问:
How do I interpret \bigwedge notation in discrete math?
I started reading Discrete Mathematics and Its Applications (Eighth Edition) by Kenneth H. Rosen and came across a problem that assumes knowledge not discussed in the book. The question reads as follows:

If $p_1, p_2,…, p_n$ are n propositions, explain why

$$\bigwedge_{i=1}^{n-1} \bigwedge_{j=i+1}^n (\neg p_i \vee \neg p_j)$$

is true if and only if at most one of $p_1, p_2, ...p_n$ is true.

From my research, it seems $\bigwedge$ denotes that I should AND each term together (what does this really mean?). I do not see why at least one of the propositions above need to be True. If someone could break down how to go about solving this I would be very appreciative. The fuller the explanation the more I would benefit as I think I am interpreting the problem incorrectly.

答:
Yes, it’s a conjunction: $\bigwedge$ bears the same relationship to $\land$ as $\sum$ bears to $+$. Thus, for example, $\bigwedge_{i=1}^3p_i$ means exactly the same thing as $p_1\land p_2\land p_3$.

In the double conjunction $\bigwedge_{i=1}^{n-1}\bigwedge_{j=i+1}^n$ each value of $i$ from $1$ through $n-1$ is paired with each value of $j$ from $i+1$ through $n$; this has the effect of running through all pairs $\langle i,j\rangle$ of indices with $1\le i<j\le n$, so

$$\bigwedge_{i=1}^{n-1}\bigwedge_{j=i+1}^n(\neg p_i\lor\neg p_j)$$

is precisely equivalent to

$$\bigwedge_{1\le i<j\le n}(\neg p_i\lor\neg p_j)\,.\tag{1}$$

If $n=4$, for instance, this is

$$\begin{align*}
&(\neg p_1\lor\neg p_2)\land(\neg p_1\lor\neg p_3)\land(\neg p_1\lor\neg p_4)\\
&\land(\neg p_2\lor\neg p_3)\land(\neg p_2\lor\neg p_4)\land(\neg p_3\lor\neg p_4)\,.
\end{align*}$$

The net effect is to take the conjunction of all pairs $\neg p_i\lor\neg p_j$ for distinct $p_i$ and $p_j$.

The conjunction $(1)$ is true precisely when all of the disjunctions $\neg p_i\lor\neg p_j$ with $1\le i<j\le n$ are true; if even one of them is false, $(1)$ is also false.

Suppose that $1\le i<j\le n$, and $p_i$ and $p_j$ are both true. Then $\neg p_i$ and $\neg p_j$ are both false, so $\neg p_i\lor\neg p_j$ is false, and therefore $(1)$ is also false. In other words, if two or more of the propositions $p_1,\ldots,p_n$ are true, then $(1)$ is false. Now suppose that at most one of these propositions is true. If $1\le i<j\le n$, then at least one of $p_i$ and $p_j$ is false, so at least one of $\neg p_i$ and $\neg p_j$ is true, and therefore $\neg p_i\lor\neg p_j$ is true. Thus, each of the disjunctions $\neg p_i\lor\neg p_j$ is true, and therefore their conjunction $(1)$ is true. Thus, we have shown that $(1)$ is true if and only if at most one of the propositions $p_1,\ldots,p_n$ is true.


问:
Congruences on the lattice $N_5$.
I am studying Universal Algebra using the book of Clifford Bergman and I encounter trouble in one of the exercises.

Task (2.12 - 3.)  Let $L$ be a copy of $N_5$ with elements $\{a,b,c,u,v\}$ and $u<a<b<v$. Let $\theta$ be a congruence on $L$. Prove that $\theta\neq O_L$ implies $(a,b)\in\theta$.

Notation Here $N_5$ is the pentagon-shaped five element lattice, where $v$ serves as 1, $u$ as 0 and we have two linear sublattices $u<a<b<v$ and  $u<c<v$. Other then this, elements are uncomparable.

By $0_L$ we mean trivial congruence where each block of equivalence is a singleton.

My approach So one can understand congruences just as partitioning elements of lattice into subsets (classes of equivalence) such that if I take two elements from one class of elements then both their meet and their join still lie in a given class of equivalence.

But if I take partition $\theta$ as $\{a\}$ and $\{b,c,u,v\}$ clearly $(a,b)\notin \theta$ but it seems to me that this is indeed a congruence.

I tried taking couples of elements, I tried thinking about it but it seems to me that if I take $x,y\in \{b,c,u,v\}$ then both their meet and join cannot be $a$

So I am clearly missing something here. Can anyone point out my mistake? Why $\theta$ is not congruence?

答:
Your definition is not quite correct. You state that an equivalence relation on a lattice is a congruence if each of its equivalence classes is a sublattice (i.e. closed under joins and meets). You need the equivalence classes to interact together as well:

if $x\mathrel{\theta}y$ and $z\mathrel{\theta}w$, then $x\wedge z\mathrel{\theta}y\wedge w$ and $x\vee z\mathrel{\theta}y\vee w$. (In particular, $x$ and $z$ do not need to belong to the same equivalence class.)

In other words, the equivalence classes need to form a lattice with well-defined operations. Can you find an example where your $\theta$ fails this condition?

ContentLicense: CC BY-SA 4.0

Title: Congruences on the lattice $N_5$.


问:
Relation between Ax=0 and Ax=b

I'm not sure how to solve this. My teacher told me that if $A\vec{x}=0$ has one unique solution (the trivial) then $A\vec{x}=\vec{b}$ has only one unique solution. But I don't know how to prove this.

答:
My teacher told me that if $A\vec x =\vec0$ has one unique solution (the trivial) then $A\vec x =\vec b$  has only one unique solution. But I don't know how to prove this.

Simply, as hinted in comments by Arturo Magidin,

if $A\vec x=\vec b$ and $A\vec y=\vec b$, then $A\vec x - A\vec y=0$, so by linearity of $A$ then $A(\vec x-\vec y)=\vec0$,

so $\vec x-\vec y=\vec 0$, so $\vec x=\vec y$, so if there is a solution to $A\vec x=\vec b$, then it is unique.

Title: Relation between Ax=0 and Ax=b


问:
Is there a smarter way to integrate?
My textbook uses the following technique to integrate. Let us take the following as an example,  evaluate the integral of:

$$f(x) = 6x(x^2 + 1)^5$$

Notice that $[x^2+1]' = 2x$, so $6x\space dx = 3 \space d (x^2+1)$

$$6x(x^2+1)^5 \space dx = \dfrac{6}{2} (x^2+1)^5 \space d(x^2+1) = 3u^5 \space du = d \dfrac{1}{2} u^6 = d \dfrac{1}{2} (x^2+1)^6$$

So $$F(x) = \dfrac{1}{2} (x^2 +1)^6 + c$$

This method seems very confusing (what is the meaning of the $d$ for example?) and besides, it seems like it would work only in textbook-cases. Is there a 'better' way of integrating these types of basic functions?

p.s. - I looked in my textbook for reference, but the toughest types of functions we have to evaluate are the integrals of for example $h(x) = \dfrac{2x-3}{x^2+6x+10}$ and $k(x) = \dfrac{\arcsin^2(x)}{\sqrt{1-x^2}}$. So the method wouldn't of course have to work for integrals tougher than these types.

答:
Precise meanings of these statements are not usually explained in calculus classes from what I've seen.

It makes more sense when you learn some differential geometry: the $d$ is just the exterior derivative of a differential ($0$-)form.

More directly, it can be seen as just a shorthand notation for integration by substitution, which usually is explained in calculus classes. And frequently, it is the most convenient way of explicit integration (well, at least for things you're supposed to integrate during calculus courses, anyway).

ContentLicense: CC BY-SA 3.0

Title: Is there a smarter way to integrate?


问:
Help with reasoning on how to set up a recurrence relation
I'm working with the following problem:

An alphabet $\Sigma$ consist of the four numeric characters 1, 2, 3, 4, and the seven alphabetic characters a,b, c, d, e, f, g. Find and solve a recurrence relation for the number of words of length n (in $\Sigma$), where there are no consecutive (identical or distinct alphabetic characters.

My problem is how to reason when I'm setting up the recurrence relation. My reasoning goes like this:

If there's a numeric character last, then the character before it can be either an numeric or an alphabetic character which there exist 4+7=11 characters of. So $11a_{n-1}$

If there's a alphabetic character last, then the character before it can only be an numeric character which there exist 4 characters of. So $4a_{n-2}$

Which gives me the recurrence relation: $a_n = 4a_{n-1} + 11a_{n-2}$

But I don't think my reasoning is sound. please help me out

答:
So I(think I) managed to solve the problem(or maybe modified @saulapatz good answer). anyway, here it is:

So basically you have two scenarios.

a numeric character is the last character.

an alphabetic character is the last character.

Scenario 1.
if a numeric character is to be the last character, then we have 4 different characters to choose from. and the character that comes before said numeric character doesn't matter. and that gives
$$4a_{n-1}$$

Scenario 2.
if an alphabetic character is to be the last character, then we have 7 different characters to choose from. But then we must have a numeric character before said alphabetic character, which there are 4 to choose from. that gives us
$$(7*4)a_{n-2}=28a_{n-2}$$
Summing it up, we get
$$a_n = 4a_{n-1} + 28a_{n-2}$$


问:
Limit involving a Differential Equation

Problem:

Consider the differential equation $$\dfrac{dy}{dx} + 10y = f(x) , \
> x>0,$$ where $f(x)$ is a continuous function such that $\displaystyle \lim_{x \to \infty} f(x) = 1$.

Find the value of $\displaystyle \lim_{x \to \infty} y(x)$.

My attempt:

Since this is a Linear Differential Equation, we can solve for $y(x)$ to get,

$$ y(x) = \dfrac{C + \int e^{10x} f(x)}{e^{10x}} $$

which gives,

$$ \require{cancel} \lim_{x \to \infty} y(x) = \cancelto{0}{\lim_{x \to \infty} \dfrac{C}{e^{10x}}} + \lim_{x \to \infty}\dfrac{{\int e^{10x} f(x)}}{e^{10x}} =  \lim_{x \to \infty}\dfrac{{\int e^{10x} f(x)}}{e^{10x}} $$

But I don't know how to solve for the last limit involving the integral.

Title: Limit involving a Differential Equation

Body:
hint

$$y=(C+\int_1^xe^{10t}f(t)dt)e^{-10x}$$

with $ f(t)=1+\epsilon(t)$


问:
Confusion on why 2 equivalence classes are either equal or disjoint
So I've just started trying to teach myself some topology and in the book I'm reading there is a proof that 2 equivalence classes are either equal or disjoint.

However I'm a bit confused on why the author just randomly states any 2 equivalence classes surely it should be any 2 equivalence classes defined by the same equivalence relation or is it that any 2 equivalence classes regardless whether or not the equivalence relations on them is equal or not are either equal or disjoint?

Thanks in advance.

答:
Yes, the equivalence classes must be defined by the same equivalence relation. Otherwise the statement is not true. For example,take the set $X=\{1,2,3\}$ ande define these equivalence relations:

for all $x,y\in X$, set $x$ and $y$ equivalent

for all $x\in X$, $x$ is equivalent only with itself and with no other elements.

In the first case there is only one equivalence class, namely the whole $X$. In the second case there are three equivalence classes, $[1]$,$[2]$ and $[3]$. As you can see the classes of the second equivalence relation are not dijoint from the only class of the first equivalence relation.


问:
using pigeonhole principle for a hand of thirteen cards
Say I shuffle and deal a hand of thirteen cards. How can I apply the pigeonhole principle in these cases:

The hand has at least four cards in the same suit

The hand has at exactly four cards in some suit

The hand has at least five cards in some suit

The hand has exactly one suit containing four or more cards

I have problems in which to assign holes and pigeons to, can anyone help?

答:
For the first question, let the four suits be the holes, and the thirteen cards be the pigeons. This is because each time you deal a card of a certain suit, it is like putting a pigeon in a certain hole. The hole fills up with each pigeon, just as the number of cards fill up with each dealt card of a certain suit. By the pigeonhole principle, you are assigning $13$ discrete objects into $4$ containers, so one container must hold at least $\lceil 13/4\rceil=4$ objects.

This makes intuitive sense, for suppose you consciously tried to assign your cards so that no suit had at least four cards in the hand. You could have at most $3$ cards of the same suit for each suit, but in doing so, you would only be able to assign $3\cdot 4=12$ cards. Then the $13^\text{th}$ card must push one of the suits over, so that you have at least $4$ four of the same suit.

For the other cases, the principle doesn't apply. Take the second option for example. You could be dealt $5$ hearts, $3$ diamonds, $3$ clubs, and $2$ spades. You can show the pigeonhole principle does apply in the other cases be finding counterexamples.


答:
I will use boldface for vectors.

We have the following:

Theorem. Let $A$ be an $m\times n$ matrix, and $\mathbf{b}$ an $m\times 1$ vector. If $\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{b}$, then every solution of $A\mathbf{x}=\mathbf{b}$ is of the form $\mathbf{y}_0+\mathbf{s}$, where $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, and every such vector is a solution to $A\mathbf{x}=\mathbf{b}$. In other words, the complete list of solutions to $A\mathbf{x}=\mathbf{b}$ is given by finding a particular solution $\mathbf{y}_0$ to $A\mathbf{x}=\mathbf{b}$, and adding the solutions to $A\mathbf{x}=\mathbf{0}$.

(This is one reason homogeneous systems play such an important role in the theory of systems of linear equations).

Proof. First, if $\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{b}$, and $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, then
$$A(\mathbf{y}_0+\mathbf{s}) = A\mathbf{y}_0 + A\mathbf{s} = \mathbf{b}+\mathbf{0}=\mathbf{b}.$$
So all such vectors are also solutions. Conversely, if $\mathbf{y}_1$ is also a solution to $A\mathbf{x}=\mathbf{b}$, then $\mathbf{s}=\mathbf{y}_1-\mathbf{y}_0$ is a solution to $A\mathbf{x}=\mathbf{0}$, since
$$A\mathbf{s} = A(\mathbf{y}_1-\mathbf{y}_0) = A\mathbf{y}_1 - A\mathbf{y}_0 = \mathbf{b}-\mathbf{b}=\mathbf{0}.$$
And $\mathbf{y}_1 = \mathbf{y}_0+\mathbf{s}$, so $\mathbf{y}_1$ is of the desired form. $\Box$

This tells you that if there is at least one solution to $A\mathbf{x}=\mathbf{b}$, then there are exactly as many solutions to $A\mathbf{x}=\mathbf{b}$ as there are to $A\mathbf{x}=\mathbf{0}$. It does not tell you that there definitely is a solution to $A\mathbf{x}=\mathbf{b}$, though. It could be that $A\mathbf{x}=\mathbf{b}$ is inconsistent, so that it has no solutions.

When you know that $A\mathbf{x}=\mathbf{0}$ has exactly one solution, you actually can say more, though you may not have the tools to justify it right now. It turns out that in this situation, the matrix $A$ must have $m\geq n$ (if $m\leq n$, more columns than rows, then $A\mathbf{x}=\mathbf{0}$ will always have more than one solution). And if $m=n$, then one can prove that the system $A\mathbf{x}=\mathbf{b}$ always has at least (and hence exactly) one solution.

And in fact you would know that the solution to $A\mathbf{x}=\mathbf{0}$ has to be $\mathbf{s}=\mathbf{0}$. If you did not realize this, note that if $\mathbf{s}$ is a solution to $A\mathbf{x}=\mathbf{0}$, then using the theorem with $\mathbf{b}=\mathbf{0}$ tells you that $\mathbf{s}+\mathbf{s}=2\mathbf{s}$ is also a solution. So the uniqueness would tell you that $\mathbf{s}=2\mathbf{s}$, and that can only hold if $\mathbf{s}=\mathbf{0}$ (assuming you are working over a field of characteristic $\neq2$... and if you don't know what that means, you are).

Title: Relation between Ax=0 and Ax=b

Body: I'm not sure how to solve this. My teacher told me that if $A\vec{x}=0$ has one unique solution (the trivial) then $A\vec{x}=\vec{b}$ has only one unique solution. But I don't know how to prove this.


Title: Limit involving a Differential Equation

Body:
Problem:

Consider the differential equation $$\dfrac{dy}{dx} + 10y = f(x) , \
> x>0,$$ where $f(x)$ is a continuous function such that $\displaystyle \lim_{x \to \infty} f(x) = 1$.

Find the value of $\displaystyle \lim_{x \to \infty} y(x)$.

My attempt:

Since this is a Linear Differential Equation, we can solve for $y(x)$ to get,

$$ y(x) = \dfrac{C + \int e^{10x} f(x)}{e^{10x}} $$

which gives,

$$ \require{cancel} \lim_{x \to \infty} y(x) = \cancelto{0}{\lim_{x \to \infty} \dfrac{C}{e^{10x}}} + \lim_{x \to \infty}\dfrac{{\int e^{10x} f(x)}}{e^{10x}} =  \lim_{x \to \infty}\dfrac{{\int e^{10x} f(x)}}{e^{10x}} $$

But I don't know how to solve for the last limit involving the integral.

Body:
$$y(x)=\frac{ \int _1^xe^{10 t} f(t)\,dt}{e^{10x}}+C e^{-10 x}$$
$$\lim_{x\to\infty}\frac{ \int _1^xe^{10 t} f(t)\,dt}{e^{10x}}\to\frac{\infty}{\infty}$$
use L'Hopital rule.
$$\frac{d}{dx}\int _1^xe^{10 t} f(t)\,dt=e^{10x}f(x)$$
so we have
$$\lim_{x\to\infty}\frac{e^{10 x} f(x)}{10e^{10x}}=\frac{1}{10}$$


问:
Is chromatic index of $G$ at most $4$?

Let $G$ be an outerplanar graph. Is chromatic index of $G$ at most $4$?

I proved that if $G$ is outerplanar then $G$ has chromatic number at most $4$ which can be proved easily using $4$ colour theorem.

However I am stuck here. Is there any relation between chromatic number and chromatic index of a graph?

Also using Vizing's Theorem we know that chromatic index of any graph $G$ is either equal to the maximum degree or maximum degree+1  of a vertex.

Can someone please help me out to prove or disprove this?

答:
There is no connection between these two numbers for outerplanar graphs.

Chromatic number
Using the 4CC to prove that the chromatic number of a simple outerplanar graph is at most $4$ is like using a sledgehammer to crack a nut. Furthermore the chromatic number can be proved to be at most $3$.

https://en.wikipedia.org/wiki/Outerplanar_graph#:~:text=According%20to%20Vizing's%20theorem%2C%20the,one%20plus%20the%20maximum%20degree.

Chromatic index

In the comments it is pointed out that there is no bound on this number for simple outerplanar graphs.

Title: Is chromatic index of $G$ at most $4$?


问:
Determine with proof if the following integral converges
Determine if the following integral converge.
$$\int^\infty_2 \ln\left(\sec\left(\pi\over x\right)\right)\ dx$$

I have a function to compare the above over the interval [2, 3], but am struggling to find one that will compare nicely over [3, $\infty$]. $1\over x^{1.1}$ can be used to show that it does converge using comparison, but is having trouble showing the function is greater than the one above over [3, $\infty$]. Any help would be appreciated!

答:
$$\sec(x)\approx1+\frac{x^2}{2}+O(x^4)...$$
$$\sec\left(\frac\pi x\right)=1+\frac{k_1}{x^2}+O(x^{-4})$$
now we know that:
$$\ln(x)\approx x-\frac{x^2}{2}+O(x^3)$$
so for large $x$ we can say that:
$$\ln\left[\sec\left(\frac\pi x\right)\right]\approx\frac{k_1}{x^2}+O(x^{-4})$$
and then by comparison test we know that $1/x^2$ integral converges


问:
construct weighted shift operator
Let $H$ be an infinite dimensional separable complex Hilbert space with orthonormal basis.

Let $W \in B(H)$ be a unilateral forward weighted shift. i.e. $We_n=w_n e_{n+1}$, where $(w_n) \in \ell^{\infty}$.

Show that there exists a unitary operator $U\in B(H)$ so that $U^*WU$ is a unilateral weighted shift with weight sequence $(|w_n|)_n$.

My work:

I have $    Ue_n=
\begin{cases}
    \frac{w_n}{|w_n|}e_n,& \text{if } w_n \ne 0\\
    e_n,              & \text{if } w_n=0
\end{cases} $

I showed that $U$ is unitary.

Now I want to show that $U^*WU$ is a unilateral weighted shift with weight sequence $(|w_n|)_n$.

So in the case $w_n \ne 0$, I have

$U^*WU e_n= U^*W(\frac{w_n}{|w_n|}e_n) = \frac{w_n}{|w_n|}U^*W(e_n)= \frac{w_n}{|w_n|}U^*w_n(e_{n+1}) = \frac{w_n}{|w_n|}w_nU^*(e_{n+1}) = \frac{w_n}{|w_n|}w_n\frac{w_{n+1}}{|w_{n+1}|}e_{n+1} = |w_n|e_{n+1}$.

I'm not sure if the last equality is correct, because I just sort of forced it in the last step...

If not, could someone let me know how to fix it?

Also, any other suggestions will be highly appreciated too!

Thank you.

答:
Let's define a sequence $u$ such that:
$$
u_0 = 0\\
u_{n+1}=u_n+\arg w_n
$$
And for the operator $U$ holds: $Ue_n = e^{iu_n}e_n$

Now let's proove that it works:$$
U^*WUe_n = e^{iu_n}U^*We_n=e^{iu_n}w_nU^*e_{n+1}=e^{iu_n}w_ne^{-iu_{n+1}}e_{n+1}= \\=\exp(iu_n+ln|w_n|+i\arg w_n-iu_{n+1})e_{n+1}=|w_n|\exp(iun+i\arg w_n - iu_n-i\arg w_n)e_{n+1}=|w_n|e_{n+1}
$$

Title: construct weighted shift operator


问:
Unit ball of $L^p$ spaces and convex combinations
I have seen, as an application of the Krein-Milman theorem, that $\text{conv}( \text{ex}B)=B$, where $B$ is the unit ball of $L^p$ (on a compact space of $\mathbb{R}^n$), $p \in (1,\infty)$, $\text{conv}$ means convex hull, $\text{ex}$ are the extremal points.

However I am confused, do we really need to apply Krein-Milman here? Indeed, $\text{ex}B=S$, the unit sphere: this holds as $||\cdot||_p$ strictly convex. Also, $\text{conv}{S}=B$ in any normed space (that is non trivial).

Is there any fallacy in my reasoning?

@JustDroppedIn

Here is a roadmap: using Banach-Alouglu, one has that $B_{X^*}$ is $w^*$ sequentially compact, convex and non empty (provided the normed space $X$ is separable).

Now,  $||x^*||:=\sum_n 2^{-n}x^*(x_n)$ is a norm on $X^*$, where $\{x_n\}_n$ form a countable dense subset of $B_X$. It has the properties that $B_{X^*}$ is compact, and that convergence of bounded sequences of functionals is equivalent to to $w^*$ convergence.

Applying Krein Milman with this new norm on $X^*$ you obtain $\overline{\text{conv}( \text{ex}B_{X^*})}=B_{X^*}$, the closure being taken in the new norm. However due to the properties above, that closure is exactly the $w^*$ closure.

And now one can substitute $X$ to $L^q$, where $q$ is the dual index of $p$.

答:
The use of Krein-Milman to prove the result in $L^p$-spaces is a quite over-powered  approach. As you show, it can be done much simpler.

I can see two applications of Krein-Milman: (1) as existence result: compact, convex sets do have extreme points, or (2) as prove of non-compactness: if $conv(ex\ B)$ is to small (e.g., empty) such that its closed (in any topology) convex hull can never be equal to $B$ then $B$ cannot be compact in any topology. This is the case for the unit balls of $L^1$, $C(K)$, $c_0$. So existence proofs that rely on compactness of bounded sets are doomed to fail in such spaces.


问:
Is $\sqrt{64}$ considered $8$? or is it $8,-8$?
Last year in Pre-Algebra we learned about square roots. I was taught then that
$\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get.
But this year, with a new school and teacher in a different state, our teacher is telling us that:
$\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is:
$+\sqrt{64}=8$

And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as
$y^2=n$ I see where he is coming from.

Upon researching this Wikipedia says:

For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$

And Wolfram MathWorld says:

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$

But on the other side, Wolfram Alpha, when given "The square root of 9" gives only 3.

So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

答:
Ultimately, it's a matter of notation. There is no right or wrong answer.

Over time, for better or for worse, mathematicians have adopted the radical sign to mean the principle (nonnegative) square root. In reality, true square roots are either positive or negative.

The first thing to understand is that when you are discussing true square roots, $ \sqrt{25} \ne 5 $. In mathematics, having equality means having one and only one distinct solution. The true $ \sqrt{25} $ has two distinct solutions, so there is nothing equal to it.

Consider the following,

$$ x^2 = 100 \Longleftrightarrow x = \sqrt{100} \Longleftrightarrow x = 10 \lor x = -10 $$

The $ x = 10 \lor x = -10 $ at the end is key here. Both $ x = 10 $ as well as $ x = -10 $ satisfy the first equation, so the assertion is true. However, it is untrue to say (in any context) that $ x = 10 \land x = -10$, since it implies that $ 10 = -10 $, which is obviously false.

Simultaneously, $ x = 10 \lor x = -10 $, but $ x \ne 10 $ (and $ x \ne -10 $). That's not a mistake. When combined, either $ x = 10 $ or $ x = -10 $ are true, but neither are true alone. Both statements directly contradict each other, therefore neither contract each other.

In the equation $ \sqrt{25} = \pm{5} $, plus-or-minus is being used to represent a value with two solutions, i.e. $ \pm{5} = 5 \lor \pm{5} = -5 $. $ \pm{5} $ is not a number, but rather, a value with two solutions; $ \pm{5} \in \{-5, 5\} $, nothing more.

Using plus-or-minus on a true square root does nothing, since the square root is already positive or negative; $ \pm{(\pm{n})} \in \{-(\pm{n}), (\pm{n})\} = \{\pm{n}, \pm{n}\} = \{\pm{n}\} \therefore \pm{(\pm{n})} = \pm{n} $. By the same token, negating a square root or making it positive have no effect.

In the quadratic formula,

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, ax^2 + bx + c = 0 $$

The $ \pm $ can be substituted with either $ + $ or $ - $ and it still holds true. The quadratic formula works regardless of whether you believe $ \sqrt{} $ means principal or true square root. Neat, huh?


问:
Showing how much bigger $|\mathbb{R}|$ is than $|\mathbb{N}|$
In one of my lectures to a group of Discrete Mathematics students (all of which are CS majors, not math), I go over the proof that the cardinality of the continuum is strictly larger in size than the cardinality of the naturals; ie the reals are uncountable.

This lesson closes out our disscussion on generalized functions, and I found students understand the proof of Cantor's argument fairly well. However, this proof only shows that there exists at least one real outside any bijective function, when in fact nearly all reals are excluded.

As stated here

Which reals could possibly be “missing” from our universe? Every real you can name—42, π, √e, even uncomputable reals like Chaitin’s Ω—has to be there, right? Yes, and there’s the rub: every real you can name. Each name is a finite string of symbols, so whatever your naming system, you can only ever name countably many reals, leaving 100% of the reals nameless.

Or did you think of only the rationals or algebraic numbers as forming a countable dust of discrete points, with numbers like π and e filling in the solid “continuum” between them? If so, then I hope you’re sitting down for this: every real number you’ve ever heard of belongs to the countable dust! The entire concept of “the continuum” is only needed for reals that don’t have names and never will.

Which is incredible, and I really want to convey this to my students as well, however I don't know of a good proof or argument that is at their level (so no measure theory) to show that "100%" of the real numbers are excluded from any possible function from $\mathbb{N}$

Clearing up my question so that it's easy to see exactly what I am asking for:

Can it be proved using only basic set theory, functions, and calculus that "100%" of all real numbers are excluded from any function $f:\mathbb{N}\rightarrow\mathbb{R}$?

答:
Although measure theory is of course needed to do this formally, perhaps there's a way to discuss this with your students informally.

You could argue like this.

List the numbers: $x_1 = f(1)$, $x_2 = f(2)$, $x_3 = r(3)$,...

Draw a sketch of the first few on a number line; it doesn't matter exactly where you put them.

Surround $x_1$ with an interval labelled $I_1$ of total length $1/2$, going from $x_1-1/4$ to $x_1 + 1/4$.

Next, surround $x_2$ with an interval labelled $I_2$ of togal length $1/4$, going from $x_2 - 1/8$ to $x_2 + 1/8$. So, the total length of $I_1$ and $I_2$ is at most $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$.

And now continue. At the $n$th stage, you'll have intervals $I_1,...,I_n$ whose lengths total up to $1 - \frac{1}{2^n}$.

And so, even if you continue letting $n$ go to infinity and take all the intervals all together, their lengths will total up to at most $1$! How can that possibly cover the entire real number line, which has infinite length?

If you wanted to, you could repeat this, except that you can start with $I_1$ having length $.00005$ instead of $.5$; the total lengths will total up to at most $.0001$!


问:
Sylow subgroup of a group generated by two subgroups
Let $G$ be a finite group and $H=\langle A ,B\rangle$ be a subgroup of $G$ generated by subgroups  $A$ and $B$ of $G$. Is it true that $H_r=\langle A_r,B_r\rangle$, where $H_r$, $A_r$ and $B_r$ are Sylow subgroups of $H,A,B$ respectively.

答:
No, take $G=H=S_3$, $A=\langle (12)\rangle$, $B=\langle (13)\rangle$.
Note: it is true when $H=AB$, see Sylow $p$-subgroup of a direct product is product of Sylow $p$-subgroups of factors


问:
Theorem about bounded functions.
Is there a theorem that says that if $f:[a,b]\rightarrow \mathbb{R} $ is a piecewise function, then it is bounded?

答:
Not without more restrictions, like continuity (which is enough).

For example, consider
$$f(x) = \begin{cases}
0,& x=a \\
\dfrac{1}{x-a},& a<x\leq b\\
\end{cases}
$$

If your function is continuous, then it is bounded since the continuous image of a compact set is compact (in $\mathbb{R}$, this means it is closed and bounded).

Title: Theorem about bounded functions.


问:
How to calculate the number of border pieces for a jigsaw puzzle?
For example I have picture which can be any size, and I want to make jigsaw puzzle with an approximate number of pieces.

So I know the picture width and height, I know the approximate number of pieces. I need to know the puzzle's border dimensions in a way where the pieces' size nearly a rectangle.

答:
Okay.  SO the picture is $W$ inches by $H$ inches.  And has $N$ pieces.

If there are $w$ pieces along the top edge (including corners) and $h$ pieces along the side (including corners) we need $w\cdot h = N$.

Assuming that these are $w,h, N$ and integers are all rows and columns have the exact same number of pieces our options will be limited by the prime divisors of $N$. If we are allowed a little nudging here or there there are an infinite number of solutions.

If we want to find the proportions of the pieces as $a\times b$ in rectangles we must have

$aw = W; bh=H; abN=WH; wh=N$.

If you know $N, W, H$ we can set any one of the others, $a$,$b$,$w$, or $h$ to anything we like and express the rest in terms of them.

example:  Let $h$ be a number we like.

Then $w = \frac Nh$

$a= \frac Ww$ and $b=\frac Hh$

(The fourth falls into place $abN= \frac {WH}{wh}N=\frac {WH}{\frac Nhh}N=WH$)

The number of edge pieces will be $2(w+h)-4$ (the $-4$ is because we don't want to count the corners twice.)


问:
find matrix such that $ Ax=(1,1,1)^t$ has exactly three distinct solutions
Does there exist a matrix $3\times 3$ order such that $ Ax=(1,1,1)^t$ has exactly three distinct solutions? If so, find $A$.

I have no idea in this question please help.

答:
No:
$Ax=\begin{pmatrix}1\\1\\1\end{pmatrix}$ is consistent implies the system possess only one or infinitely many solutions since, $x_0$ is a solution $\implies$ all solutions are obtained by adding $x_0$ to the general solution of the associated homogeneous system $Ax = 0.$


问:
Sturm-Liouville Regular problem
I'm looking at the following S-L ODE:

$\frac{d}{dx} \left[e^{ix} f'(x)\right] + e^{ix} \lambda f(x) = 0$,

where $i$ is the complex number, and $\lambda$ denotes the eigenvalues. However, this can only have real eigenvalues if the problem is regular, i.e., that somehow, one has to show that $e^{ix} > 0$ over some interval. As this is a complex function, I'm having difficulty showing this.

Any ideas?

答:
If homogenous Dirichlet boundary conditions are available, then multiplying by the complex conjugate of $f$ and integrating by parts once yields

$$\lambda = \frac{\displaystyle\int^{b}_{a} \mathrm{e}^{ix} |f'|^2 \, \mathrm{d}x}{\displaystyle\int^{b}_{a} \mathrm{e}^{ix} |f|^2 \, \mathrm{d}x}$$

For $f= \mathrm{e}^{i sx}$, which is the solution to any linear operator, the fraction above turns out to be real. Moreover, $\lambda = |s|^2$.

Title: Sturm-Liouville Regular problem


问:
Counting problem question
"How many 3 digit integers are possible, such that none of the digits appear more than twice, and none of the digits are equal to zero?"  At first glance the problem seemed easy enough:  you have 9 possible digits and 3 slots, with no two digits appearing more than twice. I assumed $9 \times 9 \times 8 = 648$ would give the solution. To my surprise, my answer was not correct. The correct answer is 720. The way they solved it was by taking all possibilities $9 \times 9 \times 9 - 9$.  (Here $-9$ expresses the break on the constraint.)  I understand what they did, but my question is why does $9 \times 9 \times 8$ fail to account for missing possibilities? I cannot wrap my head around it, $9 \times 9$ would account for any 2 digits and allowing repetition and you would have 8 more possibilities. Whats wrong with my logic?

答: If the first two digits you pick are different, there is no restriction on the third. So for the third digit there are only 8 possibilities in a few rare cases. Most of the time it's 9 there too.</p>

Title: Counting problem question


答: The factor $8$ in your calculation was based on assuming that one of the $9$ digits is unavailable for the last digit.  But that assumption is correct only if the first two digits are the same. If the first two digits are different, then any of the $9$ digits is OK for the last one.

Title: Counting problem question

Body: "How many 3 digit integers are possible, such that none of the digits appear more than twice, and none of the digits are equal to zero?"  At first glance the problem seemed easy enough:  you have 9 possible digits and 3 slots, with no two digits appearing more than twice. I assumed $9 \times 9 \times 8 = 648$ would give the solution. To my surprise, my answer was not correct. The correct answer is 720. The way they solved it was by taking all possibilities $9 \times 9 \times 9 - 9$.  (Here $-9$ expresses the break on the constraint.)  I understand what they did, but my question is why does $9 \times 9 \times 8$ fail to account for missing possibilities? I cannot wrap my head around it, $9 \times 9$ would account for any 2 digits and allowing repetition and you would have 8 more possibilities. Whats wrong with my logic?


问:
Proof regarding multiples of 3
Let $a = x^2 + y^2 + z^2$ for $a, x, y, z \in \mathbb{Z_{>0}}$. I'm trying to prove that if $a$ is divisible by 3, then $x, y, z$ are either ALL divisible by 3, or NONE of them are divisible by 3.

I am not sure where to start. I wrote down the definition, $\exists w \in \mathbb{Z_{>0}}$ such that $a = 3w$, but from there I am lost. I also know that if $x$ is divisible by 3 then so is $x^2$ but am not sure how to work that in here.

答:
Hint

$x^2$ is equal either to $0$ or to $1$ modulo $3$.

Title: Proof regarding multiples of 3


问:
What is the sum of $\sum_{j=1}^{N} \frac{j}{k+\log j}$ where $k$ is some positive constant?
What is the sum of $\sum_{j=1}^{N} \frac{j}{k+\log j}$ where $k$ is some positive constant and $\log$ is the natural log? It would even help to know what the sum of $\sum_{j=2}^{N} \frac{j}{\log j}$ is if there is one. Mathematica has provided little help on this matter, and I am not sure where to start. A strong. upper bound would also be helpful

答:
Starting from
$$  k \leq k+\ln j \leq k + \ln N  \text{,}  $$
we have
\begin{align*}
\frac{N(N+1)}{2k}
    &= \sum_{j=1}^N \frac{j}{k}  \\
    &\geq \sum_{j=1}^N \frac{j}{k+\ln j}  \\
    &\geq \sum_{j=1}^N \frac{j}{k+\ln N}  \\
    &= \frac{N(N+1)}{2(k + \ln N)}  \text{.}
\end{align*}

Those are relatively tight bounds. You don't give any hint about how tight you need, so if this isn't tight enough, the Question needs a little expansion.


问:
Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$
If $r$ and $s$ are roots of $x^3-cx+d=0$, then

$$r^3-cr+d=s^3-cs+d , $$

for $r \neq s$. Solving the equation for $c$ gives me

$$ c=r^2+rs+s^2$$

Now to find an expression for $d$

$$ r^3-cr+d=0 $$

$$ d=r(c-r^2) $$

Using the expression for $c$ gives me

$$d=sr^2+rs^2$$

Assume $t$ is a root of the second equation

$$t^3-ct+d=0$$

We can represent $c$ and $d$ in terms of $r$ and $s$ which gives us

$$s*r^2+r*s^2=t(t^2-r^2-rs-s^2)$$

Plugging this into wolfram alpha gives us $t=r+s$, which is what we wanted to show. But how can I solve for $t$ algebraically? Or perhaps there is another way to solve this question?

答:
Hint

You have to prove that

$$(r+s)^3-c(r+s)-d=0$$
and you know that, by Vietta in $x^3-cx+d=0$,

$$u+r+s=0$$


问:
Calculate $\int_{\mathcal{C}(0,3)} \frac{e^{2/z}}{z-2}\,dz$
$$\int_{\mathcal{C}(0,3)} \frac{e^{2/z}}{z-2}\,dz$$
Solution approach:
I thought of calculating that problem using the residue theorem, therefore I calculated the residue at $z=2$, which turned out to be equal to $e$ (Wolfram confirmed). However, I am not sure if I should calculate the residue at $z=0$, but I am having problem doing that. Thanks for help!

答: That pole is enclosed, so yes, we include its residue. In a neighbourhood of $z=0$, the integrand is a Laurent series,$$-\tfrac12(1+\tfrac12z+\tfrac14z^2+\cdots)(1+\tfrac{2}{z}+\tfrac{2}{z^2}+\cdots).$$Since $\tfrac1z=z^n\tfrac{1}{z^{n+1}}$, the $\tfrac1z$ coefficient is$$-\tfrac12\sum_{n\ge0}\left(\tfrac12\right)^n\tfrac{2^{n+1}}{(n+1)!}=-\sum_n\tfrac{1}{(n+1)!}=1-e.$$This makes the integral $2\pi i$. WA agrees, if you know how to ask.


问:
Find the gradient and hessian of $g(x)=f(Ax)$

Let $f(z):\mathbb R^m\rightarrow \mathbb R$ be a real-valued function
from $\mathbb R^m$ to $\mathbb R$. Let $A^{m\times n}, x^{n\times 1}$,
and let $g(x)=f(Ax)$.

Find the gradient and hessian of g(x) in terms of A, $\nabla f(x)$,
and the hessian $H(f(x))$

I tried using the chain rule $\underbrace{\nabla g(x)}_{n\times 1}=\underbrace{A}_{m\times n} \underbrace{\nabla f(Ax)}_{n\times 1}$

But it seems as if the dimensions don't work out. Is the dimension of $\nabla g(x)$ actually mx1? I thought it should be nx1 because there are n  elements of x. If that is the case, my application of chain rule is probably wrong.

答:
Let $h : \mathbb{R}^n \to \mathbb{R}^m$ be the map $h(x)=Ax$. Then $h$ is linear and $\mathrm{d}h(x) = A$. Remark that $g = f\circ h$.  The chain rule then says
$$
\forall x \in \mathbb{R}^n,~ \mathrm{d}g(x) = \mathrm{d}f(h(x))\circ \mathrm{d}h(x)
$$
Their gradients are defined thanks to the euclidean metric :
\begin{align}
\forall x\in \mathbb{R}^n, \forall v \in \mathbb{R}^n,~ \langle\nabla g(x),v \rangle_{\mathbb{R}^n} &=\mathrm{d}g(x)v \\
&= \mathrm{d}f(h(x))\circ \mathrm{d}h(x)v \\
&= \mathrm{d}f(Ax) Av \\
&= \langle \nabla f(Ax),Av\rangle_{\mathbb{R}^m} \\
&= \langle A^T\nabla f(Ax),v\rangle_{\mathbb{R}^n}
\end{align}
the last equality resulting of the fact that for all $u\in \mathbb{R}^n,\forall v\in \mathbb{R}^m,~ \langle Au,v \rangle_{\mathbb{R}^m} = \langle u,A^Tv\rangle_{\mathbb{R}^n}$. It follows that $\nabla g(x) = A^T\nabla f(Ax) \in \mathbb{R}^n$


答: Your $9\cdot9\cdot8$ is the number of ways to pick any non-zero first digit, any non-zero second digit, and a non-zero third digit different from the first digit. Thus, it fails to count such allowable numbers as $121$, $848$, and so on. There are $9\cdot8=72$ such numbers: $9$ choices for the first and third digits, and then $8$ choices for the middle digit. And sure enough, $648+72=720$, so we’ve accounted for all of the numbers missed by your original calculation.

Title: Counting problem question


问:
The quotient space of a closed orientable surface by a $\Bbb Z_2$-action
Let $M_g$ be a closed orientable surface, embedded in $\Bbb R^3$ such that the $y$-axis intersects it in $2g + 2$ points and $M_g$ is invariant under the 180$^\circ$ rotation around the $y$-axis. Thus, the 180$^\circ$ rotation around the $y$-axis defines a $\Bbb Z_2$-action on $M_g$ with $2g+2$ fixed points. I want to show that the quotient space $M_g/\Bbb Z_2$ is homeomorphic to the $2$-sphere $S^2$. But, since this action is not free, I cannot apply some well-known theorems about quotient manifolds. How can we check this?

答:
Draw a convincing picture.

This image shows the involution you are describing. What is a fundamental domain for it? Well it looks like the top/bottom half of the image as an open subset, and then these have to be glued together along the intersection of the image with the horizontal slice coming into and out of your screen. This slice cuts the handles as well, so when the gluing is carried out, each is sewn shut into a disk. Since the top piece is a disk, and every handle in the surface has been sewn shut, this space is a topological sphere.


问:
Definition of orientation-preserving homeomorphism
What is the definition of an orientation-preserving homeomorphism for a topological manifold $M$? If no such notion exists, then what is the definition of an orientation-preserving homeomorphism of $\mathbb{R}^n$?

I have seen these terms pop-up, but I cannot seem to find an actual definition of them.

答:
On any topological $n$-manifold $M$, define an orientation of $M$ to be a function $\mu$ defined on $M$ such that for each input $x \in M$, the output $\mu(x)$ is one of the two generators of the infinite cyclic group $H_n(M,M-x)$, and the following property holds: for each embedded open $n$-ball $B \subset M$ there exists a generator $\mu_B$ of the infinite cyclic group $H_n(M,M-B)$ such that for each $x \in B$ the inclusion induced homomorphism $H_n(M,M-B) \mapsto H_n(M,M-x)$ maps $\mu_B$ to $\mu(x)$.

Now one proves

Theorem (and definition): For every topological manifold $M$ exactly one of two possibilities holds: either $M$ has exactly two orientations, in which case we say that $M$ is orientable; or $M$ has no orientations, in which case we say that $M$ is nonorientable.

This theorem is one of the preliminary steps to the proof of Poincare Duality; see for example Hatcher's book "Algebraic Topology". In fact, proving that the relative homology groups $H_n(M,M-B)$ and $H_n(M,M-x)$ are infinite cyclic is also one of the preliminary steps.

Now to your question.

Let $M$ be a connected topological manifold.

If $M$ is nonorientable, then it makes no sense to ask whether a homeomorphism of $f$ preserves orientation, and the whole concept of "preserving orientation" is undefined for $M$.

If on the other hand $M$ is orientable then to say that a homeomorphism $f : M \to M$ is orientation preserving means that for either of the two orientations $\mu$ of $M$, and for any $x \in M$, the induced isomorphism $f : H_n(M,M-x) \to H_n(M,M-f(x))$ takes $\mu(x)$ to $\mu(f(x))$.


问:
Find all functions such that $f(xf(y)+y)=f(xy)+f(y)$
Find all functions, $f:\mathbb{R}\to \mathbb{R}$ such that $$f(xf(y)+y)=f(xy)+f(y)$$

What i have got till now is

Let $$P(x,y): f(xf(y)+y)=f(xy)+f(y)$$Then
\begin{align*}
&P(0,0):  f(0)=f(0)+f(0)\implies f(0)=0\\
&P\Bigg(\frac{y}{y-f(y)},y\Bigg): f(y)=0
\end{align*}
Hence $f(x)=0$ is a solution.

But we can see $f(x)=x$ is also a solution but I can't find that.

答:
For all $y$ such that $f(y) \ne y$, replace $(x,y) = (\frac{y}{y-f(y)},y)$ in the equation, we have that $f(y) = 0$. Indeed,
$$f(\frac{y}{y-f(y)}f(y)+y) = f(\frac{y}{y-f(y)}y)+f(y)$$
$$\iff f(\frac{y^2}{y-f(y)}) = f(\frac{y}{y-f(y)}y)+f(y)$$
$$\iff 0 = f(y)$$

So, we have for all $x$, $f(x) = 0$ or $f(x) =x$.

If $\exists y_0 \ne 0$ such that $f(y_0) = 0$

Then we have, for all $x$
$$f(xf(y_0)+y_0)=f(xy_0)+f(y_0)$$
$$\iff f(x.0+y_0)=f(xy_0)+0$$
$$\iff 0=f(xy_0)$$
Hence, for all $x\in \mathbb{R}$: $f(x)=0$

If $\not\exists y_0 \ne 0$ such that $f(y_0) = 0$

Then for all $x\ne0$, $f(x) \ne 0$. Because $f(x)$ can only receive value in $\{0,x \}$, then $f(x) = x$ for all $x$

Conclusion

There are 2 solutions: $f(x) = 0$ or $f(x) = x$.


问:
How do I solve $\min_x \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$ for $\lVert x \rVert_2 = 1$.
Let $f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$.
where $x, c_1, c_2, \dots, c_k \in \mathbb R^n$.
What fast iterative methods are available for finding the (approximate) min of $f$ with the constraint $\lVert x \rVert_2 = 1$?

Notes:

$f$ is convex and and non negative, that is, $c_i$ positively span $\mathbb R^n$

For my use case $k \gg n$ and (rougly) $3 \le n \le 50$ and $100 \le k \le 10000$.
I tried projected subgradients (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.

I've also tried approximating $f$ with a smooth maximum LogSumExp and doing gradient descent with projection. I also tried unconstrained gradient descent with a quadratic penalty. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in SQP because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.

Edit:
I believe this problem is equivalent to a linear programming problem with quadratic equality constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a SDP method but I'm not sure.

答:
EDIT: I have included all of the MATLAB code here: https://gitlab.com/TheRTGuy/point-cloud-support-function-minimization. There I also have other evaluations.

Given the assumption that $f$ is always nonnegative, then this implies that the convex hull of $C=\{c_1,\ldots,c_k\}$ contains the origin. Were this not the case, then $f$ is always negative, and the origin is separable from the point cloud. In this case, minimizing $f$ just amounts to finding the supporting hyperplane that maximizes the distance from the origin, a problem that is solved easily using convex optimization (e.g. solving the dual of the primal problem).

The problem, as some pointed out in the comments, boils down to enumerating the facets of the convex hull of $C$, and finding the one closest to the origin. This is a very hard problem, see https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node22.html#polytope:Vredundancy, since the number of facets is exp. in $n$ and $k$. So, forget about an "efficient" exact solution.

However, in my research I needed a way to compute a tight overapproximation of the convex hull of a point cloud. Here is a simplified adaptation, of the approach I used, to OP's problem:

The approach:

Generate vertices of a regular simplex, $\{h_1,\ldots,h_{n+1}\}$, and compute $$b_0 = \begin{pmatrix}\max_{c\in C} h_1^\top c \\ \vdots \\ \max_{c\in C}h_{n+1}^\top c\end{pmatrix}.$$

Form the new simplex $S_0 = \{x\in\mathbb{R}^n~\vert~ H_0x \leq b_0\} = \operatorname{Conv}(\{s^1_0,\ldots,s^{n+1}_0\})$, where $$H_0 = \begin{pmatrix}h_1^\top\\\vdots \\h_{n+1}^\top\end{pmatrix},$$
and $s^i_0$ is the $i$-th vertex of the simplex. This simplex tightly bounds the set $C$.

Optionally, perturb the simplex by a random rotation.

Set $i=1$, and $f_0 = \min_j [b_0]_j$, where $[\cdot]_j$ is the $j$-th component of a vector.

For each vertex $s^j_{i-1}$ solve the LP
$$
x_j=\arg\min_x ~ g(x)= x^\top s^j_{i-1} \quad \\s.t. \\\quad (c_l - s^j_{i-1})^\top x \leq -1, ~~~\forall l\in\{1,\ldots,k\}, \\
\quad (s^l_{i-1}- s^j_{i-1})^\top x \leq -1, ~~~\forall l\in \{1,\ldots,n+1\}\setminus\{j\},\\
\quad x^\top s^j_{i-1} \geq -1.
$$
The LP basically finds a supporting hyperplane of $C$ that separates it from $s^j_{i-1}$. This plane is often a facet of $\operatorname{Conv}(C)$.

Set
$$
H_i = \begin{pmatrix}\frac{x_1^\top}{\lVert x_1\rVert}\\\vdots\\\frac{x_{n+1}^\top}{\lVert x_1\rVert}\end{pmatrix},\qquad b_i = \begin{pmatrix}\frac{x_1^\top s^1_{i-1}-1}{\lVert x\rVert} \\ \vdots \\ \frac{x_{n+1}^\top s^{n+1}_{i-1}-1}{\lVert x\rVert} \end{pmatrix}.
$$
This forms a new simplex $S_i = \{x\in\mathbb{R}^n~\vert~ H_ix \leq b_i\} = \operatorname{Conv}(\{s^1_i,\ldots,s^{n+1}_i\})$.

Set $f_i = \min\{\min_j[b_i]_j , f_{i-1}\}$, $i:=i+1$, and repeat steps 5-7 for $N$ iterations, or until $f_i = f_{i-1}$ for some consecutive iterations.

Set set the optimal value to $f^* = f_i$.

Evaluation:
The implementation I have is for MATLAB.

I evaluated the approach for $n=2,\ldots,20$, with $k=1000$. For each $n$ I ran the algorithm 3 times on a uniformly generated set $C$, to which I also apply a random rotation, and time each run. The number of iterations is fixed to $N=5$.  In case you are curious, this is the set of MATLAB commands executed to generate the point cloud:

```<code>%Randomly rotated uniform
C = 5*rand(n,N)-2.5;
%Indexing pattern for off-diagonal entries of matrix
idx = nchoosek(1:n,2);
dU = zeros(n);
dth = pi*(2*rand(n*(n-1)/2,1) - 1); %Uniform
dU(sub2ind([n,n],idx(:,1),idx(:,2))) = dth;
dU(sub2ind([n,n],idx(:,2),idx(:,1))) = -dth;
C = expm(dU) * C;
</code>```

To evaluate how good the solution is, for each run I generate 1000000 points, $\{x_k\}$, normally distributed on the unit sphere in $\mathbb{R}^n$, and evaluate $f$ at each point. Then, I compute $d = \min_k\{f(x_k)\} - f^*$ and plot it. Positive means that the solution is good, negative means the solution is bad. Of course, if positive, this does not mean the solution is the best. Furthermore, the higher the dimension, the lower the quality of this evaluation. Ideally, we want to compare it with another algorithm, which I don't have.

Results:

The first plot shows the execution times vs. $n$. The second plot shows the difference $d$ vs $n$. The final plot is just an arbitrary run for $n=20$, which shows $f(x_k)$ and $f^*$.

<img src="https://i.stack.imgur.com/1Died.jpg" alt="Example for $n=20$" />

Conclusion:

On first glance, the approach seems to perform very well. Only a small number of iteratios is necessary to get a good suboptimal solution. Also, solving LPs is very efficient, even for large $n$ and $k$. However, there is no rigorous tight bound for optimality yet. More rigorous analysis is needed. Furthermore, evaluating the approach with normally distributed points on the unit sphere is obviously not the best way, but at least it gives a sense of optimality. This is clearly seen from the second plot where the gap is increasing. If the OP is interested in further evaluation, then I would be willing to share some code.

The intuition is simple: we try to "squeeze" the points by simplices as much as possible, by using the vertices of previous simplices as anchors. The facets of the simplices often coincide with the faces of $\operatorname{Conv}(\{c_i\})$.

Body:
EDIT: I have included all of the MATLAB code here: https://gitlab.com/TheRTGuy/point-cloud-support-function-minimization. There I also have other evaluations.

Given the assumption that $f$ is always nonnegative, then this implies that the convex hull of $C=\{c_1,\ldots,c_k\}$ contains the origin. Were this not the case, then $f$ is always negative, and the origin is separable from the point cloud. In this case, minimizing $f$ just amounts to finding the supporting hyperplane that maximizes the distance from the origin, a problem that is solved easily using convex optimization (e.g. solving the dual of the primal problem).

The problem, as some pointed out in the comments, boils down to enumerating the facets of the convex hull of $C$, and finding the one closest to the origin. This is a very hard problem, see https://www.cs.mcgill.ca/~fukuda/soft/polyfaq/node22.html#polytope:Vredundancy, since the number of facets is exp. in $n$ and $k$. So, forget about an "efficient" exact solution.

However, in my research I needed a way to compute a tight overapproximation of the convex hull of a point cloud. Here is a simplified adaptation, of the approach I used, to OP's problem:

The approach:

Generate vertices of a regular simplex, $\{h_1,\ldots,h_{n+1}\}$, and compute $$b_0 = \begin{pmatrix}\max_{c\in C} h_1^\top c \\ \vdots \\ \max_{c\in C}h_{n+1}^\top c\end{pmatrix}.$$

Form the new simplex $S_0 = \{x\in\mathbb{R}^n~\vert~ H_0x \leq b_0\} = \operatorname{Conv}(\{s^1_0,\ldots,s^{n+1}_0\})$, where $$H_0 = \begin{pmatrix}h_1^\top\\\vdots \\h_{n+1}^\top\end{pmatrix},$$
and $s^i_0$ is the $i$-th vertex of the simplex. This simplex tightly bounds the set $C$.

Optionally, perturb the simplex by a random rotation.

Set $i=1$, and $f_0 = \min_j [b_0]_j$, where $[\cdot]_j$ is the $j$-th component of a vector.

For each vertex $s^j_{i-1}$ solve the LP
$$
x_j=\arg\min_x ~ g(x)= x^\top s^j_{i-1} \quad \\s.t. \\\quad (c_l - s^j_{i-1})^\top x \leq -1, ~~~\forall l\in\{1,\ldots,k\}, \\
\quad (s^l_{i-1}- s^j_{i-1})^\top x \leq -1, ~~~\forall l\in \{1,\ldots,n+1\}\setminus\{j\},\\
\quad x^\top s^j_{i-1} \geq -1.
$$
The LP basically finds a supporting hyperplane of $C$ that separates it from $s^j_{i-1}$. This plane is often a facet of $\operatorname{Conv}(C)$.

Set
$$
H_i = \begin{pmatrix}\frac{x_1^\top}{\lVert x_1\rVert}\\\vdots\\\frac{x_{n+1}^\top}{\lVert x_1\rVert}\end{pmatrix},\qquad b_i = \begin{pmatrix}\frac{x_1^\top s^1_{i-1}-1}{\lVert x\rVert} \\ \vdots \\ \frac{x_{n+1}^\top s^{n+1}_{i-1}-1}{\lVert x\rVert} \end{pmatrix}.
$$
This forms a new simplex $S_i = \{x\in\mathbb{R}^n~\vert~ H_ix \leq b_i\} = \operatorname{Conv}(\{s^1_i,\ldots,s^{n+1}_i\})$.

Set $f_i = \min\{\min_j[b_i]_j , f_{i-1}\}$, $i:=i+1$, and repeat steps 5-7 for $N$ iterations, or until $f_i = f_{i-1}$ for some consecutive iterations.

Set set the optimal value to $f^* = f_i$.

Evaluation:
The implementation I have is for MATLAB.

I evaluated the approach for $n=2,\ldots,20$, with $k=1000$. For each $n$ I ran the algorithm 3 times on a uniformly generated set $C$, to which I also apply a random rotation, and time each run. The number of iterations is fixed to $N=5$.  In case you are curious, this is the set of MATLAB commands executed to generate the point cloud:

%Randomly rotated uniform
C = 5*rand(n,N)-2.5;
%Indexing pattern for off-diagonal entries of matrix
idx = nchoosek(1:n,2);
dU = zeros(n);
dth = pi*(2*rand(n*(n-1)/2,1) - 1); %Uniform
dU(sub2ind([n,n],idx(:,1),idx(:,2))) = dth;
dU(sub2ind([n,n],idx(:,2),idx(:,1))) = -dth;
C = expm(dU) * C;

To evaluate how good the solution is, for each run I generate 1000000 points, $\{x_k\}$, normally distributed on the unit sphere in $\mathbb{R}^n$, and evaluate $f$ at each point. Then, I compute $d = \min_k\{f(x_k)\} - f^*$ and plot it. Positive means that the solution is good, negative means the solution is bad. Of course, if positive, this does not mean the solution is the best. Furthermore, the higher the dimension, the lower the quality of this evaluation. Ideally, we want to compare it with another algorithm, which I don't have.

Results:

The first plot shows the execution times vs. $n$. The second plot shows the difference $d$ vs $n$. The final plot is just an arbitrary run for $n=20$, which shows $f(x_k)$ and $f^*$.

<img src="https://i.stack.imgur.com/1Died.jpg" alt="Example for $n=20$" />

Conclusion:

On first glance, the approach seems to perform very well. Only a small number of iteratios is necessary to get a good suboptimal solution. Also, solving LPs is very efficient, even for large $n$ and $k$. However, there is no rigorous tight bound for optimality yet. More rigorous analysis is needed. Furthermore, evaluating the approach with normally distributed points on the unit sphere is obviously not the best way, but at least it gives a sense of optimality. This is clearly seen from the second plot where the gap is increasing. If the OP is interested in further evaluation, then I would be willing to share some code.

The intuition is simple: we try to "squeeze" the points by simplices as much as possible, by using the vertices of previous simplices as anchors. The facets of the simplices often coincide with the faces of $\operatorname{Conv}(\{c_i\})$.


问:
How to find $N-60$ when it is based on the sum of its prime divisors?
The problem is as follows:

A certain number coincides with the sum of the prime positive divisors
of the number of zeros in which the product of the first $784$
integers ends positive, when expressed in base $15$ (pentadecimal).
Find the number minus $60$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{33}\\
2.&\textrm{39}\\
3.&\textrm{30}\\
4.&\textrm{28}\\
\end{array}$

The official method of solution according to my book is as follows:

$784!=a(15)^n$

Then it goes to say that

$n=194=2\times 97$

to finally say that:

Then it declares that $2+97=99$
and finally saying this:

$99-60=39$

declaring that the official answer is $39$.

But that's my problem with this explanation. How does it get the number from that factorial?

The number it seems way to big to be computed by hand needless to say that from the sole equation of

$784!=a(15)^n$

it doesn't say what is $a$ or what is $n$? What does this has to do with pentadecimal base?. I don't get it.

Can someone explain me what is intended in this question and a better argument of justification in other words a better solution method which implies not fancy calculations or too encyclopedical in nature?

This request is based that my knowledge on the subject is somewhat limited so I need all the steps necessary so I can understand.

How can this problematic number be found?. This problem belongs to a section in my book about the fundamental theorem of arithmetic although I don't see it in any way how could this be applied in this peculiar scenario.

Can someone help me here with an easy to follow and understand method of solution?

答:
To get the number of zeros $n$ with which the product of the first 784 integers ends,

when expressed in base $15$ (pentadecimal), we need to find the highest power of $15$ dividing $784!.$

$15=3\times5$, so $n$ is also the highest power of $5$ dividing $784!$.

That can be obtained using Legendre's formula:

$n=\left\lfloor\dfrac{784}{5}\right\rfloor+\left\lfloor\dfrac{784}{25}\right\rfloor+\left\lfloor\dfrac{784}{125}\right\rfloor+\left\lfloor\dfrac{784}{625}\right\rfloor=156+31+6+1=194.$


答:
Let $t$ be the $3$rd root of the equation $x^3 - cx + d = 0$

So $$t^3 - ct + d = 0 \tag{1}$$

Also $$r + s + t  = 0 \implies t = -(r+s) \tag{2}$$

Substitute the value of $t$ from $(2)$ into $(1)$ to get
$$-(r+s)^3 + c(r+s) + d = 0 \equiv (r+s)^3 - c(r+s) - d = 0$$

Hence the proof.

Title: Show that if $r$ and $s$ are positive roots to the equation $x^3-cx+d=0$, then $r+s$ is a root of $x^3-cx-d=0$

Body:
If $r$ and $s$ are roots of $x^3-cx+d=0$, then

$$r^3-cr+d=s^3-cs+d , $$

for $r \neq s$. Solving the equation for $c$ gives me

$$ c=r^2+rs+s^2$$

Now to find an expression for $d$

$$ r^3-cr+d=0 $$

$$ d=r(c-r^2) $$

Using the expression for $c$ gives me

$$d=sr^2+rs^2$$

Assume $t$ is a root of the second equation

$$t^3-ct+d=0$$

We can represent $c$ and $d$ in terms of $r$ and $s$ which gives us

$$s*r^2+r*s^2=t(t^2-r^2-rs-s^2)$$

Plugging this into wolfram alpha gives us $t=r+s$, which is what we wanted to show. But how can I solve for $t$ algebraically? Or perhaps there is another way to solve this question?


问:
Why is the following not a primitive recursive function?
Consider the following set of four symbols $M=\{S,\square,\circ,-\}$.  Take $S\colon \mathbb{N}\to \mathbb{N}$ to be the successor function $x\mapsto x+1$.  Given $f,g\colon \mathbb{N}\to \mathbb{N}$, let $f\circ g$ denote the usual composition operator on two functions $x\mapsto f(g(x))$, let $f-g$ be the function where $$
x\mapsto \begin{cases}f(x)-g(x) & \text{if $f(x)\geq g(x)$}\\ 0 & \text{ otherwise,}\end{cases}
$$
and let $f^{\square}$ be the "iteration" of $f$ defined recursively by the conditions $f^{\square}(0)=0$ and
$$
f^{\square}(x+1)= f(f^{\square}(x)).
$$

Instead of writing expressions as above, I will instead use Polish notation.  So, for example $\circ \square SS$ is the Polish notation for $S^{\square}\circ S$, whereas $f- \square f$ is not a well-formed expression.  There is a very easy algorithm for determining whether any string of symbols from $M$ is a well-formed expression or not.

It is known, due to work of Daniel E. Severin, that every unary primitive recursive function is given by such a well-formed expression from the set $M$.

Now, consider the following pseudo-code (that I could easily implement on a computer):

Step 1: Take $n$ as input from the user.

Step 2: Create the ordered list of all strings of length $\leq n+1$ from the alphabet $M$, in the shortlex order (where we put $S<\square < \circ < -$).

Step 3: Run through this list and remove any string that does not yield a well-formed expression.

Step 4: Take the $n+1$st expression, $e$, from the remaining expressions in the list.

Step 5: Construct the code for the function $f$ represented by $e$.

Step 6: Print $f(n)+1$ to the screen, and terminate.

This program cannot describe a unary primitive recursive function, because it disagrees with them all (due to the "$+1$" part of Step 6).  I'm trying to figure out exactly where this fails primitive recursion.  I'm guessing it is in Step #5, where we replace the expression with the actual function it expresses.

However, I've been taught that primitive recursive functions are those computer programs that don't use unbounded searches/while loops, and I'm not seeing an unbounded search here.  I expect that an expert will immediately see the issue, and so I want to express thanks in advance for any comments and thoughts.

答:
It's not step $5$, but step $6$. Or more precisely (as Andreas Blass comments below), it's the implicit "step $5.5$" which is needed to make sense of step $6$: computing $f(n)$ in the first place.

The issue is that there is no "universal compiler" for primitive recursive functions which is itself primitive recursive. The issue, intuitively, is that each individual p.r. function is "bounded complexity" in a particular sense but there is no single "bound on the bounds."

In fact there's nothing specific to primitive recursion here: by simple diagonalization, if $F=(f_i)_{i\in\mathbb{N}}$ is any indexed family of total computable functions then there is no $F$-universal function in $F$ itself. (Totality plays a crucial role here of course.)


问:
How to calculate the absolute extrema for this equation
Basically, I am trying to find the absolute extrema for this problem, but I am stuck after a few steps.

$$y=\frac{4}{x}+\tan\left(\frac{\pi x}{8}\right),\quad x\in[1,2].$$

I have figured out the derivative, which is

$$y'=-\frac{4}{x^2}+\sec^2\left(\frac{\pi x}{8}\right)\cdot\frac{\pi}{8}.$$

However, after reaching this step, I am stuck. I have tried setting the derivative equal to zero, but I do not get anywhere as I don't know what to do next. How can I solve this problem? Any help is appreciated. Thanks.

答:
Hint: You can use a difference of squares with the derivative:

$$ \frac{\pi}{8} \sec^2 \left( \frac{\pi x}{8} \right) - \frac{4}{x^2} = 0 $$

$$ \Rightarrow \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) + \frac{2}{x}\right) \left(\sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right) - \frac{2}{x}\right) = 0 $$

Update:

$$ \Rightarrow \pm \frac{2}{x} = \sqrt{\frac{\pi}{8}} \sec \left( \frac{\pi x}{8} \right)$$

$$ \Rightarrow \pm \frac{x}{4}\sqrt{\frac{\pi}{2}} = \cos \left( \frac{\pi x}{8} \right) $$

Let $f_1(x) = \frac{x}{4}\sqrt{\frac{\pi}{2}}$, $f_2(x) = -\frac{x}{4}\sqrt{\frac{\pi}{2}}$, and $g(x) = \cos \left( \frac{\pi x}{8} \right) $.

We'll start with $f_1(x) = g(x)$. First, note $f_1(0) = 0$, and $g(0) = 1$. The period of $g(x)$ is $16$, so the next time it reaches its maximum value of $1$ is at $x = 16$. At the same time, $f_1(x) = 1$ at $x = \frac{4\sqrt{2}}{\pi} < 16 $. Since $f_1$ strictly increasing, $f_1$ and $g$ intersect exactly once on $x\ge 0$, somewhere on $x \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.

For $x \le 0$, $f_1(x) = -1$ at $x = -\frac{4\sqrt{2}}{\pi} > -4 $. Again, since $f_1$ is strictly increasing, $f_1$ is below the minimum value of $g$ for all $x < -\frac{4\sqrt{2}}{\pi} $ . Since $g(x) > 0$ for $x \in (-4,0] $, we have $f_1(x) < g(x) $ for all $x \le 0$.

Combining these, we have $f_1$ and $g$ intersect exactly once, let's say at $x = r_1$, where $r_1 \in \left[0,\frac{4\sqrt{2}}{\pi}\right] $.

Then, since $f_2(x) = f_1(-x)$ and $g(x) = g(-x)$, $f_2$ and $g$ intersect exactly once at $x = r_2 = -r_1$.

Further, note that there must be a sign change in $y'$ at $r_1$ because $f_1$ is strictly increasing for all $x$, and $g$ is decreasing on $[0,8]$, implying that $g$ is decreasing at $r_1$. The same logic implies a sign change at $r_2$ as well.

These two solutions are the unique roots of the factored form of the derivative we had above.

Then, we have

$$ y'|_{x=1} = -4 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{8} \right) $$
$$ = -4 + \frac{\pi}{8}\frac{4}{2+\sqrt{2}} $$
$$ = -4 + \frac{\pi}{4+2\sqrt{2}} $$
$$ < 0 . $$

Similarly,

$$ y'|_{x=2} = -1 + \frac{\pi}{8}\sec^2 \left( \frac{\pi}{4} \right) $$
$$ = -4 + \frac{\pi}{8}*2 $$
$$ = -4 + \frac{\pi}{4} $$
$$ < 0 . $$

Since $y' < 0$ at both $x = 1$ and $x = 2$, we know that $r_1 \notin [1,2]$, since there is no sign change in $y'$ and there is exactly one root (which has a corresponding sign change) for $x \ge 0$.

Then, the extreme values of $y$ must be at $x = 1$ and $x = 2$.


问:
Why is $\langle \operatorname{grad} f, X\rangle_g$ independent of the metric on a Riemannian manifold?
Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as well as $X = X^i \partial_{x^i}$. Now I have computed that $\langle\operatorname{grad} f, X \rangle_g =  X^i\frac{\partial f}{\partial x^i}$ as follows. If we let $g^{ij}$ be the inverse matrix of $g_{ij}$ We have
$$\begin{eqnarray*} \langle\operatorname{grad} f, X \rangle_g &=&  \left\langle g^{ij} \frac{\partial f}{\partial x^i}\partial_{x^j} , X^k\partial_{x^k} \right\rangle_g\\
&=& g^{ij} \frac{\partial f}{\partial x^i}X^k\left\langle \partial_{x^j},\partial_{x^k} \right\rangle_g\\
&=& g_{jk}g^{ij}  \frac{\partial f}{\partial x^i}X^k \\
&=& g_{kj}g^{ji}\frac{\partial f}{\partial x^i}X^k \hspace{1cm} (\text{by using that $g_{ij}$ is a symmetric matrix})\\
&=& X^i\frac{\partial f}{\partial x^i}. \end{eqnarray*}$$

My question is: Is there any geometric reason as to why $\langle \operatorname{grad} f, X \rangle_g$ should be independent of the metric $g$?

答:
Yes, the reason is simple. By definition, $\operatorname{grad}f$ is the vector field (metric-dependent) so that
$\langle\operatorname{grad}f,X\rangle_g = df(X)$, and the right-hand side is independent of the metric. But the vector field itself does depend on the metric. :)


问:
How can I find the fundamental period of the following signal?
How to find the fundamental period if it is periodic?
$$
x[n] = \cos((\pi/2)n)\cos((\pi/4)n)
$$
I have simplified to the following but not sure what to do next:
$$
\cos((\pi/2)n)\cos((\pi/4)n) = 1/2[\cos((3\pi/4)n)+\cos((\pi/4)n)]
$$
Any help is appreciated!

答:
Hint:

Determine the period of $\cos\frac{3\pi}4n$ and of $\cos\frac{\pi}4n$.

The fundamental period of their sum is the l.c.m. of both periods.


Body:
Let $p(x) = x^3-xc-d$ then $p(s)=p(r)=0$.

 This is the part you already made
 $$s^3-cs =d = r^3-rc$$ so
 $$ (s-r)(r^2+rs+s^2)= (s-r)c$$ so
 $$r^2+rs+s^2=c$$

Let $q(x)=x^3-cx+d$.

Now we have \begin{align}q(s+r)&=(s+r)^3-(r+s)c+d\\
&= p(s)+p(r)+3d +3sr(r+s)\\
&= 0+0+3d-3s(c-s^2)\\
&= 3p(s)\\
&=0
\end{align}


Title: How to find $N-60$ when it is based on the sum of its prime divisors?

Body:
The problem is as follows:

A certain number coincides with the sum of the prime positive divisors
of the number of zeros in which the product of the first $784$
integers ends positive, when expressed in base $15$ (pentadecimal).
Find the number minus $60$.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{33}\\
2.&\textrm{39}\\
3.&\textrm{30}\\
4.&\textrm{28}\\
\end{array}$

The official method of solution according to my book is as follows:

$784!=a(15)^n$

Then it goes to say that

$n=194=2\times 97$

to finally say that:

Then it declares that $2+97=99$
and finally saying this:

$99-60=39$

declaring that the official answer is $39$.

But that's my problem with this explanation. How does it get the number from that factorial?

The number it seems way to big to be computed by hand needless to say that from the sole equation of

$784!=a(15)^n$

it doesn't say what is $a$ or what is $n$? What does this has to do with pentadecimal base?. I don't get it.

Can someone explain me what is intended in this question and a better argument of justification in other words a better solution method which implies not fancy calculations or too encyclopedical in nature?

This request is based that my knowledge on the subject is somewhat limited so I need all the steps necessary so I can understand.

How can this problematic number be found?. This problem belongs to a section in my book about the fundamental theorem of arithmetic although I don't see it in any way how could this be applied in this peculiar scenario.

Can someone help me here with an easy to follow and understand method of solution?

Body:
$784!$ has $1931$ decimal digits. The last $194$ are zero because of the $194$ factors $5$ in its factorization.

$194$ factors $=5$ are given by all multiples of $5$ which are $156=784$ divided by $5$.

Plus $5^2$ and its $31$ multiples $$25,50,75,100,125,150,175,200,225,250,275,300,325,350,375,400,425,\\450,475,500,525,550,575,600,625,650,675,700,725,750,775$$

$5^3$ and its $6$ multiples $$125,250,375,500,625,750$$

and $5^4=625$. Finally $156+31+6+1=194$

Therefore the factorization of $784!$ goes as:
$$784!=2^{781}\cdot 3^{390}\cdot 5^{194}\cdot\ldots$$
Prime factors of $194$ are $194=2\cdot 97$ and $2+97=99$.

So the answer is $99-60=39$


问:
Find the first few coefficients in the function represented as a power series
I'm missing something about finding the first few coefficients of a function represented as a power series. I can get the series correct, I just can't for the life of me figure out what to use to get the first few coefficients. I'm given that

$$f(x) =\frac 8{(1-6x)^2}$$

I'm able to find the radius of convergence as $R = 1/6$ and I end up with the $C_n$ as $6^{2n}$ and also $X^{2n}$ for $X^n$. I'm supposed to get the first five coefficients, which I have

$C_0 = 8$ for $(8\cdot6^{2\cdot0})$ (this one is correct)

$C_1 = 288$ for $(8\cdot6^{2\cdot1})$

$C_2 = 1728$ for $(8\cdot6^{2\cdot2})$

$C_3 = 373248$ for$(8\cdot6^{2\cdot3})$

$C_4 = 13436928$ for$( 8\cdot6^{2\cdot4})$

$C_1$ to $C_4$ are all wrong. This is plugging in $n = 0$ through $n = 4$
directly into the Cn of the series. I also tried plugging into the original
function, which was also incorrect. Where are the additional coefficients found
if not by plugging in directly? I can't seem to find anything on that. X isn't a
part of my answer, so I can't do $8\cdot6^{2n}\cdot{x^{2n}}$ and have it submit
successfully. My math course uses a website called Webwork and it won't allow X
as a part of the answer.

Sorry for my poor formatting, I don't know how to format for math notation.

Thank you in advance!
-Frank

答:
Don't need to overthink it, just use the definition and start cranking out derivatives since you know $C_k = f^{(k)}(0)/k!$:

$C_0 = 8$ since
$f(0) = 8$.

$f'(x) = 96/(1 - 6x)^{3}$ so $C_1 = 96$.

$f''(x) = 1728/(1 - 6x)^4$ and $C_2 = f''(0)/2!$

and so forth.


问:
How to find the number of tables to accomodate men and women?
The problem is as follows:

On the occasion of Bob's grand opening of his new restaurant he
noticed that he only has $10$ VIP tables. These will be used to
accomodate $48$ men and $36$ women. He wants to place them in such a
way that, using the largest possible number of tables, each table has
the same number of men and the same number of women. Find the
difference between the number of men and women at each table.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{2}\\
2.&\textrm{1}\\
3.&\textrm{3}\\
4.&\textrm{10}\\
\end{array}$

The official solution from my book is to use the quantity of divisors for 45 and for 36.

It alludes that the number of people to be seated in each table is the same but it doesn't necessarily imply that the number of men is the same as the number of women per table.

The notation $QD$ indicates the quantity or the number of divisors from that number.

It states the following:

$\operatorname{factor}(48)=2^4\cdot 3 $

$\operatorname{QD}(45)=(4+1)(1+1)=10$

$\operatorname{factor}(36)=3^2\cdot 2^2$

$\operatorname{QD}(36)=(2+1)(2+1)=9$

Then it goes to say that:

$48=6(8)$

$46=6(6)$

From this it concludes that there are $8$ men to be seated in the VIP table and $6$ women would be seated in the VIP table.

And since it is being asked the difference, it declares that:

$8-6=2$ hence the answer is the first option.

But I'm stuck. I don't know the logic for this process. Why the author did that?.

There are skipped steps here to my understanding. What does this has to do with the quantity of divisors for $48$ and $36$?

Can someone help me here?. Does it exist a better way to solve this problem perhaps using a similar approach or an easier one?.

Does this problem require the use of least common multiple or greatest common divisor?. Which one and why?

I'm lost here, it would really help me if someone could include a step by step explanation on how to solve this without much fuss, but more importantly without skipping any steps because I'm lost.

答:
Let $T$ tables be used. Since there are equal numbers of men on each table, $T$ must divide $48$. Similarly $T$ must divide $36$.

Therefore $T$ divides the gcd $(48,36)$ which is $12$ . Hence, since we only have $10$ tables, there are $6$ tables.

You have clearly understood the rest of the argument.


问:
Convergence of Gauss-Seidel for a matrix where the transposed matrix is strictly diagonally dominant
In this post, it is shown that for a matrix $A\in\mathbb R^{n\times n}$, the Gauss-Seidel iteration method for the solution of linear systems converges if $A$ is strictly diagonally dominant, i.e. if $$|a_{ii}|>\sum_{j=1\\j\neq i}^n|a_{ij}|$$ for all $i=1,\dots,n$. (One has to show that for $T:=-(D+L)^{-1} U$, given the splitting of $A$ in lower, diagonal, and upper triangular matrices, $A=L+D+U$, the spectral radius satisfies $\rho(T)<1$.)

A similar result is true if $A^\intercal$ is strictly diagonally dominant, i.e. if $$|a_{ii}|>\sum_{j=1\\j\neq i}^n|a_{ji}|$$ for all $i=1,\dots,n$. Is there any easy way to use the first result to derive the second?

答:
Yes, it is possible to apply the result that has already been established.

Let $A = M - N$ be the splitting of $A$ where $M = D + L$ is lower triangular and $N = -U$ is strictly upper triangular and let $G = M^{-1}N$ be the matrix driving the functional iteration
$$
M x_{n+1} = Nx_n + f
$$
If $A$ is strictly diagonally dominant by rows, then we already know that the spectral radius of $G$ satisfies $\rho(G) < 1$. We will now assume that the matrix $A$ is strictly diagonally dominant by columns. We now define an auxiliary sequence $\{y_n\}$ by $$y_n = M x_n.$$ The sequence $\{x_n\}$ is convergent if and only the sequence $\{y_n\}$ is convergent because $M$ is nonsingular. By definition, $y_n$ satisfies the functional iteration
$$ y_{n+1} = N M^{-1} y_n + f$$
We will know show that this sequence is convergent. Let $H = NM^{-1}$ denote the central matrix. We claim that $\rho(H) < 1$. It is enough to show that $$\rho(H^T) < 1.$$
By assumption, $A^T$ is strictly diagonally dominant by rows and $A^T = M^T - N^T$ is immediate. It follows that the matrix $(M^T)^{-1} N^T$ satisfies $$\rho((M^T)^{-1} N^T) < 1.$$ Howewer $$(M^T)^{-1} N^T = (M^{-1})^T N^T = (N M^{-1})^T = H^T.$$ This shows that $\rho(H) < 1$. It follows that $\{y_n\}$ is convergent and the limit $y$  satisfies $$y = NM^{-1} y + f.$$ It follows that $\{x_n\}$ is convergent with limit $x = M^{-1} y$ that satisfies $$Mx = Nx + f.$$
This completes the analysis.

The Jacobi iteration is discussed in here

https://math.stackexchange.com/a/2813489/307944


问:
Sampling Distribution: Difference of Normal Random Variables
This is a self-study question. I am trying to complete this textbook problem.

Suppose that $X_{1}, X_{2}, \dots, X_{m}$ and $Y_{1}, Y_{2}, \dots, Y_{n}$ are independent random samples with $X_{i}$ and $Y_{i}$ being normally distributed with means $\mu_{1}$ and $\mu_{2}$ and variances $\sigma^2_{1}$ and $\sigma^2_{2}$, respectively. The difference between the sample means, $\bar{X} - \bar{Y}$, is then a linear combination of $m+n$ normally distributed random variables is also normally distributed. If $\sigma^2_{1} = 2, \sigma^2_{2} = 2.5$, and $m=n$, find the sample sizes so that $(\bar{X} - \bar{Y})$ will be within 1 unit of $(\mu_{1} - \mu_{2})$ with probability $0.95$.

I started by saying that $P(|\bar{X} - \bar{Y} - (\mu_{1} - \mu_{2})| \le 1) = 0.95$. I think I have to standardize the variables, but I am unsure on how to do that since I am working with two random variables at once.

The answer is $n = 17.29 \rightarrow 18$, but I am unsure how to get here.

答:
Given independent random samples $X_{1}, X_{2}, \dots, X_{m}$ taken from a normal random variable with mean $\mu$ and variance  $\sigma^2$,  it is a standard result that the sample mean $\bar{X}$ is normally distributed with mean $\mu$ and variance  $\sigma^2/m$. It follows that in your example $\bar{X}$ is normally distributed with mean $\mu_{1}$ and variance  $\sigma^2_{1}/m$ and $\bar{Y}$ is normally distributed with mean $\mu_{2}$ and variance  $\sigma^2_{2}/n$. It then follows that the difference
$\bar{X} - \bar{Y}$ is normal with mean $(\mu_{1} -\mu_{2})$ and variance $(\sigma^2_{1}/m + \sigma^2_{2}/n)$. As $m=n$ and with the given values we find that $\bar{X} - \bar{Y}$ has variance $(4.5/n)$. After standardizing $P(-1<(\bar{X} - \bar{Y})-(\mu_{1} -\mu_{2})<1)=0.95$ we get $P((-1/\sqrt(4.5/n))- Z-(1/\sqrt(4.5/n)) = 0.95$ where $Z$ is the standard normal variable. The rest should be clear I hope.


问:
Inequalities for numerical radius of complex Hilbert space operator

Let $A$ be a (complex) Hilbert space operator. Show that $\omega(A) \leq ||A|| \leq 2 \omega(A)$, where $\omega(A)$ is the numerical radius of $A$ given by $\omega(A) := \displaystyle{\sup_{\|x\|=1}|\langle Ax,x\rangle|}$.

I already showed $||A|| \leq 2 \omega(A)$ by using the polarization identity. I am stuck on showing $\omega(A) \leq ||A||$. I think I am missing something rather obvious.

答:
It is obvious when you use the Cauchy-Schwarz inequality:
$$
\omega(A) = \sup_{\lVert x \rVert = 1} \lvert \langle Ax, x \rangle \rvert \leq \sup_{\lVert x \rVert = 1} \lVert Ax \rVert \lVert x \rVert  \leq \sup_{\lVert x \rVert = 1} \lVert Ax \rVert \sup_{\lVert y \rVert = 1} \lVert y \rVert = \sup_{\lVert x \rVert = 1} \lVert Ax \rVert = \lVert A \rVert
$$


问:
Is $( E^c)^o =(E^o)^c ?$ true/false
Given $X= \mathbb{R}$ and    $E \subset \mathbb{R}$

Now my question is that

Is $( E^c)^o =(E^o)^c ?$

where  $o$ denote the interior and $c$ denote the complement

I think this is not true

take $E= \emptyset$

Now $\emptyset^c= \mathbb{R}$ and $\mathbb{R}^0=\mathbb{R}$

This implies $( E^c)^o \neq (E^o)^c$

答:
This is not true in general.

Consider $E = (0,1)$. Then $E^{c} = (-\infty,0]\cup[1,+\infty)$.

Hence $\text{int}(E^{c}) = (-\infty,0)\cup(1,+\infty)$ and $(\text{int}(E))^{c} = (-\infty,0]\cup[1,+\infty)$.

Hopefully this helps.

Title: Is $( E^c)^o =(E^o)^c ?$ true/false


问:
Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares. What is the smallest possible value of $b$?
Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares.
What is the smallest possible value of $b$?

Let, $2a-b=k^2, a-2b=p^2, a+b=q^2$.

$k^2=p^2+q^2$ after adding any of the two equations.

How to proceed further?

Body:
The first few quintuples are
$$(a,b,a+b,2a-b,a-2b)\\\in\big\{
(6,3,9,9,0 ),\quad
(24,12,36,36,0 ),\quad
(54,27,81,81,0 ),\quad
(96,48,144,144,0 ),\\
\mathbf{(123,21,144,225,81 )},\quad
(150,75,225,225,0 ),\quad
(216,108,324,324,0 ),\\
(294,147,441,441,0 ),\quad
(384,192,576,576,0 ),\quad
(486,243,729,729,0 ),\\
\mathbf{(492,84,576,900,324 )},\quad
(600,300,900,900,0 ),\quad
(726,363,1089,1089,0 ),\\
(864,432,1296,1296,0 ),\quad
\mathbf{(939,357,1296,1521,225 )},\quad  \cdots\big\}
$$
The lowest value of $\space b\space $ that works is either $\space 3\space $ or $\space 21\space $ depending on whether or not you allow $\space (a-2b)=0\space $.


答:
If you simplify further expanding with the formula $\cos 3\alpha=\cos ^3(\alpha )-3 \sin ^2(\alpha ) \cos (\alpha )$ you get
$$\frac{1}{2} \cos ^3\left(\frac{\pi  n}{4}\right)+\frac{1}{2} \cos \left(\frac{\pi  n}{4}\right)-\frac{3}{2} \sin ^2\left(\frac{\pi  n}{4}\right) \cos \left(\frac{\pi  n}{4}\right)$$
So it's much easier to see that the period is $8$.

Title: How can I find the fundamental period of the following signal?

Body:
How to find the fundamental period if it is periodic?
$$
x[n] = \cos((\pi/2)n)\cos((\pi/4)n)
$$
I have simplified to the following but not sure what to do next:
$$
\cos((\pi/2)n)\cos((\pi/4)n) = 1/2[\cos((3\pi/4)n)+\cos((\pi/4)n)]
$$
Any help is appreciated!


问:
Prove that: $\lim_{t\to\infty}\frac{1}{t_n}\int_{1}^{t}{f(s)g(s)}\,ds=0$
Maybe someone can help me out with the following question:

Let $f \in L(E)$ (Lebesgue integrable in E) and g $\in M(E)$ (measurable in $E$). If the function $h(t)=g(t)/t$ is bounded, prove that:
$$\lim_{t\to\infty}\frac{1}{t}\int_{1}^{t}{f(s)g(s)}\,ds=0$$
I´ve advanced until:

I´ve taken a sequence {$t_n$} / {$t_n$}$\to\infty$. As

$\lim_{t_n\to\infty}\frac{1}{t_n}\int_{1}^{t_n}{f(s)g(s)}\,ds=0$$\iff$$\lim_{t\to\infty}\frac{1}{t}\int_{1}^{t}{f(s)g(s)}\,ds=0$.

Therefore:

$$\begin{align*}
\lim_{t_n\to\infty} \frac{1}{t_n} \int_{1}^{t_n}{f(s)g(s)}\,ds &= \lim_{t_n\to\infty} \frac{1}{t_n} \int_{1}^{t_n}\frac{f(s)g(s)s}{s}\,ds\\
&\leq \lim_{t_n\to\infty}\frac{M}{t_n}\int_{1}^{t_n}f(s)s=\lim_{t_n\to\infty}\frac{M}
{t_n}\int_{1}^{\infty}\chi_{[1,t]}f(s)s
\end{align*}$$

But I can´t continue, I might need to use the Dominated convergence theorem but I don´t know how to use it. Can someone help me?

答:
Define $F(t)=\frac{1}{t}\int_{1}^{t}f(x)g(x)dx$. To prove that $\lim_{t\rightarrow\infty}F(t)=0$,
it suffices that for each sequence $(t_{n})$ in $[1,\infty)$, if
$t_{n}\rightarrow\infty$, then $F(t_{n})\rightarrow0$ (a result
due to Heine).

Choose $M>0$ such that $\left|\frac{g(t)}{t}\right|\leq M$ for each
$t\in[1,\infty)$. Let $(t_{n})$ be an arbitrary sequence in $[1,\infty)$
such that $t_{n}\rightarrow\infty$. Observe that for each $n$ and
each $x\in[1,\infty)$, we have
\begin{eqnarray*}
 &  & \left|\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)\right|\\
 & \leq & \left|\frac{x}{t_{n}}1_{[1,t_{n}]}(x)f(x)\cdot\frac{g(x)}{x}\right|\\
 & \leq & M\left|\frac{x}{t_{n}}1_{[1,t_{n}]}(x)f(x)\right|\\
 & \leq & M\left|1_{[1,t_{n}]}(x)f(x)\right|\\
 & \leq & M|f(x)|.
\end{eqnarray*}
Moreover, for each fixed $x\in[1,\infty)$, $\lim_{n\rightarrow\infty}\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)=0$
because $\frac{1}{t_{n}}\rightarrow0$. By Lebesgue Dominated Convergence
Theorem, we have that $\lim_{n\rightarrow\infty}\int\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)dx=0$.
Note that $\int\frac{1}{t_{n}}1_{[1,t_{n}]}(x)f(x)g(x)dx=\frac{1}{t_{n}}\int_{1}^{t_{n}}f(x)g(x)dx=F(t_{n})$.
That is, $F(t_{n})\rightarrow0$.


问:
Proof about Signed Complete Graph
Let a complete graph $K_n$ in which nodes represent people. For each pair of people, we label their relationship + or -. + indicates the two people are friends while - indicates they are enemies. A labeling is balanced if it has the property that for every set of three people, if we consider the three relationships between them, either all three are labeled +, or exactly one is labeled +.

Prove that in any balanced labeling either everyone is friends with everyone, or people can be partitioned into two groups, A and B, such that all pairs in A are BFFs, all pairs in B are BFFs, and everyone in A is an enemy of everyone in B.

The first case is trivial. For the latter case, consider a - edge. I should show that one end leads to a set, and the other end leads to another set. However, I am not sure how to begin. Any help is appreciated. Thank you.

答:
The balanced labeling condition implies that friendship is transitive. It is also inherently symmetric, so if we consider everyone to be friends with themselves, it is an equivalence relation.

Therefore the people can be partitioned into equivalence classes: several groups such that everyone is friends with everyone in their own group, and nobody else.

Finally, if we could take three people from different groups, the triangle they induce would not be balanced; therefore there can be at most two groups.

Alternatively, to complete the proof in the way you suggested, suppose $v$ and $w$ are a pair of enemies. First, show that every other person $x$ is friends with exactly one of $v$ or $w$, by considering the triple $\{v,w,x\}$. We can now let $A$ be the set of all people friends with $v$ and let $B$ be the set of all people friends with $w$, to get a partition. To show that this partition is a split into two enemy factions, we must check that:

If $x, y \in A$, then $x$ is friends with $y$.

If $x, y \in B$, then $x$ is friends with $y$.

If $x \in A$ and $y \in B$, then $x$ and $y$ are enemies.

Each of the three can be done by looking at the triple $\{v, x, y\}$, in which we already know the status of the pairs $vx$ and $vy$, and can therefore infer the status of the third pair $xy$.

Title: Proof about Signed Complete Graph


问:
Why does an integral turn negative when its limits are switched?
I’m trying to find the tension at the middle of a rotating rod in a gravity-free space. To do that, I had done some math and then I integrated with the limits $L/2$ to $L$.

This sounds like a math question but I’m not sure and I want to know: Is there a significance to the negative sign in the result if I take the limits the other way around?

Does this negative sign indicate direction or do I just suck at math?

答:
It’s a basic characteristic of definite integrals that flipping the limits of integration will negate the result:

$$\int_a^b f(x)\ dx = -\int_b^a f(x)\ dx. $$

This can be seen by recalling the definition of an integral:

$$
\int_{x_{1}}^{x_{n}} f(x)\ dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\ \Delta x.
$$

If we think of this as the sum of the areas of very narrow rectangles underneath the curve of $f(x)$, then $f(x_k)$ is the height of the rectangle at the point $x_k$ and $\Delta x$ is its width.

If there are $n$ rectangles between $x_{1}$ and $x_{n}$, then

$$\Delta x = \frac{x_{n} - x_{1}}{n},\label{delx}$$

the $k$th rectangle is found at

$$ x_k = x_1 + k\Delta x, \tag{1}\label{xk}$$

and the height of the $k$th rectangle is

$$ f(x_k) = f(x_{1} + k\Delta x) = f\left(x_{1} + \frac{x_{n} - x_{1}}{n} k\right)$$

which makes the integral

$$
\int_{x_{1}}^{x_{n}} f(x)\ dx = \lim_{n \to \infty} \sum_{k=1}^{n} f\left(x_{1} + \frac{x_{n} - x_{1}}{n} k\right)\cdot \left(\frac{x_{n} - x_{1}}{n}\right).
$$

Because $x_{1}$, $x_{n}$, and $n$ are constants, we can move them outside the summation:

\begin{equation}\tag{2}\label{normal}
\int_{x_{1}}^{x_{n}} f(x)\ dx = \left(x_{n} - x_{1}\right) \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(x_{1} + \frac{x_{n} - x_{1}}{n} k\right).
\end{equation}

It’s important to note that the order in which we sum the values of $f(x)$ is unimportant due to the commutative property of addition. So instead of starting at $x=x_1$ and going to $x_n$, we could just as well start at the end and add backwards:

$$
\begin{align*}
\int_{x_{1}}^{x_{n}} f(x)\ dx &= \lim_{n \to \infty} \sum_{k=1}^{n} f(x_{n-k})\ \Delta x\\
&= \left(x_{n} - x_{1}\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f(x_{n - k}) \\
&= \left(x_{n} - x_{1}\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(x_1 + \frac{x_n - x_1}{n}(n-k)\right) \\
&= \left(x_{n} - x_{1}\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(x_n - \frac{x_n - x_1}{n} k\right)\tag{3}\label{reverse}
\end{align*}
$$

using \eqref{xk} to express $x_{n-k}$.

If your limits of integration are $\bf{x_{1} = a}$ and $\bf{x_{n} = b}$ then your integral when summing from start to end \eqref{normal} is

$$
\int_a^b f(x)\ dx = (b - a)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(a + \frac{b - a}{n} k\right) \tag{4}\label{normalab}
$$

and when summing from end to start \eqref{reverse} is

$$
\int_a^b f(x)\ dx = \left(b - a\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(b - \frac{b - a}{n} k \right).
$$

Now using the last equation let’s reverse the limits so that $\bf{x_{1} = b}$and $\bf{x_{n} = a}$ (i.e., swap all $a$s with $b$s and vice versa):

$$
\begin{align*}
\int_b^a f(x)\ dx &= \left(a - b\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(a - \frac{a - b}{n} k \right) \\
& = \left(a - b\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(a + \frac{b - a}{n} k \right) \\
&= - \left(b - a\right)\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} f\left(a + \frac{b - a}{n} k \right)
\end{align*}
$$

which is identical to the negative value of the unflipped integral \eqref{normalab}.

OwnerDisplayName: Glowingbluejuicebox

OwnerDisplayName: pion


问:
What is the factor group $\mathbb{Z}/5\mathbb Z$?
I am trying to understand the concept of factor group. The definition of factor group I know is the following: Let $G$ be a group and $H$ be a subgroup of $G$. Then the group of cosets denoted by $G/H$ is called the factor group of $G$ by $H$. Now I am looking at an example: Let $G=\mathbb{Z}$ and $H=5\mathbb{Z}$. Then I find the cosets of $5\mathbb{Z}$. Here they are:

Cosets of $5\mathbb{Z}$ containing $0$, $\{\ldots,-10,-5,0,5,10,\ldots\}$

Cosets of $5\mathbb{Z}$ containing $1$, $\{\ldots,-9,-4,1,6,11,\ldots\}$

Cosets of $5\mathbb{Z}$ containing $2$, $\{\ldots,-8,-3,2,7,12,\ldots\}$

Cosets of $5\mathbb{Z}$ containing $3$, $\{\ldots,-7,-2,3,8,13,\ldots\}$

Cosets of $5\mathbb{Z}$ containing $4$, $\{\ldots,-6,-1,4,9,14,\ldots\}$

My question is, is the factor group $\mathbb{Z}/5\mathbb{Z}$ the union of all these $5$ cosets? Then I think $G/5\mathbb{Z} =\mathbb{Z}$ since the union of these cosets are equal to $\mathbb{Z}$. Can anyone help me with this?

Thanks.

答:
The factor group is not the union of cosets, it's the set of cosets, so
$${\bf Z}/5{\bf Z}=\{5{\bf Z},1+5{\bf Z},2+5{\bf Z},3+5{\bf Z},4+5{\bf Z}\}$$

The union of cosets is the original group. And  $H$ has to be a normal subroup for the set of cosets to be a group (this is trivial if $G$ is abelian, of course).


问:
The center of the circumcircle lies on a side of a triangle
Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.

 $$AB=\dfrac{144\sqrt5}{5}$$

By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!

答:
Observe that $\angle OPA=\angle PAO=\frac\alpha2=\angle BAP\implies OP\parallel AB$. Thus $$\frac{CO}{OA}=\frac{CP}{PB}\iff \frac dR=\frac{20}{16}=\frac54$$ Also, Power of a point yields $$\begin{align*}\text{Pow}(C)_{(APB)}=\lvert d^2-R^2\rvert&=20\cdot 36\\\iff \left\lvert\left(\frac54R\right)^2-R^2\right\rvert&=720\\\iff \frac9{16}R^2&=720\\\iff R&= 16\sqrt{5}  \end{align*}$$ Use $OP\parallel AB$ again in order to infer

$$AB=R\cdot \frac{36}{20}=16\sqrt5\cdot \frac95=\frac{144\sqrt5}{5}$$


问:
Reference request: Category Theory for Theoretical Physics
I have seen that there are already reference requests questions for category theory. I would like to ask for resources and textbooks that are more oriented towards theoretical physicsists. I'm looking for an introduction to the subject, but would greatly appreciate material that demonstrates the possible use of category theory and its applications in theoretical physics.</p>

答:
I would maybe recommend reading this paper, IF, you already have some background information in algebra, or topology, and things like that. https://math.berkeley.edu/~erabin/The%20Categorical%20Language%20of%20Physics.pdf

There is also a book;https://link.springer.com/chapter/10.1007/3-540-53763-5_52.

There is also a post on mathoverflow;https://mathoverflow.net/questions/34861/how-is-category-theory-actually-useful-in-actual-physics

And a whole wikipedia page; https://en.wikipedia.org/wiki/Categorical_quantum_mechanics


答:
$\angle POC = \angle A$, so $\triangle COP \sim \triangle CAB$

Hence $\frac{R}{AB} = \frac{20}{36} = \frac{5}{9}$ ...(i)

$\angle APB = \frac{1}{2} \angle AOB = 90^0 - \angle A$

$AB = 2R \sin \angle APB = 2R \cos A$

So from (i), $\frac{R}{2R \cos A} = \frac{5}{9}$

$\cos A = \frac{9}{10} = 1 - 2 \sin^2{\frac{A}{2}} \implies \sin \frac{A}{2} = \frac{1}{2\sqrt5} $

$16 = 2R \sin \frac{A}{2} \implies R = 16\sqrt5 $

$AB = 2R \cos A = 32\sqrt5 \times \frac{9}{10} = \frac{144}{\sqrt5}$

Title: The center of the circumcircle lies on a side of a triangle

Body:
Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.

 $$AB=\dfrac{144\sqrt5}{5}$$

By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!


问:
Differential equations and linear algebra way
I am a beginner at differential equations. Never learned this before but do know calculus.

I am reading Section 6.3 (Systems of Differential Equations) in Introduction to Linear Algebra by Gilbert Strang.

At the beginning of the section:

So, e^ut is equal e^lambdatx. I don't understand this part.

So, here are what I understand:

$$
\frac{\mathrm{d}u }{\mathrm{d} t} = u(t) = {e^{ut}}
$$

And in the linear algebra:

It turns out,

$$
u(t) = e^{\lambda t}x
$$

Can anyone help to explain or let me know the key word to find the relation?

答:
One way is separation of variables. One can prove that the solution with $u(0) = x$ (in your notation) of $\frac{\mathrm{d}}{\mathrm{d}t}u(t) = u(t)$ fulfills
$$
\int^{u(t)}_x \frac{1}{g(s)}~\mathrm{d}s= \int^t_0 h(s)~\mathrm{d}s,
$$
where $g(s) = s$ and $h(s) = 1$. This comes from the ODE being in the form
$$
\frac{\mathrm{d}}{\mathrm{d}t} u(t) = g(u)h(t).
$$
I'm sure you can find the solution yourself.


问:
Distribute stones into buckets with limits
You have $s$ different stones and $b$ different buckets. How many ways $w(s, b)$ can you distribute the stones into the buckets if no bucket is allowed to contain more than $L$ stones?

Clarification: The stones are all different from each other. So, for example, if you have 3 stones, 2 buckets, and at most 2 stones per bucket, there are 6 ways to distribute the stones: (A, BC), (AB, C), (AC, B), (B, AC), (BC, A), (C, AB).

I noticed that you can express the answer for $b$ buckets in terms of the answers for $b - 1$ buckets:

$$
w(s, b) = \sum_{i=0}^{\min(s, L)} {s \choose i} w(s - i, b - 1)
$$

Also,
$$
w(s, 1) = \left\{ \begin{array}{rl}
1 & \mbox{if }s <= L\\
0 & \mbox{otherwise}
\end{array} \right.
$$

Based on those observations I came up with this Python function:

```<code>def countWays(numStones, numBuckets, L):
    # Precompute C(n, r) table
    C = [[1]]
    for n in xrange(numStones):
        C.append([1] + [C[-1][r] + C[-1][r+1] for r in xrange(n)] + [1])
    # Loop through number of buckets
    ways = [1] * (L + 1) + [0] * (numStones - L)
    for b in xrange(2, numBuckets + 1):
        ways = [sum([C[s][i] * ways[s - i] for i in xrange(min(s, L) + 1)])
                for s in xrange(numStones + 1)]
    return ways[numStones]
</code>```

Some sample output:

```<code>>>> countWays(1, 2, 1)
2
>>> countWays(3, 2, 2)
6
>>> countWays(3, 3, 2)
24
>>> countWays(10, 10, 5)
9985309740L
>>> countWays(100, 100, 5)
94759885461565034681687026066885761311439165668901259822785622795797870660611943632316455279091491854154551818337317753294675330799396303953809575897498333285450197379975598951694414085442292940800000L
</code>```

Is there a simpler solution, or closed form? Some basic principle of combinatorics I'm missing?

答:
Assuming that your stones are indistinguishable and the buckets are distinguishable, we can use generating functions to solve this.

Any given bucket can hold anywhere from 0 to L stones. This is represented by the polynomial $1+x+ x^2+\ldots x^L$. For b buckets, the total number of ways of distributing s stones is the coefficient of the term $x^s$ in the expansion $(1+x+ x^2+\ldots x^L)^b$.

Sanity Check Example: if you have two buckets, 4 stones and each bucket can hold at most 2 stones, there is only one way to do it. The coefficient of $x^4$ in $(1+x+x^2)^2$ is 1. If you have 3 buckets instead of 2, you need the coefficient of $x^4$ in $(1+x+x^2)^3$ which is 6 (you can enumerate the 6 possibilities easily).

EDIT: If the stones are distinguishable, you can still use generating functions to solve the problem. Now, the answer will be the coefficient of $x^s$ in $s!\left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots +\frac{x^L}{L!}\right)^b$. In general, I don't think this expression is easily simplified.

The idea here is that you are partitioning $s$ stones into $b$ bins. What matters is the way you divide the stones among the bins, but not the way stones inside a bin are arranged. So, if you split 4 stones in two bins, the number corresponds to $\frac{4!}{2!2!} = 6$.

In the generating function I just pulled out the factor of $s!$ outside the polynomial term and divide each $x^k$ term by $k!$ to adjust the count. I checked for the case when $s = 4$, $b=3$ and $L=2$ and I get the answer 54.