问:
How would you go about learning the combination of coins the man has?
I am trying to identify the branch of math that would help to solve the following problem:

A man has picked $10$ coins out of a bag and has laid them in a row. You cannot see them for yourself, and the heads/tails ratio is unknown. Your goal is to identify the correct heads/tails sequence of the coins by repeatedly picking $10$ coins of your own from the bag, with the man telling you how many match the position of his each time. You cannot decide whether the coins are heads or tails (for the sake of the problem assume they are the same on both sides) and your sequence depends on the order you remove them from the bag. Note that the man only selects his coins once and his never change.

For example, if the man has $HTTHHHHTTT$ and you propose $TTTHTHTHHT$ he will let you know that $5$ are a match, however he will not tell you which ones. If on your second try you pull out $THHTTTHTHT$ he will inform you that $3$ match. Without trying every possible combination $(2^{10})$, how would you go about learning the combination of coins the man has?

Comparing the two sample results I see that they share $3$ in common, but I'm not sure if or how that is useful information since I don't know which ones actually match.

I suspect it might require statistics/combinatorics, and I was told it could be solved with "either high school or college math". This isn't a homework problem so it's possible I'm just being trolled by the person. Any information you can provide for approaching this type of problem is much appreciated. Thank you.

答:
Here's a Python program that implements the algorithm mentioned by saulspatz in the question comments, and discussed in hardmath's answer. I assume that the probability of picking a heads is 0.5.

In each round of the game, we first create a pool containing all 1024 possible 10 bit strings. Next, the target bit string is generated. Then we enter the outer loop, where we generate a guess bit string, and count the number of "bulls" (positions where the target & guess bit strings match). We then loop over all the bit strings in the pool, retaining only the strings that have the same match count when tested against the current guess. We exit the outer loop when only one bit string remains in the pool, which must match the target.

```<code>""" Random binary string guessing game
    See https://math.stackexchange.com/q/4064108
    Written by PM 2Ring 2021-03-17
"""

from random import seed, getrandbits
seed(42)

def bits(n):
    " Convert int n to a bit string "
    return f"{n:010b}"

def mcount(a, b):
    " Count matches between sequences a & b "
    return sum(u==v for u,v in zip(a, b))

def test(verbose=True):
    " Play 1 round of the game "
    pool = [bits(u) for u in range(1<<10)]
    target = bits(getrandbits(10))
    if verbose:
        print("\n T", target)
    for i in range(1, 100):
        guess = bits(getrandbits(10))
        bulls = mcount(target, guess)
        pool = [u for u in pool if mcount(u, guess) == bulls]
        if verbose:
            print(f"{i:2}", guess, bulls, len(pool))
        if len(pool) == 1:
            break
    if verbose:
        print(" R", pool[0], pool[0] == target)
    return i

@interact
def main(runs=10, verbose=True, auto_update=False):
    k = 0
    for _ in range(runs):
        k += test(verbose)
    print("Mean:", float(k / runs))
</code>```

Here's the output for 10 runs, using 42 as the randomizer seed.

In each run, the target bit string is printed, followed by a number of guesses. Each guess line consists of its index, the guess bit string, the number of bits that match the target, and the number of remaining bit strings in the pool that are consistent with this guess. Finally, the only string remaining in the pool is printed, and we verify that it's identical to the target.

At the end of all runs, we print the mean number of guesses required per run.

```<code> T 1010001110
 1 0001110010 3 120
 2 0000011001 4 50
 3 1011110111 5 26
 4 0100011001 3 4
 5 0011111010 5 3
 6 0011100100 5 1
 R 1010001110 True

 T 0010001110
 1 1011110010 4 210
 2 0001101000 5 100
 3 1010110100 5 48
 4 1011110110 5 30
 5 1110010001 3 12
 6 1000101110 7 9
 7 0001011001 4 5
 8 1001011100 5 3
 9 0110110000 4 2
10 0000100000 5 1
 R 0010001110 True

 T 0000011110
 1 0001011111 8 45
 2 0011011111 7 36
 3 0011101110 6 18
 4 1000000101 5 10
 5 1001101000 4 5
 6 0000011011 8 2
 7 1000111110 8 2
 8 0011001011 5 2
 9 1011011101 5 2
10 1010011001 5 2
11 1011001110 6 1
 R 0000011110 True

 T 1000101110
 1 0110101101 5 252
 2 0011100001 3 60
 3 0111001011 3 18
 4 1001011011 5 10
 5 0100011100 5 3
 6 1100111100 7 3
 7 1101111010 6 3
 8 0000000110 7 1
 R 1000101110 True

 T 1100001001
 1 1100111001 8 45
 2 0010100011 4 24
 3 1011001010 5 12
 4 0110110000 4 6
 5 0101011100 5 6
 6 0100011100 6 3
 7 0010011111 4 1
 R 1100001001 True

 T 0011011100
 1 1111010100 7 120
 2 1100001101 4 50
 3 0101011000 7 15
 4 0001101000 6 9
 5 0001011110 8 2
 6 0110000101 5 2
 7 0001100011 3 1
 R 0011011100 True

 T 0101101111
 1 1101100011 7 120
 2 0101100000 6 63
 3 1001101010 6 30
 4 0100001110 7 6
 5 1100111010 5 3
 6 0000101100 6 1
 R 0101101111 True

 T 1011101011
 1 0111010110 3 120
 2 1000100101 5 56
 3 0001111111 6 24
 4 1111100100 5 9
 5 1110110000 4 6
 6 0110000011 5 4
 7 0001010000 3 2
 8 1000110101 4 2
 9 0100101100 3 1
 R 1011101011 True

 T 1101010001
 1 1010000011 5 252
 2 1001111001 7 60
 3 1110001010 4 30
 4 1101110010 7 8
 5 0101110010 6 8
 6 1001001111 5 4
 7 0011000100 4 2
 8 1011010001 8 1
 R 1101010001 True

 T 0001000111
 1 0000101110 6 210
 2 1010100101 5 100
 3 0011101001 5 46
 4 1100010111 6 18
 5 0100101000 3 5
 6 1111110000 2 3
 7 0001010001 7 1
 R 0001000111 True
Mean: 7.9
</code>```

It appears that the mean number of guesses required is a little over 7.

Here is a live version of the program, running on the SageMathCell server. If you set the runs parameter to 10000, it will take about one minute in non-verbose mode. I haven't tested higher numbers, but if a computation takes too long the server will abort the job.

Body:
Here's a Python program that implements the algorithm mentioned by saulspatz in the question comments, and discussed in hardmath's answer. I assume that the probability of picking a heads is 0.5.

In each round of the game, we first create a pool containing all 1024 possible 10 bit strings. Next, the target bit string is generated. Then we enter the outer loop, where we generate a guess bit string, and count the number of "bulls" (positions where the target & guess bit strings match). We then loop over all the bit strings in the pool, retaining only the strings that have the same match count when tested against the current guess. We exit the outer loop when only one bit string remains in the pool, which must match the target.

""" Random binary string guessing game
    See https://math.stackexchange.com/q/4064108
    Written by PM 2Ring 2021-03-17
"""

from random import seed, getrandbits
seed(42)

def bits(n):
    " Convert int n to a bit string "
    return f"{n:010b}"

def mcount(a, b):
    " Count matches between sequences a & b "
    return sum(u==v for u,v in zip(a, b))

def test(verbose=True):
    " Play 1 round of the game "
    pool = [bits(u) for u in range(1<<10)]
    target = bits(getrandbits(10))
    if verbose:
        print("\n T", target)
    for i in range(1, 100):
        guess = bits(getrandbits(10))
        bulls = mcount(target, guess)
        pool = [u for u in pool if mcount(u, guess) == bulls]
        if verbose:
            print(f"{i:2}", guess, bulls, len(pool))
        if len(pool) == 1:
            break
    if verbose:
        print(" R", pool[0], pool[0] == target)
    return i

@interact
def main(runs=10, verbose=True, auto_update=False):
    k = 0
    for _ in range(runs):
        k += test(verbose)
    print("Mean:", float(k / runs))

Here's the output for 10 runs, using 42 as the randomizer seed.

In each run, the target bit string is printed, followed by a number of guesses. Each guess line consists of its index, the guess bit string, the number of bits that match the target, and the number of remaining bit strings in the pool that are consistent with this guess. Finally, the only string remaining in the pool is printed, and we verify that it's identical to the target.

At the end of all runs, we print the mean number of guesses required per run.

 T 1010001110
 1 0001110010 3 120
 2 0000011001 4 50
 3 1011110111 5 26
 4 0100011001 3 4
 5 0011111010 5 3
 6 0011100100 5 1
 R 1010001110 True

 T 0010001110
 1 1011110010 4 210
 2 0001101000 5 100
 3 1010110100 5 48
 4 1011110110 5 30
 5 1110010001 3 12
 6 1000101110 7 9
 7 0001011001 4 5
 8 1001011100 5 3
 9 0110110000 4 2
10 0000100000 5 1
 R 0010001110 True

 T 0000011110
 1 0001011111 8 45
 2 0011011111 7 36
 3 0011101110 6 18
 4 1000000101 5 10
 5 1001101000 4 5
 6 0000011011 8 2
 7 1000111110 8 2
 8 0011001011 5 2
 9 1011011101 5 2
10 1010011001 5 2
11 1011001110 6 1
 R 0000011110 True

 T 1000101110
 1 0110101101 5 252
 2 0011100001 3 60
 3 0111001011 3 18
 4 1001011011 5 10
 5 0100011100 5 3
 6 1100111100 7 3
 7 1101111010 6 3
 8 0000000110 7 1
 R 1000101110 True

 T 1100001001
 1 1100111001 8 45
 2 0010100011 4 24
 3 1011001010 5 12
 4 0110110000 4 6
 5 0101011100 5 6
 6 0100011100 6 3
 7 0010011111 4 1
 R 1100001001 True

 T 0011011100
 1 1111010100 7 120
 2 1100001101 4 50
 3 0101011000 7 15
 4 0001101000 6 9
 5 0001011110 8 2
 6 0110000101 5 2
 7 0001100011 3 1
 R 0011011100 True

 T 0101101111
 1 1101100011 7 120
 2 0101100000 6 63
 3 1001101010 6 30
 4 0100001110 7 6
 5 1100111010 5 3
 6 0000101100 6 1
 R 0101101111 True

 T 1011101011
 1 0111010110 3 120
 2 1000100101 5 56
 3 0001111111 6 24
 4 1111100100 5 9
 5 1110110000 4 6
 6 0110000011 5 4
 7 0001010000 3 2
 8 1000110101 4 2
 9 0100101100 3 1
 R 1011101011 True

 T 1101010001
 1 1010000011 5 252
 2 1001111001 7 60
 3 1110001010 4 30
 4 1101110010 7 8
 5 0101110010 6 8
 6 1001001111 5 4
 7 0011000100 4 2
 8 1011010001 8 1
 R 1101010001 True

 T 0001000111
 1 0000101110 6 210
 2 1010100101 5 100
 3 0011101001 5 46
 4 1100010111 6 18
 5 0100101000 3 5
 6 1111110000 2 3
 7 0001010001 7 1
 R 0001000111 True
Mean: 7.9

It appears that the mean number of guesses required is a little over 7.

Here is a <a href="https://sagecell.sagemath.org/?z=eJyFUltr2zAUfvevOGgwZOY2cha2EWoYDPZWKFlhD10osnOciNhSqku2bPS_VxfHcWFsehKH73YuhBBYcblRPdRCcn0CY7WQW9g6NCZ-eI8Z-PcNEXbWHsxyNuu53V0by5s9_mp2XG7xulH97Gm2YB8WJfsUCd-1sBYl1Ce4u4X5KqjN2by8Yu-vyo8ZISTLWu2ddQog-oPSFgzipoAt2lCuhTXZm1Cii3meZRtsIdSozJfRhMAXJY_oeUJakGAV8IA490EiSqN1WkJL_sglK1n9TJJU3ygnLeUF1BM9XwLfYbNDAzXan-ibMPjkUDa-wuEt1K91jeupq6ojtEqDK44-C_wWh6Q7pLZoLPVBa2WwutcOR8O7jp-gBO19N6A8codx6oPHQakOKniIbbs8eQQHHeZOy5ubkuXrCLVc-7l5cMRORkg9JI8Q0cIQItlHBz8pS8kPCfekGEQSOniJiVcBJWP5hRmv5J9-4dWu6wJqmHYyKBL5gjr36S4dxpIY1-TOHKiqJLoe2X_r69KbX7xYzp_JIFAkdgEdShpMJmG90FgNPuVrvVoj3_9vkrDyTkHhga3HTxCbzna4HZFlnz0JNW9sukkuJNVOmqpkBUwvpgDurHp0hw23WH3lnTkf0d4Pjo0be7xsLOhM9rWHd9WrS0xRhti3yOXSJ287xS3dwwwiPX8Bd-04PQ==&lang=sage" rel="nofollow noreferrer">live version of the program, running on the SageMathCell server. If you set the runs parameter to 10000, it will take about one minute in non-verbose mode. I haven't tested higher numbers, but if a computation takes too long the server will abort the job.


问:
$ \int _ {- 1} ^ {3} [| x ^ 2-6x | + \text{sgn} (x-2)]\, \text {d} x $

Calculate the integral $$\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x .$$

We know that the two functions are defined as follows,
\begin{align*}
f (x) &= | x ^ 2-6x |\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x^2-6x>0; \\
6x-x^2, & \text{ if } x^2-6x<0.
\end{cases}\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x\in(-\infty, 0)\cup (6,+\infty); \\
6x-x^2, & \text{ if } x\in(0,6).
\end{cases}.
\end{align*}
And, $$\text{sgn}(x-2)=\begin{cases}
1, & \text{ if } x>2; \\
-1, & \text{ if } x<2.
\end{cases}$$

I know that the integral gives $ 58 / $ 3. But someone can explain to me the step by step, how to find that result.

I considered doing it with the graph of the function which is the following,

Calculating, the area of the triangle on the left from $ -1 $ to $ 0 $, and then calculating the two on the right and adding it, but I don't know how to calculate the area on the right.

答:
Let's first work on getting a piecewise equation for the whole thing.  There are three points where the pieces can change, namely $x = 0, 2, 6$.  So, we will consider each interval separately when writing down the piecewise equation.  For example, for $x<0$, we need to consider the following:
$$\begin{cases}
\color{red}{x^2-6x}, & \color{red}{x\in(-\infty, 0)\cup (6,+\infty)} \\
6x-x^2, & x\in(0,6)
\end{cases} + \begin{cases}
1, & x>2; \\
\color{red}{-1}, & \color{red}{x<2}.
\end{cases} = \boxed{x^2 - 6x - 1,\quad x<0 \ }$$

In the same way we can obtain the rest of the cases:
$$\boxed{| x ^ 2-6x | + \operatorname{sgn} (x-2) = \begin{cases} x^2 - 6x -1, & x<0\\ 6x - x^2 -1, &0<x<2\\ 6x - x^2 + 1, &x>2 \end{cases}}$$
So, this tells us that the integral can be split up as
$$\boxed{\int_{-1}^{0}\bigr[ x^2 - 6x - 1\bigl]\, \text {d}x + \int_{0}^{2}\bigl[6x - x^2 -1\bigr]\, \text {d}x  + \int_{2}^{3}\bigl[6x - x^2 + 1\bigr] \, \text {d}x \ }$$

ContentLicense: CC BY-SA 4.0


问:
Show that $\sum_{n\geq 1} \frac{a_n}{n^z}$ defines a holomorphic function in $\{ \operatorname{Re}z > 1\}$
I am trying to make the following demonstration: let $(a_n)_n\subset \mathbb{C}$ be a sequence fulfilling that for all $\delta>0$,
$$\sup_n \dfrac{|a_n|}{n^\delta}<+\infty. $$
Show that the series $\displaystyle \sum_{n\geq 1} \dfrac{a_n}{n^z}$, $\textrm{Re}\, z>1 $, defines a holomorphic function on the half-plane $\textrm{Re}\, z> 1 $.

What I have done is: I first show that $\displaystyle \sum_{n\geq 1} \dfrac{a_n}{n^z}$ converges uniformly in every compact rectangle $R=[a, b] \times [c, d] \subset \mathbb {C} $ with $ a> 1 $. For this I will see that the sequence of partial sums is a uniformly Cauchy sequence. Let $ \beta> \alpha $, and $ z = x + iy \in R $ with $ x> 1 $

\begin{align*}
      \left| \sum_{n=1}^\beta \dfrac{a_n}{n^z}-\sum_{n=1}^\alpha \dfrac{a_n}{n^z}\right| &=\left| \sum_{n=\alpha +1}^\beta \dfrac{a_n}{n^z}\right|\leq \sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{|n^z|}=\sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{n^x} \\
      & \leq \sum_{n=\alpha+1}^\beta \dfrac{|a_n|}{n^a} \leq \sum_{n=\alpha+1}^\beta M=M(\beta -\alpha)
  \end{align*}

where $M=\displaystyle \sup_{n\in \{ \alpha+1,\ldots, \beta\}} \dfrac{|a_n|}{n^a}$.

If we see that there exists an integer $m$ such that if $m$ exists, then $m\to 0$ then we will have that the succession of partial sums is Cauchy uniform. This is true because as we increase $n$, $n^a$ will increase.

Therefore $\displaystyle \sum_{n \geq 1} \dfrac{a_n}{n^z}$ converges uniformly on every compact subset in the interior of the half-plane of convergence $\textrm{Re} \, z> 1 $ .

Then, $ f (z) = \displaystyle \sum_ {n \geq 1} \dfrac{a_n}{n ^ z} $ is holomorphic in the half-plane $ \textrm {Re} \, z> 1 $ and the sequence of derivatives $ (f'_n) _n $ where $ f_n (z) = \dfrac {a_n} {n ^ z} $ converges uniformly on each compact subset of the half-plane of convergence to the derivative $ f '$ obtained in deriving term by term .

I don't know if it is well resolved. What I don't know how to justify very well is the step that $M$ goes to $0$ for a sufficiently large $N$.

答:
Your estimate
$$
\left| \sum_{n=1}^\beta \frac{a_n}{n^z}-\sum_{n=1}^\alpha \frac{a_n}{n^z}\right| \le M(\beta - \alpha)
$$
is not good enough:

$M=\sup_{\alpha+1 \le n \le \beta} \frac{|a_n|}{n^a}$ does not necessarily become small for $\alpha \to \infty$, and

the factor $\beta - \alpha$ is not bounded independently of $\beta$.

One can proceed as follows: For $z = x+iy$ with $x > 1+2 \delta$ and $\delta > 0$ is
$$
 M = \sup_n \frac{|a_n|}{n^{\delta}} < \infty
$$
and therefore
$$
 \left| \frac{a_n}{n^z}\right| = \frac{|a_n|}{n^x} \le \frac{|a_n|}{n^{1+2\delta}}
 \le M \frac{1}{n^{1+\delta}} \, .
$$
Since $\sum_{n=1}^\infty 1/n^{1+\delta}$ is convergent, it follows with the Weierstrass M-test that the series
$$
 \sum_{n=1}^\infty \frac{a_n}{n^z}
$$
is absolutely and uniformly convergent, and therefore defines a holomorphic function in the halfplane $\{ z \mid \operatorname{Re} z > 1 + 2 \delta \}$.

This holds for all $\delta > 0$, so that the desired conclusion follows.


答:
Basic strategy in these type of problems involving Piece-wise functions is that Firstly you have to look at Limits of your integral and then check if Point of discontinuity(Here $0$ ,$2$ and $6$) lies in limit of your integral. If They lies You have to use Linearity of integral.

$$I=\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x $$

You have already determine Points where we have to use linearity of integral.

$$I=\int _ {- 1} ^ {0} (x ^ 2-6x  -1 )dx + \int _ {0} ^ {2} (-x ^ 2+6x  -1) dx +\int _ {2} ^ {3} (-x ^ 2+6x  +1) dx$$

Now, You can evaluate integrals easily.

Title: $ \int _ {- 1} ^ {3} [| x ^ 2-6x | + \text{sgn} (x-2)]\, \text {d} x $

Body:
Calculate the integral $$\int _ {- 1} ^ {3} \bigl[| x ^ 2-6x | + \text{sgn} (x-2)\bigr]\, \text {d} x .$$

We know that the two functions are defined as follows,
\begin{align*}
f (x) &= | x ^ 2-6x |\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x^2-6x>0; \\
6x-x^2, & \text{ if } x^2-6x<0.
\end{cases}\\[5pt]
&=\begin{cases}
x^2-6x, & \text{ if } x\in(-\infty, 0)\cup (6,+\infty); \\
6x-x^2, & \text{ if } x\in(0,6).
\end{cases}.
\end{align*}
And, $$\text{sgn}(x-2)=\begin{cases}
1, & \text{ if } x>2; \\
-1, & \text{ if } x<2.
\end{cases}$$

I know that the integral gives $ 58 / $ 3. But someone can explain to me the step by step, how to find that result.

I considered doing it with the graph of the function which is the following,

Calculating, the area of the triangle on the left from $ -1 $ to $ 0 $, and then calculating the two on the right and adding it, but I don't know how to calculate the area on the right.


问:
Calculus the extreme value of the function $f(x,y)$
Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$

pleasee help me.

ContentLicense: CC BY-SA 3.0

Body: min{$f(x, y) = x^2+y^2+x y+1/x+1/y$}~~4.32675 at (x, y)~~(0.693361, 0.693361)


问:
What does "closed under subsets" mean?
I understand what it means for a set to be closed under an operation, like addition or multiplication. But what does it mean for a set $T$ to be closed under subsets? Is subset an operation? I thought subsets were just sets with a certain property...

答: If $T$ is a collection of sets, $T$ is closed under subsets (or closed under taking subsets) if it has the following property: if $s\subseteq t\in T$, then $s\in T$. That is, every subset of a member of $T$ is also a member of $T$.

Title: What does "closed under subsets" mean?


问:
Dirac Delta expansion over $SU(2)_{q}$
It is known that the Dirac $\delta(U)$ distribution can be expanded in representations for compact groups.
 For example concerning the $U(1)$
$$ \delta(\phi)= \frac{1}{2\pi}\sum_{n} e^{in\phi}$$
Or another example is $SU(2)$ where the $\delta$ is expanded over the group like
$$\delta(U) = \sum_{j} (2j+1) TrD^{(j)}(U)$$
where $D^{(j)}(U)$ are the the Wigner matrices.

There exist a similar expansion-like formula for the $\delta$ over the quantum group $SU(2)_{q}$ ?

答:
Honestly, I've never worked with Wigner matrices, so I am not sure if I understand your question/motivation correctly and therefore whether my answer will be anyhow useful. But let's try:

First of all, I think that the object you are refering to as Dirac distribution is what is in the quantum group framework called the counit and denoted by $\varepsilon$. Classically, it is the functional $C(G)\to\mathbb{C}$ mapping $f\mapsto f(e)$, where $e$ is the unit element of the group. Since with quantum groups, we work already by definition with functions rather than points, the counit actually stands kind of somewhere in the centre of the whole story, so I am not sure if it useful to try to expand it somehow, but maybe yes, I don't know.

Your examples are written in a slightly informal language, so let's formalize the first one, to make sure we understand each other (as I said, I have no practice with the second one). So, take $G=U(1)$ to be the circle. Let's denote by $\phi\in G$ its elements, where $\phi$ denotes the angle on the circle. This group has the fundamental representation $\rho\colon\phi\mapsto e^{i\phi}$ and all other irreducibles are given by $\mathop{\rm Irr}G=\{\rho^k\mid k\in\mathbb{Z}\}$. Since all the irreducibles are one-dimensional (which is because $G$ is commutative), we actually have that those are just functions $G\to\mathbb{C}$, so $\rho^k\in C(G)$ and they map $\phi\mapsto e^{ik\phi}$. Those functions even form a basis of $C(G)$ (or $L^2(G)$ or whatever) by expanding to the Fourier series. Now each of those functions induce canonically a functional $\omega_k\in C(G)^*$ mapping $\omega_k(f)=\langle\rho^k,f\rangle={1\over 2\pi}\int e^{ik\phi}f(\phi)\,{\rm d}\phi$. In particular $\omega_k(\rho^l)=\delta_{kl}$. Now, you claim that
$$\varepsilon=\sum_{k\in\mathbb{Z}}\omega_k.$$

Now, how this extends to the quantum group case: Let $G$ be a compact matrix quantum group with fundamental representation $u$. Denote by $O(G)\subset C(G)$ the coordinate algebra, that is, the $*$-algebra generated by the entries $u_{ij}$ of the fundamental representation. Denote by $\mathop{\rm Irr}G$ again the set of irreducibles of $G$ (up to equivalence). For a given $\alpha\in\mathop{\rm Irr}G$, we denote by $u^\alpha$ the corresponding representation and by $n_\alpha$ its size. It holds again that the entries of the irreducibles, that is the set $\{u_{ij}^\alpha\}$ wiht $\alpha\in\mathop{\rm Irr}G$, $i,j=1,...,n_\alpha$ forms a vector space basis of $O(G)$. Therefore, any functional $\omega\in O(G)^*$ is determined by the numbers $\omega^\alpha_{ij}:=\omega(u_{ij}^\alpha)$. Consequently, we can write
$$O(G)^*\simeq\prod_{\alpha\in\mathop{\rm Irr}G}M_{n_\alpha}(\mathbb{C})$$
Now, we can take kind of the identity matrix for each of the factor. More precisely, for every $\alpha\in\mathop{\rm Irr}G$, define $\omega_\alpha\in O(G)^*$ by $\omega_\alpha(u_{ij}^\beta)=\delta_{ij}\delta_{\alpha\beta}$. Then finally, we can write
$$\varepsilon=\sum_{\alpha\in\mathop{\rm Irr}G}\omega_\alpha.$$
(Now, as I am thinking of it, I am not sure, what topology you should choose in order to make this work. But at least pointwise it's true.)

I am not sure, however, if this is what you are looking for.

EDIT: As I understand it, the Wigner matrices are the irreducible representations of $SU(2)$. That is, $D^{(\alpha)}=u^\alpha$ here. So you could write something like $\omega_\alpha\approx\mathop{\rm Tr} D^{(\alpha)}$ and you get (almost) exactly the formula you've mentioned. But formally this is incorrect, because you should use the dual forms $\omega_\alpha$ instead of the representations $D^{(\alpha)}$ (and I suppose that the $(2j+1)$ factor is probably some normalization that appears when passing to this dual).

Nevertheless, I think that if you were really seriously working with quantum groups, then instead of those sums with questionable meaning, you would probably just use the simple equality $\varepsilon(u_{ij}^{\alpha})=\delta_{ij}$ (and maybe the fact that those $u_{ij}^\alpha$ form a basis of your algebra). But it depends on what you are working on I guess.

Title: Dirac Delta expansion over $SU(2)_{q}$


问:
How to calculate probability that I will encounter full barbershop when I go for a cut?
Let's say for example that you want to get a fresh cut. I am eeger to know how could one calculate probability that barbershop will be full? What are the parameters that should be looked. I don't need anything concrete, just wanted to know what are the possible approaches? I was thinking of maybe Poisson stochastic process, but I think it doesn't fully fit the story.</p>

答:
First of all, you would have to define a sample of possible barbershops. Maybe calculate the number of barbershops within walking distance. You would also need to specify the maximum waiting time you would be willing to wait. Then I would suggest a calculating the probability that a barbershop will be full given a whole range of other parameters such as:

its capacity

average time for a person to have his hair cut,

average number of people who cut their hair in a period (could also be number of people in an area and at which intervals they go to the barbershop)

Then you could add more parameters such as the hour of the day for your search and operation hours of a barbershop. I am sure that they do not have the same load during let's say 8a.m. and 16:00. Also this could be done for different days of the week (e.g. Saturday vs Tuesday).

I would also look to somehow model a barbershop's popularity into my calculations. But if we take for granted that supply and demand for barbershops has reached an equilibrium this might not be necessary.

You could also use a Bayesian approach where as you update the aforementioned model with observations.


问:
Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$
Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$.

I tried to solve this question as follows:

$x-y=x^2+y^2+49+2xy-14x-14y$

$x+y=x^2+y^2+1-2xy-2x+2y$

Here I realized that I can't transform one equation into $x=...$ and then replace it into the other, as that would give a polynomial of too high a degree. Hence I attempted to combine them:

$-2y=48+4xy-12x-16y$

$48+4xy-12x-14y=0$

And I got stuck here.

Could you please explain to me how to solve the question, as I feel I'm very close?

答:
Try $1≤a=x-y, 7≤b=x+y$

$$\begin{cases} a=(b-7)^2 \\ b=(a-1)^2 \end{cases} \implies a=((a-1)^2-7)^2$$

$$a^4 - 4 a^3 - 8 a^2 + 23 a + 36 = 0$$

$$(a - 4) (a^3 - 8 a - 9) = 0$$

$$ \implies a=4≥1$$

$$\implies b=(a-1)^2=9≥7.$$

Then you can find $x$ and $y.$

The last cubic equation doesn't have rational roots.


答:
We see there are $(x+y)$ and $(x-y)$ in the equations and both are positive (since under square root). So we make the transformations
$$u^2=x-y \quad , \quad v^2=x+y$$

The new equations are
$$u=v^2-7 \quad , \quad v=u^2-1$$
Eliminating $v$, we get
$$u^4-2u^2-u-6=0$$
which has an easy root $u=2$. Can you now finish?

Body:
Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$.

I tried to solve this question as follows:

$x-y=x^2+y^2+49+2xy-14x-14y$

$x+y=x^2+y^2+1-2xy-2x+2y$

Here I realized that I can't transform one equation into $x=...$ and then replace it into the other, as that would give a polynomial of too high a degree. Hence I attempted to combine them:

$-2y=48+4xy-12x-16y$

$48+4xy-12x-14y=0$

And I got stuck here.

Could you please explain to me how to solve the question, as I feel I'm very close?


问:
Quotient Group question about order of a group
How do I prove that when an element of $G/K$ (where $K$ is normal subgroup of $G$) has an order $n$, there exists an element in $G$ having order $n$?
I could only do as the order of elements are $n$. The total number of elements in $G$ are $nm$. How to proceed?

答:
Let the element of $G/K$ of order $n$ be say $rK$. $(rK)^{|r|} = K$, so $n$ divides $|r|$, say $|r| = nk$. Then consider $r^k$. Clearly $|r^k| =n$.

(Edit: Showing $|r^k|=n$: Suppose by way of contradiction $|r^k|\neq n$. Then $|r^k|$ is, say, $l$, and we know $l$ divides $n$. Hence $l<n$. Then $(r^k)^l=1$ so $|r|\leq lk<nk$, a contradiction).


答:
Use the normal derivative test. First order conditions
$$\frac{\partial f}{\partial x}=2x+y-\frac 1{x^2}=0$$
$$\frac{\partial f}{\partial y}=2y+x-\frac 1{y^2}=0$$
If $x,y\in \mathbb R$ the the solution is $x=y=\frac 1{3^{1/3}}$</p>

Title: Calculus the extreme value of the function $f(x,y)$

Body:
Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$

pleasee help me.


问:
I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$
I state $x=2^{2002}$. How many integers are there between $\sqrt{x^2+2x+4}$ and $\sqrt{4x^2+2x+1}$?

I tried to solve it as follows:

I state $a\in\mathbb{N}$, so that $\sqrt{x^2+2x+4}<a<\sqrt{4x^2+2x+1}$

Hence $x^2+2x+4<a^2<4x^2+2x+1$

$4x^2+2x+1-x^2-2x-4=3x^2-3=3*2^{4004}-3$.

Here is where I got stuck. Could you please explain to me how to solve this question?

答:
With the lower limit,

$$\begin{equation}\begin{aligned}
& (x + 1)^2 \lt x^2 + 2x + 4 \lt (x + 2)^2 \\
& x + 1 \lt \sqrt{x^2 + 2x + 4} \lt x + 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This means, with $a \in \mathbb{N}$ as you stated,

$$a \ge x + 2 \tag{2}\label{eq2A}$$

With the upper limit,

$$\begin{equation}\begin{aligned}
& (2x)^2 \lt 4x^2 + 2x + 1 \lt (2x + 1)^2 \\
& 2x \lt \sqrt{4x^2 + 2x + 1} \lt 2x + 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

This thus means

$$a \le 2x \tag{4}\label{eq4A}$$

Combining \eqref{eq2A} and \eqref{eq4A} gives

$$x + 2 \le a \le 2x \tag{5}\label{eq5A}$$

Thus, as zwim's hints state, the number of integers in between the $2$ square root limits would be

$$2x - (x + 2) + 1 = x - 1 = 2^{2002} - 1 \tag{6}\label{eq6A}$$


问:
A question about ideal of a polynomial and its vanishing set
I'm trying to show the following, but not sure how to proceed:

Let $f(x,y) = x^2 + y^2 - 1 \in \mathbb{C}[x,y]$. Suppose that $g,h \in \mathbb{C}[x,y]$ s.t. for all $(a,b) \in V(f)$ we have $g(a,b)=h(a,b)$. Then the polynomial $g(x,y)-h(x,y) \in \langle x^2 + y^2 -1 \rangle$.

Now I know that on the set of points in $V(f)$ $g(a,b)-h(a,b) =f(a,b)$. Then intuitively I would think that $g(x,y)-h(x,y) = f(x,y)$ but this doesn't need to be the case I guess right? Otherwise the problem would be trivial.

答:
Set $F(x,y)=g(x,y)-h(x,y)$. From the fact that $F(a,b)=0$, for every $(a,b)$ such that $a^2+b^2=1$ you'd like to prove that $F(x,y)=G(x,y)(x^2+y^2-1)$ for some polynomial $G$.

Let's consider $F$ as an element of $\mathbb{C}(x)[y]$, where we can perform the division
$$
F(x,y)=Q(x,y)(y^2+x^2-1)+U(x)y+V(x)
$$
where $Q\in\mathbb{C}(x)[y]$ and $U,V\in\mathbb{C}(x)$. If $(a,b)\in V(f)$, then
$$
0=U(a)b+V(a)
$$
Writing $U=U_0/W,V=V_0/W$, where $U_0,V_0,W\in\mathbb{C}[x]$, we obtain $U_0(a)b+V_0(a)=0$. In particular $U_0(a)^2b^2=V_0(a)^2$, which yields
$$
U_0(a)^2(1-a^2)-V_0(a)^2=0
$$
Thus the polynomial $U_0(x)^2(1-x^2)-V_0(x)^2$ has infinitely many roots and therefore is the zero polynomial. This is a contradiction, because $1+x$ divides $U_0(x)^2(1-x^2)$ an odd number of times, but $V_0(x)^2$ an even number of times.


问:
What does this notation mean for a squared matrix?
A is a square matrix

$A^n = A \cdot A\cdot$...$n$ times

What's $A^{2n}$?

$2A^n$ or $A \cdot A\cdot$...$2n$ times?

答: It means $A\cdot A\cdot \dotsb$ $2n$ times, it's just a convenient way of making it clear that the power is even.


问:
A change in a countable set of points does not affect the integral
I know that a change in a finite number of points of an integrable function does not change the integral (Riemman integral). Recently I started to learn about Fourier transform, and in my book they define the norm $ L^2 $.

In order for $L^2$ to be a norm, we need to define an equivalence relation, and the relation is defined as the following:

If $f$, $g$ are functions suxh that $f(x)=g(x)$ **almost for any $ x \in \mathbb{R}$ ** then we say that $f$ is equivalent to $g$. Where "almost everywhere" means that the set of points of which $ f(x) \neq g(x) $ is of measure zero.

As I understood, if 2 functions are equal up to a set of points of measure zero, then the integrals of the function would remain the same.

For a finite number of points I can totally understand. But even for a countably many points I cant see how this is true. For example, if we take the 0 function in the segment $ [0,1] $ and change the rational points to be $ 1 $, we get Dirichlet's function which is not integrable.

So where is this all come from? And what is the right claim?

Thanks in advance.

答:
The book is using the Lebesgue integral. Otherwise, you are correct that changing a countable number of values of a function could turn a Riemann integrable function into one which is not Riemann integrable.

It is a theorem that whenever the (proper) Riemann integral of a function exists, it equals the Lebesgue integral. So changing a countable number of function values will result in the same Riemann integral, provided the Riemann integral exists in both cases.

For improper integrals, if the Riemann and Lebesgue integrals both exist, they are equal. But it may be that both don't exist - consider the integral of $\frac{\sin(x)}{x}$ over $(0, \infty)$ for example, where the Riemann improper integral exists but the Lebesgue integral is undefined.


问:
Focusing Property of Parabolas (Historical Question)
I'm curious about the historical development of the idea that parabolas redirect incoming light to the focus.  I'm curious how this was discovered, and whether it was discovered mathematically or experimentally.  While it is fairly easy to show that a parabola has this property, going the other way (i.e., figuring out that such a shape would be a parabola) seems more difficult. I was able to derive a parabolic equation (i.e., a quadratic) from the meaning of "reflecting all incoming rays towards a focus", but it wound up utilizing some differential equations that I'm sure were not available when this property of parabolas was first noticed in antiquity.

Anyway, I'm curious about the history of this idea, and how it was discovered/established prior to calculus.

答: Diocles was the first prove this in "On burning mirrors". There are no surviving records but fragments were found in Eutocius's notes on Archimedes "Spheres and Cylinders". Conic sections were well-studied in antiquity including parabolas but records were mostly kept by Arabic mathematicians. Ibn Sahl has a surviving proof which likely follows Diocles' methods.</p>


问:
Attempt at solving $\frac{dy}{dx} = x^2 + y^2$
I read in my maths textbook that an analytic solution to $\frac{dy}{dx} = x^2 + y^2$ is not known. Nevertheless, as an exercise I tried to do it, to see where the difficulty lies.

In the same textbook, I learnt of a general method for solving ones of the ordinary linear form: $\frac{dy}{dx} + Py = Q$
where $P, Q$ are purely functions of $x$, or constant. The method involved finding $R = e^{\smallint P \space dx} $, resulting in $R\cdot\frac{dy}{dx} + RPy \equiv \frac{d}{dx}[Ry]$ and then finding $y = \frac{1}{R}\smallint RQ \space dx$. I attempted to form this for powers of two, with the following being the key equation:
\begin{equation} 2y \cdot \frac{dy}{dx} \cdot R + RPy^2 \equiv \frac{d}{dx}[Ry^2]
\end{equation}
I thought then, that finding $R$ in the same way would result in a very valid method for solving simple quadratic differential equations. I attempted this for $\frac{dy}{dx} = x^2 + y^2$:
\begin{equation}\frac{dy}{dx} - y^2 = x^2\end{equation}
\begin{equation}
2y \cdot \frac{dy}{dx} - 2y \cdot y^2 = 2y \cdot x^2, P = -2y
\end{equation}
\begin{equation} R = e^{\int -2y \space dx} = \frac{1}{e^{2yx}} \cdot C, C = 1 \text{ for simplicity.}
\end{equation}
\begin{equation}
2y \cdot \frac{dy}{dx} \cdot \frac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \frac{1}{e^{2yx}} = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}}
\end{equation}
\begin{equation}
\frac{d}{dx}[\frac{1}{e^{2yx}} \cdot y^2] = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}}
\end{equation}
\begin{equation}
\frac{1}{e^{2yx}} \cdot y^2 = \int 2y \cdot x^2 \cdot \frac{1}{e^{2yx}} \space dx
\end{equation}
which equals what, exactly? I am at a loss as to how to solve this integral. I pose four questions:

(1) What is the result of that final integral?

(2) Was my attempted method sound?

(3) What is the better approach to solving differential equations with a second power? I assume my method is not the best - if it works at all!

(4) If that final integral does have a result, then why did my textbook claim no solution exists?

答: Note that $\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{y^2}{e^{2xy}} \right)= 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}\cdot \color{red}{\dfrac{\mathrm dy}{\mathrm dx}}\ne 2y \cdot \dfrac{\mathrm dy}{\mathrm dx} \cdot \dfrac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \dfrac{1}{e^{2yx}}$


问:
Solve $\sqrt[3]{3-2x} + \sqrt{3x^2+4} = \frac{x}{2} + 2$
I tried letting $u$ = the cube root and $v$ for the square root and reach some easy system of equations, it did not work for me, I still got $x$ in the system of equations. I tried expressing them all as a function of $x/2 + 2$ or $3-2x$ but it did not seem like a great simplification. Algebraic solutions are preferred.

答:
You started off on the right track.

Let $v^2 = 3x^2 +4$ and $u^3 = 3-2x$. Now, we have the relation
$$x = \frac{3-u^3}{2} \quad x^2 = \frac{v^2-4}{3}$$
squaring the first equation, gives us the following relation between $u$ and $v$.
$$\frac{v^2 - 4}{3} = \left( \frac{3-u^3}{2}\right)^2$$
simplifying this gives
$$v^2 = 3\left(\frac{3-u^3}{2}\right)^2 + 4$$
On the other hand, we also have from the main equation
$$u+v = \frac{3-u^3}{4} + 2$$
solving for $v$ and squaring gives
$$v^2 = \left(\frac{11-4u - u^3}{4}\right)^2$$
So, you have two equations now for $v^2$, equate both to get
$$\left(\frac{11-4u - u^3}{4}\right)^2 = 3\left(\frac{3-u^3}{2}\right)^2 + 4$$
This is a sixth degree equation in $u$, but there wont be any odd exponents; so you can let $t = u^2$ to get a cubic equation in $t$ which should be more or less easy to solve given that you have $x=2$ already a solution as suggested in the comments (now use that and follow the right steps through $u$ and $v$ to find the $t$ corresponding to $x=2$). Now, you can use the rational root theorem to reduce to a quadratic and find the roots for $t$ using the quadratic formula. Then backtrack your steps to find all possible $u$ (basically $u = \pm\sqrt{t}$), from which you can get all possible $x$'s (use the first relation in my answer relating $x$ and $u$). From the final list, take the real $x$ values which is the answer you seek.

The last bit is a little long and painful to calculate, but isn't too hard in spirit, which is why I am not filling in all the details. All the best!


答:
I would suggest assuming $x+y=a$ and $x-y=b$. This forms a 2x2 system that given $a$ and $b$ can quickly be solved. After that you have

\begin{cases} \sqrt{b}=a-7 \\ \sqrt{a}=b-1  \end{cases}
$$\Leftrightarrow  \begin{cases} b=(a-7)^2 \\ a=(b-1)^2  \end{cases} $$
$$\Leftrightarrow  b=((b-1)^2-7)^2 $$
$$\Leftrightarrow  b=(b^2 - 2b +1 -7)^2 $$
$$\Leftrightarrow  b=(b^2 - 2b -6)^2 $$
$$\Leftrightarrow  b=b^4 -4b^3 -8b^2 +24b +36 $$
$$\Leftrightarrow  b^4 -4b^3 -8b^2 +23b +36 =0 $$
$$\Leftrightarrow  (b-4)(b^3-8b-9) =0 $$

which has two solutions. One is obviously 4 and the other on can be approximated numerically or computed using the cubic formula found here: https://math.vanderbilt.edu/schectex/courses/cubic/

After computing the two possible solutions you can substitute to find a here  $a=(b-1)^2$  and then solve the system.

Moreover the solutions for b are $b=4$ and $b \approx 3.27803\dots $ . For b =4 we have that $a=(4-1)^2 = 9$ so
\begin{cases} x+y=9 \\ \ x-y=4  \end{cases}
$$\Leftrightarrow  \begin{cases} x=13/2 \\ y=5/2  \end{cases} $$

And you do the same for $b \approx 3.27803\dots $

Title: Find the solutions $(x, y)$ of the system $\sqrt{x-y}=x+y-7$ and $\sqrt{x+y}=x-y-1$


问:
If an exact form vanishes on a submanifold, can I find a primitive that also does?
Let $M$ be a closed smooth manifold and $Q\subset M$ a closed embedded submanifold.
Furthermore, let $\omega$ be an exact differential form $\omega\in\Omega^k(M)$ and vanishing identically on $Q$ (i.e. $\omega_q=0$ for any $q\in\ Q$).

Can we always find a primitive $\alpha\in\Omega^{k-1}(M)$ (i.e. such that $d\alpha=\omega$), whose restriction to $Q$ also vanishes?

This question came up during class and maybe it is obvious but I can't even seem to convince myself whether it is true or not, so any help is greatly appreciated.

Edit: counterexamples are given in the comments for $k=\dim Q+1$ and for the case of $\omega$ being $1$-form with $Q$ disconnected. In the context of the class, we were specifically considering $\omega$ to be a $2$-form, but I am also interested in the general case.

Edit2: There is an answer dealing with the condition of the pullback of $\omega$ to $Q$ being $0$. However, I meant that $\omega$ itself vanishes identically in points that belong to the submanifold $Q$.

答:
Here is an honest example of the failure. Take $\omega = dx_1\wedge dx_2 + dx_3\wedge dx_4$ to be the standard symplectic form on $M=\Bbb R^4=\Bbb R^2\times (\Bbb R^2)'$. Let $Q$ be the torus $\{x_1^2+x_2^2=x_3^2+x_4^2=1\}$. Then of course the restriction of  $\omega$ to $Q$ is identically $0$. Let
\begin{align*}
C&=\{x_1^2+x_2^2=1, x_3=x_4=0\}=\partial D\subset\Bbb R^2 \quad\text{and} \\ C'&=\{x_1=x_2=0, x_3^2+x_4^2=1\}=\partial D'\subset (\Bbb R^2)'.
\end{align*}
Then note that if $\omega=d\alpha$, we have $\int_C \alpha = \int_D d\alpha = \int_D \omega\ne 0$ (and similarly for $C'$). Thus, the restriction of $\alpha$ to $Q$ cannot be identically $0$.


问:
Volume of solid of revolution by revolving the region $y=x^2$,$x=0$,$y=9$
Find the volume generated by revolving the region shown below:

I have done as follows:
Revolve the rectangle formed by the vertices $(0,0),(3,0),(3,9),(0,9)$ about $X$ axis.
The required volume generated is the volume of cylinder of radius $9$ and height $3$ which is $$V_1=\pi (9)^2(3)=243\pi$$

Now revolve the region formed by $y=x^2$,$x=3$ and $X$ axis about the $X$ axis.
By Disk method the volume is:
$$V_2=\pi \int_{0}^{3}(x^2)^2dx=\frac{243\pi}{5}$$
So the required volume is
$$V_1-V_2=\frac{972\pi}{5}$$
Is there any other approach?

答:
We could set it up by integrating volume of cylindrical shells as well (shell method). As we are rotating the region around x-axis,

At distance $y$ from x-axis, the shell width = $ \sqrt y \ $ and $0 \leq y \leq 9$.

So $ \ V = \displaystyle \int_0^9 2 \pi y \ \sqrt y \ dy = \frac{972 \pi}{5}$

We could also set up using the disk method as (similar to what you did),

$\displaystyle \int_0^3 \int_{x^2}^{9} 2 \pi y \ dy \ dx = \frac{972 \pi}{5}$


问:
How to prove $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$?
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.</p>

答: How about applying the triangle inequality to $\parallel x - y + y \parallel$? </p>


问:
Is the primitive of a density function Hölder continuous?
Let $I=[-1,1]$ and let $f\in L^1(I)$.

Is it true that the function $F:I\to\mathbb{R}^+$ defined as
$$F(s)=\int_{-1}^s |f(x)|dx$$
belongs to $C^{0,\alpha}(I)$ for a certain $\alpha>0$?

答:
For $j=1,2\dots$ let $\delta_j=2^{-j^2}$. Define $$c_j=\delta_j^{-1+\frac1j},$$
$$I_j=(0,\delta_j)$$ and $$f=\sum_1^\infty c_j\chi_{I_j}.$$ Since all the terms are non-negative, for every $\alpha>0$ we have $$F(\delta_j)-F(0)=F(\delta_j)\ge\int_0^{\delta_j}c_j=\delta_j^{1/j}\ne O(\delta_j^\alpha).$$


问:
If $V$ is irreducible then dim $Hom(V, W)^G$ is the multiplicity of $V$ in $W$?
From Fulton and Harris, p. 16,

If $V$ is irreducible then dim $Hom(V, W)^G$ is the multiplicity of $V$ in $W$ by Schur's lemma.

The above is for a finite group $G$.

I tried to prove this but can't quite finish it.

For a map $\varphi \in Hom(V, W)^G$, we know that $Ker(V)$ is trivial, because V is irreducible.

Therefore from Schur's lemma $\varphi(V)$  is an isomorphism onto its image and $\varphi = \lambda I$.

So, we have now that each $\varphi$ maps a 1-dimensional irreducible rep, $V$ onto itself in $W$, and the dimension of $Hom(V, W)^G$ tells us how many different one dimensional subspaces of $W$ it maps into, or the multiplicity of $V$ in $W$.

Is my argument basically correct? Also, does this say anything about the dimension of W? I think not, since W may not just be a direct sum of representations of $V$.

答:
Using the hints that runway44@'s suggested, I think the argument is roughly like so.

From a corollary of Schur's lemma, for finite groups a representation $V$ is decomposable into a direct sum of irreducible representations.

So we can take, $$W = V^1 \oplus ... V^i \oplus U^1\oplus ...U^j$$
where the $V^i$ are isomorphic representations of $V$ and the $U^j$ are any other representations.

Now, as mentioned in the comments above,
$$Hom(V,V^1 \oplus ... V^i \oplus U^1\oplus ...U^j) \cong Hom(V,V^1) \oplus
...Hom(V,V^i) \oplus Hom(V,U^1) \oplus...\oplus Hom(V,U^j)$$

We get this via projection maps $\pi_V^i$ and $\pi_U^j$. This follows from the fact that  for $w \in W$, $w = v^1+...v^i+u^1+...u^j$, uniquely.

So, $\varphi_{V^i} = \pi_{V^i} \circ \varphi$.

Now if we restrict to $Hom(V,W)^G$, the subspace of $G$-linear maps,
$V$ is irreducible, therefore, $\varphi_{V^i}$ is an isomorphism onto its image or the trivial map.
And if it is an isomorphism, $\varphi_{V^i} = \lambda I$ from Schur's lemma.
And $\varphi_{U^i}$ are all the trivial rep.

Therefore, each map in $Hom(V,W)^G$, consists of choosing a $\lambda^i$ for each $V^i$ in the decomposition of $W$. Hence, the dimension of $Hom(V,W)^G$, is just the multiplicity of the $V^i$.


问:
how big can be a rectangle inscribed into a square if...
Please forgive my English. In a square with side AB, there is inscribed a rectangle, whose area is $\frac{3}{8}$ the square area, if the measure of AB  is 28, how long can be $AP$, where $P$ is the vertex of the rectangle on the side $AB$ of the square, please have any hint?

答:
For simplicity let's suppose AP=AQ=x, we have:

$PQ^2=2x^2\Rightarrow PQ=x\sqrt2$

$PS^2=2(28-x)^2 \Rightarrow PS=(28-x)\sqrt 2$

$S_{PQRS}=\frac 38 (28)^2=194$

Therefore:

$PS\times PQ=2x(28-x)=294$

Or:

$x^2-28x+147=0$

Which gives $x=7$


问:
Find all the numbers that are equal to one quarter of the sum of their own digits

The number $1.5$ is special because it is equal to one quarter of the sum of its digits, as $1+5=6$ and $\frac{6}{4}=1.5$   .Find all the numbers that are equal to one quarter of the sum of their own digits.

I was puzzling over this question for a while, but I couldn't find a formula without using the $\sum$ , but I can't really solve generalizations, only come up with them. The only thing I could come up with was to brute-force it, but I can't really come up with any 'special' numbers. Any help?

答:
My first draft was messy.

With hindsight.

Let $N = \frac {K}4 = M\frac i4$ where $K$ is the sum of the digits. $M$ is the quotient integer of dividing $K$ by $4$ and $i=0,1,2,3$ is the fractional remainder.

Note: $K$ is the sum of the digits of $N$ which is also the sum of the digits of $M$ plus the sum of the digits of $\frac i4$.

So $4N = 4M + i =K$.  If the digits of $M$ are $d_k$ then

$4N = 4M + i=4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0  + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}$

While $K =d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$ so

$4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0  + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}=d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$

$d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}= \text{sum of the digits of }\frac i4$

But $\text{sum of the digits of }\frac i4 =\begin{cases}0\\2+5=7\\5\\7+5=12\end{cases}$

So we have $d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 = \begin{cases}0-0=0\\2+5-1=6\\5-2=3\\7+5-3=9\end{cases}$

But the RHS is less than $10$ so none of the $d_{k_{k> 0}}$ can be none zero and $M$ is a single digit, $d_0$.

And we have

$3d_0 = \begin{cases}0\\6\\3\\9\end{cases}$

So $M=d_0 = \begin{cases}0\\2\\1\\3\end{cases}$

ANd $N = M +\frac i4 = \begin{cases}0\\2.25\\1.5\\3.75\end{cases}$

==== first answer below (more thought and scrabble and not as slick-- a "rough draft")=====

So the sum of the digits is an integer.  And and integer divided by $4$ will result in four possible cases.

$\frac {integer}4 = \begin{cases}n.00\\n.25\\n.5\\n.75\end{cases}$.

So the least power of $10$ in the number can be $-2$.

So let $N = \sum_{k=-2}^m a_k\times 10^m$ (where $a_m \ne 0$ but the other $a_k$ may be).

$ \sum_{k=-2}^m a_k\times 10^m = \frac {\sum_{k=-2}a_k}4< \frac{9\times (m+2)}4$ but even more so $a_{-1}+ a_{-2} \le 7+5 = 12$ so $N < 9m + 12$ and $a10^m \le N < \frac {9m + 12}4<3m + 3$ can give us an upper limit of $m$.

If $m \ge 2$ then $3m + 3 < 10^m$ obviously so $m\le 1$.  If $m =1$ then $10a_1 < 6$ is impossible.  SO $m\le 0$

So we have four cases:

$N = a.00$ and so $a = \frac 14 a$ and $a = 0$ and $N = 0$.  That's a solution.

$N = a.25$ and $a +\frac 14 = \frac 14(a + 7)$ so $4a+1=a+7$ so $3a=6$ so $a=2$ and $N = 2.25$ yield $2.25 =\frac 14 \times 9$.  That's a solution.

$N = a.5$ and $a + \frac 12 = \frac 14(a+5)$ so $4a + 2 = a+5$ so $3a = 3$ and $a = 1$ and $N=1.5$ yields your solution.

$N = a.75$ and $a + \frac 34 =\frac 14(a+12)$ so $4a + 3=(a+12)$ so $3a =9$ and $a = 3$ and $N=3.75 = 3\frac 34 = \frac {15}4 = \frac {3+7+5}4$ is a solution.

Pretty cool and pretty cute.


答: Use triangle inequality and norm properties to show that $$\lVert x\rVert-\lVert y\rVert\le\lVert x-y\rVert$$ and $$\lVert y\rVert-\lVert x\rVert\le\lVert x-y\rVert$$</p>

Title: How to prove $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$?

Body: I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.</p>


问:
$3$ Boys and $3$ Girls sit at a circular table. What is the probability the boys sit together?
Three boys and three girls sit at a circular table. No boy or girl is more likely to sit on a particular chair. What is the probability the three boys all sit together?

I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving $5!$ possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy ($\times2$) and the next boy can sit to the right or left of him ($\times2$).

And as the $3$ boys can be arranged in $3!$ ways this gives $2\times2\times3!$ possibilities for the boys out of $5!$ overall. However this doesn't seem to work as the answer is $3/10$. This would fit with $\frac{3\times2\times3!}{5!}$ but I cant see why to change one of these twos into the required three. Am I missing something?

Many thanks for any help - I am sure it is something obvious I am overlooking so feel free to say so

答:
Call the first boy A, second boy B, and third boy C. Your method doesn't allow for the ordering ACB.

The correct logic is as follows: make the left-most boy sit in a fixed chair - so we have BBB GGG. Then there are a total of $(3!)^2$ ways of making such an arrangement. Divide this by $5!$ and you get $3 \over 10$.


问:
Find the image of line that go through zero under the exponential function
I am trying to find the image of $f(z)=e^z$ while $ z \in \mathbb{C}$. z is a line that go through the origin.

I know that $z=x+iy$ while $x,y$ real numbers. Writing $$e^{z} = e^x(\cos y + i\sin y),$$

now the meaning of the set is that $y=nx \quad n \in \mathbb{R}$

so I substitute this condition and get
$$e^x(\cos nx + i\sin nx)= e^x(\cos x + i\sin x)^n.$$

Is my method correct?

What is this set? can I say that $e^x(\cos nx + i\sin nx)$ is a circle with radius $e^x$? how does it look on the plane?

答:
I don't think your parameterization is quite right. You have $y = nx$, $n\in\mathbb{R}$, so your set is $\{(x,nx): n\in \mathbb{R}\}\subset\mathbb{R}^2$. But you're paramterizing $z$ a line through the origin in the complex plane. Your line through the origin in the real plane is a different set.

You want to find the image of a line through the origin in the complex plane. We can parametrize this as $z(r) = re^{i\theta}$, where $r\in \mathbb{R}$ and $\theta$ is constant. So $$f(z(r)) = e^{re^{i\theta}} = (e^{e^{i\theta}})^r.$$ Since $\theta$ is fixed $e^{e^{i\theta}}$ is a constant complex number, denote it with say $c$. Then $f(z(r)) = \{c^r: r\in \mathbb{R}\}$.


问:
Showing a set is closed in the real metric if it is also closed in $d_2$ metric

Above is the exercise. Showing that $S$ is bounded is straightforward by $A$ bounded and the triangle inequality, but I thought that to show S is closed, I would do the usual thing of assuming $w \in \bar S$, then by definition, there is a sequence $\{s_n\}$ with elements in $S$ so that $s_n \to x$ w.r.t. absolute metric.

The problem is now I am stuck and can't seem to use the fact that $A$ is closed to conclude that $w \in S$. Is this approach even possible? And is there a better approach, perhaps showing that $\mathbb{R^2} \backslash A$ is open is easier? Thanks!

Note: Please do not use compactness as I have not covered it yet!

(I have an exam soon and so I would like exposure to as many problems as possible. Sorry if I haven't shown too much work before posting.)

答:
I know you don't want to use compactness, but I think that solving this without compactness will result in an inefficient and/or tedious proof. Simply because the proof is almost trivial using compactness:

$f(x,y) = x + y$ is a linear transformation so it is continuous. Since $A$ is closed and bounded it is compact, so its image under $f$ is compact. Hence $S$ is closed and bounded.


问:
Confused by step in calculating fundamental groups of cell complexes using Seifert-Van Kampen theorem
I understand the statement of the Seifert-Van Kampen theorem - given an open cover ${A,B}$ of a space ${X}$ such that ${A\cap B}$ is non-empty and path-connected, then
$$
\pi_1(A\cup B) = \frac{\pi_1(A)*\pi_1(B)}{N}
$$
where ${*}$ is the free product, and $N$ is the normal subgroup of ${\pi_1(A)*\pi_1(B)}$ given by
$$
N = \{f_*(c)g_*^{-1}(c)\ |\ c \in \pi_1(A\cap B)\}
$$
where ${f_*}$ and ${g_*}$ are induced from the inclusions ${f : A\cap B\to A}$ and ${g : A\cap B\to B}$ respectively. I understand this is essentially to stop us from "overcounting" loops essentially.

Applying this to the torus, looking at the fundamental polygon for it, - we can use the open cover ${A=T^2\setminus \{x\}}$ for $x$ in the interior of the square (so the punctured torus) and ${B =}$ interior of the square of the torus. We would then have ${A,B}$ are open, and ${A\cap B}$ is clearly path-connected and ${A\cap B\simeq S^1}$. This gives us that
$$
\pi_1(T^2) = \frac{(\mathbb{Z}*\mathbb{Z})*\{0\}}{N} = \frac{\mathbb{Z}*\mathbb{Z}}{N}
$$
but then I am confused about how we can find ${N}$. I understand that for every $c$, we have ${g_*(c)=1}$, hence we essentially just have $N$ being generated by ${f_*(c)}$. But how would we find ${f_*(c)}$?

答:
I'll think of $\mathbb{T}^{2}$ as $[0,1]^{2}/\sim$, where $(0,t)\sim(1,t)$ and $(t,0)\sim(t,1)$.

Let $a$ be the loop $[t,0]$, $b$ the loop $[0,t]$ and $c$ a counterclockwise simple circle around $(\frac{1}{2},\frac{1}{2})$.

Notice that $A=\mathbb{T}^{2}\setminus\{[\frac{1}{2},\frac{1}{2}]\}$ deformation retracts into $\partial[0,1]^{2}/\sim$ (the 'border' of the square), which is homeomorphic to a wedge of two circles. Hence, $\pi_{1}(A)=\mathbb{Z}*\mathbb{Z}$ and a system of generators is given by $a$ and $b$.

Now, $B=(0,1)^{2}/\sim$ is contractible, so $\pi_{1}(B)=0$.

Lastly, $A\cap B$ deformation retracts into a circle, so $\pi_{1}(A\cap B)=\langle c\rangle$ is infinite cyclic.

The Van Kampen Theorem tells us that $\pi_{1}(\mathbb{T}^{2})$ is therefore the quotient of $\pi_{1}(A)*\pi_{1}(B)=\mathbb{Z}*\mathbb{Z}$, with generators $a,b$ and some relations.

I believe that it is easier to describe the relations as "The value of $c$ (generator of $\pi_{1}(A\cap B)$) as a loop in $A$"="The value of $c$ as a loop in $B$". We need to see what $c$ is path-homotopic to in both cases.

In $B$, since $B$ has trivial fundamental group, clearly $c=1$.

In $A$, by choosing as a base point, for example, $(\frac{3}{4},\frac{1}{2})$, and $c$ to be the circle of radius $\frac{1}{4}$ parametrized counter-clockwise, one can see (if you wish, I can elaborate on this part), that "bringing $c$ to the edge" gives $c=ba^{-1}b^{-1}a$.

Therefore, the resulting equation is $ba^{-1}b^{-1}a=1$, which is the same as $ab=ba$.

In conclusion,

$$
\pi_{1}(\mathbb{T}^{2})=\frac{\langle a,b \rangle}{\langle aba^{-1}b^{-1} \rangle}=\mathbb{Z}\oplus\mathbb{Z}.
$$

Hope this helps!

EDIT: on the "bringing $c$ to the edge part", I'm adding some (rather horrible) pictures:

First, I'll adjust the generators of our fundamental groups to the base point $p$.

<img src="https://i.stack.imgur.com/jNInV.jpg" alt="The red lines represent $a$, and the numbers indicate the order in which you must travel along $a$" />

<img src="https://i.stack.imgur.com/SOVHH.png" alt="The blue lines are also $b$, same deal as before" />

Our generator for the intersection is

We wish to see what $c$ is homotopic to in $A$. We just stretch $c$ to the edge (like when we try to stretch pizza dough to a tray!), and we get something like this:

You can certainly do this in the square (without the identifications), and applying the projection map to the homotopy (in the square), you get a homotopy (in the torus).

Now, we break up $c$ into 4 parts (I'm implicitly adding new paths, but since I travel along them in both directions, they cancel out, so I'm not cheating):

Finally, use the identifications in the torus (and the directions in which $a$ and $b$ are traveled), to see that (modulo homotopy), $c_{1}=b$, $c_{2}=a^{-1}$, $c_{3}=b^{-1}$, $c_{4}=a$, so $c=ba^{-1}b^{-1}a$.


问:
Prove area of Koch snowflake by mathematical induction
Here is an interesting construction of a geometric object known as the Koch snowflake.
Define a sequence of polygons $S_0$, $S_1$ recursively, starting with $S_0$ equal to an equilateral triangle with unit sides. We construct $S_{n+1}$ by removing the middle third of each edge of $S_n$ and replacing it with two line segments of the same length.

Let $a_n$ be the area of $S_n$. Observe that $a_0$ is just the area of the unit equilateral triangle which by elementary geometry is $\frac{\sqrt{3}}{4}$

Prove by induction that for all $n \geq 0$, the area of the nth snowflake is given by:
$$a_n = a_0\left(\frac{8}{5} - \frac{3}{5} \left(\frac{4}{9}\right)^n\right).$$

答:
Let $e_n$ be the number of edges of $S_n$, so that $e_0=3$. At each stage you replace each edge by $4$ edges, so $e_{n+1}=4e_n$, and it’s not hard to see that $e_n=3\cdot4^n$ for each $n$.

Let $t_n$ be the area of one of the triangles added added to get $S_n$; $t_0=a_0=\frac{\sqrt3}4$, and a little thought shows that $t_{n+1}=\frac19t_n$ and hence that $t_n=\frac{a_0}{9^n}$.

Clearly $a_{n+1}=a_n+e_nt_{n+1}$; can you carry out the induction step now?


问:
Sufficient conditions for a mobius transformation to map the unit circle to itself.

Find necessary and sufficient conditions for a Mobius transformation $T(z)=\frac{az+b}{cz+d}$ to map the unit circle to itself. So if $\gamma$ is a circle, $T(\gamma)=\gamma$.

I've worked out the necessary conditions. Namely, if $T(\gamma)=\gamma$, then

$|a|^2+|b|^2=|c|^2+|d|^2$

$a\bar{b}=\bar{d}c$

$\bar{a}b=d\bar{c}$

Source: Conway's Complex Functions of One Variable

How does one go about showing sufficiency? Should I simply assume conditions 1),2) and 3) and try to prove that $T(\gamma)=\gamma$? If so I can simply claim that all the implications I used to get these conditions also work backwards. Or just show that $|\frac{az+b}{cz+d}|=1$ by these conditions, which is rather simple. Is that all there is to this? I just wish the whole "necessary and sufficient" language was scrapped for some direct notation.

As an aside, I'm wondering if I'm using the words "necessary" and "sufficient" correctly in this context. Is what I showed in the first part the necessary conditions (that's what makes sense to me semantically, because they are necessary once I've assumed the map), or are they the sufficient conditions?

答: Consider the mapping of the complex plane G(w) =2w^2 on itself. What happens to the unit circle (consisting of complex numbers such that |z|=1)? How many preimages does one point have with this mapping?</p>


问:
Having $S_n$ being a sum of i.i.d. Bernoulli($p$) random variable, where does $S_n/\sqrt{n}$ converge to?
In our lecture, we learned about the CLT and we were shown the following application:

Let $X_1, X_2, \dots$ ~ $B(p)$, meaning that $P(X_i = 1) = p \in (0,1)$. Additionally, let $X_1, X_2, \dots$ be independent and identically distributed. Let $S_n = \sum_{i=1}^nX_i$, then (knowing the moment-conditions are fulfilled), the following convergence holds:

$\frac{S_n - np}{\sqrt{np(1-p)}} \longrightarrow N(0,1)$

Since $E[S_n] = np$ and $Var(S_n)=np(1-p)$ with $N(0,1)$ denoting a  standard-normal.

Since also the following equation holds:

$\frac{S_n - np}{\sqrt{n}} \longrightarrow N(0,p(1-p))$

I was wondering if (or why?) $\frac{S_n}{\sqrt{n}} \longrightarrow N(p,p(1-p))$  is also a right expression.

Having $E[\frac{S_n}{\sqrt{n}}] = \sqrt{n}p$, I do not understand why the expectation of the asymptotic normal distribution should be simply p (which is what my lecturer told us).

Can anyone please help out?

Thanks.

答:
Since $\frac{S_n-np}{\sqrt{n}} =\frac{S_n}{\sqrt{n}} - \sqrt{n}p$ converges in distribution to $\mathcal N\left(0,p(1-p)\right)$ as $n$ increases,

and $\sqrt{n}p$ increases without bounds as $n$ increases,

you will have    $\frac{S_n}{\sqrt{n}}$ diverging to $+\infty$.

What your lecturer may have said is that $\frac{S_n}{n} \to p$ in probability and almost surely.  This is the law of large numbers; note there is no square root in the denominator.


问:
Is every fixed point in an attractor stable?
Let $\Omega$ be a topological space and $\tau:\Omega\to\Omega$. We say that a fixed point $x_0\in\Omega$ of $\tau$ is stable if for every neighborhood $V$ of $x$, there is a neigheiborhood $U$ of $x$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag1.$$

On the other hand, $A\subseteq\Omega$ is called stable if for every neighborhood $V$ of $A$, there is a neigheiborhood $U$ of $A$ with $$\tau^n(U)\subseteq V\;\;\;\text{for all }n\in\mathbb N_0\tag2.$$

If $A$ is stable, can we conclude that every fixed point of $\tau$ in $A$ is stable?

If not, does the implication hold if $A$ is an attractor, i.e. $A$ is

forward invariant, i.e. $A=\tau(A)$;

stable;

there is a neighborhood $U_0$ of $A$ so that every $x\in U_0$ is attracted to $A$, i.e. $x\in E(A)$,

where $E(A)$ is the set of all $x\in\Omega$ for which the orbit $$\operatorname{orb}x:\mathbb N_0\to\Omega\;,\;\;\;n\mapsto\tau^n(x)$$ is eventually in every neighborhood of $A$?

答:
We cannot conclude that if $A$ is stable, every fixed point of $A$ is stable. Consider $\tau : \mathbb{R} \to \mathbb{R}$ defined by $\tau(x) = 2x$. Then $0$ is not stable, but $\mathbb{R}$ is stable.

In fact, we see that $\mathbb{R}$ is an attractor. So we can't conclude the statement even with the stronger condition of $A$ being an attractor.


问:
Countable product of completely metrizable spaces
Willard General topology

Here is the Teorem for the countable product of completely metrizable spaces. In the first way, How does this construction give us a isometry? Is there easy solution to show that first way?

答:
Note that the author is assuming (likely a convention throughout the book) that $\prod X_i$ is equipped with the product topology. In the product topology, for any fixed $(a_i)_i$ with $a_i \in X_i$ for each $i$, for each $n$, the map $\phi_n(x) := (a_1, \ldots, a_{n-1}, x, a_{n+1}, \ldots)$ is a homeomorphism onto its image $$X_n^* : = \{(x_1,\ldots) \in \prod X_i | x_i = a_i \text{ except for $i=n$ }\}.$$ Thus, any metric $\rho^*$ on $\prod X_i$ generating the product topology naturally defines a metric $\rho_n$ on any $X_n$ by $\rho_n(x,y) := \rho^*(\phi_n(x), \phi_n(y))$, and it is with respect to this specific metric that we can say that $(X_n^*, \rho^*|_{X_n^*})$ is isometric to $(X_n, \rho_n)$, and the function $\phi_n$ is an isometry from $X_n$ to $X_n^*$.

Edit to show $\phi_n$ is an isometry: Note that $\phi_n$ is an isometry by construction. Recall that a map $f$ is an isometry between $(X, d_X)$ and $(Y, d_Y)$ if $d_Y(f(a), f(b)) = d_X(a,b)$ for all $a,b \in X$. Let's just plug in for our situation, the condition becomes that $\rho^*(\phi_n(x), \phi_n(y)) = \rho_n(x, y)$ for all $x,y \in X_n$. But this is the definition of $\rho_n$, so of course it is true.

Note that the theorem in question always talks about "metrizable" spaces, not "metric" spaces. This means that the topology for the spaces is fixed, but any metric generating that topology can be used, so for the $\implies$ direction, the author is free to take a complete metric $\rho^*$ on $\prod X_i$ and then for each $n$ use the $\rho_n$ defined above as a compatible metric for $X_n$ that happens to be complete.


问:
Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. Find the length of $AB$.
Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. The line $OC$ is tangent to $(M, R)$ and intersects $(K,R)$ at $O$. Find the length of $AB$ in terms of $R$.

I attempted to solve it as follows:

$OC^2=(OD-2R)^2\cdot OD^2$

$MO^2+MC^2=OC^2$

$OA\cdot (OA+AB)=(OD-4R)^2+(OD-2R)^2$

I then attempted to combine the equations above, but they didn't give anything useful. Could you please explain to me how to solve the question?

答:
Hint: Draw perpendicular from $L$ and $M$ to $OC$. Say perp from $L$ meets $OC$ at $H$. Then $AH = HB$.

Now observe that $\triangle OLH \sim \triangle OMC$. Find length of $LH$ using similarity. Then use Pythagoras in $\triangle LHA$.

Find $AH$ and then $AB$.


答:
Drop $LP \perp AB$ with $P$ on $AB$. Then $\triangle OPL \sim \triangle OCM $ with similarity ratio $3/5$. Why?

Then $LPA$ is a $3-4-5$ right triangle whence $AB=8R/5$.

(Intermediate steps are left for the reader.)

Title: Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. Find the length of $AB$.

Body:
Given three circles of radius $R$, which touch only at one point and there centers belong to $OD$. The line $OC$ is tangent to $(M, R)$ and intersects $(K,R)$ at $O$. Find the length of $AB$ in terms of $R$.

I attempted to solve it as follows:

$OC^2=(OD-2R)^2\cdot OD^2$

$MO^2+MC^2=OC^2$

$OA\cdot (OA+AB)=(OD-4R)^2+(OD-2R)^2$

I then attempted to combine the equations above, but they didn't give anything useful. Could you please explain to me how to solve the question?


问:
Meaning of degenerate PDE
I am trying to understand the concept of degenerate PDE. The definitions I found depend on the type of equation but let say for elliptic (see here). Let's take an example:
$$x\partial_x^4u + \partial_y^4u + \partial_z^4 + u_x = 0 \qquad (1)$$
then $(1)$ is degenerate if $\exists \xi\in\mathbb R^3, \xi\neq 0$ such that
$$\sum_{\lvert \alpha \rvert = 4} \frac{\partial F(x,Du)}{\partial q_\alpha}\xi^\alpha=0.$$
So I have two questions:

I don't understand the definition of the polynomial (what is $\frac{\partial}{\partial q_\alpha}$?), could someone explain how to apply it to equation $(1)$?

What does it imply that an equation is degenerate? I have seen that the usual theory does not apply, but what parts exactly? Does it have consequences on the numerical solution?

答:
First, your equation is linear, and the definition of degenerate you cite is for a nonlinear PDE.

So let's start with linear: A linear pde on a domain in $\mathbb{R}^n$ is of the form
$$ \sum_{|\alpha|\le k} A_\alpha(x)\partial^\alpha u f. $$
Its principal symbol is defined to be
$$
\sigma(x,\xi) = \sum_{|\alpha|=k} A_\alpha(x)\xi^\alpha
$$
The PDE is elliptic if $\sigma(x,\xi) > 0$ for all $x$ and nonzero $\xi$. It is degenerate elliptic if $\sigma(x,\xi) \ge 0$ for all $x$ and $\xi$. You can use this definition to verify that your question is in fact degenerate elliptic.

A nonlinear PDE is of the form
$$ F(x,\partial^\alpha u) = 0 $$
In other words, it is given by a function $F(x,q^\alpha)$, where you set $q^\alpha = \partial^\alpha u$.
Its linearization at a function $u$ is defined to be the linear differential operator (acting on $v$)
$$ F'(x,\partial^\alpha u)(v) = \left.\frac{d}{dt}F(x,\partial^\alpha (u+tv))\right|_{t=0} = \frac{\partial F}{\partial q^\alpha}(x,\partial^\alpha u)\partial^\alpha v $$
The nonlinear PDE $F(x,\partial^\alpha u)$ is said to be (degenerate) elliptic at $u$, if the linearized operator is (degenerate) elliptic.

Title: Meaning of degenerate PDE


问:
(Simple) Examples on Non-Commutative Rings
Looks like it is easier to find example of commutative rings rather than
non-commutative rings. Prabably the easiest examples of the former are
$\mathbb{Z}$ and $\mathbb{Z}_n$. We can find elaborations on these two commutative rings in various
literatures including here and here. These are quite simple and easy to comprehend.

However, the examples on simple (non-commutative kind) are not that easy.
One example is found here and it has been mentioned as "one of the simplest examples of a non-commutative ring". This is also on the easier side.

EDIT: The following two are also commutative rings.

Other examples are given in this enumeration. But as you can see, examples like Gaussian integers or Eisenstein integers are difficult for starters to comprehend.

Do you think you can give one or two simple examples on non-commutative rings, based on every day numbers?

If it is that difficult, perhaps some insight comments why this is difficult would be welcome.

答:
Given an abelian group $M$, let $\operatorname{End}(M)$ denote the set of all homomorphisms $M \to M$ (i.e endomorphisms).  This set becomes a ring under pointwise addition and composition.

To see that $\operatorname{End}(M)$ may not be a commutative ring, choose another noncommutative ring $R$ (you already know one).   Left multiplication by elements of $R$ are endomorphisms of the underlying abelian group.  Since there are elements $a, b \in R$ such that $ab \ne ba$, $\operatorname{End}(R, +)$ is noncommutative.


问:
Reference Request: Gelfond Schneider Theorem
I want to learn the background for the Gelfond Schneider Theorem:

https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem

I want to learn the background needed for the proof of this theorem, as well as the proof itself. But I don't know what kind of textbook or class this would normally be covered in. Please recommend what I can study.

I want to be able to prove results on transcendence of specific numbers, which I know can be difficult in general. I also want to learn more about the subject in general, since my knowledge is very limited.

Thank you!!

答:
The standard introductory books I know about are Irrational Numbers by Ivan Niven (1956; Gelfond-Schneider is the last chapter, Chapter X on pages 134-150) and Making Transcendence Transparent by Edward B. Burger and Rubert Tubbs (2004; Gelfond-Schneider is Chapter 5 on pages 113-146).

An expository paper on the proof is Gelfond's solution of Hilbert's seventh problem by Carl Einar Hille in American
Mathematical Monthly 49 #10 (December 1942), pp. 654-661.

Less elementary is Transcendental Number Theory by Alan Baker. However, instead of Baker's book, perhaps better for you would be the 1972 Master of Science thesis Transcendental Numbers and a Theorem of A. Baker by Cameron Leigh Stewart.


问:
Intuitive proof for $(a, b) = (a', b')$.
If $p,q,r,s$ are integers such that $ps - qr = \pm 1$, and
$a,b,a',b' $ are integers such that $a' = pa +qb, b' = ra + sb$, prove that $(a, b) = (a', b')$.

Here, the particular value of $ps - qr = \pm 1$
enables us to get integer values.

Can there be an algebraic or intuitive explanation for that?

Have two explanations. My explanation is given for both sequentially. The first explanation is offered as a geometrical / intuitive one.

While the second one uses number theoretic approach, and linear system of equalities.

Request vetting & improvement for the first one

First, needs linear algebra, and orthogonal matrices.

$$\begin{bmatrix}
        a'\\
        b' \\
        \end{bmatrix} =  \begin{bmatrix}
        p & q\\
        r & s \\
        \end{bmatrix}
\begin{bmatrix}
        a\\
        b\\
        \end{bmatrix}
$$

The condition $ps - qr = \pm 1$, states two things about matrix
$$\begin{bmatrix}
        p & q\\
        r & s \\
        \end{bmatrix}
$$
#1. determinant is non-zero; hence invertible matrix.

#2. Also, is an orthogonal matrix.

A function is invertible iff it is bijective. A matrix is only a convenient representation of linear mapping function between two sets, wrt a basis of the domain and co-domain of the function.

#1: So, for linear maps, the determinant of corresponding matrix as non-zero means the mapping is possible in both directions without collapsing of any dimension (row/column).

#2: the area (min. possible contained entity, as order $2$ is the minimum dimension; similarly, for order $3$ matrix, it is volume) is unit as for orthogonal matrix of order $n$ 'that' is given by: $I^n = 1$.

Hence, $(a,b) = (a', b')$.

Think that need some geometrical insight in orthogonal matrices.

Edit ::  in case of orthogonal matrices, they practically arise in rotation or reflection, where the determinant is $\pm 1$.
Can this explanation be modified to take into account that?

Second explanation is equivalent, though needs to ignore the matrix representation. But, needs the linear system of equalities to use the Bezout's lemma. This is a number theoretic approach.

$$a' = pa + qb$$
$$b' = ra + sb$$

any common divisor of $a, b$ will divide both $a', b'$, so $(a,b) ≤ (a',b')$.

Solving for $a, b$ in terms of $a', b'$ gives

$$a = (s/(ps-qr))a' -(q/(ps-qr))b'$$

$$b = (-r/(ps-qr))a' +(p/(ps-qr))b'$$

Since $ps-qr =\pm1$, all the coefficients in this system are integers, so $(a',b')$ is a common divisor of $a$ and of $b$, therefore, $(a',b') \le (a,b)$.

答:
As for your first explanation, the equations
$$ \begin{bmatrix} a' \\ b' \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix}$$ imply that any common divisor of $a$ and $b$ is also a common divisor of $a'$ and $b'$. Since $\gcd(a,b)$ thus divides both $a'$ and $b'$ you have
$$\gcd(a,b) \le \gcd(a',b').$$

Since $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ is invertible with integer entries you have also that
$$ \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} p' & q' \\ r' & s' \end{bmatrix} \begin{bmatrix} a' \\ b' \end{bmatrix}$$
for some integers $p'$, $q'$, $r'$, and $s'$, and thus for the same reason you may conclude
$$\gcd(a',b') \le \gcd(a,b).$$


答:
Suppose that $u\in\operatorname{cl}S$; then there is a sequence $\langle p_n:n\in\Bbb N\rangle$ in $A$ such that if $p_n=\langle x_n,y_n\rangle$ for each $n\in\Bbb N$, then $\langle x_n+y_n:n\in\Bbb N\rangle$ converges to $u$. Since $A$ is bounded, the sequence $\langle x_n:n\in\Bbb N\rangle$ is also bounded, so it has a subsequence $\langle x_{n_k}:k\in\Bbb N\rangle$ converging to some $x\in\Bbb R$. Similarly, $\langle y_{n_k}:k\in\Bbb N\rangle$ is bounded and therefore has a subsequence $\langle y_{n_{k_i}}:i\in\Bbb N\rangle$ converging to some $y\in\Bbb R$. Clearly $\langle x_{n_{k_i}}+y_{n_{k_i}}:i\in\Bbb N\rangle$ still converges to $u$, so we might as well replace $\langle p_n:n\in\Bbb N\rangle$ by $\langle p_{n_{k_i}}:i\in\Bbb N\rangle$ and assume from the start that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ and $\langle y_n:n\in\Bbb N\rangle$ converges to $y$.

For future reference you might want to note that what I just did here is the long version of the useful trick that is often abbreviated to something like this: By passing to a subsequence if necessary we may assume that $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ converges to $x$ and $y$, respectively.

$A$ is closed, so $\langle x,y\rangle\in A$, and you need only verify that $x+y=u$.

Title: Showing a set is closed in the real metric if it is also closed in $d_2$ metric

Body:
Above is the exercise. Showing that $S$ is bounded is straightforward by $A$ bounded and the triangle inequality, but I thought that to show S is closed, I would do the usual thing of assuming $w \in \bar S$, then by definition, there is a sequence $\{s_n\}$ with elements in $S$ so that $s_n \to x$ w.r.t. absolute metric.

The problem is now I am stuck and can't seem to use the fact that $A$ is closed to conclude that $w \in S$. Is this approach even possible? And is there a better approach, perhaps showing that $\mathbb{R^2} \backslash A$ is open is easier? Thanks!

Note: Please do not use compactness as I have not covered it yet!

(I have an exam soon and so I would like exposure to as many problems as possible. Sorry if I haven't shown too much work before posting.)


问:
Correspondence between hamiltonian flows
Suppose I have an hamiltonian function $H$ that is fiberwise homogenenous of degree 2 , i.e, $H(q,sp)=s^2H(q,p)$,in $(T^*M,\omega)$ the cotangent manifold with the canonical symplectic structure and a path $\gamma(t)=(q(t),p(t))$ such that $\dot \gamma=X_H(\gamma)$, I want to see that we define $\gamma_c(t)=(q(t),\frac{1}{c}p(t))$ we will have that $\dot \gamma_c=X_{cH}(\gamma_c)$. Now in my attempt to do this first we have that $X_{cH}$ is defined such that $\omega_{c\gamma}(X_{cH},.)=d_{c\gamma}cH(.)=2cH(q(t),\frac{1}{c}p(t))=\frac{2}{c}H(q(t),p(t))$ since $H$ is fiberwise homogeneous of degree $2$ and using euler's identity.

Now I tried to check that $\omega_{c\gamma}(\frac{d}{dt}( c\gamma),.)=\frac{2}{c}H(q(t),p(t))$ but I got nowhere .Maybe I am making some silly mistake but I can't seem to be able to prove this.

Any help is appreciated. Thanks in advance.

答:
I hope the following helps.

Suppose you have Hamiltonians $H(q,p)$ and $\widetilde H(q,p):=c^2 H(q,p)$. Then the integral curve $(\widetilde q(t),\widetilde p(t))$ of Hamilton's equations for $\widetilde H$ with $\widetilde q(t=0)=q_0,\widetilde p(t=0)=p_0$ is $(q(c^2t),p(c^2t))$ where $(q(t),p(t))$ is the integral curve of Hamilton's equations for $H$ with initial conditions $q(t=0)=q_0,p(t=0)=p_0$.

Suppose you have Hamiltonians $H(q,p)$ and $\widetilde H(q,p):=H(q,cp)$. Then the integral curve $(\widetilde q(t),\widetilde p(t))$ of Hamilton's equations for $\widetilde H$ with $\widetilde q(t=0)=q_0,\widetilde p(t=0)=p_0$ is $(q(ct),p(ct)/c)$ where $(q(t),p(t))$ is the integral curve of Hamilton's equations for $H$ with initial conditions $q(t=0)=q_0,p(t=0)=cp_0$.

Combining the previous points, suppose that $H(q,cp)=c^2 H(q,p)$, for $c>0$. Then the integral curves $(q(t;q_0,p_0),p(t;q_0,p_0))$ of Hamilton's equations, now denoting also the dependence on initial data, have the following invariance
$$ (q(c^2t;q_0,p_0),p(c^2t;q_0,p_0)) =(q(ct;q_0,cp_0),p(ct;q_0,cp_0)/c) ,\qquad c>0, $$
or, just by setting $t\mapsto t/c$ in this identity,
$$
(q(ct;q_0,p_0),p(ct;q_0,p_0)) =(q(t;q_0,cp_0),p(t;q_0,cp_0)/c) ,\qquad c>0.
$$


问:
Attempt at solving $\frac{dy}{dx} = x^2 + y^2$
I read in my maths textbook that an analytic solution to $\frac{dy}{dx} = x^2 + y^2$ is not known. Nevertheless, as an exercise I tried to do it, to see where the difficulty lies.

In the same textbook, I learnt of a general method for solving ones of the ordinary linear form: $\frac{dy}{dx} + Py = Q$
where $P, Q$ are purely functions of $x$, or constant. The method involved finding $R = e^{\smallint P \space dx} $, resulting in $R\cdot\frac{dy}{dx} + RPy \equiv \frac{d}{dx}[Ry]$ and then finding $y = \frac{1}{R}\smallint RQ \space dx$. I attempted to form this for powers of two, with the following being the key equation:
\begin{equation} 2y \cdot \frac{dy}{dx} \cdot R + RPy^2 \equiv \frac{d}{dx}[Ry^2]
\end{equation}
I thought then, that finding $R$ in the same way would result in a very valid method for solving simple quadratic differential equations. I attempted this for $\frac{dy}{dx} = x^2 + y^2$:
\begin{equation}\frac{dy}{dx} - y^2 = x^2\end{equation}
\begin{equation}
2y \cdot \frac{dy}{dx} - 2y \cdot y^2 = 2y \cdot x^2, P = -2y
\end{equation}
\begin{equation} R = e^{\int -2y \space dx} = \frac{1}{e^{2yx}} \cdot C, C = 1 \text{ for simplicity.}
\end{equation}
\begin{equation}
2y \cdot \frac{dy}{dx} \cdot \frac{1}{e^{2yx}} - 2y \cdot y^2 \cdot \frac{1}{e^{2yx}} = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}}
\end{equation}
\begin{equation}
\frac{d}{dx}[\frac{1}{e^{2yx}} \cdot y^2] = 2y \cdot x^2 \cdot \frac{1}{e^{2yx}}
\end{equation}
\begin{equation}
\frac{1}{e^{2yx}} \cdot y^2 = \int 2y \cdot x^2 \cdot \frac{1}{e^{2yx}} \space dx
\end{equation}
which equals what, exactly? I am at a loss as to how to solve this integral. I pose four questions:

(1) What is the result of that final integral?

(2) Was my attempted method sound?

(3) What is the better approach to solving differential equations with a second power? I assume my method is not the best - if it works at all!

(4) If that final integral does have a result, then why did my textbook claim no solution exists?

答:
If that's what it really says, your textbook is wrong.  The general solution of $\dfrac{dy}{dx} = x^2 + y^2$ is

$$ y(x) = - \frac{x \left(c J_{-{3}/{4}}\! \left({x^{2}}/{2}\right)  +Y_{-{3}/{4}}\! \left({x^{2}}/{2}\right)\right)}{c J_{{1}/{4}}\! \left({x^{2}}/{2}\right)+Y_{{1}/{4}}\! \left({x^{2}}/{2}\right)}$$

where $J_\nu$ and $Y_\nu$ are Bessel functions of the first and second kinds, and $c$ is an arbitrary constant.


问:
laplace equation in a rectangle with boundary condition
$u_{xx}+u_{yy}=0 \quad in \quad  the \quad  rectangle  \quad 0<x \quad 0<y<1$

$u=0  \quad on \quad y=1$

$u=j(y) \quad on \quad x=0 $

$u_y +u=0  \quad on \quad y=0$

$u_x=0  \quad on \quad x=a$

I tried variable separation method. And the result is that this problem is not solved by this method.

答:
Variable separation works... the  Sturm–Liouville problem in $y$-variable has solutions

$$\phi_n(y)=\sin \lambda_n y -\lambda_n \cos \lambda_n y$$
where $\lambda_n>0$ are the roots of the equation  $\tan \lambda=\lambda$.

The solution is obtained in the form $$u(x,y)=\sum_n c_n \sinh [\lambda_n (x-a)]\, \phi_n(y)$$  where the coefficients $c_n$ come from expanding the function $j$ in the basis $(\phi_n)$.


问:
$3$ Boys and $3$ Girls sit at a circular table. What is the probability the boys sit together?
Three boys and three girls sit at a circular table. No boy or girl is more likely to sit on a particular chair. What is the probability the three boys all sit together?

I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving $5!$ possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy ($\times2$) and the next boy can sit to the right or left of him ($\times2$).

And as the $3$ boys can be arranged in $3!$ ways this gives $2\times2\times3!$ possibilities for the boys out of $5!$ overall. However this doesn't seem to work as the answer is $3/10$. This would fit with $\frac{3\times2\times3!}{5!}$ but I cant see why to change one of these twos into the required three. Am I missing something?

Many thanks for any help - I am sure it is something obvious I am overlooking so feel free to say so

答:
I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving 5! possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy (×2) and the next boy can sit to the right or left of him (×2).

The first problem, is that this only counts the number of solutions where the first two boys are sitting next to each other.

The second problem, is that you are going between (a) counting how many ways to place the boys in the numerator and (b) counting how many ways to place the everyone, boys and girls, in the denominator.  Either is fine, but you have to pick one and stick with it.

So if we only count how to place boys: there's $4 \times 5$ ways total to place the boys, without them needing to be together.  With all 3 boys together:  there's $4$ ways to place them with first and second next to each other and only $2$ ways to place them so that first and second are not next to each other.  So $(4+2)/(5 \times 4) = 3/10$.

If we count how to place everyone, that's $5!$ ways in general.  There's $4$ ways to place the first two boys next to each other like before, but then $3!$ ways to place the remaining girls.  There's $2$ ways to place the first two boys not next to each other, like before, but still $3!$ ways to place the remaining girls.  So $(4\times 3! + 2\times 3!)/5! = 3/10$.


问:
How would you go about learning the combination of coins the man has?
I am trying to identify the branch of math that would help to solve the following problem:

A man has picked $10$ coins out of a bag and has laid them in a row. You cannot see them for yourself, and the heads/tails ratio is unknown. Your goal is to identify the correct heads/tails sequence of the coins by repeatedly picking $10$ coins of your own from the bag, with the man telling you how many match the position of his each time. You cannot decide whether the coins are heads or tails (for the sake of the problem assume they are the same on both sides) and your sequence depends on the order you remove them from the bag. Note that the man only selects his coins once and his never change.

For example, if the man has $HTTHHHHTTT$ and you propose $TTTHTHTHHT$ he will let you know that $5$ are a match, however he will not tell you which ones. If on your second try you pull out $THHTTTHTHT$ he will inform you that $3$ match. Without trying every possible combination $(2^{10})$, how would you go about learning the combination of coins the man has?

Comparing the two sample results I see that they share $3$ in common, but I'm not sure if or how that is useful information since I don't know which ones actually match.

I suspect it might require statistics/combinatorics, and I was told it could be solved with "either high school or college math". This isn't a homework problem so it's possible I'm just being trolled by the person. Any information you can provide for approaching this type of problem is much appreciated. Thank you.

答:
My attempt is to visualize, simplify and extract maths from all the statements

\*lets assume that for every instant i begin to pick, the bag would contain exactly $10$ coins of which $5$ are heads and $5$ tails, also we'll assume the coins are distinguishable, so that they have same face on both sides and also i'll assume my coins display were not hidden from me, and my attempt was infinte plus that i made note of every of my attempt and it's information *\

$$\begin{matrix}
\cdot &1&2&3&4&5&6&7&8&9&10\\
1&H&H&H&H&H&H&H&H&H&H\\
2&T&T&T&T&T&T&T&T&T&T\\
\end{matrix}$$
this is the row of the coins, in all possibility $2^{10}$
$$\text{the man coins} \rightarrow \{ x_{1} , x_{2} , x_{3} ,x_{4} ,x_{5} ,x_{6} ,x_{7} ,x_{8} ,x_{9} ,x_{10} \}$$
the set $X$ is the man's pick, and $x_{n}$ is either $H$ or $T$ depending on how it appeared but our goal is to find the ratio $\frac{\text{Heads}}{\text{Tails}}$ for his pick, say ratio of $\text{Heads}$ to $\text{Tails}$ is a function $P()$, where $P(X) = k$
$$ \{ H \rightarrow 10, (H \rightarrow 9, T \rightarrow 1), (H \rightarrow 8, T \rightarrow 2), (H \rightarrow 7, T \rightarrow 3), (H \rightarrow 6, T \rightarrow 4),(H \rightarrow 5, T \rightarrow 5) , (H \rightarrow 4, T \rightarrow 6), (H \rightarrow 3, T \rightarrow 7), (H \rightarrow 2, T \rightarrow 8), (H \rightarrow 1, T \rightarrow 9), T \rightarrow 10 \} $$

$ P() \in \{ 0,\frac{1}{9} ,\frac{2}{8} ,\frac{3}{7} ,\frac{4}{6},1 ,\frac{6}{4} ,\frac{7}{3} ,\frac{8}{2} ,\frac{9}{1} , \text{nil} \}$, so we are attempting to find the value $k$

$$\text{all my attempts} = \{ y_{1}, y_{2}, y_{3}, y_{4}, \dots \dots\}$$

say $\text{compare()}$ is a function, that produces the number of match between the set of coins when comparing them, now in the example you gave if i compare the man's pick with my first attempt

$$\begin{matrix}
H&T&T&H&H&H&H&T&T&T\\
T&T&T&H&T&H&T&H&H&T\\
&\cdot&\cdot&\cdot&&\cdot&&&&\cdot&\\
\end{matrix}$$
there are $5$ match, also notice here that both contain same $\frac{\text{Heads}}{\text{Tails}}$ ,now in the second case you gave, there is $3$ match here
$$\begin{matrix}
H&T&T&H&H&H&H&T&T&T\\
T&H&H&T&T&T&H&T&H&T\\
&&&&&&\cdot & \cdot && \cdot\\
\end{matrix}$$
when i compare the cases of my first two attempt, remember i don't know the man's pick but  only the number of match my attempt makes, they'll also make matches then...... i can compare all my attempts in order to know the correct position of the man's pick
$$\begin{matrix}
T&T&T&H&T&H&T&H&H&T\\
T&H&H&T&T&T&H&T&H&T\\
\cdot &&&& \cdot &&&&&\cdot \\
\end{matrix}$$
from here i can see that the true value is actually just one, so $\text{compare(y_{1}, y_{2} ) } =  y_{1} \bigcap y_{2} $ therefore you can get the man's pick in a possible chance or even before that,know the position of some of the man's coin if not even all,by simply comparing all your attempts and noting the number of matches

$$\text{all my attempts} = \{ y_{1}, y_{2}, y_{3}, y_{4}, \dots \dots\}$$
my attempts can be grouped in $11$ sections based on the value for the $P()$ function
$ P() \in \{ 0,\frac{1}{9} ,\frac{2}{8} ,\frac{3}{7} ,\frac{4}{6},1 ,\frac{6}{4} ,\frac{7}{3} ,\frac{8}{2} ,\frac{9}{1} , \text{nil} \}$

so we would take all the particular case when $P() = 0$ of my attempts and compare with the man's pick
$\text{compare(X,y_{1})} = \dots \dots$, $\text{compare(X,y_{2})} = \dots \dots$, $\text{compare(X,y_{3})} = \dots \dots$, $\text{compare(X,y_{4})} = \dots \dots$, $\text{compare(X,y_{5})} = \dots \dots$,
$\dots \dots \dots \dots $
and then compare all the matches of my attempt in twos $\text{compare(y_{1},y_{2})} = \dots \dots$, $\text{compare(y_{1},y_{3})} = \dots \dots$, $\text{compare(y_{1},y_{4})} = \dots \dots$, $\dots $ , $\text{compare(y_{2},y_{3})} = \dots \dots$, $\dots \dots \dots \dots$

and then also take all the particular case when $P() = \frac{1}{9}$ of my attempts and compare with the man's pick and when we compare all the matches in twos
$\text{compare(X,y_{1})} = \dots \dots$, $\text{compare(X,y_{2})} = \dots \dots$, $\text{compare(X,y_{3})} = \dots \dots$, $\text{compare(X,y_{4})} = \dots \dots$, $\text{compare(X,y_{5})} = \dots \dots$, $\dots \dots \dots \dots  $ and when we compare all the matches of my attempt in twos  $\text{compare(y_{1},y_{2})} = \dots \dots$, $\text{compare(y_{1},y_{3})} = \dots \dots$, $\text{compare(y_{1},y_{4})} = \dots \dots$, $\dots $ , $\text{compare(y_{2},y_{3})} = \dots \dots$, $\dots \dots \dots \dots $

so this process is repeated, infinitly until scenarios appear where $\text{compare()}$ has a higher value w.r.t the man's pick and my attempts , also when $\text{compare()}$ has a low value, $1$ if i'm comparing two of my attempt w.r.t man's pick


答:
The boys can be placed in ${6\choose3}$ ways. There are exactly $6$ seatings where the boys sit together. Your probability therefore is
$${6\over{6\choose3}}={3\over10}\ .$$

Title: $3$ Boys and $3$ Girls sit at a circular table. What is the probability the boys sit together?

Body:
Three boys and three girls sit at a circular table. No boy or girl is more likely to sit on a particular chair. What is the probability the three boys all sit together?

I have been looking at this problem. One of the approaches I took is to think of one boy taking a chair - leaving $5!$ possible combinations for the remaining sitting positions of the group. Then the next boy can sit to the left or right of that boy ($\times2$) and the next boy can sit to the right or left of him ($\times2$).

And as the $3$ boys can be arranged in $3!$ ways this gives $2\times2\times3!$ possibilities for the boys out of $5!$ overall. However this doesn't seem to work as the answer is $3/10$. This would fit with $\frac{3\times2\times3!}{5!}$ but I cant see why to change one of these twos into the required three. Am I missing something?

Many thanks for any help - I am sure it is something obvious I am overlooking so feel free to say so


问:
Shared area of $2$ circles (polar integral)
Given $x^2+y^2=4y$ and $x^2+y^2=4x$ find the shared area  of the 2 circles

What I tried was :

so first I transformed them to polar coordinates
I got $r=4cos\theta$ for $x^2+y^2=4x$ and $r=4sin\theta$ for $x^2+y^2=4y$

After that I did $4cos\theta=4sin\theta$ then $\theta=\frac{\pi}{4}$ I am not sure but according to this I believe that $\frac{\pi}{4}≤\theta≤\frac{\pi}{2}$ lastly I did the integral

$\int_\frac{\pi}{4}^\frac{\pi}{2}\frac{1}{2}((4cos\theta)^2-(4sin\theta)^2) d\theta $
after integrating I got $-4$ which is a wrong answer.

what am I doing wrong? thanks for any help and tips!
Edit: I forgot to mention that I havent studied double integrals yet , I learned polar system and this formula $\int_\alpha^\beta\frac{1}{2}(r(\theta))^2 d\theta $

答:
Please note that for $0 \leq \theta \leq \frac{\pi}{4}$, you are bound by circle $4 \sin\theta $ and for $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$, you are bound by circle $4 \cos\theta$.

So the area is given by,  $A =  \displaystyle 2 \int_0^{\pi/4} \int_0^{4\sin\theta} r \ dr \ d\theta$

or $ \ \displaystyle 2 \int_0^{\pi/4} \frac{1}{2}(4\sin\theta)^2 \ d\theta$

You can see that the area between $0 \leq \theta \leq \frac{\pi}{4}$ and between $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$ are same. So we multiply by $2$ or use the fact that $\sin\theta = \cos (\frac{\pi}{2} - \theta)$.


问:
A trapezium's diagonals' point of intersection lies on the line passing through its parallel sides' mid-points

Prove, using a vector method, that a trapezium's diagonals' point of intersection lies on the line passing through its parallel sides' mid-points.

My Attempt:

Let the trapezium be $OABC,$ $O$ be the origin, and the position vectors of $A,B,C$ be $\vec{a},\vec{b},\vec{c}.$

Then the equation of diagonal $OB$ is $\vec{r}=\vec{0}+\lambda \vec{b}.\tag1$
And the equation of diagonal $AC$ is $\vec{r}=\vec{a}+\mu(\vec{c}-\vec{a}).\tag2$
And the equation of the line joining the mid points of $OA$ and $BC$ is $\vec{r}=\frac{\vec{a}}{2}+t(\frac{\vec{b}+\vec{c}}{2}-\frac{\vec{a}}{2}).\tag3$
Here, $\lambda,\mu,t$ are scalars.

I do not know how to solve $(1)$ and $(2)$ and $(3)$ to obtain the desired result.

答:
The trapezium's diagonals intersect at $X.$ We want to show that $B,X,\text{ and }R$ are collinear.$$ $$

Triangle $XOP$ is similar to triangle $XQA.$ So $$\frac{XO}{XQ}=\frac{OP}{QA}\\=\frac1\lambda.$$

Now, $$\vec{BX}=\vec{OX}-\vec{OB}\\=\frac1{\lambda+1}\vec{OQ}-\vec{OB}\\=\frac1{\lambda+1}\left(\mathbf{a}+2\lambda\mathbf{b}\right)-\mathbf{b}\\=\frac1{\lambda+1}\left[\mathbf{a}+\left(\lambda-1\right)\mathbf{b}\right],$$ while $$\vec{BR}=\vec{BO}+\vec{OA}+\vec{AR}\\=-\mathbf{b}+\mathbf{a}+\lambda\mathbf{b}\\=\mathbf{a}+\left(\lambda-1\right)\mathbf{b}.$$
Therefore $$\vec{BR}=(\lambda+1)\vec{BX}.$$ Thus $B,X,\text{ and }R$ are collinear.


问:
How to prove $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$?
I'm trying to show that $\lvert \lVert x \rVert - \lVert y \rVert \rvert \overset{\heartsuit}{\leq} \lVert x-y \rVert$. A hint would be nice.</p>

答:
Observe that

$\lVert x \rVert  = \lVert (x -y) +y \rVert \leq \lVert (x -y) \rVert + \lVert y \rVert$

which gives

$\lVert x \rVert - \lVert y \rVert \leq \lVert x -y \rVert$ ... $(1)$

Further,

$-(\lVert x \rVert - \lVert y \rVert ) \leq \lVert (y -x) \rVert = \lVert (x -y) \rVert $... $(2)$

From $(1)$ and $(2)$ result follows.


答:
HINT:

Let the radius be 1, then...

Body:
HINT:

Let the radius be 1, then...