问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
Around 1911 the group theorist William Burnside proved his famous $pq$-Theorem:
 Let $p$ and $q$ be primes and $G$ a group of order $p^aq^b$. Then G is solvable.
Burnside used linear representations over $\mathbb{C}$, in particular character theory. Although his proof was not extremely difficult, in the late 60s one began to look for a less sophisticated, that is a non-character proof, relying on new developments in group theory. In fact, David Goldschmidt was able to apply techniques that had been developed by his thesis adviser John Thompson and proved “by elementary means” the case where $p$ and $q$ are both odd. Not long after, Bender and Matsumaya found proofs for the remaining case and the combination of their arguments led to an overall simpler proof than the original one given by Goldschmidt.


问:
Why is a compact codomained function with compact fibers closed?
Let $Y$ be a compact topological space and $f:X \rightarrow Y$ a function such that
\begin{equation}
f^{-1}(y) \text{ is compact in } X \quad \forall y \in Y,
\end{equation}

i.e. fibers are compact. Then $f$ is a closed function.

Why?

Source of the statement is theorem 4.44 from Marco Manetti's "Topologia" coursebook.

EDIT: the theorem reported there is false, and it has been fixed in a later version
of the book (ISBN 8847007569,9788847007567).

答: The result is false. Let $X=[0,1]$ with the discrete topology, let $Y=[0,1]$ with the usual topology, and let $f$ be the identity map. Then $f$ has compact fibres, $Y$ is compact, and $f$ is certainly not closed, since, for instance, $(0,1)$ is closed in $X$ but not in $Y$.

ContentLicense: CC BY-SA 4.0


问:
$16$ kids dividing candies between their neighbors, prove that eventually a cycle will occur.
16 kids are playing a game. They sit on a 4 by 4 grid. Each kid has initial number of candies. Every turn, if a kid have more candies than the number of their neighbors (depending on the position of the kid, he will obviously have 2-4 neighbors), he will give each neighbor one candy. If the kid has fewer candies than the number of their neighbors, he will not give out any candies, but will continue to receive candies from his neighbors.

If we call how many candies each kid has a "state" of the game, prove that no matter what the initial state is, the states of the game will eventually recycle. For example if there is only one candy among 16 kids, in the next state this candy stays with the same kid, and the state of the game "recycles".

I wrote a python program to validate the claim. It seems indeed legit. I have tried to formulate this problem into either an Eulerian path, or a matrix multiplication, but so far has failed..

答:
Note that the total number of candies on the board doesn't change after each step! Let $N$ be the initial number of candies.

Let $X$ be the set of states in which each square has $\leq N$ candies. There are $(N+1)^{16}$ such states, so $\lvert X \rvert = (N+1)^{16}$.

Now let the sequence of states be $I_0, I_1, \ldots$ so that each state $I_n$ is an element of $X$. Then consider the first $(N+1)^{16}+1$ states,

$$I_0, I_1, \dots, I_{(N+1)^{16}}$$

This is a list of $(N+1)^{16} + 1$ elements of $X$, but $X$ has only $(N+1)^{16}$ elements, so by the pigeonhole principle, at least two of these states are the same! In other words, $I_i = I_j$ for some $i < j \leq (N+1)^{16}$, and the game repeats with period dividing $j-i$.


问:
Maximum product consecutive subsequence
I need find the maximum product of a consecutive subsequence in sequence of $n$ integers.

Example:

In: $3$, $1$, $-2$, $4$. Out: $4$.

In $2$, $5$, $-1$, $-2$, $-4$. Out: $20$. $(2*5*(-1)*(-2))$.

I've done the algorithm in $O(n^2)$, but now i want to do in $O(n)$.

I know that its possible.

Can anyone help me?

答:
A few observations you can use to devise the $O(n)$ algorithm (working under the assumption the elements of the sequence are integers):

Taking a zero is a good idea only if none of the elements is positive.

Extending the sequence by a positive element never hurts...

... nor does extending it by two negative ones.

ContentLicense: CC BY-SA 3.0

Title: Maximum product consecutive subsequence


问:
Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval

Show $\frac{(n-1)S^2}{\theta}$ is pivotal quantity of random sample $Y_1,...,Y_n$ from $N(\theta,\theta)$ then derive confidence interval $(1-\alpha)100%$ confidence interval for $\theta$

So I can show that $\frac{(n-1)S^2}{\theta}\sim \chi^2_{n-1}$ since $S^2=\sum_{k=1}^n \frac{(Y_i-\bar{Y})}{n-1}$

But I'm not sure how to find the confidence interval?

I believe it is supposed to look like $P(\chi^2_{n-1}\leq \frac{(n-1)S^2}{\theta}\leq \chi^2_{n-1})=1-\alpha$

And then solving for $\theta$ in the middle. But I know $\chi^2_{n-1}$ is wrong, I don't understand what the bounds are supposed to be.

答:
Suppose $\theta = \mu = \sigma^2,$ data are a random sample for a normal distribution with these parameters,
$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i = \bar X)^2,$ and the pivotal quantity $Q = \frac{(n-1)S^2}{\theta} \sim \mathsf{Chisq}(\nu = n-1).$

Then choose $L, U$ to cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$
$$P\left(L < \frac{(n-1)S^2}{\theta} <R\right)
= P\left(\frac{(n-1)S^2}{U}<\theta<\frac{(n-1)S^2}{L}\right)= 0.95.$$
Thus a 95% confidence interval for $\theta$ is of the form
$\left(\frac{(n-1)S^2}{U},\,\frac{(n-1)S^2}{L}\right).$

For example, if $n = 10, \theta = 7$ then data might be as simulated in R below:

```<code>set.seed(2021)
x = rnorm(10, 7, sqrt(7))
summary(x);  var(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.913   7.200   7.937   7.803   9.147  11.577
[1] 6.263132  # sample variance
</code>```

And the 95% CI for $\theta$ is $(2.96, 20.97),$

```<code>9*var(x)/qchisq(c(.975,.025), 9)
[1]  2.963196 20.874096
</code>```

Note: It seems worthwhile looking at this a little more extensively. The 95% t interval $(6.01,9.59)$ for $\theta$ is noticeably shorter than the CI you propose:

```<code>set.seed(2021)
x = rnorm(10, 7, sqrt(7))
t.test(x)$conf.int
[1] 6.012665 9.593205
attr(,"conf.level")
[1] 0.95
</code>```

Body:
Suppose $\theta = \mu = \sigma^2,$ data are a random sample for a normal distribution with these parameters,
$S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i = \bar X)^2,$ and the pivotal quantity $Q = \frac{(n-1)S^2}{\theta} \sim \mathsf{Chisq}(\nu = n-1).$

Then choose $L, U$ to cut probability $0.025$ from the lower and upper tails, respectively, of $\mathsf{Chisq}(n-1).$
$$P\left(L < \frac{(n-1)S^2}{\theta} <R\right)
= P\left(\frac{(n-1)S^2}{U}<\theta<\frac{(n-1)S^2}{L}\right)= 0.95.$$
Thus a 95% confidence interval for $\theta$ is of the form
$\left(\frac{(n-1)S^2}{U},\,\frac{(n-1)S^2}{L}\right).$

For example, if $n = 10, \theta = 7$ then data might be as simulated in R below:

set.seed(2021)
x = rnorm(10, 7, sqrt(7))
summary(x);  var(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  1.913   7.200   7.937   7.803   9.147  11.577
[1] 6.263132  # sample variance

And the 95% CI for $\theta$ is $(2.96, 20.97),$

9*var(x)/qchisq(c(.975,.025), 9)
[1]  2.963196 20.874096

Note: It seems worthwhile looking at this a little more extensively. The 95% t interval $(6.01,9.59)$ for $\theta$ is noticeably shorter than the CI you propose:

set.seed(2021)
x = rnorm(10, 7, sqrt(7))
t.test(x)$conf.int
[1] 6.012665 9.593205
attr(,"conf.level")
[1] 0.95


问:
Determining sample proportion probability using population proportion and a sample of size
Suppose 40% of all seniors have a computer at home and a sample of 64 is taken. What is the probability that more than 30 of those in the sample have a computer at home?"

My attempt:

n=64

0.4x64=25.6

p=?

x=??

A>30=??

Don't have an idea of what equation would be appropriate to determine proportion so as to use s=root p(1−p)/n

答:
Given an event $E$ with probability $p$ of occurring, where $0 < p < 1$

and $q = (1-p)$, and given $n$ independent trials of the event,

the probability
that the event $E$ occurs exactly $k$ times out of $n$ is

$\binom{n}{k}p^k q^{(n-k)}.$

This means that the probability of at least $r$ successes in $n$ trials is

$$\sum_{k=r}^n \binom{n}{k}p^k q^{(n-k)}.\tag1$$

All that you have to do is plug in the appropriate values for the variables in
equation (1) above:

$$r = 31, n = 64, p = 0.4.$$


问:
College Linear algebra: Given two determinants with 2 columns unknown solve for another determinant
Hey I have been stuck on this question for quite some time:

Given: $$A=
\begin{bmatrix}
 a  & 1 & d \\
 b & 1 & e \\
c &1  & f
\end{bmatrix}\qquad  \text{ and }\qquad\det(A)= -4,
$$

$$B=
\begin{bmatrix}
 a  & 1 & d \\
 b & 2 & e \\
c &3  & f
\end{bmatrix}
\qquad  \text{ and }\qquad\det(B)= -1,
$$

$$C=
\begin{bmatrix}
 a  & 3 & d \\
 b & 3 & e \\
c &3  & f
\end{bmatrix},\qquad\qquad
D=
\begin{bmatrix}
 a  & -1 & d \\
 b & 0 & e \\
c &1  & f
\end{bmatrix}.
$$

What is $\det C$? $\det D$?

Right from the go, I can see that $C$ is a scalar of $A$ by $3$ and I know that $\det C= k\,\det A$ but I dont know if I stand corrected.

Furthermore I tried to expand A into :

$$-1 \begin{bmatrix}
b &e \\
c&f
\end{bmatrix} +1\begin{vmatrix}
a &d \\
c&f
\end{vmatrix} -1\begin{bmatrix}
a &d \\
b&e
\end{bmatrix} = -2$$

and B into

$$ 1 \begin{bmatrix}
b &e \\
c&f
\end{bmatrix} -2\begin{vmatrix}
a &d \\
c&f
\end{vmatrix} +3\begin{bmatrix}
a &d \\
b&e
\end{bmatrix} = -1$$

Which results in A being in :
$$-1(bf-ce)+1(af-cd)-1(ae-bd) = -2$$
and B being in

$$1(bf-ce)-2(af-cd)+3(ae-bd) = -1$$

I have no idea what to do from here.. can anyone help?

Thanks

答:
Yes,
$$
\det C=3\det A=-12.
$$
For $D$, they want you to use that the determinant is linear on a fixed column (or row, but you need column here). Since
$$
\begin{bmatrix} -1\\0\\1\end{bmatrix} =-2\begin{bmatrix} 1\\1\\1\end{bmatrix} +\begin{bmatrix} 1\\2\\3\end{bmatrix},
$$
you have that
$$
\det D=-2\det A+\det B=7.
$$


问:
Need help simplifying a set of equations (and understanding how to solve it)
i have three algebraic expressions, each using the others. in these equations a, b, c and t are known and plugged in later:

$x = a^{-1}(t + y + z)$

$y = b^{-1}(t + x + z)$

$z = c^{-1}(t + x + y)$

i have managed to successfully solve the equations for two statements using substitution:

$x = a^{-1}(t + y)$

$y = b^{-1}(t + x)$

which substitutes and simplifies as:

$x = a^{-1}t + a^{-1}b^{-1}t + a^{-1}b^{-1}x$

$x - a^{-1}b^{-1}x = a^{-1}t + a^{-1}b^{-1}t$

$x(1 - a^{-1}b^{-1}) = a^{-1}t(b^{-1} + 1)$

therefore:

$x = a^{-1}t(1+b^{-1}) / (1 - a^{-1}b^{-1})$

$y = b^{-1}t(1+a^{-1}) / (1 - a^{-1}b^{-1})$

but introducing a third term has me stumped, as i keep getting a nonsense result when substituting leaving me with $x = x$ when i simplify the equation. (this is obviously correct, but also useless.) I'm hoping that someone would be so kind as to explain how i go about substituting more than one term and solving x, y and z in terms of t, a, b and c. does the same principle apply as with two terms?

for completeness, here is my (incorrect) work:

$x = a^{-1}t + a^{-1}y + a^{-1}z$

$a^{-1}z = x - (a^{-1}t + a^{-1}y)$

$a^{-1}z = x - a^{-1}(t + y)$

$z = ax - (t + y)$  (multiply both sides by a)

therefore, to find y i substitute z:

$x = a^{-1}t + a^{-1}y + a^{-1}(ax - (t + y))$

$x = a^{-1}t + a^{-1}y + x - a^{-1}t - a^{-1}y$

$x = (a^{-1}t - a^{-1}t) + (a^{-1}y - a^{-1}y) + x$

$x = 0 + 0 + x$

$x = x$

please forgive me if my terminology isn't correct. i don't exactly have a maths background and english isn't my first language.

答:
You can write your system as

$$\left\{\begin{align}
ax = (y+z) + t\\
by = (x+z) + t\\
cz = (x+y) + t
\end{align}\right.$$

and then in matrix form you'd get

$$\begin{align}
&\pmatrix{a&0&0\\0&b&0\\0&0&c}\pmatrix{x\\y\\z}
=
\pmatrix{0&1&1\\1&0&1\\1&1&0}\pmatrix{x\\y\\z}
+
t\pmatrix{1\\1\\1}\\
\implies
&\underbrace{\pmatrix{a&-1&-1\\-1&b&-1\\-1&-1&c}}_{M}\pmatrix{x\\y\\z}
=
t\pmatrix{1\\1\\1}.
\end{align}$$

Now, $M$ has determinant $\det(M) = abc - a -b -c -2$, so provided $\det(M)\neq 0$ your system admits a unique solution.
In this case, we have

$$M^{-1} = \frac1{\det(M)}\pmatrix{bc-1&c+1&b+1\\c+1&ac-1&a+1\\b+1&a+1&ab-1}$$

and

$$\pmatrix{x\\y\\z} = tM^{-1}\pmatrix{1\\1\\1}.$$

We can explicitly compute and obtain

$$\begin{align}
\pmatrix{x\\y\\z}
&= \frac t{abc-a-b-c-2}\pmatrix{b+c+bc+1\\a+c+ac+1\\a+b+ab+1}
\\
&= \frac t{abc-a-b-c-2}\pmatrix{(b+1)(c+1)\\(a+1)(c+1)\\(a+1)(b+1)}
.\end{align}$$


问:
Show that vector field is $C^\infty$ and calculate its Jacobian matrix at $(0,1,1)$
From an introductory course on calculus on manifolds:

Show that $f : U \subset \Bbb R^3 \to \Bbb R^2$ given by $$f(x,y,z) = \begin{bmatrix} \sin(x+z) + \log(y z^2) \\ e^{x+z} + y z \end{bmatrix}$$ is $C^\infty$ and calculate its Jacobian matrix at $(0,1,1)$ where $$U = \left\{ (x, y, z) \in \Bbb R^3 \mid y, z > 0 \right\}$$

I already showed that $f$ is continuous and that it is $C^1$, but I don't really get how to use induction from here. Any help?

答: Hint: Only noticing that all functions involved are well defined and $C^\infty$, should be enough. For the jacobian you need to find the partial derivatives and form the following matrix evaluated at the given point $$J(0,1,1)=\left[\begin{array}{ccc} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \end{array} \right]_{(0,1,1)} $$ where $u=\sin(x+z)+\log(yz^2)$ and $v=e^{x+z} +yz$.


问:
Comparing $2^n$ to $n!$
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?

答:
\begin{align*}
2^n &= 2  \times  2  \times  \dots \times 2  \times  2 \times 2 \\
n! &= 1 \times  2  \times  \dots  \times  (n-2)  \times  (n-1) \times n
\end{align*}

Both of these expressions are $n$ numbers multiplied together. So for $n < 4$, $2^n > n!$ because the 1 in $n!$ adds nothing. However, for $n \geq 4$, $n!$ will always be larger than $2^n$, because the 4 in $4!$ is $2 \times 2$ so it compensates for the extra 2 in $2^n$, and all other terms $3,5,6,7,\dots$ are all larger than $2$.

Title: Comparing $2^n$ to $n!$


问:
Number of necklaces given 3 beads that can take 4 colours
Find thee number of necklaces given 3 beads that can take 4 colours, two necklaces are considered the same if the colours can be matched by rotation or flipping them.

My approach is by using Burnside's lemma, and so I consider the group $D_3$. To calculate the identity $|F(r^0)|$ we have 4 colours in 3 possible spot, $|F(r^0)|$=4^3=64. I imagine it as a string with 3 spots before it's combined to form a necklace.

Burnsides lemma: $|X/G|=1/|G|*\sum_{g\in G}|F(g)|$, where $G=$$D_3$, $|X|=|F(r^0)|=64$, $|G|=6$

If $g \in G$ is a flip by one of the three flip-symmetries, two of the beads with $4^2$ choices together will force the color of the third bead. And there are 3 flips i.e $|F(g)|=3*4^2$ for the flips.

If $g \in G$ is a rotation (not identity) there are only 4*2, since both of these rotations only count when all beds are of the same colour.

And so I get: $|X/G|=1/6(4^3+3*4^2+2*4)=20$ distinct necklaces, do you agree?

Thank you in advance!

答:
Yes, that's right.
But you have so few beads that you can do without Burnside's lemma:
the number of necklaces of beads of one color is 4; the number of two-color necklaces is 12; the number of three-color necklaces is 4. The total is 20.</p>


问:
Find all the solutions to the following system of equations.
\begin{cases} x=1.5x(1-x)-0.7xy \\ y=y+2xy-0.6y\\ \end{cases} Where (x,y)= ( , )has the smallest x-value (x,y)=( , ) and (x,y)= ( , ) has the largest x-value.

The furthest I got was the second equation simplifies to y=0 and x=0.3. I plugged y=0 into the first equation and found (x,y)= ($\frac{1}{3}$,0). I tried to plug in the x value into the first equation but the (x,y) I get is incorrect. I tried to find the smallest x and largest x combinations but I cannot figure it out.

答:
The second equation gives two answers: $y = 0$ OR $x = 0.3$, not "and".  so with $y = 0$ the first equation becomes $x = 1.5x(1-x)$ which has $x=0$ as one solution and $x = 1/3$.  So two solutions are $(0,0)$ and $(1/3, 0)$.

If instead you take $x = 0.3$ from the second equation, you get an equation for $y$.  Since $x ≠ 0$ you can divide it out to get $1 = 1.5(0.7) - (0.7)y$ which gives $y = -.05/-0.7$ or $y = 1/14$.  So another solution is $(0.3, 1/14)$.  Those three are all the solutions.


答:
The second equations two solutions are $y=0$ and $x=0.3$ are separate solutions.  Plug either into the first equation to get possible solutions.

For $y=0$, the first equation leads to $x=0$ or $x=\frac{1}{3}$.

For $x=0.3$, the first equation leads to $y=\frac{1}{14}$

To answer the original question $(0,0)$ and $(\frac{1}{3},0)$ have the smallest and largest $x$ values.

Title: Find all the solutions to the following system of equations.

Body:
\begin{cases} x=1.5x(1-x)-0.7xy \\ y=y+2xy-0.6y\\ \end{cases} Where (x,y)= ( , )has the smallest x-value (x,y)=( , ) and (x,y)= ( , ) has the largest x-value.

The furthest I got was the second equation simplifies to y=0 and x=0.3. I plugged y=0 into the first equation and found (x,y)= ($\frac{1}{3}$,0). I tried to plug in the x value into the first equation but the (x,y) I get is incorrect. I tried to find the smallest x and largest x combinations but I cannot figure it out.


答: By induction, $(n+1)!=(n+1) \cdot n! \geq (n+1) \cdot 2^n \geq 2 \cdot 2^n=2^{n+1}$ if $n! \geq 2^n$. Therefore, because $4$ is the smallest integer such that $2^n \leq n!$ is true, you deduce that $n! \geq 2^n$ for all $n \geq 4$.</p>

Title: Comparing $2^n$ to $n!$

Body:
Comparing $\;2^n\;$ and $\;n!\;$

I need to rewrite one or the other so they can be more easily compared (i.e. have similar form). Because of the factorial, I'm a little lost as to how to compare the two functions. Normally, I would take the logarithm when trying to re-express a power like $2^n$, but I don't think this will work, since I don't know how to take the logarithm of a factorial, if at all.

Might the series expansions of either functions be the right approach? If not could, someone point me in the right direction?


问:
Generalized inverse matrix rules

I would like to ask how free you are when you are calculating generalized inverses? I know, that it is said that there are infinite many of them, but we always usually choose submatrix such that elements are picked in square shape so it is easy to put them back in original matrix.

For example would it really be possible to get next matrix and if not why not and if yes how?:
$\begin{bmatrix}1 &0 &0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 1\\0 & 0 & 0 & 0 & 0\end{bmatrix}$

答:
Given a rank-deficient matrix $A$, here is a method to calculate a generalized inverse $A^-$ such that
$$AA^-A = A$$
Eliminating columns/rows $\big($i.e. cols $\#2,4$ and row $\#4\big)$ yields a full-rank reduced matrix
$$A_r=\begin{bmatrix}1 &0 &0 \\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$$
Calculate $A_r^{-1}\,$ then re-insert zero rows/cols $\big($rows $\#2,4$ and col $\#4\big)$ to obtain
$$A^- = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}$$
This process does not necessarily generate the Moore-Penrose inverse $A^+$ (however for this particular matrix it does). See page 9 of these
course notes for some examples.

Title: Generalized inverse matrix rules


问:
Product of symplectic manifold is symplectic
Assume ($M_1,\omega_1 $) and ($M_2,\omega_2)$ are symplectic manifolds. Consider $M_1\times M_2$. I want to show this is also symplectic manifold.

Let $p_1, p_2$ denote the projections. Then what I need is $p_1^*(\omega_1)^n \wedge p_2^*(\omega_2)^n$ is non-zero. I am stuck here. I guess $p_1^*(\omega_1 ^n)$ and $p_2^*(\omega_2 ^n)$ are both nonzero since $\omega_i ^n$ is nonzero by non degeneracy of $\omega_i$ and $p_i$ is submersion. But how do I know $p_1^*(\omega_1)^n \wedge p_2^*(\omega_2)^n$ is non-zero?

答:
If $(M_1,\omega_1)$ and $(M_2,\omega_2)$ are symplectic manifolds then
$$
\omega:=p_1^\ast\omega_1+p_2^\ast\omega_2
$$
is a symplectic form on $M_1\times M_2$.  Then $\omega$ is closed since $d$ commutes with $p_1^\ast$ and $p_2^\ast$ and $\omega_1$ and $\omega_2$ are closed.

To see that $\omega$ is nondegenerate, let $x:=(x_1,x_2)\in M_1\times M_2$ and let $(u_1,u_2)\in T_x(M_1\times M_2)$ be a nonzero element.  WLOG assume $u_1\neq 0$.  Then there exists $v_1\in T_{x_1}M_1$ such that $\omega_1(u_1,v_1)\neq 0$.  So
$$
\omega((u_1,u_2),(v_1,0))=\omega_1(u_1,v_1)+\omega(u_2,0)=\omega_1(u_1,v_1)\neq 0
$$


问:
Expected distance from origin of random walk on Z^d using multivariate CLT
I am trying to prove that $E[D(X_n, X_0)]$ is roughly equals $ \sqrt{n}~$ using Multivariate Central limit theorem. $$$$
So I represent a d-dimensional simple symmetric random walk $~S~$ such that $~S_n = X_1 + ... + X_n$ where $X_k~'s$ are i.i.d random vectors for all $k~ {\geq}~ 0$ which are uniformly distributed on $~~~~~~~~~~~~~~~~~~~~~$ {$\pm{e_j,~ j= 1,...,d}$} where $~e_1,...,e_d$ are the canonical basis of $Z^d$.
Obviously the expected value is zero component-wise, I know $$\frac{S_n}{\sqrt{n}} \rightarrow N(\overrightarrow{0}, \Sigma)$$
here comes my struggle to apply multivariate CLT, and then to arrive to $E(D) \sim \sqrt{n} \\$ . $~~~~~~~~~~~~~~~~~~$Can you help me with that?, Thank you in advance

答:
If I understand your question correctly, the missing ingredient is the continuous mapping theorem. The assertion
$$
\frac{S_n}{\sqrt n}\stackrel d\to Z,\tag1
$$
where $Z$ has multivariate normal $N(\vec0,\Sigma)$ distribution, is already the multivariate CLT, so apply the continuous mapping theorem to (1) using the function $g(t):=\|t\|:=(t_1^2+t_2^2+\cdots + t_d^2)^{1/2}$ (i.e., $g$ is the $d$-dimensional Euclidean norm) to conclude
$$
\left\|\frac{S_n}{\sqrt n}\right\|\stackrel d\to \| Z\|.\tag2$$

The LHS of (2) equals $D/\sqrt n$. This may be enough for you to assert (heuristically) that $E(D)$ is roughly $\sqrt n$. In fact, you can argue further that
$$E\left(\left\|\frac{S_n}{\sqrt n}\right\|\right)\to E(\| Z\|),\tag3
$$
an assertion that does not follow immediately from (2). A rigorous proof of (3) would note that the LHS of (2) has constant variance independent of $n$, and therefore is uniformly bounded in $L^2$, which implies uniform integrability and justifies (3).


问:
If $E\subset Y\subset X$ and $X$ is a metric space then give the example & prove the Theorem stated below.
Suppose $E\subset Y\subset X$, where $X$ is a metric space.

Give an example such that $E$ is closed relative to $Y$, but not closed relative to $X$.

Show that $E$ is closed relative to $Y$ iff $E=Y\cap F$ for some closed set $F$ of $X$.

This question came to me when I was reading Sec. 2.32 (Compact Sets) from Baby Rudin. The questions are not explicitly stated in the book. So if any of them is false, then sorry and provide proof. But I suspect both the questions are true.

答:
2.30 Theorem:  Suppose $E\subset Y\subset X$. $E$ is open relative to $Y$ iff $E=Y\cap G$ for some open subset $G$ of $X$.

Proof of 2:
Suppose $E$ is closed relative to $Y$. Then $Y\setminus E$ is open relative to $Y$ and $Y\setminus E=Y\cap G$ for some open subset $G$ of $X$. $p\in E\Rightarrow p\notin Y\setminus E\Rightarrow p\in Y$ but $p\notin G\Rightarrow p\in Y$ and $p\in X\setminus G$. So $E\subset Y\cap X\setminus G$. Again $q\in Y\cap X\setminus G\Rightarrow q\in Y$  but $q\notin G\Rightarrow q\in Y $ but $q\notin Y\cap G=Y\setminus E\Rightarrow q\in E$. Thus $Y\cap F=E$, where $F=X\setminus G$ is a closed subset of X.

Conversely, let $F$ is closed in $X$ and $E=Y\cap F$. Now $t\in Y\setminus E\Rightarrow t\notin F\Rightarrow V_t\cap F=\emptyset$ for some open ball $V_t$ in $X\Rightarrow (Y\cap V_t)\cap (Y\cap F)=\emptyset\Rightarrow (V_t\cap Y)\cap E=\emptyset$.

So $t$ is not a limit point of $E$ relative to $Y$ and $E$ is closed relative to $Y$.


问:
Find $\binom{80}{40}\bmod 2000$

Find $\displaystyle\binom{80}{40}\bmod 2000$.

So far, I've found that $\displaystyle\binom{80}{40}$ is divisible by $2^2$ and $5^1$, so the answer isn't $0$. Usually, with smaller numbers, I would split the $\bmod 2000$ into $2^4$ and $5^3$, find the answer through brute force for each of those, then use the Chinese Remainder theorem. But $80$ and $40$ are too big for this.

答:
Since $${80 \choose 40} = \frac{80\cdot 79 \cdot 78 \cdot 77 \cdot ...\cdot 41}{40 \cdot 39 \cdot 38 \cdot 37 \cdot ... \cdot 1}$$ we can cancel the terms in the denominator with terms in the numerator leaving a factor of $2$, so $2^{40}$ is a factor of the answer. It only remains to see how many factors of 5 remain.  $75, 65, 55$, and $45$ are all factors, so at least $5^4$ is a factor.  As a result, the expression is definitely a multiple of $2000$, so modulo $2000$ the answer is $0$.

EDIT: A few minutes later I realized my mistake, and returned to find several people pointing it out: only the numbers down to 21 in the denominator cancel.  That still leaves $$\frac{2 \cdot 79 \cdot 2 \cdot 77 \cdot 2 \cdot 75 \cdot...\cdot 2 \cdot 41}{ 20 \cdot 19 \cdot 18 \cdot...\cdot 3 \cdot 2 \cdot 1}$$ which gives $20$ factors of $2$ in the numerator and $10 + 5 + 2 + 1 = 18$ factors of $2$ in the denominator, so only $2^2$ is a factor. For the $5's$, we have $5$ factors in the numerator and $4$ in the denominator, so it is a multiple of $5$ but not $5^2$.

So now we have $$\frac{79 \cdot 77 \cdot 3 \cdot 73 \cdot...\cdot 43 \cdot 41}{ 19 \cdot 9 \cdot 17 \cdot 3 \cdot 7...\cdot 6 \cdot 3}$$. Continuing to cancel factors, we find $3^{8}$ in the numerator and $3^8$ in the denominator, and so forth. We end up computing ${80 \choose 40}$ and then can find the answer by inspection.

Title: Find $\binom{80}{40}\bmod 2000$


问:
Most natural symplectic structure?
Suppose I have 2-dimensional manifold embedded in $\mathbb{R}^3$.
It's clear that the most natural Riemannian metric is the one induced by the
usual inner product.

What about symplectic forms? Is there a canonical symplectic form I can put on this manifold? If not, then I ask for something weaker, an example of a typical form that one would expect to see on such a manifold, and can do explicit computations with.

答: If the 2-dimensional manifold is non-orientable (e.g. Klein bottle), then it will never admit a symplectic structure since the symplectic structure would be a volume form and a manifold is orientable iff it has a volume form.</p>

Title: Most natural symplectic structure?


问:
Is the intersection of a descending chain of finitely generated groups finitely generated
Given a strictly descending chain of finitely generated groups $$ H_1 > H_2 > H_3 > \cdots,$$
is it always the case that $H = \cap_{i=1}^{\infty} H_i$ is finitely generated?

I think the answer is no, but I can't think of a counter-example.

答: Here is a more concrete example. Let $F={\rm gr}(x,y)$ be a free group of rank two and let $F'$ be the commutator subgroup  of F. I think it is clear that $F'$ has infinite rank. Let $H_n$ be a subgroup generated by the set $\{x^{2^n},y^{2^n},F'\}$. Then $H_n$ has a finite index in F and is therefore a finitely generated group for any n and $F=H_0>H_1>\ldots$. However, the intersection of all $H_n$ coincides with $F'$.


问:
What is the boundary at infinity in the hyperboloid model of the hyperbolic plane?
The boundary of the upper half-plane $\mathbb{H}$ model of the hyperbolic plane is the extended real line $\overline{\mathbb{R}}= \mathbb{R} \cup \{ \infty \}$.

What is the boundary of the hyperboloid model of the hyperbolic plane? Is it the positive light cone $L^+$?

答: The boundary is not the light cone $L_+$ itself, but the set of directions of the light cone, or equivalently the projectivization $L_+ / \mathbb R_+ \cong S^{n-1}$.


问:
Are these simple statements about polygons true?
Let $P$ be an $n$-gon with $n \gt 3$. I'm looking for proofs or counterexamples for the following statements:

there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$

there exist consecutive vertices $A,B,C$ of $P$ such that $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$ and the interior angle of $P$ at $B$ is convex

there exist different vertices $X,Y$ of $P$ such that $\overline{XY} \cap \partial P = \{X,Y\}$

For 1. and 2., if there is an offending vertex (must be if any other point is) in the triangle, maybe it could be put in some relation with the considered vertex, forming a finite dag which must have a maximal element, suitable as $B$.

For 3. it's actually easy (see my answer which I'm going to add in a moment) without an additional assumption, so let me also require $\overline{XY}\subset P$. Then, maybe decomposition into polygons (one a subset, one with disjoint interior) with fewer vertices by the external diagonal and applying induction would help?

答:
I believe I have all the solutions now (though there may well be better ones, which of course are still welcome).

Let's start with proving the version of 3. from my previous answer: in any $n$-gon $P$ with $n>3$ there exist different vertices $X,Y$ of $P$ such that $\overline{XY} = \overline{XY} \cap P^\mathrm{o} \cup \{X,Y\}$.

Let $T$ be a vertex of $P$ at which the inner angle of $P$ is convex and $R,S$ be its neighbours. If $\overline{RS}$ satisfies the conclusion, we're done. Otherwise, take $T$ as $X$ and let's look for a suitable $Y$. If there is at least one vertex of $P$ on $\overline{RS}$ other than $R,S$, but none in $\triangle RST ^\mathrm{o}$, take any such vertex as $Y$. Otherwise, let $C$ be the convex hull of $(\triangle RST ^\mathrm{o} \cap \partial P)^⎯$. $C$ is a convex polygon.

Let $N$ be the point of $C$ nearest to $T$. If $N$ is a vertex of $P$, take $N$ as $Y$. Otherwise, $N$ is a boundary point of $C$ but not its vertex or one of its two vertices lying on $\overline{RS}$. Then take as $Y$ any end (there are 1 or 2) of a side of $C$ containing $N$ which belongs to $\triangle RST ^\mathrm{o}$ (let's call the straight line containing this side $d$). It is surely a vertex of $P$ and, given that $C$ is convex and contains a point (e.g. a vertex of $C$ lying on $RS$ but not on $d$) on the other side of $d$ than $T$, $C$ is contained in a half-plane to which none of the points of $\overline{TY}$ belongs except $Y$.

Since 2. is stronger than 1., let's turn to it next. I'll prove a strengthened statement: if $P$ is a polygon and $U,V$ are some adjacent vertices of $P$, there exist consecutive vertices $A,B,C$ of $P$ such that $B \notin \{U,V\}$, $\triangle ABC \cap \partial P = \overline{AB} \cup \overline{BC}$ and the interior angle of $P$ at $B$ is convex.

If $P$ is a triangle, take as $B$ the vertex which is neither $U$, nor $V$. Otherwise, $P$ must have more than 3 vertices, so the previous theorem applies. Let $(X,Y)$ be a pair of points whose existence it states. $\overline{XY}$ divides $P$ into two polygons with fewer vertices, one of which (call it $R$) hasn't got $\overline{UV}$ – the forbidden segment – as a side. The result follows by mathematical induction from considering $R$ with $\overline{XY}$ as the forbidden segment.

Additionally, observe that a quadrilateral has at least two such sequences $A,B,C$ of adjacent segments, disregarding permutations of $\{A,C\}$. The above proof can probably be slightly restructured to improve the conclusion and also find two in any $P$.

Edit

Indeed it can, albeit not slightly. Forget about forbidden segments and let's find $\{A,B,C\}$ with desired properties in both polygons resulting from the division. For a polygon thus obtained, if it's a triangle, the case is trivial, otherwise, start by determining whether there exists some $Z$ that the points $X,Y,Z$ or $Z,X,Y$ satisfy the conclusion. If so, cut $\triangle XYZ$ off. Continue cutting off with the new side (previously a diagonal) in the role of $\overline{XY}$ until it becomes impossible or only a triangle remains. Then invoke the inductive hypothesis. The cutting process ensures that $\triangle ABC$ obtained in the inductive step does share two sides with the original $P$.


问:
How to prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$ diverges?

Let $u_n>u_{n+1}>0$ for all $n \in \Bbb N$, and suppose that $u_2+u_4+u_8+u_{16}+\dots$ diverges. Prove that $\sum_{n=1}^{\infty}\frac{u_n}{n}$diverges.

Please provide me a hint or a full solution.

Body: Suppose $ u_n = (\log n)^{-1}$


问:
How to solve a cauchy euler equation with multiple variables?
For this question I'm having trouble solving this equation. Here is what I have so far. Can anyone please help me out?

$ u'' + 2ku' + k^2u = e^{-kt} lnt$

$ u = k^m$

$ u' = mk^{m-1}$

$ u'' = m(m-1)k^{m-2}$

$ m(m-1)k^{m-2} + 2kmk^{m-1} + k^2k^m = 0$

答:
$$u'' + 2ku' + k^2u = e^{-kt} \ln t$$
This is not a Cauchy-Euler's differential equation. I suppose $k \ne 0$:
$$e^{kt}(u'' + 2ku' + k^2u) =  \ln t$$
$$e^{kt}(u'' + ku')+ke^{kt}(u'+ ku) =  \ln t$$
$$(e^{kt}u')'+k(e^{kt}u)' =  \ln t$$
$$(e^{kt}u'+ke^{kt}u)' =  \ln t$$
$$(e^{kt}u)'' =  \ln t$$
Integrate twice.

Note that:
$$\int \ln t dt=t \ln t -t+C$$


问:
thom space associated to a fibration
let $p:S\rightarrow B$ be a fibration wiht fiber a rational $r$-homology sphere
$\Sigma^r$, i.e., $H_*(\Sigma^r;\mathbb Q)=H_*(S^r;\mathbb Q)$.
to such a fibration we associate its Thom space
$$MS=(S\times I \cup_p B)/S\times\{1\}$$
where we consider $p$ as a map $S\times\{0\}\rightarrow B$.

1) isn't $MS $ the same as the mapping cone of $p$?

2) how does this relate to the special case of the thom space of a vector bundle?

3) to each $b\in B$ we define
$$M\Sigma^r_b:=\Sigma^r_b\times I /(\Sigma^r_b\times \{0\}\cup \Sigma^r_b\times \{1\})$$
what is $M\Sigma^r_b$ and how can we interpret it?

答:
1) Yes, you can define Thom spaces as mapping cones.

2) Certainly if your sphere bundle comes from a vector bundle then you can construct the Thom space in either way.  I don't know much about rational homology spheres, but of course there are sphere bundles (and hence rational sphere bundles) that do not come from vector bundles.  This is easy to state in the language of classifying spaces (or equivalently in the language of transition functions, if you happen to be working with actual fiber bundles): the obvious inclusion $O(n)\rightarrow Homeo(S^{n-1})$ is a homotopy equivalence iff $n\leq 3$ (I think, maybe Homeo should be Diffeo; in any case, the point still stands).

3) It looks like this is supposed to be the "restriction" of the Thom space over the point $b$.  I put "restriction" in quotes since the Thom space no longer has a projection to the base space, since there's nowhere to put the new base point.  Except when you write the gluing relation as you have, this is usually interpreted to mean that all the points in $\Sigma^r_b \times \partial I$ are identified, whereas I think you just want to crunch $\Sigma^r_b \times \{0\}$ to a point and $\Sigma^r_b \times \{1\}$ to a point.  You could write this more suggestively (w/r/t your question 1) as $(\Sigma^r_b \times I\cup_{p|_{p^{-1}(b)}}\{b\})/(\Sigma^r_b \times \{1\})$.

Just as a little pointer, from what I've seen, perhaps the most interesting thing about Thom spaces is the Thom isomorphism theorem.  Note that when you take the Thom space of a trivial vector bundle and apply the isomorphism theorem, you get back the usual suspension isomorphism!  So you can think of the Thom isomorphism as being a generalization: all Thom spaces have cohomology groups isomorphic to those of the base but shifted by the dimension of the fiber.  Of course, different bundles have different Thom classes (not that they could necessarily be compared anyways, since they live in different places), so it's not like the Thom isomorphism theorem is saying that singular cohomology is blind to twistedness...

Title: thom space associated to a fibration


问:
Major doubt in Mathematical Induction
I am doing problems in Induction. What i know is:

First we need to test Base case, that is $P(n_0)$ should be verified.

Second we need to assume that $P(k)$ is True which is the inductive Hypothesis

Third we need to prove $P(k+1)$ is True using the above Hypothesis which gives the conclusion.

But the below problem made me to fall in a confusion.

When we assume $P(k)$ is True, does it mean all of $P(1),P(2),...P(k-1)$ are also True?

Here is the problem:

Prove using induction that $10^n-(5+\sqrt{17})^n-(5-\sqrt{17})^n$ is divisible by $2^{n+1}, \:\forall n \in N$

My try:

Obviously $P(1)$ is True.

I assumed that $P(k)$ is True:
So we have:

$$10^{k}-(5+\sqrt{17})^{k}-(5-\sqrt{17})^{k}=2^{k+1} Q \to (1)$$
for some $Q \in N$

Now let
$$P(k+1)=10^{k+1}-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1}$$
Using $(1)$ we get

$$\begin{array}{l}
P(k+1)=10\left(10^{k}\right)-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1} \\
=10\left(2^{k+1} Q+(5+\sqrt{17})^{k}+(5-\sqrt{17})^{k}\right)-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1} \\
=5\left(2^{k+2}\right) Q+(5+\sqrt{17})^{k}(5-\sqrt{17})+(5-\sqrt{17})^{k}(5+\sqrt{17})\\
=5(2^{k+2})Q+(5+\sqrt{17})^{k-1}(8)+(5-\sqrt{17})^{k-1}(8)\\
=5(2^{k+2})Q+8P(k-1)
\end{array}$$

Now according to my book, the author has taken $P(k-1)$ as True and proved the result.
So my question is:
When we assume $P(k)$ is True, does it mean all of $P(1),P(2),...P(k-1)$ are also True?

答:
It is not true that $P(k)$ being true implies $P(i)$ is true for $i<k$. A usual induction proof does the following after the base case:
$$P(k) \implies P(k+1)$$

But, the technique the author has used is often referred to as strong induction. It works according to the following:
$$P(i) \space (\forall \space i \leqslant k) \implies P(k+1)$$
It can be used when you need more than just the previous case to solve a problem. The reason why it works is similar to the working of induction.

Essentially, you assume that all the previous cases are true, and use some (or all) of them to show that the next case is true. Now, the case you have proved becomes part of the list of initial cases you have proved and so on (similar to the domino effect of induction).

Title: Major doubt in Mathematical Induction


问:
Explicit form for a difficult sequence
Let us define infinite sequence $x_n$ as such: the first term of the sequence is $x_1=1$. Then, the next $x_1$ terms are 2. The next term is 1. The next $x_2$ terms are 2. The next term is 1. The next $x_3$ terms are 2. The next term is 1. And so and so forth...

Can anyone provide explicit formula for the $n$th term of this sequence? Or please provide any insight on any patterns that may exist.

For reference, here is how the sequence goes for the first several terms:

1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 1 2 2 1 2 2 1 2 1 2 2 1 2 2 1

答:
This is $A001468$ in the OEIS.
The formula is:

$$a_n = \lfloor(n+1)\varphi\rfloor - \lfloor n\varphi\rfloor,$$

where $\varphi = \frac{1+\sqrt5}2$ is the golden ratio.

Title: Explicit form for a difficult sequence


问:
Filter generated by a collection of basic neighborhoods
Exercise:

Let $x$ be an element of a topological space $X$, and let $B_x$ be a collection of basic neighborhoods of $x$. Prove that the filter on $X$ generated by $B_x$ is the collection of neighborhoods of $x$.

Proof: First we want to show that $B_x$ is non empty.

Since $x$ is an element of $U$ for all $U \in B_x$ , $B_x$ is non empty.

Next we want to show that if $U \in B_x$ with $U \subseteq V$, then $V \in B_x$.

Let $U \subseteq V$ be an element of $B_x$. Since $U \in B_x$, $x$ is an element of $U \subseteq V$. Hence $x \in V$ and thus $V \in B_x$.

Lastly we will show that for $U, V \in B_x$, $U \cap V \in B_x$.

If $U$ and $V$ are elements of $B_x$,  then x is an element of $U$ and $V$. Hence $x \in U \cap V$ and thus $U \cap V \in B_x$.

Therefore $B_x$ is a filter on X.

Is this a sufficient proof?

答:
You cannot show that $B_x$ is non-empty: that should have been part of the hypothesis. The specific problem with the argument that you used is that the fact that $x\in U$ for all $U\in B_x$ does not imply that $B_x\ne\varnothing$. If $B_x=\varnothing$, then it is vacuously true that if $U\in B_x$, then $x\in U$: there is no $U\in B_x$, so there is certainly no $U\in B_x$ such that $x\notin U$.

You are trying to prove the wrong thing, something that in general simply isn’t true: a local nbhd base $B_x$ at $x$ is not in general a filter. It is not in general true that if $U\in B_x$, and $U\subseteq V$, then $V\in B_x$. Suppose, for instance, that $X$ is the real line with the usual topology, $x=0$, and

$$B_0=\{(-q,q):q\in\Bbb Q^+\}\,;$$

$B_0$ is a local base at $0$. $(-1,1)\in B_0$, and $(-1,1)\subseteq(-\pi,\pi)$, but $(\pi,\pi)\notin B_x$, since $\pi$ is not rational. This $B_0$ is not a filter.

The filter in this question is the one generated by $B_x$, which in general is a superset of $B_x$. If $\mathscr{F}$ is the filter generated by $B_x$, and $\mathscr{N}$ is the collection of all nbhds of $x$, you need to show that $\mathscr{F}=\mathscr{N}$. Thus, you need to know what the filter generated by $B_x$ is. Let

$$\mathscr{F}_0=\left\{\bigcap\mathscr{A}:\mathscr{A}\text{ is a finite subset of }B_x\right\}\,,$$

and let $\mathscr{F}=\{N\subseteq X:\exists A\in\mathscr{F}_0(A\subseteq N)\}$; by definition $\mathscr{F}$ is the filter generated by $B_x$, provided that $\mathscr{F}$ actually is a filter. In words, a subset $N$ of $X$ belongs to $\mathscr{F}$ if and only if $N$ contains the intersection of finitely many members of $B_x$.

To show that $\mathscr{F}$ is a filter on $X$, you have to check that $\varnothing\notin\mathscr{F}$, that if $N\in\mathscr{F}$ and $N\subseteq M$, then $M\in\mathscr{F}$, and that if $M,N\in\mathscr{F}$, then $M\cap N\in\mathscr{F}$. The second of these follows very easily from the definition of $\mathscr{F}$ from $\mathscr{F}_0$; for the first and third you will need to use the hypothesis that $B_x$ is a local nbhd base at $x$.

To prove that $\mathscr{F}=\mathscr{N}$, you need to do two things.

You need to show that if $N$ is any nbhd of $x$, there is a finite $\mathscr{A}\subseteq B_x$ such that $\bigcap\mathscr{A}\subseteq N$; this will show that $N\in\mathscr{F}$ and hence that $\mathscr{N}\subseteq\mathscr{F}$.

You need to show that if $N\in\mathscr{F}$, then $N$ is a nbhd of $x$, meaning that there is a $V\in B_x$ such that $V\subseteq N$. This will show that $\mathscr{F}\subseteq\mathscr{N}$.


问:
Super-level sets of Hardy–Littlewood maximal function are open?
I am working on the book Measure, Integration and Real Analysis by Sheldon Axler. I am stuck on Problem 9 of Section 4A. For $h: \mathbb{R} \to \mathbb{R}$ Lebesgue measurable, $h^*$ is defined as follows:

$$ h^*(x)=\sup_{r>0}\frac{1}{|B(x,r)|} \int_{B(x,r)} |h(y)|dy.$$

Suppose $h: \mathbb{R} \to \mathbb{R}$ is Lebesgue measurable. Prove that
$$\left\{ b \in  \mathbb{R} : h^{\ast}(b) > c\right\}$$
is an open subset of $\mathbb{R}$ for every $c \in \mathbb{R}$.

My Idea is like this: Let $A=\left\{ b \in  \mathbb{R} : h^{\ast}(b) > c\right\}$. Pick $b \in A.$ I want to show that there is a ball $B(b,\epsilon)$ contained inside of $A$. But I don't know how to get to this.
Any help?

答:
Fix $c$. Let $b$ be such that $\sup_r \frac1{|B(b,r)|}\int_{B(b,r)}|h|>c$. Then there exists $r>0$ and $\alpha>1$  such that
$$\frac1{|B(b,r)|}\int_{B(b,r)}|h|> \alpha c.$$
Let $\epsilon>0$ and $b'\in B(b,\epsilon)$. Then
$$ B(b',r+\epsilon)\supset B(b,r)$$
and
$$c< \frac1{\alpha|B(b,r)|}\int_{B(b,r)}|h| \le \frac1{\alpha|B(b,r)|}\int_{B(b',r+\epsilon)}|h| = \frac{|B(b',r+\epsilon)|}{\alpha |B(b,r)|}\frac1{|B(b',r+\epsilon)|}\int_{B(b',r+\epsilon)}|h|$$
since (independently of $b$ and $b'$)
$$ \frac{|B(b',r+\epsilon)|}{\alpha |B(b,r)|}\xrightarrow[\epsilon\to 0]{} \frac1\alpha < 1$$
there exists $\epsilon>0$ sufficiently small such that for all $b'\in B(b,\epsilon)$,
$$c< \frac1{|B(b',r+\epsilon)|}\int_{B(b',r+\epsilon)}|h|$$
and hence $c < h^*(b')$ as needed.


问:
Radius of Convergence of $(1-x)^{1/4}$
This is my first post so I am very new to MathJax formatting - i apologize in advance for the messy formatting

This equation becomes $$\sum_{n=2}^\infty \frac{(4n-5) x^n}{(4^n)n!}$$

The textbook says the ratio test would be:

$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(4n-5)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)n!}\right\rvert$$

But I thought it would be:
$$\lim_{n\to \infty} \left\lvert\frac{(4n-1)(x^{n+1})(4^n)n!}{(4n-5)(x^n)(4^{n+1})(n+1)!}\right\rvert$$

So no extra $(4n-5)$ - but that would result in an addition $(n+1)$ in denominator (because of the expanded factorial) and the radius of convergence would now be infinite (and not $|x| < 1$ as above fraction suggests).

Thanks

答: Your series as written obviously has an infinite radius of converges, since the denominator grows much faster than the numerator. However, this is because there is a mistake - the numerator should be $(4n-5)!.$

Title: Radius of Convergence of $(1-x)^{1/4}$


问:
Find the points on the graph of $f(x) = 12(x + 9) − (x + 9)^3$ where the tangent line is horizontal.
I cannot figure out how to get started on this question. Would I First simplify, and then take the derivative? Please help! </p>

答:
Notice that $x+9$ appears twice in $f(x)$, which strongly suggests we should consider $f(x - 9) = 12x - x^3$, which is the graph of $f(x)$ translated right by $9$ units.

Thus $f'(x-9) = 0 \Rightarrow 12 - 3x^2 = 0 \Rightarrow x = ±2$, so the critical points are $(2, 16)$ and $(-2,-16)$. Now translate left by $9$ units to find the critical points of the original function: $(-7, 16)$ and $(-11, -16)$.


问:
Defining a mapping from a set to a set
How to define a map that takes a finite binary string of length $L$ to a real number in $[a,b]$
where length $L$ is the number of bits constituting the string

$x=0101$ is a binary string of length $L=4$

I am looking for something simple ...

thanks

答: A binary string $S$ of length at most $L$, if interpreted as a binary number, satisfies $0\leq S \leq 2^L-1$.  So you can just map the interval $[0, 2^L-1]$ linearly into $[a,b]$.  $$S\mapsto a+(b-a)\frac S{2^L-1}$$

Title: Defining a mapping from a set to a set


问:
For the Lagrange Multiplier Theorem, why must the number of constraints be less than the number of variables?
In Fitzpatrick's "Advanced Calculus", the beginning of the General Lagrange Multiplier Theorem is given by:

Let $U$ be an open subset of $\mathbb{R}^n$ and suppose that $f: U \rightarrow \mathbb{R}$ is continuously differentiable. Let $k$ be a positive integer less than $n$ and suppose that the mapping $G: U \rightarrow \mathbb{R}^k$ is also continuously differentiable."

The function $G$ is a vector valued function, where each component function is a constraint. Why must $k < n$, that is, why must the number of constraints be less than the number of variables?

答:
I guess here the constraints are equalities given by $G(x)=0$. As a concrete example to @amsmath's answer, just think about the case where $G$ is a linear mapping, then the constraints are $k$ linear equations in $n$ variables. If $k>n$, then some constraints are linearly dependent, therefore, are redundant and can be ignored (of course when the problem is feasible).

The method of Lagrangian multiplier also applies to the inequality case, where the constraints involve inequalities such as $G(x)\le 0$, then $k$ definitely does not have to be less than $n$. See, for example, the KKT condition in nonlinear optimization https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions


问:
Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$

Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$?

So I started by saying that by the geometric series test where $a=x^2$ and $|r| = |\frac{1}{e^x}| \leq 1$, the series converges pointwise.

But how do I exactly prove that it converges uniformly? I am quite sure it is by weistrass test but I can not find an upper bound to compare it to! Any direction would be appreciated!

答:
Let
$$f_n(x)=x^2e^{-nx}$$
then we have
$$f'_n(x)=e^{-nx}\left(2x-nx^2\right)=0\iff x=0\ \text{or}\ x=\frac{2}{n}$$
so
$$||f_n||_\infty=f_n\left(\frac{2}{n}\right)=\frac{4}{n^2}e^{-2}$$
hence the series $\displaystyle \sum_{n=1}^\infty ||f_n||_\infty$ is convergent and then the series  $\displaystyle \sum_{n=1}^\infty f_n$ is uniformly convergent.</p>


问:
Show that $F_g$ is $C^1$
Given:

$g: U ⊆ E → ℝ$, of class $C^{k+1}$ (where $k ≥ 0$ is a non-negative integer);

$E$ a real normed space;

$U$ open in $E$.

$T = {α ∈ C^k([0, 1], E) : α[0, 1] ⊆ U}$.

Show that, for $k=0$:

$F_g : α ∈ T → g ◦ α ∈ C([0, 1], ℝ)$ is $C^1$.

Book exercise on Introduction to Manifolds. Just not clicking for me.

I do know that composition of $C^k$ functions is itself $C^k$, it isn't related to that, is it? As for $k=0$ we would have a $g$ $C^1$ and α $C^0$... Any tips?

答: You can do it from first principles, but the statement is wrong: If $\alpha \in C^0,$ then $F_g$ may not be in $C^1$ (make $E=\mathbb{R},$ then set $g(x) = x$ and $\alpha$ is a $C^0$ function which is not $C^1.$ If the $C^k$ is meant to be $C^{k+1}$ then the proof follows by using the standard $\epsilon, \delta$ argument with the definition of derivative.

Title: Show that $F_g$ is $C^1$


问:
Where is my error in this combinations word problem?

A student has to answer $10$ out of $13$ questions in an examination. The number of ways in which he can answer if he must answer at least $3$ of the first five questions is:

The answer is not $^5C_3 \times {^{10}C_7}$. Why?????

$C$ here is combination.

答:
This is similar to the Textbook question in India if you recollect. First, you divide the number of questions into two categories- the first $5$ questions and the remaining $8$ questions.

Now pay attention to the words of the question. it says

" he must answer atleast 3 questions..."

and of course

student has to answer 10 questions

So the student may any $n$ number of questions among the first 5 and any $m$ number of questions among the last 8. The only conditions given are

n is greater than 3 and at most 5

$m+n=10$

You can work out the other number of possibilities by yourself and add them up

(this was supposed to be a comment)


问:
A sufficient and necessary condition of Taylor series
Let $f(x)$ be a $C^{\infty}$ function on $(-R,R)$. Prove that
$f(x)$ can be expanded as its Taylor series at the point $x=0$ over the interval $(-R,R)$ if and only if there exists a positive function $M(r)$ $(0 \le r < R)$, such that
$$
|f^{(n)}(x)| \le \frac{M(r)n!}{(r-|x|)^{n+1}}
$$
for $n \ge 0$ and $|x| < r < R$.

Letˊs try the sufficiency.Suppose $f(x)$ can be expanded as itˊs Taylor series at $x=0$, that is $$f(y)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}y^n$$
, the series converges uniformly and absolutely for $|y-x| \lt R-|x|$, hence $$f(y)=\sum_{n=0}^\infty \frac{f^{(n)}(x)}{n!}(y-x)^n$$ converges uniformly and absolutely for $|y-x|<R-|x|$. Then
$$ |f^{(n)}(x)| \le \frac{n!}{|r-x|^{n+1}}\sum_{k \ne n}\frac{|f^{(k)}(r)|}{k!}|r-x|^{k+1} \le \frac{M(r)n!}{(r-|x|)^{n+1}}$$, where
$$ M(r):=\sum_{k \ne n}\frac{|f^{(k)}(r)|}{k!}|r-x|^{k+1} $$.

But the necessity seems hard to prove.

答:
It is suffice to show $\forall r'>0$, the Taylor series converges to $f(r')$. Find $r$ such that $0 <r'<r<R$. The intergral remainder $$T(n,r')=\int_{0}^{r'}(r'-t)^{n}\frac{f^{n+1}(t)}{n!}dt$$
Together with the inequality $|f^{(n+1)}(t)| \le \frac{M(r)(n+1)!}{(r-|t|)^{n+2}}$,
$$|T(n,r')|\leq(n+1)M(r)\int_{0}^{r'}\frac{(r'-t)^n}{(r-t)^{n+2}}dt\leq\frac{M(r)(n+1)}{(1+p)^n}C, p>0$$
Thus $$\lim_{n\to\infty}T(n,r')=0$$


问:
Verification of solution: $\lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) < \infty$

$A_1$, $A_2$ .... $\subset \mathbb{R}^n$ given through $$ A_k := \{
 (x,y)\in \mathbb{R}^n:\text{  } k\leq x^2+y^2 \leq k+\frac{1}{2k^2}\}
 $$ Show:

a) $\bigcup\limits_{k=1}^\infty A_k$ is Borel-measurable

b) $\lambda \left(\bigcup\limits_{k=1}^\infty A_k\right) < \infty $

I do understand how a) works.
But b) is a bit weird in my opinion.

Edit: $\lambda$ is Lebesgue measure.

My solution:

I defined $r:= x^2+y^2$. Then $r\in [k,k+\frac{1}{2k^2}]$ for $A_k$. I also see $A_k \cap A_{k+1} = \varnothing$.
Then it is $$ \lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) \leq \sum\limits_{k=1}^\infty \lambda(A_k) = \sum\limits_{k=1}^\infty \left(k+\frac{1}{2k^2}-k\right) = \sum\limits_{k=1}^\infty \frac{1}{2k^2} < \infty $$

Official solution:

In the offical solution they use open balls $\overline{B_{r_k}}$ and $B_{\rho_k}$ with $r_k = \sqrt{k+\frac{1}{2k^2}}$ and $\rho_k = \sqrt{k}$ $\Rightarrow A_k = \overline{B_{r_k}}\setminus B_{\rho_k}$. Then $\sum\limits_{k=1}^\infty \lambda(A_k) \leq \sum\limits_{k=1}^\infty \frac{\pi}{2k^2} < \infty$

My question:

Is my way to prove this also right? And if it isn't: Why? I understand the official solution but it didnt come to my mind at first and I am unsure if this is the only right way to prove b).

答:
Let start defining $c = \lambda (\{(x,y):  x^{2}+y^{2} \leq 1\}$ (it can be proved that $c= \pi$, but we don't need it for this solution). Let us prove the following lemma:

Lemma:

If $s \geqslant 0$, then $\lambda (\{(x,y):  x^{2}+y^{2} \leq s\}= cs$.

If $0 \leqslant r \leqslant R$, then $\lambda (\{(x,y): r \leq x^{2}+y^{2} \leq R\})= c(R-r)$.

Proof:  Item 1.

\begin{align*}
\lambda (\{(x,y):  x^{2}+y^{2} \leq s\} & = \lambda (\{(\sqrt{s}x,\sqrt{s}y):  x^{2}+y^{2} \leq 1 \}= \\
& = \lambda (\sqrt{s}\{(x,y):  x^{2}+y^{2} \leq 1 \}= \\
& =(\sqrt{s})^2  \lambda (\{(x,y):  x^{2}+y^{2} \leq 1\}= \\
& = sc=cs
\end{align*}
where we used that, for any $E$ measurable and $\alpha \geqslant 0$, $\lambda(\alpha E) = \alpha^2 \lambda(E)$.

Item 2.
Using the fact that  $\lambda (\{(x,y): x^{2}+y^{2} = r\})=0$, we have
\begin{align*}
\lambda (\{(x,y): r \leq x^{2}+y^{2} \leq R\}) &= \lambda (\{(x,y): x^{2}+y^{2} \leq R\}) - \lambda (\{(x,y):  x^{2}+y^{2} < r\}) =\\
&= \lambda (\{(x,y): x^{2}+y^{2} \leq R\}) - \lambda (\{(x,y):  x^{2}+y^{2} \leq r\}) =\\
&= cR-cr = c(R-r)
\end{align*}
This completes the proof of the lemma. $\square$

Now, using the lemma above and since, for all $i , j \in \{ 1, \dots, \} $, $A_i$ and $A_j$  are disjoint,
$$ \lambda\left(\bigcup\limits_{k=1}^\infty A_k\right) \leq \sum\limits_{k=1}^\infty \lambda(A_k) = \sum\limits_{k=1}^\infty c\left(k+\frac{1}{2k^2}-k\right) = \sum\limits_{k=1}^\infty \frac{c}{2k^2} < \infty $$

Remark: As you see, your answer is almost correct, it is missing some details and a factor $c$ (whose actual value is $\pi$).


问:
Eliminate $\alpha,\beta,\gamma$ from the system of equations
Eliminate $\alpha,\beta,\gamma$ from the following system of equations.

$$a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma)=0$$
$$a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma)=0$$
$$a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma)=0$$

My try:

Squaring and adding first two equations, we get:

$$\cos(\alpha-\beta)=\frac{c^2-a^2-b^2}{2ab}$$

$$\cos(\gamma-\beta)=\frac{a^2-c^2-b^2}{2cb}$$

$$\cos(\alpha-\gamma)=\frac{b^2-a^2-c^2}{2ac}$$

Now the RHS of above all looks like negative cosines of triangle $\Delta ABC$.

But I am not sure whether it will help.

This question is taken from plane trigonometry part 1 by SL Loney book. Page number 264, question number 176.

答:
We have the following system of equations
\begin{align}
a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma) &= 0, \tag{1}\\
a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma) &= 0, \tag{2}\\
a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma) &= 0. \tag{3}
\end{align}

From (1) and (2), we have
$$a^2 + b^2 + 2ab \cos \alpha\cos \beta + 2ab \sin \alpha \sin \beta = c^2$$
and thus
$$(2ab)^2\sin^2 \alpha \sin^2\beta = (c^2 - a^2 - b^2 - 2ab\cos \alpha\cos \beta)^2. \tag{4}$$

From (1) and (3), we have
$$a^2 + b^2 + ab\frac{\cos \alpha}{\cos \beta} + ab \frac{\cos \beta}{\cos \alpha} = c^2$$
and thus
$$(2ab)^2(\cos^2 \alpha + \cos^2\beta) = 2\cdot (c^2 - a^2 - b^2)\cdot 2ab \cos \alpha \cos \beta. \tag{5}$$

$(4) + (5)$ gives
$$(2ab)^2(1 + \cos^2\alpha \cos^2\beta ) = (c^2 - a^2 - b^2)^2 + (2ab)^2 \cos^2 \alpha \cos^2 \beta$$
and thus
$$(2ab)^2 = (c^2 - a^2 - b^2)^2$$
which can be written as
$$- a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 = 0$$
or
$$(a + b + c)(a + b - c)(b + c - a)(c + a - b) = 0.$$

We are done.


答:
Welcome to MSE!

Given the first five questions:
$ 1_q \:\: 2_q \:\: 3_q \:\: 4_q \:\: 5_q $

these are all the $10$ possibilities of choosing $3$ out of the $5$:

$\{1_q,2_q,3_q\}, \{1_q,2_q,4_q\}, \{1_q,2_q,5_q\} , \{1_q,3_q,4_q\} ,
\{1_q, 3_q, 5_q \}, \{ 1_q, 4_q, 5_q \} $

$\{2_q, 3_q, 4_q\} , \{2_q, 3_q, 5_q \} , \{2_q , 4_q, 5_q\}$

$\{3_q,4_q,5_q\}$

say the student answered the first $3$ out of the $5$ : $\{1_q,2_q,3_q\}$

‘then there are still’ $10$ questions left to choose from $\{4_q,5_q,...,13_q\}$ and the student answered the $7$ questions $4_q, 6_q, 7_q, 8_q, 9_q,10_q, 11_q$,

now say the student answered questions $1_q,3_q,4_q$ out of the $5$ first,
and then answered the $7$ more questions $2_q, 6_q, 7_q, 8_q, 9_q,10_q,11_q$,

in essence the student answered the same set of questions $\{1_q,2_q,3_q,4_q,6_q,7_q,8_q,9_q,10_q,11_q\}=\{1_q,3_q,4_q,2_q,6_q,7_q,8_q,9_q,10_q,11_q\}$
but just in a different order, but that is double counting nonetheless.

The key here is at least $3$ of the first $5$ questions, the student must answer $3$, any $3$, but they may answer more, and you could want to break it up into exactitudes, the student answers exactly $3$ of the first $5$, exactly $4$ of the first $5$, all $5$ of the first $5$ to avoid overlapping/double-counting.

Say the student answers exactly $3$ of the first five questions:

$$\binom{5}{3}\binom{8}{7}$$

The student may also answer exactly $4$ of the first five questions:

$$\binom{5}{4}\binom{8}{6}$$

and the student may also answer all $5$ of the first five questions:

$$\binom{5}{5}\binom{8}{5}$$

all possibilities is the sum of all these $\binom{5}{3}\binom{8}{7} +\binom{5}{4}\binom{8}{6}+\binom{5}{5}\binom{8}{5}=276$.

Title: Where is my error in this combinations word problem?

Body:
A student has to answer $10$ out of $13$ questions in an examination. The number of ways in which he can answer if he must answer at least $3$ of the first five questions is:

The answer is not $^5C_3 \times {^{10}C_7}$. Why?????

$C$ here is combination.


问:
Why is this the trick to calculate $m\, 9$s $\times n\, 9$s?
Here is what observed for the results of $m\, 9$s $\times n\, 9$s ($m \leq n$)

$9\times9=81, 9\times99=891, 9\times999=8991$

$99\times99=9801, 99\times 999=98901, 99\times9999=989901$

$999\times999=998001,999\times 9999=9989001, 999\times99999=99899001$

Summarize the following trick  to get the result fast:

For example to get $999\times99999 =99899001$

Write down $n\,$$9$s $(n=5)$ first : $\to 99999$

Replace the $m\,$th $(m=3) 9\,$s with $8: \to 99899$

Attach $ (m-1)\, 0$s $(m-1=2): \to 99899 00$

Attach last  $1: \to  99899 00 1$

My question is how to prove this is true with algebra?

答:
This is an answer to a question that includes your question.

Just as you can cast out $9$s, you can also cast out $99$s, $999$s, etc.

For example.

$98765 \equiv 9+8+7+6+5 \equiv 35 \equiv 3+5 \equiv 8 \pmod 9$

$98765 \equiv 09+87+65 \equiv 161 \equiv 01 + 61 \equiv 62\pmod{99}$

$98765 \equiv 098 + 765 \equiv 863 \pmod{999}$

To multiply a single digit number, $n$ times $9$:

The first digit is $n-1$ and the second digit is $9 - (n-1)$.

For example, $3 \times 9 = 27 \qquad (2+7=9)$

For example, $9 \times 9 = 81 \qquad (8+1=9)$

To multiply a two-digit digit number, $ab$ times $99$:

The first two digits are $ab-1$ and the second two digits are $99 - (ab-1)$.

For example, $32 \times 99 = 3168 \qquad (31+68=99)$

For example, $40 \times 99 = 3960 \qquad (39+60=99)$

For example, $99 \times 99 = 9801 \qquad (98+01=99)$

To multiply a three-digit digit number, $abc$ times $999$:

The first three digits are $abc-1$ and the second three digits is $999 - (abc-1)$.

For example, $342 \times 999 = 341658 \qquad (341+658=999)$

For example, $44 \times 999 = 043956  \qquad (043+956=999)$

For example, $999 \times 999 = 998001 \qquad (998+001=999)$

One more example:

$123456 \times 999999 = 123455876544$


问:
Characterizing Sobolev spaces using Fourier transformation
I have a pretty basic doubt regarding Sobolev Space.

Suppose $v \in H^k(\mathbb{R}^n) := \{ v \in L^2(\mathbb{R}^n), D^{\alpha}(v) \in L^2(\mathbb{R}^n) \text{ for } \alpha \leq k \}$, where the derivative is taken in the distribution sense, so $v$ needn't be differentiable. Then how can we conclude using Plancherel that $||D^\alpha v||_{L^2} = || \zeta^\alpha \hat{v}||_{L^2}$?

I am aware that for $C^k \cap L^2$ functions we have $\widehat{D^\alpha v} = (2 \pi i \zeta)^\alpha \hat{v}$, and using Plancherel then we have $||\hat{v}||_{L^2} = ||v||_{L^2}$ since the Fourier transformation is an isometry on $L^2$, but my confusion stems from the fact that $v$ is not necessarily differentiable.

答:
Let $v\in H^k$ and let $v_n = \phi_n*v$ where $\phi_t(x)=t^{-n}\phi(x/t)$ is a $C^\infty_c$ mollifier with $\int \phi=\int \phi_t = 1$. Then it is "standard" that $v_t \in C^\infty \cap H^k$ , and $v_t \to v $ in $H^k$ as $t\to 0$, i.e. $D^\alpha v_t \to D^\alpha v$ in $L^2$ norm.

Now by what you know,
$$ \|D^\alpha v_n \|_{L^2} =  \|\zeta^\alpha \hat v_n\|_{L^2}= \|\zeta^\alpha \widehat{\phi_t}\hat v \|_{L^2}$$
Observe that $\widehat{\phi_t}(\zeta) = \hat \phi(t\zeta)\to \hat \phi(0) = 1 $ pointwise, with the dominating function
$$|\zeta^\alpha \widehat{\phi_t}\hat v |^2 \le  \|\widehat{\phi}\|^2_{L^\infty} |\zeta|^{2\alpha} |\hat v|^2 \in L^1.$$
So DCT allows us to send $t\to 0$ and obtain the result.


问:
Lagrange Multiplier Problem - Distance from point With a Circle Constraint
Given a point $P:(x_0,y_0)$ in $\mathbb{R}^2$, and a constraint function
$$x^2+y^2=R^2$$where $R$ is the radius of the circle. The distance from $P$ to any point on the circle is to be minimized using the method of Lagrange Multiplier. The distance $d$ is given by
$$d(x,y)=\sqrt{(x-x_0)^2+(y-y_0)^2}.$$
Then the Lagrange function becomes
$$\mathcal{L}(x,y,\lambda)=d(x,y)+\lambda\ (x^2+y^2-R^2)$$
with the optimality conditions $\partial_x\mathcal{L}=0$, $\partial_y\mathcal{L}=0$ and $\partial_\lambda\mathcal{L}=0$. These conditions yield:
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(x-x_0)+2\ \lambda\ x=0$$
$$\left[(x-x_0)^2+(y-y_0)^2\right]^{-\frac{1}{2}}(y-y_0)+2\ \lambda\ y=0$$
$$x^2+y^2-R^2=0$$
How should I continue from this point?</p>

答:
I thought it might be worth remarking on the geometric interpretation of the result, since we have two extrema for the distance function of points on the circle measured from the external point $ \ (x_0 \ , \ y_0) \ . $

I made the Lagrange calculation in a fashion somewhere between that of chenbai and lab bhattacharjee , extremizing the "distance-squared" function:

$$ f(x,y) \ = \ (x - x_0)^2 \ + \ (y - y_0)^2 \ \ , \ \ g(x,y) \ = \ x^2 \ + \ y^2 \ - \ R^2 $$

$$ \Rightarrow \ \ 2·(x - x_0) \ \ = \ \ \lambda · 2x \ \ \ , \ \ \ \ 2·(y - y_0) \ \ = \ \ \lambda · 2y $$

$$ \Rightarrow \ \ \lambda \ \ = \ \ \frac{x - x_0}{x} \ \ = \ \ \frac{y - y_0}{y} \ \ \Rightarrow \ \ xy \ - \ x_0y \ \ = \ \ xy \ - \ y_0x $$
$$ \Rightarrow \ \ \frac{y}{x} \  =  \ \frac{y_0}{x_0} \ \ \ \text{or} \ \ \ y \ \ = \ \ \frac{y_0}{x_0} \ x . $$

Something we notice immediately is that the  line through the center of the circle to $ \ (x_0 \ , \ y_0) \  $ is the line which contains the solution points on the circle that are closest and farthest from the external point.  [chebai comments to the OP concerning this.]  This is reasonable because that line contains the radii from the origin to the solution points, so it is perpendicular to the circumference of the circle (hence, in the direction of the gradient $ \ \nabla g \ $ ) .

Inserting this result into the constraint equation, we obtain

$$ x^2 \ + \ \left(\frac{y_0}{x_0} \ x \right)^2 \  = \  R^2 \ \ \Rightarrow \ \ x^2 \ \ = \ \ \left( \frac{x_0^2}{x_0^2 \ + \ y_0^2} \right) \ R^2 $$
$$ \Rightarrow \ \  x \ \ = \ \ \pm  \left( \frac{R}{\sqrt{x_0^2 \ + \ y_0^2}} \right) \ x_0 \ \ = \ \ \pm   \frac{R}{\rho}  \ x_0 \ \ \ , \ \ \ y \ \ =  \ \ \pm   \frac{R}{\rho}  \ y_0 \ \ ,  $$

where $ \ \rho \ = \ \sqrt{x_0^2 \ + \ y_0^2} \ $ is the distance of the external point from the center of the circle.  The extremal points are symmetrically positioned in opposite quadrants on a diameter of the circle.  [Note that this suggests how the coordinates of the extremal points could be generalized for a circle not centered on the origin, without needing to re-work the Lagrangian calculation.]

The extremal distances of $ \ (x_0 \ , \ y_0) \ $ from the circle are thus found from

$$ D^2 \ \ = \ \ \left(\pm   \frac{R}{\rho}  \ x_0  \ - \ x_0 \right)^2 \ + \ \left(\pm   \frac{R}{\rho}  \ y_0  \ - \ y_0 \right)^2  \ \ = \ \  \left(\pm   \frac{R}{\rho}  \  - \ 1 \right)^2 \ · \ (x_0^2 \ + \ y_0^2) \ \ = \ \ \left( \frac{\rho \ \mp  \  R}{\rho}   \right)^2 \ · \ \rho^2    $$
$$ \Rightarrow \ \ \ D \ \ = \ \ | \ \rho \ \mp \ R  \ | \ \ , $$

as we would expect from the geometrical arrangement.

Finally, there is almost no restriction in this calculation on the location of the point $ \ (x_0 \ , \ y_0) \ $ relative to the origin; the point could as easily be within the circle to use these results.  If this point is on the circle, we have $ \ \rho \ = \ R \ \ \rightarrow \ \ x_0 \  =   \pm    x_0 \  \ , \ \  y \  =    \pm     \ y_0 \  , \ D \ = \ 0 \ , \ 2R \ \ . $  There is no difficulty with a point $ \ (x_0 \ , \ 0) \  $ ; if the point lies on the $ \ y-$ axis, we do not use the ratio, but return to
$$  \ x_0y \ = \ y_0x \ \rightarrow \
0·y \ = \ y_0·x \ \ \Rightarrow \ \ x \ = \ 0 \ \ , \ \ y \ = \  \pm  \left( \frac{R}{\sqrt{ y_0^2}} \right) \ y_0 \ \ = \  \pm  R \ \ .  $$
It is only if we place the point at the origin that the formulas fail, since we have $ \ \frac{y}{x} \  =  \ \frac{0}{0} \  $ and $ \ \rho \ = \ 0 \ . $  We can interpret the ambiguity of the ratio as meaning that extremal points on the circle lie in every direction at the distance $ \ D \ \ = \ \ | \ 0 \ \mp \ R  \ | \ = \ R \ $ from this point.

This discussion can be readily extended to a point and $ \ n- $ sphere in $ \ \mathbb{R}^n \ \ . $  We find that the Lagrange method confirms our geometrical intuition.


问:
Write an NFA with only four states
Let $\Sigma = \{0, 1\}$, consider the language $L = \{111\}$, i.e., $L$ contains a single string with three $1$’s.  Give an NFA with $4$ states that recognizes $L$...

I am kind of stuck since I can only think of having one with $5$ states, with the last state being a garbage or catch all state that makes sure four or more quantity of $1$'s are not accepted. any intuition would be much appreciated.

答:
To expand Babado's comment, the automaton
$$
\rightarrow (1) \xrightarrow{1} (2) \xrightarrow{1} (3) \xrightarrow{1} (4) \rightarrow
$$
is indeed an automaton accepting the language $\{111\}$. This automaton is an incomplete deterministic automaton and hence also a non-deterministic automaton.
It is actually the minimal incomplete deterministic automaton accepting $\{111\}$.
The minimal complete deterministic automaton of $\{111\}$ would be the $5$-state automaton ${\cal A} = (Q, \{0,1\}, \cdot, i, F)$ with $Q = \{0, 1, 2, 3, 4\}$, $i = 1$, $F = \{4\}$ and the transitions given by the following table
\begin{array}{c|c|}
q & 1 & 2 & 3 &4 & 0\\
\hline
q \cdot 0 &0 &0 &0 &0 &0\\
\hline
q \cdot 1 &1 &1 &1 &1&0\\
\hline
\end{array}
Note that $0$ is a sink state of $\cal A$.

Title: Write an NFA with only four states


答:
You can compute the remainder term explicitly, using the formula for the geometric series:
$$ r_N(x) = \sum_{n=N}^\infty x^2 e^{-nx} = x^2e^{-Nx} \sum_{j = 0}^\infty e^{-jx} =   \dots
$$
Now find the maximum of $r_N$ on $[0, \infty)$. (It's obviously a positive function.)  </p>

<p>If this maximum tends to $0$ as $N \to \infty$, then the convergence is uniform. Otherwise it isn't. </p>

Body:
Is $\sum_{n=1}^{\infty} {x^2 e^{-nx}}$ uniformly convergent in $[0,\infty)$?

So I started by saying that by the geometric series test where $a=x^2$ and $|r| = |\frac{1}{e^x}| \leq 1$, the series converges pointwise.

But how do I exactly prove that it converges uniformly? I am quite sure it is by weistrass test but I can not find an upper bound to compare it to! Any direction would be appreciated!

Body:
You can compute the remainder term explicitly, using the formula for the geometric series:
$$ r_N(x) = \sum_{n=N}^\infty x^2 e^{-nx} = x^2e^{-Nx} \sum_{j = 0}^\infty e^{-jx} =   \dots
$$
Now find the maximum of $r_N$ on $[0, \infty)$. (It's obviously a positive function.)

If this maximum tends to $0$ as $N \to \infty$, then the convergence is uniform. Otherwise it isn't.


问:
Inscribed circle of an isosceles triangle
In a triangle $ABC$, $AC = BC = 24$ and a circle with center $J$ is inscribed. If $CH$ is altitude $(CH\perp AB,H\in AB)$ and $CJ:CH=12:17$, then find the length of $AB$.

Triangle $ABC$ is isosceles so $H$ is also the midpoint of $AB.$ So we can just find $AH$ (or $BH$) and double it to find $AB$. Let $CJ=12x, CH=17x\Rightarrow JH=5x$. I am not sure what to do from here. Thank you in advance!

答:
Since $J$ is incenter, $AJ$ is angle bisector in $\triangle HAC$. $AJ$ divides $CH$ in ratio of adjacent sides.
$$CJ/JH = 12/5 = AC/AH \Rightarrow AH = 10 \Rightarrow AB=20$$


问:
Prove the closed-form solution of low-rank linear regression.
Given matrices $X \in \mathbb{R}^{n \times d}$ and $Y \in \mathbb{R}^{n \times p}$, consider the linear regression problem in $W \in \mathbb{R}^{d \times p}$

$$\min_{W \in \mathbb{R}^{d \times p}}~\|Y-XW\|_{F}^{2}$$

In general, the closed-form solution to this problem is given by

$$\tilde{W} = X^\dagger Y$$

where $\cdot^\dagger$ denotes the pseudo-inverse.

If we impose a low-rank constraint on $W$, then the linear regression problem becomes

$$\begin{align}\min_{W}~\|Y-XW\|_{F}^{2} \\ \text{s.t.}~\text{rank}(W)\leq R\end{align}$$

How to show that the solution of this problem is $\tilde{W}P$ where $P$ is the matrix of the orthogonal projection onto the space spanned by the top-$R$ right singular vectors of $Y$?

Thank you for your help!

答:
By referring to Optimal Exact Least Squares Rank Minimization (or see DOI), I obtained the following derivation. But to be honest, I am not sure it is correct or not. If you have any new idea about this question, please let me know. Thank you!

Suppose the SVD of $Y$ is $Y=UDV^\top$ where $U\in\mathbb{R}^{n\times r},D\in\mathbb{R}^{r\times r},V\in\mathbb{R}^{p\times r},r=\min\{n,p\}$, then

$$\begin{align}\min_{W}~\|X^\dagger Y-W\|_{F}^{2} \\ \text{s.t.}~\text{rank}(W)\leq R \end{align}$$
can be written as
$$\begin{align}\min_{W}~\|X^\dagger Y-W\|_{F}^{2} \\ \text{s.t.}~\text{rank}(W)\leq R \end{align}$$
$$\Rightarrow\begin{align}\min_{W}~&\|X^\dagger YV-WV\|_{F}^{2} \\ \text{s.t.}~&\text{rank}(W)\leq R \end{align}$$

Define $Q=WV$, recall that Frobenius-norm is invariant under any orthogonal transformation, we have
$$\begin{align}\min_{Q}~&\|X^\dagger YV-Q\|_{F}^{2} \\ \text{s.t.}~&\text{rank}(Q)\leq R \end{align}\quad\Rightarrow\quad \min_{Q}~\|X^\dagger YV_{R}-Q\|_{F}^{2}$$
where $V_{R}$ consists of the top-$R$ right singular vectors of $Y$.

Therefore, the solution of this problem is $X^\dagger YP$ where $P$ is the matrix of the orthogonal projection onto the space spanned by the top-$R$ right singular vectors of $Y$.


问:
Nullity of a linear transformation does not depend on the chosen basis
Given a linear transformation $T: V \rightarrow W$ where $V$ and $W$ are finite dimensional, then is it true that nullity($T$) = nullity($[T]_\beta$) for any basis $\beta$ of $V$?

A little embarrassing that I'm 20 weeks into studying linear algebra and I've forgotten how to prove this (if I ever knew how). I think this is true, and if it is, I feel like the reasoning is extremely simple, but I can't seem to rigorously spell it out.

Is there a rigorous (and I'm expecting pretty simple) explanation for why this is true?

Thank you!

答:
Of course!  But to get a matrix for $T$, you actually need two bases, one for $V$ and one for $W$.  Then if you know the two theorems that the rank of a matrix is well-defined, and the rank-nullity theorem, you get that the nullity is equal to $n-\rm{rk}T$, where $\rm{dim}V=n$.  These are two of the in some sense biggest, best, and most basic theorems in the subject.

Every time you change bases, you conjugate by isomorphisms, so the rank, and hence nullity, are preserved.

Hmm, still not satisfied with this answer.  The kernel of $T$ is a subspace of $V$.  As such it has a dimension which as we know is an invariant (namely the number of elements in any basis).  And that dimension is called the nullity.

Similar matrices describe the same linear transformation, furthermore.


答:
Expanding on Bendit's comment:

Let $x \in N(T)$(null space of T) $\iff T(x) = 0$(the value 0) $\iff [T(x)]_{\beta} = [T]_{\beta} [x]_{\beta}=0$ (the zero matrix) $\iff [x]_{\beta} \in N([T]_{\beta})$.

As a vector's representation in a basis in unique, this implies that the vector spaces $N(T)$ and $N([T]_{\beta})$ are isomorphic and hence their dimension are the same(the nullity of $T$ = nullity of $[T]_{\beta}$).

Title: Nullity of a linear transformation does not depend on the chosen basis

Body:
Given a linear transformation $T: V \rightarrow W$ where $V$ and $W$ are finite dimensional, then is it true that nullity($T$) = nullity($[T]_\beta$) for any basis $\beta$ of $V$?

A little embarrassing that I'm 20 weeks into studying linear algebra and I've forgotten how to prove this (if I ever knew how). I think this is true, and if it is, I feel like the reasoning is extremely simple, but I can't seem to rigorously spell it out.

Is there a rigorous (and I'm expecting pretty simple) explanation for why this is true?

Thank you!


问:
Map Two Polar Disks To Four Dimensional Spherical Coordinates ($\left(\mathbb{R}^2\right)^2\to\mathbb{R}^4$)
We have two points $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$ with polar parameterizations $(r_1, \theta_1), (r_2, \theta_2)$ (i.e., $x_1=r_1 cos(\theta)$ and so on.) How would the point $(x_1, y_1, x_2, y_2) \in \mathbb{R}^4$ be represented with four-dimensional spherical coordinates $(r, \phi_1, \phi_2, \phi_3)$?

Please correct my proposed answer if I have made a mistake or upvote it if you concur.

Finally, how would one inductively generalize equivalency to $\mathbb{R}^{2^n}$ or $\mathbb{R}^{2n}$?

答:
We begin by equating the components of the two numbers in polar coordinates and the single number in spherical coordinates. Here, $r_3 = \sqrt{r_1^2 + r_2^2}$.

$$r_1 \cos \left( \theta_1 \right) = r_3 \cos \left( \phi_1 \right)$$
$$r_1 \sin \left( \theta_1 \right) = r_3 \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)$$
$$r_2 \cos \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right)$$
$$r_2 \sin \left( \theta_2 \right) = r_3 \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \sin \left( \phi_3 \right)$$

Next, the standard definition of $arccos$ does not include the full input plane, so we form a continuous extension to $\mathbb{R}^2 \setminus \mathbb{R}^+$. (We exclude the point intersecting $\mathbb{R}^+$ with $\theta = 0$ or $\theta = 2 \pi$). With an abuse of notation, we will still refer to this continuous extension as $arccos$. An interesting application of this definition of $arccos$ is that $\mathbb{R}^3 \setminus 0$ is not homologically trivial, which is an exercise in Munkre's Analysis on Manifolds.

$$\phi_1 = \arccos \left( r_1 r_3^{-1} \cos \left( \theta_1 \right) \right)$$
$$\phi_2 = \arccos \left( r_1 r_3^{-1}  \frac{ sin (\theta_1) }{sin (\phi_1)} \right)$$
$$\phi_3 = \arccos \left( r_2 r_3^{-1} \frac{ \cos \left( \theta \right) }{ \sin \left( \phi_1 \right) \sin \left( \phi_2 \right)   }  \right)$$

Fortunately, the other direction is simpler to compute.

$$\theta_{1} = \arccos(r_3 r_1^{-1} \cos \left( \phi_1 \right))=\arcsin \left( r_3 r_1^{-1} \sin \left( \phi_1 \right) \cos \left( \phi_2 \right) \right)$$

$$\theta_2 = \arccos \left( r_3 r_2^{-1} \sin \left( \phi_1 \right) \sin \left( \phi_2 \right) \cos \left( \phi_3 \right) \right)=...$$

$$\frac{ r_1 \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } = \frac{ r_1 \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)   }$$

$r_1 = \left|\frac{ \cos \left( \theta_1 \right) }{ \cos \left( \phi_1 \right) } - \frac{ \sin \left( \theta_1 \right) }{ \sin \left( \phi_1 \right) \cos \left( \phi_2 \right)   }\right|^{-1}$ provided that $cos(\phi_1)\neq0$, $cos(\phi_2)\neq 0$, and $sin(\phi_1)\neq 0$.

$r_2 = \left|\frac{1}{sin(\phi_1) sin(\phi_2)} \left( \frac{cos (\theta_2) }{cos(\phi_3)} - \frac{sin(\theta_2)}{sin(\phi_3)} \right)\right| ^{-1}$ provided that $sin(\phi_1)\neq 0$, $sin (\phi_2) \neq 0$, $cos(\phi_3)\neq 0$, and $sin(\phi_3) \neq 0$.

I wonder if there would be an easier formula for $r_1$ and $r_2$ by taking the components of $r_3$ along two axes. (In this case, $r_1$, $r_2$, and $r_3$ would all be tangent vectors on the same plane, though finding a parameterization or normal vector for the plane is not trivial; one cannot simply compute $\vec{r_1} \times \vec{r_2}$.)

By stipulating $r_3>0$ and $\theta_1, \theta_2, \theta_3 \in [0, 2 \pi)$, proof of injection (being a one-to-one map) should follow immediately. Likewise, every pair of $(x_1, y_1)=(r_1, \theta_1)$ and $(x_2, y_2) = (r_2, \theta_2)$ should produce a value for $(r, \phi_1, \phi_2, \phi_3)$, allowing surjection to follow as well. $f(r_1, \theta_1, r_2, \theta_2)= (r, \phi_1, \phi_2, \phi_3)$ should then satisfy the criteria of an isomorphic map. If feel that I have missed something. Please correct me in the comments or my answer directly if you have sufficient reputation.


问:
Continuously applying an isometry operator on to a Hilbert space
While looking at the proof of https://en.wikipedia.org/wiki/Wold%27s_decomposition#A_sequence_of_isometries, I encounted that if $V \in B(H)$ is an isometry on a Hilbert space $H$.

Then we have $V^n H \subset V^{n-1}H \subset \cdots \subset VH  \subset H$.

This part is not super clear to me. Seems like it's from the definition of an Isometry.
Could someone help me with this?

Thank you!

答:
This holds for every operator. Usually, one writes $\operatorname{ran}T = TH$ (the range of $T$). It is obvious that $\operatorname{ran}T^{n+1}\subset \operatorname{ran}T^n$ for all $n\in\mathbb N$. Indeed, if $x\in\operatorname{ran}T^{n+1}$, then $x = T^{n+1}u$ for some $u$. So, $x = T^n(Tu)\in\operatorname{ran}T^n$. So, these guys are descending.

Similarly (but the other way around) you have $\ker T^n\subset\ker T^{n+1}$. The kernels thus are ascending. If you are interested in this, look for the notions of ascent and descent.


答:
To get a grasp of the idea, an example of this is the unilateral shift operator $S$ on a separable Hilbert space $H$.

We get that $SH \subset H$, then repeated application gives you the desired $$S^{n}H \subset S^{n-1}H \subset ... \subset SH\subset H.$$

Title: Continuously applying an isometry operator on to a Hilbert space

Body:
While looking at the proof of https://en.wikipedia.org/wiki/Wold%27s_decomposition#A_sequence_of_isometries, I encounted that if $V \in B(H)$ is an isometry on a Hilbert space $H$.

Then we have $V^n H \subset V^{n-1}H \subset \cdots \subset VH  \subset H$.

This part is not super clear to me. Seems like it's from the definition of an Isometry.
Could someone help me with this?

Thank you!


问:
Given $s(x) = \int_a^x \sqrt{1 + [f'(t)]^2} \ dt$. If $9x^2 = y(y-3)^2$ and $y = f(x)$, find $\frac{ds}{dx}$ if it exists.
So here is a problem that I am having a bit of trouble with.

Given $s(x) = \int_a^x \sqrt{1 + [f'(t)]^2} \ dt$, and $y = f(x)$, if $9x^2 = y(y - 3)^2$ find $\frac{ds}{dx}$, if it exists.

My attempt to this question was to start by applying FTC1 to $s(x)$, so
\begin{align*}
s'(x) &= \sqrt{1 + [f'(x)]^2}
\end{align*}

With the function $s'(x)$ we can also say $\frac{ds}{dx}$.
\begin{align*}
\frac{ds}{dx} &= \sqrt{1 + [f'(x)]^2}
\end{align*}

Then with $y = f(x)$, we can say that $\frac{dy}{dx} = f'(x)$. With our equation, we want to determine $\frac{dy}{dx}$. We can do so by implicit differentiation:
\begin{align*}
9x^2 &= y(y - 3)^2 \\
9x^2 &= y^3 - 6y^2 + 9y \\
18x &= 3y^2 \cdot \frac{dy}{dx} - 12y \cdot \frac{dy}{dx} + 9 \cdot \frac{dy}{dx} \\
18x &= \frac{dy}{dx}(3y^2-12y+9) \\
\frac{dy}{dx} &= \frac{18x}{3y^2 - 12y + 9} \\
\frac{dy}{dx} &= \frac{6x}{y^2 - 4y + 3}
\end{align*}

So now I found $\frac{dy}{dx}$, we can square it so we can have $(\frac{dy}{dx})^2 = [f'(x)]^2$
\begin{align*}
\left(\frac{dy}{dx}\right)^2 &= \frac{36x^2}{(y^2 - 4y + 3)^2}
\end{align*}

Now putting what we know into $s'(x) = \frac{ds}{dx}$,
\begin{align*}
\frac{ds}{dx} &= \sqrt{1 + \frac{36x^2}{(y^2 - 4y + 3)^2}} \\
&= \sqrt{\frac{(y^2 - 4y + 3)^2 + 36x^2}{(y^2 - 4y + 3)^2}}
\end{align*}

After this part, I am not even sure if I am doing the question right anymore. If possible would like some advice or clarification. Thanks

Body:
Given: $$9x^2=y(y-3)^2~~~(1)$$
We get $y'=\frac{6x}{y^2-4y+3}$.
Next by Lebnitz, we get
$$\frac{ds}{dx}=\sqrt{1+y'^2(x)}= \frac{\sqrt{4y(y-3)^2+(y-3)^2(y-1)^2}}{(y-3)(y-1)} =\frac{y(x)+1}{y(x)-1},$$
when  $y\ne 1,3.$


问:
$K_0$ of an inner automorphism
An exercise in a book I am studying says

$K_0(\alpha) =id$ for every inner automorphism $\alpha$.

I am not sure why this is true, I suppose it has to do with the fact that inner automorphisms look like $Ad \; u$ for some unitary $u\in A$.

答:
If $\alpha = Ad_u$, then for any projection $p$ we have $\alpha(p) = upu^*$. The projection $upu^*$ and $p$ are unitarily equivalent. Since unitarily equivalent projections define the same classes in $K$-theory $$[p] = [upu^*] = [\alpha(p)] = K_0(\alpha)([p]).$$

It now follows that $K_0(\alpha)$ is just the identity map on $K_0(A)$ since $K_0(\alpha)([p] - [q]) = [p] - [q]$.

Title: $K_0$ of an inner automorphism


问:
Proof of perpendicularity and points equality in a triangle with inequality on area
We have three points P, R, and S in a -dimensional plane with integer coordinates. Let  be the area of
triangle PRS,  be the length of the line segment PR and  be the length of the line segment PS.
We have the condition that $$( + )^2  <  + $$
I need to prove that  =  and that PR and PS are perpendicular.
I tried to approach the problem by making some drawings and seeing how to profit from the inequality condition and the integer constraint on the coordinates but with no result.

Any suggestions??

答:
As can be seen in figure ,if PRS is right angled triangle then:

$S_{PRS}=Q=\frac{PR\cdot PS}2$

But: $PR=PS=PH\sqrt 2$

This gives maximum possible area of PRS because $PR=PS$ and we have:

$S_{PRS}=Q=\frac{(2PH)\cdot PH}2=PH^2\Rightarrow 8Q+1=8PH^2+1 $

And:

$(PR+PS)^2=(2PH\sqrt 2)^2=8PH^2$

Which suffice the inequality condition.


问:
A property of radical ideals
Let $A$ be a commutative ring with $1 \neq 0$.

Theorem (Atiyah-MacDonald 1.13 (v)). Let $\mathfrak{a, b} \subseteq A$ be ideals. Then $\sqrt{\mathfrak{a + b}} = \mathfrak{\sqrt{\sqrt{a} + \sqrt{b}}}$.

Question. Is the following generalization true?

For any finite collection of ideals $\mathfrak{a}_{1}, \dots, \mathfrak{a}_{n} \subseteq A$, we have $$\sqrt{\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}} = \sqrt{\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}}.$$

I'm pretty sure that it is true since if $x_{j} \in \sqrt{\mathfrak{a}_{j}}$ so that $x_{j}^{n_{j}} \in \mathfrak{a}_{j}$ for chosen $n_{j} \geq 1$, each term of $(x_{1} + \cdots + x_{n})^{N}$ is $x_{1}^{r_{1}} \cdots x_{n}^{r_{n}}$ and not all $r_{j} < n_{j}$ if we take $N$ large. (This is a recycled argument of Proposition 1.7 in Atiyah-MacDonald.) I just want to make sure that this statement is correct as I don't have a reference for it.

I am interested in this statement because I'd like to show that given finitely many elements $f_{j} \in A$, we have $\sqrt{(f_{1}, \dots, f_{n})} = \sqrt{0}$ if and only if $f_{j} \in \sqrt{0}$ for all $j$.

It is true when $n = 1, 2$, as follows.

Lemma. Let $f \in A$. We have $\sqrt{(f)} = \sqrt{0}$ if and only if $f \in \sqrt{0}$.

Proof. Since $f \in \sqrt{(f)}$, we only need to show one implication.

Let $f \in \sqrt{0}$. Then $(f) \subseteq \sqrt{0}$. Since taking radical does not change the order of inclusion, we have $\sqrt{(f)} \subseteq \sqrt{\sqrt{0}} = \sqrt{0} \subseteq \sqrt{(f)}$. We showed that $f \in \sqrt{0}$ implies $\sqrt{(f)} = \sqrt{0}$. Q.E.D.

Corollary. Let $f, g \in A$. We have $\sqrt{(f, g)} = \sqrt{0}$ if and only if $f, g \in \sqrt{0}$.

Proof. Since $f, g \in \sqrt{(f, g)}$, we only need to show one implication.

Suppose that $f, g \in \sqrt{0}$. Then by Lemma, we have $\sqrt{(f)} = \sqrt{(g)} = \sqrt{0}$, so $\sqrt{(f, g)} = \sqrt{(f) + (g)} = \sqrt{\sqrt{(f)} + \sqrt{(g)}} = \sqrt{0}$. Q.E.D.

答:
Summarizing what YACP wrote:

$V(\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}) = V(\cup_{i} \mathfrak{a}_{i}) = V(\mathfrak{a}_{1}) \cap \cdots \cap V(\mathfrak{a}_{n}) = V(\sqrt{\mathfrak{a}_{1}}) \cap \cdots \cap V(\sqrt{\mathfrak{a}_{n}}) = V(\cup_{i}\sqrt{\mathfrak{a}_{i}}) = V(\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}).$

Title: A property of radical ideals


问:
How to remember the derivatives of inverse trigonometry function?
I remember the derivatives of trig functions  by naming 3x basic right triangles in a specific way and using ONE simple multiplication.  Just wondering if there are  similar approaches  to remember the derivatives of inverse trig function with the assisting of some basic triangles?

Procedures $\to$

Draw a right triangle (1, $\cos x,sinx$).Label its sides accordingly as shown in Fig.1.

Put the hypotenuse label ("1" in this case)  in side the triangle , name this triangle as "1" triangle.

Here is the only rule:  |X'| = |Central value  *   Y|, where X and Y are the  two functions on the NONE "1" sides.  In this case, the central value is "1", and the two functions on the NONE "1" sides are $sinx$ and $cos x$.
$           $|$sinx'| =|1* cos x$| ,  $|cosx'|= |1*sinx|$ (Fig.1)

Similarly, divide each side of "1"triangle by $sinx$ and $cosx$ to obtain "$sec"  and  “$cscx" triangles in Fig. 2 and Fig.3.

$|secx'| =|secx*tanx|$,  $|tanx'|= sec*secx$  (Fig.2)

$|cscx'| =|cscx*cotx|$,  $|cotx'|= cscx*cscx$  (Fig.3)

(Need to apply negative signs to get values for $cosx', cotx' $and $cscx'$).

答:
Here's a slight re-packaging of your mnemonic.

We start with the figure I call the Fundamental Trigonograph (the inspiration for my avatar!), whose segment-lengths correspond to the trig values associated with (acute) angle $\theta$.

Note that the "$1$" segment separates "ordinary" segments ($\sin$, $\tan$, $\sec$) from "complementary" segments ($\cos$, $\cot$, $\csc$).

Then, with no motivation whatsoever, we augment the figure with two right triangles having legs $\sec$ and $\csc$, and hypotenuses parallel to the "$1$" segment. (I've suppressed the "$1$" label to reduce visual clutter, but extended it to highlight the separation between "ordinaries" and "complementaries".)

One readily determines the new side lengths to be $\sec\tan$, $\sec^2$, $\csc\cot$, $\csc^2$; by interesting coincidence, lengths $\sin$ and $\cos$ appear in the lower-left quadrant in an obvious way. I've attached negative signs to the new segments on the "complementary" side of the augmented figure ... um ... just because. :)

With this, the rule is something like:
$$\text{A function and its derivative are "natural" perpendicular pairs on the same side of "$1$".}$$

That is,

$\tan$ "naturally" pairs with $\sec^2$, and $\sec$ with $\sec\tan$. The pairing of the original $\sin$ segment with the augmenting $\cos$ segment is slightly contrived, but it works.

Likewise, $\cot$, $\csc$, and $\cos$ "naturally" pair with $-\csc^2$, $-\csc\cot$, and $-\sin$.

Is this the best presentation of the mnemonic? I'm not sure ---it's just what occurred to me in the moment--- but I kinda like it. I especially like that it builds on the Fundamental Trigonograph, which absolutely belongs in every student's tool belt. (See, for instance, my still-drafty note "(Almost) Everything You Need to Remember about Trig, in One Simple Diagram" (PDF link via tricochet.com). I may need to make an addition!)


答:
While the other answer would take you to the standard spherical coordinates, the formula gets very unwieldy especially for higher dimensions. An alternative will get you the same radial coordinate but different angles

$$\phi_1=\theta_1$$
$$\phi_2=\theta_2$$
$$r_3\cos\phi = r_1$$
$$r_3\sin\phi = r_2$$

The interesting thing about these coordinates is instead of having the angles be maps $[0,2\pi]\times[0,\pi]^{2n-2}\to S^{2n-1}$, these are maps $[0,2\pi]^{n}\times\left[0,\frac{\pi}{2}\right]^{n-1}\to S^{2n-1}$ (embedded in $\Bbb{R}^{2n}$). The generalization to $\Bbb{R}^{2n}$ is simply setting all of the radial coordinates to the standard spherical coordinates in $\Bbb{R}^n$ but restricted to the first quadrant/octant/etc..

Title: Map Two Polar Disks To Four Dimensional Spherical Coordinates ($\left(\mathbb{R}^2\right)^2\to\mathbb{R}^4$)

Body:
We have two points $(x_1, y_1), (x_2, y_2) \in \mathbb{R}^2$ with polar parameterizations $(r_1, \theta_1), (r_2, \theta_2)$ (i.e., $x_1=r_1 cos(\theta)$ and so on.) How would the point $(x_1, y_1, x_2, y_2) \in \mathbb{R}^4$ be represented with four-dimensional spherical coordinates $(r, \phi_1, \phi_2, \phi_3)$?

Please correct my proposed answer if I have made a mistake or upvote it if you concur.

Finally, how would one inductively generalize equivalency to $\mathbb{R}^{2^n}$ or $\mathbb{R}^{2n}$?


问:
Find $\binom{80}{40}\bmod 2000$

Find $\displaystyle\binom{80}{40}\bmod 2000$.

So far, I've found that $\displaystyle\binom{80}{40}$ is divisible by $2^2$ and $5^1$, so the answer isn't $0$. Usually, with smaller numbers, I would split the $\bmod 2000$ into $2^4$ and $5^3$, find the answer through brute force for each of those, then use the Chinese Remainder theorem. But $80$ and $40$ are too big for this.

答:
I built Pascal's triangle in a spreadsheet program, but always reducing mod 2000. There's never a need to keep more than the result mod $2000$. The result I get is $1620$.

Cell A2: 0

Cell A3: =A2+1 and carry down to Cell A82 with value $80$.

Cell B1: 0

Cell C1: =B1+1 and carry rightward to Cell AP1 with value $40$.

Now these are the indices of Pascal's triangle. For $\binom{n}{k}$, $n$ is in column A and $k$ is in row 1.

Cell B2: 1 and carry all the way down.

Cell C3: =MOD(B2+C2,2000), carry all the way rightward, and then all the way down, filling out the table.

Now Cell AP82 is $\binom{80}{40}$ mod $2000$. It gives $1620$.

To do this by hand, at first it may seem like you would calculate about $80\cdot40=3200$ cell entries mod $2000$. But actually you would only need to compute about $\frac14$ of those entries, as represented in blue here:

The red is all $0$. The orange is not needed to reach the $\binom{80}{40}$ cell. And the green can be written using symmetry from the blue. So this would take about $800$ additions mod $2000$ plus some other less signiicant labor.

But we an do better, because there is the identity $$\binom{80}{40}=\sum_{n=0}^{40}\binom{40}{n}^2=\binom{40}{20}+2\sum_{n=0}^{19}\binom{40}{n}^2$$

So actually if we can get Pascal's triangle down to the first half of the 40th row, then we can square mod $2000$ about 20 times, double, sum the results mod $20$ and have our answer.

A similar analysis reveals that you would need to calculate about $400$ of the entries of Pascal's triangle mod $2000$ for this approach.

Title: Find $\binom{80}{40}\bmod 2000$


问:
a simple doubt in graph theory
I was trying to find the distance between two vertices in the $Lexicographic\ Product$ of graphs. To find the distance between two vertices  $(x_1, x_2, . . . , x_k)$and $(y_1, y_2, . . . , y_k)$ a formula is given as follows.....

$d(x,y) := \begin{cases}
 d_{G_i}(x_i, y_i)   & \text{if $d_{G_{l}}(x_l) = 0$ for all $1\leq l<i$}\\
 \min\{d_{G}(x_i, y_i), 2\} & \text{if $d_{G_{l}}(x_l) \neq 0$ for all $1\leq l<i$}
\end{cases}$

where $i$ is the smallest index such that $x_i$$\neq$$y_i$ and $d_{G_{l}}(x_l)$ denotes the degree of $x_l$ in $G_l$.

I am taking the reference of my post doubt about basic definiton. AND understanding the basic definiton

While calculating the distance after drawing the graph, I am getting  $distance$ as $3$,  between $(a,x,1)$  and $(d,y,2)$ but acc. to above formula its coming out to be $2$.

I am so confused. Please correct me where I am going wrong. Thanks a lot for help.

note: I am considering that no graph has isolated vertex.

答:
The problem is that your expression for $d(x,y)$ isn’t quite right.

Added: It actually is right; I just misread it. I’ll match it up with mine below. It’s for the case in which $x\ne y$, so my first clause is unnecessary. My second clause is the special case of its first clause with $i=1$. My third clause is its second clause. And my fourth clause covers the rest of its first clause, i.e., the case of $1<i\le k$. This is easy to check so long as you remember that a vertex is isolated if and only if its degree is $0$.

Let $G=G_1\circ\ldots\circ G_n$, and let $x=\langle x_1,\dots,x_n\rangle$ and $y=\langle y_1,\dots,y_n\rangle$ be vertices of $G$. If $x_1\ne y_1$, $d(x,y)=d_{G_1}(x_1,y_1)$.

Now suppose that $1<m\le n$, $x_m\ne y_m$, and $x_k=y_k$ for $1\le k<m$: $x$ and $y$ agree on the first $m-1$ coordinates and disagree on the $m$-th coordinate. It’s not hard to see that $d(x,y)$ cannot be any larger than $d_{G_m}(x_m,y_m)$. However, it can be smaller. Suppose that there is some $k$ with $1\le k<m$ such that $G_k$ has a vertex $v$ adjacent to $x_k$; then in $G$ the vertices $x$ and $y$ are both adjacent to the vertex $z=\langle x_1,\dots,x_{k-1},v,x_{k+1},\dots,x_n\rangle$. To see this, note that $x$ and $y$ both first disagree with $z$ in the $k$-th coordinate, and $x_k=y_k$ is adjacent to $v$ in $G_k$, so $x$ and $y$ are adjacent to $z$ in $G$. In this case, therefore, $d(x,y)=2$, even if $d_{G_m}(x_m,y_m)>2$.

The remaining possibility is that there is an index $m$ as in the previous paragraph, but each of the vertices $x_1,\dots,x_{m-1}$ is isolated; i.e., for each $k$ with $1\le k<m$, the vertex $x_k$ has no neighbor in $G_k$. In this case $d(x,y)=d_{G_m}(x_m,y_m)$.

Thus,

$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
d_{G_1}(x_1,y_1),&\text{if }x_1\ne y_1\\
\min\{d_{G_m}(x_m,y_m),2\},&\text{if there is an }m>1\text{ such that }x_m\ne y_m\\
&\text{and }x_k=y_k\text{ for }1\le k<m\\
&\text{and }x_k\text{ is not isolated in }G_k\text{ for some }k\\
&\text{with }1\le k<m\\
d_{G_m}(x_m,y_m),&\text{if there is an }m>1\text{ such that }x_m\ne y_m\\
&\text{and }x_k=y_k\text{ for }1\le k<m\\
&\text{and }x_k\text{ is isolated in }G_k\text{ for all }k\text{ with}\\
&1\le k<m\;.
\end{cases}\tag{1}$$

Let $p=\langle a,x,1\rangle$ and $q=\langle d,y,2\rangle$ be the two vertices of interest in $P_4\circ K_2\circ K_2$, where the three component graphs are realized as shown below:

```<code>     o---o---o---o      o---o      o---o
     a   b   c   d      x   y      1   2
</code>```

Call these three graphs $G_1,G_2$, and $G_3$ respectively.

Now $a\ne d$ in the first coordinates of $p$ and $q$, so $d(p,q)$ is calculated by the second clause of $(1)$ as $d(p,q)=d_{G_1}(a,d)=3$.

Title: a simple doubt in graph theory

Body:
The problem is that your expression for $d(x,y)$ isn’t quite right.

Added: It actually is right; I just misread it. I’ll match it up with mine below. It’s for the case in which $x\ne y$, so my first clause is unnecessary. My second clause is the special case of its first clause with $i=1$. My third clause is its second clause. And my fourth clause covers the rest of its first clause, i.e., the case of $1<i\le k$. This is easy to check so long as you remember that a vertex is isolated if and only if its degree is $0$.

Let $G=G_1\circ\ldots\circ G_n$, and let $x=\langle x_1,\dots,x_n\rangle$ and $y=\langle y_1,\dots,y_n\rangle$ be vertices of $G$. If $x_1\ne y_1$, $d(x,y)=d_{G_1}(x_1,y_1)$.

Now suppose that $1<m\le n$, $x_m\ne y_m$, and $x_k=y_k$ for $1\le k<m$: $x$ and $y$ agree on the first $m-1$ coordinates and disagree on the $m$-th coordinate. It’s not hard to see that $d(x,y)$ cannot be any larger than $d_{G_m}(x_m,y_m)$. However, it can be smaller. Suppose that there is some $k$ with $1\le k<m$ such that $G_k$ has a vertex $v$ adjacent to $x_k$; then in $G$ the vertices $x$ and $y$ are both adjacent to the vertex $z=\langle x_1,\dots,x_{k-1},v,x_{k+1},\dots,x_n\rangle$. To see this, note that $x$ and $y$ both first disagree with $z$ in the $k$-th coordinate, and $x_k=y_k$ is adjacent to $v$ in $G_k$, so $x$ and $y$ are adjacent to $z$ in $G$. In this case, therefore, $d(x,y)=2$, even if $d_{G_m}(x_m,y_m)>2$.

The remaining possibility is that there is an index $m$ as in the previous paragraph, but each of the vertices $x_1,\dots,x_{m-1}$ is isolated; i.e., for each $k$ with $1\le k<m$, the vertex $x_k$ has no neighbor in $G_k$. In this case $d(x,y)=d_{G_m}(x_m,y_m)$.

Thus,

$$d(x,y)=\begin{cases}
0,&\text{if }x=y\\
d_{G_1}(x_1,y_1),&\text{if }x_1\ne y_1\\
\min\{d_{G_m}(x_m,y_m),2\},&\text{if there is an }m>1\text{ such that }x_m\ne y_m\\
&\text{and }x_k=y_k\text{ for }1\le k<m\\
&\text{and }x_k\text{ is not isolated in }G_k\text{ for some }k\\
&\text{with }1\le k<m\\
d_{G_m}(x_m,y_m),&\text{if there is an }m>1\text{ such that }x_m\ne y_m\\
&\text{and }x_k=y_k\text{ for }1\le k<m\\
&\text{and }x_k\text{ is isolated in }G_k\text{ for all }k\text{ with}\\
&1\le k<m\;.
\end{cases}\tag{1}$$

Let $p=\langle a,x,1\rangle$ and $q=\langle d,y,2\rangle$ be the two vertices of interest in $P_4\circ K_2\circ K_2$, where the three component graphs are realized as shown below:

     o---o---o---o      o---o      o---o
     a   b   c   d      x   y      1   2

Call these three graphs $G_1,G_2$, and $G_3$ respectively.

Now $a\ne d$ in the first coordinates of $p$ and $q$, so $d(p,q)$ is calculated by the second clause of $(1)$ as $d(p,q)=d_{G_1}(a,d)=3$.