答:
If $O$ is the center of the coin, the locus of $O$ inside a single $8$-$15$-$17$ right triangle $\triangle ABC$ such that the coin does not intersect any of the sides, is another triangle, say $\triangle A'B'C'$, that is similar in shape.  Moreover, $\triangle A'B'C'$ shares the same incenter as $\triangle ABC$ because corresponding pairs of vertices, e.g. $A$ and $A'$, lie on the same angle bisectors.

Note that $$|\triangle ABC| = rs$$ where $r$ is the inradius and $s = (a+b+c)/2$ is the semiperimeter.  Similarly, $$|\triangle A'B'C'| = r's'$$ where $r' = r-1$ is the inradius of the smaller triangle.  And because of similarity, $s'/s = r'/r$; consequently $$\frac{|\triangle A'B'C'|}{|\triangle ABC|} = \frac{r's'}{rs} = \left(\frac{r'}{r}\right)^2 = \left(1 - \frac{1}{r}\right)^2.$$  Since $|\triangle ABC| = (8)(15)/2 = 60$ and $s = (8+15+17)/2 = 20$, we have $r = 3$ and the desired probability is $$(1 - 1/3)^2 = \frac{4}{9}.$$

Title: Coin that lands in a plane tessellated by triangles..

Body:
A coin of radius 1 cm is tossed onto a plane surface that has been tessellated by right triangles whose sides are 8 cm, 15 cm, and 17 cm long. Find the probability that the coin lands within a triangle.

I know that this has to do with similarity because the inner triangle that is formed by the area where the coin can land is similar to the outer triangle. Therefore, I know the angles of this triangle, but I am not sure how to find one side of this triangle.


问:
Proof on showing $\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f \leq (b-a) f(\frac{a+b}{2})$ for class $C^2$ function $f$
The task is as follows:

Given:

(a) function $f \in C^2$

(b) $f \geq 0$ and  (c) $f'' \leq 0$ on $[a,b]$

Goal:

Show

$$\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f \leq (b-a) f(\frac{a+b}{2})$$

To get an understanding of the problem, I tried specific function $f(x) = \sqrt x$ on interval $[ 1, 4 ]$

(1) For the first area (triangle with base $b-a$):

$\frac{(b-a)}{2}(f(a) + f(b))$ = $\frac{(4-1)}{2}(f(1) + f(4))$ = $\frac{9}{2}$

(2) For the second area (integral):

$\int_1^4 \sqrt x$ = $\frac{14}{3}$

I also tried adding up areas of sub-rectangles for this one, using right rectangles.

(3) For the third area = rectangle with length $b-a$ and width $f(\frac{a+b}{2})$ = $4.7$ (approximately)

So the conclusion clearly holds for this specific case.

But I have issue on how to generalize my example >_<

Well, by given information, I break function $f$ into 3 cases:

Case 1: If $f = 0$ i.e: zero function

Then there is nothing to prove, since area is always 0

Case 2: If $f = c$ i.e: constant function

Then the proof is quite easy, since all the 3 areas "shrink" down to be the area of the "big rectangle" with base $b-a$ and width $c$

Case 3: $f$ is convex or concave

This is the part that I don't know how to generalize what I found from my example.

My thoughts:

When I do the first area, I'm dealing with a triangle, thus I'm going below (or exactly on) the function $f$

When I do the second area, I'm thinking about the upper Darboux sum. Thus the sub-rectangles exceed the original curve by some little fractional area, namely the upper left of the rectangles

When I do the third area, I'm also exceed the original curve by some fractional area, but I think this extra part is a bit more than the fractional areas formed by the sub-rectangles.  Or thinking another way, if I double up this rectangle, I get an area which is way bigger than the other two areas.

But then... how should I generalize all these ideas ?

Would someone please help me on this question?

Thank you in advance ^^

答:
You already know that function is concave down ($f'' \leq 0$), therefore area of trapezoid is less than the  integral.

$$\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f$$

Now the second part:
You can write the integral as:
$$\int _{ a }^{ b }{ f } =\int _{ a }^{ (a+b)/2 }{ f } +\int _{ (a+b)/2 }^{ b }{ f }$$

The function is increasing ($f'\ge0$), therefore
$$ \int _{ (a+b)/2 }^{ b }{ f } \ge \int _{ a }^{ (a+b)/2 }{ f } $$

In the first middle the rectangle has bigger area:
$$ \\ \int _{ a }^{ (a+b)/2 }{ f } \le \left( \frac { a+b }{ 2 } -a \right) f\left( \frac { a+b }{ 2 }  \right) $$
As a result:
$$\\ 2\int _{ a }^{ (a+b)/2 }{ f } \le \int _{ a }^{ b }{ f } \le \left( b-a \right) f\left( \frac { a+b }{ 2 }  \right) $$

ContentLicense: CC BY-SA 3.0


问:
meaning of notation of the set [a,b]×[c,d]?
I am trying to understand what the set  [a,b]×[c,d] means I tried searching it on google but there is nothing I can find.can someone please explain it to me using this example  A = ([−1, 1] \ {0}) × [−1, 1].</p>

答:
$[a,b]\times [c,d]$ means all of the points $(x,y) \in \mathbb{R}^2 $ such that $a \leq x\leq b$ and $c\leq y \leq d$.
Using your example, it would be the set $\left\{ (x,y) \in \mathbb{R}^2 | -1\leq x \leq 1, -1\leq y \leq 1 , x\neq 0\right\}$.

ContentLicense: CC BY-SA 4.0


答:
If $A$ and $B$ are sets, then the Cartesian product of $A$ and $B$ is defined as,

$$A \times B:=  \{ (a,b): a \in A \text{ and } b \in B \}.$$

Therefore,

$$[a,b] \times [c,d] = \{\text{ pairs } (x,y): x \in [a,b] \text{ and } y \in [c,d] \}.$$

Title: meaning of notation of the set [a,b]×[c,d]?

Body: I am trying to understand what the set  [a,b]×[c,d] means I tried searching it on google but there is nothing I can find.can someone please explain it to me using this example  A = ([−1, 1] \ {0}) × [−1, 1].</p>


问:
a semi-hard problem on combinatory
I ran into a nice interview question.

the problem is as follows:

We have array of $n$ integers. for $1 \leq i \leq j \leq n$. we want to set $c_{ij}$= Sum of all values in the range $i$ to $j$ of this array. we want to finding average of all possible $c_{ij}$ in this array. if 4 basic operation in math $(+,-,*,/)$ can be done in $O(1)$. the best algorithm can be works in $O(n)$.

And this is the solution here:

Suppose that some entry of the array has value $k$. What is its contribution to the average of the $c_{ij}$? It is counted in some $c_{ij}$ iff $i \leq k \leq j$. Out of the $\binom{n+1}{2}$ many $c_{ij}$'s, there are exactly $k(n+1-k)$ in which the entry is counted, therefore its contribution to the average is
$$
\frac{k(n+1-k)}{\binom{n+1}{2}} \cdot k = \frac{2k^2(n+1-k)}{(n+1)n}.
$$
Summing this over all values in the array, we obtain the desired $O(n)$ algorithm.

Anyone can describe me very easily how this formula is archived?  small example or any intuitive and simple idea?

答:
In order for the $k$ entry to be counted in $c_{ij}$, we must have $1\le i\le k$ and $k\le j\le n$. There are $k$ integers in the set $\{1,2,\ldots,k\}$ and $n-(k-1)=n+1-k$ integers in the set $\{k,k+1,\ldots,n\}$, so there are $k(n+1-k)$ pairs $\langle i,j\rangle$ such that the $k$ entry is counted in $c_{ij}$. It contributes $k$ to each of those sums $c_{ij}$, so altogether it contributes

$$k\cdot k(n+1-k)=k^2(n+1-k)$$

to the sum of all of the partial sums $c_{ij}$. Dividing this by the number of sums $c_{ij}$ gives us the contribution of $k$ entries to the sum of the partial sums $c_{ij}$.

There are $\binom{n}2$ partial sums $c_{ij}$ with $1\le i<j\le n$, and there are $n$ partial sums $c_{ii}$ with $1\le i\le n$, so there are altogether

$$\binom{n}2+n=\frac{n(n-1)}2+n=\frac{n(n+1)}2=\binom{n+1}2$$

partial sums $c_{ij}$.

Added example: Suppose that $n=4$, and the array is $\langle 3,2,4,1\rangle$. The $3$ contributes to $c_{ij}$ when $i\le 3\le j$, so the possible choices for $i$ are $1,2$, and $3$, and the possible choices for $j$ are $3$ and $4$. Thus, there are $k=3$ choices for $i$ and $4-(3-1)=2$ choices for $j$, for a total of $3\cdot 2=6$ possible pairs of $i$ and $j$: the $3$ contributes to $c_{13},c_{14},c_{23},c_{24},c_{33}$, and $c_{34}$. The $4$, on the other hand, contributes to $c_{14},c_{24},c_{34}$, and $c_{44}$; there are $4\cdot1=4$ of these, because there are $4$ choices for $i$ ($1,2,3$, and $4$) and only one choice for $j$ ($4$).

Title: a semi-hard problem on combinatory


问:
How many solutions are possible to this equation?
Given
$$A+2B+3C=N
$$
where $N$ is a given positive integer.

$A ,B,C\in\mathbb{N}$ vary from $0$ to $\infty$.

How many solutions will be there to this equation?

答: The result will be equal to the coeeficient of $x^{n}$ in $(x+x^{2}+x^{3}+\dots)(x^{2}+x^{4}+\dots)(x^{3}+x^{6}+\dots)$</p>


问:
Second order shape funtions on first order element
In a Finite Element context, say I have a 1D finite element with 2 nodes, $x=0$ and $x=1$. The typically used shape funcions are

$N_1(x) = 1 - x$

$N_2(x) = x$

Would it make sense to use second order polynomials on that same element (without introducing an extra node at $x=0.5$) as described below?

$N_1(x) = 1 - 0.5x - 0.5x^2$

$N_2(x) = 0.5x + 0.5x^2$

Should I expect better accuracy? Is there something wrong in using those shape functions?

答:
If $A_1(x)=1−0.5x−0.5x^2$ and $A_2(x)=0.5x+0.5x^2$. The interpolation fails with $u(x)=\sin(x)$ since the error estimation $$\|u(x)−u(0)A_1(x)−u(1)A_2(x)\|_{\infty}≈0.1644.$$
The linear interpolation $L_1(x)=1-x$ and $L_2(x)=x$ yields $$\|u(x)−u(0)L_1(x)−u(1)L_2(x)\|_{\infty} \approx 0.06$$

The usual quadratic interpolation $N_1(x)=2(x-1)(x-1/2)$, $N_2(x)=-4(x-1)x$, $N_3(x)=2x(x-1/2)$ gives,
$$\|u(x)−u(0)N1(x)−u(1/2)N2−u(1)N3\|_{\infty}≈0.0072$$ Even doesn't interpolate well $u(x)=x^2$ where the usual quadratic interpolation is exact.


问:
You roll a fair, six-sided die as part of a game. If you roll a 5, you will win the game
e. Your friend will pay you $4$ dollars if you win the game. You owe your friend $1$ dollar
if you lose the game. What are your expected winnings?

f. What is the variance of your expected winnings?

My work:

For e) I know that $E(X) = p = \cfrac{1}{6}$
Using this information, I can calculate the expected winnings $E(X) = 4 \cdot \cfrac{1}{6} + -1 \cdot \cfrac{5}{6} = \cfrac{-1}{6}$ is the expected winnings

f) I need help with this one. I know for a Bernoulli random variable, $Var(X) = pq = \cfrac{1}{6} \cdot \cfrac{5}{6} = \cfrac{5}{36}$ But how can I use this information to calculate the variance of the expected winnings?

Can someone give me a clue?

Thank You!

答: Calculate $E(X^2)$ as $4^2\cdot\frac16+(-1)^2\cdot\frac56=\frac{21}6$. Then the variance is $E(X^2)-E(X)^2=\frac{21}6-\left(-\frac16\right)^2=\frac{125}{36}$.


答:
Runso answered perfectly.

I just want to add, if you graphed in on an $xy$ plane it would be all the points inside a rectangle (centered at the origin and extending for $(-1,1)$ in the upper left to $(1,-1)$ in the lower right), but with the line $x=0$ (a vertical line contained in the $y$ axis) removed from the rectangle.


问:
Computing Expected Value using the Double Expectation Formula

A fair die is rolled repeatedly.  Let $X$ be the number of rolls needed to obtain a 5 and $Y$ the number of rolls needed to obtain a 6.  Calculate $E[X \mid Y=2]$.

This problem has been discussed here and here, but none of the solutions in those posts use the double expectation formula. I want to solve the problem above using the Double Expectation formula because I think that will be more efficient. If nothing else, I'll gt more practice with the formula. The formula is $E[Y] = E_X[E_Y[Y\mid X]]$ for jointly distributed random variables $X$ and $Y$. I think both $X$ and $Y$ follow the geometric distribution with mean $1/6$. So, we can write

$P(X=x) = (5/6)^{x-1} \cdot (1/6)$

$P(Y=y) = (5/6)^{y-1} \cdot (1/6)$

Assuming $X$ and $Y$ to be independent, we have $f(x, y) = (5/6)^{x-1} \cdot (1/6)^2 \cdot (5/6)^{y-1}$. This is where I get stuck. Can someone please explain how this solution can be completed from here?

Body:
$f(x,y)=f(x|y)f(y)$

$f(x|y)=
\begin{cases}
(4/5)^{x-1}(1/5), & x<y \\
0, & x = y \\
(4/5)^{y-1}(5/6)^{x-y-1}(1/6), & x>y
\end{cases}
=
\begin{cases}
(4/5)^{x-1}(1/5), & x<y \\
0, & x = y \\
(24/25)^{y}(5/6)^{x}(1/4), & x>y
\end{cases}
$

It doesn't seem to give an elegant solution.


问:
Counting the number of surjections.
How many functions from set $\{1,2,3,\ldots,n\}$ to $\{A,B,C\}$ are surjections? $n \geq 3$

Attempt

I was hoping to count the number of surjections by treating $A,B,C$ like bins, and counting the number of ways to fill them with $1,2,3,\ldots,n$ such that no bins are empty. This is equivalent to putting two "separator" bars in between the $n$ numbers or $n-1 \choose 2$.

However, I think I am missing a step here: prior to putting in the separator bars, I should permutate the $n$ elements. So, multiply by $n!$.

The order of elements in the same bins do not matter. So I should be divided by the number of permutations of the elements in each bins. This is where I am stuck.

Additional Info

I made the question up myself. Feel free to modify the question into a more solvable form if it is too difficult/complicated.

答:
Using PIE one can find a more generalized form of the number of surjections.
The number of surjections of a set containing $m$ elements onto a set containing $n$ elements ($m>n$) is

$$\sum\limits_{i=0}^{n}(-1)^{i}\dbinom{n}{i}(n-i)^{m}$$

In the given question $n=3$ and $m=n$.

Title: Counting the number of surjections.


答:
Each row in the array below represents a set of terms in one of the sums. So $n=4$ in this example, and the first row contains the term of $c_{11}$, the second row the terms of $c_{12}$, and so on until the last row which contains the term of $c_{44}$.
$$
\begin{array}{ccccc}
a_1\\
a_1 & a_2\\
a_1 & a_2 & a_3\\
a_1 & a_2 & a_3 & a_4\\
\hline
    & a_2\\
    & a_2 & a_3\\
    & a_2 & a_3 & a_4\\
\hline
    &     & a_3\\
    &     & a_3 & a_4\\
\hline
    &     &     & a_4
\end{array}
$$
The best way to do the sum is by columns, not rows. So the first column has $4$ copies of $a_1$, which we can think of as $1$ group of $4$. The second column has six copies of $a_2$, which we can think of as $2$ groups of $3$: a group where $a_1$ is the element of lowest index and a group where $a_2$ is the element of lowest index. Similarly, the third column has $3$ groups of $2$ copies of $a_3$: a group with $a_1$ as the element of lowest index, a group with $a_2$ as the element of lowest index, and a group with $a_3$ as the element of lowest index. Finally, the fourth column has $4$ groups each with $1$ copy of $a_4$.

Therefore you get
$$
1\cdot4\cdot a_1+2\cdot3\cdot a_2+3\cdot2\cdot a_3+4\cdot1\cdot a_4.
$$

Added: I may have misinterpreted the question. I understood that there was an array, $[a_1, a_2, \ldots, a_n]$ and that the goal was to find the average of the sums
\begin{align}
&c_{1,1}=a_1\\
&c_{1,2}=a_1+a_2\\
&c_{1,3}=a_1+a_2+a_3\\
&\vdots\\
&c_{n-1,n-1}=a_{n-1}\\
&c_{n-1,n}=a_{n-1}+a_n\\
&c_{n,n}=a_n.
\end{align}
There are $\binom{n+1}{2}$ such sums, so the average would be
\begin{align}
\frac{1}{\binom{n+1}{2}}\sum_{1\le i\le j\le n}c_{ij}&=\frac{1}{\binom{n+1}{2}}\sum_{i=1}^n\sum_{j=i}^n\sum_{\ell=i}^j a_\ell\\
&=\frac{1}{\binom{n+1}{2}}\sum_{i=1}^n\sum_{\ell=i}^na_\ell\sum_{j=\ell}^n1\\
&=\frac{1}{\binom{n+1}{2}}\sum_{i=1}^n\sum_{\ell=i}^na_\ell(n+1-\ell)\\
&=\frac{1}{\binom{n+1}{2}}\sum_{\ell=1}^na_\ell(n+1-\ell)\sum_{i=1}^\ell 1\\
&=\frac{1}{\binom{n+1}{2}}\sum_{\ell=1}^na_\ell(n+1-\ell)\ell,
\end{align}
which is an $O(n)$ computation. These rearrangements of summation symbols are essentially the "sum by columns rather than by rows" idea in the example above.

It seems, however, that what was actually meant is that there is an array, $[b_1,b_2,\ldots,b_n]$ and that $c_{ij}$ is defined by
$$
c_{ij}=\sum_{\ell=1}^n b_\ell\theta(b_\ell-i)\theta(j-b_\ell),
$$
where
$$
\theta(m)=\begin{cases}0 & \text{if $m<0$}\\ 1 & \text{if $m\ge0$.}\end{cases}
$$
This could also be written as
$$
c_{ij}=\sum_{\ell=1}^nf(i,j,b_\ell),
$$
where
$$
f(i,j,k)=\begin{cases}k & \text{if $i\le k\le j$}\\ 0 & \text{otherwise.}\end{cases}
$$
This problem can be converted to the problem I solved in $O(n)$ time by initializing the array $[a_1,a_2,\ldots,a_n]$ so that every element is $0$. Then for each $b$ in $[b_1,b_2,\ldots,b_n]$ such that $1\le b\le n$, add $b$ to $a_b$. For example, if $n=4$ and $[b_1,b_2,b_3,b_4]=[6,4,2,2]$ then $[a_1,a_2,a_3,a_4]$ would, at the end of the process, equal $[0,4,0,4]=[0,2,0,1]\circ[1,2,3,4]$, where $\circ$ is the element-wise product. In general, at the end of the process we have
$$
[a_1,a_2,\ldots,a_n]=[m_1,m_2,\ldots,m_n]\circ[1,2,\ldots,n],
$$
where $m_k$ is the number of times $k$ appears in $[b_1,b_2,\ldots,b_n]$.
Now proceed using the previous algorithm, which, as was said, is also $O(n)$.

If we combine this with the previous formula, we get that the average is
$$
\frac{1}{\binom{n+1}{2}}\sum_{\ell=1}^nm_\ell\cdot\ell\cdot(n+1-\ell)\ell.
$$
It's not necessary to use the multiplicities $m_\ell$; you can instead sum over the individual $b_k$, using the function $f$ to omit those that might be out of the interval $[1,n]$:
$$
\frac{1}{\binom{n+1}{2}}\sum_{k=1}^nf(1,n,b_k)\cdot(n+1-b_k)b_k.
$$
This is manifestly an $O(n)$ calculation.

Title: a semi-hard problem on combinatory

Body:
I ran into a nice interview question.

the problem is as follows:

We have array of $n$ integers. for $1 \leq i \leq j \leq n$. we want to set $c_{ij}$= Sum of all values in the range $i$ to $j$ of this array. we want to finding average of all possible $c_{ij}$ in this array. if 4 basic operation in math $(+,-,*,/)$ can be done in $O(1)$. the best algorithm can be works in $O(n)$.

And this is the solution here:

Suppose that some entry of the array has value $k$. What is its contribution to the average of the $c_{ij}$? It is counted in some $c_{ij}$ iff $i \leq k \leq j$. Out of the $\binom{n+1}{2}$ many $c_{ij}$'s, there are exactly $k(n+1-k)$ in which the entry is counted, therefore its contribution to the average is
$$
\frac{k(n+1-k)}{\binom{n+1}{2}} \cdot k = \frac{2k^2(n+1-k)}{(n+1)n}.
$$
Summing this over all values in the array, we obtain the desired $O(n)$ algorithm.

Anyone can describe me very easily how this formula is archived?  small example or any intuitive and simple idea?


问:
When does function iteration converge to a fixed point?
After watching a Youtube video by blackpenredpen (only 54 seconds long) showing how to write 5 as an infinite nested square root it got me thinking.

This is what he showed in the video:

$
\begin{align}
x  &= 5
\\x-5& = 0
\\(x-5)(x+4)&=0
\\x^2-x-20&=0
\\x^2&=20+x
\\x&=\sqrt{20+x}
\\x&=\sqrt{20+\sqrt{20+x}}
\\x&=\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+\sqrt{20+x}}}}}
\end{align}
$

$x$ converges to 5 with the above infinite square root. Nothing weird about that, I tried multiplying by other equations instead of $(x+4)$ and got other fun infinite square roots. $x=\sqrt{10+3\sqrt{10+3x...}}$ and $x = \sqrt{30-\sqrt{30-\sqrt{30-x...}}}$. Both converge to 5 too, all good.

However I wanted to see if it would work without using square roots everywhere. So this is what I tried:

$
\begin{align}
x^2&=20+x
\\x&=x^2-20
\\x&=(x^2-20)^2-20
\end{align}
$

Both $x=x^2-20$ and $x=(x^2-20)^2-20$ diverge (tested by iterating with code), and I have no clue why and I want to know why. My guess would be that it has something to do with the derivative. The derivative at the interesting point, $x=5$, is $\frac{d}{dx}(x^2-20)=2x=10$. And it does not converge. The derivative of one that does converge is $\frac{d}{dx}\sqrt{20+x}=\frac{1}{2\sqrt{20+x}}=0.1$, that is two orders of magnitude smaller than the diverging one. Am I onto the right track? Is there some kind of theorem that states that the derivative has to be less than 1?

I know about Newton's method, a proven technique for finding the zeros to functions. I am curious about why using equations on the form ($x=\sqrt{x}$) converges and why using $x=x^2$ diverges.

答: You are exactly right. There is a simple rule coming from chaos theory that tells you whether certain fixed points are sinks or sources for a function being iterated. Basically, suppose $f$ is some continuously differentiable function such that $f(x_0)=x_0$. If $|f'(x_0)|>1$ then $x_0$ is a source. If $|f'(x_0)|<1$ then $x_0$ is a sink. If $|f'(x_0)|=1$ then it can go either way.


问:
If $p$ is a prime and $m,n\in\mathbb{N}$, prove that $\binom{pm}{pn}\equiv\binom{m}{n}$ mod $p$.
Hint: Compare the binomial expansions of $(1+x)^{pm}$ and of $(1+x^m)^p$ in $\mathbb{F}_p[x]$.

If $R$ is a commutative ring, then the set of all polynomials with coefficients in $R$ is denoted by $R[x]$.

$\mathbb{I}_p[m]$ is the integers mod $m$.

When $p$ is a prime, we will usually denote the field $\mathbb{I}_p[x]$ by $\mathbb{F}_p[x]$.

My question: What's the meaning of the hint?

答: for $a,b\in \Bbb F_{p}$ and $n\in \Bbb N$ you can write $(a+b)^n = \sum_{k=0}^n {n\choose k} a^kb^{n-k}$. The hint is to compare $(1+x)^{pm} = \sum_{n=0}^{pm}{pm  \choose n}x^n$ to $(1+x^m)^p= \sum_{n=0}^p {p \choose n}(x^m)^n$.


问:
Line passing through the origin and incident to two other lines
I came across this silly exercise but it's giving me a headache, because I cannot figure whether or not I'm not doing it right.

I have two lines of cartesian equations given

$$
r : \left\{
\begin{array}{l}
x = 2z - 1 \\
y = z + 1
\end{array}\right. \;\;\;
s : \left\{
\begin{array}{l}
z = 2 \\
x + y - 1 = 0
\end{array}\right.
$$

And I need to find the line passing through origin $O = (0,0,0)$ and intersecting both $r$ and $s$.

I don't think the exercise has solution.

My attempt: I wrote both lines in parametric form (I don't think this was necessary but it made the equations a bit easier to read)

$$
r : \left\{
\begin{array}{l}
x = 2t - 1 \\
y = t + 1 \\
z = t
\end{array}\right. \;\;\;
s : \left\{
\begin{array}{l}
x = -s + 1 \\
y = s \\
z = 2
\end{array}\right.
$$

Then I imposed the collinearity among the origin, generic point for $r$ and generic point for $s$ which is equivalent to impose

$$
\text{rank}
\begin{pmatrix}
2t - 1 & 1 - s \\
t + 1 & s \\
t & 2
\end{pmatrix} = 1
$$

which gave me the system (non linear but easy)

$$
\left\{
\begin{array}{l}
2t + 2 - ts = 0 \\
3t - 2 + ts = 0
\end{array}
\right.
$$

Subtracting the second equation from the first gives me $t = 0$, but substituting this value in any of the two equations gives me a contradiction $1 = 0$, which makes me say there's no such line.

Am I missing anything?

答:
I agree.  I worked it slightly differently.

$$x_1 -2z_1 = -1$$
$$y_1 - z_1 = 1$$

$$x_2 + y_2 = 1$$
$$z_2 = 2$$

And some scaling of a solution to the first set equals the 2nd set so that we get a line through the origin. $(x_1,y_1,z_1) = c (x_2,y_2,z_2)$.

So substituting:

$$z_1/c = 2$$
$$x_1/c + y_1/c = 1$$
$$z_1 - 2c = 0$$
$$x_1+y_1 -c =0$$

Using the first 2 and last 2 equations I got the solution $(x_1,y_1,z_1,c) = (-1,1,0,0)$

Since $z_1 = c = 0$ there is no scaling that makes a solution.


问:
Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$
Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$

The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$

Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$
in the numerator using reciprocal identities.

How would I do that?
Here is what the assignment looked like:

答:
Continuing from what you got:

$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$

and since $\sec \theta \cos \theta = 1, \cot \theta \tan \theta = 1$, expand the brackets:

$$\sec\theta \cot\theta = \frac{\cot \theta + \sec \theta}{\cos \theta + \tan \theta}$$

There is no need to rewrite $\sec \theta \cot \theta$.


问:
Can an algebraic structure have indistinguishable elements?
Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.

Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.

答: $i$ and $-i$ are algebraically indistinguishable points of $\mathbb C$ because conjugation is a field automorphism over $\mathbb R$.</p>


问:
Complement of closed convex set in Hilbert space
Let $H$ be a real Hilbert space and $C$ be a closed convex subset of $H$.

Is it true that if $x$ is a boundary point of $C$ then there exists $y\in H\setminus C$ such that the set $\{x+t(y-x);\;t>0\}$ lies in $H\setminus C$ ?

Thank you for any hint.

答:
Here is a somewhat opportunistic answer since it is not much more than putting together the ideas of @daw and @gerw,  but I
think it is still justified because it is a self contained proof.

Without loss of generality suppose that the boundary point $x$ referred to in the statement is the orgin.  As observed by @gerw,  assuming by
contradiction that the result is false,  for every $y$ in $H$,  there exists $t>0$ such
that $ty\in  C$.  This is equivalent to the fact that
$$
  H = \bigcup_{n\in \mathbb N} nC,
  $$
so Baire's Theorem implies that $C$ has a nonempty interior which is what one needs to make  @daw's proof work.


问:
For which $n$ are there primitive Pythagorean triples with legs of lengths $a$ and $a+n$?
For which n can $a^{2}+(a+n)^{2}=c^{2}$ be solved, where $a,b,c,n$ are positive integers?
I have found solutions for $n=1,7,17,23,31,41,47,79,89$ and for multiples of $7,17,23$...
Are there infinitely many prime $n$ for which it is solvable? </p>

答:
There are infinitely many Pythagorean triples where the difference between two legs is either $1$ or any of the infinite prime numbers $P\equiv\pm 1 \mod 8\quad$ taken to any non-zero power.

Under $100\quad
P\in \{1,7, 17, 23, 31, 41, 47, 49, 71, 73, 79, 89, 97\}.\quad $
To find these, we begin with Euclid's formula
$ \quad A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$ solve the $\quad (B-A)$ difference equation for $m$ and generate the $(m,k)$ values needed to feed Euclid's formula. Each iteration uses a seed $k$ which is either given or is the $m$-value of the previous iteration. Here is the formula for $P=1$ with a seed $k=1$. Note that $(m,k)$ are Pell numbers.

\begin{equation}
\quad m=k+\sqrt{2k^2+(-1)^k}
\end{equation}
\begin{align*}
k=1\space &\implies m=(1+\sqrt{2(1)^2+(-1)^1}\space)\big)=2\quad & F(2,1)=(3,4,5)\\
k=2\space  &\implies m=(2+\sqrt{2(2)^2+(-1)^2}\space)\big)=5\quad & F(5,2)=(21,20,29)\\
k=5\space  &\implies m=(5+\sqrt{2(5)^2+(-1)^5}\space)\big)=12\quad & F(12,5)=(119,120,169)\\
k=12\space  &\implies m=(12+\sqrt{2(12)^2 (-1)^{12}}\space)\big)=29\quad  & F(29,12)=(697,696,985)
 \end{align*}

For $P>1$, the formula is

\begin{equation}
 m=k+\sqrt{2k^2\pm P}
 \end{equation}
and there is more than one seed $(s)$ in a set $(k_1)$ each of which generates a subset of the entire set of triples for that difference. There is one more seed than the power $n$ of the prime number that is P. For example, $1$ is a zero power so there is $1$ seed.
$$P=1\implies k_1=\{s1\}=\{1\}\quad
P=7^1\implies k_1=\{s1,s2\}=\{2,1\}\\
P=343=7^3\implies k_1=\{s1,s2,s3, s4\}=\{14,16,3,7\}$$

The elements of $k_1$ are generated as follows.

\begin{equation}
\end{equation}
\begin{align*}
\text{where}\quad s_i=\sqrt{\frac{j_i^2 + P}{2}}
\\  &\text{for}\space\space\space 1 \le i \le \bigg\lceil\frac{(n+1)}{2}\bigg\rceil \space
  \land \space \space                                      1 \le j_i                \le  \lfloor\sqrt{X}\rfloor\\
\text{where}\space s_i=\sqrt{\frac{j_i^2-P}{2}}
\\ &\text{for}\space    \bigg\lceil\frac{(n+1)}{2}\bigg\rceil + 1 \le i \le (n+1)\space
  \land \space    \lfloor\sqrt{P} +1 \rfloor \le j_i \le  \lfloor\sqrt{2P}\rfloor
\end{align*}

For example, we have $n=3, P=7^3=343\\$
\begin{align*}
 1 \le   & i \le   2
 \space             1 \le     & j \le \lfloor\sqrt{343}\rfloor=18
 \space  \\
s_1=\sqrt{\frac{7^2+343}{2}}=14\space\quad  &  s_2=\sqrt{\frac{13^2+343}{2}}=16  \\
3 \le  & i \le  4
\space  19 \le    &  j  \le \lfloor\sqrt{2(343)}\rfloor=26
 \space   \\
s_3= \sqrt{\frac{19^2-343}{2}}=3 \quad
&  s_4=\sqrt{\frac{21^2-343}{2}}=7  \
 \end{align*}

\begin{align*}
s_1= 14\quad
     &14+\sqrt{2(14)^2-343}=21\quad   &F(21,14)=(245,588,637)\\
     &21+\sqrt{2(21)^2+343}=56\quad   &F(56,21)=(2695,2352,3577)\\
s_2=16 \quad
      &16 +\sqrt{2(16)^2-343}=29\quad  &F(29,16)=(585,928,1097)\\
      &29 +\sqrt{2(29)^2-343}=74\quad  &F(74,29)=(4635,4292,6317)\\
s_3=3 \quad
      &3+\sqrt{2(3)^2+343}=22\quad  &F(22,3)=(475,132,493)\\
      &22+\sqrt{2(22)^2+343}=47\quad  &F(47,22)=(1725,2068,2693)\\
s_4=7 \quad
      &7 +\sqrt{2(7)^2+343}=28\quad  &F(28,7)=(735,392,833)\\
      &28+\sqrt{2(28)^2+343}=63\quad  &F(63,28)=(3185,3528,4753)\\
 \end{align*}

For $P=1$
These $(m,k)$-values
may be generated directly using a Pell-related equation as follows.
\begin{equation}
 m_n= \frac{(1 + \sqrt{2})^{n+1} - (1 - \sqrt{2})^{n+1}}{2\sqrt{2}}\qquad \qquad\qquad
 k_n= \frac{(1 + \sqrt{2})^n - (1 - \sqrt{2})^n}{2\sqrt{2}}
 \end{equation}
For example

{\small
\begin{align*}
\frac{(1 + \sqrt{2})^{2} - (1 - \sqrt{2})^{2}}{2\sqrt{2}}=2 \quad
     \frac{(1 + \sqrt{2})^1 - (1 - \sqrt{2})^1}{2\sqrt{2}}=1 \quad&  F(2,1)=(3,4,5)\\
 \frac{(1 + \sqrt{2})^{3} - (1 - \sqrt{2})^{3}}{2\sqrt{2}}=5 \quad
     \frac{(1 + \sqrt{2})^2 - (1 - \sqrt{2})^2}{2\sqrt{2}}=2 \quad&  F(5,2)=(21,20,29)\\
\frac{(1 + \sqrt{2})^{4} - (1 - \sqrt{2})^{4}}{2\sqrt{2}}=12 \quad
      \frac{(1 + \sqrt{2})^3 - (1 - \sqrt{2})^3}{2\sqrt{2}}=5 \quad& F(12,5)=(119,120,169)\\
\frac{(1 + \sqrt{2})^{5} - (1 - \sqrt{2})^{5}}{2\sqrt{2}}=29 \quad
     \frac{(1 + \sqrt{2})^4 - (1 - \sqrt{2})^4}{2\sqrt{2}}=12 \quad& F(29,12)=(697,696,985)
 \end{align*}

}


问:
How comes that integrating factor = $1/(x\,y)$
I'm just curious how the integrating factor m(x,y) = $\frac{1}{x\,y}$ of $\frac{\mathrm{d}}{\mathrm{d}x}f(x,y) = x+y-\frac{x^2}{y}\,y'(x)$ was determined. It was just given. When I was introduced to this topic this week, my fist impression was that integrating factor just can depend on x or y. Because determinations like $m(x) = \exp(\int \alpha(x)\mathrm{d}x)$ whereas $\alpha(x) = \frac{\partial_y\partial_x(f(x,y))-\partial_x\partial_y(f(x,y))}{\partial_xf(x,y)}$ or such just seem to work this way. So loosely speaking: Are there other methods?

答:
You may have to guess, but in this case you can make it an educated guess.

If you multiply by $dx$ and thus rewrite the differential as

$M dx + N dy=(x+y)dx - (x^2/y)dy$

you see that if $x$ and $y$ were multiplied by a constant $\lambda$, you would get just the same differential multiplied by a constant:

$(\lambda x+\lambda y)d(\lambda x) - [(\lambda x)^2/\lambda y]d(\lambda y) = (\lambda^2)[(x+y)dx - (x^2/y)dy]$

A differential with this property is called homogeneous.

Homogeneous differentials tend to lend themselves to integrating factors having the form $x^ay^b$, which is itself homogeneous.  With that in mind let us try this type of integrating factor.  We then have the differential

$P dx +Q dy=x^ay^b(x+y)dx - x^ay^b(x^2/y)dy=(x^{a+1}y^b+x^ay^{b+1})dx - x^{a+2}y^{b-1}dy$

Applying the exactness criterion we have

$\partial P/\partial y = \partial Q/\partial x$

$bx^{a+1}y^{b-1}+(b+1)x^ay^b=-(a+2)x^{a+1}y^{b-1}$

Combining like terms reduces the exactness criterion to

$(a+b+2)x^{a+1}y^{b-1}+(b+1)x^ay^b=0$

So we must have $a+b+2=0,b+1=0$.  From this we get $b=-1$ and then $a=-1$, so we find the integrating factor $x^ay^b=1/(xy)$.


问:
function with disconnected dense graph
Assume we have a function $f$ with dense graph, that is, $$\{\langle x, f(x)\rangle\colon x\in\mathbb R\}\cap (U\times V)\neq\emptyset$$ for every open set $U,V\subset\mathbb R.$ let $\ell_a(x)=ax$ where $a\in\mathbb R\setminus\{0\}.$ Let $I$ be an open interval,if $$f(x)\neq\ell_a(x)$$ for every  $x\in I,$ then $f$ has a disconnected graph. I think it is correct since $f$ has a dense graph so it will have points from both side of the line. So, we can have  separates that make the graph disconnected. Is it true ? Any help will be appreciated greatly.

答:
You have to work a little harder than that. There must be $u,v\in I$ such that $u<v$, and $\langle u,f(u)\rangle$ and $\langle v,f(v)\rangle$ are on the same side of $\ell_a$. Without loss of generality assume that $f(u)>au$ and $f(v)>av$. Let

$$U=\left\{\langle x,y\rangle\in\Bbb R^2:u<x<v\text{ and }y<ax\right\}\,,$$

the set of points between the lines $x=u$ and $x=v$ and below $\ell_a$. Let $G$ be the graph of $f$. Then $U\cap G\ne\varnothing\ne G\setminus U$, and $U\cap G$ is a relatively open subset of $G$, so it only remains to show that $G\setminus U$ is relatively open in $G$.

We can do this by showing that $U\cap G$ is relatively closed in $G$ which is easy: we’ve carefully chosen $U$ so that

$$\operatorname{cl}U=\left\{\langle x,y\rangle\in\Bbb R^2:u\le x\le v\text{ and }y\le ax\right\}=U\,,$$

and $U\cap G$ is therefore clopen in $G$.

Title: function with disconnected dense graph


问:
Find the Lyapunov function for the nonlinear second order system.
Consider the system
$$
\begin{aligned}
\dot{x}_1 &= -x_1+x_2,\\
\dot{x}_2 &= -x_2^3.
\end{aligned}
$$
The origin is obviously globally asymptotically stable. Indeed, $x_2\to 0$ and the dynamics of $x_1$ is linear and thus converges to zero for the converging input $x_2$.

However, I got stuck trying to find a Lyapunov function for this system. Could you please help?

答:
When $x_2$ gets close to the origin its derivative very quickly becomes very small, mean while the dynamics of $x_1$ drives $x_1$ closer to $x_2$. So $x_1$ should be driven to some manifold defined by $x_2$. In order to also utilize this behavior for the Lyapunov function one could define it as

$$
V(x) = \frac{1}{2} (x_1 - f(x_2))^2 + \frac{1}{2} x_2^2. \tag{1}
$$

Taking the time time derivative of this yields

$$
\dot{V}(x) = (x_1 - f(x_2)) \left(-x_1 + x_2 + \frac{d\,f(x_2)}{d\,x_2} x_2^3\right) - x_2^4. \tag{2}
$$

If there exists a function $f(x)$ such that

$$
x + \frac{d\,f(x)}{d\,x} x^3 = f(x), \tag{3}
$$

then $(2)$ can also be written as

$$
\dot{V}(x) = (x_1 - f(x_2)) \left(-x_1 + f(x_2)\right) - x_2^4 = -(x_1 - f(x_2))^2 - x_2^4, \tag{4}
$$

which would be negative definite.

The equation from $(3)$ is equivalent to the following differential equation in $f(x)$

$$
\frac{d\,f(x)}{d\,x} = \frac{f(x) - x}{x^3}, \tag{5}
$$

which can be written into a form that is easier to solve by using an integrating factor of $e^{1/(2\,x^2)}$, yielding

\begin{align}
\frac{d}{d\,x}\left(e^{\frac{1}{2\,x^2}}\,f(x)\right) &= -\frac{e^{\frac{1}{2\,x^2}}}{x^2}, \tag{6a} \\
e^{\frac{1}{2\,x^2}}\,f(x) &= -\int \frac{e^{\frac{1}{2\,x^2}}}{x^2} dx. \tag{6b}
\end{align}

The integral from $(6b)$ can be shown to equal

$$
\int \frac{e^{\frac{1}{2\,x^2}}}{x^2} dx = -\sqrt{\frac{\pi}{2}}\, \text{erfi}\!\left(\frac{1}{\sqrt{2}\,x}\right) + c. \tag{7}
$$

Therefore, the general solution for $f(x)$ can be expressed as

$$
f(x) = e^{\frac{-1}{2\,x^2}} \left(\sqrt{\frac{\pi}{2}}\, \text{erfi}\!\left(\frac{1}{\sqrt{2}\,x}\right) + c\right). \tag{8}
$$

Since $(8)$ contains expressions which are not very common (mainly the imaginary error function)/ Therefore, I also tried if I could approximate it with a "simpler" expression and I got a decent approximation for $c=0$ using

$$
f(x) \approx \frac{x + 5.506\,x^3 + 1.012\,x^5}{1 + 2.175\,x^2 + 6.107\,x^4 + x^6}. \tag{9}
$$

The expression from $(8)$ is not well defined near $x=0$ (although the limit does exist) but the expression from $(9)$ does not suffer from this. However, I have not yet been able to show that the Lyapunov function is also negative definite using this expression for $f(x)$.


问:
Prove that $d_2=\sum_{j=1}^{k}\left|{x_j-y_j}\right|$ is a complete metric on $\mathbb{R}^k$.
I am trying to prove that $d_2\left({x,y}\right)=\sum_{j=1}^{k}\left|{x_j-y_j}\right|$ is a complete metric on $\mathbb{R}^k$. Here is my reasoning for why it is, but I am unsure about its correctness.

The first three properties of a metric are omitted for brevity, where as the triangle inequality property can be shown as follows. Since $\left|{x_j-y_j}\right| \leq \left|{x_j-z_j}\right| + \left|{z_j-y_j}\right|$, summing up this inequality for $j=1,2,...,k$ gives $\sum_{j=1}^{k}\left|{x_j-y_j}\right| \leq \sum_{j=1}^{k}\left|{x_j-z_j}\right|  + \sum_{j=1}^{k}\left|{z_j-y_j}\right|$. Hence, $d_2\left({x,y}\right) \leq d_2\left({x,z}\right) + d_2\left({z,y}\right)$ and is a metric on $\mathbb{R}^k$.

To show that $d_2\left({x,y}\right)$ is complete, I must show that a Cauchy sequence $\left({x_n}\right)$ converges to an element of $\mathbb{R}^k$, i.e. that $x_j^n$ for each $j=1,2,...,k$ converges to an element of $\mathbb{R}$. So I must determine whether there exists $N$ such that $n,m>N\implies \left|{x_j^n-x_j^m}\right|<\epsilon$. Since $d\left({x,y}\right)$ is a complete metric on $\mathbb{R}$, and for any $j$ we see that $\left|{x_j^n-x_j^m}\right|=d\left({x_j^n,x_j^m}\right)$ it follows that $x_j^n$ converges in $\mathbb{R}$ for each $j=1,2,...,k$ and thus $\left({x_n}\right)$ converges in $\mathbb{R}^k$.

答:
Almost there! You're a bit confused on the exposition.

It's true that $|x_j-y_j|\leq d(x,y)$, but saying $|x_j-y_j|$ converges makes no sense, since this is not even a sequence; rather, this is the difference in the $j$-th components of $x$ and $y$. Go back to your original sequence $(x_n)$ and apply what you've just said. You want to say that the the components of the $x_n$-s are Cauchy sequences. From this, you can actually extract the components of the limit of your sequence $(x_n)$. How?


问:
Compute triple integral $ \iiint\limits_D \sin(xy) - \sin(xz) + \sin(yz) \, dx \, dy \, dz $
Let $D = \{ (x,y,z): 1 \leq x^2 + y^2 + z^2 \leq 4 \}$ then compute

$$
\iiint\limits_D \sin(xy) - \sin(xz) + \sin(yz) \, dx \, dy \, dz
$$

I tried to use spherical coordinates but then the integrand becomes super ugly if I try to use cartesian coordinates then the domain gets ugly, it looks like there must be something better to do here.

答:
Due to symmetry

$$
\iiint\limits_D \sin(xy)\, dx \, dy \, dz
= \iiint\limits_D \sin(xz) dx \, dy \, dz
= \iiint\limits_D \sin(yz) \, dx \, dy \, dz
$$
Then
$$\iiint\limits_D \sin(xy) - \sin(xz) + \sin(yz) \, dx \, dy \, dz
= \iiint\limits_D \sin(xy)\, dx \, dy \, dz=0
$$
which vanishes per the symmetry  with respect to $z$


问:
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$
For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
$$ (a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2.
$$
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.

Can someone help me [rove this? It's USAMO 2009 .I thought of using cauchy.

答:
Update:

Just realized we can prove directly without taking derivative.

WLOG assume $a_1 \ge a_2 \ge \cdots \ge a_n$, and $a_1 > 4a_n$ then

$$\begin{split} & & \left(\sum a_i \right) \left( \sum \frac{1}{a_i} \right) \\ & = & n + \left(\frac{a_1}{a_n}+\frac{a_n}{a_1}\right) + \sum_{1 < i<j<n} \left( \frac{a_i}{a_j} + \frac{a_j}{a_i}\right) + \sum_{i=2}^{n-1} \left( \frac{a_i}{a_1} + \frac{a_n}{a_i} \right) +
\sum_{i=2}^{n-1} \left( \frac{a_i}{a_n} + \frac{a_1}{a_i} \right)\\
& > & n+\left(4+\frac 14\right) + 2 \binom{n-2}{2} + (n-2)\left(2\sqrt{\frac{a_n}{a_1}}+ 2\sqrt{\frac{a_1}{a_n}}\right)\\
& > & n + 4 + \frac 14 + (n-2)(n-3) + 2(n-2) \left( 2 + \frac 12 \right)\\
& = & n + 4 + \frac 14 + (n-2)(n+2) = n^2 + n + \frac 14 \left(n+\frac 12\right)^2 \\
\end{split}$$

Original proof:

We prove by induction that if $\max(a_i) > 4 \min (a_i)$ then $\sum a_i \sum \frac{1}{a_i} > \left(n+\frac 12\right)^2$

When $n=2$ it's true.

Suppose it's true for $n$. WLOG we assume $a_1 \ge a_2 \ge \cdots \ge a_n$ and $a_{n+1}$ is between $a_n$ and $a_1$.  Denote

$$f(x)=\frac{a_1}{x} + \frac{x}{a_1} + \frac{a_n}{x}+\frac{x}{a_n}$$

$$f^{(2)}(x) > 0,f'(x)=\frac{1}{a_1} + \frac{1}{a_n} - \frac{a_1+a_n}{x^2}$$

It's easy to show that $f(x)$ obtains its minimum at $x=\sqrt{a_1 a_n}$, and $$f(\sqrt{a_1a_n})=2 \left(\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}} \right) > 2(2+\frac 12)=5$$

Define $S_n = \sum_{i=1}^n a_n, T_n = \sum_{i=1}^n 1/a_n$. Then
$$
\sum_{i=1}^{n+1} a_i \sum_{i=1}^{n+1} \frac{1}{a_i} = (S_n+a_{n+1})\left(T_n+\frac{1}{a_{n+1}}\right) \\
= S_n T_n + 1 + \sum_{i=2}^{n-1} \left(\frac{a_{n+1}}{a_i} + \frac{a_i}{a_{n+1}}\right)+f(a_{n+1})\\
> \left(n+\frac 12\right)^2 + 1 + 2(n-2)+5=\left(n+1+\frac 12\right)^2.\blacksquare
$$


问:
Why are the Left and Right Riemann sums for $\int_a^b f(x)\ dx$ equal?

For finding the area under $\operatorname{f}\left(x\right) = x$ from $a$ to $b$,

if a right Riemann sum is
$$
\lim_{n \to \infty}  \frac{b - a}{n}
\sum_{i = 1}^{n}
\operatorname{f}\left(a + i\,\frac{b - a}{n}\right)
\quad\mbox{and}
$$

if a left Riemann sum is
$$
\lim_{n \to \infty}\frac{b - a}{n}
\sum_{i = 0}^{n - 1}
\operatorname{f}\left(a + i\,\frac{b - a}{n}\right),
$$

then why would they be equal $?$.

Is it because the limits as $n\to \infty$ make it so the summation's indexing doesn't matter when it moves down by one $?$.

答:
The difference between the two terms is

$$d_n=\frac{b-a}{n}(f(b)-f(a)).$$

As a function Riemann integrable is bounded, let say $\vert f(x) \vert \le M$ for $x \in [a,b]$, we get

$$\vert d_n \vert \le \frac{2M(b-a)}{n}$$ and

$$\lim\limits_{n \to \infty} d_n=0.$$

Indeed the two sums are converging to the Riemann integral $\int_a^b f$.


答:
Note that the difference between the two sums is  $$(b-a)(f(b)-f(a))/n$$ which approaches zero as $n$ goes to infinity.

Thus the two limits are the same provided that they exist.

Title: Why are the Left and Right Riemann sums for $\int_a^b f(x)\ dx$ equal?

Body:
For finding the area under $\operatorname{f}\left(x\right) = x$ from $a$ to $b$,

if a right Riemann sum is
$$
\lim_{n \to \infty}  \frac{b - a}{n}
\sum_{i = 1}^{n}
\operatorname{f}\left(a + i\,\frac{b - a}{n}\right)
\quad\mbox{and}
$$

if a left Riemann sum is
$$
\lim_{n \to \infty}\frac{b - a}{n}
\sum_{i = 0}^{n - 1}
\operatorname{f}\left(a + i\,\frac{b - a}{n}\right),
$$

then why would they be equal $?$.

Is it because the limits as $n\to \infty$ make it so the summation's indexing doesn't matter when it moves down by one $?$.


答:
W.L.O.G. Assume $a_{1}$ is the minimum and $a_{2}$ is the maximumm
if we have $x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n} \in \mathbb R $ then by cauchy-Schwars we have
$(x_{1}y_{1}+...+x_{n}y_{n})^2\leq(x_{1}^2+...+x_{n}^2)(y_{1}^2+...+y_{n}^2)$, we need an equation relating $a_{1}$ to $a_{2}$, if we put  $\sqrt a_{2}, \sqrt a_{1}, \sqrt a_{3},....,\sqrt a_{n}$ in place of $x_{1},....x_{n}$ and $\frac{1}{\sqrt a_{1}},....,\frac{1}{\sqrt a_{n}}$ in place of $y_{1},...,y_{n}$ this gives $(\sqrt \frac{a_{2}}{a_{1}}+\sqrt \frac{a_{1}}{a_{2}}+n-2)^2 \leq (a_{2}+a_{1}+...+a_{n})(\frac{1}{a_{1}}+\frac{1}{a_{2}}+....+\frac{1}{a_{n}})$ combining this with the above inequality it will be much easier to proceed :) .

Body:
For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
$$ (a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2.
$$
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.

Can someone help me [rove this? It's USAMO 2009 .I thought of using cauchy.


问:
Find all proper coloring of paths, cycles and wheels if a fix number of color in allowed
Recall that a proper vertex coloring is an assignment of a color to each vertex in a graph such that
adjacent vertices receive different colors. For a fix n and fix color choices k, find the number of proper vertex colorings for the following graph:

A path with n vertices (and n - 1 edges).

A cycle with n vertices (and n edges).

A wheel graph which has n + l vertices.

Now, 1 is rather simpler. I'm trying to attempt this problem by considering a recurrence relation: if we have already known the proper coloring of a path with (n-1) vertices, we need only multiply it with (k-1) colors to get all the proper coloring with n vertices.

But 2 and 3 seems to need a little bit more thinking.
Are there any theorems related to it?

答:
Now I proceed to answer 2 and 3.

By the power of deletion-contraction theorem, note that the proper coloring of a cycle (with n vertices) is reduced to the difference between proper coloring of a path (whose proper coloring we have calculated in 1), and the graph that we merge two arbitrary vertices together, which, when n larger than 3, is also a cycle.

We then apply deletion-contraction theorem again on the smaller cycle to derive a path of (n-1) vertices and a cycle of (n-2) vertices. Repeat this to derive the formula $ P_n - P_{n-1} + P_{n-2} + ... + (-1)^n K_3 $, where P denote the coloring of the paths and K denote the coloring of the complete graph.

For the proper coloring of the wheel graph, notice that a wheel graph is a cycle with an additional vertex connecting to every other vertice. We can color the cycle first, and then take a look at the vertex left. If the colors we had in hand are more than the vertex needs to be coloring, then we can freely choose the unused colors. It follows that the proper coloring of a wheel graph with (n+1) vertices is $(k-n)*C_{n}$


问:
Proof of Power Rule for Limits using epsilon-delta
The power rule for limits is,

$\lim\limits_{x \to a} \left[f(x)^n\right]=L^n$

It seems that the typical approach to proving this rule is by using induction in combination with the product law for limits. But is there a way to prove the power rule using the epsilon-delta style of proofing? I can't find a single resources online that discusses this.

答:
If $\lim\limits_{x\to a}f(x)=L$, then there is a $\delta_1>0$ such that if $0<|x-a|<\delta_1$ then $|f(x)-L| <1$, and hence $|f(x)|<1+|L|$ (the point here is that if a function has a limit, then it is bounded in some small interval around that point).

Now, let $\epsilon>0$, and choose $\delta_2>0$ such that if $0<|x-a|<\delta_2$, then $|f(x)-L|<\epsilon$. Define $\delta:= \min(\delta_1,\delta_2)$; I claim this works. So, suppose $0<|x-a|<\delta$ now. Then,
\begin{align}
|f(x)^n-L^n|&=\left|[f(x)-L]\sum_{k=1}^{n} f(x)^{n-k}L^{k-1}\right|\\
&\leq |f(x)-l|\sum_{k=1}^n|f(x)|^{n-k}L^{k-1}\\
&\leq \epsilon \cdot \sum_{k=1}^n (1+|L|)^{n-k}L^{k-1}
\end{align}
Now, that big ugly summation term, let's call it $M$. Whatever it is, it doesn't matter; it is just some finite positive number depending only on $L$ (the key point is it doesn't depend on $\epsilon$).

Thus, what we have shown is that there exists an $M\geq 0$ (namely $M=\sum_{k=1}^n(1+|L|)^{n-k}L^{k-1}$) such that for every $\epsilon>0$, there exists a $\delta>0$ such that if $0<|x-a|<\delta$ then $|f(x)^n-L^n|\leq M\epsilon$. This is entirely equivalent to proving $\lim\limits_{x\to a}f(x)^n=L^n$.

If you have the "desire" to finish off proofs with an $\epsilon$ in the end, then I invite you to reword this proof to make that happen.

The idea of the proof is to use the familiar formula to factor $f(x)^n-L^n$ as a product of $f(x)-L$ and some function which is bounded. Since $|f(x)-L|$ can be made small, if you multiply by something bounded (i.e you have control over how big it is), the result is still "small" (in the grand scheme of things). Hence, $|f(x)^n-L^n|$ is also small. In other words, the crux of this proof is that
\begin{align}
\text{small}\cdot \text{bounded} =\text{small}.
\end{align}


问:
Properties of a a group $G$ of order $2021$.

Let $G$ be a group of order $2021$. Then which of the following statements are true:

(1) $G$ is a cyclic group.

(2) $G$ is non-abelian.

(3) All subgroups of $G$ are normal.

(4) Number of non-trivial subgroups is 3

(5) $G$ is a simple group.

(6) $G\simeq \mathbb Z_{2021}$

(7) If $a_1+a_2+...+a_{2021}$ is class equation of $G$ then $a_i=a_j\;\forall i,j\in \{1,2,...,2021\}$

(8) If $H\le G$ then $G/H$ is always cyclic.

(9) Non-trivial subgroup of $G$ may not be cyclic.

Assuming $G$ is a cyclic group I got:
(1) True;(2) False;(3) True;

(4) If by trivial subgroup you mean only $\{e_G\}$ then True, but if $\{e_G\}$ and $G$ are both considered as trivial subgroups then False;

(5) False (6) True;(7) True;(8) True;(9) Same argument as question (4), since $G$ is not simple.

But does there exist a non-cyclic group of order $2021$. If yes then which of the statements will be True.

答:
Every group $G$ of order $2021$ is cyclic. To see this, observe that $2021=43\cdot 47$, so any group of order $2021$ has a normal subgroup $H_1$ of order $43$ and another normal subgroup $H_2$ of order $47$. Since $H_1$ acts on $H_2$ by conjugation, there is a homomorphism $H_1\to \text{Aut}(H_2) \cong \mathbb Z/46\mathbb Z$. But $43$ is coprime to $46$, so this homomorphism has trivial image, and we conclude that $H_1$ commutes with $H_2$.

Thus $G$ is a direct product $H_1\times H_2 \cong \mathbb Z/43\mathbb Z \times \mathbb Z/47\mathbb Z \cong \mathbb Z/2021\mathbb Z$.

This is a special case of the general setup of semidirect products.

For your other question, the trivial subgroup is $\{e\}$, and all other subgroups (including $G$ itself) are nontrivial.


问:
Associativity of compositions of relations
Prove that given relations

$R_1 \subseteq A \times B$, $R_2 \subseteq B \times C$, $R_3 \subseteq C \times D$

then

$(R_1 \circ R_2) \circ R_3 = R_1 \circ (R_2 \circ R_3)$

I don't know where exactly to start? What does it mean for something to be in $(R_1 \circ R_2) \circ R_3$?

Here is what I know.

I know that I have to show that $(R_1 \circ R_2) \circ R_3 \subseteq R_1 \circ (R_2 \circ R_3)$ and $R_1 \circ (R_2 \circ R_3) \subseteq (R_1 \circ R_2) \circ R_3$.

答:
Preparations:
$\newcommand{\op}[2]{ \left\langle #1 ,\, #2 \right\rangle }$
$\newcommand{\rel}[1]{ \mathcal{#1} }$
$\newcommand{\dom}[1]{ \mathrm{dom} \ {#1} }$
$\newcommand{\ran}[1]{ \mathrm{ran} \ {#1} }$

Define ordered pair:

$$\op{x}{y} = \left\{\left\{x\right\},\,\left\{x,\,y\right\}\right\}$$

What to prove:

If $\rel{R}$, $\rel{S}$, $\rel{T}$ are relations, such that $\left(\rel{R}\circ\rel{S}\right)\circ\rel{T}$ is well defined, then $\rel{R}\circ\left(\rel{S}\circ\rel{T}\right)$ is also well defined and equals $\left(\rel{R}\circ\rel{S}\right)\circ\rel{T}$.

Proof:

Suppose $\rel{R} \subseteq A \times B,\; \rel{S} \subseteq B \times C,\; \rel{T} \subseteq C \times D$, while $A,\ B,\ C,\ D$ are arbitrary sets.

\begin{align}
\op{a}{d} \in \left(\rel{R}\circ\rel{S}\right)\circ\rel{T}
&\iff
\exists c \in C :\: \left\{ \begin{array}{l} \op{a}{c}\in\rel{R}\circ\rel{S}\\\op{c}{d}\in\rel{T} \end{array} \right. \\
&\iff
\exists b \in B,\; \exists c \in C :\: \left\{ \begin{array}{l} \op{a}{b} \in \rel{R} \\ \op{b}{c} \in \rel{S} \\ \op{c}{d} \in \rel{T} \end{array} \right. \\
&\iff
\exists b \in B :\: \left\{ \begin{array}{l} \op{a}{b} \in \rel{R} \\ \op{b}{d} \in \rel{S}\circ\rel{T} \end{array} \right. \\
&\iff
\op{a}{d} \in \rel{R}\circ\left(\rel{S}\circ\rel{T}\right)
\end{align}

Q.E.D.


答:
Consider the field $\Bbb Q(t,s)$ where $t,s$ are two algebraically independent transcendental numbers.

Then these two numbers are completely inseparable by a first-order formula in the language of fields.

Generally speaking, if $\cal L$ is some first-order language of some structure, then there are at most $\aleph_0\cdot|\cal L|$ definable elements in any given structure. If by "indistinguishable" we mean "inseparable by a first-order formula with limited parameters$^1$", then any sufficiently large structure will invariably contain a lot of indistinguishable elements.

One good place to learn about these things is model theory, and in particular the concept of "type".

Edit: To your last edit, about $(\Bbb N,f)$ note that $0$ is a definable element of the structure with the formula $x=f(x)$. And since we don't have any other symbols in the language it's really impossible to express anything else. Therefore it's very easy to see that over the empty set, every two non-zero elements satisfy the same formulas with one free variable.

(To see that we can't express anything else, at least without parameters, note that if $m,n$ are non-zero then there is an automorphism which exchanges between the two. Therefore every two non-zero elements must satisfy the same formulas [in one free variable].)

Footnotes:

Of course if we allow any parameter then $\varphi(x,y)$ defined as $\lnot(x=y)$ is sufficient to distinguish between any two members. But if, like in the first example, we allow no parameters - or parameters from a small substructure - then if the universe of the structure is large enough, there will be many indistinguishable elements.

Title: Can an algebraic structure have indistinguishable elements?

Body:
Sometimes, a topological space has indistinguishable points - we call those spaces non-$T_0$. But given such a space, we can always identify indistinguishable points, thereby yielding a $T_0$ space. (Technically, we've taken the Kolomogorov quotient).

Does this sort of thing ever happen in abstract algebra?

Here's two more examples.

A preordered set can have comparable, distinct points - in other words it can fail to be antisymmetric. But that's cool, we can identify comparable points to obtain a partially ordered set.

Sometimes a pseudometric space has distinct points that are zero distance apart. But that's okay, we can identify zero-distance points to obtain a metric space.

Edit: It would be nice to see a definition of 'indistinguishable' for the elements of arbitrary structures. It would then be a consequence of this more general definition that for an arbitrary preordered set $X$ (order relation $\leq$) it holds that $x,y \in X$ are indistinguishable iff $x \leq y$ and $y \leq x$.

Here's an example. Consider the function $f : \mathbb{N} \rightarrow \mathbb{N}$, with $f(n)=0$ for all $n \in \mathbb{N}$. The associated notion of indistinguishability for the structure $(\mathbb{N},f)$ should probably be the relation $\sim$ such that $a \sim b$ iff both $a$ and $b$ equal $0$, or both $a$ and $b$ are distinct from $0$.

Edit2: On the other hand, perhaps it does not make sense to speak of 'the natural notion of indistinguishability in a structure $X$' without first situating that structure in a category. After all, if we're going to quotient out by the indistinguishability relation, epimorphisms will probably show up at some point.


问:
Exercise 4 Conway, a course in functional analysis.
I am trying to test Conway problem 4 page 49. I'm having trouble proving an implication.

Corollary 5.4. If $T$ is a compact self-adjoint operator, then there is a sequence $\left\{\mu_n\right\}$ of real numbers and an orthonormal basis $\left\{e_n\right\}$ for $(ker(T))^{\perp}$ such that for all $h$, $Th=\sum_{n=1}^{\infty} \mu_n \langle h,e_n\rangle e_n$.

Exercise 4. If $T$ is a compact self-adjoint operator and $\left\{e_n\right\}$ and $\left\{\mu\right\}$ are as in (5.4) and if $h$ is a given vector in $\mathcal{H}$, show that there is a vector $f\in \mathcal{H}$ such that $Tf=h$ if and only if $h\perp ker(T)$ and $\sum_{n}\mu_n^{-1}|\langle h,e_n\rangle|^2<\infty.$ Find the form of the general vector $f$ such that $Tf=h$.

I have this:

$(\Rightarrow)$ Let $f\in\mathcal{H}:Tf=h$. Let $g\in ker(T)$ then $T(g)=0$.
\begin{eqnarray}
\langle h,g\rangle&=&\langle T(f),g\rangle\\
&=&\langle f,T^*(g)\rangle\\
&=&\langle f,T(g)\rangle\\
&=&\langle f,0\rangle\\
&=&0
\end{eqnarray}
Therefore $h\perp ker(T)$.

On the other hand,
\begin{eqnarray}
\langle h,e_n\rangle&=&\langle T(f),e_n\rangle\\
&=&\langle \sum_{n=1}^{\infty} \mu_n \langle f,e_n\rangle e_n,e_n\rangle\\
&=&\sum_{n=1}^{\infty}\mu_m \langle f,e_m\rangle \langle e_m,e_n\rangle\\
&=&\mu_n\langle f,e_n\rangle\\
\end{eqnarray}
Therefore $\displaystyle \frac{\langle h,e_n\rangle}{\mu_n}=\langle f,e_n\rangle$
Now, $\displaystyle \sum_{n=1}^{\infty} |\frac{\langle h,e_n\rangle}{\mu_n}|^2=\sum_{n=1}^{\infty} |\langle f,e_n\rangle |^2\leq \|f\|^2$ (Bessel Inequality)
Therefore $\displaystyle \frac{ |\langle h,e_n\rangle|}{\mu_n}\in l^2$

($\Leftarrow$) (I don't know how to prove this implication)

Let $h\in ker(T)$ and $\displaystyle \frac{ |\langle h,e_n\rangle|}{\mu_n}\in l^2(\mathbb{N})$

Formally, I think the $f$ to find is $\displaystyle f=\sum_{n=1}^{\infty}\frac{\langle h,e_n\rangle}{\mu_n}e_n$ because $T(f)=\sum_{n=1}^{\infty} \langle h,e_n\rangle e_n$ and   $h=\sum_{n=1}^{\infty} \langle h,e_n\rangle e_n$ but I'm not sure if this is correct.
How proves this implicance?

答:
Observe that, as a direct result of the corollary, $Te_n=\mu_ne_n$. Since $h\in Ker(T)^\perp,$ you can expand it in the basis: $$h=\sum\limits_{n=1}^\infty\langle h,e_n\rangle e_n.$$ You also know that $$Th= \sum\limits_{n=1}^\infty\mu_n\langle h,e_n\rangle e_n.$$ As you recommended, let's call $$f=\sum\limits_{n=1}^\infty\frac{1}{\mu_n}\langle h,e_n\rangle e_n.$$ This makes sense due to the fact that $\mu_n^{-1}|\langle h,e_n\rangle|\in\ell^2.$ Then, applying $T$ gives $$Tf=\sum\limits_{n=1}^\infty \frac{1}{\mu_n}\langle h,e_n\rangle Te_n=\sum\limits_{n=1}^\infty \frac{1}{\mu_n}\langle h,e_n\rangle \mu_n e_n=\sum\limits_{n=1}^\infty\langle h,e_n\rangle e_n=h.$$ We were able to pull the infinite sum out as a consequence of linearity and continuity.

EDIT: Let's justify the first equality. Call $a_n=\mu_n^{-1}\langle h,e_n\rangle$. The missing work is showing that $$T\left(\sum\limits_{n=1}^\infty a_n e_n\right)=\sum\limits_{n=1}^\infty T(a_n e_n).$$ If we show this, then we just use linearity to conclude that it equals $\sum\limits_{n=1}^\infty a_n Te_n.$

First, since $T$ is linear, $$T\left(\sum\limits_{n=1}^m a_n e_n\right)=\sum\limits_{n=1}^m T(a_n e_n).$$ Taking the limit as $m\rightarrow\infty$, $$\lim_{m\rightarrow\infty}T\left(\sum\limits_{n=1}^m a_n e_n\right)=\sum\limits_{n=1}^\infty T(a_n e_n).$$ The convergence of these limits should be understood as occurring in the $\mathcal{H}$-norm.

We're done if we can justify moving the limit inside of $T$. The follows from continuity: Since $T$ is continuous, it sends convergent sequences to convergence sequences. In particular, if we call
$b_n=\sum\limits_{n=1}^m a_n e_n$, then $b_n\rightarrow b:=\sum\limits_{n=1}^\infty a_n e_n$ in $\mathcal{H}$, so $T(b_n)\rightarrow T(b)$ in $\mathcal{H}.$ That is, $$\lim_{m\rightarrow\infty}T\left(\sum\limits_{n=1}^m a_n e_n\right)=T\left(\sum\limits_{n=1}^\infty a_n e_n\right).$$


问:
Is it necessarily true that$P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$
I've been trying my luck at the seemingly trivial and intuitive statement that $P(A\lvert B) = 1 \implies P(A\lvert C)=1$ where $C\subseteq B$ but I have struggled to prove it.

My attempts:

$$P(A\lvert C)= \frac{P(A\cap C)}{P(C)}\geq\frac{P(A\cap C)}{P(C)}\times \frac{P(C)}{P(B)}=\frac{P(A\cap C)}{P(C)}\times\frac{P(C\cap B)}{P(B)}=\frac{P(A\cap C)}{P(B)}\times\frac{P(C\cap B)}{P(C)}\\ $$

$$=\frac{P(A\cap C)}{P(B)}\times P(B\lvert C)$$

I do not know where I am going with this... any hints?

答:
Algebraically, consider two conditional probabilities:

$$\begin{align*}
P(A\mid C) &= \frac{P(A\cap C)}{P(C)}\\
P(A\mid(B\setminus C)) &= \frac{P(A\cap (B\setminus C))}{P(B\setminus C)}
\end{align*}$$

The mediant of $P(A\mid C)$ and $P(A\mid(B\setminus C))$ (in the above fractional forms) is

$$\frac{P(A\cap C) + P(A\cap(B\setminus C))}{P(C) + P(B\setminus C)} = \frac{P(A\cap B)}{P(B)} = P(A\mid B)$$

using $C\subseteq B$ and therefore $C \cup (B\setminus C) = B$.

If (assuming the contrary) that $P(A\mid C) \ne P(A\mid (B\setminus C))$, then by the mediant inequality,

$$\begin{align*}
\min(P(A\mid C), P(A\mid (B\setminus C))) < P(A\mid B) &< \max(P(A\mid C), P(A\mid (B\setminus C)))\\
1 &< \max(P(A\mid C), P(A\mid (B\setminus C)))
\end{align*}$$

which contradicts that $P(A\mid C)$ and $P(A\mid (B\setminus C))$ are probabilities.

Therefore the opposite holds,

$$P(A\mid C) = P(A\mid (B\setminus C)) = P(A\mid B) = 1$$

(unless $C=\emptyset$)


问:
Growth analysis: fractional power functions dominate polylogarithmic?
In big-O notation for algorithmic time-complexity analysis, given real-valued functions $f$ and $g$, $f(x)$ is $O(g(x))$ if there are constants $C$ and $k$ such that $$ |f(x)| \leq C |g(x)| \quad \forall x > k $$

I'm struggling with the case $f(x) = x^c$ where $0 < c < 1$, and $g(x) = \log^b(x)$ where $b > 1$. Which grows faster, polylogarithmic $g(x)$ or fractional power $f(x)$?

Plotting an example suggests that polylogarithmic dominates, but how can I formally prove this?

答:
In a formal big-O proof, you need to pick two values $k$ and $C$ and show that $0 \leq f(x) \leq C g(x)$ for every $k \geq x$.

If you're not sure how to pick you can go backwards and see how you can solve for $x$. Start with:

$$C x^c \geq \log^b(x)$$ (as the comment says a positive power of $x$ is going to grow bigger than any power of $\log x$, but even if you get it wrong here you can go back and fix it).
Then do some algebraic manipulation. Since $0<c<1$ you can do
$C x \geq \log^{b/c}(x)$ and here it seems like maybe taking the log of both sides might be helpful.

Then you get $\log(C x)  \geq b/c \log\log(x)$. Next, move all the $x$ terms to one side and get

$\frac{\log(Cx)}{\log \log(x)} \geq \frac{b}{c}$. We think $\log x$ grows faster than $\log \log x$ but we can figure out some constants that will work for us. Since $b/c$ is a constant set before, we can just say there's some value $\ell$ for which $10^\ell/\ell \geq b/c$, and $C = 1$. I might try $x =10^{10^\ell}$ (you can try another value as well).

Then I have $$\frac{\log (10^{10^\ell})}{ \log\log(10^{10^\ell}) } = \frac{10^\ell}{\ell}.$$

At this point, you go back and write the proof saying I set $C=10$ and $k= 10^{10^\ell}$ and work through the steps backwards to show the inequality holds. As a last step, you'd have to convince yourself that $\log x$ grows faster than $\log \log x$.


问:
Proving convergence of a Hilbert modular theta function $\vartheta(z):= \sum\limits_{x \in \mathcal{O}_F} e^{\pi i \operatorname{Tr}(x^2 z)}$
I'm trying to understand a somewhat sketchy proof that I found online of the convergence of the analog of Jacobi's theta function $\displaystyle{\theta(\tau) := \sum_{n = -\infty}^{\infty} e^{2 \pi i n^2 \tau } = 1 + 2 \sum_{n = 1}^{\infty}e^{2 \pi i n^2 \tau } }$ in the context of Hilbert modular forms for real quadratic fields. It is well known that this  function is holomorphic on the upper half plane $\mathbb{H}$, and I made sure to understand well and wrote down a detailed proof of this before asking this question.

The context of my question is as follows

Let $F = \mathbb{Q}(\sqrt{d})$ be a real quadratic field, with $d > 0$ a positive square-free integer such that $d \equiv 2, 3 \pmod{4}$. Assume also that $F$ has class number $1$. Also its ring of integers is $\mathcal{O}_F = \mathbb{Z}[\sqrt{d}] = \mathbb{Z} \oplus\sqrt{d} \, \mathbb{Z} = \{ n + m\sqrt{d} \mid n, m \in \mathbb{Z} \}$. Then there are two real embeddings $\sigma_i : F \hookrightarrow \mathbb{R}$ given by  $\sigma_1(a + b\sqrt{d}) = a + b\sqrt{d}$ and $\sigma_2(a + b\sqrt{d}) = a - b\sqrt{d}$, so that $\sigma_1$ is just the identity map, and $\sigma_2$ gives the "conjugate" of an element.

Now for $z = (z_1, z_2) \in \mathbb{H} \times \mathbb{H}$ and $\alpha \in F$, we define the "trace" as follows $\operatorname{Tr}(\alpha z) := \sigma_1(\alpha) z_1 + \sigma_2(\alpha) z_2 = \alpha z_1 + \alpha' z_2$, where we wrote $\alpha'$ for the conjugate of $\alpha$.

Finally we define the holomorphic Hilbert modular theta function $\vartheta: \mathbb{H} \times \mathbb{H} \rightarrow \mathbb{C}$ by

$$
\vartheta(z) := \vartheta(z_1, z_2) := \sum_{x \in \mathcal{O}_F} e^{\pi i \operatorname{Tr}(x^2 z)}
$$

I'm trying to prove the absolute convergence of this infinite series.

The argument I found online
To prove the convergence of the series, it is compared to the Jacobi theta function defined above. To be precise, the author observes that

$$
|\vartheta(z)| = \left| \sum_{x \in \mathcal{O}_F} e^{\pi i \operatorname{Tr}(x^2 z)} \right| = \left| \sum_{x \in \mathcal{O}_F} e^{\pi i x^2 z_1} e^{\pi i x'^2 z_2} \right| \leq \left| \sum_{x \in \mathcal{O}_F} e^{2 \pi i x^2 z_1} \right|^{1/2}  \left| \sum_{x \in \mathcal{O}_F} e^{2 \pi i x'^2 z_2} \right|^{1/2}
$$

where apparently the Cauchy-Schwarz inequality has been used in the last step, although I'm not convinced that it is true as stated. In any case, then it is noted that since the Jacobi theta function defined above $\displaystyle{\theta(\tau) = \sum_{n = -\infty}^{\infty} e^{2 \pi i n^2 \tau }}$ is holomorphic on $\mathbb{H}$, then each of the two series on the right hand side of the above inequality are absolutely convergent on compact sets in $\mathbb{H}$, so that the original series $\vartheta(z_1, z_2)$ converges absolutely on compact subests of $\mathbb{H} \times \mathbb{H}$.

Questions

Can someone please help me understand how does the convergence of each of the series on the right hand side of the above inequality follows from the corresponding fact for the Jacobi theta function? I mean, since each of the two sums on the RHS of the inequality are over all algebraic integers $x \in \mathcal{O}_F = \mathbb{Z} \oplus \sqrt{d} \, \mathbb{Z}$, then the sums are actually double sums, so it doesn't seem to follow trivially.

Is the argument where the Cauchy-Schwarz inequality is used correct? If I'm not mistaken, I think that the absolute value should be inside the infinite series, like this
$$
\left | \sum x_n y_n  \right| \leq \left( \sum |x_n|^2 \right ) ^{1/2} \left( \sum |y_n|^2 \right ) ^{1/2}
$$

Thank you very much for any help.

答:
The convergence of the sum on the right-hand side, essentially is

absolute convergence of
$$
\sum_{m,n\in \mathbb{Z}} e^{-am^2-bmn-cn^2}$$
where $ax^2+bxy+cy^2$ is a positive-definite quadratic form.

Because of absolute convergence, there is no problem with using C-S.

Note that absolute convergence implies convergence.

The actual point where you used C-S, need to be replaced by absolute values.


问:
Find least value of $9\tan\phi+4\cot\phi+3$ without derivatives
The problem is as follows:

In a pharmaceutical lab in Daegu, a group of scientists need to find
the power of a visible blue light lamp to get through a sucrose
crystal in order to measure its purity for a certain medication.

The power for such light is given by:

$$J=\frac{1}{H}$$

Assuming that $H$ has to be the minimum value of the equation from
below:

$$H=9\tan\phi+4\cot\phi+3$$

$\theta$ is an acute angle which lies in the first quadrant.

What would be the value of $J$ so that the scientists could set up
their lamp?

The alternatives given in my book are:

$\begin{array}{ll}
1.&\frac{1}{5}\\
2.&\textrm{1}\\
3.&\textrm{15}\\
4.&\frac{1}{15}\\
\end{array}$

This problem is making me feel confused. Should derivatives be used here?

I've attempted to do so and derivate the equation, by doing so this yields:

$H'=9\sec^2\phi-4\csc^2\phi$

But equating this to zero doesn't seem to be the intended approach.

Then I came up with the idea of adding the intervals from where $\phi$ in the first quadrant lies.

$0\leq \cot \phi < +\infty$

Tangent is also in the same interval:

$0\leq \tan \phi < +\infty$

Then multiplying by $9$ and by $4$ and adding $3$ to the second equation.

This results into:

$3\leq 9\tan\phi+4\cot\phi+3 < +\infty$

But without any further constrain there isn't any way to tell, which would be the least value?

If it would had been the integer value, wouldn't had been $4$?

Running out of idead, what it came to my mind was to equate both functions:

$\tan\phi=\cot\phi$

This corresponds to $\phi=\frac{\pi}{4}$

Assuming this: I'm guessing that the result would be

$9\tan\phi=1$

$4\cot\phi=1$

then $1+1+3=5$

But this is more like guessing and not a serious way to justify the answer. I cannot be sure if that's the answer. Can someone help me here and explain me why derivatives doesn't seem to work?

How to solve this without derivative?. Help me please, I'm confused.

答:
If $x^2 = \tan \phi$

$$ 0 \leq \left(3x - \frac{2}{x} \right)^2 = 9 x^2 - 12 + \frac{4}{x^2}  $$

$$ 12 \leq   9 x^2  + \frac{4}{x^2}  $$
$$ 12 \leq   9 \tan \phi + 4 \cot \phi $$

https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality


问:
does not has continuous extension of $S^1 \to S^1$ to $\Bbb{R}^2 \to S^1$
Let $S^1$ be the smooth manifold then the identity map is smooth $i:S^1 \to S^1$,prove there exist not continuous extension from $i:\Bbb{R}^2 \to S^1$

I try to construct some connected set in $\Bbb{R}^2$ to $S^1$ that fail to be connected,for example removing two point that still keeps the domain connected but not in $S^1$ is my idea correct?

答: Suppose we may extend $i$ to $r$. That is $r:\mathbb{R}^2\rightarrow \mathbb{S}^1$ is such that $r|_{\mathbb{S}^1}=i$ and is continuous. So in particular, $r$ is a retraction from $\mathbb{R}^2$ to $\mathbb{S}^1$. Thus $\mathbb{S}^1$ is a retract of a simply connected space, which implies that $\mathbb{S}^1$ is simply connected. Contradiction.


答:
Using drivatives works fine.

$H' = 9\sec^2\phi - 4\csc^2\phi =0\\
9\sin^2\phi = 4\cos^2\phi\\
\tan^2\phi = \frac {4}{9}\\
\phi = \arctan {\frac 23}$

Plugging this into the original equation gives:

$9(\frac 23) + 4(\frac 32) + 3 = 15$

Title: Find least value of $9\tan\phi+4\cot\phi+3$ without derivatives

Body:
The problem is as follows:

In a pharmaceutical lab in Daegu, a group of scientists need to find
the power of a visible blue light lamp to get through a sucrose
crystal in order to measure its purity for a certain medication.

The power for such light is given by:

$$J=\frac{1}{H}$$

Assuming that $H$ has to be the minimum value of the equation from
below:

$$H=9\tan\phi+4\cot\phi+3$$

$\theta$ is an acute angle which lies in the first quadrant.

What would be the value of $J$ so that the scientists could set up
their lamp?

The alternatives given in my book are:

$\begin{array}{ll}
1.&\frac{1}{5}\\
2.&\textrm{1}\\
3.&\textrm{15}\\
4.&\frac{1}{15}\\
\end{array}$

This problem is making me feel confused. Should derivatives be used here?

I've attempted to do so and derivate the equation, by doing so this yields:

$H'=9\sec^2\phi-4\csc^2\phi$

But equating this to zero doesn't seem to be the intended approach.

Then I came up with the idea of adding the intervals from where $\phi$ in the first quadrant lies.

$0\leq \cot \phi < +\infty$

Tangent is also in the same interval:

$0\leq \tan \phi < +\infty$

Then multiplying by $9$ and by $4$ and adding $3$ to the second equation.

This results into:

$3\leq 9\tan\phi+4\cot\phi+3 < +\infty$

But without any further constrain there isn't any way to tell, which would be the least value?

If it would had been the integer value, wouldn't had been $4$?

Running out of idead, what it came to my mind was to equate both functions:

$\tan\phi=\cot\phi$

This corresponds to $\phi=\frac{\pi}{4}$

Assuming this: I'm guessing that the result would be

$9\tan\phi=1$

$4\cot\phi=1$

then $1+1+3=5$

But this is more like guessing and not a serious way to justify the answer. I cannot be sure if that's the answer. Can someone help me here and explain me why derivatives doesn't seem to work?

How to solve this without derivative?. Help me please, I'm confused.


问:
For the discrete topology, why must the singletons be in every basis?
I know the set of all singletons are a basis for a discrete topology, but if we had another basis, why must it also include all the singletons?</p>

答:
This follows from the following lemma:

Lemma: Let $X$ be a space and $\mathbb{B}$ a basis for $X$. $U\subseteq X$ is open if and only if for each $p\in U$, there exists some $B\in \mathbb{B}$ such that $p\in B\subseteq U$.

Indeed, if $X$ is a topological space endowed with the discrete topology, then every singleton is open. Hence by the preceding lemma, any basis for the topology of $X$ must contain all the singletons. A consequence of this is, if $X$ is uncountable and discrete then $X$ is not second countable.


问:
Numerical Methods - Fixed Point Iteration
We were set the following question before we had learned about systems of non-linear equations and I'm confused as to how we're meant to answer it using the fixed point method theorems that we have been provided with.

Consider the fixed point iteration $\underline{x}^{n+1} = \underline{\Psi}(\underline{x}^{n})$ where

$\Psi(\underline{x}) = \Psi((x_1,x_2,x_3)^T))$ =

Analyse the convergence of the method to the fixed point $\alpha$ = $(0,\frac{1}{3},0)^T$.

Are we meant to show that $|\Psi(\underline{x})|<1$ and continously differentiable? And if so, do we do this using a Jacobian matrix?

答:
When $x$ is near $\alpha$, the map $\Psi$ can be approximated by the first-order Taylor series
$$ \Psi(x) \approx \Psi(\alpha) + J(\alpha) (x - \alpha) = \alpha + J(\alpha) (x - \alpha) $$
where $J(\alpha)$ is the Jacobian matrix evaluated at $x=\alpha$.  Note that we have used the fact that $\alpha$ is a fixed point, i.e., $\Psi(\alpha)=\alpha$.
From this you can show that that the recursion
$$ x^{n+1} = \Psi(x^n) $$
is equivalent to
$$ x^{n+1} - \alpha = J(\alpha) (x^n - \alpha) $$
So if you define $y^n = x^n - \alpha$ then
$$ y^{n+1} = J(\alpha) y^n $$
so
$$ y^n = J(\alpha)^n y^0 $$
where the superscript on $J(\alpha)$ is an exponent rather than a time index.  Now the critical question is whether and how quickly $J(\alpha)^n$ shrinks with $n$.  This depends on the eigenvalues of $J(\alpha)$.

Basically, the fixed point is stable if all of the eigenvalues of $J(\alpha)$ are less than 1.  The convergence error is roughly $\lambda_1^n$, where $\lambda_1$ is the largest eigenvalue.


问:
Immersion and Submersion between Manifolds
I am reading "An introduction to Manifolds" by Loring Tu. There they've first defined immersion and submersion between manifolds and then gave an example. In the last line of the example they've written This example shows in particular that a submersion need not be onto. And i am unable to understand that statement. For reference the following is the way they've defined immersion and submersion:
A $C^\infty$ map $F:N\to M$ is said to be an immersion at $p\in N$ if its differential $F_{*,p}:T_pN\to T_{F(p)M}$ is injective and a submersion at p if $F_{*,p}$ is surjective.We call F an immersion if it is an immersion at every $p\in N$ and a submersion if it is a submersion at every $p\in N$. I am attaching the screenshot of the example they've given and there i have highlighted the statement that i could not understand. My doubt is that surjective by definition means onto so how can a submersion need not be onto?

答: If $\iota: U\hookrightarrow M$ denotes the inclusion, where $U\neq M$ is open in $M$, then $\iota_{*}:T_pM\rightarrow T_pM$ is both, injective and surjective. Hence $\iota$ is an immersion and a submersion. But $\iota$ clearly need not be surjective.


问:
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$
Show that exists $x \in \mathbb{R} $ such that $x^3+x=3$

I did:

I take $S=\{x\in \mathbb{R} \mid x^3+x<3\}$

So, this set is bounded and by completeness of the real numbers exists $a=\sup(S)$

Then I want to prove that $a^3+a=3$

And then in order to get a contradiction I suppose that $a^3+a>3$

and how $a $ is the supremum then $\forall\epsilon\in\mathbb{R} $ exists $b \in S$ such that
$a-\epsilon<b$ then $(a-\epsilon)^3+b<b^3+b<3$

But I'm stuck from here. I'm not able to use derivatives, or continuity so I can't use IVT

答:
If you cannot use the IVT or continuity, then the simplest method in my mind is to explicitly solve the polynomial equation. It is difficult to show that the following real number

$$x= \sqrt[3]{\frac{1}{18}(27+\sqrt{741})}-\sqrt[3]{\frac{2}{3(27+\sqrt{741})}}$$

is a solution to the equation $x^3+x-3=0$ (at least by hand). However, it is possible (again, not as easy as using the IVT) and suffices as a direct proof of the original claim.


问:
Find the poles of a function defined by series
Let $(z_n)$ be a sequence of distinct complex numbers such that $\sum_n \left| z_n \right|^{-3}$ converges, define

$$
f(z) = \sum_n \frac{1}{(z - z_n)^2} - \frac{1}{z_n^2}
$$

Prove that the function is meromorphic and find it's poles.

So, my idea is to show that the function converges absolutely for any $z \not= z_n$, if we do the sum we get that the series is the same as

$$
f(z) = \sum_n \big( -z^2 \frac{1}{(z - z_n)^2 z_n^2} + 2z \frac{1}{(z-z_n)^2 z_n} \big)
$$

I think that there exists constants c_1 and c_2 such that

$$
\left| \frac{1}{(z - z_n)^2 z_n^2}  \right| \leq C_1 \frac{1}{\left|z_n \right|^4}
$$

and

$$
\left| \frac{1}{(z - z_n)^2 z_n}  \right| \leq C_2 \frac{1}{\left|z_n \right|^3}
$$

for all $z$ outside of the sequence so it is dominated by a sum of absolutely convergent series.

To show it is analytic, if those constants are independent from $z$ then Weierstrass M-test can be applied and the convergence is uniform. Then we can use Morera's theorem to show that it is analytic.

The problem is to find those constants, if they exist.

答:
$$f(z) = \sum_n \frac{1}{(z - z_n)^2} - \frac{1}{z_n^2}= \sum_n \frac{2z_n z-z^2}{z_n^2(z - z_n)^2}$$

For $U$ a disk of finite radius with no $z_n$ on the boundary then $$f(z)-\sum_{z_n\in U}( \frac{1}{(z - z_n)^2} - \frac{1}{z_n^2})=  \sum_{z_n\not \in U} \frac{2z_n z-z^2}{z_n^2(z - z_n)^2} \tag{1}$$
Since $|z_n|\to \infty$ then $$\lim_{n\to \infty}\frac{z_n^3}{2 z}\frac{2z_n z-z^2}{z_n^2(z - z_n)^2} = 1$$ So $(1)$ converges uniformly on $U$ and it is analytic by Morera's theorem.

(alternatively it is mostly obvious that $(1)$ is holomorphic on $U$ since the series of termwise  derivatives converges uniformly too)


问:
Question about quadratic maximum/minimum word problem with linear modeling
I have the below problem from a textbook:

It goes through the solution. I don't get this part here:

This is a simple part, as this is just describing a linear equation. I thought I knew how to model these. Why is it not Q = 84,000 - 2,500p? In English, I can say we are starting at 84,000 subscribers, and deceasing 2,500 subscribers for every $1 we increase the price. It seems so easy! But in this problem they make me solve for b. I've done other linear equation problems where I didn't have to do this, so I'm a little confused.

*I guess what I'm confused about is, when reading a word problem, how do I know something is a starting value or a regular y/output value? knowing that will help me determine where I need to plug in these figures.

答:
Currently, $Q=84000$, $p=30$.  Try substituting these values into your proposed equation $$Q=84000-2500p$$ and see if you get a true statement.

To say that $Q$ is a linear function of $p$ is to say $$Q=ap+b$$ for some constants $a$ and $b$.  Equation of a straight line, right?  Now, we are told that $a=-2500$, so it remains to find $b$.


答:
Note

\begin{align}
9\tan\phi+4\cot\phi+3&=( 9\tan\phi+4\cot\phi-12)+15\\
&=(3\sqrt{\tan \phi}- 2\sqrt{\cot \phi})^2+15\ge 15
\end{align}


答:
A group of order $pq$, where $p\lt q$ and $p\not|(q-1)$ is always cyclic. This is well-known.

Thus we have a cyclic group and the answers are all fairly trivial.  The class equation is $\underbrace{1+\dots+1}_{\text{2021-times}}$.  Any quotient or subgroup of a cyclic group is cyclic.

Title: Properties of a a group $G$ of order $2021$.

Body:
Let $G$ be a group of order $2021$. Then which of the following statements are true:

(1) $G$ is a cyclic group.

(2) $G$ is non-abelian.

(3) All subgroups of $G$ are normal.

(4) Number of non-trivial subgroups is 3

(5) $G$ is a simple group.

(6) $G\simeq \mathbb Z_{2021}$

(7) If $a_1+a_2+...+a_{2021}$ is class equation of $G$ then $a_i=a_j\;\forall i,j\in \{1,2,...,2021\}$

(8) If $H\le G$ then $G/H$ is always cyclic.

(9) Non-trivial subgroup of $G$ may not be cyclic.

Assuming $G$ is a cyclic group I got:
(1) True;(2) False;(3) True;

(4) If by trivial subgroup you mean only $\{e_G\}$ then True, but if $\{e_G\}$ and $G$ are both considered as trivial subgroups then False;

(5) False (6) True;(7) True;(8) True;(9) Same argument as question (4), since $G$ is not simple.

But does there exist a non-cyclic group of order $2021$. If yes then which of the statements will be True.


问:
Epsilon-delta proof for $\lim\limits_{x \to a} \left[f(x)^2\right]=L^2$
As the title suggests, I'm looking for help with how to formulate an epsilon-delta proof for the square of a limit. That is,

Prove $\lim\limits_{x \to a} \left[f(x)^2\right]=L^2$ using the precise definition of a limit.

I know an easy proof using the product rule,

Suppose $\lim\limits_{x \to a} \left[f(x)\right]=L$ then

$\lim\limits_{x \to a} \left[f(x)^2\right]=\lim\limits_{x \to a} \left[f(x)f(x)\right]=\lim\limits_{x \to a}\left[f(x)\right]\cdot\lim\limits_{x \to a} \left[f(x)\right]=L\cdot L=L^2$

But I am required to  know how to perform the proof using epsilon-delta. I know the precise definiton of a limit but I'm quite horrendous at applying it. Could I recieve some guidance on how to set it up?

答:
Suppose $L\neq 0$. (I will leave the case $L=0$ for you). Let $\epsilon>0$. As $\lim_{x\rightarrow a}f(x)=L$, there exists $\delta>0$ such that whenever $0<|x-a|<\delta$, we have $|f(x)-L|<\epsilon$.

Hence whenever $0<|x-a|<\delta$,

$|f(x)^2-L^2|=|f(x)-L||f(x)+L|=$

$|f(x)-L||f(x)-L+L+L|\leq|f(x)-L|(|f(x)-L|+2|L|)<\epsilon(\epsilon+2|L|)$


问:
How to calculate the surface area of a cone using cylindrical coordinates?
In cylindrical coordinates, the infinitesimal surface area is $dA=sd\theta dz$.

In order to find the surface area of the curved portion of a cone,with radius R and height h, I compute the integral:

$$A  = \int_{\theta=0}^{2\pi}\int_{z=0}^{h}dA = \int_{\theta=0}^{2\pi}\int_{z=0}^{h} sd\theta dz$$

Using the straight line equation which gives $s = \frac{R}{h}(h-z)$, I obtain $A = \pi R h$.

This however does not give the literature solution, $ A = \pi R (R + \sqrt{R^2 + h^2})$. Have I gone wrong somewhere in my calculations?

答:
Parameterize your surface.

$S = (r\cos\theta, r\sin\theta, r\frac {H}{R})$

Find $dS$ and $\|dS\|$

$dS = (\cos\theta, \sin\theta, \frac {H}{R})\times(-r\sin\theta, r\cos\theta, 0)\ dr\ d\theta\\
dS = (-\frac {H}{R}r\cos\theta, -\frac {H}{R}r\sin\theta, r)\\
\|dS\| = r\sqrt {\frac {H^2}{R^2} + 1}$

Giving us the integral.

$\int_0^{2\pi}\int_0^R r\sqrt {\frac {H^2}{R^2} + 1}\ dr\ d\theta$

And resolve the integral:

$2\pi (\frac 12R^2)(\sqrt {\frac {H^2}{R^2} + 1})\\
\pi R\sqrt{H^2 + R^2}$


问:
Differential equation on a manifold in terms of the flow?
On an $n$-dimensional differentiable manifold $\mathcal{M}$, select a smooth tangent vector field $v : \mathcal{M} \to T\mathcal{M}$. Let $\varphi : U \to \mathbb{R}^n$ be a chart on an open subset $U \subseteq \mathcal{M}$. A curve $\gamma : \mathbb{R} \to U$ is an integral curve of $v$ if $\ \frac{d}{dt}\big{(}\varphi \circ \gamma\big{)}(t) = v \circ \gamma(t)\ \ \forall t \in \mathbb{R}$. The intuition is that, as you follow the curve, the tangent vector to the curve where you currently are must match what $v$ specifies for where you currently are.

Denote $\gamma_x$ as the unique integral curve to $v$ with $\gamma_x(0)=x$. Then the flow $\Phi : \mathcal{M} \times \mathbb{R} \to \mathcal{M}$ of $v$ can be defined as $\Phi(x,t) = \gamma_x(t)$. The intuition is that $\Phi(x,t)$ selects the integral curve initialized at $x$, follows it for a duration $t$, and tells you where on the manifold it landed. It is essentially an assemblage of integral curves with a "selector" argument.

Because $\gamma_x$ is an integral curve, it of course satisfies $\ \frac{d}{dt}\big{(}\varphi \circ \gamma_x\big{)}(0) = v \circ \gamma_x(0) = v(x)$. This leads me to an alternative definition for the flow of $v$ on $\mathcal{M}$ as the map $\Phi : \mathcal{M} \times \mathbb{R} \to \mathcal{M}$ such that,
$$
\frac{d}{dt}\big{(}\varphi_x \circ \Phi(x,t)\big{)}\big{|}_{t=0} = v(x)\ \ \ \forall x\in\mathcal{M} \tag{1}
$$

where $\varphi_x$ is a chart on $\mathcal{M}$ with $x$ in its domain. What I like about this perspective is that there is one condition for each $x \in \mathcal{M}$, rather than one condition for each $t \in \mathbb{R}$ along every integral curve. Though not proved here, it is well known that integral curves cannot cross and thus these conditions appear one-to-one.

The usual equation $\ \frac{d}{dt}\big{(}\varphi \circ \gamma\big{)} = v \circ \gamma\ $ is a general autonomous ODE. For a Euclidean manifold with an identity chart we write $\dot{\gamma} = v(\gamma)$. I am trying to interpret Eq.1 in a similar light as $\ \dot{\Phi}(x,0) = v(x)\ $. But something is weird about that, because it seems to be trivially satisfied by $\Phi(x,t) = tv(x)$.

Does this mean that Eq.1 is somehow not a sufficient condition for $\Phi$ to be the flow? Am I just misinterpreting the equation? Or is there a deeper flaw in my reasoning above?

答:
$(1)$ does not uniquely determine the map $\Phi$, as your counterexample suggests. Indeed, there are a great varienty of distinct maps $\mathcal{M}\times\mathbb{R}\to\mathcal{M}$ whose derivatives all agree on $\mathcal{M}\times\{0\}$. The missing piece of information is that the ODE determined by a fixed vector fields is autonomous, in that it satisfies
$$
\Phi(x,t_1)=\Phi(y,t_2)\implies\frac{\partial\Phi}{\partial t}(x,t_1)=\frac{\partial\Phi}{\partial t}(y,t_2)
$$
Put another way, the vector field $\left.\frac{\partial\Phi}{\partial t}\right|_{t=t_0}$ does not depend on $t_0$. One can show that this condition is equivalent to $\Phi$ being an $\mathbb{R}$-action, satisfying
$$
\Phi(\Phi(x,t_1),t_2)=\Phi(x,t_1+t_2)
$$
Flows are typically defined as maps having this property. In particular, given a vector field $v$ there is a unique flow satisfying $(1)$.

To describe a greater variety of maps $\mathcal{M}\times\mathbb{R}\to\mathcal{M}$, one often generalizes to time-dependent flows which need not be autonomous, and are generated by time-dependent vector fields. However, as the nomenclature suggests, to specify such a map one must specify its derivatives for all $t\in\mathbb{R}$.


问:
Number of Pythagorean triplets
How can I find the number of Pythagorean triplets $(a, b, c)$ such that:

$1\le a \le b \le c \le N$,

$
a, b, c, N  \in  \mathbb {N}
$

$
1 \le N\le10^6
$

答:
The number of primitive Pythagorean triples for $A\le N$ is over $50\%$ because there are one-or-more triples for every odd number greater than one. As $A$ increases, the chance of having more than one [primitive] triple with the same side increases, e.g.
$$(15,8,17)\quad (15,112,113)\quad (21,20,29)\quad (21,220,221)\quad (33,56,65)\\ (33,544,545)\quad (35,12,37)\quad (35,612,613)\quad (39,80,89)\quad (39,760,761)\\ (45,28,53)\qquad (45,1012,1013)\qquad (51,140,149)\qquad (51,1300,1301)$$

If you wanted to consider all Pythagorean triple then the number of them are greater than $N$ because, for example with $N=100$ there are well more than $100$ triples, about $170$ by my estimate.

$(3,4,5),\space (6,8,20),\cdots,(96,40,104),\space (96,110,146),\space
 (96,280,296),\space(96,1150,1154)$

By the way, half of all Pythagorean triples have $A<B$ as I confirmed in my recent question here.

Title: Number of Pythagorean triplets


问:
Why is the integral of harmonically related sinusoids either $\frac{T}{2}$ or $0$?
I'm taking a signals processing class and am trying to wrap my head around why the following is true.

$$\int_{t_0}^{t_0 +T}cos(\frac{2\pi}{T}kt)cos(\frac{2\pi}{T}mt) = \begin{cases}
                                   \frac{T}{2} & \text{if $k = m$} \\
                                   0 & \text{if $k \neq m$}
   \end{cases}$$

What does it mean(physically) for two sinusoids when they are multiplied with each other? Is there an easy way of visualizing that? What is the significance of $\frac{T}{2}$ when $k = m$?

Why is the above equation true physically/intuitively for signals?

答:
OP Here -

Thanks to Semiclassical's answer I've decided to answer my own questions.

(Also for more insight on the trig identities check out this)

So we have $$\int_{t_0}^{t_0 +T}\cos(\frac{2\pi}{T}kt)\cos(\frac{2\pi}{T}mt)$$

We will use the following trigonometric formula $$\cos(A)\cos(B) = \frac{1}{2}\left(cos(A+B) + cos(A - B)\right)$$

Applying that identity
$$\int_{t_0}^{t_0 +T}\cos(\frac{2\pi}{T}kt)\cos(\frac{2\pi}{T}mt)dt = \frac{1}{2}\int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m)) + \cos(\frac{2\pi t}{T}(k-m))\right)dt$$
$$ =\frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m))\right)dt + \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k-m))\right)dt \right)$$

Case 1: $k = m$

$$ = \frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(2k))\right)dt + \int_{t_0}^{t_0 +T}\cos(0)dt \right)$$

Notice that the first term contributes $0$ because we are integrating cosine over an entire period and the "positive areas" and "negative areas" negate each other, while the second term is integrating $1$ over the period. Thus,
$$ = \frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(2k))\right)dt + \int_{t_0}^{t_0 +T}\cos(0)dt \right) \\ = \frac{1}{2} \int_{t_0}^{t_0 +T}dt = \frac{T}{2}$$

Case 2: $k \neq m$

From
$$=\frac{1}{2} \left( \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k+m))\right)dt + \int_{t_0}^{t_0 +T}\left(\cos(\frac{2\pi t}{T}(k-m))\right)dt \right)$$

Both terms contribute $0$ since they are integrating a cosine wave over a full period with "negative" and "positive" areas

And of course, a similar argument can be made for sine waves.


问:
Faithfullness of Weyl group action
Let $I$ be a finite indexing set, and $A \in \operatorname{Mat}_I(\mathbb{Z})$ be a generalised Cartan matrix, i.e.

$a_{ii} = 2$,

$a_{ij} \leq 0$ for $i \neq j$, and

$a_{ij} = 0 \iff a_{ji} = 0$.

Then there is an abstract Weyl group $W_A$ associated to the matrix $A$, defined on the generators $\{s_i \mid i \in I\}$ subject only to the following relations:

$s_i^2 = 1$ for all $i \in I$, and

$(s_i s_j)^{m_{ij}} = 1$ for $i \neq j$, where $m_{ij} = 2, 3, 4, 6$ or $\infty$ depending on whether $a_{ij} a_{ij} = 0, 1, 2, 3$ or $\geq 4$ respectively. (If $m_{ij} = \infty$, we mean that $s_i, s_j$ generate an infinite dihedral group).

A based root datum of type $A$ is the data $R = (X, X^\vee, \langle - , - \rangle, (\alpha_i)_{i \in I}, (\alpha_i^\vee)_{i \in I})$ of:

Two $\mathbb{Z}$-modules $X, X^\vee$ of free finite rank in a perfect pairing $\langle -, - \rangle \colon X \times X^\vee \to \mathbb{Z}$, and

Choices of elements $\alpha_i \in X$ and $\alpha_i^\vee \in X^\vee$ such that $\langle \alpha_j, \alpha_i^\vee \rangle = a_{ij}$ for all $i, j \in I$.

Any time we have such data, we may define the simple reflections

$$ r_i \colon X \to X, \quad r_i(\lambda) = \lambda - \langle \lambda, \alpha_i^\vee \rangle \alpha_i, $$
$$ r_i^\vee \colon X^\vee \to X^\vee, \quad r_i^\vee(\mu) = \mu - \langle \alpha_i, \mu \rangle \alpha_i^\vee $$

and with a little work it is possible to show that the map $W_I \to \operatorname{GL}_{\mathbb{Z}}(X)$ defined by $s_i \mapsto r_i$ is in fact a group homomorphism, and similarly for $W_A \to \operatorname{GL}_{\mathbb{Z}}(X^\vee)$. Note that the perfect pairing is $W_A$-invariant, meaning that $\langle r_i \lambda, r_i^\vee \mu \rangle = \langle \lambda, \mu \rangle$, and so the $W_A$ action on $X$ is faithful iff the action on $X^\vee$ is faithful.

Question: What are necessary and sufficient conditions on the based root datum $R$ such that the abstract Weyl group acts faithfully on $X$ and $X^\vee$?

A sufficient condition is that both the sets $\{\alpha_i\} \subseteq X$ and $\{\alpha_i^\vee\} \subseteq X^\vee$ are linearly independent. Then one may do some chamber geometry on $X \otimes \mathbb{R}$ to show that $\langle r_i \mid i \in I \rangle$ is an abstract Coxeter group, as is done in Kac' book Infinite dimensional Lie algebras. However, this condition is not necessary, as one can see by taking the affine $\mathfrak{sl}_2$ Cartan matrix
$$\begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix}$$
and taking a root datum where $X \cong \mathbb{Z}^2 \cong X^\vee$ and let $\alpha_1^\vee, \alpha_2^\vee$ be a basis of $X^\vee$, forcing $\alpha_1 + \alpha_2 = 0$ to be linearly dependent roots. It is not hard to see that $r_1$ and $r_2$ then have no relation, so the action here is faithful.

On the other hand, we could also take a rank-one realisation of that same Cartan matrix by choosing any $\alpha_1 \in \mathbb{Z}$ and $\alpha_1^\vee \in \mathbb{Z}$ satisfying $\langle \alpha_1, \alpha_1^\vee \rangle = 2$ and then setting $\alpha_2 = -\alpha_1$ and $\alpha_2^\vee = - \alpha_1^\vee$. In this case, both $r_1$ and $r_2$ are just multiplication by $-1$, and so we have the relation $r_1 r_2 = 1$.

答: The progress I made on this shortly afterwards is that a sufficient condition is for either the simple roots or coroots to be linearly independent. Assuming the simple roots are linearly independent, one finds during the proof of (rirj)mij=1 that mij is the smallest integer for which this holds, meaning that the Coxeter matrix coincides with the matrix [ord(rirj)]ij. Together with the fact that the ri satisfy the exchange condition (3.11 in Kac), the ri define the same Coxeter group as the si. – Joppy</p>

Title: Faithfullness of Weyl group action


问:
Show that $8 \mid (a^2-b^2)$ for $a$ and $b$ both odd

If $a,b \in \mathbb{Z}$ and odd, show $8 \mid (a^2-b^2)$.

Let $a=2k+1$ and $b=2j+1$. I tried to get $8\mid (a^2-b^2)$ in to some equivalent form involving congruences and I started with
$$a^2\equiv b^2 \mod{8} \Rightarrow 4k^2+4k \equiv 4j^2+4j \mod{8}$$
$$\Rightarrow k^2+k-j^2-j=2m$$
for some $m \in \mathbb{Z}$ but I am not sure this is heading anywhere that I can tell.

Second attempt: Use Euler's Theorem and as $\gcd(a,8)=\gcd(b,8)=1$ and $\phi(8)=4$, $a^4 \equiv b^4 \equiv 1 \mod 8$ so $a^4-b^4\equiv 0 \mod{8}$.

I haven't gotten too much further are there any hints?

答:
Hint:

$$a^2 \equiv b^2 \equiv 1 \pmod8$$

When both $a$ and $b$ are odd.

$(2k+1)^2=4k^2+1+4k=4k(k+1)+1$, here either $k$ or $k+1$ is even.

Therefore $4k(k+1) \equiv 0 \pmod 8$


问:
Shifting Summations and Integrals
When adding summations, for example, $\sum_{n = 1}^{\infty} 3x^{n - 1} + \sum_{n = 1}^{\infty} nx^n$, a variable change such as $m = n - 1$ is often made, in this case yielding $\sum_{m = 0}^{\infty} 3x^m + \sum_{n = 1}^{\infty} nx^n$.  Although the motivation to do so depends on this question having been answered, I'll break a term out of the first summation to match the indices, yielding $3 + \sum_{m = 1}^{\infty} 3x^m + \sum_{n = 1}^{\infty} nx^n$.

The next step is to rename $m$ as $n$, and proceed to combine and factor the expression into $3 + \sum_{n = 1}^{\infty} (3 + n)x^n$.  This is the part that seems very unnatural to me.  The explanations I've heard to justify renaming $m$ as $n$ are "What does the name of a variable really matter? It's just a name!" and "$m$ and $n$ are just iteration variables that won't be present in the final expression."  Neither of these is very satisfying.  I get that it doesn't matter what name a variable has.  There is a very natural and obvious bijection between $r + 2 = 6$ and $y + 2 = 6$.  I even get that a variable name can be used multiple times in a problem.  You can use $A$ in an ODE system to represent both the coefficient matrix and an arbitrary constant, provided you keep the uses straight in your mind.  The problem here is that factoring step doesn't seem to keep the uses straight.  It treats old $n$ and new $n$ as the same variable.  The fact that $m$ and $n$ are iteration variables also doesn't seem to imply that they're the same variable; if values of $n$ represent days in a week ($n = 1$ is Sunday, $n = 2$ is Monday, etc.), then equivalent values of $m$ (or new $n$) represent different days (shifted by $1$).

I am not looking for an answer that expands a summation and its shifted version to prove that they match.  Also, I can do the algebra for a summation shift quickly with no problem, and it makes perfect sense at that level of abstraction.  It's only when I slow down and actually make the variable changes that things start to feel wrong.

答:
This is really just explaining Greg Martin's comment:

Let $f$ and $g$ be functions, and for each $n$, let $A_n={\displaystyle \sum_{a=1}^n} f(a)$, $B_n={\displaystyle \sum_{b=1}^n} g(b)$, and $C_n={\displaystyle \sum_{c=1}^n} \left(f(c)+g(c)\right)$.

Note that $C_n=(f(1)+g(1))+(f(2)+g(2))+\cdots+(f(n)+g(n))$, so that by associativity and commutativity of (finite) addition, $C_n=A_n+B_n$. Then by limit laws, ${\displaystyle \lim_{n\to\infty}}A_n+{\displaystyle \lim_{n\to\infty}}B_n={\displaystyle \lim_{n\to\infty}}C_n$ whenever you know at least two of the limits exist (it's a sum law if you know the $A$ and $B$ limits exist, and it's a difference law otherwise).

However, by the definition of the sum of a series, this means that we have

$$\boxed{{\displaystyle \sum_{a=1}^\infty} f(a)+{\displaystyle \sum_{b=1}^\infty} g(b)={\displaystyle \sum_{c=1}^\infty} \left(f(c)+g(c)\right)\text{if two series converge}}\tag{$1$}$$

Writing $a$ as $m$ and $b$ as $n$ and $c$ as $n$, but keeping the uses straight (each $n$ appears in its own sum on its own side of the equation), we obtain

$$\boxed{{\displaystyle \sum_{m=1}^\infty} f(m)+{\displaystyle \sum_{n=1}^\infty} g(n)={\displaystyle \sum_{n=1}^\infty} \left(f(n)+g(n)\right)\text{if two series converge}}\tag{$2$}$$

But this says exactly that the step of "rename $m$ to $n$ and then combine the two series, treating the two $n$s the same" is mathematically valid*.

*But this is only guaranteed to work under the assumption that either: 1. both of the original $m$ and $n$ series converge. or 2. one of the $m$ and $n$ series converges, and the end result converges too.

For example, if we choose $f(m)=\dfrac{1}{2m-1}$ and $g(n)=-\dfrac{1}{2n}$, then the left side is undefined (it looks like the indeterminate form $\infty-\infty$) but the right side is $\ln2$.

For this sort of reason, you were right to be cautious. But, in practice, this issue doesn't come up often in calculations.

Title: Shifting Summations and Integrals