问:
Solving Coin-Weighing Problem (81 Coins, 1 Fake) Using Information Theory
So I have a coin-weighing puzzle under these situations:

There are 80 real coins and 1 fake coin (total of 81 coins).

The real coins are all the same weight, and the weight of the fake coin is different from the real coin.

The real and fake coins cannot be distinguished, other than weight.

A balance is used to identify the fake coin.

When using the balance, the same number of coins is placed on either plate. The result is either "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

According to prior research, this problem cannot be solved by using the balance 4 times or less. I tried to show this by using information theory as follows, but I feel like I am missing something here.

Label the coins 1,...,81, and let $C$ be the number of the fake coin. Also, let $W$ be a random variable defined so that $W=1$ when the fake coin is heavier than the real coin, and $W=0$ when lighter. At the initial state, $C$ and $W$ can be regarded as independent random variables following a uniform distribution.

Then, define random variable $R_k$ so that $R_k = 0$, $R_k = 1$, $R_k = 2$, when, the result after using the balance for the $k$th time is, respectively, "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

Then, $R_k$ can be regarded as uniquely determined by $R_1, ..., R_k,$ and the real values of $C$ and $W$. (Do I have to add additional proof that this is true?)

Assume that the fake coin can be surely identified by four measurements using the balance, then this means that the value of $C$ is determined when $R_1,...,R_4$ have been determined. Thus, using the entropy function and its chain rule, \begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(C) + H(R_1 | C) + H(R_2 | R_1, C) + H(R_3 | R_2, R_1, C) + H(R_4|R_3, R_2, R_1, C) + H (W| R_4, R_3, R_2, R_1, C)
\end{eqnarray*}

And, we also have\begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(W) + H(C|W) + H(R_1 | C, W) + H(R_2 | R_1, C, W) + H(R_3 | R_2, R_1, C, W) + H(R_4|R_3, R_2, R_1, C, W)
\end{eqnarray*}

Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get
$H(W) < H(W | R_1, ..., R_4, C)$

But we know that $H(W | R_1, ..., R_4, C) < H(W)$, so there is a contradiction.

OK. So I somehow arrived at a contradiction, showing that the problem cannot be solved by using the balance for times. But then I realized that the same logic applies when using the balance for any number of times, so apparently there is something wrong...

What sort of logic am I missing? I would appreciate any help.

答: You have 162 possible outcomes.  You can't do light and heavy in just four trinary weighs.  The usual means is 12 in three weighs or 120 in five.  This gives 24 or 240 outcomes.</p>


答:
The conclusion in this statement is wrong:

Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get ${\color{red}{H(W) < H(W | R_1, ..., R_4, C)}}$

The conclusion would be $H(W) > H(W|R_1,\ldots, R_4,C)$.

Your logic would be like saying "If $a+b=c+d$ and $b < c$, then $a < d$"  The inequality is backwards.

ContentLicense: CC BY-SA 4.0

Title: Solving Coin-Weighing Problem (81 Coins, 1 Fake) Using Information Theory

Body:
So I have a coin-weighing puzzle under these situations:

There are 80 real coins and 1 fake coin (total of 81 coins).

The real coins are all the same weight, and the weight of the fake coin is different from the real coin.

The real and fake coins cannot be distinguished, other than weight.

A balance is used to identify the fake coin.

When using the balance, the same number of coins is placed on either plate. The result is either "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

According to prior research, this problem cannot be solved by using the balance 4 times or less. I tried to show this by using information theory as follows, but I feel like I am missing something here.

Label the coins 1,...,81, and let $C$ be the number of the fake coin. Also, let $W$ be a random variable defined so that $W=1$ when the fake coin is heavier than the real coin, and $W=0$ when lighter. At the initial state, $C$ and $W$ can be regarded as independent random variables following a uniform distribution.

Then, define random variable $R_k$ so that $R_k = 0$, $R_k = 1$, $R_k = 2$, when, the result after using the balance for the $k$th time is, respectively, "the left plate is heavier," "the right plate is heavier," or "the two plates are the same weight."

Then, $R_k$ can be regarded as uniquely determined by $R_1, ..., R_k,$ and the real values of $C$ and $W$. (Do I have to add additional proof that this is true?)

Assume that the fake coin can be surely identified by four measurements using the balance, then this means that the value of $C$ is determined when $R_1,...,R_4$ have been determined. Thus, using the entropy function and its chain rule, \begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(C) + H(R_1 | C) + H(R_2 | R_1, C) + H(R_3 | R_2, R_1, C) + H(R_4|R_3, R_2, R_1, C) + H (W| R_4, R_3, R_2, R_1, C)
\end{eqnarray*}

And, we also have\begin{eqnarray*}
H(C, W, R_1, ..., R_4) &=& H(W) + H(C|W) + H(R_1 | C, W) + H(R_2 | R_1, C, W) + H(R_3 | R_2, R_1, C, W) + H(R_4|R_3, R_2, R_1, C, W)
\end{eqnarray*}

Then we notice that $H(C|W) = H(C)$, as $C$ and $W$ are independent variables. Also, $H(R_1 | W, C) < H(R_1 | C), H(R_2 | R_1, C, W)  < H(R_2 | R_1, C) $, and so on. Thus, from the two equations, we get
$H(W) < H(W | R_1, ..., R_4, C)$

But we know that $H(W | R_1, ..., R_4, C) < H(W)$, so there is a contradiction.

OK. So I somehow arrived at a contradiction, showing that the problem cannot be solved by using the balance for times. But then I realized that the same logic applies when using the balance for any number of times, so apparently there is something wrong...

What sort of logic am I missing? I would appreciate any help.


问:
How are these cdf partitioned?

In the solution, it says for $x \leq 0$, it's $P(X \leq x)$, but shouldn't it be $P(X \leq 0)$ or $P(x \leq X \leq 0)$? Also, on that same line, how did it go from $e^{-|y|}$ to $e^{y}$?

And for the second line, why is $F_X(0)$ there, but it wasn't there for  $x \leq 0$?

答:
In the solution, it says for $x \leq 0$, it's $P(X \leq x)$, but shouldn't it be $P(X \leq 0)$ or $P(x \leq X \leq 0)$?

As sudeep mentioned you are missing the definition of CDF.

The cumulative distribution function is the probability that the variable takes a value less than or equal to x. That is

$F_X(x)=P(X \leq x)$

Also, on that same line, how did it go from $e^{-|y|}$ to $e^{y}$?

Definition of absolute function says

$|y|= \begin{cases}
      -y & -\infty < y < 0 \\
      \ \  y& \ \  0 \leq y < \infty \\
   \end{cases}$

substituting $-y$ gives $e^{y}$

And for the second line, why is $F_X(0)$ there, but it wasn't there for  $x \leq 0$?

Again using the definition of CDF which gives probability less than or equals $x$. When you are computing CDF for $x>0$ you have to add the probability for $x<0$. When we were calculating $P(X<x)$ for $-\infty<x<0$ there was nothing to add because  $F(-\infty)=0$.

Does it help?

Title: How are these cdf partitioned?


问:
Why does the proof of Theorem 3.10 in Apostol (1976) use generalized convolutions from previous chapter when there could be a simpler proof?
The proof of Theorem 3.10 in Introduction to Analytic Number Theory by Apostol goes like this:

Theorem 3.10 If $ h = f * g $, let
$$
H(x) = \sum_{n \le x} h(n),
F(x) = \sum_{n \le x} f(n), \text{ and }
G(x) = \sum_{n \le x} g(n). $$
Then we have
$$
H(x)
= \sum_{n \le x} f(n) G \left( \frac{x}{n} \right)
= \sum_{n \le x} g(n) F \left( \frac{x}{n} \right). $$

PROOF. We make use of the associative law (Theorem 2.21) which relates the operations $ \circ $ and $ * $. Let
$$
U(x) = \begin{cases} 0 & \text{ if } 0 < x < 1, \\
1 & \text{ if } x \ge 1. \end{cases}
$$
Then $ F = f \circ U $, $ G = g \circ U $, and we have
$$ f \circ G = f \circ (g \circ U) = (f * g) \circ U = H, $$
$$ g \circ F = g \circ (f \circ U) = (g * f) \circ U = H. $$
This completes the proof.

I am curious to understand why use generalized convolution and Dirichlet multiplication here at all.

Since section 3.5, the book has been relying on the following reordering of summation at various places,
$$
\sum_{n \le x} \sum_{d | n} f(d)
= \sum_{\substack{q, d \\ qd \le x}} f(d)
= \sum_{d \le x} \sum_{q \le \frac{x}{d}} f(q).
$$
Can we not arrive at a simpler proof with a similar reordering of summation like this:
\begin{align*}
H(x)
& = \sum_{n \le x} h(n)
= \sum_{n \le x} (f * g)(n)
= \sum_{n \le x} \sum_{d | n} f(d) g\left(\frac{n}{d}\right)
= \sum_{\substack{q, d \\ qd \le x}} f(d) g\left(q\right) \\
& = \sum_{d \le x} \sum_{q \le \frac{x}{d}} f(d) g(q)
= \sum_{d \le x} f(d) \sum_{q \le \frac{x}{d}} g(q)
= \sum_{n \le x} f(n) \sum_{q \le \frac{x}{n}} g(q)
= \sum_{n \le x} f(n) G\left(\frac{x}{n}\right).
\end{align*}
I am trying to understand if a simpler proof using basic arithmetic and reordering of summations is correct or if I am missing something and there is a specific reason why Apostol went for a proof using generalized convolution and Dirichlet multiplication.

答:
The other proof is arguably simpler (it is actually shorter and requires less manipulation). Also, the first few chapters of Apostol make a big deal about using convolutions.

There is a reason for this. As you move to more complicated content, with much "messier" expressions, techniques like this can be incredibly useful, since they allow for much more efficient notation. If anything, I think this proof shows how nice convolutions can be, since it solves the problem without writing down a single sum.

Hence, although it might seem like a more difficult proof if you are not familiarized with convolutions, it is actually a good choice from an educational point of view, since it will get you used to using this kind of technique for more difficult problems. I always find changing the order of summation a bit fiddly (I have to draw the curves and convince myself I have not made any mistakes) whereas convolutions are pretty much automatic


问:
Associated Prime notation
Let $I$ be an ideal of ring $R$ with $I\neq R$. Prime ideal $P$ of $R$ is said to be an associated prime of $I$ in the sense of Bourbaki, if $P=(I :_R x)$ for some $x\in R$. In this case we say that $P$ is a $ B$-prime of $I$.

Can anyone explain meaning of notation $(I :_R x)$?

答: Presumably it is $(I:_Rx):=\{r\in R:rx\in I\}$. Typically we suppress the subscript $R$.

Title: Associated Prime notation


问:
How to solve this analytical geometry problem?-parable inscribed within a square
This problem appeared on the network, and although it looks simple I am not sure of the result.

The polygon $ABCD$ is a square with side $4$ cm and the curve inscribed inside the square is a parabola, determine the length of the segment $AP$, if $QC$ is $3$ cm.

problem

答:
Hope you know differentiation to find maximum/minimum of standard parabola curve of form $y = ax^2+ bx + c $.

Taking B as origin since it passes through (0,0), we get c=0

$$ y =a x^2+bx \tag1 $$

Differentiate to get max tangent point

$$ 2 ax +b=0\to x_{t} =- b/(2a)  \tag 2$$

plug into (1) and simplify

$$ y_t= -b^2/(4a) = 4 \text{  (given)} \tag3$$

Point $(4,3)$ is given lying on the parabola

$$ 3= 16 a +4 b \tag4 $$

Eliminate $a$ between (3) and (4) and simplify resulting in the quadratic equation

$$ b^2-4b+3=0,\; (b-1)(b-3)=0 ; b=(1,3);\;\tag5 $$

From(3) corresponding $a$ values are

$$ \left(\dfrac{-1}{16},\dfrac{-9}{16} \right) \tag 6$$

Plug $(a,b) $ values into (2) we get two values of tangent point

$$ x_t=(8,8/3) \tag7$$

The first value lies outside interval $0<x_t<4$ so is discarded, second value is taken. The parabola has equation

$$y=3 x\left(1-\dfrac{3x}{16} \right) \tag 8$$

whose graph is the one  given in your question.


问:
Why does the constant of integration always come with the independent variable whilst integrating differential equations?
Say, we had the following differential equation:

$$\frac{\mathrm{d}y}{\mathrm{d}x}x = 1+y$$

When we re-arrange the equation and integrate both sides, why will the constant of integration ONLY come with the independent variable, $x$?

My assumption is that it's because we can think of a function as having an input and an output. The output is the dependent variable, and input is the independent variable. So, all the work is done on the input, to get a desired output. So, the constant should only come with the independent variable. Am I wrong?

答:
It doesn't matter what side of the equation we put it on. Observe

$$\frac{\mathrm{d}y}{\mathrm{d}x}x=y+1$$

$$\frac{\mathrm{d}y}{y+1}=\frac{\mathrm{d}x}{x}$$

$$\int\frac{\mathrm{d}y}{y+1}=\int\frac{\mathrm{d}x}{x}$$

$$\ln{\left(y+1\right)}+C_y=\ln{x}+C_x$$

Now, here is the reason it doesn't matter: just bring both constants over to either side

$$\ln{\left(y+1\right)}+\left(C_y-C_x\right)=\ln{x}$$

$$\left(y+1\right)\,\mathrm{e}^{C_y-C_x}=x$$

If we want $y$ in terms of $x$, then we can rearrange to get

$$y=x\,\mathrm{e}^{C_x-C_y}-1$$

But, you ask, aren't there two constants now? No, the expression $\mathrm{e}^{C_x-C_y}$ acts as a single constant. If you include an initial condition, then you would solve for this entire expression to get the final solution.


问:
Prove that $\ln(u)$ is subharmonic
Assume that $u \in C^2(\Omega)$ is strictly positive such that the function $v=u \cdot e^{\sum \limits_{i=1}^{N} a_{i}x_{i}}$ is subharmonic, for any choice of the constants $a_{1},\ldots,a_{N}$.

Prove that $\ln(u)$ is subharmonic in $\Omega$.

Of course I calculated $$\Delta v=e^{\sum \limits_{i=1}^{N} a_{i}x_{i}} \left(\Delta u + 2\sum \limits_{i=1}^{N} a_{i} \frac{\partial u}{\partial x_{i}} + \sum \limits_{i=1}^{N} a_{i}^2 u \right).$$
It is clear that $\Delta v \geq 0$ because of subharmonic.

There is a hint saying that: "It follows that $u \Delta u - |\nabla u|^2 \geq 0$, hence $\ln(u)$ is subharmonic".

Please explain me, step by step, from where it follows and because we can say sth about $\ln(u)$ from that.

答:
Note that $\Delta( \log{u}) = u^{-2} (u \Delta u - | \nabla u|^2)$, so it suffices to show that $u \Delta u - |\nabla u|^2 \geq 0$, as was hinted before.

You have computed that
$$
\Delta v = e^{a \cdot x} \left( \Delta u + 2 a \cdot \nabla u + u |a|^2 \right),
$$
and by assumption this is $\geq 0$. This holds pointwise (that is, for every $x \in \Omega$) as well as for every $a \in \mathbb{R}^N$. Therefore also
$$
\Delta u(x) + 2 a \cdot \nabla u(x) + u(x) |a|^2 \geq 0 ,
$$
where I've added the dependency on $x$ to be explicit. Now, let us focus our attention on any given $x_0 \in \Omega$. Then the above is still true, and furthermore we are allowed to choose any vector $a$, so choose $a = -\nabla(x_0)/u(x_0)$. Then, substituting this into the above,
$$
\Delta u(x_0) - 2 |\nabla u(x_0)|^2/ u(x_0) + |\nabla u(x_0)|^2 / u(x_0) \geq 0.
$$
Simplifying and multiplying through by $u(x_0)$ yields
$$
u(x_0) \Delta u(x_0) - | \nabla u(x_0)|^2 \geq 0.
$$
Since this holds for arbitrary $x_0$, then the inequality is proven.

Title: Prove that $\ln(u)$ is subharmonic


问:
The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$
The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____

My approach is as follow

Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below

$z = \cos \theta  + i\sin \theta ;\theta  \notin \left\{ 0 \right\}$ because $z\ne1$

$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta  - i\sin \theta }}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

How do we get the required answer which if $\frac{\pi}{2}$

答:
$$\left| {\arg\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$$ $$= \left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right) - \arg\left(2\sin \frac{\theta }{2}\right)\right|$$

$$=\left|\arg \left(\operatorname{cis}\left(\frac{\pi - \theta}{2}\right)\right)\right| = \left|\frac{\pi-\theta}{2} \right| \leq \frac{\pi}{2}$$

Here, $\operatorname{cis}(x) = \cos x + i\sin x$.


问:
Liminf of Pointwise Norms of a Weakly Convergent Sequence
Let $X_1, X_2, \cdots$ be a sequence of $p$-integrable $\mathbb{R^d}$ valued random variables. Assume that $X_n$ converges $0$ weakly, then can we say that $r(\omega) = liminf\{ |X_1 (\omega)|, |X_2 (\omega)|, \cdots\}$ is $0$ for almost every $\omega ?$

The classical example of a weakly convergent but not strongly convergent sequence is those of orthogonal basis, yet it is not a counterexample for the above claim. I could not prove the claim but intuitively I believe it holds, at least for $\mathbb{R^d}$ valued random vectors.

If needed one can assume that the sequence $(X_n)_n$ is bounded in $p$-norm, I do not feel like this is necessary.

答: Weak convergence together with boundedness of second moments implies convergence of expectations. By Fatou's Lemma we get $E \lim \inf |X_n| \leq \lim \inf E|X_n|=0$ since $E|X_n| \to 0$. This implies that $ \lim \inf |X_n|=0$ almost surely.


问:
Conditional joint probability and independence
Let's have a joint probability of three events, $\mathbf{P}(X,A,B)$. If $\mathbf{P}(X|A) = \mathbf{P}(X)$, can we show that $\mathbf{P}(X|A,B) = \mathbf{P}(X|B)$? If so, how?</p>

答:
Toss a fair coin twice, and let $A$ be the event that the first toss is a head and $B$ the event that the second toss is a head.

Now let $X$ be the event that the result of the two tosses is the same.

$P(X|A) = P(X|B)=P(X)=\dfrac 12$ but $P(X|A,B)=1$.

ContentLicense: CC BY-SA 3.0


答:
Hint:

Arg$\left(\dfrac12+i\cdot\dfrac{\cot\dfrac\theta2}2\right)=$Arg$\left(\cot\dfrac\theta2\right)=$Arg$\left(\tan\left(\dfrac{\pi-\theta}2\right)\right)$

Use this

Body:
The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____

My approach is as follow

Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below

$z = \cos \theta  + i\sin \theta ;\theta  \notin \left\{ 0 \right\}$ because $z\ne1$

$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta  - i\sin \theta }}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$

How do we get the required answer which if $\frac{\pi}{2}$


问:
Find roots of a specific discontinuous trigonometric function
Given the following function $f(x)$ and an arbitrary positive non-prime integer $N$

$ \{ f(x) = \sin(N\pi/x) + \cos(2N\pi/x) + |sin(3N\pi/x)| + |cos(5N\pi/x)| + 1, N \in \mathbb{Z}, x \in \mathbb{R} \}  $

I am looking for any root in the range $2 <= x < N$, I am pretty sure all $f(x)$ are greater than or equal to zero, so the roots are global minima. All the roots are also points of discontinuity and their $x \in \mathbb{Z} $.

Question

Is there any, even numerical way, to go about finding these roots for any arbitrary non-prime N? Or even just point me in the right direction like an algorithm name, I'm a software engineer, and familiar with arbitrary precision libraries like GMP, so I can implement numerical methods if I can understand the math.

Here is a plot of $f(x)$ for $N=517$. It isn't very good, its hard to choose a zoom level which will demonstrate all the features. But you can see I marked a zero at $x=22$, you will notice it looks like there is a zero at $x \approx 26.5$ but it is actually $ f(x) \approx 0.0011 $ if one were to zoom in.

There are many of these "close" to zero points, but none are actually zero, in the range I'm interested in ($ 2<=x<N $), unless 2 conditions I have found are met:

$x \in \mathbb{Z} $

$GCD(x,N) > 1 $

My example $N=517$ the roots are $ x \in \{-2,22,94\} $, we don't care about $-2$ and $GCD(22,517)=11$ and $GCD(94,517) = 47$.

I don't think the GCD observation will help in finding the roots, but I mention it because I think its interesting. And if anyone has an insight as to why this GCD relationship exists but not the main question still feel free to answer.

Here is another plot, in response to @poetasis comment below. I zoomed out to show a little better that all values of $f(x)$ are greater than zero except 3 points which are equal to zero where $ x \in \{-2, 22, 94 \} $ though its only really easy to see $ f(94) = 0 $ at this zoom level.

Also notice the period is not regular or an integer, it gets larger as we move right. The zeroes occur at those M-shaped downward "spikes". The spikes occurring where x is an integer actually touch the zero line, like at $ x = 94 $, the others occurring off an integer like the one around $ x = 147.7 $ can be close but not zero $ f(147.7) = 0.00859 $.

I put in some $x$ and $f(x)$ values of the M-shaped spikes so you can see below. There are other spikes, they all occur on non-integer $x$'s as well and have non-zero $f(x)$

In case its too hard to read, for N=517

$f(22)=0$

$f(38.3)=0.0767$

$f(68.96)=0.0480$

$f(94)=0$

$f(147.7)=0.00859$

答:
This will be easier if you substitute $t=\frac{1}{x}$ and graph it as a function of $t$.

Looking at the zeros of the $t$ function, you will see that for each $N$, they form an arithmetic progression whose terms are easily discernible.

For example, here is the graph of $f(t)$ for $N=4$.

It is not difficult to work out that the first zero is always

$$ t_0(N)=\frac{3}{2N} $$

and that the common difference of the sequence of zeroes is

$$ d=\frac{2}{N} $$

So the $k$th zero for $N$ will be

$$ t_k(N)= \frac{3+4k}{2N} $$

The reciprocals of these will be the zeros of $f(x)$.

$$ x_k(N)=\frac{2N}{3+4k} $$

Note that $x_{11}(517)=22$.


问:
What is the last digit of $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$
$2005$ $7's$ are used in the following number ,so what is its last digit   :

Number: $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$

As i mentioned in the introduction , this number is written using $2005$ times $7's$ and i want to find its last digit.

My work: Its last digit is wanted so i used $\mod(10)$.

After that i formed the expression into $\bigl((7)^{7^{2004}}\bigr)$ , then i decided to use Euler's totient function.

${7^{\phi(10)=4}\equiv 1 \ (mod10)}$

In the next step , i said that ${7^{4k}\equiv 1 \ (mod10)}$ and  i checked over $7^{2004}(mod4)$.Then ,I found that $7^{2004}(mod4)=1$

As a result ,  $\bigl((7)^{7^{2004}}\bigr) \equiv 7^1 (mod 10)=7$

My first question is that is my solution correct ?

Secondly , if my solution is not correct , can you help me to write the correct ?

Thirdly , if my solution is correct but it is long ,can you share any shortcut or beneficial information to make it elegant.

$EDIT$: I want to find its last digit and it is an exponential tower where $n=2005$

Body:
$7^{2004}=\big(7^2\big)^{1002}=49^{1002}\equiv1^{1002}=1\pmod 4$

As $\phi(10)=4$,

$7^{7^{2004}}\equiv7^1=7\pmod{10}$

Your solution is correct.


问:
Computing homology classes
Problem statement with diagram: https://drive.google.com/file/d/1GLSEP919BMVwWckaJKsjb_Wi8TKVWMYj/view?usp=drivesdk

I'm completely stuck on exercise 2.15. I'm self studying a bit of topology during winter break and this is the first time I've had a problem like that and there are no example problem in the book.

The way the book defined homology on 2d spaces, two curves are homologous if and only if there is a surface connecting them. Algebraically, I believe that means there must be a continuous invertible function between them. Earlier, homology was defined as an equivalence relation and I'm good with that. So my understanding is that I need to find the equivalence class of paths in R³/(B ∪ C) that can all be "smoothly" transformed into A. But I don't have a clue how to do that. The book explained the idea of homology classes, but not how to specify a specific one, let alone how to do it for two different spaces. I've been Googling trying to find examples of how to specify/define homology classes, but I couldn't find any example problems. Can someone please help?

答:
See that $A$ is homologous to the union of the two little red loops going around the blue circle. Then note the two little loops go around the blue circle in opposite directions, so they cancel to get the $0$ homology class.

Now $A$ (the red circle above left) minus the red cycle (the two little red loops above right) is the boundary of a surface that does not intersect $B, C$, as see below left; finally the two little red loops is trivial in homology since they together form the boundary of a cylinder that does not intersect $B, C$, see below right.

Title: Computing homology classes


问:
Find variables from a multiplication of geometric sequences

Given $$f(x) = \sum_{i=0}^\infty a_ix^i$$
And $$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}$$
Find the values of $a_0$, $a_1$, and $a_2$.

I try to expand them but it doesn't seem to lead me to any solution.
How can I isolate a0, a1, and a2 to find them?

Body:
Tip:

Notice $1+2x+2x^2+x^3=\left(1+x+x^2\right)(1+x)$

Then $$\begin{split}f(x)&=\frac1{(1-x)^3(1+x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{(1-x)(1+x)}\cdot\frac1{(1-x)\left(1+x+x^2\right)}\\&=\frac1{1-x}\cdot\frac1{1-x^2}\cdot\frac1{1-x^3}\\&=\left(1+x+x^2+x^3+...\right)\left(1+x^2+x^4+x^6+...\right)\left(1+x^3+x^6+x^9+...\right)\end{split}$$

Then expansion finishes it all. Try it!


Title: Find variables from a multiplication of geometric sequences

Body:
Given $$f(x) = \sum_{i=0}^\infty a_ix^i$$
And $$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}$$
Find the values of $a_0$, $a_1$, and $a_2$.

I try to expand them but it doesn't seem to lead me to any solution.
How can I isolate a0, a1, and a2 to find them?

Body:
From the second condition we get
$$f(x) = \frac{1}{(1-x)^3(1+2x+2x^2+x^3)} \tag 1$$
From the first condition we get
$$f(x)=\sum_{i=0}^\infty a_ix^i = a_0 + a_1x+a_2x^2+a_3x^3 + ... \tag 2$$
Hence, $$a_0 = f(0) \overset {(1)}= \dfrac{1}{1\cdot 1} = 1 \tag 3$$
To find $a_1$, differentiate $(2)$
$$f'(x) = a_1 + a_2x + ...$$ Hence, $$a_1 = f'(0)$$ wich you can find from $(1)$
$$f'(x)=-\frac{-3(1-x)^2(1+2x+2x^2+x^3) + (1-x)^3(2+4x+3x^2)}{\text{denominator}^2}$$ which gives $$ a_1 = f'(0) = -\frac{-3+2}{1} = 1$$
Similarly, you can find  $$a_2 = f''(0)$$


问:
Solve $y=xy'-((y')^5+40y'+\sin(y'))$
I tried to take derivative each side but at the end,

I ended up with $x=5(y')^4+40+\cos(y')$

because $y''$ were cancelled out. I m stuck here.

答:
This is called Clairaut ODE. You actually get two ODEs. One is $y''=0 \implies y'=C$. Put thois in the original ODE to get the general solution as
$$y=Cx-C^5-40 C-\sin C,~~~~~(1)$$
which a familly of lines when the parameter $C$ is varied. The solution of
$$x=5y'^4+40+\cos(y') ~~~~(2)$$ will be a fixed curve (without a constant) which will be envelope of the family of lines given by (1). in fact for any value of $C$ the line woll touch this fixed curve. But (2) ids difficult to solve by hand.

I may get back again.

Title: Solve $y=xy'-((y')^5+40y'+\sin(y'))$


答:
I think the answer is right because I turned up with the same one with this approach:

$7^2\equiv -1\ \left(\text{mod}10\right)$ (Here -1 corresponds to 9)

$7^6\equiv \left(-1\right)^3=-1\ \left(\text{mod}10\right)$

$7^7\equiv 3\ \left(\text{mod}10\right)$

i.e.$7^7\equiv -7\ \left(\text{mod}10\right)$

similarly continuing, we get

$7^{7^7}\equiv \left(-7\right)^7\equiv \left(-7\right)\cdot \left(-7\right)^6\equiv -7\cdot \left(7\right)^6\equiv -7\cdot -1\equiv 7\ \left(\text{mod}10\right)$
and as 2004 is a multiple of 7, all will get cancelled to leave the remainder of 7.

Hence  the last digit is 7.

Title: What is the last digit of $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$

Body:
$2005$ $7's$ are used in the following number ,so what is its last digit   :

Number: $.....\Biggl(\biggl(\Bigl(\bigl((7)^7\bigr)^7\Bigr)^7\biggr)^7\Biggr)^7 ......$

As i mentioned in the introduction , this number is written using $2005$ times $7's$ and i want to find its last digit.

My work: Its last digit is wanted so i used $\mod(10)$.

After that i formed the expression into $\bigl((7)^{7^{2004}}\bigr)$ , then i decided to use Euler's totient function.

${7^{\phi(10)=4}\equiv 1 \ (mod10)}$

In the next step , i said that ${7^{4k}\equiv 1 \ (mod10)}$ and  i checked over $7^{2004}(mod4)$.Then ,I found that $7^{2004}(mod4)=1$

As a result ,  $\bigl((7)^{7^{2004}}\bigr) \equiv 7^1 (mod 10)=7$

My first question is that is my solution correct ?

Secondly , if my solution is not correct , can you help me to write the correct ?

Thirdly , if my solution is correct but it is long ,can you share any shortcut or beneficial information to make it elegant.

$EDIT$: I want to find its last digit and it is an exponential tower where $n=2005$


问:
Triple integral: cylinder inside a sphere
I'm trying to solve the following exercise about triple integrals, but I need a check.

Let $V \subset \mathbb{R}^3$ which is made by one sphere of radius $2$ centered at $(0,0,0)$ and a cylinder with base at $(0,0,0)$ and radius $1$. Formally, this is

$$V=\{x^2+y^2+z^4 \leq 4 \} \cup \{(x,y,z): x^2 + y^2 \leq 1, z \in [0,4] \}$$

Goal: Compute the Volume of $V$

My attempt:

$|V|=|\text{Sphere}| + |\text{Cylinder}| - |\text{Portion of Sphere inside Cylinder}|$

The first two terms are straightforward: $|\text{Sphere}|=\frac{16}{3} \pi $, $|\text{Cylinder}|=4 \pi$

The tricky part for me is $|\text{Portion of Sphere inside Cylinder}|$

To  do that, I consider the $Oxz$ plane: here I can see that cylinder intersects the sphere at $(1,\sqrt{3})$ and the angle $\varphi = \frac{\pi}{6}$ ($\varphi$ is intended with this convenction)

Now I want to use spherical coordinates to compute the volume of that region: to this aim, I let the $\rho$ of the spherical coordinates range in $(0,1)$, in order to be inside the cylinder. All in all, the triple integral do to is the following:

$$\int_{0}^{2 \pi} \int_0^{\frac{\pi}{6}} \int_{0}^{1} \rho^2 \sin(\varphi) d \rho d \varphi d \theta$$

which can easily be solved.

答:
Your approach is fine but the integral does not give the volume of the sphere inside the cylinder - i) the way polar angle works, it will leave out parts of the sphere inside the cylinder on the sides unless you specifically address it in your integral. ii) you take $0 \leq \rho \leq 1$ but that will not take out the correct volume for this sphere.

As you are trying to find the volume of the combined region between $0 \leq z \leq 4$, here is what I would suggest.

You have already found that the cylinder and sphere intersect at $z = \sqrt3$. So we can add the volume of the hemisphere ($V_1$) AND the volume of the cylinder between $\sqrt3 \leq z \leq 4 \, (V_2)$. We have to then just subtract from the hemisphere the volume of the spherical cap that is inside the cylinder above $z = \sqrt3 \,, \, (V_3)$.

$V_3$ can be calculated with the below integral,

$V_3 = \displaystyle \int_{0}^{2 \pi} \int_0^{\pi /6} \int_{(\sqrt3 / \cos \phi)}^{2} \, \rho^2 \sin(\phi) \, d \rho \, d \phi \, d \theta$

$V_1 = \frac{16 \pi}{3} \,$ as you mentioned

$V_2 = \pi \times 1^2 \times (4 - \sqrt3) = (4 - \sqrt3) \pi$

Total volume of the combined solid is $V_1 + V_2 - V_3$.


问:
Sublinear functional and norm.
Let $X$ be a normed linear space. Let $C$ be a convex absorbing subset of $X.$ Define the function $p_C : X \longrightarrow \Bbb R_{+}$ by $p_C (x) = \inf \left \{t \gt 0\ |\ x \in tC \right \}.$ Then I have proved that $p_C$ is a sublinear functional. Can $p_C$ be a norm on $X\ $? It is clear that $p_C (0) = 0.$ What about the converse? Suppose $p_C(x) = 0.$ Can we say that $x = 0\ $?

Can anybody please help me in this regard? Thanks for your time.

答:
$p_C$ is  a norm if $C$ is bounded. $C=X$ is convex and absorbing but $p_C(x)=0$ for all $x \in X$ in this case!

Suppose $C$ is bounded and $p_C(x)=0$. Then theer exists a sequence $(t_n)$ decreasing to $0$ such that $x \in t_nC$ for all $n$. Hence, $\|x\| \leq t_n M$ where $M=\sup \{\|y\|: y \in C\}$. Hence $x=0$.

Title: Sublinear functional and norm.


问:
Why do you run into problems in counting with distinct sets combined with indistinct sets?
Say you want to arrange $3$ indistinct students into $8$ seats such that no student is next to each other.

My method:

x_x_x_x_x

let the _ denote a seat and $x$ a possible student. There are $5C3$ ways of slotting in the students. Why is this wrong?

答:
Via Stars and Bars you want the number of non-negative integer solutions to

$$x_1 + x_2 + x_3 + x_4 = (8-3) = 5$$

where

$$x_2, x_3 ~\text{must be}~ \geq 1.$$

This bijects to

$$x_1 + x_2 + x_3 + x_4 = (5-2) = 3$$

where

all 4 variables must be non-negative.

By Stars and Bars, the enumeration is

$$\binom{3 + [4-1]}{[4-1]} = \binom{6}{3} = 20.$$

As comment/questioned, $\binom{5}{2} + \binom{5}{3} = \binom{6}{3}.$


Body:
$f(x)= a_0+a_1x+a_2x^2+O(x^3)$
$f(x)(1+2x+2x^2+x^3)$
$=a_0+(2a_0+a_1)x+(2a_0+2a_1+a_2) x^2
+O(x^3)$
$$\textrm{Again } f(x)(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}=1+3x+6x^2+O(x^3)$$
Comparing the like powers of x we get:$$a_0=1,a_1=1,a_2=2$$


Body:
$$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}\implies $$ $$(1-x-x^2+x^4+x^5-x^6)\,f(x)=1$$

Now, let $f(x)=a+b x+c x^2$ and ignore the terms $x^n$ if $n>3$. This gives
$$a+ (b-a)x- (a+b-c)x^2=1$$
Comparing the terms, $a=1$, $b=a=1$, $c=a+b=2$.


答:
$a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$
That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe
$$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}{a^2-a}-500)^5 \\&\overset 1= (\color{blue}{502} - 500)^5 \\&= 32 \end{align}$$

Body:
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{2009}}{2}\Bigg] - 500\Bigg[\frac{\sqrt{2009} - 1}{2}\Bigg]\Bigg)^5$$
$$= \Bigg[\Bigg(\frac{1010\sqrt{2009} + 3018}{2}\Bigg)\Bigg] - 250(\sqrt{2009} - 1)\Bigg]^5$$
$$=(505\sqrt{2009} + 1509 - 250\sqrt{2009} - 250)^5$$
$$= (250\sqrt{2009} - 1259)^5$$

However, the answer given is $32$, so there could have been more simplifications.

As a question, where did I go wrong? Also can anyone give me some simpler way of solving this?


答:
Also possible as:

Thanks for all of your help guys! :)


问:
Constrained $L^2([0,1],\mathbb{R})$ space
The space $L^2([0,1],\mathbb{R})$ admits an orthonormal basis:

$1, \sqrt{2} cos(2\pi n x), \sqrt{2} sin(2\pi n x), \quad n= 1,2,...$

What happens to $L^2([0,1],\mathbb{R})$ if we add a constrain, say $u(x)=0 , x \in [0,\epsilon]$ for every function $u \in L^2([0,1],\mathbb{R})$, with $\epsilon$ a small positive number?

Clearly the above trigonometric basis, can not be a basis for the new constrained space. So what is a new basis for this space ? Is the a rigorous way to determine the new basis of the constrained space ?

Also, it is clear that the dimensionality of the space is still infinite. But the space has changed, so is there any notion that can capture this change ?

Thanks a lot

答:
One orthonormal basis of the subspace $M$ where $u=u\chi_{(\epsilon,1]}$ consists of functions that vanish on $[0,\epsilon]$ and equal the following on $(\epsilon,1]$:
$$
       c_n(x)=C_n\cos(2n\pi (x-\epsilon)/(1-\epsilon)),\;\; n\ge 0, \\
       s_n(x)=D_n\sin(2n\pi (x-\epsilon)/(1-\epsilon)),\;\; n \ge 1.
$$
The constants $C_n,D_n$ are chosen so that $\|c_n\|=1$ and $\|d_n\|=1$.


答:
Take out $3$ chairs for the students.

$5$ chairs remain with $6$ gaps where they can be inserted back.

$-C-C-C-C-C-$

Reinsert them in the $6$ gaps in $\binom6 3$ ways

Title: Why do you run into problems in counting with distinct sets combined with indistinct sets?

Body:
Say you want to arrange $3$ indistinct students into $8$ seats such that no student is next to each other.

My method:

x_x_x_x_x

let the _ denote a seat and $x$ a possible student. There are $5C3$ ways of slotting in the students. Why is this wrong?


答:
If this is indeed a tetration, rather than a repeated taking the seventh power, the answer is different.

Tetration: $7^{\large 7^{\large 7^{\dots}}}$

Repeated power: $((7^7)^7)^7\cdots$

For tetration, the modulus of even a short tower of exponents rapidly settles down to a single value, unaffected by extending the tower, under the influence of the reducing cycles of higher exponents.

Here $7^k \equiv 7^\ell \bmod 10$ when $k \equiv \ell \bmod 4$ since $\lambda(10) = 4$

Then $7^{7^m} \equiv 7^{3^m} \equiv 7^{-1^m} \bmod 10$, directly giving a $2$-cycle, and since we can see that in the given case (of tetration) that $m$ is odd, we have

$$ 7^{\large 7^{\large 7^{\large 7^{\dots}}}} \equiv 7^{\large 7^{\text{[odd #]}}} \equiv 7^{-1} \equiv 3 \bmod 10$$


Body:
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

I see no reason for elegance.  Since one of the factors in the numerator is $1$, computing $a^3$ is is straightforward.

$$a^3 = \left(\frac{1}{8}\right) \times
\left[
1 + 3\sqrt{2009} + 3(2009) + 2009\sqrt{2009}
\right]$$
$$=~
\left(\frac{1}{8}\right) \times
\left[6028 + 2012\sqrt{2009}\right]
~=~
\frac{1507 + 503\sqrt{2009}}{2}.
$$

Therefore,

$$(a^3 - 503a - 500)$$

$$=~ \frac{1507 + 503\sqrt{2009}}{2} ~-~
\frac{503 + 503\sqrt{2009}}{2} - \frac{1000}{2}
~=~ \frac{4}{2} \implies $$

$$(a^3 - 503a - 500)^5 = 2^5 = 32.$$


问:
What is the Shortest Set of Moves Needed to Solve a Rubik's Cube?
The question is asking what is the shortest sequence of moves that will have the cube solved at some point during the sequence. The most obvious such sequence would be one that has every solution to every position followed immediately by its inverse. This would be incredibly long, somewhere in the hundreds of quadrillions of moves. Is it even possible to find a shorter sequence?</p>

答:
You are looking for what is sometimes called the "devil's algorithm", which is a Hamiltonian cycle over cube states.

People have wondered about such a thing for some time, and there are explicit constructions; see here for example.


问:
unable to prove the property that uniform metric is actually a metric
Definition of uniform Topology: Let $\Gamma$ be a non empty set and let $\bar{d}$ be the standard bounded metric on $\mathbb{R}$ . Define a metric $\bar{\mathbb{\rho}}$ on $\mathbb{R}^{\Gamma}$ by  $\bar{\mathbb{\rho}}(x,y)$= {lub${\bar{d}(x_{\alpha} ,y_{\alpha})}: \alpha \in \Gamma$}.Then $ \bar{\mathbb{\rho}}$ is a metric called uniform metric and topology induced by $\bar{\mathbb{\rho}}$ is called uniform topology on $\mathbb{R}^{\Gamma}$.

I am unable to prove triangle inequality for $\bar{\rho}$,

I have proved it till here:$\frac{ d(x_{\alpha} , y_{\alpha})} { 1+ d(x_{\alpha}, y_{\alpha})} \leq \frac{ d(x_{\alpha} , z_{\alpha})} { 1+ d(x_{\alpha}, y_{\alpha})} +  \frac{ d(z_{\alpha} , y_{\alpha})} { 1+ d(x_{\alpha}, y_{\alpha})}$.

Can you please help me in completing the proof of triangle inequality.

Thanks!

答:
It’s easiest to do by contradiction.

Suppose that $\bar\rho(x,y)>\bar\rho(x,z)+\bar\rho(z,y)$; then by the definition of least upper bound there is an $\alpha\in\Gamma$ such that

$$\bar d(x_\alpha,y_\alpha)>\bar\rho(x,z)+\bar\rho(z,y)\,.$$

But $\bar\rho(x,z)\ge\bar d(x_\alpha,z_\alpha)$, and $\bar\rho(z,y)\ge\bar d(z_\alpha,y_\alpha)$, so

$$\bar d(x_\alpha,y_\alpha)>\bar d(x_\alpha,z_\alpha)+\bar d(z_\alpha,y_\alpha)\,,$$

contradicting the fact that $\bar d$ is a metric on $\Bbb R$.


Title: Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .

Body:
Your working leads to the answer. Here is the correction -

$a = \frac{1 + \sqrt{2009}}{2}$

$a^2 - 3 = \frac{999 + \sqrt{2009}}{2}$

$a (a^2 - 3) = 752 + 250 \sqrt {2009}$

$500 (a + 1) = 250 ( \sqrt {2009} + 3)$

$a(a^2-3) - 500(a+1) = 2$

So $(a^3 - 503a - 500)^5 = 32$


问:
Taking derivative of $\frac{\log x}{\sqrt{x}}$
Is it correct to write
$$
\frac{d}{dx}\frac{\log x}{\sqrt{x}} = \frac{2 - \log x}{2x^{{3/2}}}?
$$ I tried applying the quotient rule,
$$
\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}
$$
but I'm confused on how to get the numerator equal to $2 - \log x$ and the denominator equal to $2x^{3/2}$.

答: $\frac{d}{dx}\frac{\ln x}{\sqrt{x}}= \frac{\frac 1x×\sqrt{x}-\ln x\frac{1}{2\sqrt{}x}}{x} =\frac{2-\ln x}{2x^{3/2}}$


问:
Homemorphism between Mat($n,m$) and $\mathbb{R}^{nm}$
Defining matrices Lie groups I have noticed that we use the clear identification of Mat($n,m$) and $\mathbb{R}^{nm}$ given by $(a_{i,j})_{i,j} \to (a_{1,1},\cdots a_{n,m})$.

This seems quite important for computation of tangent space for example, everytime we use a limit to compute a differential in term of the coefficients of the matrix. What I was asking was, if we put on Mat($n,m$) the operator norm $||A|| := \text{max}\left\lbrace \frac{||Ax||}{||x||} : x \in \mathbb{R}^{n} -\textbf{0}\right\rbrace$, the induced topology should turn the identification above between  Mat($n,m$) and $\mathbb{R}^{nm}$ an homeomorphism.

There is a nice way to see it explicitly ? How could I prove it ? My difficulty relies on the fact that I'm not confortable using the topology induced by the operator norm to prove the continuity of the map.

答: On a finite dimensional real vector space, all norms are equivalent. So, since that map is a homeomorphism if we consider the norm$$\bigl\|(a_{ij})_{1\leqslant i,j\leqslant n}\bigr\|=\sqrt{\sum_{i=1}^n\sum_{j=1}^ma_{ij}^{\,2}}$$on $\operatorname{Mat}(n,m)$ (it's actually an isometry then), it will still be a homeomorphism if we use the operator norm there.


问:
Cauchy-Riemann Equations
I have a question.
If I have a complex function and I know that for $r=1$, $f(z=e^{iθ})=\cos(θ)\sin(θ)$
And I know that $f(z)$ is entire.
Now, I need to calculate $f'(z)$ for every $z$ in $r=1$, meaning $z=e^{iθ}$.
I tried to define $u(r,θ)= \cos(θ)\sin(θ)$,
$v(r,θ)= 0$
And use Cauchy-Riemann Equations in polar coordinates..and then I get that the derivative is zero, but I am not sure cause I can calculate the derivative by definition (derivative by $θ$) and then I got something which is a function of $θ$ so I am a bit confused..why is that? What is the right way to calculate it?

答:
$f(z)=\frac 1 {4i} (z^{2}-\frac 1 {z^{2}})$  whenever $|z|=1$. By Identity Theorem we must have $f(z)=\frac 1 {4i} (z^{2}-\frac 1 {z^{2}})$  for all $z \neq 0$, so $f$ cannot be continuous at $0$. Hence, there is no such entire function.

If you consider $f$ as an analytic function on $\mathbb C \setminus \{0\}$ then we can compute $f'(z)$ easily: $f'(z)=\frac 1 {4i} (2z+\frac 2 {z^{3}})$. This is valid for all $z \neq 0$, in particular for $|z|=1$.

Title: Cauchy-Riemann Equations


问:
The number of integer solution $x_1+x_2+\cdots+x_k=n$ such that $x_i\in \{1,2\}$
I am trying to find a combinatorial interpretation for the number of integer solution of the equation
$$x_1+x_2+\cdots+x_k=n$$ such that $x_i\in \{1,2\}$.

I know that the number of this solution is ${k \choose n-k}$. Also
$$\sum_{k=0}^n {k\choose n-k}=F_{n+1},$$
where $F_n$ is $n^{th}$-Fibonacci number.

Is it possible to find the solution for this system with the expression of Balls In Bins With Limited Capacity?

答:
It’s an especially easy version of that problem.

First put one ball in each of the $k$ bins; you now have $n-k$ balls left to distribute amongst the $k$ bins. You can put at most one of them into any bin, so you just have to choose which $n-k$ bins will get these extra balls, and you can do this in $\binom{k}{n-k}$ different ways. Each of those ways corresponds to a solution of the equation, and vice versa.


答:
Each of the $k$ values will either be $1$ or $2$. Let's say there are $m$ values equal to $2$, and the rest, $k-m$, are equal to $1$.

This means that $x_1+\cdots + x_k = m\cdot 2 + (k-m) \cdot 1 = k+m$, and since you also know that $x_1+\cdots x_k = n$, you have $k+m = n$ and so $m = n-k$.

Every solution to your problem matches directly with one selection of $m$ out of the $k$ values (i.e., you choose which of the $k$ values will be equal to $2$, and the rest will be equal to $1$), and therefore, there are ${k\choose m}$ of them.

Title: The number of integer solution $x_1+x_2+\cdots+x_k=n$ such that $x_i\in \{1,2\}$

Body:
I am trying to find a combinatorial interpretation for the number of integer solution of the equation
$$x_1+x_2+\cdots+x_k=n$$ such that $x_i\in \{1,2\}$.

I know that the number of this solution is ${k \choose n-k}$. Also
$$\sum_{k=0}^n {k\choose n-k}=F_{n+1},$$
where $F_n$ is $n^{th}$-Fibonacci number.

Is it possible to find the solution for this system with the expression of Balls In Bins With Limited Capacity?


问:
$\prod_{k=1}^{m}\sin(kt)$ related identity
Investigating the product $$\prod_{k=1}^{m}\sin(kt)$$
And became to the following enough trivial identity, but have a feeling that I am missing something here. Any hint about the following is appreciated.

\begin{align}
\sum _{k=1}^m \log \sin (k t) &=
  \sum _{k=1}^m \left(-\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2\right)=\\
  &=-\sum _{n=1}^{\infty } \sum _{k=1}^m \frac{\cos (2 k n t)}{n}-m \log 2=\\
  &=-\sum _{n=1}^{\infty } \frac{\csc (n t) \sin (m n t) \cos ((m+1) n t)}{n}-m \log 2
\end{align}

So the final result would be:
$$\prod_{k=1}^{m}\sin(kt)=2^{-m} \prod _{k=1}^{\infty } \exp\left(-\frac{\csc (k t) \sin (k m t) \cos (k (m+1) t)}{k}\right)$$

答:
Your final answer is true for some intervals of the real line, but not for all $t \in \mathbb{R}$. The problem is that $\ln(\sin(kt)) \not= -\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2$ when $\sin(kt) < 0$. What is always true is $\ln\left|\sin(kt)\right| = -\sum _{n=1}^{\infty } \frac{\cos (2 k n t)}{n}-\log 2$. You can also see that the LHS of the product would be negative for some $t$, while the RHS would always be positive.

The identity would be right if it was instead $$\prod_{k=1}^{m}\left|\sin(kt)\right|=\left|\prod_{k=1}^{m}\sin(kt)\right|=2^{-m} \prod _{k=1}^{\infty } \exp\left(-\frac{\csc (k t) \sin (k m t) \cos (k (m+1) t)}{k}\right)$$


答:
Let $y_i:=x_k-1\in\{0,1\}$. The sum of the $y_i$ must be $n-k$ so you must pick $n-k$ ones among $k$.

$$\binom k{n-k}.$$


问:
Show that this family is equicontinuous at $0$
Let $E$ be a normed vector space, $$p_K(\varphi):=\sup_{x\in E}|\varphi(x)|\;\;\;\text{for }\varphi\in E'$$ for compact $K\subseteq E$ and $\sigma_c(E',E)$ denote the initial topology with respect to $(p_K,K\subseteq E\text{ is compact})$, i.e. the subspace topology on $E'$ inherited from the topology of compact convergence on $C(K)$.

Let $\mathcal C\subseteq C(E')$ be uniformly $\sigma_c(E',E)$-equicontinuous.

Why can we conclude that $$\forall\varepsilon>0:\exists\delta>0:\forall\varphi\in E':\left\|\varphi\right\|_{E'}<\delta\Rightarrow\sup_{f\in\mathcal C}\left|f(0)-f(\varphi)\right|<\varepsilon?\tag1$$

Most probably the desired claim is trivial to obtain, but I'm not able to see how due to the rather complicated setting.

$(1)$ is obviously some kind of equicontinuity at $0$. I'm not sure if it's relevant, but by the Banach-Alaoglu theorem $\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is $\sigma_c(E',E)$-compact for all $\delta>0$.

答:
If I'm not mistaken, this should be an instance of a more general result: Let

$(X,\tau)$ be a topological space;

$Y$ be a normed $\mathbb R$-vector space;

$$\overline p(f):=1\wedge\sup_{x\in X}\left\|f(x)\right\|\;\;\;\text{for }f\in C(X,\tau;Y);$$

$$p_K(f):=\sup_{x\in K}\left\|f(x)\right\|_Y\;\;\;\text{for }f\in C(X,\tau;Y)$$ for $\tau$-compact $K\subseteq X$ and $$P:=\{p_K:K\subseteq X\text{ is }\tau\text{-compact}\}.$$

$(Z,d)$ be a metric space;

$F:C(X,\tau;Y)\to Z$ be continuous with respect to the locally convex topology on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$.

Then we easily see that $f$ is continuous with respect to the norm $\overline p$ on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$: Let $f\in C(X,\tau;Y)$ and $\varepsilon>0$. By the continuity assumption on $F$, there is a $P$-neighborhood $N$ of $f$ with $$d(F(f),F(g))<\varepsilon\;\;\;\text{for all }g\in N\tag1.$$ Let $U_p$ denote the open unit ball in $$C(X,\tau;Y)$$ with respect to $p\in P$. We can write $N=f+N_0$ for some $P$-neighborhood $N_0$ of $0$. Moreover, there are $k\in\mathbb N_0$, $\tau$-compact $K_1,\ldots,K_k\subseteq X$ and $\delta_0>0$ with $$B_0:=\delta_0\bigcap_{i=1}^kU_{p_{K_i}}\subseteq N_0\tag2.$$ Now let $\delta\in(0,1)$ with $\delta\le\delta_0$. Then, $$\delta U_{\overline p}\subseteq B_0\tag3$$ and hence $$d(F(f),F(g))<\varepsilon\;\;\;\text{for all }g\in f+\delta U_{\overline p}\tag4;$$ i.e. $f$ is continuous at $f$ with respect to the locally convex topology on $C(X,\tau;Y)$ generated by $P$ and the metric $d$ on $Z$.

Alternatively, the result would have been followed immediately by noting that the topology generated by $P$ is coarser than the topology generated by $\overline p$, as discussed here.

Now, if $X$ is a normed $\mathbb R$-vector space and $\tau$ is topology generated by $\left\|\;\dot\;\right\|_X$, then $$\left\|A\right\|=1\wedge\sup_{x\in X}\left\|Ax\right\|_Y\le\left\|A\right\|_{\mathfrak L(X,Y)}\tag5\;\;\;\text{for all }A\in\mathfrak L(X,Y)$$ and hence the topology generated by $\left\|\;\cdot\;\right\|$ is coarser than the uniform operator topology (i.e. the topology generated by $\left\|\;\cdot\;\right\|_{\mathfrak L(X,Y)}$). So, we immediately obtain that $F$ is continuous with respect to the topology generated by $\left\|\;\cdot\;\right\|_{\mathfrak L(X,Y)}$ and the metric $d$ on $Z$ as well.


问:
Could $\sum e^{a_i}$ be simplified? Does it have an identity?
$\sum_{i=1}^n e^{a_i}$ (where $a_i \in \mathbb R$) is expensive for large $n$ (a sum and $n$ exponential operations). I was wondering if there is any way for simplifying this?</p>

答:
One possible way to reduce the cost of the exponentiation operation may be to replace exponentiation involving large exponents with multiplications. This is what I mean:

Assume that $\{a_i\}$ is sorted in ascending order (if not, sort them first), so that $a_{i+1}\geq a_i$. Now, let's use the notation
$$
A(n) = \sum_{i=1}^n e^{a_i}
$$
Obviously, $A(1)=e^{a_1}$. Now, we can write
$$
A(2)=A(1)+e^{a_2}
$$
But we could also write this as
$$
A(2) = A(1)+b_1e^{a_2-a_1}
$$
where $b_1=e^{a_1}$. Similarly, we have
$$
A(3)=A(2)+b_2e^{a_3-a_2}
$$
where $b_2=e^{a_2}=b_1e^{a_2-a_1}$. If we define the sequence
$$
b_{i+1} = e^{a_{i+1}-a_i}b_i
$$
with $b_1=e^{a_1}$, then we can write
$$
A(n+1)=A(n)+b_{n+1}
$$
The advantage of this is that, while $a_n$ may be very large, and thus require a lot of computation, it may be much quicker to calculate the various $b_i$. It requires just as many exponentiation operations, but reduces the size of the exponent significantly, which may save on operation cost.

However, it is subject to the exponentiation cost relationship. If the cost of exponentiation remains almost constant with size of the exponent, then this will actually be slower than simply evaluating the exponentials directly, due to the added multiplications.


问:
Inverse permutation
Can anyone help me with this question.
((14638)(259))^-1

I understand the normal way of multiplying permutation with cycle notation but here is an inverse sign so I don’t know what to do?

Many thanks.

答:
Hint: The cycles are disjoint and so the inverse of  $\pi\rho= (14638)(259)$ is $(\pi\rho)^{-1} = \rho^{-1}\pi^{-1}$. So you have to form the inverse of cycles, which is quite easy:

The inverse permutation of $(x_1x_2\ldots x_n)$ is $(x_1x_n\ldots x_2)$.

Title: Inverse permutation


答: In a group, $(xy)^{-1}=y^{-1}x^{-1}.$ And the inverse of a cycle can be obtained by reversing the order of entries, e.g. $(acbd)^{-1}=(dbca).$

Title: Inverse permutation

Body:
Can anyone help me with this question.
((14638)(259))^-1

I understand the normal way of multiplying permutation with cycle notation but here is an inverse sign so I don’t know what to do?

Many thanks.


问:
Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$
Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.

We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$

Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?

答:
After your 3rd line, you are actually done, without realizing it; you simply went down the wrong path.

The RHS

$$= \frac{\sin(a) + \cos(a)}{\sin(a) - \cos(a)}.$$

Edit

See the comments following this answer.

A case can be made that my analyis is flawed, since I didn't bother to prove that the denominator above is always positive in the interval $(45^\circ, 90^\circ)$.

Addendum

Responding to

You can prove what the OP wants using pure geometry. Only using the definition of cosine and sine.

Okay: Imagine that you have a unit circle (i.e. of radius $= 1$) centered at the origin, with any point $(x,y)$ that is on the unit circle representing $(\cos[a],\sin[a])$, where $a$ is the angle formed by the two line segments $\overline{(0,0),(1,0)}$ and $\overline{(0,0),(x,y)}$.

Clearly, at $a = 45^\circ$, you have that $(x,y) = \left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right),$ since in a $45^\circ - 45^\circ - 90^\circ$ triangle, the two legs are equal.

Further, you can see geometrically, that for $a$ equal to any angle in $(45^\circ,90^\circ)$, for the corresponding point on the unit circle, the corresponding $y$ coordinate will have increased from $\frac{1}{\sqrt{2}}$, and the corresponding $x$ coordinate will have decreased from $\frac{1}{\sqrt{2}}$.

Therefore, since the $y$ coordinate represents $\sin(a)$ and the $x$ coordinate represents $\cos(a)$, you have that
$\sin(a) > \frac{1}{\sqrt{2}} > \cos(a),$ for $a$ equal to any angle in $(45^\circ,90^\circ)$.

A variation on the above argument is to notice that for $a$ equal to any angle in $(45^\circ,90^\circ)$, the slope of the line segment $\overline{(0,0),(x,y)}$ is $> 1$, which (alternatively) implies that $y = \sin(a) > x = \cos(a).$


问:
Exercise 1.5.7 (b) in Abbott’s Understanding Analysis
In the second edition, this exercise is as follows:

Find a 1-1 mapping from $S$, the open unit square $\{(x,y) : 0<x,y<1\}$, to the open interval $(0,1)$. Use the fact that each real number has a decimal expansion.

The answer I’ve been seeing the most, and is also the one I thought of, was to consider the decimal expansions of $x$ and $y$, $0.x_1x_2x_3...$ and $0.y_1y_2y_3...$, respectively. Map $(x,y)$ to the number in $(0,1)$ of the form $0.x_1y_1x_2y_2...$

Since numbers of the form $0.a_1a_2...a_n$ are equal to $0.a_1a_2...a_{n-1}(a_n-1)99999...$, I noticed that this creates the following issue:

Suppose I have $(.2,.7)$ which is equal to the pair $(.1999...,.6999...)$. Upon applying the previous mapping to both, they map to $.27$ and $.16999...$, respectively. Since $.16999...$ is the same as $.17$, this means the ordered pairs $(.1,.7)$ and $(.2,.7)$ map to the same element of $(0,1)$. So it isn’t 1-1? Is this problem avoided by just referring to $x$ and $y$‘s finite decimal expansion if it exists?

答:
More importantly, this tells us that the mapping is not well-defined and so is not a function. (This mapping is required to be a function in the problem statement in my copy of this book).

That is, the pairs $(.2,.7)$,$(.1999...,.6999...)$ each map to distinct numbers, even though the pairs are the same.

Perhaps someone else can shed some light on this as the proofs I have seen for this do the same thing as you, but never check that the mapping is well-defined in the first place, which would make the argument invalid.


问:
Proving an inequality with n positive real numbers
Is there any alternative  way of proving
$$\sum_{k=1}^n\frac{a_k-1}{a_k+1}\geq\frac{\prod_{k=1}^n a_k -1}{\prod_{k=1}^na_k+1}$$ for any positive real numbers $a_1,a_2,\cdots,a_n,$  greater than or equal to 1, other than the one   given in one of the the answers; An inequality associated with sum of n real numbers?

Body:
We write the inequality as
$$\sum_{i=1}^n\frac{1}{a_i+1}\leq \frac{n-1}{2}+\frac{1}{a_1a_2\cdots a_n+1}. \quad (1)$$
For $n=1$ the inequality become
$$\frac{1}{a_1+1}+\frac{1}{a_2+1} \leqslant \frac{1}{2} + \frac{1}{a_1a_2+1}, \quad (2)$$
or
$$\frac{(a_1-1)(a_2-1)(a_1a_2-1)}{2(a_1+1)(a_2+1)(a_1a_2+1)} \geqslant 0.$$
Which is true.

Now, let $x=a_1a_2 \cdots a_k,$ suppose $(1)$ is true for $n=k,$ we get
$$\sum_{i=1}^k\frac{1}{a_i+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1}. \quad (3)$$
We will show that this is also true for $n=k+1.$

Indeed, setting $y = a_{k+1},$ use $(3)$ and $(2)$ we have
$$\sum_{i=1}^{k+1}\frac{1}{a_i+1} = \sum_{i=1}^k\frac{1}{a_i+1} + \frac{1}{y+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1} + \frac{1}{y+1}$$
$$\leqslant \frac{k-1}{2}+\frac{1}{2}+\frac{1}{xy+1} = \frac{k}{2}+\frac{1}{x_1x_2\cdots x_{k+1}+1}.$$
Done.


答:
By the definition of cosine as the x-coordinate of a circle, cosine is negative in the interval $(90^{\circ},180^{\circ})$

$\cos  2x =2\cos^2x-1=1-2\sin^2x$

$\cos  2x =(\sqrt{2} \cos x-1)(\sqrt{2} \cos x+1)=-(\sqrt{2} \sin x-1)(\sqrt{2} \sin x+1)$

Using the fact that $\cos 2x$ is -ve in the interval $x \in (45^{\circ},90^{\circ})$

Prove the fact that $\cos x < \frac{1}{\sqrt{2}}< \sin x$ in the interval $x \in (45^{\circ},90^{\circ})$

Proof 2:
There is 1 more way people define $\cos x = \frac{\text{adjacent}}{\text{hypotenuse}}$

Let the angles be $x,90-x,90$

Use the fact that the side opposite to the greater angle is greater.

Therefore, adjacent < opposite (for $x \in (45^{\circ},90^{\circ})$)

Therefore, $\cos x < \sin x$ for $x \in (45^{\circ},90^{\circ})$

Title: Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$

Body:
Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.

We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$

Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?


问:
Non-zero prime ideals in the ring of all algebraic integers
Let $\mathcal{O}$ be the ring of all algebraic integers: elements of $\mathbb{C}$ which occur as zeros of monic polynomials with coefficients in $\mathbb{Z}$.

It is known that $\mathcal{O}$ is a Bezout domain: any finitely generated ideal is a principal ideal.

In addition, $\mathcal{O}$ has no irreducible elements, since any $x \in \mathcal{O}$ which is not a unit can be written as $x = \sqrt{x}\cdot\sqrt{x}$, where $\sqrt{x}$ is also not a unit in $\mathcal{O}$.

My question is:

Does $\mathcal{O}$ have any prime ideal other than $(0)$?

答:
We don't need the axiom of choice. We can write down a maximal ideal.

First, we observe that $R$ is countable: For every monic $p \in \mathbb{Z}[x]$ let $V(p)$ be the set of its complex roots. It has a canonical enumeration, when we order the roots using the lexicographic order on $\mathbb{R} \times \mathbb{R}$. For every $n \geq 1$ the set of these $p$ of degree $n$ identifies with $\mathbb{Z}^n$ and is therefore countable, with an explicit enumeration. It follows that also $\mathcal{O}_n = \cup_{\mathrm{deg}(p)=n} V(p)$ has an explicit enumeration, and then also $\mathcal{O}= \cup_n \mathcal{O}_n$. This enumeration is complicated, but it is computable.

Now let $R \neq 0$ be any countable ring. Any enumeration $R=\{a_0,a_1,\dotsc\}$ produces a maximal ideal as follows: Define an increasing chain of proper ideals $I_k$ as follows: Let $I_0=0$. If $I_k + \langle a_k \rangle$ is proper, let $I_{k+1} = I_k + \langle a_k \rangle$. If not, call $a_k$ bad, and let $I_{k+1}=I_k$. Then $I:=\cup_k I_k$ is a maximal ideal: By construction $1 \notin I$. Now let $a_k \in R \setminus I$. Then $a_k$ has to be bad (otherwise $a_k \in I_{k+1} \subseteq I$), so that $I_k + \langle a_k \rangle = R$ and hence $I+\langle a_k \rangle = R$. $\square$

More generally, let $R$ be a ring $\neq 0$ whose underlying set is well-orderable. Every enumeration $R=\{a_{\alpha} : \alpha < \kappa\}$ with some limit ordinal $\kappa$ produces a maximal ideal (without using the axiom of choice): Let $I_0=0$, and construct $I_{\alpha+1}$ from $I_{\alpha}$ as above. For limit ordinals $\lambda<\kappa$, let $I_{\lambda}=\cup_{\alpha<\lambda} I_{\alpha}$. Then $I=\cup_{\alpha<\kappa} I_{\alpha}$ is maximal: If $\alpha<\kappa$ and $a_{\alpha} \notin I$, then $a_{\alpha}$ is bad (otherwise $\alpha+1<\kappa$, $a_{\alpha} \in I_{\alpha+1} \subseteq I$), hence $I_{\alpha}+\langle a_{\alpha}  \rangle = R$ and $I+\langle a_{\alpha} \rangle = R$.

Even more generally, the proof of "well ordering principle $\Rightarrow$ Zorn's Lemma" in ZF actually shows that every partial order, in which every chain has an upper bound, and whose underlying set is well-orderable, has a maximal element.


答:
$\sin \alpha + \cos \alpha$ and $\sin \alpha - \cos \alpha$ are both positive in the given domain, so their quotient is also positive, and $|x| = x$ when $x \in \mathbb R^+$. $\sin \alpha + \cos \alpha > 0$ as $\sin \alpha, \cos \alpha$ are positive in the domain. Hence we can focus our attention to just proving $\sin \alpha > \cos \alpha$.

Divide both sides by $\cos \alpha$. Since it is positive in the given interval, the inequality does not change sign.

Thus you have $\tan \alpha > 1$, which is true because $\tan 45º = 1$ and $\tan x$ is a strictly increasing function in the given range $(45º, 90º)$. This is already sufficient as a justification, but on some insight as to why this is, $\tan x = \frac{\sin x}{\cos x}$, and $\sin x$ is increasing while $\cos x$ is decreasing in the given interval, which both increase the value of the function. Anything more rigorous has to involve a geometric argument with the unit circle or calculus.


问:
How to analyze the following complex number?
$f(T) = \int_{0}^{1}e^{-iT\lambda(s)}ds$, where $\lambda(s)$ increases monotonously. I want to show that  $|f(T)|$ has an upper bound that decreases with $T$ for $T \gg 1$.

For example, if $\lambda(s) = s$ then $|f(T)| \le 2/T$.

This question comes from numerical analysis of my research. It's the expression of error analysis in adiabatic evolution. More specifically, the digital error has expression : $\sum_{k=1}^{L}e^{-iT\sum_{j<k}\lambda_{j}}/L$, where $\lambda_{j}$ equals $\lambda_{j} = \lambda(j/L)$. From numerical result I am sure it has an upper which bound decays with $T$. However, I don't know how to prove that. I wonder if there is related formulas in complex analysis and fourier series.

答:
It suffices to prove that $f(T)$ tends to $0$ when $T$ tends to infinity, because this is equivalent to having an upper bound that decreases to $0$, namely $U(T) = \sup_{T'\ge T}|f(T')|$.

Let us suppose that $\lambda(s)$ is differentiable and that $\lambda'(s)$ never vanishes. The substitution $x = \lambda(s)$ gives
\begin{equation}
f(T) = \int_{\lambda^{-1}([0,1])}e^{-iTx}\frac{1}{|\lambda'(\lambda^{-1}(x))|} d x
\end{equation}
The Riemann-Lebesgue lemma now tells us that $\lim_{T\to\infty}f(T) = 0$ because $\chi_{x\in\lambda^{-1}([0,1])} \frac{1}{|\lambda'(\lambda^{-1}(x))|}\in L^1({\mathbb R})$. Indeed, its integral is $f(0) = 1$.

I'm pretty confident the differentiability assumption can be replaced by a weaker condition with some work.


问:
Group action on the set of irreducible characters are orthogonal.
I am studying representation theory and I came across the following problem:

Let $\mathbb{C}[G]$ be the set of $\mathbb{C}$-valued functions on a finite group $G$ and $S(\subset \mathbb{C}[G])$ be the set of irreducible characters of $G$. Let $\mathbb{C}_{\chi}[G]$ be the subspace of $\mathbb{C}[G]$ generated by the set $\{g* \chi \; : \; g \in G \} $ where $g*\chi$ is the action of $G$ on $\chi$ defined as $(g*\chi)(x) = \chi (g^{-1}x) \; \forall x \in G$.

Let us define the scalar product $<|>: \mathbb{C}[G] \times \mathbb{C}[G] \rightarrow \mathbb{C} $ s.t. $<\phi|\varphi> = \frac{1}{|G|} \sum_{g \in G} \phi(g) \overline{\varphi(g)}$.

Questions:

How do we show that if $\chi_1 \neq \chi_2$, then $\mathbb{C}_{\chi_1}[G]$ and $\mathbb{C}_{\chi_2}[G]$ are othogonal w.r.t. above scalar product?

Show that the subspaces $\mathbb{C}_{\chi}[G]$ are $\mathbb{C}[G \times G]$-submodules of $\mathbb{C}[G]$ irreducible by two non-isomporphs. (the action of $G \times G$ on $\mathbb{C}[G]$ is defined as $((g,h)*f)(x) = f (g^{-1}xh) $).

My attempt:

We have the following:
\begin{align*}\label{key}
\mathbb{C}_{\chi_1}[G] &= \text{span} \{ g*  \chi_1 \; : \; g \in G \} \\[2pt]
\mathbb{C}_{\chi_2}[G] &= \text{span} \{ g*  \chi_2 \; : \; g \in G \}
\end{align*}

The dot product $ \langle \phi,\varphi \rangle $can be written as
\begin{align*}
\langle \phi,\varphi \rangle &= \langle \sum_{g \in G} \lambda_g \left( g* \pi \right) , \sum_{h \in G} \lambda_{h}^{'} \left( h* \pi'  \right)  \rangle \\[2pt]
\text{After simplifying, we get} \\
& =  \sum_{g \in G} \sum_{h \in G} \lambda_g \overline{\lambda_{h}^{'}} \langle  \left( g* \pi \right) , \left( h* \pi'  \right)  \rangle
\end{align*}

If $g=h$, then the product will be $0$ but what if $g \neq h$?

(2) I don't even understand the statement.

答:
I am going to use the notation $\mathbb{C}^G$ rather than $\mathbb{C}[G]$.

(1) It suffices to check the basis elements $g\ast\chi$ of $\mathbb{C}^G_\chi$ are orthogonal to the basis elements $h\ast\psi$ of $\mathbb{C}_\psi^G$:

$$ \langle g\ast\chi,h\ast\psi\rangle=\frac{1}{|G|}\sum_{x\in G} \chi(g^{-1}x)\overline{\psi(h^{-1}x)}. \tag{$\circ$}$$

Suppose $\chi$ and $\psi$ are characters of $V$ and $W$. Note $\chi(g^{-1}x)\overline{\psi(h^{-1}x)}$ is then the trace of the linear map $T\mapsto (g^{-1}x)T(h^{-1}x)^{-1}$ on $V\otimes W^\ast \cong\hom(W,V)$. Then $(\circ)$ is the trace of

$$ T\mapsto g^{-1}\left(\frac{1}{|G|}\sum_{x\in G} xTx^{-1}\right)h, $$

on $\hom(W,V)$ but by Schur's lemma the expression in parentheses is zero (it is the projection of $T$ onto $\hom_G(W,V)$), so this is the trace of the zero map, which is zero.

(2) Saying subspaces are irreducible "by two non-isomporphs" does not seem like gramamatical English, so perhaps there is a typo or you transcribed/translated incorrectly.

But we can show $\mathbb{C}_\chi^G$ is an irrep of $G\times G$. The key, at least in the following, is in understanding how the Artin-Wedderburn theorem says the group algebra $\mathbb{C}[G]$ acts on irreps $V$. Recall a rep is an algebra homomorphism $\mathbb{C}[G]\to\mathrm{End}(V)$, and AW says $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ is an isomorphism. Keep in mind, also, that $\mathrm{End}(V)\cong V\otimes V^\ast$ as a $G\times G$-rep (where $(g,h)T:=gTh^{-1}$, which is an irrep since $V$ is. (Recall that if $V$ is a $G$-irrep and $W$ is an $H$-irrep then $V\otimes W$ is a $G\times H$-irrep.)

Suppose $\chi$ is the character of an irrep $\rho_V$. Then any $f\in\mathbb{C}^G_\chi$ is expressible as

$$ f_u(x)=\mathrm{tr}\rho(ux) $$

for some $u\in\mathbb{C}[G]$ (the group algebra). There is some redundancy in representing functions this way, though. Indeed the map $\mathbb{C}[G]\to\mathbb{C}^G_\chi$ given by $u\mapsto f_u$ annihilates every summand of $\mathbb{C}[G]$'s AW decomposition except $\mathrm{End}(V)$. Note also $(g,h)\ast f_u=f_{hug^{-1}}$, so in fact we have an onto $G$-morphism from $\mathrm{End}(V)$ to $\mathbb{C}^G_\chi$. (I am sure there is a better way to explain this, but this is how I think about the problem.)


问:
Lack of analytic solution for a second order non-linear differential equation in the form $f''+c_1 f^2 = c_2$
The following problem is from Bird Transport phenomena.

I have solved this problem and I know the solution. But one part got me interested, which I cannot prove the statement and that is the first sentence of part c.

Can you help about the proof of non-existence of analytical solution for the differential equation 3.K-1?

答:
Splution of $$f''+af^2=b$$ multiply by $2f'$ both sides to get
$$2f'f''(x)=-2af'f^2+2f'b \implies (f'^2)'=(-2f^2+2b) f'.$$
Integrate both sides w.r.t. $x$, to get
$$\int\frac{df'^2}{dx} dx=\int(-2af^2+2b)\frac{df}{dx} dx \implies f'^2=-\frac{2a}{3}f^3+2bf+A$$
$$\implies f'= \pm \sqrt{-2af^3+2bf+A} \implies \int \frac{df}{\sqrt{-2af^3+2bf+A}}=\pm\int dx+B.$$
The integral on the left relates to Elliptic integrals.


问:
Reference Needed: $m$-faced dice with $n$ options
I would like to investigate the following question and I am not sure if similar ideas have been investigated. I cannot really find a detail discussion about this matter on the Internet. It will be great if you can tell me what domain I should look into, or you may also give your view on the following questions.

Suppose we have one fair dice. We would like to choose an option randomly from 6 possible options. We roll the dice once then we can decide which option we should choose.

Suppose we now have 8 options. To choose an option randomly, different algorithms can be used.
For example, roll the dice once first.

If the result is odd, we focus on option 1 to 4. Then roll the dice again, if we get 1 to 4, we know which option to use. If it is a 5 or 6, repeat the second step again.

If the result is even, we focus on option 5 to 8. Then roll the dice again, if we get $n$, which $1\leq n\leq 4$, we choose option $n+4$. If it is a 5 or 6, repeat the second step again.

One can check the the probability of choosing each option is exactly $\frac{1}{8}$.

My question now is to investigate if it is possible that for whatever number of options we have, we can always use a dice to choose an option randomly. And is there a general algorithm to follow in order to choose an option?

My second question is more advance. Suppose we have a fair $m$-faced dice and we have $n$ total options. Can we choose an option randomly by using the dice for any $m$ and $n$?

Once again, you don't have to answer the above two questions. I am looking for hints or domain that I should look into. Thank you.

答:
A further tweak to mathmandan's scheme is as follows.

For fixed $m,n,\ell$, allocate sequences of $\ell$ numbers from $1,\ldots,m$ to outcomes as before, i.e. choose the largest $k$ such that $kn\leq m^{\ell}$ and allocate $k$ of the smallest (lexicographically) $kn$ sequences to each of $1,\ldots,n$ and the others to "reroll". However, instead of rolling a whole batch $\ell$ and then rerolling if necessary, roll up to $\ell$ dice one by one and stop as soon as you know you will need to reroll. As well as improving the expectation, this can actually change which choice of $\ell$ is best.

For example, with $m=6,n=20$, mathmandan's approach gets $3.6$ with $\ell=2$ and $3.24$ with $\ell=3$, so prefers the latter. However, with $\ell=2$ the reroll combinations are $(4,3),(4,4),\ldots,(6,6)$, so if the first die of the first pair is 5 or 6, you don't need to roll the other one. The expected number of rolls with this modification, $E$, satisfies
$$E=\frac{1}{3}(1+E)+\frac{4}{36}(2+E)+\frac{20}{36}\times2,$$
which gives $E=3$.

(This tweak will also improve the expectation for $\ell=3$ slightly, since you can stop and reroll if your first two rolls are $6,5$ or $6,6$, but it will clearly be worse than $E=3$ for $\ell=2$.)


问:
Property of a divisor function, $\lim_{n\to\infty} \frac{\rho(n, n)}{n^2}=1-{\pi^2\over12}$ where $\rho(n, k):=\sum_{i=1}^k{(n\mod i)}$
Let $$\rho(n, k):=\sum_{i=1}^k{(n\mod i)}$$ where $(n \mod i)$ is a remainder of $n$ divided by $k$. Also let $\sigma(n)$ be the sum of all factors of $n$, or $$\sigma(n):=\sum_{d|n} d$$ i.e. a division function of order 1. I found the relation between $\rho(n, n)$ and $\rho(n+1, n+1)$, which was $$\rho(n+1, n+1)=\rho(n, n)+2n+1-\sigma(n+1)$$ This follows because, for $i=1, 2, \cdots, n$, $((n+1)\mod i) \ \ \ \  -(n\ \  \mod i)$ is 1 if $i|n+1$ and $1-i$ otherwise, which in turn $$\rho(n+1, n+1)-\rho(n, n)=\sum_{ i\nmid n+1} 1+\sum_{i\mid n+1}{(1-i)}=\sum_{i\le n}1-\left(\sum_{i \mid n}i - (n+1)\right)\\=2n+1-\sigma(n+1)$$ Through this, I got the closed-form of $\rho(n, n)$, which was $$\rho(n, n)=n^2-1-\sum_{k=2}^n \sigma(k)=n^2-\sum_{k=1}^n \sigma(k)$$ Now, the real problem is, I have to show that $$\lim_{n\to\infty} \frac{\rho(n, n)}{n^2}=1-{\pi^2\over12}$$ and it gives a well-known property $\zeta(2)={\pi^2\over6}$ as a hint. It seems to prove $$\lim_{n\to\infty} \frac{\sum_{k=1}^n \sigma(k)}{n^2}={1\over2}\sum_{k=1}^\infty{1\over k^2}$$ but I have no idea. Any hints are going to be appreciated. Thanks.

答:
For the sum of divisors, we have $$\sigma(m)=\sum_{kl=m}l,$$ where we sum over all pairs $(k,l)$ with $kl=m$. Thus, $$\sum^n_{m=1}\sigma(m)=\sum_{kl\le n}l,$$ the sum over all pairs of positive integers $(k,l)$ with $kl\le n$, i.e. under the hyperbola defined by $kl=n$. It's a two-dimensional sum, and if you sum over $k$, first, you get
$$\sum_{kl\le m}l=\sum^n_{l=1}l\,\sum^{\lfloor n/l\rfloor}_{k=1}1=\sum^n_{l=1}l\,\lfloor n/l\rfloor,$$ interesting, but not helpful in this case. BTW, it's another proof of your result above, because $n\bmod l=n-l\lfloor n/l\rfloor$.

If you sum over $l$, first, you get $$\sum_{kl\le m}l=\sum^n_{k=1}\,\sum^{\lfloor n/k\rfloor}_{l=1}l=\sum^n_{k=1}\,\frac{\lfloor n/k\rfloor(\lfloor n/k\rfloor+1)}2.$$


问:
Determine whether a point $c$ is in a wake $W_P$.
One can draw wakes in the parameter plane of the Mandelbrot set, by tracing external rays inwards from near $\infty$ (with Newton's method or other algorithm) to get polygonal outlines which can be filled.  But reading the following paper I think there should be a way to plot them implicitly, starting from each pixel's $c$ coordinate to a binary classification:

Definition. Let $O = \{z_1, \ldots, z_p\}$ be a periodic orbit for $f$ . Suppose that there is some rational angle $t \in \mathbb{Q}/\mathbb{Z}$ so that the dynamic ray $R^{K(f)}_t$ lands at a point of $O$. Then for each $z_i \in O$ the collection $A_i$ consisting of all angles of dynamic rays which land at the point $z_i$ is a finite and non-vacuous subset of $\mathbb{Q}/\mathbb{Z}$. The collection $\{A_1, \ldots, A_p \}$ will be called the orbit portrait $P = P(O)$.

Theorem 1.2. The Wake $W_P$. The two corresponding parameter rays $R^M_{t\pm}$ land at a single point $r_P$ of the parameter plane. These rays, together with their landing point, cut the plane into two open subsets $W_P$ and $\mathbb{C} - W_P$ with the following property: A quadratic map $f_c(z) = z^2 + c$ has a repelling orbit with portrait $P$ if and only if $c \in W_P$, and has a parabolic orbit with
portrait $P$ if and only if $c = r_P$.

-- Periodic Orbits, Externals Rays and the Mandelbrot Set: An Expository Account, John W. Milnor https://arxiv.org/abs/math/9905169

However, I haven't yet deciphered if the paper provides a constructive algorithm for checking, or if it is just an existence proof.

Question: given an orbit portrait $P$, provide a constructive algorithm to determine if a point $c \in \mathbb{C}$ is in the wake $W_P$.

答:
One way of drawing them implicitly is by tracing the $2$ rays corresponding to the wake's characteristic angles in the dynamical plane and seeing if they land together, for each pixel $c$ of the parameter plane:

Theorem 2.5 (Ray portraits are born in wakes). Every cycle of periodic dynamic rays has two characteristic addresses/angles $s_+ , s_−$ ; the parameter rays of addresses/angles $s_+ , s_−$ land together at a parabolic parameter and hence partition the plane into two regions, one of which does not contain the unique period one hyperbolic component and which is called a parabolic wake $W_{s_− ,s_+}$. The two dynamic rays of address/angle $s_+ , s_−$ land together for a map $f_c \in F$ if and only if $c \in W_{s_− ,s_+}$.

-- A survey on MLC, Rigidity and related topics, Anna Miriam Benini https://arxiv.org/abs/1709.09869

The boundaries are a bit rough (I don't know why), and convergence near the parabolic point is slow (which I expected), but it seems to give reasonable images in a brief test:


问:
Integral of a rational expression involving quadratics.
I want to solve this integral but I have some problems...

$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$

I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x^2+2x+1)}$, but I wasn't capable of finding any antiderivative.

Thanks for responding!

答:
Hints

Rewrite the numerator as $(x+1)^2 - 4x$.

Ask yourself, what is the $\frac{d}{dx} \log[f(x)]$?

What is the anti-derivative of $\frac{1}{(x+1)^2}$?


Title: Integral of a rational expression involving quadratics.

Body:
I want to solve this integral but I have some problems...

$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$

I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x^2+2x+1)}$, but I wasn't capable of finding any antiderivative.

Thanks for responding!

Body:
\begin{gather*}
I=\int ^{3}_{2}\frac{x^{2} -2x+1}{x^{2} +2x+1} dx\\
\text{Decompose $x^{2} -2x+1$ into partial fractions of $( x+1)$}\\
I=\int ^{3}_{2} 1-\frac{4x+4-4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{x^{2} +2x+1}\\
I=\int ^{3}_{2} 1-\frac{4}{x+1} +\frac{4}{( x+1)^{2}} dx\\
\end{gather*}Can you take it from here?


Body: $$ \frac{x^2-2x+1}{x^2+2x+1}=1-\frac {4x+4-4}{(x+1)^2}=1-4\frac {x+1}{(x+1)^2}+\frac 4{(x+1)^2}$$


问:
Is $X:=\{a\}$ Hausdorff space?
I wonder if $X:=\{a\}$ is a Hausdorff space.

Definition;

$A$ is a Hausdorff space $\iff
\forall x,y \in A(x\neq y);
\exists \text{ open set } U,V s.t. x\in U,y\in V, U\cap V=\phi.$

But we cannot pick up two elements from $X=\{a\}.$

How should I judge whether $X$ is a Hausdorff space or not?

答: It is a Hausdorff space since that condition is trivially true, by the reason that you have mentioned: there are no two distinct elements on $X$. By the same argument, every function whose domain is a singleton is injective.

Title: Is $X:=\{a\}$ Hausdorff space?


问:
Why is this inequality true?
I was solving an inequality and got stuck at this part:
$$abc+abd+acd+bcd\le a^3+b^3+c^3+d^3$$
Why is this true? I think it has a similar solution as $ab+ba\le a^2+b^2$, because in both cases the left side is rearranged.</p>

Title: Why is this inequality true?

Body: Hint $a^3+b^3+c^3\ge 3abc$ for $a, b, c \ge 0$.


问:
Solving Binomial probability question- what's wrong with my solution?

A company establishes a fund of $120$ from which it wants to pay an amount, $C$, to any of its $20$ employees who achieve a high-performance level during the coming year. each employee has a $2\%$ chance of achieving a high-performance level during the coming year, independent of any other employees. Determine the maximum value of $C$ for which the probability is less than $1\%$ that the fund will be inadequate to cover all payments for high performance.

Multiple-Choice Options: $24, 30, 40, 60, 120$.

My attempt:

Let $X$ be the RV denoting the number of employees that need to be paid the amount $C$. $X \sim Bin(20, 0.02)$. We need to find $P(X>120/C)<0.01$.

Note that $P(X= 0) = 0.667$, $P(X= 1) = 0.272$, $P(X= 2) = 0.052$, $P(X= 3) = 0.006$. Since $P(X= 3)<0.01$, we obtain $$3>\frac{120}{C} \implies C>40$$

I chose the option for $40$, however, the correct answer is $60$. Now, since my answer said $C>40$, I should have chosen $60$ or $120$, but I suspect that the solution has major flaws. Can someone please review my solution and let me know where I went wrong? Thanks.

答:
From your  $P(X= 0) = 0.668$, $P(X= 1) = 0.272$, $P(X= 2) = 0.053$, $P(X= 3) = 0.006$ (with slightly better rounding)

you can find $P(X> 0) = 0.332$, $P(X> 1) = 0.060$, $P(X> 2) = 0.007$, $P(X> 3) = 0.001$.

So there is a probability of less than $1\%$ that there are more than $2$ high-performers, but more than $1\%$ that there is more than $1$ high-performer,

and thus you need the fund of $120$ to be able to pay out to up to two high performers at a potential cost of $2C$.

$2C \le 120$ implies $C \le 60$.


问:
The Lebesgue Dominated Convergence Theorem: Example.
The following question is from Real Analysis by Royden (4th Edition). More precisely, it is problem 36 of chapter 4. Further, the problem follows right after the section over The Lebesgue Dominated Convergence Theorem, hence, that is what should be used. Below is the problem followed by my attempted solution which I have a question about.

Problem:

Let $f$ be a real-valued function of two variables $\left(x,y\right)$ that is defined on the following unit square
$Q:=\{\left(x,y\right): 0\leq x\leq 1, 0\leq y\leq 1\}$ and is a measurable function of $x$ for each fixed value of
$y$. For each $\left(x,y\right)\in Q$ let the partial derivative a $\partial f/\partial y$ exist. Suppose there is a function $g$ that is integrable over $[0,1]$ and such that,

$$\left\vert\frac{\partial f}{\partial y}\left(x,y\right)\right\vert \leq g(x) \text{ for all } \left(x,y\right)\in Q.\tag{*}$$

Prove that

$$\frac{d}{dy}\left[\int_0^1 f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx, \text{ for all } y\in[0,1].$$

(My) Proof:

Fix $y\in[0,1]$ and consider the sequence of functions $f_n\colon[0,1]\to\mathbb{R}$ defined by,

$$f_n(x):=\frac{f(x,y+1/n) - f(x,y)}{1/n}.$$

Then, $\{f_n\}$ is a sequence of measurable functions ($f$ is a measurable function of $x$ for each fixed value of $y$) dominated by $g(x)$ that converge pointwise almost everywhere to $\partial f/\partial y$ on $[0,1]$ (for each $(x,y)\in Q$ the partial derivative $\partial f/\partial y$ exists). Therefore, by The Lebesgue Dominated Convergence Theorem,

$$\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$$
$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx =
\int_0^1\frac{\partial f}{\partial y}(x,y)dx.\tag{1}$$

Notice,

$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx =
\lim_{h\to 0}\frac{\int_0^1f(x,y+h)dx-\int_0^1f(x,y)dx}{h}=
\frac{d}{dy}\left[\int_0^1f(x,y)dx\right].\tag{2}$$

Thus, combining (1) and (2),

$$\frac{d}{dy}\left[\int_0^1f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx,$$

which was to be shown.

My main concern with my proof is that, I am not sure how the sequence $\{f_n\}$ are dominated by $g(x)$. I know by (*) that the limit of the sequence $\{f_n\}$ is bounded by $g(x)$. Also, if there are any mistakes in the proof, please let me know!

Edit:

Does the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence?

答:
Another way to prove that $f_n$ is bounded by $g$ is
\begin{align}
|f_n(x)| = & \bigg \vert \frac{ f(x, y+1/n)-f(x,y)}{1/n}\bigg \vert\\
         = & \bigg \vert \frac{ \int_y^{y+1/n} \partial_y f(x,y) dy }{1/n}\bigg \vert\\
         \leq & \frac{ \int_y^{y+1/n} |\partial_y f(x,y)| dy }{1/n}\\
         \leq & g(x).
\end{align}


问:
$f:(1,-\infty) \times \mathbb{R} \to \mathbb{R}:f(x,y)=(\ln(x)^{y})$ is Borel measurable.
I'm trying to prove that the function $f:(1,\infty) \times \mathbb{R} \to \mathbb{R}:f(x,y)=(\ln(x)^{y})$ is a Borel measurable function .

First I thought that the function is continuous so it needed to be Borel measurable, but I think that's to easy.

As followed I tried to prove that $f^{-1}(A)$ is a Borelsubset for every Borelsubset $A \subset [0,\infty]$. My intentions were to prove it for $A \in B$ with $B$ the smalest $\sigma$-algebra of $[0,\infty]$

I don't get anything usefull on paper to prove it correctly.

EDIT: my first thought was correct. Because the function is continuous it is enough so you can see it's a Borrel measurable function.

答: my first thought was correct. Because the function is continuous it is enough so you can see it's a Borrel measurable function</p>


问:
Expectation of stochastic integral, martingale
I have to show that
$$\mathbb{E}\left[ \int_0^t a B_s^{a-1} \ dB_s \right] = 0$$
I started with the fact that, if I show that the inner integral $\int_0^t a B_s^{a-1} \ dB_s$ is a martingale I get the wanted result.

One way to show that the stochastic integral is a martingale is to show that:
$$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] < \infty$$
which is basically saying $aB_s^{a-1} \in \Lambda^2(t)$ (to have this we also need the process to be progressive, but it follows by BM properties I guess).

Using Fubini I can interchange expectation and integral, bring outside constants and write:
$$\mathbb{E}\left[ \int_0^t (a B_s^{a-1})^2 \ ds \right] =  \int_0^t a^2 \ \ \mathbb{E}\left[ B_s^{\ 2(a-1)} \right] ds< \infty$$
any hint to show this? I should not use results on Brownian motion moments.

NOTE: edited after @surb comment!!

答:
Here is a proof for $\int_0^{t} aEB_s^{a-1}ds <\infty$ which you wanted . You need to modify this based on the comment of Surb.

$B_s \sim \sqrt s X$ where $X$ has standard  normal distribution. Hence, your integral $\int_0^{t} aEB_s^{a-1}ds <\infty$ is a constant times $\int_0^{t} s^{(a-1)/2}ds$ and this last integral equals $\frac {t^{(a+1)/2}} {(a+1)/2}$ if $a>-1$.


问:
Minimum Absolute Difference in a List of Integers
Finding the minimum absolute difference in a list of integers by comparing each possible pair in the list is inefficient due to the number of comparisons required. If we sort the list first, then apparently the minimum can be found by simply comparing each integer with its neighbour.

Why is it the case that after sorting, the minimum absolute difference will be found comparing each item only with its neighbour please?

答:
Let $S = \{s_1. s_2, \ldots, s_n\}$ be your set of integers ordered so that $s_1 \le s_2\le \cdots \le s_n$.

Let $k$ be arbitrary.

Then $(s_k - s_{k - 1}) + (s_{k-1} - s_{k-2}) = s_k - s_{k-2}$.

So $s_k - s_{k - 1} \le s_k - s_{k-2}$

As $k$ is arbitrary, it follows (after a few moments thought) that the difference between $s_k$ and its adjacent elements is never greater than the difference between $s_k$ and any other elements of $S$.

So the minimum distance is bound to be one of those differences between adjacent elements.


问:
Proof: Kish's effective sample size is no greater than the number of samples
Given a dataset $X=\{x_1,x_2,\dots,x_N\}$, and two densities $p(x)$ and $q(x)$ with $p(x)$ being absolutely continuous with respect to $q(x)$, the effective sample size is defined as the number of samples from $p(x)$ that would provide an estimator with a performance equal to that of the importance sampling estimator $w(x)=p(x)/q(x)$ with $N$ samples. The Kish's effective sample size is defined to be

$$
N_{ESS}={\Vert w(X)\Vert_1^2\over \Vert w(X)\Vert_2^2}={\left(\sum_xw(x)\right)^2\over(\sum_xw(x)^2)}
$$

How can I prove that $N_{ESS}\le N$?

Update

I've come up with a solution that utilizes the nonnegativity of variance. I regard $w(x)$ as a random variable $Y$. Then, $\Vert w(X)\Vert_1^2=N^2\mathbb E[Y]^2$ and $\Vert w(X)\Vert_2^2=N\mathbb E[Y^2]$. Because $\mathbb E[Y^2]-\mathbb E[Y]^2=Var(Y)\ge0$, ${\mathbb E[Y]^2\over\mathbb E[Y^2]}\le 1$ and $N_{ESS}\le N$. I'm still open to any analytical solution.

答:
Hint:

For a given sum $S=\sum_x w(x)$, you can show that you minimise $\sum_xw(x)^2$ when the $w(x)$ are all equal, i.e. all $w(x)=\frac1NS$, the average of the weights

That $\sum_xw(x)^2$ will then give an upper bound on $N_{ESS}$


问:
Two terms that I want to understand: weakest topology and jointly continuous (in the following context).
I was reading an article online, please help me to understand the following lines (in bold letters). -
Topological structure:
If (V, ‖·‖) is a normed vector space, the norm ‖·‖ induces a metric and therefore a topology on V.the distance between two vectors u and v is given by ‖u−v‖.

This topology is precisely the weakest topology which makes ‖·‖ continuous and which is compatible with the linear structure of V in the following sense:

1.The vector addition + : V × V → V is jointly continuous with respect to this topology. This follows directly from the triangle inequality.

2.The scalar multiplication · : K × V → V, where K is the underlying scalar field of V, is jointly continuous. This follows from the triangle inequality and homogeneity of the norm.

Please explain me the Weak topology and how does it makes norm ‖·‖ continuous. what does it mean by "Addition + : V × V → V is jointly continuous with respect to this topology"

Thank you so much in advance.

答:
There is already a very nice answer, but I thought I could give an example of a function which is discontinuous as a function of two variables, but continuous in each variable. Consider the function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$,

$$
f(x,y) = \begin{cases}
      \frac{xy}{x^2 + y^2} & (x,y) \neq (0,0) \\
      0 & (x,y) = (0,0) \\
   \end{cases}
$$

This is discontinuous as it is 1/2 arbitrarily close to $(0,0)$ when $x=y$, but it is easily seen to be continuous in each variable.


问:
Prove that dim(M∩N)=dim(N)−1
Let L be a vector space of dimension $ n $ and  $ M,N \subset $ L linear subspaces such that $ dim(M)=n-1 $ and $ N \not \subset M $.

Prove that $ dim( M \, \cap N ) = dim(N) -1 $

My attempt:

Let $ (v_1,...,v_{n-1}) $ be a basis for $ M $ , we'll expand the basis of $ M $ to basis of $ M + N $ by the vectors $ (v_n,...,v_t) $ so that the basis of $ M + N $ is $ (v_1,...,v_t) $.
Since $ M + N \subseteq V $ then $ dim(M + N) \leq dim(V) $ , so that
$ n-1 \leq dim(M+N) = t \leq n $.
[ I dont know how to continue from here ]

Basically I want to use the dimension theorem for subspaces $ dim(M+N) = dim(M) + dim(N) - dim(M \cap N)$ But I got stuck and don't know how to continue, can you please help?

Edit:

As shown in the answers below, after showing that $ dim( M + N ) = n $ then, using the dimension theorem,
$ dim(M+N) = dim(M) + dim(N) - dim(M \cap N) \iff $ $ dim(M \cap N) = dim(M) + dim(N) - dim(M+N) \iff $ $ dim(M \cap N) = n-1 + dim(N) - n \iff dim(M \cap N) = dim(N) - 1  $ . And we've finished the proof.

答:
If $\dim(M+N) = \dim(M)$, then since $M \subset M+N$, we have $M = M+N \supset N$ (contradicting $N \not\subset M$)

Therefore $\dim(M+N) > \dim(M) = n-1,$ which along with $\dim(M+N) \leq n$ implies $\dim(M+N) = n$.

Now finish by the dimension theorem.

Title: Prove that dim(M∩N)=dim(N)−1


答: You are almost there! Take a basis $(v_1, \dots, v_{n-1})$ of $M$. Since $N$ is not included in $M$, then you can take a vector $v_n \in N \backslash M$. This vector does not belong to the span of $(v_1, \dots, v_{n-1})$ (which is $M$), so the family $(v_1, \dots, v_n)$ is linearly independent. This gives a linearly independent family of size $n$ in $M+N$, and therefore $\dim(M+N) \geq n$, and it cannot be more than the dimension of the space, so $\dim(M+N) = n$. Finally, conclude with the dimension formula.

Title: Prove that dim(M∩N)=dim(N)−1

Body:
Let L be a vector space of dimension $ n $ and  $ M,N \subset $ L linear subspaces such that $ dim(M)=n-1 $ and $ N \not \subset M $.

Prove that $ dim( M \, \cap N ) = dim(N) -1 $

My attempt:

Let $ (v_1,...,v_{n-1}) $ be a basis for $ M $ , we'll expand the basis of $ M $ to basis of $ M + N $ by the vectors $ (v_n,...,v_t) $ so that the basis of $ M + N $ is $ (v_1,...,v_t) $.
Since $ M + N \subseteq V $ then $ dim(M + N) \leq dim(V) $ , so that
$ n-1 \leq dim(M+N) = t \leq n $.
[ I dont know how to continue from here ]

Basically I want to use the dimension theorem for subspaces $ dim(M+N) = dim(M) + dim(N) - dim(M \cap N)$ But I got stuck and don't know how to continue, can you please help?

Edit:

As shown in the answers below, after showing that $ dim( M + N ) = n $ then, using the dimension theorem,
$ dim(M+N) = dim(M) + dim(N) - dim(M \cap N) \iff $ $ dim(M \cap N) = dim(M) + dim(N) - dim(M+N) \iff $ $ dim(M \cap N) = n-1 + dim(N) - n \iff dim(M \cap N) = dim(N) - 1  $ . And we've finished the proof.


问:
At least 20 tries to get 3 lefties
A problem is as follows: 15% of the population are left-handed, if people are stopped randomly on the street, what is the probability that it takes at least 20 tries to get 3 lefties?

The solution that the author has proposed is:

$P(2 \ lefties \ in \ 19 \ tries) + P(1 \ lefty \ in \ 19 \ tries) + P(no \ lefty \ in \ 19 \ tries)$

The solution that I think would be correct is:

$P(2 \ lefties \ in \ 19 \ tries \cup 1 \ lefty \ in \ 18 \ tries \cup no \ lefty \ in \ 17 \ tries) = $

$P(2 \ lefties \ in \ 19 \ tries) + P(1 \ lefty \ in \ 18 \ tries) + P(no \ lefty \ in \ 17 \ tries) - P(1 \ lefty \ in \ 18 \ tries \ and \ 19th \ is \ a \ lefty) - P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ is \ a \ lefty) - P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ and \ 19th \ are \ lefties) + P(no \ lefty \ in \ 17 \ tries \ and \ 18th \ and \ 19th \ are \ lefties)$

If the author is right, why? For exaxmple, how could 1 lefty in 19 tries mean "at least 20 tries to get 3 lefties"? Let's say the 19th one is a lefty, then to get 3 lefties, the most compact way would be "19th, 20th and 21st are lefties", which is not "at least 20 tries", it is "at least 21 tries". Also there seems to be intersections in that solution.

(UPDATE: The book is "The Probability Tutoring Book by Carol Ash" in case it matters.)

答:
At least 20 means in particular at least 21. The question is not asking the probability that it takes exactly 20 tries.

Think of it is way: saying that it takes at least 20 tries is exactly like saying that out of the first 19 tries, there are strictly less than 3 lefties. This leaves the disjoint possibilities 0, 1, or 2 lefties out of 20. Do you know how to compute the final probability?

EDIT. Your solution is technically correct, but the issue is that writing the event $E =$ "it takes at least 20 tries" as the union of $A =$ "2 in 19", $B =$ "1 in 18", and $C=$ "0 in 17" makes it hard to compute. This is because these three events are not disjoint, and thus as you say, there are intersections to take into account. However, it is also true that $E$ is the union of the events $A$, $B' =$ "1 in 19", and $C'=$ "0 in 19". Now, these events are disjoint, so
$$
P(E) = P(A \cup B' \cup C') = P(A) + P(B') + P(C').
$$
The added advantage is that the probability of $A, B', C'$ is easy to compute (how?).

Title: At least 20 tries to get 3 lefties


问:
What is the idea behind sheaves of rings on distinguished open sets
In the book on algebraic geometry of Mumford (which can be found here), he said :

We want to enlarge the ring $R$ into a whole sheaves of rings on $SpecR$, written $\mathcal{O}_{SpecR}=\mathcal{O}_{X}$.

So he need to define $\mathcal{O}_{X}(U)$ for $U$ is an open set in $X$, in particular he need to define $\mathcal{O}_{X}(X_f)$ for some distinguished open subset $X_f$. And he said

We want to define : $\mathcal{O}_{X}(X_f)=R_f$=localization of the ring $R$ w.r.t multiplicative system $\lbrace 1,f,f^2,..\rbrace$

My question is : Why do we want to define the sheaf $\mathcal{O}_{X}(X_f)$ to be $R_f$ ?

Back to case of affine variety $V$, $V\setminus V(f)$ is a quasi-affine variety with the ring of regular function $k[V]_f$. So can we think about the sections of $\mathcal{O}_{X}(X_f)$ as the regular function on some spaces ?

答:
You can think of $X_f$ as the the set of those points where $f$ does not vanish (this is true verbatim in the case of classical varieties, but also for schemes when you use the structure sheaf and think of $f$ as a function valued in the residue fields). Thus $f$ should be invertible on $X_f$, and therefore $r/f^k$ should be a regular function on $X_f$, for every regular function $r$ on $X$. There is no reason to impose more regular functions. So we just define $\mathcal{O}(X_f) := R_f$.

In the end, it is just a definition, and when you work with it you will see that it is the right one. You have already mentioned that this fits perfectly to the case of classical varieties.