问:
How do you solve this type of definite integral?
I'd really appreciate some help with solving this problem (specifically parts b, c, and d)! I'm not quite sure how to solve this problem; I do not really understand the rules for solving definite integrals.. I'd be grateful if someone could help me with this!

答:
The integral defines a linear operator, so you can split the total integral as
$$\int_0^7f(x)dx=\int_0^5f(x)dx+\int_5^7f(x)dx=20+12=32.$$
In general, given $a,b\in\mathbb R$ with $a<c<b$ , then
$$\int_a^b f(x)dx=\int_a^cf(x)dx+\int_c^bf(x)dx.$$

Try to think about the other requests using the property of linearity of the integral.


问:
Express the following power series as a raional function
Consider the following power series:

$f(x) = \sum\limits_{i>=1} 2^{i-1}x^{3i}$ = $\ x^3 + 2x^6 + 4x^9 + ...$

$g(x) = \sum\limits_{i=2}^{20} f(x)^{i}$

Express both f(x) and g(x) as rational functions, ie p(x)/q(x) where P(x) and q(x) are polynomials.

How would I go about doing this? Is there some kind of algorithm or pattern I can follow?

ContentLicense: CC BY-SA 3.0

Body:
Hints:

$$2^{i-1}x^{3i}=\frac12\left(2x^3\right)^i$$

So assuming

$$2|x|^3<1\;\;\text{(why?)}\;,\;\;\text{we get}\;\;\sum_{k=1}^\infty2^{i-1}x^{3i}=\frac12\frac{2x^3}{1-2x^3}$$


问:
Prove that the lower Riemann integral of $f$ on $[a, b]$
Prove that the lower Riemann integral of $f$ on $[a, b]$ is $\sup\{\int_a^b\psi: \psi \text{ is a step function on } [a, b],\text{ such that } \psi\leq f \text{ on }[a, b]\}$.

The definitions that I am working with are:

For a partition $P = \{x_0, ..., x_n\}$ of $[a,b]$ with $a = x_0 < x_1 < ... < x_n = b$. Then
$\int_a^b\psi = \sum_{j=1}^{n}\psi_{j}(x_j - x_{j-1})$ where $\psi_i$ is the constant value of $\psi$ on the interval $\psi$ on $[x_{j-1}, x_j]$.

The Riemann integral of $f$ is defined to be (using the partition $P$ above)
$\int_a^b f = L(f,[a,b]) = U(f,[a,b])$ where
$L(f,[a,b]) = \sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f~~,
U(f,[a,b]) = \sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f
 ~$are the lower and upper Riemann sums respectively.

I was trying to show that the supremum over the step functions has to be equal to the lower sum/ the constant value $\psi_j = \inf f$ on any of our sub intervals, but going from here to proving that the sums are equal I was running to to troubles keeping my epsilons straight.

Basically, I wanted to show that for our step functions we need that the step function value needs to be equal to the $\inf f$ on any of the sub intervals and from there we'd hopefully get the desired equality.

Any help on if this is the correct direction to go/ what a clean proof looks like is appreciated. Thanks =)

答:
You want to show two things:

For every step function $\psi$ with $\psi \le f$ we have $\int \psi \le \underline{\int} f$. And to do this, simply define a partition so that $\psi$ is constant on each interval $(x_i, x_{i+1})$ (we don't need to worry about the endpoints of the intervals but do check this).

For every $\varepsilon$, we can find a step function $\psi$ such that $\underline{\int} f \le \int \psi + \varepsilon$. And to do this, you can do exactly as you say: define $\psi_i = \inf_{[x_i, x_{i+1}]} f(x)$ and define $\psi$ piecewise using the constants $\psi_i$.

For 1. you want to check that $\int \psi \le L(f, \mathcal{P})$ for any sufficiently small partition (refining the partition given by $\psi$). It follows then that $\int \psi \le \sup_{\mathcal{P}} L(f, \mathcal{P}) = \underline{\int} f$.

For 2. you can check that $\int \psi = L(f, \mathcal{P})$ by definition and by definition of $\underline{\int} f = \sup_{\mathcal P} L(f, \mathcal{P})$ there exists a partition $\mathcal{P}$ such that $\underline{\int} f \le L(f, \mathcal{P}) + \varepsilon$ (that's the definition of $\sup$).

This is a common strategy: if you want to show $A = B$ you can show that $A \le B$ and $B \le A + \varepsilon$ for every $\varepsilon > 0$.

ContentLicense: CC BY-SA 4.0


答:
For (b), use that $$\int_a^b f(x)\ dx = - \int_b^a f(x) \ dx$$ and think about why this is true!

For (c), you are integrating over a single point $\{5\}$. Geometrically, what does integration represent? So what must the answer be in this case?

For (d), you can take the constant factor out by linearity of the integral.

Title: How do you solve this type of definite integral?

Body: I'd really appreciate some help with solving this problem (specifically parts b, c, and d)! I'm not quite sure how to solve this problem; I do not really understand the rules for solving definite integrals.. I'd be grateful if someone could help me with this!


答:
There are several relevant properties of integrals that you may find useful:

$$\int_{a}^{b}f(x)\,dx + \int_{b}^{c}f(x)\,dx = \int_{a}^{c}f(x)\,dx \tag{1}$$

$$\int_{a}^{b} \left(f(x) \pm g(x)\right)\,dx = \int_{a}^{b}f(x)\,dx \pm \int_{a}^{b}g(x)\,dx\tag{2}$$

$$\int_{a}^{a}f(x)\,dx = 0\tag{3}$$

$$\int_{b}^{a}f(x)\,dx = -\int_{a}^{b}f(x)\,dx\tag{4}$$

$$\int_{a}^{b}kf(x)\,dx = k\int_{a}^{b}f(x)\,dx\tag{5}$$

You should be able to solve all your problems with a combination of these.


问:
Can a knight visit every field on a chessboard?
I was doing excercises about graphs theory and I came across a quite interesting excercise (which probably has something to do with Hamiltonian Cycle):
"Is it possible to step on every field of a 4x4 or 5x5 chessboard just once and return to the starting point using a knight?"
Does anyone have any idea how to tackle this problem? I am more interested in a outline of how to do it or just some hints.</p>

答:
HINT:

$$\begin{array}{|c|c|c|c|} \hline
\cdot&&&\\ \hline
&&\cdot&\\ \hline
&\cdot&&\\ \hline
&&&\cdot\\ \hline
\end{array}$$


问:
theory of equation
let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
 ( c+d)^{-1}+ (c+d)^{-1}$$

my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck

答:
If you combine your result with that of B. Goddard, you have

$$\frac{1}{a+b}+\frac{1}{c+d} = \frac{3}{(a+b)(c+d)}$$

Now suppose $a=1+w, b=1+w^2, c=1+w^3, d=1+w^4$ where $w$ is one of the imaginary roots of $w^5=1$, then

$$(a+b)(c+d) = (2+w+w^2)(2+w^3+w^4) \\= 4 + 2(w+w^2+w^3+w^4) + (w^4 + 1 + 1 + w)=4+w+w^4$$
Similarly,
$$(b+c)(a+d)=1$$
$$(a+c)(b+d)=4+w^2+w^3$$

That's probably easier since both $w+w^4$ and $w^2+w^3$ are real. You will need $\cos \left(\frac{2\pi}{5}\right)=\frac{\sqrt 5 -1}{4}$

Edit: Actually you don't need the cosine. Just add $1/(4+w+w^4)$ and $1/(4+w^2+w^3)$ like the above it becomes very clean.

$$(4+w^2+w^3)(4+w+w^4)\\=16+4(w+w^2+w^3+w^4)+(w^3+w+w^4+w^2)\\=16-4-1=11$$

Can you end it now?

Title: theory of equation


问:
Let $M:=(\Bbb{Z}_{23},+,\cdot)$ the ring of integers modulo $23$. Is there a subset $T\subseteq M$ with $|T|=22$ such that $(T,\cdot)$ forms a group?
Let $M:=(\mathbb{Z_{23}},+,\cdot)$ the ring of integers modulo $23$. Is there a subset $T \subseteq M$ with $|T|=22$ such that $(T, \cdot)$ forms a group?

So I know what is asked here, but how to show that this exists (or I think that this does not exist?) because to show the existence one would have to prove the group axioms for each combination of the 22 elements?

答: Hint: Use Euler's Theorem and note that, since $23$ is prime, $\varphi (23)=23-1=22$.


问:
Indexing problem in array
Given an array A=[a1 a2 a3 a4 a5 ... aN], I am looking for an algorithm that enables to find all potential combinations of subarrays of size at least 2 without changing the sequence of intervals (consecutive partitioning of the original array). The concatenation of all sub-arrays shall give the original array.

For example
A=[1 3 4 5 6], I am looking for an algorithm that will give the following answers:

(1 array combinations)
a1: [1 3 4 5 6]
(2 array combinations)
a2: [1 3] [4 5 6]
a3: [1 3 4] [5 6]

Another example A=[1 3 4 5 6 8]

(1 array combinations)
a1: [1 3 4 5 6 8]
(2 array combinations)
a2: [1 3] [4 5 6 8]
a3: [1 3 4] [5 6 8]
a4: [1 3 4 5] [6 8]
(3 array combinations)
a5: [1 3] [4 5] [6 8]

What the should be the state space size (complexity of the search algorithm)?

答:
Let $F(n)$ be the number of partitions for an array of $n$ elements. My claim is that $F(1) = 0$, $F(2) = 1$, and$F(n) = F(n-1) + F(n-2)$ when $n > 2$. It is trivial to see that the values for $F(0)$ and $F(1)$ are correct so the tricky part is proving the recursive formula. To see that it is true note that the partitions of an array $[a_1, a_2, ..., a_n]$ come in two different forms:

The cases where $[a_1, a_2]$ is part of the partition

The cases where $[a_1, a_2]$ is not part of the partition

You can get all of the partitions in case (1) by generating the partitions for the smaller array $[a_3, a_4, ..., a_n]$ and adding in $[a_1, a_2]$ for each partition. This means there are $F(n-2)$ partitions like this.

You can get all of partitions in case (2) by generating the partitions for the array with $a_1$ missing ($[a_2, a_3, ..., a_n]$) and then prepending $a_1$ to the sub-array containing $a_2$ of the partitions. This means we have $F(n-1)$ such partitions.

The total number of partitions would then be the sum of these two cases which is $F(n) = F(n-1) + F(n-2)$. This is exactly the fibonacci sequence except shifted over (normally $F(1) = 1$) which has the closed solution:

$$ F(n)=\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}- \frac{1}{\sqrt{5}}\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}$$

This means your complexity is in fact exponential.

Title: Indexing problem in array


问:
Solving a linear system of equations use row operations
Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?

$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$

Every time I solve this it feels like it could have been done simpler... Thanks!

My attempt:

I first exchanged row 1 and row 2  to get:

$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$  $$3x_1-2x_2+3x_3=8$$

Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:

$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$

But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!

答: What you can do is multiply rows by nonzero constants. For instance $5R_2 \to R_2$ and $2R_3 \to R_3$. Then you can cancel the $x_2$ term in the last equation without getting a fraction in the coefficients. Of course, if there are fractions in the right-hand side, we cannot avoid those.


问:
Find all $x$ such that $|4x^2 - 12x - 27|$ is prime
Find all integers $x$ such that $|4x^2 - 12x - 27|$ is prime.

I first factored $|4x^2 - 12x - 27|$ as $(2x+3)(2x-9),$ but I was unsure where to go from there.

答:
Hint:

$|4x^2 - 12x - 27| = |(2x+3) \ (2x-9)|$

For it to be a prime number, one of the factors has to be $\pm1$ and absolute value of the other factor has to be the prime number.

So find $x$ when,

$2x+3 = \pm1$ or $2x - 9 = \pm1$ and check the value of the other factor.

There are only $4$ possible values of $x$ to check.


答:
You say: "because to show the existence one would have to prove the group axioms for each combination of the 22 elements"

This is correct. However it is not as bad as it sounds. Sometimes the key to solving a problem like this is just starting.

Suppose you first check if the group axioms are satisfied for 'all elements except the (congruence class of the) number 7'.

You will find [SPOILER ALERT] that it does not satisfy the axioms. But at that point you also see that the reasons that this set does not satisfy the group axioms also apply to a lot of the others and you can quickly cross 19 or 20 or 21 more subsets off of your list.

Then there are just a few left to check and the task isn't so daunting anymore.

So the lesson is: don't conclude too early that something is too much work. First start doing it and see if in the process you discover a way to speed it up. It does not always work but it works more often than you would expect.

Title: Let $M:=(\Bbb{Z}_{23},+,\cdot)$ the ring of integers modulo $23$. Is there a subset $T\subseteq M$ with $|T|=22$ such that $(T,\cdot)$ forms a group?

Body:
Let $M:=(\mathbb{Z_{23}},+,\cdot)$ the ring of integers modulo $23$. Is there a subset $T \subseteq M$ with $|T|=22$ such that $(T, \cdot)$ forms a group?

So I know what is asked here, but how to show that this exists (or I think that this does not exist?) because to show the existence one would have to prove the group axioms for each combination of the 22 elements?


问:
Proof that vector spaces $k^M$ and $k^{(M)}$ are not isomorphic
Assume $k$ is a division ring and $M$ an infinite set. There are two vector spaces over $k$ to consider: $k^M$ and $k^{(M)}$ (the latter is the subset of $k^M$ containing all infinite sequence with only a finite number of nonzero components).

Obviously $|k^{(M)}|\leq |k^M|$. I know that $k^{(M)}$ has a canonical basis $(e_\mu)_{\mu\in M}$. This means that $(e_\mu)_{\mu\in M}$ is a family of linearly independent elements of $k^M$. How can I show that the $e_\mu$'s don't generate $k^M$? I have to find an element of $k^M$ that isn't a linear combination of the $e_\mu$. What is a good choice for this supposed element? (Or is a proof by contradiction more appropriate?)

答:
Finding such an element would not be enough, as it would simply show that the inclusion $k^{(M)}\to k^M$ isn't an isomorphism, but that doesn't mean that there is no isomorphism.

For the record, the constant family with value $1$ is not in the span of the $e_\mu$.

The usual way to prove that they are not isomorphic is to prove that any $M$-indexed family does not span $k^M$ - in fact that would be equivalent because one can extract a basis from any such family, and it must then have cardinality $\geq |M|$ and $\leq |M|$ so $=|M|$. But for some reason I always forget the standard argument for this - it's a sort of diagonal argument where you build a new element not in the span of any $M$-indexed family of elements.

So another way to proceed is to prove that $k^{(M)}$ has too many linearly independent linear forms. Indeed, $(k^{(M)})^* \cong k^M$ (canonically), and so these linearly independent linear forms contradict the existence of an isomorphism.

For this I will use the following well-known set-theoretic lemma:

If $M$ is an infinite set, there is a bijection between the set of (nonempty) finite subsets of $M$, and $M$.

The proof then goes as follows (hidden, to leave some suspense):

 It follows from this lemma - actually, a variation on that lemma that I'll let you figure out- that $k^{(M)}$ is isomorphic, as a $k$-vector space, to $k[(X_\mu)_{\mu\in M}]$, the polynomial algebra in $M$-many indeterminates. Now for any $z\in k^M$, $ev_z$,"evaluation at $z$", defined by sending a polynomial $P$ to $P(z)$, defines a linear form on $k[(X_{\mu})_{\mu\in M}]$.

The claim is that these are linearly independent. But this is classical polynomial algebra: if you have a finite family $(z_i)_i \in k^M$, then I can find a finite subset $J\subset M$ such that the restrictions of these families to $J$ are distinct - and now in a finite number of variables, I can of course design a polynomial that takes any value I like on a finite list of points, in particular I can make the values leave any hyperplane. This is left as an exercise.

 The coup-de-grâce is in claiming that these are too many. But Cantor's theorem ensures that $|k^M|> |M|$, as $k$ is a field and so $|k|\geq 2$, so we are done.


问:
Number of students in school club : $4$ Circles Venn Diagram Puzzle
There are $200$ students in a class. Each student is a member of at least on of the clubs Eagles, Falcons, Tigers and Pythons.

If each club has $100$ students, then the number of students who are members of all the four clubs is at the most .....

If each club has $120$ students, then the number of students who are members of at the most two clubs  is at the least .....

If each club has $140$ students, then the number of students who are members of at least three clubs is at least ....

I came across this question and I tried to solve it using the $4$ circles Venn diagram but it is a very chaotic way to solve this (Too many variables come into picture and it all got messed up) and even I was only able to solve the first question.

Any way or method to simplify the solution approach for this question? Any help is appreciated.

Thanks!

答:
The set $A$ is the set of all students that are member of the first club, the set $B$ is the set of all students that are member of the second club.
The sets $C,D$ are the sets of students of the third and fourth club.

The set of all student can be partitioned to 16 different sets with regards to A,B,C,D. There is not student that is not member of any club. So the set $ A'\cap  B' \cap  C' \cap  D'$ is empty. The symbol $X'$  is the complement of the set $X.$
We define the set $P_{\mathit{BD}}=A'\cap B \cap C' \cap D$, and other members of the partion,

$P_A, P_B, P_C, P_D, P_\mathit{AB}, P_\mathit{AC}, P_\mathit{AD}, P_\mathit{BC}, P_\mathit{BD}, P_\mathit{CD}, P_\mathit{ABC}, P_\mathit{ABD}, P_\mathit{ACD}, P_\mathit{BCD}, P_\mathit{ABCD}$ in a similar way.

We define the variable $\mathit{bd}=|P_{BD}|$ and the other fifteen
nonegative  integer variables $a, b, c, d, \mathit{ab}, \mathit{ac}, \mathit{ad}, \mathit{bc}, \mathit{bd}, \mathit{cd}, \mathit{abc}, \mathit{abd}, \mathit{acd}, \mathit{bcd}, \mathit{abcd}$ are defined in a similar way.

The number of students that are in the first club is
$$|A|=a+\mathit{ab}+\mathit{ac}+\mathit{ad}+\mathit{abc}+\mathit{abd}+\mathit{acd}+\mathit{abcd}$$
and this number is $S$, where $S=100$ in your first question, and  $S=120$ and $S=140$ in your other questions. But we do our reasoning for arbitrary $S.$
For $|B|$, $|C|$, $|D|$ are defined in a similar way. So we get the four equations

$$a+\mathit{ab}+\mathit{ac}+\mathit{ad}+\mathit{abc}+\mathit{abd}+\mathit{acd}+\mathit{abcd}=S\\
b+\mathit{ab}+\mathit{bc}+\mathit{bd}+\mathit{abc}+\mathit{abd}+\mathit{bcd}+\mathit{abcd}=S\\
c+\mathit{ac}+\mathit{bc}+\mathit{cd}+\mathit{abc}+\mathit{acd}+\mathit{bcd}+\mathit{abcd}=S\\
d+\mathit{ad}+\mathit{bd}+\mathit{cd}+\mathit{abd}+\mathit{acd}+\mathit{bcd}+\mathit{abcd}=S$$

Because the number of a students is $200$ additionally  the following equation holds
$$a+b+c+d+\mathit{ab}+\mathit{ac}+\mathit{ad}+\mathit{bc}+\mathit{bd}+\mathit{cd}+\mathit{abc}+\mathit{abd}+\mathit{acd}+\mathit{bcd}+\mathit{abcd}=200
$$

The number $\mathit{abcd}$ is the number  of students that are member of all four clubs. This is the number we want to maximize. So this is an integer optimization problem that we have to solve.

Now let us assume that we have already found a assignments of the students to the clubs that these five equations hold but $abcd$ may not be as large as possible.
The following i such an assignment:
$$a=3,b=10,c=13,d=72,\\\mathit{ab}=7,\mathit{ac}=5,\mathit{ad}=2,\mathit{bc}=0,\mathit{bd}=1,\mathit{cd}=4,\\\mathit{abc}=62,\mathit{abd}=5,\mathit{acd}=1,\mathit{bcd}=0,\\\mathit{abcd}=15 \tag{x.1}$$

We want to try to change the assignment of some students to the clubs so that still all clubs have $S$ members but the number $\mathit{abcd}$ increases.

Here is an example:

Assume there is a student $x$ that is member of the set $P_\mathit{ABC}$ but not $D.$ So $x$ contributes $1$ to the number $\mathit{abc}$ and nothing to any of the other fourteen variables. Also assume there is another student $y$ that is member of the sets $A,B,D$ but not member of $C.$ So $y \in P_\mathit{ABD}.$ The student $y$ contributes $1$ to the variable $\mathit{abd}$ and nothing to the other variables.

Now we will resolve the membership of $x$ of the club $C$, that is , we will remove $x$ from $C$. But we make $y$ a member of $C$. Now $x$ is only a member of $A$  and $B$ and $y$ is a member of all clubs now. The club $C$ lost a member but another student joined the club, so the club has still $S$ members. But the number of students that are member of all clubs increase by one.
All in all the following happend:
the variables $\mathit{abc}$ and $\mathit{abd}$ decreased by $1$, the variables $\mathit{abcd}$ and  $\mathit{ab}$ increased by one.

We can apply this operation on our example $(x.1)$ and this results in

$$a=3,b=10,c=13,d=72,\\\mathit{ab}=8,\mathit{ac}=5,\mathit{ad}=2,\mathit{bc}=0,\mathit{bd}=1,\mathit{cd}=4,\\\mathit{abc}=61,\mathit{abd}=4,\mathit{acd}=1,\mathit{bcd}=0,\\\mathit{abcd}=16 \tag{x.2}$$

we can repeat this action 4 times and get
$$a=3,b=10,c=13,d=72,\\\mathit{ab}=12,\mathit{ac}=5,\mathit{ad}=2,\mathit{bc}=0,\mathit{bd}=1,\mathit{cd}=4,\\\mathit{abc}=57,\mathit{abd}=0,\mathit{acd}=1,\mathit{bcd}=0,\\\mathit{abcd}=16 \tag{x.3}$$

There are a lot of such operations that enables us to decrease one or more of the variables $\mathit{ab}, \mathit{ac}, \mathit{ad}, \mathit{bc}, \mathit{bd}, \mathit{cd}, \mathit{abc}, \mathit{abd}, \mathit{acd}, \mathit{bcd}$
by changing the club membership of one or more studens and keppeng the numbers $|A|$,$|B|$,$|C|$,$|D|$ constant and never decreases $\mathit{abcd}.$

There are other way to change the membership of the elements to the sets A,B,C,D such that the size of A,B,C and D remain the same and the size of  $\mathit{abcd}$ does not increase. If we do the correct changes after a finite number of steps we arrive at a system (maybe after changin the names) such that

the size of A, B, C and D did not change

the sets $P_\mathit{AC}, P_\mathit{AD}, P_\mathit{BC}, P_\mathit{BD}, P_\mathit{CD}, P_\mathit{ABD}, P_\mathit{ACD}, P_\mathit{BCD}$ are empty and so $\mathit{ac}=\mathit{ad}=\mathit{bc}=\mathit{bd}=\mathit{cd}=\mathit{abd}=\mathit{acd}=\mathit{bcd}=0$

Either $P_\mathit{AB}$ is empty and $\mathit{abc} \in \{0,1,2\}$ or $P_\mathit{ABC}$ is empty and $\mathit{ab} \in \{0,1\}$

$\mathit{abcd}$ is maximal

Conclusion

(The meaning of the variable $\mathit{abcd}$ and the other variables consisting of the letters $a,b,c,d$ is defined above)

If $m$ is the maximum possible size of $\mathit{abcd}$ then $m$ satisfies at least one of these systems of equations of nonnegative integer variables for $(u,v)\in \{(0,0), (1,0), (2,0), (0,1), (0,2)\}$:

$$
ab=u\\
abc=v\\
a+ab+abc+m=S\\
b+ab+abc+m=S\\
c+abc+m=S\\
d+m=S\\
a+b+c+d+m=200
$$

For $S=100$ only the system with $(u,v)=(1,0)$ has an integer solutions, and we get $m=66$, and $a=b


答: Yes.  $\Bbb Z_{23}$ is actually a special kind of ring known as a field.  Then it follows that the set of nonzero elements forms a (cyclic) multiplicative group.  It is denoted $M^*$.  This agrees with the group of units, $M^\times$, in this case.


问:
Local isometry which is not covering map
The theorem of Ambrose establishes that given a local isometry between two connected Riemannian manifolds $(M, g)$ and $(N,h)$ if $M$ is complete then $f$ is a covering.

I was wondering why completeness is necessary and I was trying to construct a local isometry which is not a covering until I realized that I do not know of many examples of local isometries, particularly on non complete manifolds such as the punctured plane.

Does anyone know of a few examples?

Any help would be welcome, thanks in advance.

答:
I believe that the local isometry condition tells you that your map has the sort of local covering property that is really at the heart of covering spaces. The problem is that the map may not be surjective. For instance, take the inclusion map from the punctured plane into the plane. This is clearly a local isometry, but it is not surjective. Thus the map does not have the covering property near the origin $(0,0)$ in the codomain.

So really, you should think of local isometries as being like covering spaces. After all, if you were just to complete the domain and codomain and uniquely extend the map $f$, you would have a covering space.


问:
$\lim_{ x \to 0 } \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$
I have to evaluate this limit:
$$\lim_{ x \to 0 }  \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$$

I have an indeterminated form $[\frac{0}{0}]$.
So I applied the Hopital rule
$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \sim $$
$$-e^{\frac{\log(1+\arctan x)}{x}} \cdot \frac{\frac{x}{(1+\arctan x)(1+x^2)}-\log(1+\arctan x)}{x^2} \sim$$
$$ -e \cdot \frac{1}{x^2} \cdot \left(\frac{x}{(1+x)(1+x^2)}-x\right)
\sim +e $$

According to my book and wolfram alpha the final result should be $\frac{e}{2}$ and I don't know where I made mistakes.

答:
We look at the expression :
$$
-e^{\frac{\log(1+\arctan x)}{x}} \cdot \left( \frac{1}{x(1+\arctan x)(1+x^2)} - \frac{\log(1+\arctan x)}{x^2}\right)
$$

which is correctly derived.

We focus on the first term's limit : consider the limit of the exponent $\lim_{x \to 0} \frac{\log(1+\arctan x)}{x}$, where we note that we can use L'Hopital to get that this is equal to the limit of $\frac{1}{(x^2+1)(1+\arctan x)}$ which equals $1$ at $x=0$. So $\lim_{x \to 0} -e^{\frac{\log(1+\arctan x)}{x}} =-e^1= -e$ by continuity of the exponential.

For the second part, we do what you did, and get a common denominator so we can use L'Hopital. That gives :
$$
\frac{\frac{x}{(1+\arctan x)(1+x^2)} - \log(1+\arctan x)}{x^2}
$$

Now, I think the issue with your working is what has happened on the next line i.e. on the list of $\sim$s, one is wrong. The one I think is wrong, is that $-x$ should be replaced with $-x \over 2$. Let me explain.

What you tried to do, is separate the bottom $x^2$ so we get $$
\frac 1{x^2}\left(\frac{x}{(1+x^2)(1+\arctan x)} - \log(1+\arctan x)\right)
$$

up till here, everything is fine.

What you did wrong

And then you said : $-\frac{\log(1+\arctan x)}{-x}$ goes to $1$ as $x \to 0$, so replacing $-\log(1+\arctan x)$ with $-x$ in this, will still give the same answer. Similarly, $\arctan x$ looks like $x$ as $x\to 0$ so I can again replace $\arctan x$ by $x$ in the first part of the bracket and that limit will be the same. That led to :
$$
\frac 1{x^2}\left(\frac{x}{(1+x^2)(1+x)} - x\right)
$$

Now there's a problem here , and J.G. pointed it out in his now deleted answer. It is a subtlety one has to be careful of.

If you have two separate functions $f(x)$ and $g(x)$, and you are studying $\lim_{x \to 0} f(x) - g(x)$. You notice that $f(x) \sim h(x)$ and $g(x) \sim j(x)$ as $x\to 0$, and then are tempted to write $\lim_{x \to 0} f(x) -g(x) = \lim_{x \to 0} h(x) - j(x)$. Not True.

The reason why this is not true, is actually in the "hidden" terms. I will say it roughly : a "dominating" term for $f(x)$ is the power of $x$ (with coefficient) that $f$ "behaves like" as $x \to 0$. So for example, $f(x) = \log(1+x)$ would behave like $x$ as $x \to 0$. Anything that kind of lies beneath the dominating term, we refer to as the "hidden" terms. For example, $\log(1+x) \sim x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots$ (found by Taylor, only for illustrative purposes), and while $x$ is the dominating term, the rest are hidden terms.

The hidden terms will come into play when you take a limit by removing the dominating : for example, $\frac{\log(1+x)  - x}{x^2} = \frac 12$. (found from Taylor series, but can be done by L'Hopital too)

Now, $f(x) \sim h(x)$ means that $f(x)$ and $h(x)$ have the same dominating term as $x\to 0$, these kind of cancel each other out when we take a ratio and the "hidden terms" don't come into play because they get "divided out" (try this with some functions) so the limit becomes $1$.

Now, imagine I take $f(x) - g(x)$. This takes the dominating terms out of the equation, sure, but the hidden terms don't get canceled out. These guys remain in the whole equation, and the problem is (like the $\frac 1{x^2}$ sitting silently on the side in your answer) if there are other terms like a $\frac 1{x^2}$ or something, these hidden terms kind of contribute. For example, $\lim_{x \to 0} x = 0$, but if $x$ was part of a huge expression and there was also a $\frac 1x$ sitting somewhere on the side I did not see, then replacing $x$ with $0$ is wrong, since with the $\frac 1x$, we get $\lim_{x \to 0} x \times \frac 1x = 1$, so $x$ "contributes" to the limit.

That is the idea : taking the difference of functions exposes the "hidden" terms, and because asymptoticity doesn't say anything about the hidden terms, we can't really say anything about comparing $f(x)-g(x)$  and $h(x) - j(x)$.

In your question, replacing $x$ with $\arctan x$ in both the terms has the problem that $x$ and $\log(1+\arctan x)$ are asymptotic as $x \to 0$, but the latter has the hidden terms (found from Taylor series, only for illustration) $\log(1+\arctan x) = x - \color{blue}{\frac{x^2}{2}} + ...$ and this one hidden term comes into prominence because there's a $\frac 1{x^2}$ in the expression as well. So that cancels with the $x^2$ giving a $-\frac 12$, which is exactly the (additive for the inner bracket)
factor you are missing.

Back to Question

To solve the question, we can still do L'Hopital, and be cleverer about our choice of split ,so we can L'Hopital each part separately! The point is, I want to finish each part off with a separate L'Hopital and then combine the two answers.

For example, if we write :
$$
\left( \frac{1}{x(1+\arctan x)(1+x^2)} - \frac{\log(1+\arctan x)}{x^2}\right) \\= \left( \frac{1}{x(1+\arctan x)(1+x^2)} - \frac 1x - \frac{\log(1+\arctan x)}{x^2} + \frac 1x\right) \\ = \left( \frac{1}{x(1+\arctan x)(1+x^2)} - \frac 1x\right) -\left( \frac{\log(1+\arctan x) - x}{x^2} \right)
$$

where I deliberately added and subtracted $\frac 1x$ so that each of these brackets can be evaluated separately. Now each of these is L'Hospitable (?) but we still need to be very clever. Your problem is extremely hard, as you've already seen.

The first one first :

$$
\left( \frac{1}{x(1+\arctan x)(1+x^2)} - \frac 1x\right) = \frac {(1- (1+\arctan x)(1+x^2))}{x(1+\arctan x)(1+x^2)}
$$

Apply L'Hopital, and it turns out both derivative of top and bottom have non-zero limits ($-1$ for top and $1$ for bottom), which leads to $-1$ at the end of it. The expressions are messy, but simplified by the fact that we are looking at $x=0$ so things simplify. (I am leaving this for you)

As for the second term, indeed it is again L'Hopital twice.

Do it once :
$$
\frac{\log(1+\arctan x) - x}{x^2} \to \frac{\frac 1{(x^2+1)(1+\arctan x) } - 1}{2x} \\ \to \frac{1}{(x^2+1)(1+\arctan x)} \frac{1 - (x^2+1)(1+\arctan x)}{2x}
$$

The first one is $1$ at $x=0$. The second, well L'Hopital and substitute $x=0$, you get the non-zero answer $\frac 12$.

Combining everything, the answer is :
$$
-e(-1 + \frac 12) = \frac e2
$$

and so the missing factor was just the $\frac 12$ in the end.


答:
If the polynomial has roots $a,b,c,d$ then we can write it as:
$$(x-a)(x-b)(x-c)(x-d)=0$$
which if we expand out we get:
$$x^4-(a+b+c+d)x^3+(cd+bc+bd+ad+ac+ab)x^2-(abc+abd+acd+bcd)x+abcd$$

From this we can take values from our polynomial and so:
$$a+b+c+d=3$$
$$cd+bc+bd+ad+ac+ab=4$$
$$abc+abd+acd+bcd=2$$
$$abcd=1$$

Now try and get a single expression from the desired product and you should be able to represent it in terms of the expressions above, giving you a value :)

Title: theory of equation

Body:
let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
 ( c+d)^{-1}+ (c+d)^{-1}$$

my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck


问:
show Banach Isomorphism theorem using Closed Graph theorem
Closed Graph Theorem: let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear map with closed graph. Then $T$ is continuous.

Definition: let $X$ and $Y$ be two normed spaces and let $T : X \to Y$ be a mapping. We say that $T$ is an isomorphism if it is continuous, opened and bijective.

Banach Isomorphism Theorem: let $X$ and $Y$ be Banach spaces and let $T : X \to Y$ be a linear bijective continuous map. Then $T$ is isomorphism.

Is it possible to show the Banach Isomorphism theorem using Closed Graph theorem? it should be sufficient if we had the graph from $T^{- 1}$, $G(T^{-1}) = \{(x , y) \in X \times Y : x = T^{- 1}(y)\}$ is closed in $X \times Y$, but I don't think that it is as easy.

答:
If
$$G=\{(y,x) \in Y \times X: x=T^{-1}(y)\}$$
and we take a sequence $\{(y_n,x_n)\} \subset G$ such that is convergent to some point $(y,x) \in Y \times X$ then we have that $x_n=T^{-1}(y_n)$ and as $T$ is bijective, $y_n=T(x_n)$.

So, on the one hand $T(x_n) \rightarrow y$ but on the other hand, as $T$ is continuous, $T(x_n) \rightarrow T(x)$, so $T(x)=y$ and then $x=T^{-1}(y)$. We conclude then that $(y,x) \in G$ and $G$ is closed.


问:
A question based on a n irreducible polynomial dividing $x^{q^n}-x$ in finite fields
This question was part of my assignment in field theory and I need help in solving it.

If $\vert K \vert= q$ and $f\in K[x] $  is irreducible, then $f$ divides $x^{{q^n}}-x$ iff $\deg f \mid n$.

I assumed  $\deg f = a$   and $f \mid x^{q^n}-x$ , but there is no progress.

For converse, as expected I let that degree $f \mid n$ but How to use it?

I am really sorry but I am blank on this question and would really like to have some hints given.

I am following Algebra By I N Hernstein.

答:
I worked through this exercise myself. I apologize if I gave "too much" away, the proof of one of the directions in particular has a few somewhat intricate steps, so it would be hard to give merely an illuminating hint.

Let us set $b$ to be the degree of $f$.

The result $b|n \Rightarrow f|(x^{q^n}-x)$ is the more straightforward direction but this is the proof for sake of completeness. Let $f$ be an irreducible polynomial of degree $b$ in $K[x]$. Then $K[x]/f(x)K[x]$ is a field with $q^b$ elements. Then for each nonzero element $y(x) \in K[x]/f(x)K[x]$ the equation $$y^{q^b-1}(x) \equiv 1 \mod f(x)$$ holds, which gives $$y^{q^b}(x) - y(x) \equiv 0 \mod f(x).$$ So taking $y(x) = x$ this gives $$x^{q^b}-x \equiv 0 \mod f(x)$$ or $$x^{q^b} \equiv x \mod f(x).$$ So raising each side to the $q^{b}$-th power gives

$$x^{q^{2b}} =  (x^{q^b})^{q^b} \equiv (x)^{q^b} \equiv x \mod f(x).$$

In fact, repeating this $j$ times gives
$$x^{q^{jb}} \equiv x^{q^{(j-1)b}} \equiv \ldots \equiv x^{q^b} \equiv x \mod f(x)$$
which gives $$x^{q^{jb}}-x \equiv 0 \mod f(x) \quad \forall j \in \mathbb{Z}^+.$$
So take $j = n/b$ to get $$x^{q^n}-x \equiv 0 \mod f(x)$$ or equivalently,
$$f(x)|(x^{q^{n}}-x).$$ So indeed, $b|n \Rightarrow f|(x^{q^n}-x)$. $\surd$

The other direction $f(x) | (x^{p^n}-x) \Rightarrow b | n$ is somewhat trickier. Let us suppose that $ n \mod b = r \not = 0$. We show the other direction by showing that $f$ cannot divide $x^{p^n}-x$.
Write $n= jb+r$.  Then repeating the reasoning from the other direction gives
$$x^{p^n} = (x^{p^{jb}})^{p^r} \equiv (x^{p^{(j-1)b}})^{p^r} \equiv \ldots  \equiv (x^{p^{b}})^{p^r} \equiv x^{p^r} \mod f(x),$$ or more concisely, $x^{p^n} \equiv x^{p^r} \mod f(x)$. Suppose that $f$ divides $x^{p^n}-x$. Then  gives $$x^{p^n} \equiv x \mod f(x),$$ and so putting that together with the relation $x^{p^n} \equiv x^{p^r} \mod f(x)$ gives $$x^{p^r}-x \equiv 0 \mod f(x),$$ or equivalently, $f(x)|(x^{p^r}-x)$. We show next that this cannot be.

Now, on the one hand, as $K[x]/K[x]f(x)$ is a field, the polynomial $X^{p^r}-X$ has at most $p^r$ roots in $K[x]/K[x]f(x)$ equivalently

(A) there are at most $p^r$ polynomials $y(x) \in K[x]/K[x]f(x)$ that satisfy $y^{p^r}(x)-y(x) \equiv 0 \mod f(x)$.

However, one can also check that for any polynomial $Y \in K[X]$, the following holds because $K$ is characteristic $p$: $Y^{p^r}(X) = Y(X^{p^r})$. So if the relation $x^{p^r} \equiv x \mod f(x)$ holds, then $y^{p^r}(x) = y(x^{p^r}) \equiv y(x) \mod f(x)$ for each  $y(x) \in K[x]/K[x]f(x)$, which gives that each of the $p^b$ polynomials $y(x) \in K[x]/K[x]f(x)$ satisfy $y^{p^r}(x)-y(x) \equiv 0 \mod f(x)$. However, as $r = n \mod b$ it follows that $p^b>p^r$, and so this contradicts (A). Thus
$f(x)$ cannot divide $x^{p^r}-x$ after all, if deg$(f) = b$ does not divide $n$. $\surd$


问:
Express $n(\pi/2)^{n-1} -n(n-1)(n-2)(\pi/2)^{n-3} + ...$ as a sum
I have obtained that
$$
\mathbb{S}_n =\int_0^{\frac{\pi}{2}} x^n \sin(x)dx = n\left(\frac{\pi}{2}\right)^{n-1} -n(n-1)(n-2)\left(\frac{\pi}{2}\right)^{n-3} + n...(n-4)\left(\frac{\pi}{2}\right)^{n-5} ...
$$
Or:
$$
\mathbb{S}_n = \frac{n!}{(n-1)!}\left(\frac{\pi}{2}\right)^{n-1} -\frac{n!}{(n-3)!}\left(\frac{\pi}{2}\right)^{n-3} + \frac{n!}{(n-5)!}\left(\frac{\pi}{2}\right)^{n-5} ...
$$
How can I express this using $\sum$ notation. And if there is more elegant way to express this how should it be done. Thank you.

答:
Let $ n\in\mathbb{N} $, using double integration by parts, we have : \begin{aligned}\int_{0}^{\frac{\pi}{2}}{x^{n}\sin{x}\,\mathrm{d}x}&=\left[-x^{n}\cos{x}\right]_{0}^{\frac{\pi}{2}}+n\int_{0}^{\frac{\pi}{2}}{x^{n-1}\cos{x}\,\mathrm{d}x}\\ &=n\left[x^{n-1}\sin{x}\right]_{0}^{\frac{\pi}{2}}-n\left(n-1\right)\int_{0}^{\frac{\pi}{2}}{x^{n-2}\sin{x}\,\mathrm{d}x}\\ I_{n}&=n\left(\frac{\pi}{2}\right)^{n-1}-n\left(n-1\right)I_{n-2}\end{aligned}

If $ n=2p $, such that $ p\in\mathbb{N} $, then, for all $ k\leq p $, we have : \begin{aligned}I_{2k}&=2k\left(\frac{\pi}{2}\right)^{2k-1}-2k\left(2k-1\right)I_{2k-2}\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\left(-1\right)^{k}}{\prod\limits_{j=1}^{k}{2j\left(2j-1\right)}}I_{2k}&=\frac{2\left(-1\right)^{k}k}{\prod\limits_{j=1}^{k}{2j\left(2j-1\right)}}\left(\frac{\pi}{2}\right)^{2k-1}+\frac{\left(-1\right)^{k-1}}{\prod\limits_{j=1}^{k-1}{2j\left(2j-1\right)}}I_{2k-2}\\ \Longrightarrow\sum_{k=1}^{p}{\left(\frac{\left(-1\right)^{k}}{\prod\limits_{j=1}^{k}{2j\left(2j-1\right)}}I_{2k}-\frac{\left(-1\right)^{k-1}}{\prod\limits_{j=1}^{k-1}{2j\left(2j-1\right)}}I_{2k-2}\right)}&=2\sum_{k=1}^{p}{\frac{\left(-1\right)^{k}k}{\prod\limits_{j=1}^{k}{2j\left(2j-1\right)}}\left(\frac{\pi}{2}\right)^{2k-1}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\left(-1\right)^{p}}{\prod\limits_{j=1}^{p}{2j\left(2j-1\right)}}I_{2p}-I_{0}&=2\sum_{k=1}^{p}{\frac{\left(-1\right)^{k}k}{\prod\limits_{j=1}^{k}{2j\left(2j-1\right)}}\left(\frac{\pi}{2}\right)^{2k-1}}\end{aligned}

Now since, for all $ k\in\mathbb{N} $, $ \prod\limits_{j=1}^{k}{2j\left(2j-1\right)}=\left(2k\right)! $, we get : $$ \fbox{$\begin{array}{rcl}\displaystyle I_{2p}=\left(-1\right)^{p}\left(2p\right)!\left(1+\sum_{k=1}^{p}{\frac{\left(-1\right)^{k}}{\left(2k-1\right)!}\left(\frac{\pi}{2}\right)^{2k-1}}\right)\end{array}$} $$

Now we can do the same if we suppose $ n=2p+1 $, $ p\in\mathbb{N} $, then : $$ \fbox{$\begin{array}{rcl}\displaystyle I_{2p+1}=\left(2p+1\right)!\sum_{k=0}^{p}{\frac{\left(-1\right)^{p-k}}{\left(2k\right)!}\left(\frac{\pi}{2}\right)^{2k}}\end{array}$} $$


问:
By making the substitution $x=\pi-t$, show that: $ \int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx $
I have a question regarding the following problem:

By making the substitution $x=\pi-t$, show that:

$$
\int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx
$$

This is my work so far:

$$
x=\pi-t
\\dx=-dt
$$

$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}(t-\pi)f(\sin(\pi-t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}tf(\sin(t))dt-\pi\int_{0}^{\pi}f(\sin(t))dt
$$

I noticed that
$$\int_{0}^{\pi}xf(\sin{x})dx$$
and
$$\int_{0}^{\pi}tf(\sin(t))dt$$
are very similar and
$$\pi\int_{0}^{\pi}f(\sin(t))dt$$
is also close to what I want to find.

However, I am unsure how to make further progress.

Any help would be much appreciated

EDIT:
As the comment mentioned, I forgot to change the limits of integration. After changing the limits, I got:
$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}tf(\sin(t))dt-\pi\int_{\pi}^{0}f(\sin(t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt
$$

I can see that this is very close to the final answer. However, I still don't know how I can eliminate the variable $t$

答:
You have made a mistake - your last equation is missing a negative sign. If you let $I=\displaystyle \int_0^{\pi} x f(\sin x)\mathrm dx$ , then the correct result is $$\int_0^{\pi} xf(\sin x)\text dx=\pi \int_0^{\pi}f(\sin t) \text d t-\int_0^{\pi} tf(\sin t)\text d t$$ But since $$\displaystyle \int_0^{\pi} xf(\sin x)\text dx=\int_0^{\pi }tf(\sin t) \text d t$$ because $t$ is just a dummy variable,  we get $\displaystyle I=\dfrac{\color{red}\pi}{2} \int_0^{\pi} f(\sin x) \text{d} x$.

Edit: This was answered in context to original question before editing.


问:
Suppose that $P[|X| \geq \lambda ] \leq Ce^{-c\lambda^{2}}$ then $E[e^{c'|X|^{2}}] \leq C'$
The following assertion was made in Terry Tao's book on random matrix theory:
Let $X$ be a real valued random variable

Suppose that $P[|X| \geq \lambda ]  \leq Ce^{-c\lambda^{2}}$ then $E[e^{c'|X|^{2}}] \leq C'$

where $c,c',C,C'$ are positive real numbers.

I was having trouble actually verifying this and was looking for some help in doing so.

答:
The layer cake formula gives

\begin{align*}
E[e^{c'|X|^2}] &= \int_0^\infty P(e^{c'|X|^2} \ge \lambda)d\lambda \\
&= \int_0^\infty P\left(|X| \ge \sqrt{\frac{1}{c'} \ln \lambda}\right)d\lambda \\
&\le 1 + \int_1^\infty P\left(|X| \ge \sqrt{\frac{1}{c'} \ln \lambda}\right)d\lambda \\
&\le 1 + C\int_1^\infty e^{-\frac{c}{c'} \ln \lambda} d\lambda \\
&= 1 + C \int_1^\infty \lambda^{-c/c'}d\lambda,
\end{align*}

so choose $c' < c$ and $C'$ to be the final line above.


问:
Prove: if a $\sigma$-algebra over $X$ contains one unique atom $A$, then $A=X$.
Let $\mathcal{F}$ be a $\sigma$-algebra over some non-empty $X$. Given that $A$ is the only atom of $\mathcal{F}$ (recall $A \in \mathcal{F}$ is an atom of $\mathcal{F}$ if it's non-empty and no proper subset of $A$ other than the empty set is in $\mathcal{F}$),  show that $A=X$.

My attempt: It's clear $A \subset X$. To show $X \subset A$, suppose for contradiction there's some $x \in X$ and $x \in A^c$. Let
$$
B = \bigcap_{S \in \mathcal{F}: x \in S}  S
$$
(which is clearly non-empty as $x \in A^c \in \mathcal{F}$). My idea is to show that $B$ is also an atom of $\mathcal{F}$ (basically using this idea), and $A\cap B = \emptyset \implies A \neq B$, contradicting that $A$ is the unique atom. However, I'm stuck on showing $B \in \mathcal{F}$, because the intersection defining $B$ might not be countable. It's intuitively clear that if $\mathcal{F}$ has $n \in \mathbb{N}$ distinct atoms, then $ | \mathcal{F}| = 2^n$ (or at least $\mathcal{F}$ should be finite), but I have yet to prove this (it's actually the last part of this problem, and I'm stuck on part 1)...

答:
As I understand it, the problem as stated is false.

Let $Y$ be a set with an atomless $\sigma$-algebra $\mathcal{G}$ (see here for examples). Let $X = Y \sqcup \{a\}$ for some $a$. Define a $\sigma$-algebra $\mathcal{F}$ on $X$ by $E \in \mathcal{F}$ if and only if $E \backslash \{a\} \in \mathcal{G}$. In other words, $E \in \mathcal{F}$ exactly when $E \in \mathcal{G}$ or $E$ is the union of a set in $\mathcal{G}$ with $\{a\}$.

In this scenario, we have that $\{a\}$ is an atom. On the other hand, every other nonempty set $E \in \mathcal{F}$ satisfies $E \backslash \{a\} \neq \emptyset$, and since $\mathcal{G}$ is atomless we can find $\emptyset \neq F \subsetneq E$ with $F \in \mathcal{G}$, which also means $F \in \mathcal{F}$. Thus, $E$ contains a proper nonempty subset in $\mathcal{F}$, so is not an atom.

As noted in the comments below, the problem works out if the $\sigma$-algebra is countably generated. In this case, we can follow the idea in the question, making just one modification.

Suppose $\mathcal{F}$ is generated by some countable collection $\mathcal{E}$. We might as well assume $\mathcal{E}$ is an algebra, since the algebra generated by a countable collection is still countable. With this in hand, we can restrict our intersection defining B to sets $S \in \mathcal{E}$, and B should still be an atom.


答:
We are given
$$\begin{pmatrix}
2 & 4 & -1\\
1 & 1 & -1\\
3 & -2 & 3
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
0\\
8
\end{pmatrix}$$

Take $R_1 - 2R_2 \to R_2$ and $-3R_1+2R_3 \to R_3$ to form
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & -16 & 9
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
-5
\end{pmatrix}$$
Then let $8R_2 + R_3 \to R_3$ to obtain
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & 0 & 17
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
51
\end{pmatrix}$$
We then find that $x_1=1,x_2=2,x_3=3$.

Title: Solving a linear system of equations use row operations

Body:
Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?

$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$

Every time I solve this it feels like it could have been done simpler... Thanks!

My attempt:

I first exchanged row 1 and row 2  to get:

$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$  $$3x_1-2x_2+3x_3=8$$

Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:

$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$

But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!


答:
Just a few sign flips away from it.
$$\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}(\pi-t)f(\sin(\pi-t))(-dt)\\
=\int_{0}^{\pi}(\pi-t)f(\sin(\pi-t))dt\\
=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt
$$

Clearly, the value of a definite integral is independent of the dummy variable we use to describe it.  So we can do an algebraic manipulation to get

$$2\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt$$
$$\int_{0}^{\pi}xf(\sin{x})dx=\frac\pi2\int_{0}^{\pi}f(\sin(t))dt$$

Title: By making the substitution $x=\pi-t$, show that: $ \int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx $

Body:
I have a question regarding the following problem:

By making the substitution $x=\pi-t$, show that:

$$
\int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx
$$

This is my work so far:

$$
x=\pi-t
\\dx=-dt
$$

$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}(t-\pi)f(\sin(\pi-t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}tf(\sin(t))dt-\pi\int_{0}^{\pi}f(\sin(t))dt
$$

I noticed that
$$\int_{0}^{\pi}xf(\sin{x})dx$$
and
$$\int_{0}^{\pi}tf(\sin(t))dt$$
are very similar and
$$\pi\int_{0}^{\pi}f(\sin(t))dt$$
is also close to what I want to find.

However, I am unsure how to make further progress.

Any help would be much appreciated

EDIT:
As the comment mentioned, I forgot to change the limits of integration. After changing the limits, I got:
$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}tf(\sin(t))dt-\pi\int_{\pi}^{0}f(\sin(t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt
$$

I can see that this is very close to the final answer. However, I still don't know how I can eliminate the variable $t$


问:
Solving a stochastic differential equation using Ito’s formula
I was wondering wether anyone could help me solve this question.

Let $B=(B_t)t\geq0$ be a standard Brownian motion started at zero. Consider the stochastic differential equation:

$dX_t = (1+ 3X_t)dt + (1+5X_t)dB_t$,

for a stochastic process $X=(X_t)_{t\geq0}$ where $X_0=1$.

Verify by Ito’s formula that this solution is given by

$X_t = Y_t(1-4\int_0^t\frac{1}{Y_s}ds+ \int_0^t\frac{1}{Y_s}dB_s)$ for $t\geq0$,
where the process $Y=(Y_t)_{t\geq0}$ solves

$dY_t = 3Y_tdt + 5Y_tdB_t$, with $Y_0=1$

I’m not sure how to apply Ito’s formula to obtain the solution the question is after? Any help would be great!

答:
Let $A_t := 1 - 4\int_0^t \frac 1{Y_s}ds + \int_0^t \frac 1{Y_s}dB_s$, so $dA_t = -4 \frac{1}{Y_t}dt + \frac 1{Y_t}dB_t.$  Note that the equality for $dA_t$ is just notation for how we defined $A_t$.  Then Ito's product formula applied to $X_t = Y_t A_t$ gives

\begin{align*}
dX_t &= Y_t dA_t + A_t dY_t + dA_tdY_t \\
&= (-4dt + dB_t) + (3Y_tA_t dt + 5Y_tA_t dB_t) + 5dt \\
&= (1+3Y_tA_t)dt + (1+5Y_tA_t)dB_t \\
&= (1+3X_t)dt + (1+5X_t)dB_t.
\end{align*}


问:
Can a set with an isolated point have a boundary?
If I have a set formed by the union of $x^2+y^2\le4$ and the point $(5,5)$, does the set have a boundary in the form of a circumference centred at the point $(0,0)$ of radius $2$?

What I am asking is whether the point $(5,5)$ affects in any way the existence of the boundary given by $x^2+y^2\le 4$.

Thank you.

答:
Your set is
$$
S=\{x^2+y^2\le 4\}\cup \{(5,5)\}
$$

The interior of $S$ is the open disk $\{x^2+y^2< 4\}$. Since $S$ is closed, its boundary is given by
$$
\partial S
=\overline{S}\setminus \textrm{int}(S)
=S\setminus \textrm{int}(S)=\{x^2+y^2=4\}\cup \{(5,5)\}.
$$


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
This one is quick and quite surprising application of linearity of expectation.

Let $G=(V, E)$ be a graph with $n$ vertices and $e$ edges. Then it contains a bipartite subgraph of at least $\frac{e}{2}$ edges.

Let $T \subseteq V$ be a random subset given by $\mathbb{P}[x\in T] = \frac{1}{2}$. An edge $\{x, y\}$ is a crossing if exactly one of $x, y$ are in $T$.

Denote by $X_{xy}$ the random indicator for ${x, y}$ being a crossing, and set
$$
X = \sum_{\{x, y\} \in E} X_{xy}.
$$

Since $\mathbb{E}[X_{xy}] = \frac{1}{2}$, by linearity of expectation we have
$$
\mathbb{E}[X] = \sum_{\{x, y\} \in E} \mathbb{E}[X_{xy}] = \frac{e}{2}.
$$
Consequently, there exists a choice of $T$ with at least $\frac{e}{2}$ crossing edges, which form a bipartite graph.

This beautiful proof is from Noga Alon and James Spencer's excellent book 'The Probabilistic Method'. I know it can be directly proved by induction, but the above probabilistic argument is way more elegant.


问:
Prove or disprove: if $f(0)=0$ and $f'(0)=3$ then $\sum_{n=0}^\infty f(\frac {1}{n})$ converges.
Just learned about series recently and been trying to solve this question and I got stuck, the derivative that is taking me to taylor's polynomial, but I have no idea how to link between it and my series.
 Any help is appreciated.

答:
One counterexample suffices, the obvious one being $f(x)=3x$, but the general result that all $f$ satisfying the given conditions result in a divergent series can be obtained directly from the definition of the derivative. Since $f(0)=0$, you know that

$$3=f'(0)=\lim_{h\to 0}\frac{f(h)}h\,,$$

so there is an $\epsilon>0$ such that $2<\frac{f(h)}h<4$ whenever $|h|<\epsilon$. Thus, for $n>\frac1\epsilon$ we must have

$$2<\frac{f\left(\frac1n\right)}{\frac1n}<4$$

and hence

$$\frac2n<f\left(\frac1n\right)<\frac4n\,.$$

But then $\sum\limits_{n\ge 1}f\left(\frac1n\right)$ diverges by comparison with $\sum\limits_{n\ge 1}\frac2n$.


问:
Examples of convergence of random variables
First, let's recall the definitions of 4 different types of convergence:almost surely, in $r$th mean, in probability and in distribution:

$X_n\xrightarrow{a.s.}X$ if $\{\omega \in \Omega:X_n(\omega)\rightarrow X(\omega),$ as $n\rightarrow\infty\}$ is an event with probability 1.

$X_n\xrightarrow{L^r} X$ if $\forall r\geq1,\mathbb{E}[X_n^r]<\infty$, $\mathbb{E}[|X_n-X|^r]\rightarrow 0$, as $n\rightarrow \infty$

$X_n\xrightarrow{p}X$ if $\forall \epsilon >0,\mathbb{P}(|X_n-X|>\epsilon)\rightarrow 0,$ as $n\rightarrow \infty$

$X_n\xrightarrow{D}X$ if $\mathbb{P}(X_n\leq x)\rightarrow \mathbb{P}(X\leq x)$ as $n\rightarrow \infty$

In Wikipedia I got the following implication relations:

If $s>r\geq1,(X_n\xrightarrow{L^s} X) \Rightarrow (X_n\xrightarrow{L^r}X)$

Convergence almost surely and convergence in mean both imply convergence in probabilty.

$(X_n\xrightarrow{p}X)\Rightarrow (X_n\xrightarrow{D}X)$ (Convergence in probability implies convergence in distribution)

So, what I want to ask here is that if somebody can give me some simple examples to briefly explain why the implication works and some counter examples why it doesn't work conversely(the other direction of the implication arrow), because all those definitions look so similar to me, especially, for example, why convergence in probability doesn't imply convergence almost surely? For the definitions of them are really the same thing.

I'll really appreciate if you can help me out. Thanks!

答:
why convergence in probability doesn't imply convergence almost surely? For the definitions of them are really the same thing.

No, they are not. For a counterexample, assume $(X_n)_{n\geqslant1}$ are independent Bernoulli with $P[X_n=1]=1/n$ and $P[X_n=0]=1-1/n$. Then $X_n\to0$ in probability (why?) while the event $[X_n\to0]$ has probability zero (hint: Borel-Cantelli).


问:
The Fourier Transform of a Bessel function: $J_a(be^{-x}), a,b>0$
Currently we require the Fourier transform  $g(p)$ of  $f(x)=J_a(be^{-x}), a,b>0$
$$g(p)=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{ipx} f(x) dx$$
We could not find it in Tables and Handbooks. Mathematica gives it in terms of Gamma functions and Gauss Hyper geometric function. A closer look suggested that Hypergeometric function was actually a hoax. As we were finally interested in $|g(p)|^2$, it was a pleasant surprise to get
$$|g(p)|^2=\frac{1}{2\pi(p^2+a^2)},~~~~~(1)$$
which is independent of $b$ !, even the verification of (1) by

Mathematica in analytic mode  for various sets of $a,b,p$ has been time consuming.

The question is how to prove (1) and get $g(p)$ by hand?

EDIT:
By Integrate of  Mathematica, I finally get
$$g(p)=\frac{1}{2\sqrt{2\pi}} \left(\frac{2}{b}\right)^{ip} \frac{\Gamma[(a+ip)/2]}{\Gamma[1+(a-ip)/2]}$$
EDIT:
Here is the Mathematica out put, see the pre=factor of hypergeometric function is 0.

g(p)=1/2 Gamma[
1/2 (a + I p)] ((2^(I p) (1/b^2)^((I p)/2))/
Gamma[1/2 (2 + a - I p)] +
2^-a (-(1/b^2)^(-a/2) + b^
a) HypergeometricPFQRegularized[{1/2 (a + I p)}, {1 + a,
1/2 (2 + a + I p)}, -(b^2/4)])

答:
The Fourier transform has a closed-form, but my derivation lacks rigor.
Eliminate the b-dependence by translation: $x \to x+\log{(1/b)}.$ Thus
$$ g(p):= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ipx}J_a(be^{-x})dx=$$
$$=\frac{(\log{(1/b)})^{ip}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{ipx}J_a(e^{-x})dx = \frac{(\log{(1/b)})^{ip}}{\sqrt{2\pi}}G(a,p) $$
where $G(a,p)$ has been defined so that we can deal with a 2-variable integral now.
Use the known expansion (Gradshteyn & Rhyzik 8.411.10)
$$J_a(z) = \frac{(z/2)^a}{\Gamma(a+1/2)\Gamma(1/2)} \int_{-1}^1e^{izt}(1-t)^{a-1/2}dt$$
Insert this into $G(a,p).$ Interchange integrals.  Do the inner integral explicitly,
$$ \int_{-\infty}^\infty e^{ipx}e^{-ax}e^{ie^{-x}} = (-it)^{-a+ip}\Gamma(a-ip)$$
Here, though, the second non-rigorous step has been taken (the first is that the intergrals can be intergchanged).  Mathematica says the previous eq. is true for t with a non-zero imaginary part.  I'm assuming that someone good with contour integration can justify assuming completely real t is O.K.  Mathematica can do the remaining integral as well.  Collecting,
$$G(a,p):=\int_{-\infty}^\infty e^{ipx}J_a(e^{-x})dx=$$
$$=\frac{ 2^{-a} (-i)^{-a+ip}\ \Gamma(1/2+(ip-a)/2)}{2\Gamma(1/2)\Gamma(1+(ip+a)/2)}  (-1)^{-a}\big((-1)^{a}+e^{-p\pi})\big)\Gamma(a-ip).$$
(Hopefully there are no typos.)
As a numerical check, I let a=3/10 and p=6/10.  For the integral I chopped the integral limits at $\pm$ 1200, used 30 digits precision, and MinRecursion of 5 in Mathematica's NIntegrate.  I got 14 digits of agreement from the integral, and the explicit form presented.

Out of time; hope this helps.


问:
true or false: if fn(x) is a decreasing sequence of functions and converges point wisely to f(x) then f(x) is a decreasing function
i tried to use Dini's theorem to prove it converges uniformly but the question didn't say anything about its continuity, so f(x) might be discontinuous.
then i tried to find a counter example but all of them were decreasing
can you help me ? thanks</p>

答:
False.

Let $f_n(x)=(1-x)^n$ for $x$ in $(0,1)$ and $n=1, 2, 3, ...$.

$f_n$ converges pointwise to the constant function $f(x)=0$.


问:
Fresnel Diffraction and Fourier Transforms
I'm reading through Paganin's Coherent X-Ray optics textbook and could use a hand with Fourier Transform (FT) manipulation. For the duration of this post, $\mathcal{F}$ is the forward FT and $\mathcal{F}^{-1}$ is the inverse transform. I define the conventions at the bottom of this post.

In the text, Paganin defines the angular-spectrum representation of a propagated wave-field as

\begin{equation}
\psi_\omega (x, y, z=\Delta) \approx \exp(ik\Delta) \mathcal{F}^{-1} \exp\bigg[\dfrac{-i\Delta (k_x^2 + k_y^2)}{2k} \bigg]\mathcal{F} \psi_\omega (x, y, z=0)
\end{equation}

Here, $\Delta$ is a spatial coordinate and $\Delta > 0$. In the discussion of the convolution formulation of the Fresnel Diffraction section, he defines the following form of convolution:
\begin{equation}
f(x,y) *g(x,y) = 2\pi \mathcal{F}^{-1}\big( \{ \mathcal{F} [f(x, y)]\} \times \{ \mathcal{F}[g(x,y)] \} \big).
\end{equation}

I'm a little confused with what Paganin does after this. He states that we can rewrite the angular-spectrum representation of the wavefield in the following suggestive form:
\begin{align}
\psi_\omega (x, y, z=\Delta)
&=
2\pi \mathcal{F}^{-1} \bigg( \dfrac{\exp(ik\Delta)}{2\pi} \exp\bigg[ \dfrac{-ik\Delta(k_x^2 + k_y^2)}{2k} \bigg]  \times \{ \mathcal{F} [\psi_\omega (x, y, z=0] \} \bigg)
\\[.5em]
&=
2\pi \mathcal{F}^{-1} \bigg( \bigg\{ \mathcal{F}\mathcal{F}^{-1} \dfrac{\exp(ik\Delta)}{2\pi} \exp\bigg[\dfrac{-i\Delta(k_x^2 + k_y^2)}{2k} \bigg] \bigg\} \times \{ \mathcal{F} [\psi_\omega (x, y, z=0)] \} \bigg)
\end{align}

I've tried to work out how he does this, but I haven't had any luck. Does anyone have any insight into how we go from the first equation listed to the one just before this sentence?

Since there are a million ways to write the FT, I'll write the definition Paginin uses.
\begin{align}
f(x,y)
&=
\frac{1}{2\pi} \iint_{-\infty}^\infty \tilde{f}(k_x,k_y)\exp[i(k_x x + k_y y)] \, dk_x \, dk_y
\\[.5em]
\tilde{f}(k_x,k_y)
&=
\frac{1}{2\pi} \iint_{-\infty}^\infty f(x,y)\exp[-i(k_x x + k_y y)] \, dx \, dy
\end{align}
Here, $\tilde{f}$ is the FT of $f$. i.e., $\tilde f=\mathcal{F}[f]$ and ${f}=\mathcal{F}^{-1}[\tilde f]$.

答:
$\def\F{\mathcal{F}}
\def\Fi{\F^{-1}}
\def\p{\pi}
\def\D{\Delta}$Schematically, we have
\begin{align*}
\psi &= 2\p A\Fi(B\F C)
= 2\p\Fi(AB\F C)
= 2\p\Fi(\F(\Fi AB)\F C) \\
&= (\Fi AB)\star C
= (A\Fi B)\star C.
\end{align*}
Note that $A = e^{i k\D}/(2\p)$ is not a function of $x,y$.


问:
Smallest cardinality of generating set of maximal ideal in a ring $R$
I wonder the smallest cardinality of generating set of maximal ideal in a ring $R$ in it coincide with the Krull dimension? i.e. Let $J=\{$Maximal ideal in $R\}$, $S_{I}$ is the generating set of maximal ideal $I$,Krull dimension of $R=n$. Is $min_{I\in J}|S_{I}|=n$?

If $R$ is polynomial ring or $C([0,1])$ the cases are true, but can we generalize to any commutative $R$?

答: No. For any commutative ring $R$ and $I\leqslant R$ any ideal, denote by $\mu(I)$ the cardinality of a smallest generating set for $I$. If $R$ is Noetherian, then the best we can do in general is Krull's height theorem, which tells us that, for any $P\geqslant I$ a minimal prime ideal over $I$, we have $\text{ht}(P)\leqslant\mu(I)$. However, equality certainly does not have to hold. Let $F$ be your favorite field, and consider the ring $R:=F[x_1,\dots,x_n]\big/(x_ix_j)_{1\leqslant i,j\leqslant n}$. Then $R$ is local with $\dim R=0$ (why?), but the unique maximal ideal of $R$, $(\overline{x_1},\dots,\overline{x_n})$, is not generated by fewer than $n$ elements (why?). You can extend this example to the ring $F[x_i]_{i\in\mathbb{N}}\big/(x_ix_j)_{i,j\in\mathbb{N}}$, which is still local of Krull dimension $0$, but doesn't even have a finitely generated maximal ideal.


问:
orthogonal matrix and elementary matrix
Answer is False. But I can't think of the counter example... Could anybody have it?

Let A be an orthogonal 4 x 4 matrix such that $$ Ae_1 = e_2, Ae_2 = e_3, Ae_3 = e_1$$ Then $$Ae_4 = e_4 $$

答:
Let $$A = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & -1 \end{pmatrix}$$ Then $Ae_4 = -e_4$.

On the other side, the given three conditions and orthogonality of $A$ imply $Ae_4 \in \{\pm e_4\}$, as $Ae_4 \cdot Ae_i = e_4 \cdot e_i = 0$ for $i \in \{1,2,3\}$ hence $Ae_4 \cdot e_j = 0$ for $j \in \{1,2,3\}$. So $Ae_4 = \alpha e_4$ for some $\alpha \in \mathbb R$. Now $1 = Ae_4 \cdot Ae_4 = \alpha^2$ gives $\alpha \in\{\pm 1\}$.

Title: orthogonal matrix and elementary matrix


问:
CCM $3\times3$ into $4\times4$ (image processing)

Preface

Color Correction Matrix (a.k.a. CCM or CMX) is applied to raw image data (before de-mosaicking, usually from a standard bayer $RGGB$ or $BGGR$ pattern) to correct for the physical characteristics of the digital image sensor, most importantly for a so called "cross-talk" between sensor elements.

For reference: the most universally used bayer pattern produces pixel values with alternating "color brightness" at pre-assigned positions, like the following picture:

which are then processed to produce "real color" $RGB$ triplets per pixel with the assumption that chroma doesn't change significantly from one pixel to another.

The most common form used in pretty much every document I've found online of that CCM is a $3\times3$ matrix for input trio $(R_0, G_0, B_0)$ of pixel values used to produce a trio of "color corrected" pixel values $(R, G, B)$ like the following:

$$
\begin{pmatrix}
 R \\
 G \\
 B \\
\end{pmatrix}
=
\begin{bmatrix}
 A_{00} & A_{01} & A_{02} \\
 A_{10} & A_{11} & A_{12} \\
 A_{20} & A_{21} & A_{22} \\
\end{bmatrix}
\cdot
\begin{pmatrix}
 R_0 \\
 G_0 \\
 B_0 \\
\end{pmatrix}
$$

That assumes that values for Green are averaged out (or some other operation is done to produce 3 values from the original 4).

However the real input data I'm working with would be a stream of true 4-values in $2\times2$ blocks $\begin{vmatrix}B_0&G_{b0}\\G_{r0}&R_0\end{vmatrix}$.

Quite obviously for me to preserve the values for $G_b$ and $G_r$ I'd like to perform the color correction using a $4\times4$ matrix, not a $3\times3$. Note that the order of pixel values is now reversed from $RGB$ to $BGGR$:

$$
\begin{pmatrix}
 B \\
 G_b \\
 G_r \\
 R \\
\end{pmatrix}
=
\begin{bmatrix}
 C_{00} & C_{01} & C_{02} & C_{03} \\
 C_{10} & C_{11} & C_{12} & C_{13} \\
 C_{20} & C_{21} & C_{22} & C_{23} \\
 C_{30} & C_{31} & C_{32} & C_{33} \\
\end{bmatrix}
\cdot
\begin{pmatrix}
 B_0 \\
 G_{b0} \\
 G_{r0} \\
 R_0 \\
\end{pmatrix}
$$

Conversion

So here's my struggle: how do I go from that $3\times3$ color correction matrix to the final $4\times4$ form?

Of course the first thing to do is straight-forward: flip the matrix along the $A_{20}..A_{02}$ diagonal and then... add a row and a column?

The Green is the one that gets "split" into two independent values, one for each row (blue and red). So my thinking is to just blindly do exactly that:

$$
C =
\begin{bmatrix}
 A_{22} & A_{12} & A_{12} & A_{02} \\
 A_{21} & A_{11} & A_{11} & A_{01} \\
 A_{21} & A_{11} & A_{11} & A_{01} \\
 A_{20} & A_{10} & A_{10} & A_{00} \\
\end{bmatrix}
$$

Question

Would like to know if the above is the correct way to do such conversion? A bit of explanation beyond a "yes" or "no" would be most welcome, too.

答:
After extensive testing and some more investigation and research it became clear that this would not be the correct way to do the color correction (as stated) in a first place. Primarily due to the fact that an important dependency is missing in the proposed approach: namely the fact that $RGB$ trio depends on the specifics of the de-mosaicing.

In short - this is not a proper way to do things.


问:
Volume of bowl with inscribed cone
Suppose the volume of a bowl is defined by the following inequalities: $$z\geq x^2+y^2-1 {}, \\z\leq \sqrt{x^2+y^2}+1,\\z\geq 0$$
What is the value of its volume?

So I reasoned that the sought after volume is the volume of an elliptical paraboloid minus its bottom, (where $z<0$ ), and minus an inscribed cone. I think that the volume can be calculated with a triple integral. First with respect to $z$:

$$\int_{x^2+y^2-1}^{\sqrt{x^2+y^2}+1}dz=(\sqrt{x^2+y^2}+1)-(x^2+y^2-1)$$

Now we can integrate by x and y by first transform into polar coordinates: $$x^2+y^2=r^2\\{Jacobian}:r$$

So $$2\pi\int_1^2 (\sqrt{r^2}+2-r^2)r dr=2{\pi}[\frac{r^3}{3}+r^2-\frac{r^4}{4}]_1^2=\frac{19\pi}{6}$$

Nevertheless this is the wrong answer and I'm not quite sure where I've made a mistake. The answer is supposed to be $\frac{29\pi}{6}$.

答:
Vertex of the cone is at $z = 1$ and it forms for $z \geq 1$. The circular paraboloid has vertex at $z = -1$ and it intersects the cone at $z = 3$. We are interested in the volume of the paraboloid above $z = 0$ outside the cone.

Intersection of cone and paraboloid using cylindrical coordinates,

$r^2 - 1 = r + 1 \implies r^2 - r - 2 = 0$.

That gives us $r = 2$ and $z = 3$.

So one way to integrate is paraboloid volume between $0$ and $1$ (as there is no cone) and then between $z = 1$ and $3$.

$V = \displaystyle \int_0^{2\pi} \int_1^3 \int_{z-1}^{\sqrt{z+1}} r \ dr \ dz \ d\theta \ + \ \int_0^{2\pi} \int_0^1 \int_{0}^{\sqrt{z+1}} \ r \ dr \ dz \ d\theta$

As far as your method, that works too but you also need to add the volume for $0 \leq r \leq 1$ where $z$ is between the plane $z = 0$ and the cone.

So add the component $\displaystyle \int_0^{2\pi} \int_0^1 \int_{0}^{1+r} r \ dz \ dr \ d\theta$.

Title: Volume of bowl with inscribed cone


问:
Dimension of vector space and symmetric matrix
Why the following statement is true?
I am so frustrated that I could not have any clue on this problem.

The dimension of the vector space of all symmetric 4 by 4 matrices is 10.

Please help me.

答:
Hints:

A symmetric $\,n\times n\,$ matrix is completely characterized by the elements on its main diagonal and by those above it. How many elements are there...?


问:
In relation to the Babylonian method for computing a square root, if N/b = c, will c always be less than b?
I'm trying to wrap my head around the Babylonian method/algorithm for computing the square root of a number N. I can't seem to explain in words why, when a is too large i.e a^2 > N, then why you divide N/a to get b.

For example.

Compute the square root of 9409, where N = 9409.

If you guess a = 100, you get 10,000, which is too large as a^2 > N

so you then decide N/a = b and you say that b is an under estimate.

Why would b, be an underestimate?

In the wikipedia article for this: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method it relates to the sentence "The basic idea is that if x is an overestimate to the square root of a non-negative real number S then S/x will be an underestimate, or vice versa".

I hope I'm making sense, if not, please let me know.

If there are any resources where I may be able to get a better understanding please feel free to link them, including any practise problems.

Body: If $a^2>n$, then $\frac{n}{a^2}<1$. Let $b=\frac{n}a$; then $$b^2=\frac{n^2}{a^2}=n\cdot\frac{n}{a^2}<n\cdot1=n\,,$$ so $b<\sqrt{n}$.


问:
Counting function and its Artin L-function
I am working with the following function: given a polynomial, $P(x)$, with non-negative integer coefficients and passing through origin, we can define
$$f(d)=\{1\leq a\leq d \ | \ P(a)\equiv 0 \mod(d)\}$$
It is easy to see that $f$ is a multiplicative function. I am interested in the L-function associated with $f$, that is,
$$\sum_{n=1}^\infty \frac{f(n)}{n^s}=\prod_{p \ \text{prime}}\left(1+\sum_{k=1}^\infty\frac{f(p^k)}{p^{sk}}\right)^{-1}$$
I need to show the following that the above L-function is analytic in $\Re(s)>1$ and it has a pole of order $n$ at $s=1$ where $n$ is the degree of the polynomial $P(x)$. Any help would be much appreciated.

答:
The order of the pole at $s=1$ is the number of distinct irreducible factors.

Let $m\in \Bbb{Z}[x]$ monic and irreducible, $K=\Bbb{Q}[x]/(m(x))$.

$$f(n)=\# \{ a\bmod n, m(a)=0\bmod n\}$$

$f$ is clearly multiplicative. $m(x)\bmod p$ is separable when $p$ doesn't divide $Disc(m)$. For those primes Hensel lemma tells us that $f(p^k)=f(p)$ so that
$$F(s)=\sum_n f(n)n^{-s} = \prod_p (1+\sum_{k\ge 1} f(p^k) p^{-sk})$$ For the primes dividing the discriminant, I am thinking to the $\deg(m)$ roots of the polynomial in $\overline{\Bbb{Q}}_p$, for $k$ large enough there is a one-to-one correspondence between these and their reductions $\bmod p^k$. Not sure of the simplest argument $$=(\prod_{p\ | \ Disc(m)} (1+O(p^{-s}))\prod_{p\ \nmid\ Disc(m)} (1+f(p)\frac{p^{-s}}{1-p^{-s}})$$ $$=(\prod_{p\ | \ Disc(m)} (1+O(p^{-s}))\prod_{p\ \nmid\ Disc(m)} (1+f(p)p^{-s})(1+O(p^{-2s}))$$

For the primes not dividing the discriminant $f(p)$ is the number of prime ideals of $O_K$ with residue field $\Bbb{F}_p$ which means that
$$\prod_{P\ni p\subset O_K}\frac1{1-N(P)^{-s}}=
(1+O(p^{-2s}))\prod_{P\subset O_K,N(P)=p}\frac1{1-N(P)^{-s}}$$ $$=
 1+f(p)p^{-s}+O(p^{-2s})$$
And hence $$F(s)=\zeta_K(s)(\prod_{p\ |\ Disc(m)} (1+O(p^{-s})) (\prod_p (1+O(p^{-2s}))$$

Since $(s-1)\zeta_K(s)$ is entire then $(s-1)F(s)$ is analytic for $\Re(s)>1/2$.

For $m$ non-monic it works the same way, multiplying $Disc(m)$ by the leading coefficient of $m$.

For $m(x) = \prod_{j=1}^J m_j(x)^{e_j}$ reducible it is mostly the same as for the primes not dividing $Disc(\prod_{j=1}^J m_j)$ $$f(p^k)=\sum_jf_j(p^k),\qquad f_j(p^k)=\# \{ a\bmod n, m_j(a)=0\bmod p^k\}$$

Letting $K_j=\Bbb{Q}[x]/(m_j(x))$ then
$$F(s)=(\prod_{p\ |\ Disc(\prod_j m_j)} (1+O(p^{-s})) (\prod_p (1+O(p^{-2s})) \prod_{j=1}^J \zeta_{K_j}(s)$$ and hence $(s-1)^J F(s)$ is analytic for $\Re(s)> 1/2$.


问:
By making the substitution $x=\pi-t$, show that: $ \int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx $
I have a question regarding the following problem:

By making the substitution $x=\pi-t$, show that:

$$
\int_{0}^{\pi}xf(\sin{x})dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin{x})dx
$$

This is my work so far:

$$
x=\pi-t
\\dx=-dt
$$

$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}(t-\pi)f(\sin(\pi-t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\int_{0}^{\pi}tf(\sin(t))dt-\pi\int_{0}^{\pi}f(\sin(t))dt
$$

I noticed that
$$\int_{0}^{\pi}xf(\sin{x})dx$$
and
$$\int_{0}^{\pi}tf(\sin(t))dt$$
are very similar and
$$\pi\int_{0}^{\pi}f(\sin(t))dt$$
is also close to what I want to find.

However, I am unsure how to make further progress.

Any help would be much appreciated

EDIT:
As the comment mentioned, I forgot to change the limits of integration. After changing the limits, I got:
$$
\int_{0}^{\pi}xf(\sin{x})dx=\int_{\pi}^{0}tf(\sin(t))dt-\pi\int_{\pi}^{0}f(\sin(t))dt\\
\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\pi}f(\sin(t))dt-\int_{0}^{\pi}tf(\sin(t))dt
$$

I can see that this is very close to the final answer. However, I still don't know how I can eliminate the variable $t$

答:
hint

Let $$I=\int_0^\pi xf(\sin(x))dx$$

with $ t=\color{red}{\pi} -x$, it becomes

$$I=\int_{\color{red}{\pi}-0}^{\color{red}{\pi}-\pi}(\pi-t)f(\sin(\pi-t))(-dt)=$$

$$\int_0^\pi(\pi-t)f(\sin(t))(+dt)=$$

$$\pi\int_0^\pi f(\sin(t))dt-I$$

thus

$$2I=\pi \int_0^\pi f(\sin(t))dt$$
$$=\pi \int_0^\pi f(\sin(x))dx$$

and
$$I=\frac{\pi}{2}\int_0^\pi f(\sin(x))dx$$


问:
How to interpret the graph?
It is about vector addition! I can not see how $x$ and $y$ connect with $x+y$, intuitively,cut a small triangle near $y-$axis and transfer to $(b_1,b_2)$, please use analytic geometry to interpret it. In other words, use $xy$-plane coordinate system to interpret. I know how parallelogram law works for this graph

答:
What is being demonstrated here is that by the Parallelogram definition for vector addition, the coordinates of $x + y$ are the sums of the coordinates of $x$ and $y$.

They did not just "cut a small triangle". What they did was copy the particular triangle $\triangle PUV$

This is a right triangle with the $x$ vector as its hypotenuse. The vertical leg has length $a_2$, while the shorter leg has length $a_1$.

If you rotate this triangle by $180^\circ$, the three sides of the resultant triangle $\triangle P'U'V'$  will be parallel to their originals. The horizontal leg will still be horizontal, the vertical leg still vertical, and the hypotenuse will still be parallel to the $x$ vector.

Next translate $\triangle P'U'V'$ until $U'$ coincides with the end of the vector $y$, resulting in a third triangle $\triangle P''U''V''$. The hypotenuse is still parallel to $x$ and of the same length. But the right side of the parallelogram also has its endpoint at $y$, and is parallel to and of the same length as $x$. Thus the hypotenuse of $\triangle P''U''V''$ and the right side of the parallelogram must be the same, and $\triangle P''U''V''$ has to be as shown in the diagram, with $P'' = Q$.

The horizontal leg $\overline{U''V''}$ is horizontal and has length $a_1$. Since $U''$ is the end of $y$, it has coordinates $(b_1, b_2)$, so moving horizontally a distance of $a_1$ puts $V''$ at $(a_1 + b_1, b_2)$. The vertical leg $\overline{QV''}$ has length $a_2$, and moving vertically a distance of $a_2$ from $V''$ gives coordinates $(a_1 + b_1, a_2 + b_2)$. Thus $Q = (a_1 + b_1, a_2 + b_2)$.

Title: How to interpret the graph?


问:
Extension of Bayes' Theorem coin flipping example - finding a fair coin in a bag of biased coins
There's a fairly common example of Bayes' Theorem in which a coin is drawn from a population with a known amount of biased coins and fair coins. A coin is drawn, flipped a certain amount of times, and then one is challenged to determine the probability that the flipped coin was biased, resulting in a standard implementation of Bayes' Theorem.

I was thinking of an extension of this question, where the goal is to find a fair coin drawn from a population.

As a concrete example, suppose we have $N$ coins, of which $n$ are fair coins and the remainder will always flip heads. Our goal then is essentially to flip a tails so that we know we have a fair coin. Beforehand we decide after $m$ heads flipped from an individual coin, we will draw a new coin.

How would I approach the problem of calculating how many coins I would need to draw before reaching a certain probability threshold $p$ of having at some point flipped a tails?

In this case Bayes' Theorem tells us that the probability the first randomly chosen coin is biased after $m$ heads is given by:

$$\frac{\frac{N-n}{N}}{\frac{N-n}{N}+\frac{1}{2}^m*\frac{n}{N}}$$

My initial thought was to approach the problem in the vein of "how many coin flips do I need to have a probability $p$ of having flipped a heads?"-type problems and using Bayes' Theorem between each flip, but the examples of those questions that I've encountered require mutually independent events so I don't think that applies to this problem.

Additionally, would the question of finding an "optimal" $m$ be related to stopping time?

edit: After coming back to this question with a fresh mind, I had the thought that in the case there is only one fair coin (i.e. $n=1$) then we could say that the probability we would flip $i$ coins and not see a tails would be:

$$\frac{i}{N}*\frac{1}{2}^m+\frac{N-i}{N}$$

which would compute the probability that the fair coin was in the first $i$ coins drawn and that it was flipped $m$ times without seeing a tails, and add that to the probability that the fair coin hasn't yet been drawn. If this is valid, then I think I see how to extend it to $n$ fair coins (by summing over the different amounts of possibly drawn fair coins) as well as how to extend it to individual flips, but I'm also worried I'm violating some assumption in probability and intuiting something incorrectly. Would this work?

答:
Start with $N$ coins of which $n$ are fair and $N-n$ always show heads.

Draw $k$ coin at random and flip each of them $m$ times.

The probability of not getting any tails is:

$$\sum_{i=0}^k P[\text{draw exactly } i \text{ fair coins}]\times \left(0.5^m\right)^i=\sum_{i=0}^k \frac{{n\choose i}{{{N-n}\choose {k-i}}}}{N \choose k}\times \left(0.5^m\right)^i$$

Now, find the smallest $k$ that makes this probability less than or equal to $1-p$.  Then, the probability of seeing at least one tail will be at least $p$.


问:
Prove that there exists a $g \in N(\phi)^\perp$ such that $\phi(g)=1$
I am practising with a past exam of my functional analysis course and I am stuck on the following question:

Let $H$ be a real Hilbert space and let $\phi: H
\to \mathbb R$ be a non-zero bounded (hence continuous) linear functional. Prove that there exists a $g \in (N(\phi))^\perp$ such that $\phi(g)=1$. (Where the $N$ denotes the null space).

Attempt.  We may write that every $x\in H$ can be represented $x=n +n_c$ where $n$ is in the null space and $n_c$ in its complement. (Since $H= N(\phi) \oplus
 N (\phi)^\perp$ ). Then by linearity we get:
$$\phi(x)= \phi(n+n_c)=\phi(n)+\phi(n_c)= 0+ \phi(n_c)\neq 0 \text{ since } n_c \not \in N(\phi).$$
Since $n$ lies in the null space of $\phi$. Now we simply consider:
$$ g:= \frac{1}{\phi(n_c)} n_c \in N(\phi) ^\perp$$
which would have the desired property. Is my reasoning close?

答:
Since $N(\phi)^\perp$ is a subspace, it suffices to find some $x \in N(\phi)^\perp$ such that $\phi(x) \neq 0$: then we can just define $g = x/\phi(x)$.

Well, suppose for contradiction that this didn't happen. Then $\phi(x) = 0$ for all $x \in N(\phi)^\perp$. By definition, we also have $\phi(x) = 0$ for all $x \in N(\phi)$! As you noted, $H = N(\phi) \oplus N(\phi)^\perp$, so we conclude that $\phi = 0$. This contradicts the assumption that $\phi$ is nonzero.


问:
Is it true that if the property "being a closure of its interior" holds for a set, then it also holds for each connected component of that set?
Suppose that $A\subseteq\mathbb{R}^k$ is the closure of its interior. Is it true, then, that each connected component of $A$ is the closure of its respective interior? I.e., does it hold that

$$\tag{1} A = \mathrm{clos}(\mathrm{int}(A)) \qquad \Longrightarrow \qquad \left(\forall\, \text{connected component $C$ of $A$} \ : \ C = \mathrm{clos}(\mathrm{int}(C))\right)\quad ?$$

答:
No, it is not true. For exemple, consider $$A = \lbrace 0 \rbrace \cup \bigcup_{n \geq 1} \left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \subset \mathbb{R} \, \text{.}$$ Then we have $$\mathring{A} = \bigcup_{n \geq 1} \left( \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right) \, \text{,}$$ and hence $\overline{\mathring{A}} = A$. However, we have $$\overline{\mathring{\lbrace 0 \rbrace}} = \varnothing \neq \lbrace 0 \rbrace \, \text{.}$$

Edit: I have been asked in the comments if the property is true if one also assumes that each connected compact has nonempty interior. The answer is still no. For example, consider $$A = C \cup \bigcup_{n \geq 1} \left( \left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \times [-1, 1] \right) \subset \mathbb{R}^{2} \, \text{,}$$ where $$C = \left( [-1, 0) \times [-1, 0] \right) \cup \left( \lbrace 0 \rbrace \times [-1, 1] \right) \, \text{.}$$ We have $\overline{\mathring{A}} = A$. Moreover, the connected components of $A$ are precisely $C$ and the sets $$\left[ \frac{1}{2} \left( \frac{1}{n +1} +\frac{1}{n} \right), \frac{1}{n} \right] \times [-1, 1] \, \text{,}$$ with $n \geq 1$, which all have nonempty interior. However, we have $$\overline{\mathring{C}} = [-1, 0]^{2} \neq C \, \text{.}$$


问:
Euler method fails
Consider the differential equation $$y'=y-2e^{-x},$$
with initial value $y(0)=1$. It's not hard to see that $y=e^{-x}$ is the solution.

I am trying to solve this numerically via the forward Euler method. However, no matter how I shrink my step size $h$, the error of my numerical solution always diverges: in fact, the error at $x$ seems to be growing like $e^x$.

My question is why does the numerical solution error always diverge?

The numerical solution from the Euler method can be expressed in the form $$y_n=A(1+h)^n+Be^{-hn},$$
for appropriate constants $A,B$. I have a feeling that divergence has something to do with the fact that $|1+h|>1$, but I'm not sure how to formalize this.

As a follow-up, which numerical methods will solve this DE without the error blowing up?

答: The general solution of this ODE is $e^{-x}+C\,e^x$. With $y(0)=1$, we conclude $C=0$. But every rounding error in any numerical method will introduce a spurious $C\neq0$, and given the notorious behavior of exponential functions, this will overwhelm your numerical approximation.

Title: Euler method fails


问:
Exponential vs Erlang Distribution
I am confused as to what the primary difference is between an Exponential and an Erlang Random Variable since both are also related with a Poisson Distribution

From what I've read, Erlang refers to the xth occurences within a set time interval while Exponential also refers to the intervals.

Suppose we have the following examples:

The distribution of the arrivals of a shipment is Poisson Distributed. Let A be the time between two shipment arrivals at the company. Does A refer to an Erlang RV referring to the time it takes until the second occurence happens?

If for example we have the variable B which refers to the time between each arrival? Is this now an exponential random variable?

答:
The relationship between the exponential and Erlang distributions is exactly analogous to the relationship between a geometric and negative binomial random variable.

A geometric random variable counts the number of Bernoulli trials needed to obtain the first success.  A negative binomial random variable generalizes this idea to count the number of trials needed to obtain $r$ successes.  Thus a geometric distribution is a special case of the negative binomial with $r = 1$.  Another way to conceptualize a negative binomial random variable is that it is the sum of $r$ independent and identically distributed geometric random variables.

Similarly, an exponential random variable counts the time until the first event of interest occurs.  An Erlang random variable counts the time it takes to observe $r$ events.  And so it is also the sum of $r$ IID exponential random variables.

The following claim is incorrect:

Erlang refers to the $x^{\rm th}$ occurrences within a set time interval while Exponential also refers to the intervals.

The incorrect part of this claim is "within a set time interval."  There is no such set time interval; we are only interested in the time it takes for the $x^{\rm th}$ occurrence, no matter how long that may be.

As for your example questions:

Not quite.  If arrivals are Poisson distributed with rate $\lambda$ per unit time, then the time between two consecutive shipment arrivals is simply exponential with rate $\lambda$.  Since the exponential distribution is a special case of the Erlang with $r = 1$, it is "Erlang" with shape parameter $r = 1$ and rate $\lambda$.  What is genuinely (nontrivially) Erlang, however, is the total time until we observe two arrivals, which includes the time it takes to wait for the first arrival, plus the time between the first arrival and the second.  In this case, the total waiting time is Erlang with shape $r = 2$ and rate $\lambda$.

As already implied above, the interrarival time is exponential for a (homogeneous) Poisson process.

Title: Exponential vs Erlang Distribution


问:
What does "well defined" mean?
On page 153 of Linear Algebra Done Right the second edition, it says:

Define a linear map $S_1: \text{range}(\sqrt{T^*T} ) \to \text{range}(T)$ by:

7.43:   $S_1 (\sqrt{T^* T}v)=Tv$

First we must check that $S_1$ is well defined. To do this, suppose $v_1, v_2 \in V$ are such that $\sqrt {T^*T}v_1 = \sqrt{T^*T}v_2$. For the definition given by 7.43 to make sense, we must show that $Tv_1=T v_2$.

It is not entirely clear to me what the term 'well-defined' means here. Can someone clarify?

Thanks

答: I'll describe well-definedness in terms of a simpler function.  Let's say that I wanted to describe a function on non-zero rational numbers by $$f(p/q)=q/p$$  In this case, it falls on me to show that the value of $f$ at a certain rational number is independent of the way that the ratio is formed, for instance that $f(1/2)=f(2/4)$.  If I can show that, then I can get away with defining a function whose "input" is not a single independent variable.

Title: What does "well defined" mean?


问:
Indexing problem in array
Given an array A=[a1 a2 a3 a4 a5 ... aN], I am looking for an algorithm that enables to find all potential combinations of subarrays of size at least 2 without changing the sequence of intervals (consecutive partitioning of the original array). The concatenation of all sub-arrays shall give the original array.

For example
A=[1 3 4 5 6], I am looking for an algorithm that will give the following answers:

(1 array combinations)
a1: [1 3 4 5 6]
(2 array combinations)
a2: [1 3] [4 5 6]
a3: [1 3 4] [5 6]

Another example A=[1 3 4 5 6 8]

(1 array combinations)
a1: [1 3 4 5 6 8]
(2 array combinations)
a2: [1 3] [4 5 6 8]
a3: [1 3 4] [5 6 8]
a4: [1 3 4 5] [6 8]
(3 array combinations)
a5: [1 3] [4 5] [6 8]

What the should be the state space size (complexity of the search algorithm)?

答:
Here is pseudocode for an algorithm which yields all such partitions:

If $N<2$, yield nothing.

For each $k$ in $\{2,3,\dots,N\}$:

Recursively yield $\{A(1),A(2),\dots,A(k)\}$, appended with each of the partitions of $\{A(k+1),\dots,A(N)\}$.

Title: Indexing problem in array


问:
Seeking Proof for a Conclusion from Positive Definite Homogeneous Polynomials
I am trying to find a proof for the following proposition from multivariate real analysis:

Suppose that $p$ on $\mathbb{R}^{n}$ is a homogeneous polynomial of degree $r$ in $x$ and $p\left(x\right) > 0$ for all $x \neq 0 \in \mathbb{R}^{n}$. Show that there is a $\rho > 0$ such that $p\left(x\right) \geq \rho\lVert x \rVert^{r}$ for all $x \in \mathbb{R}^{n}$.

A homogeneous polynomial of degree $r$ is of the following form:
\begin{equation}
\begin{aligned}
\forall x \in \mathbb{R}^{n},\ p\left(x\right) = \sum_{i_{1},i_{2},\dots,i_{r}=1}^{n}a_{i_{1},i_{2},\dots,i_{r}}{x_{i_1}}{x_{i_2}}\dots {x_{i_r}},
\end{aligned}
\end{equation}
where $x_{s}$ is the $s$-th element of the vector $x \in \mathbb{R}^{n}$.

P.S. This proposition serves as the basis for proving a theorem about local extreme points in multivariate real analysis. A hint is provided by the textbook: use the fact that $p$ assumes a minimum on the set $\left\{x \vert \lVert x \rVert = 1\right\}$. I am not sure how to construct the proposition from this starting point. Does anyone have an idea?

答:
The homogeneous polynomial of order $r$ is of the following general form:
\begin{equation*}
\begin{aligned}
\forall x \in \mathbb{R}^{n},\ p\left(x\right) = \sum_{i_{1},i_{2},\dots,i_{r}=1}^{n}a_{i_{1},i_{2},\dots,i_{r}}{x_{i_1}}{x_{i_2}}\dots {x_{i_r}}.
\end{aligned}
\end{equation*}

As the set
\begin{equation*}
S = \left\{x \vert x \in \mathbb{R}^{n}, {\lVert x \rVert} = 1\right\}
\end{equation*}
is closed and bounded, and $p$ is continuous, there exists an $x_0 \in S$ ($\lVert x_{0}\rVert = 1$) such that
\begin{equation*}
\forall x \in S,\ p\left(x\right) \geq p\left(x_0\right)
\end{equation*}
Let $\rho = p\left(x_{0}\right)$. As $p$ is positive definite, we have $\rho > 0$. Then for all $x \neq 0$, we have
\begin{equation}
\begin{aligned}
p\left(\frac{x}{{\lVert x \rVert}}\right) &= \sum_{i_{1},i_{2},\dots,i_{r}=1}^{n}a_{i_{1},i_{2},\dots,i_{r}}{\frac{x_{i_1}}{{\lVert x \rVert}}}{\frac{x_{i_2}}{{\lVert x \rVert}}}\dots {\frac{x_{i_r}}{{\lVert x \rVert}}}\\
&= \frac{1}{{\lVert x \rVert}^{r}}\sum_{i_{1},i_{2},\dots,i_{r}=1}^{n}a_{i_{1},i_{2},\dots,i_{r}}{x_{i_1}}{x_{i_2}}\dots {x_{i_r}}\\
& = \frac{1}{{\lVert x \rVert}^{r}}p\left(x\right) \geq \rho,
\end{aligned}
\end{equation}
and considering the trivial case where ${\lVert x \rVert} = 0$, we have

\begin{equation*}
\forall x \in \mathbb{R}^{n}, p\left(x\right) \geq \rho {\lVert x \rVert}^{r}.
\end{equation*}
And the existential proof is constructed with such a $\rho$.


问:
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$

If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.

If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.

Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.

Any hints? I have the exam soon =/

答:
Note $\,\phi(3)=2 = \phi(4)\mid \phi(5) = \color{#c00}4,\ $ therefore,

by little Euler:  $\ (a,n)=1  \Rightarrow\ a^{\color{#c00}4} \equiv 1\pmod{n},\ \ \text{for} \ \ n\in \{3,4,5\},$

therefore   $\,\ 3,4,5\mid a^4-1\ \Rightarrow\ 3\cdot4\cdot 5\mid a^4 - 1,\ \ $ for $\, n\,$ coprime to $\,3,4,5.$

Remark: more generally see Carmichael's lamda function, $\,\lambda(m) =\,$  exponent of  $\,\Bbb Z/m^*$


问:
true or false: if fn(x) is a decreasing sequence of functions and converges point wisely to f(x) then f(x) is a decreasing function
i tried to use Dini's theorem to prove it converges uniformly but the question didn't say anything about its continuity, so f(x) might be discontinuous.
then i tried to find a counter example but all of them were decreasing
can you help me ? thanks</p>

答:
This is false

Taking $f_n(x)=x^2 +1/n$, the sequence of functions $\{f_n\}$ is decreasing and converges uniformly to $f(x)=x^2$. However $f$ is not decreasing.


问:
Trying to graph $x^{\frac12} + y^{\frac12} =1$
I am a bit rusty with math and I am not sure what the steps would be to graph the function  $x^{\frac12} + y^{\frac12} =1$. I tried doing implicit differentiation but then I wasn't sure where to go from there. Maybe I'm using the wrong approach.

Thanks, everyone.

答: Solving for $y$ we see that $y^{1/2}=1-x^{1/2}$ and so squaring both sides see that $y=(1-x^{1/2})^2 = 1 -2 x^{1/2} + x$.


问:
B-Spline interpolation method used in AutoCAD?
I have a number of input points that I would like to fit a B-Spline through using global interpolation (given a degree p). In other words, my B-Spline needs to pass directly through these points. Therefore, I need to determine the control points that will produce this B-Spline. AutoCAD has an option "Spline by Fit" which describes exactly what I want.

My problem has been in determining these control points. What formula should I use for this global interpolation? I have not found much on the internet about this, except for the document at https://pages.mtu.edu/~shene/COURSES/cs3621/NOTES/INT-APP/CURVE-INT-global.html, and have managed to successfully construct a B-Spline that does in fact pass through in the input points, however this method result in curves that wiggle a lot more than the ones in AutoCAD do, and so I am not sure if there are any alternate formulae which might produce slightly different result which AutoCAD may be using.

Help or guidance would be much appreciated.

Edit: I assigned the knot vectors using uniform distribution, with the first p + 1 knots being 0, and the last p + 1 being 1, with p being the degree of the curve. I assigned the parameters also uniformly, calculating t using t = i / n, where I is the current input point and there are n + 1 input points.

I figured that this wouldn't change the result THAT drastically, and that something had to be going wrong in my interpolation algorithm, but I'm not 100% sure. How could I improve this to make my curve less "wiggly"?

答:
For B-spline curve interpolation, the two most important factors that would affect the resulting spline's shape are (a) the parameter assigned to each data point and (b) the knot vector.

For assigning parameter to each data point, there are many different ways but the two most popular choices are chord-length parametrization and centripetal parametrization. In this link, there is a picture showing the difference for using different parametrizations.

Once the parameter for all data point are decided, the knot vector needs to be decided based on the parameters in a way so that the matrix is not singular or nearly-singular. This is known as "Schoenberg-Whitney condition". A common way to determine knot vector that meets this condition is the parameter averaging scheme, which is described below using cubic B-spline as example.

Suppose we want to use a cubic B-spline to interpolate given $N$ points, the resulting B-spline will also have $N$ control points. Therefore, there are $N+4$ knot values that need to be decided. Assuming the point parametrization is $u_0=0.0,u_1,u_2,...,u_{N-1}=1.0$, we can decide the knot vector as

Assign $0.0$ to the first 4 knots,

$knots[4]=(u_1+u_2+u_3)/3$,

$knots[5]=(u_2+u_3+u_4)/3$,

....

Assign $1.0$ to the last 4 knots.


问:
proof of associative-ring
I want to check if the structure is ring

define the structure in this way:

$R$={${\frac{m}{2n+1},+,*} $} $n,m\in \mathbb Z$

i want to prove associativity in relation to +

$(a+b)+c=a+(b+c)$

$(a+b)+c=a/(2c+1)(2b+1)$

$a+(b+c)=a(2c+1)/2b$

i'm missing something here or the group just does not associative?

答:
You need to show

$$\dfrac{a}{2b+1} + \left(\dfrac{c}{2d+1}+\dfrac{e}{2f+1}\right) = \left(\dfrac{a}{2b+1} + \dfrac{c}{2d+1}\right)+\dfrac{e}{2f+1}$$

If you know $\mathbb Q$ is a ring, then you wouldn't need to tediously show this, though.

Title: proof of associative-ring


问:
Cocartesian squares in the category of abelian groups.
Recently, I've been doing a recap of (basic) category theory and found an old exercise I seem to be unable to solve. The question is as follows.

Let $A, B$ be abelian groups, $A'<A$ and $B'<B$ subroups and $\phi:A\to B$ such that $\phi(A')\subseteq B'$. Let $\phi':A'\to B'$ and $\phi'':A/A'\to B/B'$ be the maps induced by $\phi$. Prove that the diagram

\begin{array}
AA' & \to & A \\
\downarrow{\phi'} & & \downarrow{\phi} \\
B' & \to & B
\end{array}

is a cocartisian square if and only if $\phi''$ is an isomorphism.

The exercise preceding this one looks like a dual statement which I did manage to prove (a similar square being cartesian iff $\phi'$ is an iso), yet just dualising the proof doesn't seem to work. That proof, however, suggests using the universal property of cocartesian squares for one of the implications ($\Rightarrow$), while using the explicit construction of pushouts in the category of abelian for the other one. The pushout of $A_1\overset{f_1}{\leftarrow} A_3 \overset{f_2}{\rightarrow} A_2$ in this category is given by $A_1\oplus A_2/\{(f_1(x),-f_2(x))\mid x\in A_3\}$.

Can anyone provide me with a proof, or give a sketch of proof that might point out which step I fail to see just yet?

答:
Take $A' \to A \to A/A'$, the first map is the inclusion and the second is the quotient projection. Take then $B' \to B \to B/B'$, the maps as above. These are both exact sequences, which means $\operatorname{im}(\text{inclusion}) = \ker(\text{projection})$. You should notice that $(\phi', \phi, \phi'')$ is a morphism of exact sequences, which means that this triple makes the obvious diagram commute.

Now, you may check that the sequence $B' \to P \to \operatorname{coker}(i)$ is exact, if $P$ is the pushout of the inclusion $A' \subseteq A$ along $\phi'$ and $i$ is the standard inclusion of $B'$ into the direct sum you mention. If you put the two diagrams together (the one that describes the map $(\phi', \phi, \phi'')$ and the one with $(\phi', j, k)$, where $j\colon A \to P$ is the standard inclusion and $k\colon A/A' \to \operatorname{coker}(i)$), you notice that there is also a map form $B$ to $P$ because of the universal property of $P$. As you should know the universal map is an iso, so $B \cong P$. Then you notice that $\operatorname{coker}(i)$ is iso to $B/B'$, because the inclusions $B' \subseteq B$ and $B' \subseteq P$ differ by an iso. This means that your square is in fact a pushout.

This is a succession of iff-s, so your claim is proved. This should be so much easier if you write down the diagrams I mention, which I cannot do here, since it looks like mathjax does not support xypic, and I'm not familiar with any other way of writing commutative diagrams.

Clarification: I understand this is an exercise in category theory and I didn't use any strictly categorical methods, but you should notice that the category of abelian groups is an abelian category and this is actually done in utter generality in any abelian category if, instead of quotients you take cokernels.

EDIT: Here's the diagram I was talking about.


问:
Striking applications of linearity of expectation
Linearity of expectation is a very simple and "obvious" statement, but has many non-trivial applications, e.g., to analyze randomized algorithms (for instance, the coupon collector's problem), or in some proofs where dealing with non-independent random variables would otherwise make any calculation daunting.

What are the cleanest, most elegant, or striking applications of the linearity of expectation you've encountered?

答:
If you randomly shuffle a deck of $n$ cards numbered $1, 2,...,n$ and deal them out, then the expected number of cards that are greater than all previous cards is the $n$-th partial sum of the harmonic series: That is, the expected number of cards that are the 'maximum so far' is given by $H(n)=1+\frac12+\frac13+\cdots +\frac1n$

To see this: Let $X_i=1$ if the $i$-th card is the greatest so far; and $X_i=0$ otherwise. Then $X:=\displaystyle\sum_{i=1}^n{X_i}$ gives the number of cards that are maximums so far.  But $E[X_i] = \frac1i$ since there is a $\frac1i$ probability that the greatest of the first $i$ cards is in fact the $i$-th card.