答:
This looks like a job for Cauchy's inequality. By Cauchy, we have that $(a_1^2+a_2^2+a_3^2...+a_{17}^2+a_{18}^2)(1+1+1 ... + 1 +1) \ge (a_1(1) + a_2(1) +a_3(1)...a_{17}(1) + a_{18}(1))^2$.

Note that there are $18$ ones in $(1+1+1 ... + 1 +1)$ to pair up with $18$ $a$s.

So, this inequality simplifies as $18(a_1^2+a_2^2+a_3^2...+a_{17}^2+a_{18}^2) \ge (a_1 + a_2 +a_3...a_{17} + a_{18})^2$. We know that the sum on the left is $237$, thus we have $18(a_1^2+a_2^2+a_3^2...+a_{17}^2+a_{18}^2) \ge (237)^2$. Simplifying, we get $(a_1^2+a_2^2+a_3^2...+a_{17}^2+a_{18}^2) \ge 3120.5$. By Cauchy's inequality condition, we must have all $a$ be equal, so we get $18a_1^2 \ge 3120.5$, or $a_1 \le -13.6666$  or $a_1\ge 13.16666...$. However, that means that making all of the terms $13/-13$ actually gets you to the MINIMUM. However, since Cauchy's inequality has no upper bound, that means that the maximum of the squares is INFINITE. This is because you can take ridiculously small negative integers, and ridiculously big integers, and still have a sum of $237$. However, the sum of their squares will be HUGE.

Example, the sum of $-1000,-2000,-3000,-4000,-5000,-6000,-7000,-8000,-9000,-10000,-11000,-12000,-13000,-14000,-15000,-16000,-17000, 153237$

is $237$, but the sum of the squares of this set is $2.5266578169×10^{10}$.

Title: Maximize $\sum{a_i^2}$ given $\sum{a_i}=237$

Body:
I recently got stuck with the following problem:

Consider 18 integer numbers in segment $[10; 20]$. Find the maximum possible value of
$$
\sum_{i=1}^{18} {a_i^2}
$$
given that
$$\sum_{i=1}^{18} {a_i}=237$$

I don't really know how to approach this problem. The thing is, solving it like constraint optimization problem I got the solution $a_1=a_2=...=a_n=\frac{237}{18} \approx13.2$

But we need to solve it in integers. And just taking numbers around $13.2$ doesn't make sense to me. Any ideas?


问:
Rank of matrix as a function of x
Given $B$=$\begin{pmatrix}
x & 1 & 2 \\
5 & 5 & -2 \\
3 & 3 & 6 \\
1 & x & 2
\end{pmatrix}$

Find rank of matrix $B$ as a function of $x$

For finding the rank I converted the above matrix to row reduced echelon form and got the following

$\begin{pmatrix}1 & 0 & \frac{-2}{5}\frac{x+1}{x-1}\\ 0 & 1 & \frac{12}{5}\frac{1}{x-1}\\ 0 & 0 & 1\\ 0 &0 & 0\end{pmatrix}$

When $x \neq 1$ the rank of the above matrix is 3 and when $x=1$ rank of the above matrix is 2 but when $x=1$, I am getting infinite value in the last column.

Can you please let me know if the above approach to the formulation is correct?

答: If $x=1$ the two first columns are equal and independent from the last column. Hence the rank is equal to $2$

ContentLicense: CC BY-SA 4.0

Title: Rank of matrix as a function of x


答:
When $x=1$ there are two identical columns, independent to the third column, so rank =2.

As you've seen, when $x \ne 1$, the rank is three.

Title: Rank of matrix as a function of x

Body:
Given $B$=$\begin{pmatrix}
x & 1 & 2 \\
5 & 5 & -2 \\
3 & 3 & 6 \\
1 & x & 2
\end{pmatrix}$

Find rank of matrix $B$ as a function of $x$

For finding the rank I converted the above matrix to row reduced echelon form and got the following

$\begin{pmatrix}1 & 0 & \frac{-2}{5}\frac{x+1}{x-1}\\ 0 & 1 & \frac{12}{5}\frac{1}{x-1}\\ 0 & 0 & 1\\ 0 &0 & 0\end{pmatrix}$

When $x \neq 1$ the rank of the above matrix is 3 and when $x=1$ rank of the above matrix is 2 but when $x=1$, I am getting infinite value in the last column.

Can you please let me know if the above approach to the formulation is correct?


问:
If $f(x)$ is a polynomial, is it true that $\lim_{x\to \pm\infty}\frac{1}{f(x)}$ both exist?
If $y=f(x)$ is a polynomial, then $\lim_{x\to -\infty}\frac1{f(x)}$ and $\lim_{x\to +\infty}\frac1{f(x)}$ both exist.

I need explain that why it's true or false.

答:
WLOG, assume that $f(x)$ is a monic polynomial of degree $n$.  That is, it has form $x^n + a_{(n_1)}x^{(n-1)} + \cdots + a_1x^1 + a_0.$

Attack the problem with intuition.  Suppose (for example) that $n$ is odd.  Can you prove (for example) that the $\lim_{x\to -\infty} f(x) = -\infty.$

Can you then, using a similar approach, prove that the $\lim_{x\to -\infty} \frac{1}{f(x)} = 0.$

What about the supposition that each of the coefficients

$\{a_{(n-1)}, \cdots, a_0\}$ is an element of $[-M,M]$, for some (large) fixed positive real $M$?


问:
Supremum in a dictionary ordered set.
Take double arrow space $[0,1]\times \{1,2\}$

Where topology is dictionary topology defined by linear order as $(a,b)<(c,d)$ iff $a<c$; or ;$a=c \;\&\; b<d$

open sets are in the form $$((a, b] \times\{1\}) \cup([a, b) \times\{2\})$$

I want to know that is this topology has least upper bound property?

I know that $\mathbb R$ has this axiom, by using this can I show the above topology has least upper bound property?

By doing this I want to show the above topology is compact. I have seen https://dantopology.wordpress.com/2009/10/07/the-lexicographic-order-and-the-double-arrow-space/ proves it but it seems a bit advanced and less informative.

答:
Let $S\subseteq [0,1]\times\{1,2\}$ be a nonempty set. Let
$$T=\{x\in [0,1]\mid \exists t\in\{1,2\}\text{ such that }(x,t)\in S\}.$$ Then $T$ is a nonempty subset of $[0,1]$, and hence has a least upper bound, $s$.

If $(s,2)\in S$, then $(s,2)$ is the maximum of $S$ and hence the least upper bound of $S$: indeed, if $(x,t)\in S$, then $x\in T$, so $x\leq s$; and since $t\leq 2$, we must have $(x,t)\leq (s,2)$.

If $(s,1)\in S$ and $(s,2)\notin S$, then $(s,1)$ is the maximum of $S$: if $(x,t)\in S$, then $x\leq s$; if $x\lt s$ then $(x,t)\lt (s,1)$, and if $x=s$, then it must be the case that $t=1$, so $(x,t)=(s,1)$. Thus, $(s,1)$ is the maximum of $S$, and hence the least upper bound of $S$.

Assume then that $(s,1),(s,2)\notin S$.

We know that for all $\epsilon\gt 0$ there exists $x\in T$ such that $s-\epsilon\lt x\leq s$ (in fact, $s-\epsilon\lt x\lt s$, since $x\in T$ but $s\notin T$).

Claim. $(s,1)$ is the least upper bound of $S$.

First, note that it is an upper bound for $S$: if $(x,t)\in S$, then $x\leq s$; since $(s,1)\notin S$ and $(s,2)\notin S$, then $x\lt s$, so we are done: $(x,t)\lt (s,1)$. Thus, $(s,1$ is an upper bound for $S$.

Now, let $(x,t)\lt (s,1)$. Since by assumption $x\lt s$, then there exists $x_1\in T$ such that $x\lt x_1\leq s$, and there exists $t_1\in\{1,2\}$ such that $(x_1,t_1)\in S$, so $(x,t)\lt (x_1,t_1)\leq (s,1)$. Thus, $(x,t)$ is not an upper bound for $S$. Thus, $(s,1)$ is the least upper bound for $S$.

Thus, $S$ has a least upper bound in $[0,1]\times\{1,2\}$, ordered lexicographically.

Title: Supremum in a dictionary ordered set.


问:
Differential Equations Reference Request
Currently I'm taking the Differential Equations course at college, however the problem is the book used. I'll try to make my point clear, but sorry if this question is silly or anything like that: the textbook used (William Boyce's book) seems to assume that the reader doesn't familiarity with abstract math, so it lacks that structure of presenting motivations, then definitions, then theorems and corolaries as we see in books like Spivak's Calculus or Apostol's Calculus.

I've already seem a question like this on Math Overflow, however unfortunatelly some people felt offended somehow and said: "are you saying that Boyce is easy? It doesn't matter if you know how to prove things, you must learn to compute", and my point is not that: Apostol and Spivak teaches how to compute also, however since their books are aimed to mathematicians they take care to build everything very fine and their main preocupation is indeed in the theoretical aspects.

I really don't like the approach of: well, in some strange way we found that this equation works, so memorize it, know how to compute things and everything is fine. I really want to understand what's going on, and until now I didn't find this possible with Boyce's book (certainly there are people that find this possible, but I'm used to books like Spivak's and Apostol's, so I don't really do well with books like Boyce's).

I've already seem Arnold's book on Differential Equations, but the prerequisites for reading it are bigger: he uses diffeomorphisms many times and although I'm currently also studying differential geometry, I don't feel yet comfortable on reading a book like this one.

Can someone recommend a book that covers ordinary differential equations, systems of differential equations, partial differential equations, and so on, but that can be read without much prerequisite, and that still has the structure of a book of mathematics? And when I say "structure of book of mathematics" is being like Spivak's and Apostol's books: not mixing up definitions, theorems and examples inside histories of how the theory developed. Since I'm a student of Mathematical Physics, of course motivations and examples from Physics are welcome, but not all mixed up in the text.

In truth I don't really believe that a book like the one I described exists (if a book on this topic is good in my point of view, I believe it'll have a lot of prerequisites). Anyway I hope that I don't get misunderstood in what's my doubt, and really sorry if this question is silly in someway.

答:
1). Try Verhulst's book on nonlinear differential equations. While the focus of this book is nonlinear ODEs, it does have a chapter on linear equations. It is a nice middle between the formula-computation based ODE books, and something like Arnold's ODE book.

2). For a more comprehensive but still accessible book, try the one by L. Perko. This is a gem.

ContentLicense: CC BY-SA 3.0


问:
What am I doing wrong with my probability formula?
The question is

I have a class of 10 students. The class consists of 4 statistics majors, 3 philosophy majors, and 3 sociology majors.

I choose three students at random, without replacement. (Order doesn’t matter.)

Let event A be "All three students I choose are statistics majors."

Let event B be "At least one student I choose is a philosophy major."

For P(A) I did $$4/10 \cdot 3/9 \cdot 2/8 = .033$$

For P(B) I thought it would be $$(7!\cdot 3!)/(10!) = .833$$ But that is not correct. I am a little confused on where I am going wrong, I thought it would be number of events that satisfy B/number of equally likely events some help in the right direction is all im asking for, thank you in advance.

答:
Comment:  @ ClementC's answer (+1) is correct, comprehensive, and to the point. Also,
both parts can be solved as hypergeometric probability problems,
and both answers can be easily approximated by simulation.

Using the hypergeometric PDF function phyper in R: (Exact answers to the stated number of decimal places.)

(a)

```<code>dhyper(3, 4,6, 3)
[1] 0.03333333
</code>```

(b)

```<code>1 - dhyper(0, 3,7, 3)
[1] 0.7083333
</code>```

Simulation with a million iterations can be expected to give 2 or 3 place
accuracy.

```<code># a
set.seed(201)
class = c(1,1,1,1, 2,2,2, 3,3,3)
nr.1 = replicate(10^6, sum(sample(class, 3)==1))
mean(nr.1 == 3)
[1] 0.03342

# b
set.seed(202)
nr.3 = replicate(10^6, sum(sample(class, 3)==3))
mean(nr.3 >= 1)
[1] 0.708252
</code>```

Body:
Comment:  @ ClementC's answer (+1) is correct, comprehensive, and to the point. Also,
both parts can be solved as hypergeometric probability problems,
and both answers can be easily approximated by simulation.

Using the hypergeometric PDF function phyper in R: (Exact answers to the stated number of decimal places.)

(a)

dhyper(3, 4,6, 3)
[1] 0.03333333

(b)

1 - dhyper(0, 3,7, 3)
[1] 0.7083333

Simulation with a million iterations can be expected to give 2 or 3 place
accuracy.

# a
set.seed(201)
class = c(1,1,1,1, 2,2,2, 3,3,3)
nr.1 = replicate(10^6, sum(sample(class, 3)==1))
mean(nr.1 == 3)
[1] 0.03342

# b
set.seed(202)
nr.3 = replicate(10^6, sum(sample(class, 3)==3))
mean(nr.3 >= 1)
[1] 0.708252


问:
Why does the addition of $P(A^c | B^c)$ and $P(A | B^c)$ result in 1 (Why are they complements)?
While watching Harvard's Statistics 110 on YouTube, in the lecture on conditional probability, The professor relates $P(A^c | B^c)$ to $P(A | B^c)$. It is given that $P(A^c | B^c) = 0.95$. The professor says, without explanation, that $P(A|B^C) = 1 - 0.95 = 0.05.$ So I think I'm missing something fundamental. I did some trials to prove it but, unfortunately, came up with nothing.

$$P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{P(A^c \cap B^c)}{1- P(B)} = \frac{1 - P(A \cup B)}{1 - P(B)}$$

Not sure, to be honest, about the benefit of those steps, or how can I come up with a relation between the two probabilities.

答: These probabilities are conditioned on $B^c$.  In that subset of events, the two events are mutually exclusive, and their union is the whole (sub)set of events.


问:
Is $I= \langle 5, x^2\rangle$ a principal ideal or not in $\Bbb Z[x]$?
So I know the case with $\langle 2,x\rangle$which is not a principal ideal, however I've seen other questions with stuff like $\langle 5,x^2+3\rangle$ which is a principal ideal, however what I don't get is the fact that I can make the proof from $\langle 2,x\rangle$ to my case, and it would say it's not a principal ideal, but $x^2$ is irreducible over $\Bbb Z_5$ so you can say it's principal ideal.

I'm very bad at maths by the way, I'm just trying to pass an exam, help. Also, maybe a better question would be why can't we say that $x$ is irreductible over $\Bbb Z_2$ in the case of $\langle 2,x\rangle$ so that it would come to the conclusion it is a principal ideal

答:
$\mathbb{Z}[X]/\langle 5,X\rangle \cong \mathbb{Z}_5$ is a field so $\langle 5,X\rangle$ is a prime ideal (it is even maximal). On the other hand, note that $\mathbb{Z}[X]/\langle 2, X^2\rangle \cong \mathbb{Z}_2[X]/(X^2)$ is not a domain because $X^2 = 0$ in the latter quotient ring, so $\langle 2,X^2\rangle$ is not a prime ideal.

Upon edit, you should ask yourself if we can write $\langle 5,X^2\rangle = \langle P\rangle$ for some $P \in \mathbb{Z}[X]$. In particular, this means we can find polynomials $Q,R\in \mathbb{Z}[X]$ with $5 = PQ$ and $X^2 = PR$. From $5 = PQ$, we see that both $P$ and $Q$ must be constant polynomials. Say, $P = n$ for some $n \in \mathbb{Z}$. But also $X^2 = PR = nR$ so that $n=1$ and $R=X^2$. It follows that $P =1$, but then $\langle 5,X^2\rangle = \langle 1 \rangle = \mathbb{Z}[X]$.  This is a contradiction since $X \notin \langle 5, X^2\rangle.$ Hence, the ideal is not principal.


答: If you know the event $B^c$ then you have either $A$ or $A^c$. Hence the sum of the two probabilities is equal to one.

Title: Why does the addition of $P(A^c | B^c)$ and $P(A | B^c)$ result in 1 (Why are they complements)?

Body:
While watching Harvard's Statistics 110 on YouTube, in the lecture on conditional probability, The professor relates $P(A^c | B^c)$ to $P(A | B^c)$. It is given that $P(A^c | B^c) = 0.95$. The professor says, without explanation, that $P(A|B^C) = 1 - 0.95 = 0.05.$ So I think I'm missing something fundamental. I did some trials to prove it but, unfortunately, came up with nothing.

$$P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{P(A^c \cap B^c)}{1- P(B)} = \frac{1 - P(A \cup B)}{1 - P(B)}$$

Not sure, to be honest, about the benefit of those steps, or how can I come up with a relation between the two probabilities.


问:
Proving $\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$
This is problem 4a of Spivak's Calculus, 4th Edition, Prologue Chapter 2:

Prove that $$\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$$
Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.

And this is the solution

Since $$(1+x)^n(1+x)^m=(1+x)^{n+m}$$ we have
$$\sum_{k=0}^n\binom{n}{k}x^k\;\cdot\; \sum_{j=0}^m\binom{m}{j}x^j \;=\; \sum_{\ell=0}^{n+m} \binom{n+m}{\ell}x^\ell.$$
But the coefficient of $x^\ell$ on the left is clearly
$$\sum_{k=0}^\ell\binom{n}{k}\binom{m}{\ell-k},$$
one term of the sum occurring for each pair $k$, $j=\ell-k$.

It's frustrating when a solution uses terms like "clearly", and then you look at an equation and don't see it at all.

How is the coefficient of $x^l$ "clearly" the expression shown in the solution above?

答:
The lefthand side is of the form
$$(a_0 + a_1 x + \dots + a_n x^n)(b_0 + b_1 x + \dots + b_x x^m),$$
where $a_k = \binom{n}{k}$ and $b_j = \binom{m}{j}$. When we multiply everything out, when will terms of the form $c x^\ell$ appear?

Well, when a left term $a_k x^k$ meets a right term $b_j x^j$, we get a resulting term $a_k b_j x^{k + j}$. Thus, we are looking for pairs $(a_k x^k, b_j x^j)$ where $k + j = \ell$.

P.S. I agree with your sentiment that phrases like "clearly" are frustrating. You might like this article by Francis Su on the topic.


答:
If $\mathbb P(B^\complement)>0$ then
$$
\mathbb P(A^\complement\vert B^\complement)=\frac{\mathbb P(A^\complement\cap B^\complement)}{\mathbb P(B^\complement)}=\frac{\mathbb P(B^\complement)-\mathbb P(A\cap B^\complement)}{\mathbb P(B^\complement)}=1-\frac{\mathbb P(A\cap B^\complement)}{\mathbb P(B^\complement)}=1-\mathbb P(A\vert B^\complement).
$$

For the second equality I used that $B^\complement$ is the disjoint union of $A^\complement\cap B^\complement$ and $A\cap B^\complement$, hence $\mathbb P(B^\complement)=\mathbb P(A^\complement\cap B^\complement)+\mathbb P(A\cap B^\complement)$.


答: $\binom{n+m}{l}$ counts the ways of choosing $l$ items from a collection of $n+m$ items. Think your collection as having $n$ blue items and $m$ red ones. In order to choose $l$ items from your collection, you can first choose some blue ones, say $k$ of them, and then you can choose the rest $l-k$ from the red ones.

Title: Proving $\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$

Body:
This is problem 4a of Spivak's Calculus, 4th Edition, Prologue Chapter 2:

Prove that $$\sum_{k=0}^\ell \binom{n}{k}\binom{m}{\ell-k}=\binom{n+m}{\ell}$$
Hint: Apply the binomial theorem to $(1+x)^n(1+x)^m$.

And this is the solution

Since $$(1+x)^n(1+x)^m=(1+x)^{n+m}$$ we have
$$\sum_{k=0}^n\binom{n}{k}x^k\;\cdot\; \sum_{j=0}^m\binom{m}{j}x^j \;=\; \sum_{\ell=0}^{n+m} \binom{n+m}{\ell}x^\ell.$$
But the coefficient of $x^\ell$ on the left is clearly
$$\sum_{k=0}^\ell\binom{n}{k}\binom{m}{\ell-k},$$
one term of the sum occurring for each pair $k$, $j=\ell-k$.

It's frustrating when a solution uses terms like "clearly", and then you look at an equation and don't see it at all.

How is the coefficient of $x^l$ "clearly" the expression shown in the solution above?


答:
Let $X$ be the double arrow space. Your description of the order topology on $X$ isn’t quite right. The set that you describe is the open interval in $X$ between $\langle a,1\rangle$ and $\langle b,2\rangle$, but that is not the only kind of open interval in $X$; for instance, the open interval between $\langle a,2\rangle$ and $\langle b,1\rangle$ is $(a,b)\times\{1,2\}$. And of course all unions of open intervals are open as well.

Let $\preceq$ be the lexicographic order on $X$; the linear order $\langle X,\preceq\rangle$ does have the least upper bound property. To prove this, let $\varnothing\ne A\subseteq X$. Let

$$P=\{x\in[0,1]:\exists i\in\{1,2\}\,(\langle x,i\rangle\in A)\}\,;$$

$[0,1]$ has the least upper bound property, so let $y=\sup P$, where the supremum is with respect to the usual order on $[0,1]$.

Show that if $y\notin P$, then $\langle y,1\rangle=\sup_\preceq A$.

Show that if $y\in P$, then $$\sup_\preceq A=\begin{cases}\langle y,1\rangle,&\text{if }\langle y,2\rangle\notin A\\\langle y,2\rangle,&\text{if }\langle y,2\rangle\in A\,.\end{cases}$$

Thus, in all cases $\sup_\preceq A$ exists.

Title: Supremum in a dictionary ordered set.

Body:
Take double arrow space $[0,1]\times \{1,2\}$

Where topology is dictionary topology defined by linear order as $(a,b)<(c,d)$ iff $a<c$; or ;$a=c \;\&\; b<d$

open sets are in the form $$((a, b] \times\{1\}) \cup([a, b) \times\{2\})$$

I want to know that is this topology has least upper bound property?

I know that $\mathbb R$ has this axiom, by using this can I show the above topology has least upper bound property?

By doing this I want to show the above topology is compact. I have seen https://dantopology.wordpress.com/2009/10/07/the-lexicographic-order-and-the-double-arrow-space/ proves it but it seems a bit advanced and less informative.


答:
One formal derivation of the "clear" observation using a change in the index of summation $l = k + j$, or equivalently $j = l-k$.
\begin{align}
\sum_{k=0}^n \binom nk x^k \cdot  \sum_{j=0}^m \binom mj x^j &=
\sum_{j=0}^m\sum_{k=0}^n \binom nk \binom mj x^{k + j}
\\ & = \sum_{l=0}^{m+n} \sum_{k=0}^l \binom nk \binom m{[l - k]}x^{k + [l-k]}
\\ & = \sum_{l=0}^{m+n} \sum_{k=0}^l \binom nk \binom m{l - k}x^{l}
 = \sum_{l=0}^{m+n} \left[\sum_{k=0}^l \binom m{l - k}\binom nk\right]x^{l}.
\end{align}


问:
Using matrix algebra to solve a differential vector equation
So I am solving a simple problem under the influence of gravity, with no resistance or extra forces.

Labeling the axes, let us use the downward vertical as our positive $y%-axis, with $x$ being standard.

Of course, the easiest way to actually define the position of the pendulum (or it's center of mass) is via

\begin{align*}
\text{position: }\vec{r} &= \begin{bmatrix} x\\ y\\ \end{bmatrix} = R\begin{bmatrix} \sin\theta\\\cos\theta \end{bmatrix}\\
\text{velocity: }\vec{v} &=\frac{d}{dt}(\vec{r})=\begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix}= R\begin{bmatrix}\dot{\theta}\cos\theta\\-\dot{\theta}\sin\theta\end{bmatrix}\\
\text{acceleration: }\vec{a} &=\frac{d^2}{dt^2}(\vec{r})= R\begin{bmatrix}\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta\\-\ddot{\theta}\sin\theta - \dot{\theta}^2\cos\theta\end{bmatrix}
\end{align*}

where $\theta$ is the angle formed with the vertical by the pendulum.

Normally we would write the EOM as two separate equations:

$$\begin{cases}m\ddot{x}=0\\m\ddot{y}=F_g=mg\end{cases}$$

Of course, the two equations of motion are coupled by the relationship $x^2+y^2=R^2$, where  but I wanted to try something a little different.

Instead, I wanted to try solving this system through matrix algebra, so rewriting the vector equation in polar coordinates by substituting for $\ddot{x}$ and $\ddot{y}$ yields:

$$m\vec{a} =
m  \begin{bmatrix} \ddot{x}\\\ddot{y}\end{bmatrix} =
mR \begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
                         -\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix}
=  \begin{bmatrix}0\\mg\end{bmatrix}$$

Normally we would express this system like so:

$$\begin{cases}m\ddot{x} = mR(\ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta)=0\\
m\ddot{y} = mR(-\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta)=mg\end{cases}$$

Here is my first question: Instead of solving the two equations separately, does it make sense to factor the matrix version of the equation, then employ some matrix multiplications to both sides to solve the problem?

The way I am imagining this process is as follows (beginning from Newton's second law)

\begin{align*}
m\vec{a} &= \vec{F}_{net} = \vec{F}_g\\
%
m\begin{bmatrix}\ddot{x}\\\ddot{y}\end{bmatrix} &= \begin{bmatrix}0\\ mg\end{bmatrix}\\
%
\text{Subsitute for $x,y$} \rightarrow m\begin{bmatrix}\ddot{x}\\\ddot{y}\end{bmatrix} &= mR\begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
                         -\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix}\\
%
\text{Factor} \rightarrow \begin{bmatrix} \ \ddot{\theta}\cos\theta-\dot{\theta}^2\sin\theta \\
                         -\ddot{\theta}\sin\theta-\dot{\theta}^2\cos\theta \\\end{bmatrix} &=
\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}\\
%
\text{So we have: }\rightarrow \begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}
\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix} \cos\theta&-\sin\theta \\ -\sin\theta&-\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}0\\ \frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}
\begin{bmatrix} \cos\theta&-\sin\theta \\ \sin\theta&\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}\begin{bmatrix}0\\ -\frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix} \cos^2\theta&-\sin\theta\cos\theta\\ \sin^2\theta&\sin\theta\cos\theta \\\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}0\\ -\frac{g}{R}\sin\theta\end{bmatrix}\\
%
\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix} \cos^2\theta&-\sin\theta\cos\theta\\ \sin^2\theta&\sin\theta\cos\theta \\\end{bmatrix} \begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=\begin{bmatrix}1&1\end{bmatrix}\begin{bmatrix}0\\ -\frac{g}{R}\sin\theta\end{bmatrix}\\
%
\begin{bmatrix}\cos^2\theta+\sin^2\theta&\sin\theta\cos\theta-\sin\theta\cos\theta\end{bmatrix}\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&= -\frac{g}{R}\sin\theta\\
%
\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&= -\frac{g}{R}\sin\theta\\
\ddot{\theta}&= -\frac{g}{R}\sin\theta\\
%
\end{align*}

$$\frac{d\dot{x}}{dx}\dot{x}= -\frac{g}{R}\sin\theta$$

$$\dot{x}d\dot{x}= -\frac{g}{R}\sin\theta dx$$

Is this a valid derivation of the equation of motion for a pendulum?

Edit: Here is another question. Instead of left multiplying by $\begin{bmatrix}\cos\theta&0\\0&\sin\theta\end{bmatrix}$, if we left multiply by $\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}$, we get:

\begin{align}
\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}
\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}0&\cos\theta\\-\sin\theta&0\end{bmatrix}
\begin{bmatrix}0\\-\frac{g}{R}\end{bmatrix}\\
%
\begin{bmatrix}\sin\theta\cos\theta&\cos^2\theta\\-\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}-\frac{g}{R}\cos\theta\\0\end{bmatrix}\\
%
\begin{bmatrix}1&1\end{bmatrix}
\begin{bmatrix}\sin\theta\cos\theta&\cos^2\theta\\-\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
\begin{bmatrix}1&1\end{bmatrix}
\begin{bmatrix}-\frac{g}{R}\cos\theta\\0\end{bmatrix}\\
%
\begin{bmatrix}0&1\end{bmatrix}
\begin{bmatrix}\ddot{\theta}\\ \dot{\theta}^2\end{bmatrix}&=
-\frac{g}{R}\cos\theta\\
%
\dot{\theta}^2 &= -\frac{g}{R}\cos\theta \\
%
\end{align}

If the above is valid, this should be valid too, correct?

答: The problem is that you started with a wrong assumption, that the gravity is the only force. If you just have a mass suspended at some point, and you release it, it will not oscillate, but instead it will go straight down. You need to add the tension in the suspending wire.</p>


问:
How to prove by induction $(n+1)^{1/n}$ is bounded above?
Was wondering if anyone could help with the problem above. It seems like the sequence is bounded by $2$ and I proceeded with the induction steps to show all elements are at most $2$.

Base case $n=1$ holds true because $2$ is at most $2$.

Induction assumption for some value $k$, from which I get $(k+2)^{1/(k+1)}\le 2$

But how do I proceed from here? Did I get here correctly? Do I use logarithms to proceed?

Thanks! (Sorry for the unformatted post).

Edit: sequence begins with $n=1,2,3\ldots$

答:
An idea:

Begin with $\;n<2^n\;$ forall $\;n\in\Bbb N$ :

$$\;\forall\,n\in\Bbb N\;,\;\;\;n<2^n\implies n^{1/n}<(2^n)^{1/n}=2\iff \sqrt[n]n<2\,,\,\,\forall\,n\in\Bbb N\;$$

If you want, you can carry on induction to prove $\;n<2^n\;$ for all $\;n=1,2,3,...\;$ N;$


问:
Representing sums of matrix algebras as group rings
Let $A = M_{n_1}(\mathbb R) \oplus M_{n_2}(\mathbb R) \oplus ... \oplus M_{n_m}(\mathbb R)$ be a direct sum of real matrix algebras. Under what conditions does there exist a group ring $\mathbb R[G]$ which is isomorphic to $A$?

I know every group ring is isomorphic to some such $A$ by Wedderburn's theorem, and I want to determine the extent to which the converse holds. I know that $A$ must have a $\mathbb R$-summand, corresponding to the linear span of $\sum_{g\in G} g \in \mathbb R[G]$. Is this condition sufficient as well? If not, is there a nice criterion, at least for small numbers of summands?

答:
This is similar to Qiaochu Yuan's answer (+1) but makes the claims more strongly and uses Miller's classification of groups with very few conjugacy classes.

Suppose $G$ is a finite group such that $\mathbb{R}[G]$ is the direct product of matrix rings $M_{n_i}(\mathbb{R})$ for $n_1 \leq n_2 \leq \dots \leq n_m$.

The original question claims this is true for any group, but actually this places fairly severe restrictions on the group which we can use to classify those groups with small $m$. Since all representations are real, we have that the $n_i$ are the character degrees of $G$, and so we have that $n_i$ divides $\sum n_i^2 = |G|$ and that $m$ is the number of conjugacy classes of $G$ (which must all be real).  The number of the $n_i$ that are equal to 1 is the index $[G:G']$, and in particular is at least one. Since all the representations are real, we also have that $G$ contains exactly $\sum n_i$ elements of order dividing 2, and so $|G|$ is even.

If $m=1$, there is only one degree and we know it is $n_1=1$ so that $|G|=1$ and $G$ is the trivial group.

If $m=2$, then $G$ has only two conjugacy classes, so $|G|=2$ and $G$ is cyclic of order 2 with $n_1=n_2=1$.

If $m=3$, then $G$ has only three conjugacy classes, so $|G| \in \{3,6\}$, but only $G$ the nonabelian group of order 6 has only real conjugacy classes, so $n_1= n_2=1$ and $n_3=2$.

If $m=4$, then $G$ has only four conjugacy classes, so $|G| \in \{4,10,12\}$, but only $G = C_2 \times C_2$ and $G = D_{10}$ work, giving either $n_1=n_2=n_3=n_4=1$ or $n_1=n_2=1$ and $n_3=n_4=2$.
$$\begin{array}{ccccc|c}
n_1 & n_2 & n_3 & n_4 & G \\ \hline
1 & 1 & 1 & 1 & C_2 \times C_2 \\
1 & 1 & 2 & 2 & D_{10}
\end{array}$$

If $m=5$, then we get only four possibilities:
$$\begin{array}{ccccc|c}
n_1 & n_2 & n_3 & n_4 & n_5 & G \\ \hline
1 & 1 & 1 & 1 & 2 & D_8 \\
1 & 1 & 2 & 2 & 2 & D_{14} \\
1 & 1 & 2 & 3 & 3 & S_4 \\
1 & 3 & 3 & 4 & 5 & A_5
\end{array}$$


问:
$\arg(zw)=\arg(z)+\arg(w)$ any one of $zw$, $z$, or $w$ outside of $(-\pi, \pi]$
In the solution of 3(b), it states that the solution to $1+z^3 = 0$ is $-1, -w, -w^2 (w = -1/2 + \sqrt{3}i/2)$. However, I think it should be $-w^k, k \in \mathbb{N}$.

Take $k=3$ for example, then the equation is still true.

答: Well, yes, but there are only three solutions.  The additional solutions you found are repeats.</p>


问:
pdf of product of standard normal and uniform
Even though this question has been answered somewhat here, here, here, I'd like to ask what the probability density function of $z = x y$ is, where $x \sim \mathcal{U}[-1,1]$, $y \sim \mathcal{N}(0,1)$, and $z$ are all univariate. Furthermore, what are the mean and variance of $z$? (By symmetry, it is easy to see $\mathbb{E}[z] = 0$, but I want a more rigorous proof.)

Edit: What if $x \sim \mathcal{U}[-\theta,\theta]$?

答: Since $|X|,\,|Y|$ have respective PDFs $1_{[0,\,1]}(x),\,\sqrt{\tfrac{2}{\pi}}e^{-y^2/2}1_{[0,\,\infty)}(y)$, $|Z|=|X||Y|$ has on-$[0,\,\infty)$ PDF $\int_0^1x^{-1}\sqrt{\tfrac{2}{\pi}}e^{-z^2/(2x^2)}dx=\tfrac{1}{\sqrt{2\pi}}\Gamma\left(0,\,\tfrac12z^2\right)$ in terms of the upper incomplete Gamma function. Since $X,\,Y$ have symmetric distributions, $Z$'s PDF on $\Bbb R$ is $\tfrac{1}{\sqrt{8\pi}}\Gamma\left(0,\,\tfrac12z^2\right)$.


问:
Solving Simple Mixed Fraction problem?
How do you wrap your head around mixed fraction, does anyone knows how to figure out, can someone give me an example how it can be solved?</p>

答:
$$a +\frac bc =  \frac {a\times c}c + \frac bc = \frac{(a\times c) + b}{c} $$

What we do first is like when finding a common denominator between two fractions, only in $a$ above, we have $a = \dfrac a1$:

We multiply $a = \dfrac a1$ by $\dfrac cc = 1$, to get $\dfrac a1 \times \dfrac cc = \;\dfrac{a\times c}{c}\;$ and then add that to the fraction $\dfrac bc$.

For example

$$3 \frac 58\; = \;3 + \frac 58\; = \;\frac{3 \times 8}{8} + \frac 58\; = \;\frac{(3\times 8)+ 5}{8}\; = \;\frac{24 + 5}{8} \;= \;\frac{29}{8}$$

Title: Solving Simple Mixed Fraction problem?


答:
Consider the sequence $a_n = (n + 1)^{1/n}$ defined for all integers $n \geq 1.$ Using logarithmic differentiation on the continuous function $f(x) = (x + 1)^{1/x}$ defined for all real numbers $x \geq 1,$ we can prove that $f(x)$ is decreasing, hence $f(x)$ attains an absolute maximum. Prove that $f(x)$ takes its maximum value of $2$ at $x = 1,$ hence the sequence $a_n$ is bounded above by $2.$

Once you believe this, you can proceed by induction to prove that $a_n \leq 2$ for all integers $n \geq 1,$ if you wish. For $n = 1,$ the claim holds, hence we may assume inductively that $a_n \leq 2.$ Observe that $$a_{n + 1} \leq 2 \iff \ln a_{n + 1} = \ln (n + 2)^{1 / (n + 1)} = \frac{\ln(n + 2)}{n + 1} \leq \ln 2$$ because $\ln x$ and $e^x$ are increasing functions. Consequently, it suffices to show that $\ln(n + 1) \leq (\ln 2)n$ for all integers $n \geq 0$; this can be accomplished via induction.

Ultimately, this shows that there are better ways to prove that $(n + 1)^{1/n}$ is bounded above by $2.$

Title: How to prove by induction $(n+1)^{1/n}$ is bounded above?

Body:
Was wondering if anyone could help with the problem above. It seems like the sequence is bounded by $2$ and I proceeded with the induction steps to show all elements are at most $2$.

Base case $n=1$ holds true because $2$ is at most $2$.

Induction assumption for some value $k$, from which I get $(k+2)^{1/(k+1)}\le 2$

But how do I proceed from here? Did I get here correctly? Do I use logarithms to proceed?

Thanks! (Sorry for the unformatted post).

Edit: sequence begins with $n=1,2,3\ldots$


问:
Find a formula for the $n$-th power, given $x^3=-ax$.
I am given the relation $$x^3=-ax$$, $x$,$a$ are real numbers.

Find the $x^n$

My attempt, by trying some values I get $$x^4=-ax^2$$
$$x^5=a^2x$$
$$x^6=a^2x^2$$
$$x^7=-a^3x$$
$$x^8=-a^3x^2$$
and so on

So I can see that $x^n$ looks like $$a^{\frac{n}{2}-1} x^2$$ for $n$ even
and $$a^{\frac{n-1}{2} }x$$ for $n$ odd.

However, this is not the answer since I can see that for each case, the sign alternates, but I can't seem to write a function that alternates sign every step.

In general, I know that $(-1)^n$ does exactly this, however, here I have only odd numbers in one case and only even numbers in the other, so for example, $$x^n=(-1)^n a^{\frac{n}{2}-1} x^2$$ for $n$ even, would not alternate sign.

I want to ask, what function would do that?

Body:
$$x^3 + ax = 0$$
$$x(x^2+a) = 0$$

If $a\ge 0$ then $x=0,\quad x^n=0.$

If $a<0$, then $x=0$ or $x\in \{(-a)^{1/2}, -(-a)^{1/2}\}, \quad x^n \in \{(-a)^{n/2}, (-1)^n(-a)^{n/2}\}$.

Thanks to @JoshuaWang for correcting my algebra!


答:
The right-hand side of your final answer should not contain $x$.

A brute-force hint.

If $x=0$, then $x^n=0$ for any positive integer $n$.

If $a=0$, then you must have $x=0$.

If $a>0$, and $x\ne 0$, then $x^2=-a$, which is impossible.

If $a<0$, and $x\ne 0$, then $x^2=-a$, and $x=\pm\sqrt{-a}$.

Alternatively, consider the equivalent equation $x^3+ax=0$, which can be written as $x(x^2+a)=0$.

Title: Find a formula for the $n$-th power, given $x^3=-ax$.

Body:
I am given the relation $$x^3=-ax$$, $x$,$a$ are real numbers.

Find the $x^n$

My attempt, by trying some values I get $$x^4=-ax^2$$
$$x^5=a^2x$$
$$x^6=a^2x^2$$
$$x^7=-a^3x$$
$$x^8=-a^3x^2$$
and so on

So I can see that $x^n$ looks like $$a^{\frac{n}{2}-1} x^2$$ for $n$ even
and $$a^{\frac{n-1}{2} }x$$ for $n$ odd.

However, this is not the answer since I can see that for each case, the sign alternates, but I can't seem to write a function that alternates sign every step.

In general, I know that $(-1)^n$ does exactly this, however, here I have only odd numbers in one case and only even numbers in the other, so for example, $$x^n=(-1)^n a^{\frac{n}{2}-1} x^2$$ for $n$ even, would not alternate sign.

I want to ask, what function would do that?


问:
sine equation - signal processing
I have a problem to solve in my signal processing course. The chapter of this question is "Sampling and aliasing".

Let's call $g_f(t) = \sin(2\pi.f.t) + \sin(2\pi.(f+\Delta_f).t) + \sin(2\pi.(f+2\Delta_f).t) + 2.\sin(2\pi.(f+3\Delta_f).t)$.

The first question was to visualize the frequency representation of $g_{1000}$.
The result was coherent : 4 negative Diracs in $f$, $f+\Delta_f$, $f+2\Delta_f$, $f+3\Delta_f$ and 4 positive Diracs in $-f$, $-f-\Delta_f$, $-f-2\Delta_f$, $-f-3\Delta_f$. Then I had to visualize the frequency representaion of $g_{2250}$. This time, the result was not according to theory because of aliasing ($2250 > \frac{f_e}{2}$, with $f_e$ the sempling frequency).
Here is the next question :

Find all the $f$ so $g_f(t)$ is equal to a unique sinus of another specific frequency.

Any idea about how to begin that ?

答:
Let $f = -\Delta_f$.
We'll have,
$$g_f(t) = \sin(2\pi.-\Delta_f.t) + \sin(2\pi.(-\Delta_f+\Delta_f).t) + \sin(2\pi.(-\Delta_f+2\Delta_f).t) + 2.\sin(2\pi.(-\Delta_f+3\Delta_f).t)$$
So,
$$g_f(t) = \sin(2\pi.-\Delta_f.t) + \sin(2\pi.\Delta_f.t) + 2.\sin(2\pi.(2\Delta_f).t)$$

If we think in terms of frequency, the first sine has one negative Dirac peak in $-\Delta_f$ and one positive in $\Delta_f$.
The second one has one positive in $-\Delta_f$ and one negative in $\Delta_f$.
They cancel each other.

Finally it will only result 2 peaks, one positive in $-2\Delta_f$ and one negative in $2\Delta_f$ corresponding to the last sine.

So $g_{-\Delta_f.t}$ is equivalent to a unique sine of frequency $2\Delta_f$.

Title: sine equation - signal processing


问:
Why is the angle exactly 1/2 of one of the angles in the isosceles triangle
I'm sure there's a simple theorem that answers my question, but I'm unsure as to how to formulate the problem (not easy to describe geometry with words): It's a isosceles triangle where you draw a normal line on one of the "hypotenuse" sides, and then you make a right angled triangle out of that. My question is why is the 30 degree angle exactly 1/2 of the 60 degree angles we find in the isosceles triangle?

答: You have an equilateral triangle. Bisecting angle $\beta$ and drawing the line through the line segment $AC$ will go through the midpoint. You should be able to convince yourself that angle $\sigma$ has to be 30 degrees from that.


问:
Every open interval of $\mathbb{R}$ is a countable intersection of open sets of $\mathbb{R}$.
How can I prove using topological properties that every open interval of $\mathbb{R}$ is a countable intersection of open sets of $\mathbb{R}$? Is that true that every closed interval of $\mathbb{R}$ is a countable intersection of open sets of $\mathbb{R}$?

I have seen several divergent responses for those questions and I have no idea of how to answer to these statements.

答:
$$[a,b] = \cap\{ (u,v) : u,v \in \mathbb{Q}, u < a, b < v \}$$

In a simular way you can prove the dual statement that every open interval is a countable union of closed sets.


答:
If $ a=0$ then $ x=x^n=0$.

if $ x=0 $ then $ x^n=0$

If $ x\ne 0 $ then $ x^2=-a$ and

necessarily, $  a<0 $ or $ a=-b $ with $ b>0$.

So, $ x^2=b $ and $ |x|=b^{\frac 12}$

Thus

$$x>0\implies x^n=b^{\frac n2}$$

and
$$x<0\implies x^n=(-1)^nb^{\frac n2}$$


问:
The segment joining the vertices of altitudes
A triangle $ABC$ with sides $AB=15,BC=14,AC=13$ is given. $AA_1=12$ and $BB_1$ are heights. Find $A_1B_1$.

I was thinking about the Ptolemy's theorem: $$A_1B_1\cdot CH=B_1C\cdot HA_1+B_1H\cdot CA_1,$$ where $H$ is the intersection of the heights (the orthocenter). We can use it because $HA_1CB_1$ is cyclic: $$\measuredangle HB_1C+\measuredangle HA_1C=180^\circ.$$ I was able to find that $CA_1=5$. I am not sure this is the easiest solution, because we have to find $5$ more segments. Any thoughts on the problem will be appreciated. Thank you!

Body:
Equate the area of the triangle to find $BB_1 = \frac{12 \times 14}{13}$.

$A_1C = 5$ as you mentioned.

Applying Pythagoras, $B_1C^2 = BC^2 - BB_1^2 \implies B_1C = \frac{70}{13}$

Now find $\cos C = \frac{AC^2 + BC^2 - AB^2}{2 AC \cdot BC} = \frac{5}{13}$

$A_1B_1^2 = A_1C^2 + B_1C^2 - 2 A_1C \ B_1C \cos C = 25 + (\frac{70}{13})^2 - \frac{3500}{13^2} = (\frac{75}{13})^2$

Hence $A_1B_1 = \frac{75}{13}$


问:
Inverse | Modulo | Power
Describe the inverse of $5$ modulo $18$ as a positive power of $5\pmod{18}$.

I've got that the inverse of $5$ is $11$, but is this question asking to find a $t$ such that
$$ 11=5^t\pmod{18}?$$

答:
$$5^t \equiv 11 \pmod{18}$$

Notice that $\phi(18)=6$

$$5^{6} \equiv1\pmod{18}$$

$$5^5 \cdot 5 \equiv 1\pmod{18} \implies 5x \equiv 1\pmod{18}$$

$$\implies x \equiv 11\pmod{18}$$

Therefore, $$5^5 \equiv 11\pmod{18}$$

Title: Inverse | Modulo | Power


答:
The area of the triangle is, by Heron's formula, $$|\triangle ABC| = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(6)(7)(8)} = 84,$$ thus $$AA_1 = \frac{2(84)}{14} = 12, \quad BB_1 = \frac{2(84)}{13} = \frac{168}{13},$$ and by the Pythagorean theorem, $$CA_1 = 5, \quad CB_1 = \frac{70}{13}.$$  Since by the Law of Cosines we have $$\cos \angle C = \frac{13^2 + 14^2 - 15^2}{2(13)(14)} = \frac{5}{13},$$ we obtain
$$(A_1 B_1)^2 = (CA_1)^2 + (CB_1)^2 - 2(CA_1)(CB_1) \cos \angle C = \left(\frac{75}{13}\right)^2,$$ and the desired length is $75/13$.

Title: The segment joining the vertices of altitudes

Body:
A triangle $ABC$ with sides $AB=15,BC=14,AC=13$ is given. $AA_1=12$ and $BB_1$ are heights. Find $A_1B_1$.

I was thinking about the Ptolemy's theorem: $$A_1B_1\cdot CH=B_1C\cdot HA_1+B_1H\cdot CA_1,$$ where $H$ is the intersection of the heights (the orthocenter). We can use it because $HA_1CB_1$ is cyclic: $$\measuredangle HB_1C+\measuredangle HA_1C=180^\circ.$$ I was able to find that $CA_1=5$. I am not sure this is the easiest solution, because we have to find $5$ more segments. Any thoughts on the problem will be appreciated. Thank you!


问:
Does a Line Integral that sums to a vector exist ? And is it useful?
We know that we can integrate a vector field on a curve to get a scalar sum: this is how the line integral is defined. Is there a use for another sort of integral whose sum is a vector ? I can't seem to find that defined anywhere.</p>

答:
Integrals are (limits of) sums of many small things. If the small things are vectors, then the result is a vector too.

The simplest example is probably the following: The sum of many small displacements is a displacement. More specifically, let a particle follow the path $\vec p(t)$, where $t$ is time. Then we have the velocity vector $\vec v(t) =\vec p'(t)$.

The net displacement of this particle from $t$ to $t+\Delta t$ is roughly $\vec v(t)\cdot\Delta t$. For small $\Delta t$, this is a small vector. This can be turned into an integral.

If you integrate the velocity vector with respect to time from $t=t_0$ to $t=t_1$, the end result is the net displacement $\vec p(t_1)-\vec p(t_0)$:
$$
\int_{t_0}^{t_1}\vec v(t)\,dt=\vec p(t_1)-\vec p(t_0)
$$


问:
What is a good book to learn number theory?
What would be a good book to learn basic number theory? If possible a book which also has a collection of practice problems? Thanks.</p>

答:
I would recommend the dover book $\mathit{Number\,\,Theory}$ by George R. Andrews, which is extremely friendly too beginners but also does a good job at giving an overview of all topics in the general "elementary number theory" umbrella.

Counting exercises its a short 300 page read, and cover to cover you could probably get through it in a month or two. I really can't recommend this book enough.


问:
How can we compute this quantity, constructed using motivation from Vandermonde's identity?
Vadermonde's identity states that

$$ \sum_{k=0}^{r} {m \choose k}{n \choose r-k} = {m+n \choose r} $$

I wonder if we can use this formula or otherwise to calculate:

$$ \sum_{k=0}^{r}k^l {m \choose k}{n \choose r-k}, l \in \mathbb{N} ?$$

or if not,  at least when $l=1,$ namely:

$$ \sum_{k=0}^{r}k {m \choose k}{n \choose r-k}?$$

I'd appreciate the relevant calculations or some links where I can find the above?

答:
We  have for with the dominant parameter being $p$

$$\sum_{k=0}^r k^p {m\choose k} {n\choose r-k}$$

that it is

$$p! [w^p] \sum_{k=0}^r \exp(kw) {m\choose k} {n\choose r-k}
\\ = p! [w^p] [z^r] (1+z)^n
\sum_{k=0}^r \exp(kw) {m\choose k} z^k.$$

Now the coefficient extractor enforces the upper limit of the range and
we may continue with

$$p! [w^p] [z^r] (1+z)^n
\sum_{k\ge 0} \exp(kw) {m\choose k} z^k
\\ = p! [w^p] [z^r] (1+z)^n (1+z\exp(w))^m
\\ = p! [w^p] [z^r] (1+z)^n (1+z+z(\exp(w)-1))^m
\\ = p! [w^p] [z^r] (1+z)^n
\sum_{j=0}^m {m\choose j} (1+z)^{m-j} z^j (\exp(w)-1)^j
\\ = [z^r] \sum_{j=0}^m {m\choose j} (1+z)^{m+n-j} z^j j! {p\brace j}
\\ = \sum_{j=0}^m {m\choose j} {m+n-j\choose r-j} j! {p\brace j}.$$

Note that if $m\gt p$ the values with $m\ge j\gt p$ produce a zero
Stirling number  so we may lower $m$ to $p.$ If $m\lt p$ the values with
$p\ge j\gt m$ produce a zero binomial coefficient and we may raise $m$ to
$p.$ We thus obtain

$$\bbox[5px,border:2px solid #00A000]{
\sum_{j=0}^p {m\choose j} {m+n-j\choose m+n-r} j! {p\brace j}.}$$

a sum with $p$ non-zero terms except for $p=0$, when it has one term. (We
could also use $\min(m,p)$ as the upper limit  but we want to emphasize
the dependence on $p.$) Note that in the initial sum for it to be
non-zero with non-negative $k$ we must have $m\ge k$ and $n\ge r-k$  or
$k\ge r-n$ so that $m\ge k\ge r-n$ and for the range not to be empty we
must  have $m\ge r-n$ or $m+n-r\ge 0$ which ensures that the middle
binomial  coefficient in the boxed form is non-zero and well defined.
Observe that with $p=0$ we obtain ${m+n\choose m+n-r} =  {m+n\choose r}$
which is Vandermonde. A slight variation is

$$\bbox[5px,border:2px solid #00A000]{
\sum_{j=0}^p m^\underline{j} {m+n-j\choose m+n-r} {p\brace j}.}$$

Remark. This is the same as the formula that was first to appear.

We get

$${m+n\choose r} {p\brace j} {m\choose j} {m+n\choose j}^{-1}
r^\underline{j}
\\ = {p\brace j} {m\choose j}
\frac{(m+n)!/(m+n-r)!/r!}{(m+n)!/(m+n-j)!/j!} r^\underline{j}
\\ = {p\brace j} {m\choose j} j! {m+n-j\choose m+n-r}
(r-j)! r^\underline{j} / r!
 = {p\brace j} {m\choose j} j! {m+n-j\choose m+n-r}.$$

Remark II. We may keep the ${m+n-j\choose r-j}$ if we remember that it originates with $[z^r] (1+z)^{m+n-j} z^j$ and hence is zero when $j\gt r.$


问:
Solve the equation $(2-x^3)^3+x-2=0$
I would like for you to help me solve this equation using simple factorisation.

$(2-x^3)^3+x-2=0, x \in \mathbb{R}$

I have been trying to expand the expression which now becomes $-x^9+6x^6-12x^3+x+6$ and find a way around it but still no luck. Any help would be appreciated.

Title: Solve the equation $(2-x^3)^3+x-2=0$

Body: $$\begin{aligned} (2−x^3)^3+x−2=0 \\ x^9+6x^6-12x^3+x+6=0 \\ -\left(x-1\right)\left(x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6\right)=0 \\ \mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0 \\x-1=0\quad \mathrm{or}\quad \:x^8+x^7+x^6-5x^5-5x^4-5x^3+7x^2+7x+6=0 \\x-1=0 \\ \boxed{x=1} \end{aligned}$$


问:
Fifth root of unity pentagon proof
I am stuck at an exercise that refers to the fifth root of unity with the equation $z^5-1=0$. I showed in an exercise before that 1 and $x=\frac{1}{2}(h+i\sqrt{4-h^2})$ are possible solutions for the equation.

Now the exercise says that there are $1,x,x^2,x^3,x^4$ given as edges of a pentagon and one can show that the side lengths are all equal.
I tried it by looking at:

$|x^{k+1}-x^k| = |x^k(x-1)| = |x^k||x-1| = |x|^k|x-1|  $(for integers k) but I don't know how to go on :/

Note: We haven't introduced Pi and Eulers identity in our lectures yet, so were not allowed to use them here!

Title: Fifth root of unity pentagon proof

Body: If $x^5=1$ then $|x|^5=|x|^5=1$ so $|x|=1$.


答:
Here is an example.

$$2 \frac{3}{11}=\frac{2 \cdot 11}{11}+\frac3{11}=\frac{22}{11}+\frac3{11}=\frac{25}{11}$$

Or conversely,

$$\frac{11}3=\frac{11-3 \cdot 3}{3}+\frac{3 \cdot 3}{3}=\frac{11-9}{3}+\frac{9}{3}=\frac{2}{3}+3=3 \frac{2}3$$.

In general,

$$x+\frac{y}{z}=\frac{x \cdot z}{z}+\frac{y}{z}=\frac{x \cdot z +y}z$$ and vice versa.

In most cases you'll encounter $y<z$.

Title: Solving Simple Mixed Fraction problem?

Body: How do you wrap your head around mixed fraction, does anyone knows how to figure out, can someone give me an example how it can be solved?</p>


答:
Note that $ABA_1B_1$ is cyclic, implying similar triangles $ABC $ and $A_1B_1C$. Thus,
$$ A_1B_1= \frac{CA_1}{CA}AB =\frac 5{13}\cdot15= \frac{75}{13}$$


问:
Expected value calculations
I am reading Montgomery book for DOE.
this is part of his proofs. I could figure out what he does until this last line, how did he get the second line from the first line. I know E(x) formulas for continuous and discrete functions, but I still have difficulty understanding this.

答:
There is one squared summation and one squared double summation, for which you will need the appropriate formulae.

The formulae are as follows: $\underbrace{\left(\sum_{i=1}^n a_i\right)^2}_{\text{n$^2$ terms}}=\underbrace {\sum a_i^2}_{\text{n terms}}+\underbrace {\sum_{i\lt j} a_ia_j}_{\text{$\frac{n(n-1)}2$ terms}}+\underbrace{\sum_{i>j}a_ia_j}_{\text{$\frac{n(n-1)} 2$ terms}}=\sum a_i^2+\underbrace{\sum_{i\ne j}a_ia_j}_{\text{n(n-1) terms}}$.

For the double sum, the square is $\underbrace{\left(\sum_{j=1}^m\sum_{i=1}^n a_{ij}\right)^2}_{\text{$n^2m^2$ terms}}=\underbrace{\sum_{j=1}^m\sum_{i=1}^n a_{ij}^2}_{\text{nm terms}}+\underbrace{\sum_{l=1}^m\sum_{j=1}^m\sum_{i\ne k} a_{ij}a_{kl}}_{\text{$m^2n(n-1)$ terms}}+\underbrace{\sum_{i=1}^n\sum_{j\ne l}a_{ij}a_{il}}_{\text{mn(m-1) terms}}$

To get you started, the left sum is $E\left[\sum_{i=1}^a\frac 1 {n_i}\left(\sum_{j=1}^{n_i}(\mu+\tau+\epsilon_{ij})\right)^2\right]=E\sum_{i=1}^a\frac 1 {n_i}\left[\sum_{j=1}^{n_i}(\mu+\tau_i+\epsilon_{ij})^2+\sum_{j\ne k}(\mu+\tau_i+\epsilon_{ij})(\mu+\tau_i+\epsilon_{ik})\right]=...$

Eventually you'll need to use the fact that $E(\epsilon)=0,E(\epsilon^2)=var(\epsilon)-[E(\epsilon)]^2=\sigma^2$, and further, that $\mu$ is constant, also don't know what "$\sum \tau_i$" is.

Title: Expected value calculations


问:
Determine if the Implication is True (REGARDLESS OF THE VALUE OF F(x, y))
Given $F(x,y)$ and ignoring whether or not it is true, determine if the following implications are true:

a) $\forall x \exists y F(x,y) \implies \exists y \forall x F(x,y)$

b) $\exists y \forall x F(x,y) \implies \forall x \exists y F(x,y)$

My answer:

Part a would be false because the second part of the implication wouldn't necessarily be true if all of x relates to at least 1 y variable from the first part.
Part b I am unsure but I think this is true since you are making the relation more broad in the second part of the implication.

Does my logic make sense?

答: Yes you are correct. In the second example, given $x$, one can simply choose the $y$ that works for all $x$.


答:
I am surprised no one has mentioned Elementary Number Theory by David M. Burton yet.

This book is easy to follow, has lots of illustrative examples, and also comes with a bunch of exercises/practice problems at the end of each section.

In fact, it was exactly this book that introduced me to the topic of odd perfect numbers back in the year $1999$.

Title: What is a good book to learn number theory?

Body: What would be a good book to learn basic number theory? If possible a book which also has a collection of practice problems? Thanks.</p>


问:
Calculating the improper integral $\int _{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$
I want to calculate the value of $$\int_{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$$

I was able to prove using Dirichlet's test that it does converge, but how can I calculate its value?

Body:
We have that, integrating by parts
$$
\eqalign{
  & I = \int_1^\infty  {e^{\,\sin x} {{\cos x} \over x}dx}
  = \int_1^\infty  {{1 \over x}d\left( {e^{\,\sin x} } \right)}  =   \cr
  &  = \left. {{{e^{\,\sin x} } \over x}} \right|_1^\infty
   - \int_1^\infty  {e^{\,\sin x} d\left( {{1 \over x}} \right)}
  =  - e^{\,\sin 1}  + \int_1^\infty  {{{e^{\,\sin x} } \over {x^{\,2} }}dx}  =   \cr
  &  =  - e^{\,\sin 1}  + J \cr}
$$
with the advantage of having replaced $1/x$ with $1/x^2$.

Then being $\exp(\sin x)$ periodic
$$
\eqalign{
  & J = \int_1^\infty  {{{e^{\,\sin x} } \over {x^{\,2} }}dx}
  = \int_1^{1 + 2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx}
  + \int_{1 + 2\pi }^{1 + 4\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx}  +  \cdots  =   \cr
  &  = \sum\limits_{0\, \le \,k} {\int_{1 + k2\pi }^{1 + \left( {k + 1} \right)2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} }
  = \sum\limits_{0\, \le \,k} {\int_0^{2\pi }
 {{{e^{\,\sin \left( {x + 1} \right)} } \over {\left( {x + 1 + k2\pi } \right)^{\,2} }}dx} }  =   \cr
  &  = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)}
 \left( {\sum\limits_{0\, \le \,k}
 {{1 \over {\left( {{{\left( {x + 1} \right)} \over {2\pi }} + k} \right)^{\,2} }}} } \right)dx}  =   \cr
  &  = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi }
 {e^{\,\sin \left( {x + 1} \right)} \zeta \left( {2,{{x + 1} \over {2\pi }}} \right)dx}  =   \cr
  &  = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi }
 {e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right)dx}  =   \cr
  &  = {1 \over {2\pi }}\int_0^{2\pi }
 {e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right)
d\left( {{{x + 1} \over {2\pi }}} \right)}  =   \cr
  &  = {1 \over {2\pi }}\int_{{1 \over {2\pi }}}^{{1 \over {2\pi }} + 1}
 {e^{\,\sin \left( {2\pi t} \right)} \psi ^{\left( 1 \right)} \left( t \right)d\left( t \right)}   \cr
  }
$$
where $\psi ^{\left( 1 \right)} \left( t \right)$ denotes the
Trigamma Function.

So we have reduced the improper integral to a proper integral,
much more manageable by numeric integration.

My CAS gives:
$$
\eqalign{
  & J \approx 1 .94017  \cr
  & {\rm I} \approx  - \,0.3796 \cr}
$$


问:
Prove that there is NO monomorphism of $\mathbb{Z_{6}}$ in $\mathbb{Z_{11}}$.

Prove that there is NO monomorphism of $\mathbb{Z_{6}}$ in $\mathbb{Z_{11}}$ ($\mathbb{Z_{6}}$ and $\mathbb{Z_{11}}$ are rings).

Let $f: \mathbb{Z_{6}} \rightarrow \mathbb{Z_{11}}$ be a homomorphism. If $f$ is injective, that is, it is a one-to-one function, then $f$ is said to be a monomorphism.

I'm trying to find a contradiction but can't find it for this test.

答:
$\mathbb{Z}_{11}$ is a field ($11$ is prime) so it has no zero divisors. The image of any ring homomorphism is a subring. Since $\mathbb{Z}_{6}$ has zero divisors, its image has them too. Therefore only zero homomorphism is possible.

Another approach: $\mathbb{Z}_{11}$ and $\mathbb{Z}_{6}$ are additive groups. The image of a group under any homomorphism is a subgroup. By Lagrange's theorem its order divides the order of an ambient group. But $11$ is prime.


问:
How does $f(x)= x \sin(\frac{\pi}{x})$ behave?
I think this function is increasing for $x>1$ but wanted to find the reason. So I thought about taking the derivative:

$f(x)= x \sin(\frac{\pi}{x})$

Aplying the chain an the product rule, we get:

$f'(x)= \sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x})$

The function is increasing if the derative is more than or equal to $0$, so:

$\sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x}) \ge 0$

$\sin(\frac{\pi}{x}) \ge \frac{\pi}{x} \cos (\frac{\pi}{x}) $

Since $ \cos ( x) > 0$, if $ 0< x < \pi$,  $ \cos (\frac{\pi}{x}) > 0 $, because $ 0<\frac { \pi}{x}< \pi$.

$ \tan (\frac{\pi}{x}) \ge \frac{\pi}{x}$

I get to this point and don't know how to continue. I'd like you to help me or give me a hint, or maybe see a different way of showing it. Anyway, thanks.

答:
Try the Maclaurin series for tangent:
$$\tan x = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \ldots$$
All terms are positive when $x > 0$, so $\tan x > x$. This proves that $\tan \frac{\pi}{x} > \frac{\pi}{x}$ for $0 < x < \frac{\pi}2$.

Alternatively, define $f(x) = \tan x - x$. Then $f(0) = 0$ and $f'(x) = \sec^2 x - 1 > 0$ for $0 < x < \frac{\pi}2$, so $f(x)$ is increasing, and $f(x) > 0$ on $0 < x < \frac{\pi}2$.

Now, here's the catch: remember that tangent is not defined at $x = \frac{\pi}2$. However, the argument of your tangent function is $\frac{\pi}{x}$. So, for $x > 2$, this fraction decreases from $x = 2$ to $x = \infty$, and in fact, approaches zero. Therefore, you are safe with your argument showing that the original function is increasing for $x > 2$. For $1 < x \leq 2$, you need to be a bit more careful.


问:
How to prove these triangles are similar?
I have the following setup.

A triangle $ABC$ has incentre $I$ and $AB>AC$. $E$ is a point on the circumcircle of $ABC$ such that $\angle IEA=90^\circ$. $U$ is a point on $AC$ such that $IU$ is perpendicular to $AC$, and $W$ is a point on $AB$ such that $IW$ is perpendicular to $AB$.

How do I prove that $\triangle EUC$ is similar to $\triangle EWB$?

答:
Clearly we have for triangles $EUC$ and $EWB$,
$$\angle ECU = \angle EBW$$
subtended by chord $AE$ on the circumcircle $(ABC)$.

Also $E,U,W$ lie on the circle with diameter $AI$. So
$$\angle AUE= \angle AWE$$
$$\Rightarrow \angle EUC= \angle EWB$$

Thus $\triangle EUC \sim \triangle EWB$ by angle-angle similarity criterion.


问:
Basic question on the transformation of Exponential distribution.
Why central moments coincide for random variables

$V\sim E(a,h)$  and $Y\sim E(h)$

where a=location parameter

h= scale parameter.

答: Because $E(a,h)$ is $E(h)$ translated by $a$.  Taking central moments (by subtracting the mean) removes the translation.</p>


问:
Maximize $\sum{a_i^2}$ given $\sum{a_i}=237$
I recently got stuck with the following problem:

Consider 18 integer numbers in segment $[10; 20]$. Find the maximum possible value of
$$
\sum_{i=1}^{18} {a_i^2}
$$
given that
$$\sum_{i=1}^{18} {a_i}=237$$

I don't really know how to approach this problem. The thing is, solving it like constraint optimization problem I got the solution $a_1=a_2=...=a_n=\frac{237}{18} \approx13.2$

But we need to solve it in integers. And just taking numbers around $13.2$ doesn't make sense to me. Any ideas?

答:
An optimal solution cannot have $10 < a_i \leq a_j < 20$ for some $i$, $j$ because that contradicts optimality (you could increment $a_j$ and decrement $a_i$ and get a higher objective value). Therefore, there can be at most one element that is not $10$ or $20$.

There is therefore an optimal solution for which $a_1, \ldots, a_m$ take the value $10$, $a_{m+1},\ldots,a_{17}$ take the value $20$, and $a_{18} \in [10,20] \cap \mathbb{Z}$. From the equality constraint it follows that $a_{18} = 237 - 10m - 20(17-m) = 10m-103$. The only $m$ for which $10m-103 \in [10,20]$ is $m=12$, so the problem is solved.

If you had multiple $m$, you should evaluate the objective value ($100m + 400(17-m) + a_{18}^2 = 100m^2 - 2360m + 17409$) and pick the one for which the objective is the highest.

Other method This problem is also easy to solve with an integer programming solver after you formulate it appropriately. Let $x_{10}$ denote the number of times you pick $10$, etc, then the problem is:

\begin{align}
\max_{x} \quad & \sum_{i=10}^{20} i^2 x_i \\
\text{s.t.} \quad & \sum_{i=10}^{20} i x_i = 237 \\
& \sum_{i=10}^{20} x_i = 18 \\
& x_i \in \mathbb{Z}_+
\end{align}

I have added the GAMS model below, which you can solve on the NEOS server. The optimal solution is
$$a = (10,10,10,10,10,10,10,10,10,10,10,10,17,20,20,20,20,20)$$

with an objective value of 3489.

```<code>Set
   i 'idx'    / r10*r20 /;

Parameter
   a(i) / r10 10, r11 11, r12 12, r13 13, r14 14, r15 15, r16 16, r17 17, r18 18, r19 19, r20 20 /;

Integer   Variable x(i);
Variable z;

Equation objective, budget, eighteen_numbers;
objective.. z =e= sum(i, a(i)*a(i)*x(i));
budget.. sum(i, a(i)*x(i)) =e= 237;
eighteen_numbers.. sum(i, x(i)) =e= 18;

Model knapsack / all /;
solve knapsack max z using mip;
</code>```

Body:
An optimal solution cannot have $10 < a_i \leq a_j < 20$ for some $i$, $j$ because that contradicts optimality (you could increment $a_j$ and decrement $a_i$ and get a higher objective value). Therefore, there can be at most one element that is not $10$ or $20$.

There is therefore an optimal solution for which $a_1, \ldots, a_m$ take the value $10$, $a_{m+1},\ldots,a_{17}$ take the value $20$, and $a_{18} \in [10,20] \cap \mathbb{Z}$. From the equality constraint it follows that $a_{18} = 237 - 10m - 20(17-m) = 10m-103$. The only $m$ for which $10m-103 \in [10,20]$ is $m=12$, so the problem is solved.

If you had multiple $m$, you should evaluate the objective value ($100m + 400(17-m) + a_{18}^2 = 100m^2 - 2360m + 17409$) and pick the one for which the objective is the highest.

Other method This problem is also easy to solve with an integer programming solver after you formulate it appropriately. Let $x_{10}$ denote the number of times you pick $10$, etc, then the problem is:

\begin{align}
\max_{x} \quad & \sum_{i=10}^{20} i^2 x_i \\
\text{s.t.} \quad & \sum_{i=10}^{20} i x_i = 237 \\
& \sum_{i=10}^{20} x_i = 18 \\
& x_i \in \mathbb{Z}_+
\end{align}

I have added the GAMS model below, which you can solve on the NEOS server. The optimal solution is
$$a = (10,10,10,10,10,10,10,10,10,10,10,10,17,20,20,20,20,20)$$

with an objective value of 3489.

Set
   i 'idx'    / r10*r20 /;

Parameter
   a(i) / r10 10, r11 11, r12 12, r13 13, r14 14, r15 15, r16 16, r17 17, r18 18, r19 19, r20 20 /;

Integer   Variable x(i);
Variable z;

Equation objective, budget, eighteen_numbers;
objective.. z =e= sum(i, a(i)*a(i)*x(i));
budget.. sum(i, a(i)*x(i)) =e= 237;
eighteen_numbers.. sum(i, x(i)) =e= 18;

Model knapsack / all /;
solve knapsack max z using mip;


问:
Intersection of topologies: two topologies, finite number of topologies and topologies indexed by index set - why we need distinct proofs
I'm self learning topology by reading "Topology without tears" by Sidney A. Morris. There are three exercises related to intersection of topologies:

Let $\tau_1$ and $\tau_2$ be topologies on $X$. Prove that $\tau_1 \cap \tau_2$ is a topology on X

If $\tau_1, \tau_1, ..., \tau_n$ are topologies on $X$, prove $\bigcap_{i=1}^{n} \tau_{i}$ is a topology on X

If for each $i \in I$, for some index set $I$, each $\tau_i$ is a topology on the set $X$, prove $\bigcap_{i \in I} \tau_{i}$ is a topology on X

Why do we need three distinct proofs for 2 topologies, countable number of topologies and (possibly infinite number of?) topologies indexed by index set? Why we cannot naturally extend the proof from 1. to 2. and 3? Is there other similar construct in mathematics that is true for two operands but not when taking account more ?

答: There is clearly no actual requirement for three separate theorems, since the last implies the others and is easily proved without appealing to either of the others. The reason for having all three exercises is probably pædagogical: even though the arguments for (1) and (3) are essentially the same, many students find it easier to think about two topologies than to think about an arbitrary family of topologies. As for (2), it can of course be proved by the same argument, but it can also be proved by induction on $n$, using (1) to carry out the induction step, much as one proves that the intersection of any finite family of open sets is open.


问:
Prove $ \frac{\partial f}{\partial t}\left(tx\right)=\langle \nabla f\left(tx\right),x \rangle $
let $ f :\mathbb{R}^n \to \mathbb{R} $ and prove that for any $ t \in \mathbb{R} $ the following equality holds:

$$ \frac{\partial f}{\partial t}\left(tx\right)=\langle\nabla f\left(tx\right),x\rangle.$$

I'm not sure how to start because I don't know which of the following would be the right way to right the deriavtive:

$$ \frac{\partial f}{\partial t}\left(tx\right)=\frac{\partial f}{\partial x_{1}}\left(tx\right)x_{1}+\ldots+\frac{\partial f}{\partial x_{n}}\left(tx\right)x_{n} $$

or

$$ \frac{\partial f}{\partial t}\left(tx\right)=\frac{\partial f}{\partial\left(tx_{1}\right)}\left(tx\right)x_{1}+\ldots+\frac{\partial f}{\partial\left(tx_{n}\right)}\left(tx\right)x_{n} $$

答:
While understandable, the notation $\frac{\partial f}{\partial t}(tx)$ is actually quite "bad", and confusing.

Let me rephrase your problem a bit. Consider a differentiable function $f:\mathbb{R}^n\to\mathbb{R}$. Define the function $g:\mathbb{R}\times{\mathbb{R}}^n\to\mathbb{R}$ with $g(t,x)=f(tx)$. Find
$
\frac{\partial }{\partial t}g(t,x).
$

When applying the chain rule, the Leibniz notation of derivatives could be confusing, in particular without an explicit choice of the "intermediate variable".

It is revealing to look at the case when $n=1$.

By the chain rule, you have
$$
\frac{\partial }{\partial t}g(t,x)=f'(tx)\cdot x=\frac{df}{d\color{red}{z}}\bigg|_{\color{red}{z}=tx}\cdot x
$$
where we explicitly write the variable for the function $f$ as $z$.

In the higher dimension, you have
$$
\frac{\partial }{\partial t}g(t,x)=
\frac{\partial f}{\partial\color{red}{z_1}}\bigg|_{\color{red}{z}=tx}\cdot x_1
+\cdots
+\frac{\partial f}{\partial\color{red}{z_n}}\bigg|_{\color{red}{z}=tx}\cdot x_n
=
\langle
\nabla f(tx),x\rangle$$

Sometimes, people simply write

$$
\frac{\partial }{\partial t}g(t,x)=
\partial_1f(tx)\cdot x_1
+\cdots
+\partial_nf(tx)\cdot x_n
=
\langle
\nabla f(tx),x\rangle$$


问:
Evaluating $I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy$
I came across the following double integral in a statistics problem:

For $z>0$
$$
I(z,s)=\int_0^1\int_0^1\left(1-\frac{(1-x)(1-y)}{(1-(1-z)x)(1-(1-z)y)}\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
All I was interested in was the values of $s$ for which $I$ converges $(s>0)$. That said, the integral looked interesting enough to try and evaluate.  I was successful in evaluating the special case $z=1$ (see below) but could not solve for the general solution. Any ideas on how to solve it?

Case for $z=1$:

We have
$$
I(1,s)=\int_0^1\int_0^1\left(1-(1-x)(1-y)\right)^{s-2}\,\mathrm dx\mathrm dy.
$$
Substituting $(u,v)=(1-(1-x)(1-y),x)$, the resulting integral in $v$ is easily evaluated in terms of the natural logarithm; hence,
$$
I(1,s)=-\int_0^1 u^{s-2}\log(1-u)\,\mathrm du.
$$
Observe that $-u^{s-2}\log(1-u)\sim u^{s-1}+\mathcal O(u^s)$ as $u\nearrow 0$, which tells us the integral should converge for $s>0$. Using the series expansion for the logarithm we then integrate termwise to find
$$
I(1,s)=\sum_{k=1}^\infty\frac{1}{k(k+s-1)}=\frac{H_{s-1}}{s-1},
$$
where $H_{s-1}=\psi(s)+\gamma$ is the generalized harmonic number, $\psi(z)=\partial_z\log\Gamma(z)$ is the digamma function and $\gamma=0.577\dots$ is the Euler–Mascheroni constant.

答:
This is meant to supplement the solution derived by @jwimberley. The solution derived was (after some algebraic manipulations):
$$
I(z,s)=\sum_{k=0}^{s-2}\frac{(2-s)_k}{k!}\frac{F^2(1,k;k+2;1-z)}{(k+1)^2},
$$
which held for $z>0$ and $s\in\Bbb N$ with $s\geq 2$. I would like to show that this solution in fact holds for all $s\in\Bbb C: \Re s>0$. First note that the solution above is equivalent to
$$
I(z,s)=1+\sum_{k=1}^\infty\frac{(2-s)_k}{(k+1)^2k!}F^2(1,k;k+2;1-z),
$$
since the first term is always one and $s\in\Bbb N\land s\geq 2\implies (2-s)_k=0$ for $k=s-1,s,s+1,\dots$

To show convergence for other values of $s$ we will use the integral representation of the hypergeometric function (DLMF $15.6.1$) to write
$$
F\left({1,k \atop k+2};1-z\right)=\int_0^1\frac{1}{1-(1-z) t} \frac{t^{k-1}(1-t)}{\operatorname{B}(k,2)}\,\mathrm dt,\quad k\in\Bbb N
$$
which can be interpreted as an expected value w.r.t. the $\operatorname{Beta}(k,2)$ distribution, i.e. $\mathsf E[(1-(1-z) X)^{-1}]$ with $X\sim\operatorname{Beta}(k,2)$. By Jensen's inequality we then have
$$
F^2\left({1,k \atop k+2};1-z\right)\leq \mathsf E[(1-(1-z) X)^{-2}]=\int_0^1\frac{1}{(1-(1-z) t)^2} \frac{t^{k-1}(1-t)}{\operatorname{B}(k,2)}\,\mathrm dt.
$$
Now note for $z>0$ and $t\in[0,1]$ that
$$
\min\{1,z^{-2}\}\leq\frac{1}{(1-(1-z) t)^2}\leq\max\{1,z^{-2}\}.
$$
Consequently,
$$
F^2\left({1,k \atop k+2};1-z\right)\leq\max\{1,z^{-2}\}
$$
and
$$
\tag{1}
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\sum_{k=0}^\infty\frac{(2-s)_k}{(k+1)^2k!}-1\right).
$$
By the properties of the Pochhammer symbol we now write $k+1=(2)_k/(1)_k$. Substituting this result into our inequality for $I$ gives
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\sum_{k=0}^\infty\frac{(1)_k(1)_k(2-s)_k}{(2)_k(2)_kk!}-1\right),
$$
which according to DLMF $\S 16.2(\mathrm{iii})$ is absolutely convergent if $\Re s>0$. Furthermore, if we evaluate the series we can obtain an explicit expression for the upper bound on $I$. Upon inspection of $(1)$ we see that
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(I(1,s)-1\right).
$$

Hence, for $z>0$ and $s\in\Bbb C:\Re s>0$:
$$
I(z,s)=\sum_{k=0}^\infty\frac{(2-s)_k}{k!}\frac{F^2(1,k;k+2;1-z)}{(k+1)^2}
$$
and
$$
I(z,s)\leq1+\max\{1,z^{-2}\}\left(\frac{H_{s-1}}{s-1}-1\right).
$$


答: When you refer to "the line integral," you're talking of course about the work line integral $\displaystyle\int_C \vec F\cdot d\vec r$. So this amounts to integrating a scalar function. But in physics you'll also encounter a line integral like $\displaystyle\int_C \vec F\times d\vec r$ (e.g., Biot-Savart), and so you'll be integrating a vector function along the length of the curve.

Title: Does a Line Integral that sums to a vector exist ? And is it useful?

Body: We know that we can integrate a vector field on a curve to get a scalar sum: this is how the line integral is defined. Is there a use for another sort of integral whose sum is a vector ? I can't seem to find that defined anywhere.</p>


问:
Maximizing the size of a rectangle inside a rectangle
I'm an engineer who hasn't done calculus in a while, but I know there's a way to figure this out with a summation or a differential or something.

I need to make a rectangle (length of $x$, width of $y$) inside another rectangle (length of $a$, width of $b$) such that they both have the same resulting area $\left(xy=\dfrac{ab}{2}\right)$. Also, $0<x<a$ and $0<y<b$.

How do I figure this out?

答:
Lots of ways. Pick any number $r$ between $1/2$ and $1$. Let
$x = ar$ and $y = b/2r$. Then put the smaller rectangle anywhere inside the larger.

A natural way to do this is to use $r=\sqrt{2}/2$. Then the small rectangle will be the same shape as the large one.


问:
A function and continuity problem (multiplying by a constant)
I have this problem about continuity and functions the question is:

$f(x)=$\begin{cases}-1 & x>1\\
-2 & x≤1 \end{cases}

the question has 3 parts

Is there a constant 'a' for the function $f(a)*f(x)$ that will be continuous for $x=1$?

is there a constant 'b' for the function $(b+f(x))^2$ that will be continuous for every x?

is there a constant 'c' for the function $(1+f(x))*(f(x)+c)$ that will be continuous for every x?

my attempt for the first part

We know that $f(1)=-2$ , and for it to be continuous  we need $\lim \limits_{x \to 1^+ }f(x)$$=$$\lim \limits_{x \to 1^- }f(x)$$=$$f(1)$ when we apply what we have we will get $-1=-2=-2$ which means it is not continuous   (before trying anything with the $f(a)$)

so now I tried applying $f(a)$

$\lim \limits_{x \to 1^+ }f(x)*f(a)$$=$$\lim \limits_{x \to 1^- }f(x)*f(a)$$=$$f(a)*f(x)$

after applying what we have (applying what we have about $f(x)$ (-1) for the left side of the limit and (-2) for the right side)

$\lim \limits_{x \to 1^+ }-1*f(a)$$=$$\lim \limits_{x \to 1^- }-2*f(a)$$=$$f(a)*-2$

for the last step what I did was take the 2 cases we have , for $a>1$ we get $f(a)=-2$ and for $a<1$ we get $f(a)=-1$
so lastly we get $2\ne4=4$ for when applying $f(a)=-2$ and we get $1\ne2=2$ for $f(a)=-1$ so my conclusion was that there is no such a constant but I do not know if my way or approach is even right

For the second and third part I really don't know how to even begin with, so I am asking for help and tips on how to approach these type of question and not just an answer because I seem to have a lot of them the answers are not really important but I need advice.

thanks for any tips and help on how to approach these type of questions , and sorry for the English mistakes!

答:
Your conclusion for part $1$ is correct.  I think your work is also, but I haven't checked it closely.

I'll show you how to do the second part, and leave you to attempt the third part.

If the limit exists we must have $$\lim_{x\to1^-}(b+f(x))^2=\lim_{x\to1^+}(b+f(x))^2\\
(b-2)^2=(b-1)^2\\
b=\frac32$$ and it's easy to confirm that this works.


问:
What is the maximum of the product of three numbers whose sum is 3n?
Let $x,y,z$ be three integers and their sum is $3n,$ where $n$ is a constant integer. How to find the maximum of the value $xyz$?

I have gotten through that let $$f(x,y)=xy(3n-x-y),$$

what would be the next step to find the maximum calculus-ly?

Thanks!

答:
By the well-known inequality
$$
\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3} = n
$$
and the equality is obtained iff $x = y = z$. So the maximum is $n^3$. But if they are not positive, then $xyz$ is unbounded even when the sum $x+y+z = 3n$ is fixed. Just consider the case $x = y < 0$ so $xyz = x^2z = x^2(3n-x)$.


问:
a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3
A random variable has the density  $f\left(x\right)=a+bx^{2}$ with  $0<x<1$. Determine and b so that its mean will be 2/3.

I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\right)=\int_{0}^{1}ax^{2}dx+\int_{0}^{1}bx^{4}dx$$
$$P\left(0<X<1\right)=a\int_{0}^{1}x^{2}dx+b\int_{0}^{1}x^{4}dx=a\left(\frac{x^{3}}{3}\right)_{0}^{1}+b\left(\frac{x^{5}}{5}\right)_{0}^{1}$$
$$P\left(0<X<1\right)=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=\frac{2}{3}=\frac{5a+3b}{15}$$
$$10=5a+3b$$
I'm assuming x as a continuous random variable
In this case shouldn't be a=0 b=1?

答:
You are mixing up the $P(X\in(0,1))=1$ and $E(X)=\frac 2 3$. The other post presumably showed you the expected value correctly, so $P(X\in(0,1))=\int_{-1}^1 f(x) dx=\int_0^1(a+bx^2)dx=\left[ax+\frac{bx^3}{3}\right]_{x=0}^1=a+\frac b 3=1$

Solving, you obtain $a=\frac 1 3, b=2$.

To check:

$E(X)=\int_{-\infty}^\infty xf(x)dx=\int_0^1x(\frac 1 3 + 2x^2)dx=\left[\frac{x^2}{6}+\frac{x^4}{2}\right]_{x=0}^1=\frac 2 3$


问:
Proving $\int^c_0 xf(x)\,dx=(c-1)\int^c_0 f(x)\,dx$

Let $f\colon [0,1]\to\mathbb{R}$ be a continuous function such that $$\int^1_0 xf(x)\,dx = \int^1_0 f(x)\,dx.$$
Prove that exists a $c\in(0,1)$ such that
$$\int^c_0 xf(x)\,dx=(c-1)\int^c_0 f(x)\,dx.$$

I tried to apply the Mean Value Theorem but I did no get to any point.

答:
Let $$F : x \mapsto \int_0^x f(t)dt$$

By hypothesis, one has
$$F(1)=\int_0^1 f(x) \,dx= \int_0^1 xf(x) \,dx= \int_0^1 \int_0^x f(x) \,dt \,dx = \int_0^1 \int_t^1 f(x) \,dx \,dt$$ $$= \int_0^1 F(1)-F(t) \,dt = F(1)-\int_0^1F(t)\,dt$$

(you can also make this calculation with an integration by parts on $\int_0^1 xf(x) \,dx$).

So you deduce that
$$\int_0^1 F(t) \,dt = 0$$

Now let $$G : x \mapsto \int_0^x F(t) \,dt, \quad \quad \text{and }\quad h : x \mapsto e^{-x}G(x)$$

Because $G(0)=G(1)=0$, then $h(0)=h(1)=0$, so by Rolle's theorem, there exists $c \in (0,1)$ such that $h'(c)=0$. And one has
$$h'(x)=e^{-x}\left(G'(x)-G(x) \right)$$

so you get $c \in (0,1)$ such that $G'(c)=G(c)$, i.e. such that
$$F(c)=\int_0^c F(t) \,dt$$

You deduce that
$$\int_0^c xf(x) \,dx = \int_0^c \int_0^x f(x) \,dt \,dx = \int_0^c \int_t^c f(x) \,dx \,dt = \int_0^c F(c)-F(t) \,dt =cF(c)-F(c)$$

i.e. that
$$\int_0^c xf(x) \,dx = (c-1)\int_0^c f(x)\,dx$$


问:
Prove that $2<\int^2_1ln(x^2+1)+x<3$
To show that $2<\int^2_1ln(x^2+1)+x$ , I took the derivative of the integrand and got that it's positive always in $[1,2]$, so its strictly increasing and the minimum in $[1,2]$ is at $x=1$, so $ln(2)+1<ln(x^2+1)+x$, and $ln(2)<ln(e)=1$ so: $ln(2)+1<2< ln(x^2+1)+x$ in $(1,2]$.

EDIT: I just realized that the lower bound doesn't work too.

About the upper bound I tried to substitute $2$ but it becomes $ln(5)+2>3$, so it doesn't really help.

I would appreciate any help.
 Thanks in advance.

答:
for $x\in [1.2] $. We have $ x-1< \ln(x^2+1)< x$

So :$\displaystyle \int_1^2(x-1)dx=\frac{1}{2}<\displaystyle\int_1^2\ln(x^2+1)dx<\displaystyle\int_1^2 xdx=\frac{3}{2}$

$\Rightarrow \frac{1}{2}+\displaystyle\int_1^2 xdx =2< \displaystyle\int_1^2 \ln(x^2+1)+x dx < \frac{3}{2}+\displaystyle\int_1^2 xdx =3$

Title: Prove that $2<\int^2_1ln(x^2+1)+x<3$


问:
Convergence of the expectation of a bounded random variable
I'm trying to prove the following statement. Let $(X_n)$ be a sequence of random variables such that $\lvert X_n \rvert \leq C$ almost surely. If $X_n$ converges in probability to 0, then $\mathbb{E}[\lvert X_n \rvert]$ converges to 0.

The converse is straight forward by Markov theorem. But I don't know how to start for this one. I know that I only need to prove that $\forall \epsilon > 0, \exists N \in \mathbb{N}, \forall n \geq N, \mathbb{E}[\lvert X_n \rvert] < \epsilon$.

答:
Let $\varepsilon>0$. Then
$$
\mathbb E[\vert X_n\vert]=\mathbb E[\vert X_n\vert1_{\{\vert X_n\vert\ge\varepsilon\}}]+\mathbb E[\vert X_n\vert1_{\{\vert X_n\vert<\varepsilon\}}]\le C\mathbb P(\vert X_n\vert\ge\varepsilon)+\varepsilon,
$$
hence $\limsup_{n\to+\infty}\mathbb E[\vert X_n\vert]\le\varepsilon$. For $\varepsilon\to0$ we deduce that $\lim\sup_{n\to+\infty}\mathbb E[\vert X_n\vert]=0$, or equivalently $\mathbb E[\vert X_n\vert]$ converges to $0$ as $n\to+\infty$.


问:
Distribution of the minimum of discrete Uniform R.V.s

What is the cumulative distribution function of the minimum of $n$ discrete uniform i.i.d  R.V.s?
In particular when $X_i\sim\text{Unif}\,\{1,6\}$.

For $n=2$ the problem becomes quite simple as listing all possible choices is not too  bad. This becomes fairly hard when $n$ is large. Are there any closed forms?

答:
Let's denote the cdf by $F_n$ when there are $n$ dice.  Then

$F_n(k) = P(\min(X_1, X_2, \ldots, X_n) \le k)$

Now for the minimum to be at most $k$ is a complicated event, but its complement is simpler.  So we'll rewrite this as

$F_n(k) = 1 - P(\min(X_1, X_2, \ldots, X_n) > k).$

For the minimum to be greater than $k$, all of the $X_i$ have to be greater than $k$, and then you can use the independence and identical distribution of the $X_i$.  You can get  a closed form in  terms of $k$ and $n$, and check it against the results you can get by listing out all cases when $n = 2$.


答:
Assume temporarily $x,y$ real. The function has a stationary point where

$$y(3n-x-y)-xy=x(3n-x-y)-xy=0,$$ or

$$x+y=3n\lor x=y=n\lor x=y=0.$$

The corresponding values of the function are

$$0,n^3,0.$$

Title: What is the maximum of the product of three numbers whose sum is 3n?

Body:
Let $x,y,z$ be three integers and their sum is $3n,$ where $n$ is a constant integer. How to find the maximum of the value $xyz$?

I have gotten through that let $$f(x,y)=xy(3n-x-y),$$

what would be the next step to find the maximum calculus-ly?

Thanks!


问:
Does a Line Integral that sums to a vector exist ? And is it useful?
We know that we can integrate a vector field on a curve to get a scalar sum: this is how the line integral is defined. Is there a use for another sort of integral whose sum is a vector ? I can't seem to find that defined anywhere.</p>

答:
If you have a (scalor-valued) function $f:[a,b]\to\mathbb{R}$ with "nice" properties, you can define the Riemann integral:
$\int_a^b f(x)dx$, the result of which is a scalar.

If you have a vector-valued function $f:[a,b]\to\mathbb{R}^n$, where each component  of $f=(f_1,\cdots,f_n)$ is a scalar (integrable) function, you have
$$
\int_a^b f(x)dx:=(\int_a^b f_1(x)dx,\cdots,\int_a^b f_n(x)dx) \tag{1}
$$
the result of which is a vector.

More generally, if you have a vector-valued function $f:X\to\mathbb{R}^n$ where $(X,\mu)$ is some measure space,  then you have the analog of (1) as
$$
\int_Xfd\mu:=(\int_Xf_1d\mu,\cdots,\int_Xf_nd\mu)
$$

Even more generally, if you replace $\mathbb{R}^n$ with some Banach space, you have the notion of Bochner integrals.


问:
Is it possible to find the closed form for this series $\frac{(-1)^{2n}x^{4n}}{[(2n)!]^2}$
In this thread, I made a basic mistake of expanding series for $\cos^2(x)$:

$$\cos^2(x)\neq \color{red}{\sum_{n=1}^{+\infty} \left(\frac{(-1)^nx^{2n}}{(2n!)}\right)^2=\sum_{n=1}^{+\infty} \dfrac{(-1)^{2n}x^{4n}}{[(2n)!]^2}}$$

I wonder what is the function to which the above series converge if it is not $\cos^{2}(x)$. Is it possible to find its closed form?

Thanks!

答:
This is
$$
\frac{J_0(2x)+I_0(2x)}2
$$
where $J_0(x)$ and $I_0(x)$ are Bessel and modified Bessel functions of the first kind, respectively.


问:
Product of fibrations is a fibration in a fibration category
Let $\mathcal{C}$ a category with a class of fibrations and weak equivalence such that

The category $\mathcal{C}$ has a terminal object $1$ and all maps $X\to 1$ are fibrations

Pullbacks of fibrations exist and are fibrations again

Pullbacks of acyclic fibrations are acyclic fibrations

Weak equivalences satisfy 2-out-of-3

Fibrations are closed under composition

Every map $f:X\to Y$ factorizes as a weak equivalence followed by a fibration

Isomorphisms are acyclic fibrations

I am trying to show that given to maps $f_1:X_1\to Y_1$ and $f_2:X_2\to Y_2$ are fibrations, the product $f_1\times f_2:X_1\times X_2 \to Y_1\times Y_2$ is a fibration. It suffices to show this for the product of a fibration $f:X\to Y$ with the identity $1_Z:Z\to Z$ for some $Z$. I have tried a few things but nothing worked out yet, it shouldn't be that hard I think, but I can't get it to work.

Any help is welcome!

答:
Let $p:E\rightarrow B$ be a fibration. Given any object $X$ form the product $X\times B$ as the pullback of the cospan $X\rightarrow \ast\leftarrow B$. Observe that the following square is a pullback
\begin{array}{ccc}
X\times E &\xrightarrow{pr_2} & E \\
\ {1_X\times p}\downarrow & &\ \downarrow p\\
X\times B &\xrightarrow{pr_2} &B.
\end{array}
From the axioms we get that $1_X\times p:X\times E\rightarrow X\times B$ is a fibration. Similar reasoning shows that $p\times 1_X:E\times X\rightarrow B\times X$ is a fibration.

Now let $p_i:E_i\rightarrow B_i$, $i=1,2$, be fibrations and factor $p_1\times p_2$ as
$$p_1\times p_2:E_1\times E_2\xrightarrow{1_{E_1}\times p_2}E_1\times B_2\xrightarrow{p_1\times 1_{B_2}}B_1\times B_2.$$
Since a composition of fibrations is a fibration we get that $p_1\times p_2$ is a fibration.


问:
How to solve $x^2 - ny - 2 = 0$ for integer solutions, with different coefficients of $y$?
$$x^2 - ny - 2 = 0, n \in Z$$

I put this equation in wolframalpha, and got integer solutions for random n = 7, 23, 31, 47, 49, 71, 73, 343.

for n = 7 it gave the integer solutions as $$x = 7 n + 3, y = 7 n^2 + 6 n + 1, n \in Z$$
$$x = 7 n + 4, y = 7 n^2 + 8 n + 2, n \in Z$$

and for n = 23 it gave the integer solutions as $$x = 23 n + 5, y = 23 n^2 + 10 n + 1, n \in Z $$
$$x = 23 n + 18, y = 23 n^2 + 36 n + 14, n \in Z $$

It was noteworthy that the sum of free numbers in x is always equal to n (i.e. coefficient of y). 7 had 3 and 4 as free-x-values, 23 had 5 and 18 as free-x-values, and so on. So, as we can see, those free-x-values are always smaller than n.

For some values (like even numbers) there weren't any solutions shown, So I guess there weren't any. But as most of these have only 2 integer solutions, some had 4 integer solutions, like 511 for an example, which had 2 pairs of free-x-values, adding  to 511.

My question is, how do I determine if the equation has only exactly 2 solutions, or more than 2 solutions, for a given value of n (i.e. coefficient of y)?

I'm not particularly concerned with the solutions themselves, just that the number of solutions is just 2 or more than that. I can't try values one by one, as the coefficient of y can get large. Wolfram isn't (understandably) handling that big numbers, so I'm trying to write code to solve these equations, but as I don't know what is going on behind the scenes, I can't do anything.

Also, let's say if I find the 1st (smallest) solution, and the 4th (as their sum is predetermined). Can I use them to find the other two (2nd and 3rd) solutions, if they exist?

I looked into Diophantine equations of degree two, but the examples I found had all variables of second degree in them. So, I'm not sure if it qualifies for that. If if still does, please let me know.
I tried to solve the equation as a quadratic equation, but got stumped as the coefficient of $y^2$ is 0.

My mathematical background is only upto basic highschool level. So, if I need to study some field(s), to be able to solve this, please guide me, as I'm willing to study to solve this problem.

答:
Well, if you would have $n|x^2-2$, then putting $y = \frac{x^2-2}{2}$ gives us a legitimate solution. So we only need to be worried about solutions to that divisibility.

You can easily see, that if $x$ is a solution, then $n-x$ and $n+x$ are also solutions. That yields your observation with summing up to $n$.

Now we will focus on the set $S$ of such $n$, that some solution exists.

If $k$ and $l$ are coprime numbers in $S$, then using Chinese Remainder Theorem we can find solution for $kl$. Moreover, the number of solutions for $kl$ will be equal to the product of these numbers for $k$ and $l$. We also see quite obviously that if $n \in S$, then all divisors of $n$ are in $S$.

From these two facts, we only need to be worried about $n$ equal to prime powers. It is not very easy to prove that $n=p^k$ works iff $n=2$ or $p \equiv 1,7 \pmod 8$, but you should be able to find proof of this fact in any materials about quadratic residues (here, for example, $2$ is a quadratic residue mod $n$, since $x^2 \equiv 2 \pmod n$).

So, solutions exist for $n$ iff all of its prime divisors are $2$ or primes $\equiv 1$ or $7 \pmod 8$, where $2$ can show up only once (so $n$ can't be divisible by $4$, but it can be even). Number of these solutions (or, to be exact, number of different families of sulutions modulo $n$) is equal to $2^k$, where $k$ is number of distinct odd prime divisors of $n$.


问:
Identify this graph problem.
[Edit 01/02/2021] : Now the problem has been identified, as a relaxation of "minimal spanning tree" (it is more a minimal spanning sub-graph). Are they some references I could use on this problem ?

Let consider a graph made of $N$ nodes, numeroted from $B_1$ to $B_N$.

Each nodes can or cannot be connected to the other nodes, but overall we consider a family of weighed edges : $(\nu_{ij})_{1 \leq i < j \leq N} \in \mathbb{R}^{\frac{N(N-1)}{2}}$.

The edges must be understood as followed :

$\nu_{ij}=0$ if $B_i$ and $B_j$ are not linked by an edge.

there is no edge connecting one node to himself.

I want to operate with a specific way on this graph (I want to select some specific edges), and I believe I am looking for a family of numbers $(x_{ij})_{1 \leq i < j \leq N}$ with $x_{ij}=0$ or $1$ that verifies the following discrete optimization problem :

$$\underset{(x_{ij})_{1 \leq i < j \leq N} \in (0,1)^{\frac{N(N-1)}{2}}}{\min} \sum_{i<j}\  \nu_{ij} \  x_{ij}$$
with the following constraints on the $(x_{ij})$:

$\sum_{i<j} x_{ij} = N-1$ (i.e we are only gonna keep $N-1$ edges in the graph)

$\forall i \in \{1,\dots,N\}$, we have $$\sum_{j=i+1}^N x_{ij} + \sum_{j=1}^{i-1} x_{ji}  \geq 1$$ (i.e every nodes have at least one "non-zero" edge)

My questions

Is this problem known in operation research or discrete optimization or is it closed to a known problema with a theoritical analysis ? I don't know a lot of them, except for the traveling salesman problem or the cover set problem but I am quite unfamiliar with them.

Separately from this first question, I believe (heuristically, at least) that any family $(x_{ij})_{1 \leq i < j  \leq N}$ admissible for this problem verifies the following property :

the final graph with value $(x_{ij} \nu_{ij})$ on the edge is made of cycles and of only one path

But i am not sure how to prove it. I'have been looking for counter-examples using $4/5$ nodes and cycle but couldn't find any...
Any help, references or remarks are welcomed on the problem :).

答:
(I assume here that your edges' costs are in $\mathbb R_+^* \cup \{+\infty\}$)

First, you can remove the $\Sigma x_{ij} = N+1$ constraint.
If you have a minimum-cost solution, it will be a forest covering all the nodes (since if there was a cycle, you could remove its biggest edge, and the "non-zero degree" constraint will still be preserved at a better cost).
Given this minimum-cost solution, if you want to add edges to add up to $N+1$ edges, you can use the smallest edges that aren't used yet.

So now you just need a minimum-size edge cover of the vertices, and this has a polynomial solution (see https://en.wikipedia.org/wiki/Edge_cover)
EDIT : The Wikipedia page only cites the problem with constant-valued edges. A polynomial solution also exists for the real-valued one : see [Plesník, Ján, Equivalence between the minimum covering problem and the maximum matching problem, Discrete Math. 49, 315-317 (1984). ZBL0581.05045.]

Concerning the shape of the solution, it turns out that, in this minimum-cost forest, you won't have a 3-length path. If you have a chain $u - v - w - x$ then the same forest without the $(u,v)$ edge still works for the constraints and has a lower cost. So the solution is a collection of trees whose diameter is 1 or 2 : those are stars (https://en.wikipedia.org/wiki/Star_(graph_theory)).
But then, you just fill it with the smallest edges you get, so you could have any graph shape.

Title: Identify this graph problem.


问:
Unifying the connections between the trigonometric and hyperbolic functions
There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.

The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.

Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.

The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}

In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.

Analytic connections

$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.

Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.

It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.

If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}

All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}

And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$

答:
The Jacobi elliptic functions,
in particular the sine and cosine of the amplitude are also connected to the
hyperbolic trigonometric functions using the
Gudermannian function
which relates the circular functions and hyperbolic functions without
explicitly using complex numbers.
More precisely, if the modulus $\,m=0\,$ then
$$ \text{sn}(x,0) = \sin(x)\quad \text{ and }
\quad \text{cn}(x,0) = \cos(x) $$ while if $\,m=1\,$ then
$$ \text{sn}(x,1) = \tanh(x)\quad \text{ and }
\quad \text{cn}(x,1) = \text{sech}(x). $$
The Gudermannian function links the two cases with the identity
$$ \sin(\text{gd}(x)) = \tanh(x) \quad \text{ and }
\quad \tan(\text{gd}(x)) = \sinh(x) $$
and the related identity
$$ \cos(\text{gd}(x)) = \text{sech}(x) \quad \text{ and }
\quad \sec(\text{gd}(x)) = \cosh(x). $$
In some sense, the Jacobi functions generalize the circular
and hyperbolic trigonometric functions. Note the fundamental identity
$$ \text{sn}(x,m)^2 + \text{cn}(x,m)^2 = 1. $$
Whem $\,m=0\,$ this becomes $$ \sin(x)^2+\cos(x)^2=1 $$ and
when $\,m=1\,$ this becomes $$ \tanh(x)^2+\text{sech}(x)^2=1 $$
which is equivalent to $$\cosh(x)^2-\sinh(x)^2=1. $$


问:
$\mu (A \bigtriangleup B ) = \mu(B) - \mu(A) + 2\mu(A \setminus B)$
How can I prove it?:
$\mu (A \bigtriangleup B ) = \mu(B) - \mu(A) + 2\mu(A \setminus B)$

I started with: $\mu (A \bigtriangleup B )=\mu((A \cup B)\setminus B) + \mu ((A \cup B) \setminus A) = 2\mu (A \cup B) - \mu(B) - \mu(A)=... $

Does anyone have an idea how to finish it?

答:
Hint.

Use the axioms of measures. Also,
$$
A\Delta B=(A\setminus B)\cup (B\setminus A)
$$
is a disjoint union, and you have the disjoint unions:
$$
(B\setminus A)\cup A=B\cup(A\setminus B)
$$


问:
Why is the angle exactly 1/2 of one of the angles in the isosceles triangle
I'm sure there's a simple theorem that answers my question, but I'm unsure as to how to formulate the problem (not easy to describe geometry with words): It's a isosceles triangle where you draw a normal line on one of the "hypotenuse" sides, and then you make a right angled triangle out of that. My question is why is the 30 degree angle exactly 1/2 of the 60 degree angles we find in the isosceles triangle?

答:
Because if you have a right triangle, with acute angles $\alpha,\beta$, then $(\alpha + \beta) = 90^\circ$.

When $\alpha = 60^\circ = \frac{2}{3} \times 90^\circ,$

then $\beta = 90^\circ - 60^\circ = 90^\circ\left(1 - \frac{2}{3}\right)$

$= 90^\circ\left(\frac{1}{3}\right) = \left(\frac{1}{2}\right) \times \alpha.$


问:
Solution for Summation of $\cos^2x$
Can you give me the solution for the summation
$$
\sum_{n=0}^{\infty} \cos^2(\pi n)
$$

Edit: Please give me the explanation of how it is calculated and also final answer in integers.

答:
Have you noticed out the terms behaved? $\cos(\pi n) = (-1)^n$, so $\cos^2(\pi n) = 1$. This series is just summing a bunch of ones, so it diverges.

Hope that helps,

Title: Solution for Summation of $\cos^2x$