\alpha }$?

As in the title, why is $\operatorname{Nil}(R) = \operatorname{rad}(0)$?

I - prime ideal $\iff R/I$ integral domain and $I$ - max ideal $\iff R/I$  is a field - Why? What's the proof for that?

We defined something as a multiplicative set S, which is if $1\in S, 0\notin S, a,b \in S$ then $ab \in S$. Why do we even bother to define something like that, what's the practical use and interpretation here?

Following that: If $I$ - prime ideal $\iff R/I$ multiplicative set, why?

If $I_n\subset I_{n+1}$ are all ideals, why is $\bigcup _{n\in N}\:I_n$ also an ideal?

We had such a theorem:

Let $S\subset R$ be a multiplicative set and let $J$ be a family of ideals $I$ from ring $R$ so that $I\cap S=\varnothing $. Then $\forall _{I_0\:\in \:J}\:\:\exists \,{P\:\in \:J-\text{ prime and maximal ideal}},\:\:I_o \subset P$.

I kind of understand why P is a maximal Ideal (J fulfills Zorn's lemma thus if we take a chain of ideals so that $I_a\subset I_b$ and $a\le b$ we can say that $P = \bigcup _{\alpha \in A}\:I_{\alpha }$). But why is it also a prime ideal?

We had those statements and like to know why they are true:

(a) If $f\::\:R\:\rightarrow \:T$ is an epimorphism, and $I\subset R$ is an ideal, then $f\left(I\right)$ is an ideal in T

(b) If I take an ideal $J\supset I$, and $\pi :\:R\:\rightarrow \:R/I$ is an epimorphism, then $\pi \left(J\right)$ is also an ideal

(c) The function $\pi$ is a bijection which retains the inclusion relation

(d) $\left\{J\supset I\::\:J\:\in \Max\:R\right\}\ni L\:\rightarrow \:\pi \left(L\right)\in \:\left\{M\:\subset R/I\::\:M\:\in \:Max\:R/I\right\}$ is also a bijection which retains the inclusion relation (the same goes for Spec R and Spec R/I)

(e) $I \in \Max\:R\:\Leftrightarrow \:\pi \left(I\right)=\left(0\right)\in \Max R/I \Leftrightarrow  R/I\text{ is a field}$

Sorry that it's so much but I would really really appreciate any effort here, I'm just curious and want to understand every detail and I'm not satisfied accepting something without a reason.

答:
Some answers (or questions):

What are the $I_\alpha$?

This is the definition of the nilradical. Remember the radical of an ideal $I$ of the ring $R$ is $\{x\in R\mid\exists\, n\in\mathbf N, \,x^n\in I\}$.

The definition of a prime ideal $I$ is that, for all $x,y\in R$, $xy\in I\implies x\in I\text{ or }y\in R$. Translated in terms of congruence classes modulo $I$, it becomes $\overline{xy}=\bar x\bar y=\bar x=\bar 0\implies \bar x=\bar0$ or $\bar y=\bar 0$, which corresponds to the definition of an integral domain.
As to the maximality of the ideal $I$, a field can be characterised by the property that it has only two ideals  – $\{0\}$ and the field itself.

Multiplicative sets are the basis to defines rings of fractions od a given ring. Among these are the localisations of a ring at a prime ideal, which is an essential tool in algebraic geometry.

If $I$ is a prime ideal, it is not $R/I$ which is a multiplicative set, but $R\setminus I$

$\bigcup\limits_{n\in\mathbf N}I_n$ satisfies the properties defining an ideal – the sum of two elements of the union is in the union essentially because by hypothesis, the set of ideals is totally ordered by inclusion.

A maximal ideal is also a prime ideal (because, considering the quotient $R/I$, a  field is an integral domain).


问:
how can I find a orthogonal matrix Q such that Q$x_i=y_i$
How can I find a orthogonal matrix Q such that Q$x_i=y_i$ given that $\|x_i\|=\|y_i\|$ and $\|x_i-x_j\|= \|y_i-y_j\|$.
I know that Householder reflectors would do the trick here Q=$I-2pp^T$ where p is ($x_i-y_i$)/$\|x_i-y_i\|$.  but I am unable to understand how can someone please explain this to me.

Here $x_i , y_i \in R^n$

I also read this Rotation by Householder matrices solution but couldn't understand the how part.

答:
The Householder reflection gives you a reflection across the hyperplane orthogonal to the unit vector $p$ (note that $pp^\top$ gives projection onto the line spanned by $p$, and so subtracting twice that leaves the hyperplane orthogonal to $p$ fixed and reflects across that hyperplane). That said, if you do a Householder reflection as you suggested with $p=\dfrac{x_i-y_i}{\|x_i-y_i\|}$, it switches $x_i$ and $y_i$ and leaves the hyperplane orthogonal to $x_i-y_i$ fixed. If the remaining $x_j$ and $y_j$ are not in that hyperplane, they will be messed up by this reflection. Thus, I don't see any good way to use this repeatedly with more vectors.

However, as Will Jagy pushed me to realize, the solution is easy. Let $A$ be the matrix whose columns are the $y_i$ and let $B$ be the matrix whose columns are the $x_i$. Observe that when we set $Q=AB^{-1}$, it will have property that $Qx_i = y_i$ for all $i$. The easiest way to see this is the following: If $e_j$ are the standard basis for $\Bbb R^n$, then by construction we have $Be_i=x_i$ and $Ae_i = y_i$. Therefore, $(AB^{-1})x_i = A(B^{-1}x_i) = Ae_i = y_i$, as required. Finally, as Will pointed out, the geometric hypotheses tell us that the $n$-dimensional pyramids formed by the $x_i$ and by the $y_i$ are congruent, and so $Q$ must be an isometry of $\Bbb R^n$. By a standard argument, $Q$ must be linear and, indeed, an orthogonal matrix. (For a proof, see, for example, pp. 114-115 of my differential geometry text.)


问:
Suffices to check a morphism of schemes is unramified on closed points.
I am starting to read Milne's Etale Cohomology notes and in it they mention:

A morphism $\varphi :Y \rightarrow X$ is unramified if it is of finite type and if the maps $\mathcal{O}_{X,f(y)} \rightarrow \mathcal{O}_{Y,y}$ are unramified for all $y\in Y$. It suffices to check the condition for the closed points $y$ of $Y$.

I wasn't really sure how it suffices to check that the condition holds for closed points. This is a local definition so we can check it locally: consider a homomorphism of rings $\varphi :R\rightarrow S$. Then if $\mathfrak{p}$ is a prime ideal in $S$ which is contained in the maximal ideal $\mathfrak{m}$ then we can't really compare the fields $S_{\mathfrak{p}}/{\frak{p}}$ and $S_{\frak{m}}/{\frak{m}}$. I don't think there's a canonical morphism in either direction.

I've also searched around and I couldn't find a mention of this anywhere. Is this statement true? and if so why?

答: No one answered my question yet and I did some digging. Not quite sure if this is absolutely correct or not, but again if $\varphi: Y \rightarrow X$ is unramified at $y\in Y$ then this means we can find affine opens around $y \in U = \text{Spec}(A)$ and $\varphi(y) \in V = \text{Spec}(R)$ such that $\Omega_{A/R} = 0.$ Now $\Omega_{A/R}$ is just an $A-$module  so we can check that it vanishes by checking it at the maximal ideals, and vanishing at the maximal ideals is then equivalent to checking that the map is unramified at the closed points.


问:
How many solutions are there in the following equation over the natural numbers if $x_1+x_2+x_3>x_4+x_5+x_6+x_7$

How many solutions are there in the following equation over the natural numbers such that $x_1+x_2+x_3+x_4+x_5+x_6+x_7=30$ if $x_1+x_2+x_3>x_4+x_5+x_6+x_7$  ?

I made up a combinatorics question in my mind  , but i stuck in it. What i tried is :

Firstly , i wanted to use symmetry property but , it has odd number of variables . Hence ,i could not do anything.

Secondly , i thought about whether generating functions can be used or not , but i stuck in writing the desired generating formula for it.

So , i  hope to find nice approaches to my question.

$NOTE=$Natural numbers start with zero.

答:
Using stars and bars you can write the number of solution as
$$
\sum_{i=0}^{14}\binom{i+3}3\binom {30-i+2}2,
$$
where the first factor counts the number of solutions to the equation $x_4+x_5+x_6+x_7=i $ and the second one those of $x_1+x_2+x_3=30-i $.


问:
Convergence and differentiability of a power series
I want to show that

$$\sum_{n=1}^{\infty} \left(x-n \sin\left(\frac{x}{n}\right)\right)  $$

converges uniformly and is differentiable on any bounded open interval $(a,b)$ in $\mathbb{R}$.

I don't know how to show uniform convergence here. $ |x- n \sin(\frac{x}{n})| \leq |x-n| $, but that does not provide any help. Do I use the Weierstrass $M$ test? or does it involve the Cauchy criterion?

Regarding the differentiability part, I invoked the taylor series expansion of $\sin{x}$:

$$ \sum_{n=1}^{\infty} \left(x-n \sin\left(\frac{x}{n}\right)\right) = \sum_{n=1}^{\infty} \left(x-n \sum_{m=0}^{\infty} (-1)^{m} \frac{x^{2m+1}}{(2m+1)!} \right)$$

So, can I conclude it is differentiable because the series on the right is differentiable?

答:
Hint

For the uniform converge, make a formal proof based on following arguments:

For $y$ small enough we have $\sin y \simeq y -\frac{y^3}{6}$.

Therefore $\vert y - \sin y \vert \le \vert y \vert^3$ for $y$ small enough using a Taylor formula.

Replacing $y$ by $x/n$ and multiplying by $n$ we get
$\vert x -n\sin x/n \vert \le \frac{\vert x \vert^3}{n^2} \le \frac{\max( \vert a \vert^3, \vert b\vert^3)}{n^2}$ for $n$ large enough.

This allows to conclude to the uniform converge on $[a,b]$.

Regarding differentiability, use a similar approach and a standard theorem about differentiability of series of maps.


问:
If a system of equations is consistent and the determinant of the coefficient matrix is zero does this mean that there are infinitely many solutions?
I am trying to figure out for what constant $a$ the following system has infinitely many solutions:

\begin{align*}
ax+y+z&=0 \\
x+ay+z&=0 \\
x+y+az&=0
\end{align*}

For $a=-2$ and $a = 1$ the determinant of the coefficient matrix is zero and since the rank of the coefficient matrix is the same as the rank of the augmented matrix, does this mean that for these two values there are infinitely many solutions?

More generally, what I want to ask is: does the determinant of the coefficient matrix being zero mean that there can't be unique solutions?

答: Consider the linear map associated to the matrix. The set of solutions corresponds to the kernel of this linear map, and there results from the rank-nullity formula, that the rank of the matrix is the codimension of the kernel.


问:
Definition of $T$ operator in proof of Excision Theorem on Homology Theory
I had some problems with the proof of Excision Theorem.
I use two books at the same time, one is (Algebraic Topology, Hatcher) and the other one is "An Introduction of Algebraic Topology", Rotman.
In the first book the main work is done in Proposition $2.21$ in four steps. At the second step is defined an operator $T$, that work as chain homotopy between the identity and the barycentric subdivision operator $S$ on $LC_n(Y)$, with $Y$ convex. Hatcher defines $T$ operator inductively as $ \ T_n\lambda=b_{\lambda}(\lambda-T_{n-1}\partial_n \lambda)$. Is proved that the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$ holds at degree $n$ using the same formula at degree $n-1$.

The same fact in Rotman is proved in Lemma $6.13$ using $T$ operator defined as $T_n \gamma =b.(\gamma -Sd \gamma -T_{n-1}\partial_n \gamma)$, where $b.$ is the cone operator $b_{\lambda}$ used by Hatcher, and $Sd_n$ operator is $S$ in the notation above. Also in this case the formula $\partial_{n+1} T_n +T_{n-1} \partial_n=1-Sd_n$ holds using the properties of cone operator and the formula
$\partial_n(\gamma -Sd_n \gamma-T_{n-1} \partial_n \gamma)=\partial_n \gamma - Sd_{n-1}\partial_n-(1-Sd_{n-1}-T_{n-2} \partial_{n-1}) \partial_n \gamma$

that holds by induction and properties of $Sd$ operator.

The two definitions of $T$ are not the same and differ by an $Sd_n$ term that appears in Rotman definition and in Hatcher doesn't.

$1)$ Can someone explain me the reasons (algebraic and geometric) of this difference?

$2)$ Hatcher and other authors say that the geometric interpretation of $T$ operator is a subdivision of $\Delta^n \times I$ obtained by joining simplices in $\Delta^n \times\{0\} \cup (\partial \Delta^n) \times I $ to the barycenter of $\Delta^n \times \{ 1\}$.
I don't understand this explanation, can someone be more clear?

答:
I am following Hatcher. I don't think answering so late makes much sense, but still I present my interpretation for your consideration.
We need an operator T that satisfies $\partial_{n+1} T_n +T_{n-1} \partial_n=1-S$.

Now for simplicity and motivation we look at a $2$-simplex as illustrated in Hatcher. Whenever we need to define a chain homotopy like $T$ we think of a way to decompose $\Delta^n \times I$ to some $n+1$ simplices. $T$ describes this decomposition. The decomposition should be such that, boundary $\partial T \lambda$ of  the $3$-simplices when summed up in a certain way (to be thought of next) cancel at all the faces except the ones remaining in the linear combination of $2$-simplices appearing in $(1-S)\lambda-T\partial\lambda$. This hints (though vaguely) that the decomposition should give us some $3$-simplices with the edges and vertices as those appearing in the barycentric division of the $2$-simplex. Thus the decomposition (as in the illustration in Hatcher) is in some sense not that out of the blue.
Now we try to define $T$ on $\lambda$ for the $2$-simplex case. We want $T$ to be the map which projects a particular linear combination of the $3$-simplices to the $2$-simplex  followed by composition with $\lambda$. Given the way $T\partial\lambda$ looks from Hatcher's calculation we see that $b_\lambda(\lambda - T\partial\lambda)$ is the one possible linear combination (by a brute brute force calculation), and $T$ is the map which projects this sum into the $2$-simplex followed by composition with $\lambda$ (Here $b_\lambda$ is to be identified with the barycenter of $\Delta_{n} \times \{1\}$) . But here we have assumed the definition of $T$ on $1$-simplices. In order to verify that we would have followed same the line of reasoning.We then have a form for $T$ for the $2$-simplex (verified for higher $n$, anyhow considering a 3-simplex for geometric motivation is unyielding due to obvious limitations), which works for higher n when defined as in Hatcher.

P.S. Apologies for poor presentation and vague answer

ContentLicense: CC BY-SA 3.0


问:
If each $f_n$ has $b_n$ zeros, then the number of zeros of $g$ is at most $\liminf_{n \to \infty} b_n$
My question is the following:

Let $f_n:B_1(0) \to \mathbb{C}$ be holomorphic so that $f_n \to g$  uniformly on compact sets. Assume that $g$ is non-constant and let $b_n$ be the number of zeros (with multiplicity) that $f_n$ has. Show that the number of zeros of $g$ (with multiplicity) is at most $\underset{n \to \infty}{\liminf} b_n$.

My attempt:

Let $z \in B_1(0)$ be a zero of $g$.  Because $g$ is nonconstant, the zeros are isolated. So we can find an open ball about $z$ that contains no other zeros of $g$, say $B_\delta(z)$, for some $\delta > 0$.  Note $f_n \to g$ uniformly on $\overline{B_\delta(z)}$ so $\forall \epsilon > 0 \exists N$ such that $|f_n - g | < \epsilon$, $\forall n > N$.  For $\epsilon < \min_{w \in \partial B_\delta (z)} \{|g(w)| \}$ we see that $|f_n - g| < \min_{w \in \partial B_\delta (z)} \{|g(w)| \}$ for all $n > N$.  By Rouche's theorem, the number of zeros of $g$ on $B_\delta(z)$ is equal to the number of zeros of $g + (f_n - g) = g$ for all $n > N$.

I would like to then conclude that the number of zeros of $g$ on $B_1(0)$ is at most $\liminf_{n\to \infty} b_n$, but I am having trouble justifying this...

答:
Wlog we can assume $N=\underset{n \to \infty}{\liminf} b_n$ is finite (else nothing to prove as the zeroes of $g$ are at most countable as they are discrete in the open disc) and since the number of zeroes is always an integer, it follows there is an infinite subsequence of the $f_n$'s with precisely $N$ zeroes (counting multiplicities of course).

Passing to this subsequence we can assume wlog each $f_n$ has $N$ zeroes and we need to prove that $g$ has at most $N$ zeroes which follows now by contradiction as in the original post

(Let $M$ the number of zeroes of $g$, assume $M>0$ so there is something to prove and split them into distinct numbers $z_1,..z_k$ with multiplicities $m_1,..m_k, m_1+..m_k=M$
Working on small mutually disjoint discs around each $z_q$, one gets by Rouche that there is an $n_q$ st $f_n$ has $m_q$ zeroes close to $z_q$ for all $n \ge n_q, q=1,...k$. Picking $n(M) \ge n_q$ for all $q$ we get that $f_n$ has at least $M$ zeroes for $n \ge n(M)$ so by hypothesis we have $M \le N$ and we are done)

Note that an easy example like $f_n(z)=z-1+1/n \to z-1$ shows that $f_n$ has each a zero in the open unit disc, but $f$ doesn't so the inequality can be strict


问:
Find the expression for $f(x)$ an expression for $f(g(x))$
I fail to understand something fundamental about universal statements and sets.

Suppose we have a function $f$, such that $\forall x \in \mathbb{R}\:f(\frac34x)=2x^2$

I want to find the expression for $f(x)$.

$\forall x \in \mathbb{R}\: \exists x'=\frac34x$. Substitute $x'$ into the function to get $f(x')=\frac{32x'^2}{9}$, and it would seem that I can substitute $x'$ to $x$ getting $f(x)=\frac{32x^2}{9}$, but why? With this substitution I'm replacing the whole domain of $f$ from $\mathbb{R}$ to the set $\left\{ x' | x'=\frac34x, x \in \mathbb{R} \right\}$. Should I now show that the set of all $x'$ equals $\mathbb{R}$?

答: That's a notation play. Just rewrite it: $f\left(\dfrac{3x}{4}\right) = \dfrac{32}{9}\cdot\left(\dfrac{3x}{4}\right)^2\implies f(x) = \dfrac{32}{9}x^2$.


问:
How to calculate $p$-value in a clinical trial
I'm trying to verify the $p$-value mentioned in this article:

$8\%$ ($n=\frac{1}{13}$) of mavrilimumab-treated patients progressed to mechanical ventilation by Day $28$, compared to $35\%$ ($n=\frac{9}{26}$) of control-group patients who progressed to mechanical ventilation or died ($\textbf{p=0.077}$).

How can I calculate the $p$-value for the data above?

I used $3$ different calculators and they all gave different results, none of them matches with the article.

Calculator 1: $p=0.12$

Calculator 2: $p =  0.06876$

Calculator 3: $p = 0.154$

答:
I can't explain the 'why'. However, looking at an abstract for the research, all of the $p$-values are different:

$\big[\frac{0}{13}, \frac{7}{26}\big]$ Globenewswire: $p=0.086$,
Scientific Abstracts: $\log$ rank $p=0.046$

$\big[\frac{13}{13}, \frac{17}{26}\big]$ Globenewswire: $p=0.0001$,
Scientific Abstracts: $p=0.018$

$\big[\frac{10}{11}, \frac{11}{18}\big]$ Globenewswire: $p=0.0093$,
Scientific Abstracts: $p=0.110$


问:
Interesting/Paradoxical Math Problems for High School Students
I am creating a math course for high school students (not necessarily advanced) about interesting/paradoxical math problems (think Monty Hall and Hilbert's Hotel). What are some other interesting math problems that are accessible to average high school students? So no calculus or higher required (unless the necessary machinery can be taught within 30 minutes). Thanks!</p>

答:
Zeno's paradox could be fun. I had a lively discussion with some high school students about whether $0.999999... = 1$, and we enjoyed looking at pictures such as this one:

While this topic comes up against the idea of a limit and infinite series, it doesn't have to be presented with that language.


问:
Measure theory and topology books that have solution manuals
I am trying to find a book to learn measure theory that contains complete solutions manual. Does someone know of any?
Also, I would like to know if there is a book with solutions manuals about topology.

thanks.

答:
There are two parts of this answer:

Textbooks (or problem books) that have solution manual sold along with them;

Good and classic textbook most of whose problems have solution online.

For 1, here are are some recommendations:

1a) "Principles of Real Analysis 3rd Edition" by Charalambos D. Aliprantis and Owen Burkinshw.

This is the original book, and it has a companion solution manual of all the exercises in this book:

1a') "Problems in Real Analysis, A Workbook with Solutions 2nd Edition" by Charalambos D. Aliprantis and Owen Burkinshw.

I have to say that I never had any professors or courses that used this book. This book is good if you are not the first time learner, since it introduces measure, integrable functions, etc, in a way that first-time learner may find hard to understand and to relate to other classic books like Royden, Rudin, or Stein. It is more like a Folland-approach. But if you just want to do some extra exercises and you have learnt measure theory more than once, this is good.

By the way, this book also contains some basic analysis problem, so you can review the undergraduate stuff.

Another book is a problem book:

1b) "Problems in Mathematical Analysis III --- Integration" by W.J. Kaczor and M.T. Nowak.

It has detailed solution inside the book. This is pretty user friendly if you are the first-time learner, but not really useful if learn it in a second time and in a deeper level. The exercise simply does not cover the deeper stuff. But if you learn the measure theory first time in the first semester PhD course or so, this can be pretty good book to supplement your course.

This is pretty much of it.

For 2), I will refer the books and list some links of solution I found.

2a) "Real Analysis: Modern Techniques and Their Applications" by Gerald B. Folland.

Most of Chapter 1 to Chapter 6

Most of Chapter 7

Most of Chapter 8

Q1 and Q10 of Chapter 9

Another solution set from Chapter 1 to Chapter 9

Some problems in Folland are pretty hard and technical so they don't have solution online (and they are likely in your homework if you are with a harsh professor), so go here and ask.

2b) "Real Analysis" by H.L. Royden and P.M. Fitzpartrick

A really good solution set that contains most of the exercises in the whole book

Despite being commonly used, I personally hate this book. Perhaps my first time learning experience was with this book, and the proof in this book was not readable for the first time learning. However, it does cover a pretty broad of topics, if you are not the first time, companying Folland with Royden will be a perfect choice.

2c) "Probability: Theory and Examples" by Rick Durrett.

Some solutions as solutions of homework

Never imaged a customized solution manual can be sold...

This is a really commonly used book to introduce measure theoretic probability, and MSE (and internet) contains the solutions of pretty much all the exercises. However, be careful with the question numbering. The book now is of 5th edition, despite no change in the exercise, the ordering has a drastic change, and solutions were created during the 2nd of 3rd edition of this book.

I must say that, if you are for a serious treatment (and I think this is vital) of measure theoretic probability, do NOT read this book. This book is somehow not deep enough and not standard enough. Measure theoretic probability contains an inevitable and hard experience of getting your technicality perfect. This is a "must-experience" experience. This book tries to skip them (or cannot cover them in details) or tries to taught them in a not standard way. The book becomes funny when it treats conditional probability distribution, like it goes crazy.

For a serious treatment of measure theoretic probability, here are some really good choices:

3a) Amir Dembo's note:  This book is really technical;

3b) "Probability Second Edition" by A.N. Shiryaev;

3c) "Probability - 2" by A.N. Shiryaev.

Shiryaev's book is really good and really detailed. It also covers a really wide range of topics, from basic measure theory to orthogonal random measure (so discrete spectrum theory) ect. You can not only learn measure theoretic probability, but can learn stochastic process a lot from the discrete view (from discrete to continuous, it is pretty simple).

Dembo's note didnt do a great job when it introduces Brownian motion, but it is more of a topic of stochastic process. If you need, I also have some recommendation.

But for now I think above is enough for you :)


问:
How many squares contains paint
I read a book containing competition level problems. I was unable to solve the following:

Consider an $n\times m$ grid. A paintbrush whose width is the same as the width of a grid square is used to paint a line from one corner of the grid to the opposite corner. How many squares will have paint in their interiors?

How can I solve this?

答: Hint:  how many vertical and horizontal lines does the paintbrush cross?  You need to think about the width of the brush.  Draw a picture for a few small cases.</p>

Title: How many squares contains paint


问:
How to prove this particular set is a countable set?
Suppose $f(x)$ is defined on $[0,1]$, and $\exists M$, s.t. for any finite $x_1, \cdots, x_n \in [0,1]$, we have $|f(x_1)+\cdots+f(x_n)|\le M$. Prove E = $\{x \in [0,1]:f(x)\neq 0\}$ is a countable set.

I've got a simple idea to change to $\cup\{x:|f(x)|\ge \frac{1}{n}\}$, but don't know what to do next.

答: if $f(x)\neq0$ then there exists $x_0\in[0,1]$ such that $f(x_0)\neq0$. Using the hypothesis for $x_1=x_2=...=x_n=x_0$ then $n|f(x_0)|<M$ and so $|f(x_0)|<\frac{M}{n}$ for any $n\in\mathbb{N}$, which means that $|f(x_0)|$ can be made arbitrarily small by choosing $n$ large enough. This is impossible.


问:
Machine Precision or Machine Epsilon Formula
I'm reading Heath's Scientific Computing.

The machine epsilon is denoted by $\epsilon$.
With rounding to nearest, the machine epsilon can be determined by the following formula:

$\epsilon = \frac{1}{2} \beta^{1-p}$

where $\beta$ is the base and $p$ is the precision (i.e. number of significant digits).

For $\beta = 10$ and $p = 3$ since $1 + 0.005 = 1.01 > 1$, $\epsilon = 0.005 =  \frac{1}{2} 10^{1-3}= \frac{1}{2} \beta^{1-p}$

For $\beta = 2$ and $p = 3$ since $(1 + 0.001)_2 = (1.01)_2 > 1$, $\epsilon = (0.001)_2 = \frac{1}{2} 2^{1-3} = \frac{1}{2} \beta^{1-p}$

For $\beta = 3$ and $p = 3$ since $(1 + 0.002)_3 = (1.01)_3 > 1$, $\epsilon = (0.002)_3 = \frac{2}{27}$ which does not agree with $\frac{1}{2}3^{1-3}= \frac{1}{18}$

Is the formula only applicable to bases that can represent $\frac{1}{2}$ exactly or did I misunderstand how the formula is to be applied?

答:
You are confusing real numbers with their representations.

You write $(1 + 0.002)_3 \stackrel?= (1.01)_3,$
which is an abuse of notation to begin with;
the left side is not equal to the right, it rounds to the right-hand side
when $p = 3.$

The question Heath appears to be trying to answer is,
"What is the largest relative error due to a single rounding off?"
For $\beta = 10$ and $p = 3$ we can use $1.005$ as a test case because that is the very smallest number between $1.00$ and $1.01$ that rounds up instead of down.

But when $\beta = 3$ and $p = 3,$
lots of other numbers that are less than $1.002_3$ also round off to $1.01_3$, for example $(1 + 0.001112)_3.$

In fact, the very smallest number between $1$ and $1.01_3$ that rounds up to $1.01_3$
is
$$1.001111\ldots_3 = 1 + \frac1{18}.$$


问:
Unsure of what's happening in this fraction
I can't seem to see what is happening to make this equivalence. More specifically, how the left-side's $n+1$ makes it into the right side's numerator.
$$
\frac {n(n+1)} 2  + (n+1) = \frac {(n+1)(n+2)}2
$$

答: hint: Factor the $(n+1)$ using distributive property, and note also that $\dfrac{n}{2}+1 = \dfrac{n+2}{2}$


问:
Fixed points on complex function $T(z) = (1+i)z + 3-4i$
If $T(z) = (1+i)z + 3-4i$ is a complex function, and I got this fixed point: $z=4+3i$

Is this correct? Are there more fixed points?

$(1+i)z+3-4i=z$

$3-4i=z-(1+i)z$

$3-4i=-iz$

$z=-\frac{3-4i}{i}=4+3i$

答:
Your proof is correct and there are no other fixed points. You can see this abstractly by saying that the fixed point equation is a degree one polynomial and therefore has at must one solution. More concretely, each deduction you made was "reversible" so each implication was actually an equivalence.

You had

$$
\begin{align*}
(1+i)z+3-4i=z &\implies 3-4i=z-(1+i)z\\
&\implies 3-4i=-iz\\
&\implies z=-\frac{3-4i}{i}=4+3i
\end{align*}
$$

but at each step the reverse implication holds as well. For instance, in the last step you divided by $-i$, so to go the other way you multiply by $-i$. Hence, you really have

$$
\begin{align*}
(1+i)z+3-4i=z &\Longleftrightarrow 3-4i=z-(1+i)z\\
&\Longleftrightarrow 3-4i=-iz\\
&\Longleftrightarrow z=-\frac{3-4i}{i}=4+3i.
\end{align*}
$$

In other words, $z$ is a fixed point of $T$ if and only if $z = 4+3i$. Be warned that not all manipulations you will do are reversible, however. $(-1)^2 = 1^2$ but you cannot reverse this to get $-1=1$!


问:
Probability mass function from the distribution
I have the following distribution function:

\begin{equation*}
      F(x) = \begin{cases}
      0, x<-1 \\
      \frac{1}{2}, -1 \leq x < \frac{1}{2} \\
      \frac{3}{4}, \frac{1}{2} \leq x <2 \\
      1, x \geq 2
        \end{cases}
    \end{equation*}

How can i obtain the probability mass function? My attempt was to use the discontinuites and try to evaluate the probabilities, but i was no able to confirm anything. I know that the sums off all probabilities must be one.

答: Take the differences from the DF, it will give you the pdf at the jump discontinuities.</p>


问:
Proving that a set and operation form a group

$S$ is a nonempty finite set with associative operation $∗$ such that, $$∀ a, b ∈ S, ∃ c,d ∈ S\text{ s.t. }a ∗ c = b \text{ and }d ∗ a = b$$Prove that $(S, ∗)$ is a group.

I'm stuck on a difficult problem from some PSET I found online (I'm self studying Algebra).

I have been trying to show the existence of an identity in $S$ on $*$ as I think the rest will follow, but I cannot seem to make any progress.

My work so far is not very solid, but it is an attempt to show that there is at least one $e$ exists such that

$$a ∗ e = e ∗ a = b, \text{ where }a,b\in S$$

Reasoning as follows:

From the problem statement:

$$∀ (a, b) ∈ S × S\ ∃ (c, d) ∈ S × S:\ a*c = d*a = b$$

I would like to show that at least one such $(a, b)$ has $c = d$.

I try a proof by contradiction:

Assume

$$∀ (a, b) ∈ S × S\ \exists (c, d) ∈ S × S\ c ≠ d:\ a*c = d*a = b$$

Since $S$ is finite, so let $|S| = n$.

$$\therefore|S × S| = n^2$$

But there are $n(n-1)$ pairs of the form $(c, d) ∈ S × S$ such that $c ≠ d$.

From here I am trying to use the pigeonhole principle to show that there exists $(a,x)$ and $(b,y)$ such that

$$(a = b\text{ and }x ≠ y)\text{ or }(x = y\text{ and }a ≠ b)$$
that satisfy

$$a*c = c*a = x$$

$$b*c = c*b = y$$

leading to an obvious contradiction.

I think there is a way to derive this existence, since intuitively there are $n$ "excess" pairs after relating pairs $(a, b)$ one-to-one with pairs $(c, d)$.

I would greatly appreciate a hint.

Thanks!

答:
In the end I wrote a proof far simpler than what I was trying to do above. Might be helpful to someone at some point so:

From the problem statement:

Existence of right and left identities for each element

∀  ∈, ∃ e, e' ∈:

ae = a and e'a = a

Right and left identities are universal

Now:

∀ b ∈ , ∃ x, y ∈:

ax = b and ya = b

⇒ e'ax = e'b and yae = be

⇒ ax = b = e'b and ya = b = be

In other words, e is a left identity for all elements of S, and e' is a right identity for all elements of S.

Equivalence of right and left identities

∃ q ∈:

qe = e' ⇔ q = e'

e'e = e'

Similarly, ∃ p ∈:

e'p = e ⇔ p = e

e'e = e

so:
e = e'

Uniqueness of the identity

We have shown that if a left-inverse for one element is a left-inverse for all elements, and that such a left-inverse is also a right inverse. Now suppose ∃ p ∈:

al = la = a for some a ∈

It follows that:

el = e

But el = l, since e is an identity for all elements

So l = e.

Existance of inverses

For any element a ∈,  ∃ ar-1, al-1 ∈ :

aar-1 = e

al-1a = e

⇒ al-1(aar-1) = al-1

⇒ ar-1 = al-1

So for all a ∈ , we have a-1 ∈

Uniqueness of inverses

Suppose ab = e

⇒ a-1(ab) = a-1e

⇒ b = a-1

So each element has a unique inverse.

Closure

let xy = z

⇒ x = zy-1

From the problem statement, we know there exists p in S:

x = py-1

⇒ xy = p

⇒ z = p

Hence {S, ∗} is a group

Please feel free to suggest edits, and do let me know if I made a mistake.


问:
Communicating mathematics
As a TA, my ratings have been fairly mediocre.  There's nothing to suggest I'm a problem, but I have room to improve.

What's more disturbing to me is that I even struggle to communicate math with people who know a lot more than I do.  It's no surprise that they struggle to communicate with me.  They know more, probably forgot what it's like to know less, and generally do not give the details on everything, sometimes even neglecting to mention that a particular detail needs to be filled in.  But when I communicate, I try to leave out as few details as possible and clearly enunciate when I am leaving them out, and to minimize the number of mistakes I make.  I am extremely nitpicky about this.

So there must exist other obstacles to communication other than just misrepresentations.  I suspect that what I'm lacking is a pedagogical element, which would also explain my interaction with students.  How can I go about finding out what is lacking, and then fixing it?  What are some common teaching mistakes made by people with the attitude that mathematics should be precise and airtight?  It's not that I'm such an adherent of my own philosophy that I refuse to "dumb things down."  That used to be the problem, but it appears that when I attempt to convey vague ideas, people do not understand me.

答:
It is very hard to know when precision is going to trade off with clarity and when precision is going to be necessary for clarity. I don't pretend to know. However, two defaults:

at all levels of math, you should probably go a little slower than your instincts tell you to go. Do not worry about insulting an expert by going too slow. If the expert wants you to speed up, the expert will indicate that. Then you can speed up. Similarly, most students appreciate a slow clear explanation and are not insulted (in fact are happy) when you explain something they already know.

pausing to let the other person talk is extremely helpful. I find I have to force myself to do this.

Title: Communicating mathematics


答:
If before -1 the distribution function equals 0 and after -1 it equals $\frac{1}{2}$, it means that there is a mass of probability $\frac{1}{2}$ in $x=-1$.

For the same reason, there is a mass of $\frac{3}{4} - \frac{1}{2} = \frac{1}{4}$ in $x=\frac{1}{2}$, and finally a mass of $1 - \frac{3}{4} = \frac{1}{4}$ in $x = 2$.

The sum of all probabilities is $\frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1$.

Title: Probability mass function from the distribution

Body:
I have the following distribution function:

\begin{equation*}
      F(x) = \begin{cases}
      0, x<-1 \\
      \frac{1}{2}, -1 \leq x < \frac{1}{2} \\
      \frac{3}{4}, \frac{1}{2} \leq x <2 \\
      1, x \geq 2
        \end{cases}
    \end{equation*}

How can i obtain the probability mass function? My attempt was to use the discontinuites and try to evaluate the probabilities, but i was no able to confirm anything. I know that the sums off all probabilities must be one.


问:
How to prove or disprove if $a_n \leq M$, for some number $M$, then $L \leq M$?
So here is the problem I am having a bit of trouble with. Not exactly sure how to approach this problem.

Given the sequence $\{a_n\}_{n = 1}^{\infty}$ and $\lim_{n \to \infty} a_n = L$ for some number $L$. If $a_n \leq M$ for all $n$, where $M$ is some finite number, prove or disprove $L \leq M$.

I am thinking about using a Proof by Contraposition. So, if $L > M$, then $a_n > M$.

Since we are told that $\lim_{n \to \infty} a_n = L$, then we can say that $\lim_{n \to \infty} a_n > M$. But this is where I am lost or do not how to proceed. I am not exactly sure how to get rid of the limit on the left side.

Some advice or some suggestions would be useful. Thanks

答:
Hint:Assume that $L>M$ and let $L-M=\epsilon$

than use $\epsilon-\delta$ definition of limit for given $\epsilon$


答:
An idea for you: suppose $\;L>M\implies L-M=\epsilon>0\;$ . Then there exists $\;N\
\in\Bbb N\;$ such that

$$n>N\implies |a_n-L|<\frac\epsilon2$$

But remember that if $\;a>0\;$ , then for some real number $\;x\;,\;\;|x|<a\iff -a<x<a\;$ . Use this and the above to get your contradiction.

Title: How to prove or disprove if $a_n \leq M$, for some number $M$, then $L \leq M$?

Body:
So here is the problem I am having a bit of trouble with. Not exactly sure how to approach this problem.

Given the sequence $\{a_n\}_{n = 1}^{\infty}$ and $\lim_{n \to \infty} a_n = L$ for some number $L$. If $a_n \leq M$ for all $n$, where $M$ is some finite number, prove or disprove $L \leq M$.

I am thinking about using a Proof by Contraposition. So, if $L > M$, then $a_n > M$.

Since we are told that $\lim_{n \to \infty} a_n = L$, then we can say that $\lim_{n \to \infty} a_n > M$. But this is where I am lost or do not how to proceed. I am not exactly sure how to get rid of the limit on the left side.

Some advice or some suggestions would be useful. Thanks


答: Yes, you are right. You prove: $L \le M$. Assume otherwise, then take $\epsilon = \dfrac{L - M}{2}>0$, then you will reach at some point a contradiction. Hope this helps. Note that the $2$ at the denominator can be adjusted to fit the need of the contradiction proof. So maybe you want to use $\epsilon = C(L - M)$, and you adjust $C$ to make it work.


问:
What is the difference between probability distribution and probability simplex?
I was watching a Reinforcement Learning theory course presented by Nan Jiang, and I noticed that he's using probability simplex instead of a probability distribution. So, I wonder what is the difference between probability distribution and probability simplex.</p>

答:
According to https://en.wikipedia.org/wiki/Simplex, a probability simplex is defined as
$$\{x \in \mathbb{R}^k \ : \ \sum_{i=1}^k x_i = 1, x_i \geq 0 \ \forall i\}$$
or in other words, a finite set of non-negative real numbers that sum to 1 (also called a convex combination).

For any probability space, you can partition your sigma field (set of observable events) into a probability simplex, which could be useful for optimization.


答:
hint

Assume that $ L>M$.

take $$\epsilon=L-M$$

$$\lim_{n\to +\infty}a_n=L \implies$$
for $ n $ great enough,
$$L-\epsilon<a_n \implies$$
$$L-(L-M)<a_n\implies$$

$$M<a_n$$

Which contradicts the hypothesis $ a_n\le M$.


问:
Probability $k$ bins are non-empty
The following problem arises in the analysis of Bloom filters.

Consider $m$ bins and $N=nk$ balls placed uniformly and independently at random into the bins. A query chooses $k$ bins uniformly and independently at random and asks if they are all non-empty.  The main question is as follows.

What is the probability that all $k$ bins in the query are non-empty?

It is assumed that this probability will be a function of $k$, $m$ and $n$.

The second question is:

For what value of $k$ (which is a function of $m$ and $n$) is this probability minimized?

The standard version of the analysis taught the world over and reproduced in the wikipedia page linked above contains a "now the magic occurs" step which ignores the non-independence of the bins. It gives $k \approx \frac{m}{n} \ln{2}$ as the answer to the second question.

Is there a clean and rigorous way of doing this analysis correctly?

This is a repost of this one which hasn't had any good answers yet.

Clarification.

The difficultly arises because the probability that all $k$ bins are non-empty is not $(1-(1-\frac{1}{m})^{kn})^k$. The reason is that the probability that one of the $k$ (not necessarily distinct) bins that is queried is non-empty is dependent on whether the other queries return non-empty bins.

An approximate answer or good bounds will suffice.

答:
(Note: This is the correct analysis for the $k$ queries being not necessarily distinct, but I've posted this as a separate answer because my previous one was already too long and messy.)

After the $N = nk$ balls are put in the $m$ bins, suppose a fraction $q$ of bins are empty. When we make the $k$ queries, the probability that a particular query encounters a non-empty bin is exactly $(1 - q)$, as the query might encounter any of the $m$ bins with equal probability. Further, even though the bins are not independent, the $k$ queries we make are completely independent, so the probability that they all encounter non-empty bins is exactly $(1-q)^k$.
The final answer (the probability) is got by considering all possible values for $q$, weighted by their probability: the probability of all $k$ queries seeing a nonempty bin is $$\sum_{\alpha} \Pr(q = \alpha)(1-\alpha)^k$$ where the sum is over values $\alpha \in [0, 1]$ that $q$ can take, namely values of the form $\frac{r}{m}$ for $0 \le r \le m$. No approximation here.

The question of independence of the bins comes up when we try to say what the fraction $q$ might be. The probability of a particular bin (say bin $i$) being empty is $p = \left(1 - \frac1m\right)^N$, since for this bin to be empty, each of the $N$ balls must have gone to any of the other $(m - 1)$ bins other than this one. This probability $p$ is also the expected value of the fraction $q$.

(To prove this rigorously: define the indicator variable $X_i$ to be $1$ if bin $i$ is empty, and $0$ otherwise. The total number of empty bins $X = X_1 + X_2 + \dots + X_m$ has expected value $E[X] = E[X_1 + \dots + X_m] = E[X_1] + \dots + E[X_m] = mE[X_1]$ by linearity of expectation, so the expected value of the fraction $q$ (which is the same as $X/m$) is $E[q] = E[X/m] = E[X]/m = E[X_1] = p.$)

The actual value of $q$ can vary, and be different from its expected value $p$. However, the probability of its being far from $p$ is very low: as we move away from $p$, the probability of $q$ taking that value falls off exponentially. As $p \approx e^{-N/m}$, this also means that the probability of $q$ being far from $e^{-N/m}$ is also very low. (We still need to prove this.)

So the probability of all $k$ bins being nonempty, which is
$\sum_{\alpha} \Pr(q = \alpha)(1-\alpha)^k$, is effectively the same as $\left(1 - e^{-N/m} \right)^k$, as $q$ takes on a value around $e^{-N/m}$ with overwhelmingly high probability.

Left to prove: that $q$ is unlikely to be far from $e^{-N/m}$. It's a bit cumbersome to analyze the fraction $q$, as it depends on all the bins, and they are not independent. One approach is to bound the error caused by the independence assumption. Mitzenmacher and Upfal, in their book Probability and Computing: Randomized Algorithms and Probabilistic Analysis, give an elegant technique for doing precisely this.

Let us focus on a particular bin $i$ of the $m$ bins. Consider the number of balls in bin $i$, after $N$ balls have been dropped independently and uniformly at random into the $m$ bins. For each bin $i$, this number (call it $X_i$) follows the binomial distribution: the probability that the bin has $r$ balls is

$\Pr[X_i = r] = \binom{N}{r}\left(\frac1m\right)^r\left(1-\frac1m\right)^{N-r}.$ This is approximately a Poisson distribution with parameter $\lambda = \frac{N}{m}$, or in other words $\Pr[X_i = r] \approx e^{-\lambda}(\lambda^r / r!)$.

Motivated by this, for the $m$ bins all viewed together, they introduce what they call the Poisson approximation. Consider $m$ bins with the number of balls in each bin $i$ (call this number $Y_i$) as being independently distributed, following a Poisson distribution with parameter $\lambda = \frac{N}{m}$. Of course, under this distribution, the total number of balls across the $m$ bins could vary, though it's indeed $N$ in expectation. However, it is a surprisingly straightforward exercise to prove that

The distribution of $(Y_1, \dots, Y_m)$ when conditioned on $\sum_{i=1}^{m} Y_i = N$ is the same as the distribution of $(X_1, \dots, X_m)$.

Any event that takes place with some probability $p$ in the "Poisson approximation scenario" takes place with probability at most $pe\sqrt{N}$ in the "exact scenario".

This inequality above is proved by showing that for any nonnegative function $f(x_1, \dots, x_m)$ (such as the indicator function of an event), we have
$$\begin{align}
E[f(Y_1, \dots, Y_m)]
&\ge E\left[f(Y_1, \dots, Y_m) | \sum_i Y_i = N\right]\Pr(\sum_i Y_i = N) \\
&= E[f(X_1, \dots, Y_m)] \Pr(\sum_i Y_i = N) \\
&\ge E[f(X_1, \dots, Y_m)] (1 / e\sqrt{N})
\end{align}
$$
as $\Pr(\sum_i Y_i = N) = e^{-N} (N^N / N!)$ and $N! \le e\sqrt{N} (N/e)^N$.

So, they prove something like(?) $\Pr(|q - p|) \ge \epsilon) \le 2e\sqrt{m}e^{-m\epsilon^2 / 3p}$ using this and the Chernoff bound.

In fact, they prove using martingales and the Azuma-Hoeffding inequality (12.5.3, p. 308) that $\Pr(|q - p| \ge \frac{\lambda}{m}) \le 2\exp(-2\lambda^2/m)$.

They even have an exercise (12.19, p. 313) showing that $\Pr(|q - p| \ge \frac{\lambda}{m}) \le 2\exp(-\lambda^2(2m - 1) / (m^2 - p^2m^2))$.

Title: Probability $k$ bins are non-empty


问:
What's the difference between a probability mass function and a discrete probability distribution?
The Wikipedia page on "probability mass function" says "The probability mass function is often the primary means of defining a discrete probability distribution" (emphasis added).  I was under the impression that those two terms were completely synonymous.  What's the difference between them?

答:
Note that the statement is the probability mass function can define the distribution of a discrete random variable, but the distribution is not the same thing as the distribution function

The probability mass function for a discrete random variable $X$ is $P(X = k)$. The probability distribution function often refers to the cumulative distribution function, $P(X \leq k)$.

I think when speaking or writing, people often drop the 'function' at the end, hopefully it'll be clear from context what they mean.


问:
Numeric Integration
Consider the following question :
enter image description here

Can someone please explain why it's suitable to use here all of the numeric integration techniques?
Also here :
enter image description here

Why the trapezoidal rule is the correct answer?

答: The integrand is a linear function. All three rules are exact on linear functions, Simpson's rule even on quadratic functions.</p>

Title: Numeric Integration


问:
Solve $x^2 - 5x + 6 \equiv 0 \pmod{18}$

Solve $x^2-5x+6\equiv 0 \pmod{18}$.

Here is what I did:
I rewrote the equation into $(x-2)(x-3) \equiv 0\pmod{18}$
And I know if (mod p is a prime number) $x \equiv 2\pmod{18}$, $x \equiv 3\pmod{18}$
However, it is not in this case, how do I find other solutions within the range of $1$ to $17$. ($11$ and $12$)?

答: An integer is divisible by $18$ iff divisible by each of the coprime prime powers $2,\,9$. Your function of $x$ is always divisible by $2$, but is divisible by $9$ iff at least one of its factors is, as these are also coprime. So we have two cases $9|x-2,\,9|x-3$.

Title: Solve $x^2 - 5x + 6 \equiv 0 \pmod{18}$


问:
continuation of integer-valued solution
I have a recurrence relation that is given by

$$f(n) = 2af(n-1)-a,\quad f(1) = a.$$

The solution given in Wolfram alpha is

$$f(n) = \frac{(a-1)2^na^n+a}{2a-1}.$$

(This is not a homework problem).  My question is, supposing that I want to determine the continuous solution $f(x)$, can I find it by knowing $f(n)$ - it's values at the positive integers?  If so, any suggestions on how to do this?

答: Well you could just set $f(x):= \frac{(a-1)2^xa^{x}+ a}{2a-1}$. This is continuous.... (assuming $a\neq 0.5$ and $a>0$.)

Title: continuation of integer-valued solution


答:
Here it is a proof by contradiction.

Let us suppose that $L > M$. Then, for $\varepsilon = (L - M)/3 > 0$, there corresponds $n_{\varepsilon}\in\mathbb{N}$ such that
\begin{align*}
n\geq n_{\varepsilon} \Rightarrow |a_{n} - L| < \varepsilon \Rightarrow  L - \varepsilon < a_{n} \leq M
\end{align*}

Hence we can conclude that
\begin{align*}
L - M \leq \varepsilon \Rightarrow \frac{2(L - M)}{3} \leq 0 \Rightarrow L \leq M
\end{align*}

which is a contradiction, and we are done.

In fact, for any constant $0 < k < 1$, the choice $\varepsilon = k(L - M)$ works.

Hopefully this helps!


问:
Mathematically determine ways to empty a box of toys
Consider a box with $n \geq 2$ toys. We can empty it by removing two, three or four toys. How many different ways can we do this, if order matters (i.e. taking out 2, 3, then 2 toys is different from taking out 3, 2, then 2 toys)?

Obviously, this problem can be solved programmatically by iterating through all possible arrangements for two, three, or four removals, selecting only the ones that take out all the toys, and counting permutations as we do in the MISSISSIPPI problem.

Is there a more elegant mathematical solution?

答:
Let $a_n$ be the number of ways to empty the box of $n$ toys. Suppose that $n\ge 5$. If we first remove $2$ toys, there are $a_{n-2}$ ways to finish emptying the box. If we first remove $3$ toys, there are $a_{n-3}$ ways to finish emptying the box. And if we first remove $4$ toys, there are $a_{n-4}$ ways to finish emptying the box. Thus,

$$a_n=a_{n-2}+a_{n-3}+a_{n-4}\tag{1}$$

for $n\ge 5$. In fact, the recurrence $(1)$ is valid for $n\ge 4$ if we set $a_0=1$. (This actually makes some sense: if $n=0$ we empty the box in one ‘move’ by removing nothing at all.) Solving for a closed form is going to be messy, since it will require factoring a quartic, but the recurrence itself makes it easy to calculate $a_n$ efficiently for reasonable values of $n$.


答: So you have $(x - 2) (x - 3) \equiv 0 \pmod{18}$. Thus the two factors (modulo $18$) multiply to $18 \equiv 0$, i.e., one is $1, 2, 3, 6, 9, 18$ and the other is the other factor. This gives the set of values.

Title: Solve $x^2 - 5x + 6 \equiv 0 \pmod{18}$

Body:
Solve $x^2-5x+6\equiv 0 \pmod{18}$.

Here is what I did:
I rewrote the equation into $(x-2)(x-3) \equiv 0\pmod{18}$
And I know if (mod p is a prime number) $x \equiv 2\pmod{18}$, $x \equiv 3\pmod{18}$
However, it is not in this case, how do I find other solutions within the range of $1$ to $17$. ($11$ and $12$)?


问:
Criteria for a function to be plottable
Assume that $f$ is a real function. My question is how can one decide if $f$ is plottable or not? My assumption is that $f$ must be of class $C^1$, but I am not aware of such a result. My assumption is based on the fact that there are differentiable functions that can not be plotted (e.g.,
$$
f:\mathbb{R}\rightarrow\mathbb{R},\quad f(x)=\left\{
\begin{array}
[c]{l}%
x^{2}\sin\dfrac{1}{x},~\text{if }x\neq0\\
0,~\text{if }x=0
\end{array}
\right.
$$

is not plottable in the neighborhood of $0$, since it's derivative does not have limit at $0$, hence the slope of the tangent to the graph varies in an "uncontrolable" fashion.

Edit: The problem is that I don't have an "official" definition for what "plottable" means, I only have an intuitive understanding of the concept. Also, it is clear that there are plotable functions which have points where they are not differentiable, e.g., $|x|$, so my requirement of being $C^1$ is not valid. Maybe, i should have said "picewise $C^1$".

答:
Without a precise definition of plottable, we cannot give a precise classification of the functions that are.  Some things that can get in the way:

1) Too many discontinuities:  The Dirichlet function, $1$ on the rationals and $0$ on the irrationals is an example.  Roughly speaking, I would say we can handle a finite number of discontinuities as long as they are not to close together

2) Too much local variation:  Even $\sin \frac 1x$ on $[\frac 1{100000},1]$ has so much variation that it will be a blur.  This one is $C^{\infty}$

3) To much variation of scale:  Think of taking a square wave with amplitude $1000$ and adding $\sin x$ to it.  Of course we can round off the corners to get continuity.  But to see what is going on, you need to see the square wave, and then the sine wave will be too small to see.  You can have similar problems in the $x$ direction.  Maybe you have stuff going on close to the origin, then around $x=100000$.

4) As a takeoff of 3, too much detail.  There are only so many pixels on a piece of paper.  If there is too much important stuff, some will get lost.  Think of the Mandelbrot set (not a function but it gives the idea)

5) Too much global variation.  What do you do with $e^{1000x}\sin x?$

I think the real answer is that almost no functions are plottable, but the ones we use are selected to usually be plottable.  This is like the fact that almost every function is discontinuous, but most of the ones you see in class will be continuous.

Title: Criteria for a function to be plottable


问:
Odds of getting at least one triple consecutive wins.
Given P(winning game) = p and P(losing game) = 1-p.

If you play 10 rounds of games only, what is the probability of getting at least one triple consecutive wins? in terms of p.

Where WWWLLLLLLL = one triple consecutive win.

答:
It is easier to calculate the probability of no three consecutive wins.

To solve it you should only notice that any 10 games that have no three consecutive wins must start with "L", "WL", "WWL".

I think you can now proceed using recursion.


问:
Find the expression for $f(x)$ an expression for $f(g(x))$
I fail to understand something fundamental about universal statements and sets.

Suppose we have a function $f$, such that $\forall x \in \mathbb{R}\:f(\frac34x)=2x^2$

I want to find the expression for $f(x)$.

$\forall x \in \mathbb{R}\: \exists x'=\frac34x$. Substitute $x'$ into the function to get $f(x')=\frac{32x'^2}{9}$, and it would seem that I can substitute $x'$ to $x$ getting $f(x)=\frac{32x^2}{9}$, but why? With this substitution I'm replacing the whole domain of $f$ from $\mathbb{R}$ to the set $\left\{ x' | x'=\frac34x, x \in \mathbb{R} \right\}$. Should I now show that the set of all $x'$ equals $\mathbb{R}$?

答:
Yes, that's correct.  Your work so far proves that $f(x')=32x'^2/9$ for all $x'$ such that $x'=3x/4$ for some $x\in\mathbb{R}$.  This only completely determines $f$ if you can prove that every $x'$ in the domain of $f$ (which I assume is supposed to be $\mathbb{R}$) actually can be written as $3x/4$ for some $x\in\mathbb{R}$.

The following alternative example may be instructive.  Suppose you have a function $g:\mathbb{R}\to\mathbb{R}$ such that $g(x^2)=2x^2$ for all $x\in\mathbb{R}$.  Then you cannot conclude that $g(x)=2x$ for all $x\in\mathbb{R}$.  You can only conclude this for such $x$ that are the square of a real number, i.e. for $x\geq 0$.  You have no information at all about the value of $g(x)$ if $x<0$.


问:
Logic behind addition of two intervals
I would like to know the logic behind adding two intervals $$[a, b]$$ $$[c, d]$$ together and for the sake of simplicity let's say that the numbers in the interval are just integers and not real numbers. I know that the idea is to just add the lowest and highest of the two $$[a+c, b+d]$$ and this is the result of the addition.

My question is whether this is just defined to be like this mathematically or is there a real world example that can shed some light on why it is calculated so?

答:
Here's a proof that if $r$ is in $[a+c,b+d]$ then $r=x+y$ for some $x$ in $[a,b]$ and some $y$ in $[c,d]$:

First, note that $a+c\le r\le a+d$ or $a+d\le r\le b+d$.

In the former case, there must be $y$ in $[c,d]$ such that $r=a+y$. Let $x=a$.

In the latter case, there must be $x$ in $[a,b]$ such that $r=x+d$. Let $y=d$.

Note: I'm not sure that this engages with OP's concerns. And I suspect that what I've written here has been written before on this site, probably multiple times. But it does seem to speak to some of the comments on the question.

Title: Logic behind addition of two intervals


问:
Sturm-Liouville operator completeness
Can someone please provide a reference to a proof of the completeness of the eigenvectors of a Sturm-Liouville operator, it would also be great if the aforementioned proof relies on operator theory and functional analysis. Thanks in advance.</p>

答:
The completeness of the eigenvectors of a Sturm-Lioiville operator follows immediately from the spectral theorem for compact operators.

A Sturm-Liouville operator always (when zero is not an eigenvalue, which can always be arranged) has a right inverse, the Green's function operator (also called the solution operator). This Green's function operator is a compact and self-adjoint operator.

If you want to see the concrete definition of the Green's function operator (and the proof that it is a right-inverse), you can check "Linear Functional Analysis" by Rynne and Youngson, Section $8.4$ in the $2$nd edition.

Title: Sturm-Liouville operator completeness


答:
The difference is subtle. I think it helps to realize that a (probability) distribution is a more general concept. A distribution is something intrinsic to a random variable and determines/describes the behaviour of it. In contrast, a probability mass function (PMF) is one way to concretely describe the distribution quantitatively. Things to note:

In general, a distribution does not necessarily have a PMF or a probabilty density function (PDF). However, a discrete distribution always has a PMF, and a PMF always describes a discrete distribution, and therefore the two things are basically the same.

You should think of a PMF as just one way to describe/define a dicrete distribution. We could also use the cummulative density function for example, or any other exotic description we can think of.

Maybe you can very loosely think like this: The distribution is what carries out the random experiments, and it might look at the PMF to decide what to do, or it might look at something else.

Title: What's the difference between a probability mass function and a discrete probability distribution?

Body: The Wikipedia page on "probability mass function" says "The probability mass function is often the primary means of defining a discrete probability distribution" (emphasis added).  I was under the impression that those two terms were completely synonymous.  What's the difference between them?


答:
As an experienced teacher, let me offer a few bits of advice.

By all means, do go to talks and see what makes some speakers effective (in general, perhaps not always to you) and what makes other speakers less so.

One of the arts of teaching is to anticipate where the audience will have difficulties. In large part, this comes with experience. But it's important to think about it when you're preparing your own lectures. If you've already seen that students have had issues understanding certain concepts or certain types of computations, try to address that head-on in your presentations.

Another mistake that many teachers make (sadly, even those with lots of experience) is to get mired in proofs/derivations before the students even have a context or any reason to care. Well-chosen examples done before the general theory help a lot. Indeed, I would say the same thing holds for you to understand a theorem: You really don't understand it until you can give me one or two examples (perhaps one of them trivial) to illustrate its significance and its application. This is part of the motivation to which others have referred. :)

Last, as far as teaching is concerned, letting the students know that you care about their learning and progress is huge. It's ok, once you establish rapport, to be honest about your disappointment in their test scores, etc., but they have to know that you want them to learn. And, Jeff, particularly in relatively low-level courses, they do not want to sit through your doing pedantic proofs that would satisfy you as a student. It's much more important for you to teach your precalculus and calculus students how to approach problems and how to write up understandable solutions of the problems. They don't need to know (let alone understand) the proof of the Maximum Value Theorem in Calc I; they do need to know how to give a solid application of it in the context of an applied max/min problem. They need to know how to set up functions in a logical way, think about their domains, understand why the function should have a maximum (other than just "the problem asked for it"), and then find it. (Many students these days have such weak algebra skills that those weak skills get in the way of so much more conceptually interesting stuff.)

And, last, I would concur with the folks on here encouraging you to answer more questions by way of learning to practice explaining things at different levels. Sometimes on here it's very hard to assess the level of the asker. I've noticed several graduate students (at least I think they are) giving explanations that would be appropriate for people at their level, rather than for the one who asked (who was clearly in an introductory undergraduate course).

OK, so my notion of "a few bits" is askew. :)

Title: Communicating mathematics

Body:
As a TA, my ratings have been fairly mediocre.  There's nothing to suggest I'm a problem, but I have room to improve.

What's more disturbing to me is that I even struggle to communicate math with people who know a lot more than I do.  It's no surprise that they struggle to communicate with me.  They know more, probably forgot what it's like to know less, and generally do not give the details on everything, sometimes even neglecting to mention that a particular detail needs to be filled in.  But when I communicate, I try to leave out as few details as possible and clearly enunciate when I am leaving them out, and to minimize the number of mistakes I make.  I am extremely nitpicky about this.

So there must exist other obstacles to communication other than just misrepresentations.  I suspect that what I'm lacking is a pedagogical element, which would also explain my interaction with students.  How can I go about finding out what is lacking, and then fixing it?  What are some common teaching mistakes made by people with the attitude that mathematics should be precise and airtight?  It's not that I'm such an adherent of my own philosophy that I refuse to "dumb things down."  That used to be the problem, but it appears that when I attempt to convey vague ideas, people do not understand me.


问:
Check if two rotated cuboids overlap?
Is there a general approach towards finding the intersection over union between two rotated cuboids? Or even a method that checks whether two rotated cuboids overlap?

At the moment, I sample points along each of the 12 edges of a cuboid and then check whether any of these points are inside the other cuboid. However, this is not an elegant solution and I was looking for something more mathematically straightforward. Is there any such method in math that can tell whether two rotated cuboids overlap?

答:
Describe each cuboid using its center $\vec{c}$, orthonormal basis $\hat{e}_1$, $\hat{e}_2$, $\hat{e}_3$, where
$$\hat{e}_i \cdot \hat{e}_j = \begin{cases} 1, & i = j \\ 0, & i \ne j \end{cases}$$
halved length/size $h_1$, $h_2$, and $h_3$ along each basis vector,
and circumradius $r = \sqrt{h_1^2 + h_2^2 + h_3^2}$. This means the vertices are
$$\begin{aligned}
\vec{v}_0 &= \vec{c} + h_3 \hat{e}_3 + h_2 \hat{e}_2 + h_1 \hat{e}_1 \\
\vec{v}_1 &= \vec{c} + h_3 \hat{e}_3 + h_2 \hat{e}_2 - h_1 \hat{e}_1 \\
\vec{v}_2 &= \vec{c} + h_3 \hat{e}_3 - h_2 \hat{e}_2 + h_1 \hat{e}_1 \\
\vec{v}_3 &= \vec{c} + h_3 \hat{e}_3 - h_2 \hat{e}_2 - h_1 \hat{e}_1 \\
\vec{v}_4 &= \vec{c} - h_3 \hat{e}_3 + h_2 \hat{e}_2 + h_1 \hat{e}_1 \\
\vec{v}_5 &= \vec{c} - h_3 \hat{e}_3 + h_2 \hat{e}_2 - h_1 \hat{e}_1 \\
\vec{v}_6 &= \vec{c} - h_3 \hat{e}_3 - h_2 \hat{e}_2 + h_1 \hat{e}_1 \\
\vec{v}_7 &= \vec{c} - h_3 \hat{e}_3 - h_2 \hat{e}_2 - h_1 \hat{e}_1 \\
\end{aligned}$$
and its volume is $8 h_1 h_2 h_3$.

Then, if the centers are further away than the sum of the circumradiuses,
$$\left\lVert \vec{c}_A - \vec{c}_B \right\rVert \gt r_A + r_B \tag{1a}\label{G1a}$$
which is easily calculated by squaring both sides,
$$(\vec{c}_A - \vec{c}_B) \cdot (\vec{c}_A - \vec{c}_B) \gt (r_A + r_B)^2 \tag{1b}\label{G1b}$$
the cuboids cannot intersect (overlap) at all.

Otherwise, we can check if any of the vertices of one cuboid are inside the other cuboid.  Vertex $\vec{v}$ is inside, if and only if
$$-h_i \le \hat{n}_i \cdot ( \vec{v} - \vec{c} ) \le h_i \tag{2}\label{G2}$$
is true for all $i = 1, 2, 3$.  If at least one vertex of one cuboid is inside the other cuboid, then the two do intersect.

Note that because of how the vertices are defined, and because dot product is distributive over vector addition, you only need to calculate 15 dot products (and not 24).

Otherwise, we need to examine if any edges of one cuboid intersects any face in the other cuboid.  (We do need to test it both ways; it is not sufficient to test all twelve edges of one cuboid against the six faces of the other cuboid.)

Let the edge be between $\vec{w}$ and $\vec{w} + m \hat{\epsilon}$ (of cuboid B, the other vectors and $h_k$ referring to those of cuboid A). Calculate
$$\begin{aligned}
p_k & = \hat{e}_k \cdot (\vec{w} - \vec{c}) \\
q_k & = \hat{e}_k \cdot (\vec{w} - \vec{c} + m \hat{\epsilon}) \\
\end{aligned} \quad k = 1, 2, 3 \tag{3a}\label{G3a}$$
and then test the following points if they're contained in cuboid A,
$$\begin{array}{lll}
\vec{w} - \displaystyle m \frac{h_k + p_k}{q_k - p_k} \hat{\epsilon}, & \text{if} \quad p_k \lt -h_k \lt q_k \\
\vec{w} - \displaystyle m \frac{h_k + q_k}{p_k - q_k} \hat{\epsilon}, & \text{if} \quad q_k \lt -h_k \lt p_k \\
\vec{w} + \displaystyle m \frac{h_k - p_k}{q_k - p_k} \hat{\epsilon}, & \text{if} \quad p_k \lt h_k \lt q_k \\
\vec{w} + \displaystyle m \frac{h_k - q_k}{p_k - q_k} \hat{\epsilon}, & \text{if} \quad q_k \lt h_k \lt p_k \\
\end{array} \tag{3b}\label{G3b}$$
again for $k = 1, 2, 3$, but note that the points are on the faces of cuboid A (the one(s) perpendicular to $\hat{e}_k$), so you should ignore the face pair $i = k$ in the $\eqref{G2}$ test (since that is true by definition, as the above finds the point where the edge intersects with a face plane), and only test the other two.  If true, the cuboids intersect.

Title: Check if two rotated cuboids overlap?


问:
Why is the solution to the particular of $y^{(\text{iv})}+2y′′+y=x \cos x$ equal to $x^{2} \sin(x)+x^{3} \cos(x)$
I was wondering why the cosine in this particular solution has $x$ to power 3. I don't understand why this is the case.

答:
$$y^{iv}+2y′′+y=x \cos x$$
We first try:
$$y_p=(ax+b) \cos x +(cx+d) \sin x$$

But since the homogeneous solution is:
$$r^4+2r^2+1=(r^2+1)^2=(r-1)^2(r+i)^2=0$$ $$\implies r=\pm i$$
$$y_h=c_1 \cos x +c_2 \sin x +c_3 x \cos x +c_4x\sin x$$
We need to multiply our guess by $x^2$ so that we have :
$$y_p=x^2(ax+b) \cos x +x^2(cx+d) \sin x$$
Since $y_p$ is odd $(y_p(x)=-y_p(-x))$ we can simplify the guess as:
$$y_p=ax^3 \cos x +d x^2\sin x$$
Plug this in the differential equation and find the constants $a$ and $d$.
You should find;
$$\boxed {y_p=-\dfrac {x^3}{24}\cos x +\dfrac {x^2}8 \sin x}$$


问:
Can we prove $|A| < \left|2^{2^A}\right|$ constructively?
Can we prove that there is no:

Injection $2^{2^A} \to A$

Surjection $A \to 2^{2^A}$

Bijection between $A$ and $2^{2^A}$

Constructively?

答:
You can show that there is no injection from $\mathcal{P}(\mathcal{P}(A))$ to $A$ by using Cantor's theorem $|A|<2^{|A|}$. As a consequence, there is no bijection between $\mathcal{P}(\mathcal{P}(A))$ and $A$. Since Cantor's theorem does not require any choice, we can prove the absence of injection without choice.

For the surjection part, use the following fact that is provable without choice:

If there is a surjection from $X$ to $Y$ then $2^{|Y|}\le 2^{|X|}$.

(The proof uses inverse image: if $f:X\to Y$ is a surjection, consider $f^{-1}:\mathcal{P}(Y)\to\mathcal{P}(X)$.)

If there is a surjection from $X$ to $\mathcal{P}(\mathcal{P}(X))$, then $2^{2^{2^{|X|}}}\le 2^{|X|}$, but we have $2^{2^{|X|}}<2^{2^{2^{|X|}}}$, so $2^{2^{|X|}}<2^{|X|}$, a contradiction.

As @bof pointed out, Cantor's theorem already excludes the surjection from $A$ to $\mathcal{P}(A)$, so my previous argument is verbose. Also, the OP also asked these facts hold constructively, and the answer is yes (at least over $\mathsf{IZF}$.)

Before answering, I have to point out some subtlety that came from the notation: the OP (supposedly) uses $2^A$ to denote the power set. It makes sense if we are working with classical logic. However, the power set of $A$ need not be bijective with $2^A$ constructively. In general, $\mathcal{P}(A)$ is bijective with $\Omega^A$ for $\Omega=\mathcal{P}(1)$. (Here $A^B$ be the set of all functions from $A$ to $B$.)

The usual proof of Cantor's theorem also works constructively, so $\mathsf{IZF}$ also proves there is no surjection from $A$ to $\mathcal{P}(A)$. Moreover, a slight modification of the proof of Cantor's theorem also shows there is no injection from $\mathcal{P}(A)$ to $A$: assume that there is such an injection $f$. Let
$$X=\{a\in A \mid \exists b\in\mathcal{P}(A) [f(b)=a\land a\notin b]\}.$$
We can see easily that $f(X)\in X$ implies $f(X)\notin X$. If $f(X)\notin X$, then there is no $b\in\mathcal{P}(X)$ such that $f(b)=f(X)$ and $f(X)\notin b$, but $b=X$ satisfies that condition, a contradiction.

However, the answer changes if we stick to $2^A$ instead of $\mathcal{P}(A)$: we can still show that there is no surjection between $A$ and $2^A$. The usual proof for Cantor's theorem still works. However, it is consistent with $\mathsf{IZF}$ that there are just two decidable subsets of $\mathbb{R}$. (A subset $X\subseteq A$ is decidable if, for every $a\in A$, we have either $a\in X$ or $a\notin X$.) In that case, $2^{\mathbb{R}}\simeq 2$, hence $2^{2^{\mathbb{R}}}$ is embeddable into $\mathbb{R}$.


问:
Is this a characteristic function for some random variable?
I am aware that $\phi(t) = \cos(t) $ is the ch.f. of the random variables that takes values $\pm 1 $ with probability 1/2 each.

My question is about the following function:

$$ \phi(t) = \begin{cases} \cos(t) &\text{ if } -\pi/2 \le t \le \pi/2\cr  0 &\text{ otherwise. }
\end{cases} $$

How can one conclude that this is either a ch.f. or not? I believe that a ch.f. for a discrete random variable on a lattice, this has to be periodic. If I compute the moments using my definition of $\phi(t), $ it all works, but I cannot really tell if there is some reason to say that it is not a ch.f.   Thank you for any insight.

Maurice

答: It is not: Suppose for the sake of contradiction that $\phi$ is the characteristic function of some random variable $X$. Since $\phi$ has a second derivative at $0$, $X$ has a finite second moment (see, e.g., If the characteristics function of a random variable is differentiable even times then it has finite moment of even order). However, this would imply that $\phi$ is twice-differentiable globally, which is clearly not the case. Hence such an $X$ cannot exist.


问:
$\int \frac{\ln(1-\sin(2x))}{\sin(2x)}\, dx$
$$\int \frac{\ln(1-\sin(2x))}{\sin(2x)}\,\mathrm{d}x$$

I tried to find the solution through WolframAlpha and I got the result, but I do not know how the solution is:
$$1/2 (2 \operatorname{Li}_2(1 - \tan(x)) + \operatorname{Li}_2(-i \tan(x)) + \operatorname{Li}_2(i \tan(x)) + \log(1 - i \tan(x)) \log(\tan(x)) + \log(1 + i \tan(x)) \log(\tan(x)) + \log(1 - \sin(2 x)) \log(\tan(x))) + \operatorname{const.}$$

答:
We want to prove that
$$I := {\displaystyle\int}\dfrac{\ln\left(1-2\sin\left(x\right)\right)}{\sin\left(2x\right)}\,\mathrm{d}x = -\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)-1\right)}{4}-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)\right)}{2}-\dfrac{\ln\left(1-2\sin\left(x\right)\right)\ln\left(2-2\sin\left(x\right)\right)}{4}-\dfrac{\ln\left(3\right)\ln\left(-2\sin\left(x\right)-2\right)}{4}+\dfrac{\operatorname{Li}_2\left(-\frac{-2\sin\left(x\right)-2}{3}\right)}{4}.$$

Rewrite

$$={\displaystyle\int}-\class{steps-node}{\cssId{steps-node-1}{2\cos\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{\ln\left(1-2\sin\left(x\right)\right)}{\left(-2\sin\left(x\right)-2\right)\left(2-2\sin\left(x\right)\right)\sin\left(x\right)}}}\,\mathrm{d}x$$
With the substitution $u=1-2\sin(x)$ we get to
$$-\class{steps-node}{\cssId{steps-node-3}{2}}{\displaystyle\int}\dfrac{\ln\left(u\right)}{\left(u-3\right)\left(u-1\right)\left(u+1\right)}\,\mathrm{d}u.$$
By using partial fraction decomposition
$$\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{8}}}\underbrace{\int\dfrac{\ln\left(u\right)}{u+1}\,\mathrm{d}u}_{I_1}-\class{steps-node}{\cssId{steps-node-5}{\dfrac{1}{4}}}\underbrace{{\int}\dfrac{\ln\left(u\right)}{u-1}\,\mathrm{d}u}_{I_2}+\class{steps-node}{\cssId{steps-node-6}{\dfrac{1}{8}}}\underbrace{{\int}\dfrac{\ln\left(u\right)}{u-3}\,\mathrm{d}u}_{I_3}.$$

$I_1$

Partial integration yields
$$I_1=\ln\left(u\right)\ln\left(u+1\right)-{\displaystyle\int}\dfrac{\ln\left(u+1\right)}{u}\,\mathrm{d}u$$
If you substitute $v=-u$ in the integral, this is equal to $\operatorname{Li_2}(v)$. Substituting back yields
$$I_1=\ln\left(u\right)\ln\left(u+1\right)+\operatorname{Li_2}(-u).$$

$I_2$

Substitute $v=u-1$. This will get us to a very similar integral we calculated in $I_1$. In conclusion
$$\operatorname{Li_2}(1-u)$$

$I_3$

After substituting $v=u-3$:
$$I_3={\displaystyle\int}\left(\dfrac{\ln\left(\frac{v}{3}+1\right)}{v}+\dfrac{\ln\left(3\right)}{v}\right)\mathrm{d}v$$
We can split the integral up. The first integral can be evaluated with the same methods as earlier. This will yields $-\operatorname{Li}_2\left(-\dfrac{v}{3}\right)$. The second integral easily evaluates to $\ln 3\ln v$.
$$I_3=\ln\left(3\right)\ln\left(u-3\right)-\operatorname{Li}_2\left(-\dfrac{u-3}{3}\right).$$

$I$

Putting all of the above together:
$$I=-2\left(\dfrac{\ln\left(u\right)\ln\left(u+1\right)}{8}+\dfrac{\operatorname{Li}_2\left(-u\right)}{8}+\dfrac{\ln\left(3\right)\ln\left(u-3\right)}{8}-\dfrac{\operatorname{Li}_2\left(-\frac{u-3}{3}\right)}{8}+\dfrac{\operatorname{Li}_2\left(1-u\right)}{4}\right).$$
Now remember $u=1-2\sin x$.

$$I=-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)-1\right)}{4}-\dfrac{\operatorname{Li}_2\left(2\sin\left(x\right)\right)}{2}-\dfrac{\ln\left(1-2\sin\left(x\right)\right)\ln\left(2-2\sin\left(x\right)\right)}{4}-\dfrac{\ln\left(3\right)\ln\left(-2\sin\left(x\right)-2\right)}{4}+\dfrac{\operatorname{Li}_2\left(-\frac{-2\sin\left(x\right)-2}{3}\right)}{4}.$$


问:
How to calculate $p$-value in a clinical trial
I'm trying to verify the $p$-value mentioned in this article:

$8\%$ ($n=\frac{1}{13}$) of mavrilimumab-treated patients progressed to mechanical ventilation by Day $28$, compared to $35\%$ ($n=\frac{9}{26}$) of control-group patients who progressed to mechanical ventilation or died ($\textbf{p=0.077}$).

How can I calculate the $p$-value for the data above?

I used $3$ different calculators and they all gave different results, none of them matches with the article.

Calculator 1: $p=0.12$

Calculator 2: $p =  0.06876$

Calculator 3: $p = 0.154$

答:
Without knowing which test they chose, and whether the test was stratified or adjusted by any covariates, it is unlikely that we can replicate the $p$-value.

For instance, we could use Fisher's exact test, the chi-squared test (which is the same as the two-sample independent proportion test with a pooled standard error), or a two-sample independent proportion test with an unpooled standard error, or a likelihood ratio test, and these are just different choices of statistic, not considering whether continuity correction is used, or if there was some other adjustment for other prognostic factors that are not stated in the press release.

That said, I would not be surprised if there was an error.  Small biotech companies don't always perform the analyses correctly.  But to really know for sure, one would have to read the study protocol and the statistical analysis plan, neither of which is generally available to the public.


问:
Prove $2^n - 1$ is NOT a prime number, when n is NOT a prime number also
I started with $n = 2k$ , $2^{2k} - 1 = (2^k - 1)(2^k + 1)$, where i go from there...

答:
Welcome to MSE!

Hint:

If $n = kl$ shows $n$ is composite, then we can write

$$
2^n - 1 = 2^{kl} - 1 = (2^k)^l - 1^l
$$

But do you know of a way to factor $a^l - b^l$?

I hope this helps ^_^


Title: Prove $2^n - 1$ is NOT a prime number, when n is NOT a prime number also

Body: I started with $n = 2k$ , $2^{2k} - 1 = (2^k - 1)(2^k + 1)$, where i go from there...

Body:
$ n $ not prime $ \implies $

$$\exists p,q>1 \;\;:\;\; n=pq\implies$$

$$2^n-1=(2^p)^q-1$$
$$=(2^p-1)(1+2^p+2^{2p}+...+2^{(q-1)p})$$

is not prime.


问:
When will $M$ be an $R_n$ module?
Problem: Let $n$ be an integer and let $R_n\subset \mathbb{Q}$ consist of rational numbers whose denominator is divisible by primes dividing $n$. Let $m$ be a different integer, and consider $M$, the cyclic group with $m$ elements. When will $M$ be a $R_n$-module?

Claim: $M$ is a $R_n-$module when $\gcd(m,n)=1$.

Case 1: $\gcd(m,n)=1$.
So $R_n=\mathbb{Z}[\frac{1}{n}]=\{\frac{a}{b} \mid b=p_1 p_2\cdots p_k, p_i|n \,\forall i\}$. If $\gcd(m,n)=1,$ I want to show that $\frac{a}{b}x\in M$ for any $x\in M$. $$\frac{a}{b}x=\frac{ax}{p_1 p_2\cdots p_k}.$$ Since $m, n$ are coprime, then that means the don't share any prime factors so for each $p_i\in M, \exists v_i\in M$ s.t. $p_iv_i=1\implies \frac{1}{p_i}=v_i$ in $M$. Now that scaling by $\frac{1}{n}$ is well defined in $M$, then I can rewrite  $$\frac{ax}{p_1 p_2\cdots p_k}=ax\Pi_{i=1}^{n} v_i\in \mathbb{Z}.$$ So thus $\frac{a}{b}x\in M$ since $M$ is a $\mathbb{Z}-$module.

Case 2: $\gcd(m,n)\neq 1$.

Now $\frac{a}{b}x=\frac{ax}{p_1 p_2\cdots p_k}$. There is at least one $p_i$ s.t. there is no $v_i$ in $M$ s.t. $p_i$ is a unit. So $\frac{a}{b}x\notin \mathbb{Z}\implies \frac{a}{b}x\notin M$.

I'm wondering if it's okay that I treated $M$ like $\mathbb{Z}/m\mathbb{Z}$ in this case? I was given that $M$ is a $\mathbb{Z}-$module. I'm questioning this implication: $p_iv_i=1\implies \frac{1}{p_i}=v_i$. Could someone provide feedback on my argument please?

答:
An abelian group $ M $ has a natural structure as a $ R_n $-module if and only if multiplication by $ n : M \to M $ is invertible. In particular, when $ M $ is cyclic of order $m$, multiplication by $n$ is invertible if and only if $ \operatorname{gcd}(m,n)=1 $.

The reason for the above boils down to the fact that $R_n$ is $\mathbb{Z}[1/n]$, i. e. the ring $ \mathbb{Z} $ with an inverse of the element $n$ adjoined. This ring has a universal property, namely that a homomorphism of rings $\varphi \colon \mathbb{Z} \to A $ factors through a map $ \mathbb{Z}[1/n] \to A $ if and only if $ \varphi(n) $ is invertible in $ A $. Interpreting a $\mathbb{Z}[1/n]$-module structure as a ring homomorphism $\mathbb{Z}[1/n] \to \operatorname{End}_{\mathbb{Z}}(M) $, the result is immediate.

To put this into a more general context, the process of passing from $\mathbb{Z}$ to $\mathbb{Z}[1/n]$ is known as localization. For a commutative ring $ A $ and multiplicative subset $ S \subseteq A $ one form the localized ring $ A[S^{-1}] $, informally described as the ring $ A $ with inverses of every element of $ S $ adjoined. One then has the general result that the category of $ A[S^{-1}]$-modules is equivalent to the category of $ A $-modules on which multiplication by every element of $ S $ is invertible.

Title: When will $M$ be an $R_n$ module?


问:
Prove this solution is genuine
I cannot prove that the solution to :$$y'=-\frac{x\sqrt{1-y^2}}{y\sqrt{1-x^2}}$$

is:

$$y=\sqrt{x^2-2C\sqrt{-x^2+1}-C^2},\:y=-\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}$$

By substituting the solution(s) back into the original equation. I have taken derivative and attempted to equate LHS with RHS but I just can't simplify enough to obviously justify the solution(s) $$\frac{x\left(C+\sqrt{-x^2+1}\right)}{\sqrt{-x^2+1}\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}}$$

答:
Square the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$

Taking the derivative:
$$2yy'= 2x+2C\frac{x}{\sqrt{1-x^2}}$$

So
$$yy' = x + C\frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{1-y^2}}{\sqrt{1-x^2}}$$

Assuming $x$ is not $0$, then
$$\sqrt{1-x^2} + C=\sqrt{1-y^2}$$

Squaring both sides:
$$1-x^2+2C\sqrt{1-x^2}+C^2=1-y^2$$

Then,
$$y^2=x^2-2C\sqrt{1-x^2}-C^2$$

QED

================

The solution to this ODE is by first noticing it is separable. Then you arrive at the equation:
$$-\sqrt{1-y^2}=\sqrt{1-x^2}+C$$

This gives you the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$

Title: Prove this solution is genuine


问:
A proof that if $f_n \rightarrow f$ in measure and $g_n \rightarrow g$ in measure, then $f_n\cdot g_n \rightarrow f\cdot g$.
The problem is to prove that if $f_n \rightarrow f$ in measure and $g_n \rightarrow g$ in measure, then $f_n\cdot g_n \rightarrow f\cdot g$. Where these functions are defined on a set that has a finite measure. The portion I am confused about is when they say $$\{x: |f_n(x)| > M+1\} \subset \{x: |f(x)| > M \} \cup \{x: |f(x)-f_n(x)|
\geq 1\},$$ this is on page (1) of the file. I feel like this should be simple, and it makes sense intuitively, however when trying to prove this rigorously I fail.

Link: https://www.math.arizona.edu/~friedlan/teach/523/sol.pdf

Any help or hints would be much appreciated.

Edit: Sorry I just realized is this, but is it because we are considering the case for when $|f(x)| > M$, where the definition of $M$ is given in the previous lines in the proof.

答:
This can definitely be a fiddly question but hopefully I can help you out. I do not know your mathematical level so please forgive me if anything here is patronising I shall talk through things throughly.

Let us come up with a way to show generally that:

$\{x: \text{Condition 1 holds} \} $ $ \subset \{x: \text{Condition 2 holds} \} \cup \{x: \text{Condition 3 holds} \} \ $

In general to show one set is a subset of another mathematicians tend to show if anything is in the first set then it has to belong to the other!

In our case that means that if ever Condition $1$ holds then at least one of Condition 2 or 3 must hold as union means or.

One one to do this would be via a contradiction. That is to assume Condition $1$ holds and that neither Condition $2$ or $3$ hold and then arrive at a contradiction. i.e showing it impossible for Condition $1$ to hold without Condition $2$ or $3$

In our case:

Condition $1$ is $ |f_n(x)| > M+1$

Condition $2$ is $|f(x)|>M$

Condition $3$ is $|f(x)-f_n(x)| \geq 1$

Let us proceed by assuming Condition $1$ holds and neither $2$ or $3$, hoping to find a contradiction.

Firstly we have that $|f_n(x)| = |f_n(x) -f(x) +f(x)|$

This adding then instantly subtraction $f(x)$ may seem odd but I promise it will help us out!

Via the triangle inequality, that is $|a-b| \leq |a| + |b|$ we have that:

$|f_n(x)| = |f_n(x) -f(x) +f(x)| \leq |f_n(x)-f(x)| + |f(x)| $

However as we assumed both Condition $2$ and $3$ are false we have that:

$|f(x)|\leq M$ and  $|f(x)-f_n(x)| < 1$

Hence:

$|f_n(x)| = |f_n(x) -f(x) +f(x)| \leq |f_n(x)-f(x)| + |f(x)| < M+1 $

i.e

$f_n(x) < M+1$

However this violates Condition $1$!

Hence if $ |f_n(x)| > M+1$ then either $|f(x)|>M$ or $|f(x)-f_n(x)| \geq 1$ (or both)

Hope this helps,

Oskar

Edit the tricky part of this question would be showing we have a proper subset and not $\subseteq$ that is there are some points where condition $2$ or $3$ hold without condition $1$ just on the assumption that they are measurable functions. Is this needed for you?


问:
Convex function is left-hand and right-hand differentiable.
I've come across a proof on ProofWiki (link here) which aims to show that a convex function $f$ is left-hand and right-hand differentiable on an open interval $(a,b)$. It uses that convex function inequality to show that the function $$F(h) = \frac{f(x+h) - f(x)}{h}$$ for $x \in (a, b)$ is increasing on an interval $(0, \delta)$ for some $\delta > 0$. However, the article proceeds by making the claim that from the Theorem of Limit of Increasing Function, it follows that the function $F$ has a left-hand and right-hand limit in $0$. But from my understanding of the cited theorem, this claim is not trivial because $0 \notin (0, \delta).$ The Limit of Increasing Function only claims that if $F$ is increasing on an open interval $(a,b)$, then $F$ has a a right-hand and left-hand limit for every point INSIDE the interval $(a,b)$.

My first instinct is to doubt my judgement because the proof is provided with citations and a flawed article probably wouldn't be allowed to stay on the site. So what am I missing? I would greatly appreciate if anyone could point the error in my reasoning. If I happen to be right, how should the proof be continued?
Thanks for all the help.

答:
The proof in the referenced article is correct, but does not mention explicitly that both monotony and boundedness is used to prove the existence of the one-sided limits.

The precise argument goes as follows: For $x \in (a, b)$ we can choose a $\delta > 0$ such that $(x-\delta, x+\delta) \subset(a, b)$. The convexity condition then implies that
$$
\frac{f(x+h_1) - f(x)}{h_1} \le\frac{f(x+h_2) - f(x)}{h_2} \le \frac{f(x+k_1) - f(x)}{k_1} \le \frac{f(x+k_2) - f(x)}{k_2}
$$
for $-\delta < h_1 < h_2 < 0 < k_1 < k_2 < \delta$.

This shows that $\frac{f(x+h) - f(x)}{h}$ is

increasing and bounded above on $(-\delta, 0)$, so that $\lim_{h \to 0-}\frac{f(x+h) - f(x)}{h} = f'_-(x)$ exists, and

increasing and bounded below on $(0, \delta)$,  so that $\lim_{h \to 0+}\frac{f(x+h) - f(x)}{h} = f'_+(x)$ exists.

It also shows that the limits satisfy $f'_-(x) \le f'_+(x)$.


问:
looking for combinatorial identity
Does anybody know a nicer expression for the sum $\sum_{i=0}^{k} \binom{n}{i} \binom{n-i}{k-i}$?

Title: looking for combinatorial identity

Body:
$$
\sum_{i=0}^{k} \binom{n}{i} \binom{n-i}{k-i}
=\sum_{i=0}^{k} \binom{n}{k} \binom{k}{i}
=\binom{n}{k} \sum_{i=0}^{k} \binom{k}{i}
=\binom{n}{k} 2^k
$$