问:
Pointwise convergence of sequence of functions on $[0,1]$
I'm thinking about a sequence of functions on closed interval $[0,1]$, where $f_n(x)=(-1)^nn$ for$x \in  (0,1/n]$, and $0$ elsewhere.

It's clear that $f_n$ is not uniformly convergent since the supremum doesn't go to zero. But does it converge pointwise? Intuitively, I think it does converge to $f(x)=0$, because as $n$ goes to infinity, the interval should vanish. But I have no idea how to prove it formally.

Could anyone tell me whether I'm correct, and how to prove this pointwise convergence?

答:
It does converge pointwise to $0$, for the reason you said. Basically, $f_n(x) = 0$ for sufficiently large $n$ (where "sufficiently large" depends on $x$).

To prove it formally, suppose $x \in [0, 1]$. If $x = 0$, then by definition, $f_n(0) = 0$ for all $n$, so $f_n(0) \to 0$ as $n \to \infty$.

Otherwise $x > 0$. We can then use the fact that $\frac{1}{n} \to 0$ from above as $n \to \infty$. We can find some $N$ such that
$$n \ge N \implies 0 < \frac{1}{n} < x.$$
For such $n$, we have $x \notin (0, \frac{1}{n}]$, and hence
$$n \ge N \implies f_n(x) = 0.$$
For any $\varepsilon > 0$, we can set this same $N$ to see that
$$n \ge N \implies |f_n(x) - 0| < \varepsilon,$$
i.e. $f_n(x) \to 0$ as $n \to \infty$.


问:
Where did we use local compactness and how is Urysohn's lemma used?
Notation $Ba_X$ is the sigma-algebra of Baire sets, i.e. the smallest sigma-algebra generated by the compact $G_{\delta}$ sets of $X$.

lemma : Assume that $X$ is a locally compact Hausdorff space. Then $Ba_X$ is the smallest $σ$-algebra relative to which all functions in $C_c(X)$ are measurable.

proof : We verify first that each $f∈C_c(X)$ is measurable[$Ba_X$], and we can restrict ourselves to the collection of nonnegative functions since it generates C_c(X) as a vector space. For the other direction Assume then that $f∈C_c(X)$ is nonnegative and observe that $(f≥α)=⋂_{n=1}^{\infty}(f>α−1n)$ is a compact $G_δ$ set for $α>0$, while $(f≥α)=X$ for $α<0$. Thus $f$ is measurable [BaX].

Assume that all the functions in $C_c(X)$ are measurable [S] for some σ-algebra S. We need to show that every compact $G_δ$ set belongs to S. Indeed, assume that $K=\cap_{n=1}^{\infty}G_n$ is compact,where $G_n$ is open for every $n∈N$, and use Urysohn’s lemma to produce functions $f_n∈C_c(X)$ such that $χ_K≤f_n≤χ_{G_n}$.The result follows from the equality $K=\cap_n(f_n>0)$

But in this proof I don't really understand two things:

where did we use that $X$ is locally compact?

How did we get $(f_n>0)=G_n$ since the wiki for Urysohn's lemma seems to say in general we cannot require $f(x)\neq 0$ outside the closed sets we separate?

答:
(1) We used local compactness to get that $X$ is normal, so that Urysohn's Lemma applies.

(2) The proof didn't say that $(f_n>0)=G_n$. But the definition of $f_n$ immediately gives that $K\subseteq (f_n>0)\subseteq G_n$. Taking te intersection over all $n$, we get $K\subseteq\bigcap_n(f_n>0)\subseteq K$.

ContentLicense: CC BY-SA 4.0


问:
Differential Calculus: How to find the slope of the curve $y=x(x+1)(x+2)$ at the points where it crosses the $x$-axis
Find the slope of the curve $y=x(x+1)(x+2)$ at the points where it crosses the $x$-axis.

答:
$y = x(x+1)(x+2)$

First: where does it cross the $x$-axis? When $y=0.$

$0=x(x+1)(x+2) \implies x=\{0,-1,-2\}$

Now the slope at each point of the curve is given by: $\dfrac{dy}{dy} = 3x^2+6x+2$

So slopes are: $\{2,-1,2\}$ respectively.


问:
If $a,b,c,d,e$ be real number other than $-2≤a≤b≤c≤d≤e≤2$, prove the following inequation $\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}≥4$
Question
If $a,b,c,d,e$ be real numbers $-2≤a≤b≤c≤d≤e≤2$, prove the following inequation
$$\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}≥4$$

I tried to use Cauchy-Schwarz like
$$\left(\frac{1}{b-a}+\frac{1}{c-b}+\frac{1}{d-c}+\frac{1}{e-d}\right)(b-a+c-b+d-c+e-d)≥16$$
I think I should prove it using $-2≤a≤b≤c≤d≤e≤2$, but I'm confused about how to do it. Can anyone help me?

答:
$w=b-a, x=c-b,y=d-c, z=e-a$

you have four non-negative numbers $w,x,y,z$ that sum to $e-a\leq 4,$ so
$w+x+y+z \leq 4$   and $\frac{1}{w+x+y+z} \geq \frac{1}{4}$

The harmonic mean of thew is smaller then the arithmetic mean, for
$$ \frac{16}{w+x+y+z}  \leq \frac{1}{w} + \frac{1}{x} + \frac{1}{y} +\frac{1}{z} $$

This book I have refers to an AM-HM inequality for positive real numbers, if there are $n$ numbers

(with equality only when all the numbers are equal)

$$ \left(x_1 + x_2 + \cdots + x_n \right)  \left( \frac{1}{x_1} + \frac{1}{x_2}+ \cdots +\frac{1}{x_n} \right) \geq n^2   $$


问:
Intermediate field extensions
Let $\zeta$ be a primitive 12th-root of unity. How many field extensions are there between $\mathbb{Q}(\zeta^3)$ and $\mathbb{Q}(\zeta)$?

Here is my solution:

$\mathbb{Q}(\zeta)$ is the splitting field of $X^{12}-1$, so $\mathbb{Q}(\zeta):\mathbb{Q}$ is a Galois extension. The minimal polynomial of $\zeta$ is the cyclotomic polynomial $\Phi_{12}$, whose degree is 4, therefore $G(\mathbb{Q}(\zeta):\mathbb{Q})=\mathbb{Z}_{12}^*$. Then,  every non trivial subgroup has order 2, so every intermediate extension of $\mathbb{Q}(\zeta):\mathbb{Q}$ is quadratic, which shows that there is no intermediate field extensions between $\mathbb{Q}(\zeta^3)$ et $\mathbb{Q}(\zeta)$.

I would like to know if my reasoning is right.

Thanks in advance.

答: Since $[Q(ζ):Q)]=|Z_{12}^*|=4$ and $[Q(ζ^3):Q)]=2$, then we get $[Q(ζ):Q(ζ^3))]=2$ and hence your reasons are right.

Title: Intermediate field extensions


问:
Counterexamples for $\alpha < \beta \Longrightarrow \alpha^{\gamma} < \beta^{\gamma}$, where $\alpha$, $\beta$ and $\gamma$ are nonzero ordinals
I am trying to find counterexamples for the following formula:

$\alpha < \beta \Longrightarrow \alpha^{\gamma} < \beta^{\gamma}$, where $\alpha$, $\beta$ and $\gamma$ are nonzero ordinals.

I think $\gamma$ may have to be a limit ordinal, for instance, maybe $\omega$ (another notation for $\mathbb{N}$). Can anyone provide hints?

答:
I came up with this proof thanks to hints provided. Consider the following two ordinals: $\boldsymbol{2}^{\omega}$ and $\boldsymbol{3}^{\omega}$. The two ordinals can be expanded as
\begin{equation*}
\boldsymbol{2}^{\omega} = \bigcup\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\},
\end{equation*}
and
\begin{equation*}
\boldsymbol{3}^{\omega} = \bigcup\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}.
\end{equation*}
It is obvious that $\omega$ is an upper bound for both $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$ and $\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}$.

Next assume that there exists some ordinal $\lambda$ such that $\lambda < \omega$ and that $\lambda$ is also an upper bound for $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$. Immediately we have $\lambda < \boldsymbol{2}^{\lambda} \in \left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$, contradicting that $\lambda$ is an upper bound of $\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}$. Thus, $\omega = \operatorname{lub}\left(\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\right)$. Similarly, $\omega = \operatorname{lub}\left(\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\right)$. Then we have
\begin{equation*}
\begin{aligned}
\boldsymbol{2}^{\omega} &= \bigcup\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\\
&= \operatorname{lub}\left(\left\{\boldsymbol{2}^{\theta}: \theta \in \omega\right\}\right)\\
&= \omega\\
&= \operatorname{lub}\left(\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\right)\\
&= \bigcup\left\{\boldsymbol{3}^{\theta}: \theta \in \omega\right\}\\
&= \boldsymbol{3}^{\omega}.
\end{aligned}
\end{equation*}


问:
Let $A, B,C$ be subsets of $\{1,2,...n\}$. What is $P((A\cap B)\subseteq C)$?
My reckoning is that there are totally $2^{n} \cdot 2^{n} \cdot2^{n}$ options for $3$ subsets and for $(A\cap B)\subseteq C$ ,  for each of $\sum_{k=0}^{n}\binom{n}{k}$ options for C there are $\sum_{i=0}^{k}\binom{k}{i}$ for $A$ and $\sum_{i=0}^{k}\binom{k}{i}$ for $B$ and so in total we have:

$$P((A\cap B)\subseteq C)=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 2^{k}\cdot 2^{k}}{2^{n}\cdot 2^{n}\cdot 2^{n}}=\frac{\sum_{k=0}^{n}\binom{n}{k}\cdot 4^{k}}{8^{n}}=\frac{5^{n}}{8^{n}}=(\frac{5}{8})^{n}$$

Is this reckoning right?

答:
Along the lines of your attempt, here's a corrected version . . .

Fix a positive integer $n$ and let $X=\{1,...,n\}$.

Suppose $A,B,C$ are subsets of $X$, chosen independently at random from a uniform distribution on the power set of $X$.

For each $k\in\{0,...,n\}$ there are ${\large{\binom{n}{k}}}$ choices for $C$ with $|C|=k$.

Fix some such choice of $k$ and $C$.

For each $j\in\{0,...,k\}$ there are are ${\large{\binom{k}{j}}}$ choices for $A\cap B$ with $(A\cap B)\subseteq C$ and $|A\cap B|=j$.

Fix some such choice of $j$ and $A\cap B$.

Each of the $n-j$ elements of $X\,{\large{\setminus}}(A\cap B)$ can be assigned to

$A$ but not $B$.$\\[4pt]$

$B$ but not $A$.$\\[4pt]$

neither of $A,B$.

Thus for the fixed choice of $j$ and $A\cap B$, there are $3^{n-j}$ choices for $A,B$.

Then $P\bigl((A\cap B)\subseteq C\bigl)=\dfrac{w}{8^n}$, where
\begin{align*}
w
&=
\sum_{k=0}^n
\binom{n}{k}
\sum_{j=0}^k
\binom{k}{j}3^{n-j}
\\[4pt]
&=
3^n\sum_{k=0}^n
\binom{n}{k}
\sum_{j=0}^k
\binom{k}{j}\left(\frac{1}{3}\right)^j
\\[4pt]
&=
3^n\sum_{k=0}^n
\binom{n}{k}\left(\frac{4}{3}\right)^k
\\[4pt]
&=
3^n\left(\frac{7}{3}\right)^n
\\[4pt]
&=
7^n
\end{align*}


问:
Is the infinite sum of reciprocals powers of 2 a rational number?
I am trying to prove that the sum $$\sum_{n \in J}^{\infty} 2^{-n}$$ is a rational number, where $J$ is some infinite subset of $\mathbb{N}$, but not equal $\mathbb{N}$. That is, I want to show that $1 + \frac{1}{4} + \frac{1}{32} + \cdots$ is a rational number. I tried to prove by induction that every partial sum is a rational number, which seem'd okay, but I know I can't generalize to say that if all partial sums are rational, then the infinity sum is rational. So I'm stuck... and would appreciate any help!

答:
$\sum\limits_{n \in J} 2^{-n}$ is not always rational.  For example, let $J=\{0!,1!,2!,3!,4!...\}$.

$\dfrac1{2^{0!}}+\dfrac1{2^{1!}}+\dfrac1{2^{2!}}+\dfrac1{2^{3!}}+\dfrac1{2^{4!}}+...$ is a Liouville number,

a non-repeating decimal, and thus irrational.


问:
Solving the equation $10^{-x} = 5^{2x}$ with logarithms
$$10^{-x} = 5^{2x}$$

I'm having trouble isolating $x$. I get both logs on one side and then I'm stuck because I have nothing to divide with on the other side, and I can't factor it.

Thanks

答:
If $-x\log 10 = 2x\log 5$ then $(2x\log5) + (x\log10)=0$, so $x\cdot(2\log5 -\log10)=0$.

So it reduces to $x$ times some constant equals $0$.  And then the only thing you need to know about that constant is that it's not $0$.

ContentLicense: CC BY-SA 3.0


问:
Expected discount shop owner will have to shell out in a guessing game with customers
This question is a bit of a corollary to this one: Generate a random permutation of the first $n$ numbers on the fly

I have some customers coming to my shop. I'll accept the first $n$ and then close the shop. I'll assign each customer an ID between $1$ and $n$. The customers know they're going to get an integer ID, but don't know what $n$ is. Before they're assigned their ID, they will be given a chance to guess it. If they get it right, they'll be given a $1\$$ discount on any purchase.

The customers are very smart and will maximize their chances of getting the discount. They also talk to each other and so know the previous ID's. I want to minimize the discount I end up shelling out.

It seems intuitive that the best strategy for assigning ID's is to do it in a random order. So, take the numbers $1$ to $n$ and put them in an array. Randomly permute that array using the Fisher Yates shuffle and assign the IDs based on successive indices of the array. If the customers knew what $n$ was, the first customer would guess anything from $1$ to $n$, the second one would rule out the ID of the first customer and so on, making the total discount I'd end up giving becoming on average:

$$d = \sum\limits_{i=1}^n\frac{1}{i}$$

But, what will $d$ become when they don't know $n$?

答:
Up to a term that goes to $0$ as $n \to \infty$, the expected discount is still the same. Essentially, the customers most likely to get the discount can have a pretty good idea of $n$, so they'll still get the discount with about the same probability.

In particular, suppose that the customers use the following strategy: each one guesses a random unassigned ID between $1$ and the highest ID assigned so far. (If all IDs in that range have been assigned, the customer just gives up in frustration.)

I'm going to ignore the first $\sqrt n$ customers. For each of the other customers:

With probability $\frac1k \le \frac1{\sqrt n}$, the $k^{\text{th}}$ customer will have the highest ID assigned so far, which is unguessable by this strategy.

Otherwise, the $k^{\text{th}}$ customer will have at least a $\frac1{n-k+1}$ chance of guessing their ID (possibly better).

So these customers get an expected discount of at least
$$
   d' = \left(1 - \frac{1}{\sqrt n}\right) \sum_{i=1}^{n - \sqrt n} \frac1i.
$$
Let's compare it to the original $d = \sum_{i=1}^n \frac1i \approx \ln n + \gamma$. This $d'$ loses out on the entirety of $\sum_{i=n-\sqrt n}^{n} \frac1i$, but that's asymptotically only $\ln n - \ln (n - \sqrt n) = O(\frac1{\sqrt n})$ as $n \to \infty$. For $i \le \frac n2$, $d'$ only gets $(1 - \frac1{\sqrt n})$ of the expected discount, which loses us $O(\frac{\ln n}{\sqrt n})$.

So $d' = d - O(\frac{\ln n}{\sqrt n})$, even with this suboptimal strategy.


问:
Struggling with the meaning of discrete.
Even though the set of all integers is infinite, is it still discrete?  Also, is a finite set of decimals, such as the following set of $3$ decimals $\{ .1, .2, .3\}$ discrete because it's members are finite, separated and countable?

答: Any set with the property that for every element of the set there is an open set containing a singleton element of the given set is discrete. For the set of integers, the open set containing $\left(a-\frac12,a+\frac12\right)$ contains only one integer ($a$), so it is discrete. Any finite set is discrete (take the length of interval to be minimum of the distances between the elements). Set of all rational numbers are not discrete under the usual topology (because they are dense in $\mathbb R$).


问:
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$?
How to evaluate

$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$

The integral arises in the computation of

$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as

$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$

A similar Integral

$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$

Where $ C $ = Catalan Constant

Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and  using Taylor Series as well as Mathematica.

Q = Is there a closed form for Integral $(1)$?

EDIT

$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$

Where $W$ is some value.

答:
$$I=\int_0^1 \frac{x^2\ln(1-x^4)}{1+x^4}dx=\frac{5\pi^2}{48\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(\delta_S)}{2\sqrt 2}+\frac{\ln^2(\delta_S)}{2\sqrt 2}-\frac{\ln 2 \ln(\delta_S)}{2\sqrt 2 }$$
$$+\frac{3\operatorname{Li}_2\left(-(\delta_S)^2\right)}{8\sqrt 2}-\frac{3\operatorname{Li}_2\left(-1/(\delta_S)^2\right)}{8\sqrt 2}+\frac{\operatorname{Li}_2\left(1/(\delta_S)^4\right)}{8\sqrt 2}+\frac{\operatorname{Ti}_2\left(-\delta_S\right)}{\sqrt 2}-\frac{\operatorname{Ti}_2\left(1/\delta_S\right)}{\sqrt 2}$$
Where $\delta_S=1+\sqrt 2\, $ is the silver ratio, $\operatorname{Li}_2(x)$ is the dilogarithm and $\operatorname{Ti}_{2}(x)$ is the inverse tangent integral.

To obtain this closed form we'll start by moving the bounds from $(0,1)$ to $(0,\infty)$. That's simply because afterwards the plan is to differentiate under the integral sign and usually having the bounds as $(0,\infty)$ simplifies the final result a lot when dealing with rational functions.
$$I=\frac12\int_{-1}^1\frac{x^2\ln(1-x^4)}{1+x^4}dx\overset{x\to\frac{1-x}{1+x}}=\frac12\int_0^\infty \frac{(1-x)^2}{1+6x^2+x^4}\ln\left(\frac{8x(1+x^2)}{(1+x)^4}\right)dx$$
$$=\frac{3\pi \ln 2}{4\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{2\sqrt 2}+\frac12X-2Y$$
$$X=\int_0^\infty \frac{(1-x)^2\ln\left(1+x^2\right)}{1+6x^2+x^4}dx;\quad Y=\int_0^\infty \frac{(1-x)^2\ln\left(1+x\right)}{1+6x^2+x^4}dx$$
Also one integral from above vanished via the substitution $x\to \frac{1}{x}$:
$$\int_0^\infty \frac{(1-x)^2\ln x}{1+6x^2+x^4}dx=0$$
As mentioned above, to evaluate $X$ and $Y$ we'll start by differentiating under the integral sign and even though the rational functions from above are somewhat more complicated than if we kept the original ones, we expect to have an easier time when integrating back.

$$X(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax^2\right)}{1+6x^2+x^4}dx\Rightarrow X'(a)=\int_0^\infty \frac{(1-x)^2x^2}{(1+6x^2+x^4)(1+ax^2)}dx$$
$$=\frac{2}{1-6a+a^2}\int_0^\infty\left(\frac{ax}{1+ax^2}-\frac{ax+x^3}{1+6x^2+x^4}\right)dx$$
$$+\frac{1-a}{1-6a+a^2}\int_0^\infty\frac{1}{1+ax^2}dx-\frac{1}{1-6a+a^2}\int_0^\infty\frac{1-a-(a-5)x^2}{1+6x^2+x^4}dx$$
$$=\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}+\frac{\pi}{2\sqrt a}\frac{1-a}{1-6a+a^2}+\frac{\pi}{2\sqrt 2}\frac{a-3}{1-6a+a^2}$$
$$\Rightarrow X=\int_0^1\left(\frac{\ln a}{1-6a+a^2}-\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{a-3}{1-6a+a^2}\right)da+\frac{\pi\ln 2}{2\sqrt 2}+\frac{\pi\ln (1+\sqrt 2)}{2\sqrt 2}$$

$$Y(a)=\int_0^\infty\frac{(1-x)^2\ln\left(1+ax\right)}{1+6x^2+x^4}dx\Rightarrow Y'(a)=\int_0^\infty \frac{(1-x)^2x}{(1+6x^2+x^4)(1+ax)}dx$$
$$=\int_0^\infty\left(\frac{(1+5a^2-2a^3)x+(1+a)^2x^3}{(1+6a^2+a^4)(1+6x^2+x^4)}-\frac{a(1+a)^2}{(1+6a^2+a^4)(1+ax)}\right)dx$$
$$+\int_0^\infty \left(\frac{a(1+a)^2-(2-5a-a^3)x^2}{(1+6a^2+a^4)(1+6x^2+x^4)}\right)dx$$
$$=-\frac{(1+a)^2\ln a}{1+6a^2+a^4}-\frac{a(3-a+a^2)+1}{1+6a^2+a^4}\frac{\ln(1+\sqrt 2)}{\sqrt 2}+\frac{\pi}{2\sqrt 2}\frac{a(3+a+a^2)-1}{1+6a^2+a^4}$$
$$\Rightarrow Y=-\int_0^1\frac{(1+a)^2\ln a}{1+6a^2+a^4}da-\frac{\pi\ln(1+\sqrt 2)}{8\sqrt 2}-\frac{3\ln 2\ln(1+\sqrt 2)}{4\sqrt 2}-\frac{\pi^2}{16\sqrt 2}+\frac{3\pi\ln 2}{8\sqrt 2}$$

Let's reorder things a little, just to see what we have so far (combining with the $X$ and $Y$ integrals):
$$I=\frac{\pi^2}{8\sqrt 2}+\frac{\pi\ln 2}{4\sqrt 2}+\frac{\pi\ln(1+\sqrt 2)}{2\sqrt 2}+2\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx$$
$$+\frac12\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$

So all that's left is to evaluate the remaining two integrals.
For the first one we have:

$$\int_0^1\frac{(1+x)^2\ln x}{1+6x^2+x^4}dx=\int_0^1\left(\frac{1}{2\sqrt 2}\frac{(1+\sqrt 2)-x}{(1+\sqrt 2)^2+x^2}-\frac{1}{2\sqrt 2}\frac{(1-\sqrt 2)-x}{(1-\sqrt 2)^2+x^2}\right)\ln x \, dx$$
$$\small =\frac{1}{8\sqrt 2}\left(\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)-\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)\right)+\frac{1}{2\sqrt 2}\left(\operatorname{Ti}_2\left(-(1+\sqrt 2)\right)-\operatorname{Ti}_2\left(-(1-\sqrt 2)\right)\right)$$
The result from above follows using:
$$\int_0^1 \frac{x\ln x}{a^2+x^2}dx\overset{x^2\to x}=\frac14\int_0^1 \frac{\ln x}{a^2+x}dx=\frac14\operatorname{Li}_2\left(-\frac{1}{a^2}\right)$$
$$\int_0^1 \frac{a\ln x}{a^2+x^2}dx\overset{x\to ax}=\int_0^\frac1a\frac{\ln a +\ln x}{1+x^2}dx\overset{IBP}=-\operatorname{Ti}_2\left(\frac1a\right)$$

The second one is a little troublesome since we must keep the terms togheter to avoid divergence issues. So we'll do a couple substitutions first:
$$\int_0^1\left(\frac{\ln x}{1-6x+x^2}+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{3-x}{1-6x+x^2}\right)dx$$
$$\overset{x\to\frac{1-x}{1+x}}=\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\left(\frac{2x}{2x^2-1}-\frac{1}{x+1}\right)\right)dx$$
$$=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}+\int_0^1\left(\frac12\frac{1}{2x^2-1}\ln\left(\frac{1-x}{1+x}\right)+\frac{\ln(1+\sqrt 2)}{\sqrt 2}\frac{2x}{2x^2-1}\right)dx$$
$$\overset{\sqrt 2 x\to x}=-\frac{\ln 2\ln(1+\sqrt 2)}{\sqrt 2}-\frac{1}{\sqrt 2}\int_0^\sqrt 2\left(\frac12\frac{1}{1-x^2}\ln\left(\frac{\sqrt 2 -x}{\sqrt 2 + x}\right)+\frac{x}{1-x^2}\ln(1+\sqrt 2)\right)$$
$$\overset{x\to\frac{1-x}{1+x}}=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\int_{-(1-\sqrt 2)^2}^1\ln\left(\frac{1+(1+\sqrt 2)^2x}{1+\frac{x}{(1+\sqrt 2)^2}}\right)\frac{dx}{x}$$
$$\scriptsize{=\frac{\ln^2(1+\sqrt 2)}{\sqrt 2}-\frac{\ln 2 \ln(1+\sqrt 2)}{\sqrt 2 }-\frac{1}{4\sqrt 2}\left(\frac{\pi^2}{6}-\operatorname{Li}_2\left(-(1+\sqrt 2)^2\right)+\operatorname{Li}_2\left(-(1-\sqrt 2)^2\right)-\operatorname{Li}_2\left((1-\sqrt 2)^4\right)\right)}$$
The last integral was split into two and the result follows by using the definition of the dilogarithm.

Finally putting everything togheter gives the announced closed form.


问:
Without using prime factorization, show if $m\mid n^2$ then $\gcd(m,n^2/m)\mid n$
It's easy to use prime factorization to show:

If $m\mid n^2$ then $\gcd(m,n^2/m)\mid n$.

Can anybody find some other proof - perhaps a simple reduction of some sort? Maybe solving $m^2x + n^2y=mn$, for example?

I suppose the result would follow if you could prove that $\gcd(m^2,n^2)=\gcd(m,n)^2$, since $\gcd(m,n)^2\mid mn$ since $\gcd(m,n)\mid m$ and $\gcd(m,n)\mid n$. That result doesn't even need $m\mid n^2$.

答: By here $\,\color{#0a0}{(m^2,n^2)} = (m,n)^2 = (m^2,\color{#c00}{mn},n^2)\Rightarrow  \color{#0a0}{(m^2,n^2)}\mid \color{#c00}{mn} \overset{\!\div\,m}\Rightarrow  (m,n^2/m)\mid n$


问:
Struggling to understand basics of complete residue system
I'm really struggling to understand the literal arithmetic being applied to find a complete residue system of modulo $n$. Below is the definition my textbook provides along with an example.

Let $k$ and $n$ be natural numbers. A set $\{a_1,a_2,...,a_k\}$ is called a canonical complete residue system modulo $n$ if every integer is congruent modulo $n$ to exactly one element of the set

I'm struggling to understand how to interpret this definition. Two integers, $a$ and $b$, are "congruent modulo $n$" if they have the same remainder when divided by $n$. So the set $\{a_1,a_2,...,a_k\}$ would be all integers that share a quotient with $b$ divided by $n$?

After I understand the definition, this is a simple example provided by my textbook

Find three residue systems modulo $4$: the canonical complete residue system, one containing negative numbers, and one containing no two consecutive numbers

My first point of confusion is "modulo $4$". $a{\space}mod{\space}n$ is the remainder of Euclidean division of $a$ by $n$. So what is meant by simply "modulo $4$"? What literal arithmetic do I perform to find a complete residue system using "modulo $4$"?

答:
Prelim: The definition is confusing because it is not assuming $k = n$.  You will be able to prove $k = n$ later but in mathematics we don't include anything in a definition that we can prove later.

The definition of a complete residue system is a collection of integers $\{a_j\}$ so that for any integer, that integer will be congruent (have the same remainder) with exactly one of those integers.

In Other words, and probably a much more straightforward definiton, For every possible remainder,  there will be be exactly one integer with that remainder.

For instance if $n = 7$, the easiest and most obvious complete residue system would be simply $\{0,1,2,3,4,5,6\}$. Every integer will be have remainder $0,1,2,3,4,5,6$ and those are precisely the numbers in there.

Another complete system could be $\{63, 8, 15, -4, 32, 75, 146\}$.  If an integer $b$ has remainder $0$ it is congruent to $63$.  $63$ represents all the integers with remainder $0$.  ANd if $b$ has remainder $8$ then $b$ is congruent to $8$. $8$ represents all the integers with remainder $1$.... And so on.

Every remainder is represented exactly once.

And that is what a complete residue system means.  A residue is a representation of one class of remainders (all the integers with remainder $4$ for example are represented by $32\equiv 4 \pmod 7$, for example). And a complete residue system means every residue is represented.

And as there $n$ possible remainders there will be $n$ elements in the system so if the system is $\{a_1, ...., a_k\}$ then $k = n$.  (If it were me, I wouldn't even bring up the idea this could be in doubt.  It just confuses things the first time you see the definition.)

.....

Okay. To do a completer residue system $\pmod 4$ you need to find a $\{a_1, a_2, ..... , a_k\}$ were for every integer $-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,....$ is congruent to exact one of them.

So we need and $a_1\equiv -6\pmod 4$.  Well $-6\equiv 2 \pmod 4$ so let's let $a_1 = 2$.  And we need something $\equiv -5 \pmod 4$.  We $-5\equiv 3 \pmod 4$ and $15 \equiv 3 \pmod 4$ and $15 \equiv -3 \pmod 4$ so lets use $a_2 = 15$.

And we need something $\equiv - 4 \pmod 4$.  Well $-4 \equiv 0 \equiv 28\pmod 4$ so let's use $a_3 = 28$.  And we need something $\equiv -3\pmod 4$ but $-3\equiv 1 \equiv 48321 \pmod 4$.  SO lets let $a_4 = 48321$.

And we need something $\equiv -2\pmod 4$.  But $-2 \equiv 2 = a_1$ so we already have it.  In fact it looks like we have one of each.

So $\{2, 15,28, 48321\}$ seems to be complete.

If $b$ is an integer we have either $b = 4k$ or $4k + 1$ or $4k + 2$ or $4k + 3$.

If $b = 4k$ then $b \equiv 28\pmod 4$.  ANd if $4k + 1$ then $b\equiv 48321$.  And if $b = 4k + 2$ then $b \equiv 2\pmod 4$ and if $b= 4k + 3$ then $b\equiv 15\pmod 4$.

So ....

the definition is:

$\{2, 15,28, 48321\}$  is called a canonical complete residue system modulo n if every integer is congruent modulo $4$ to exactly one element of the set

Well, that is true so  $\{2, 15,28, 48321\}$  is a complete residue system.

That's all.


问:
Isometry condition in components form
From the book of Nakahara "Geometry, Topology and Physics":

Let $(M,g)$ be a (pseudo-) Riemannian manifold. A diffeomorfism
  $f:\mathcal{M}\to \mathcal{M}$ is an isometry if it preserves the
  metric: $$ f^{*}g_{f(p)}=g_{p} $$

In components this condition becomes: $$ \frac{\partial
 y^{\alpha}}{\partial x^{\mu}}\frac{\partial y^{\beta}}{\partial
 x^{\nu}}g_{\alpha\beta}(f(p))=g_{\mu\nu}(p) $$ where $x$ and $y$ are
  the coordinates of $p$ and $f(p)$, respectively.

I don't understand how he got the component form of the isometry condition $f^{*}g_{f(p)}=g_{p}$.

It reminds me the transformation law of a tensor (in this case the metric $g$):
$$ \frac{\partial
 x'^{\alpha}}{\partial x^{\mu}}\frac{\partial x'^{\beta}}{\partial
 x^{\nu}}g'_{\alpha\beta}(x')=g_{\mu\nu}(x) $$
when we change coordinates of the point $p\in M$ from $x$ to $x'$. But here the coordinates $x$  and $x'$ are referring to the same point $p$ the manifold $M$. In the isometry condition we have two different points on $M$, $p$ with coordinates $x$ and $f(p)$ with coordinates $y$. How are these two related? Is a diffeomorfism ("active" transformation - "moving" points on the manifold) the same thing as a coordinate transformation of the same point ("passive" transformation).

答:
Recall that the pullback is given by
$$(f^\ast g)_{p}(u, v) = g_{f(p)}(df_p(u), df_p(v))$$
for $u, v \in T_p M$. So if we have local coordinates $(x^\alpha)$, the components of $f^\ast g$ at $p \in M$ are
\begin{align}
(f^\ast g)_{\alpha\beta}(p) & = f^\ast g_{p}\left( \frac{\partial}{\partial x^\alpha}, \frac{\partial}{\partial x^\beta}\right) \\
 & = g_{f(p)}\left( df_p\left(\frac{\partial}{\partial x^\alpha}\right), df_p\left(\frac{\partial}{\partial x^\beta}\right)\right) \\
 & = g_{f(p)}\left( \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial}{\partial x^\mu}, \frac{\partial f^\nu}{\partial x^\beta}\frac{\partial}{\partial x^\nu}\right) \\
 & = \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial f^\nu}{\partial x^\beta}g_{f(p)}\left( \frac{\partial}{\partial x^\mu}, \frac{\partial}{\partial x^\nu}\right) \\
 & = \frac{\partial f^\mu}{\partial x^\alpha}\frac{\partial f^\nu}{\partial x^\beta}g_{\mu\nu}(f(p)).
\end{align}

Note that we only use the coordinate system $(x^\alpha)$ at $p$ here. To get the coordinates at $f(p)$, we just pull back the coordinates at $p$, i.e. we use the coordinates $(x^\alpha \circ f^{-1})$ at $f(p)$.

Title: Isometry condition in components form


问:
Prove or disprove: If $A$ is regular and $B$ is not regular, then $A ∩ B$ is not regular.
I suspect the Myhill–Nerode theorem may come into play, but not certain. If this was a union instead of an intersect, I'm certain it's true. I'm relatively confident that this statement is false and thus, is to be disproven, but not certain how to formally prove it. I was thinking that, if my intuition is correct, a counterexample would be optimal. I know that the empty language is a regular subset of all non-regular languages (or any language), so I was thinking of doing a proof that is essentially a reordering of this post (Prove or disprove: If $A$ and $A ∩ B$ are regular, then $B$ is regular). Of course, I could be wrong and it's true. I just want to be fully confident - anything to help me better understand this would be greatly appreciated.

答:
The proof is the same as suggested in the answer of your post (Prove or disprove: If $A$ and $A \cap B$ are regular, then $B$ is regular). You can choose $A = \emptyset$ and $B$ to be any non-regular language, then $A \cap B = \emptyset$ is regular.

The reason why the answers are the same is because the two statements are logically equivalent, i.e. Let $p = A$ is regular, $q = B$ is regular, $r = A\cap B$ is regular. Your previous question is $(p \land r)\to q$, and your current question is $(p \land \neg q) \to \neg r$. We can show that $((p \land r)\to q) \equiv ((p \land \neg q) \to \neg r)$ by proof tree, or by natural deduction:

\begin{align} &\quad((p\land r)\to q) \\&\equiv (\neg (p \land r)\lor q) \\&\equiv (\neg p \lor \neg r \lor q)\\&\equiv (\neg p \lor q \lor \neg r) \\&\equiv (\neg(p\land \neg q)\lor \neg r)\\&\equiv((p\land \neg q)\to \neg r)\end{align}


问:
Find the spectrum and eigenvalues of an operator
Let $\{e_n\}$ be the orthonormal basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$.

Find $\sigma(A)$ and show that if $K \subset \mathbb{C}$, then there exists some $T \in B(\ell^2)$ such that $\sigma(T)=K$.

Find the eigenvalues of $A$,

Attempt:

Let $E$ be the closure of $\{ \alpha_n\}$.

From Spectrum of Diagonal Operator in $\ell^2$, I am guessing that $E= \sigma(A)$.

But I'm not sure if this is the case. I don't see why we have $(\alpha_n) \subset E$.

I know that for any $T \in B(\ell^2)$, $\sigma(T)$ is compact. But I'm not sure how to construct a $T \in B(\ell^2)$ such that $\sigma(T)=K$.

I know that the matrix of $A$ relative to the orthonormal basis $\{e_n \}$ is the diagonal matrix $diag(\alpha_n)$. So I think that the eigenvalues will be $\{\alpha_n \}$. But I'm not sure about this.

Any help is appreciated!

Thank you in advance.

答:
Hints:

Using only the definition, show that each $\alpha_n$ is an eigenvalue of $A$. Since $\sigma(A)$ is closed, $\overline{\{\alpha_n\}}\subseteq \sigma(A).$ Now show by direct calculation, that if $\lambda\notin \overline{\{\alpha_n\}}$, then $A-\lambda$ is invertible. Finally, if $K$ is given, choose a countable dense subset of $K$, which will be in $\ell^\infty$ since $K$ is compact (being a spectrum).


问:
Taylor expansion of a gaussian integral
I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}

So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.

My questions are:

How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)

Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?

答:
This is not a gaussian integral.

$$f(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} e^{-\frac{xy^4}{24}-\frac{y^2}{2}} \, dy=\frac{\sqrt{3}}{\sqrt{2\pi}}\frac{e^{\frac{3}{4 x}} }{\sqrt{x}}K_{\frac{1}{4}}\left(\frac{3}{4 x}\right)$$
Using the asymptotics of the Bessel function (since $x \to 0 \implies \frac 1 x \to \infty$), you should have
$$f(x)=1-\frac{x}{8}+\frac{35 x^2}{384}-\frac{385 x^3}{3072}+\frac{25025
   x^4}{98304}-\frac{1616615 x^5}{2359296}+O\left(x^6\right)  $$

If you make $x=\frac 1{10}$, this truncated series will give
$$f\left(\frac{1}{10}\right)=\frac{46634068277}{47185920000}=0.9883047$$ while the exact value is
$$f\left(\frac{1}{10}\right)=e^{15/2} \sqrt{\frac{15}{\pi }} K_{\frac{1}{4}}\left(\frac{15}{2}\right)=0.9883064$$

If you want more terms, let $x=\frac{3}{4 z}$ which makes
$$f(z)=\sqrt{\frac{2}{\pi }} e^z \sqrt{z} K_{\frac{1}{4}}(z)$$ with $z \to \infty$. Have a look at this paper (equation $(1.10)$) to have
$$f(z)=\Bigg[\sqrt{\frac{2}{\pi }} e^z \sqrt{z}\Bigg]\sqrt{\frac{\pi }{2z}} e^{-z} \sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}=\sum_{n=0}^\infty a_n\left({\frac 14}\right)\,z^{-n}$$
From equation $(1.9)$
$$ a_n\left({\frac 14}\right)=(-1)^n\frac{  \cos \left(\frac{\pi }{4}\right) \Gamma
   \left(n+\frac{1}{4}\right) \Gamma \left(n+\frac{3}{4}\right)}{\pi\, 2^n\, \Gamma (n+1)}$$

Make $z=\frac 3{4x}$ to recover the first expansion.

Edit

Without using the exact solution (as I did), you could have done the same work expanding first the integrand as a Taylor series around $x=0$
$$e^{-\frac{xy^4}{24}-\frac{y^2}{2}}=\sum_{n=0}^\infty\frac {(-1)^n}{24^n\,n!} e^{-\frac{y^2}{2}} y^{4 n} x^n$$ and use that
$$\int_{-\infty}^\infty e^{-\frac{y^2}{2}} y^{4 n}\,dy=2^{2 n+\frac{1}{2}} \Gamma \left(2 n+\frac{1}{2}\right)$$ This makes
$$f(x)= \frac{1}{\sqrt{\pi}}\sum_{n=0}^\infty\frac {(-1)^n}{6^n\,n!}  \Gamma \left(2 n+\frac{1}{2}\right) x^n$$ ad the same result.

Title: Taylor expansion of a gaussian integral


问:
Why do we assume principal root for the notation $\sqrt{}$

I'm wondering why when $n$ is even we always assume the positive root for $\sqrt[n]{}$.

For example, if we have $x = \sqrt{4}$, we always assume $x = 2$.  But if we have $x^2 = 4$, we do $x = \pm\sqrt{4} \Longrightarrow x = -2, 2$.

The problem is that if I take the 1st equation, and square both sides, I get
$$
x = \sqrt{4} \Longrightarrow x^2 = 4 \Longrightarrow x = \pm\sqrt{4} \Longrightarrow x = -2, 2
$$
My teacher says that if you introduce the radical sign, use $\pm$, but those two equations are the same under the rules about radicals she taught us, so in reality, (I think) "taking the positive root" rule has no meaning.

I get that people want to make the radical mean something without ambiguity, but making arbitrary rules (at least I thing they're arbitrary), like taking the positive answer, seems to confuse things and leads to inconsistency, like what is shown above.

There are other examples, such as solving for $x$, then substituting the original equation with the value of $x$, and not getting it to work because you can only take the positive root.

For things like the Pythagorean theorem, people like to say it's obvious to take the positive one only, but there are ways to represent the Pythagorean theorem without relying on human judgement to decipher the final answer.  Say you have a right triangle with legs $a=3$, $b=4$, and you want to find c.  So you do: $3^2 + 4^2 = c^2$, $c > 0$. and solve the systems of equations, just like any other: $c = \pm\sqrt{25} \Longrightarrow c = -5, 5$.  $c = -5, 5$ intersects $c > 0$ at $c = 5$, so the answer is $c = 5$.  I don't see why you would need to redefine square root: $\pm\sqrt{}$ to mean principal square root: $\sqrt{}$ to find the correct answer.

Does it have something to do with imaginary numbers, or am I missing something?  Thanks for any help.

Edit:  I am not asking if $\sqrt{}$ means positive, I'm asking why.

答:
Mathematicians use the principal square root more often.

We would rather use $ \pm \sqrt x $ in a few cases then have to write $ |\sqrt x|$ all the time.

That's pretty much the only reason why.

$ \sqrt \cdot $ is usually how we notate a principal square root. It is not an inverse operation.

$ \pm \sqrt \cdot $ is usually how we notate a (true) square root. Is is the inverse operation of $ \cdot^2 $.


问:
Prove every irreducible polynomial over a field of characteristic of $0$ is separable.
I want to prove that every irreducible polynomial over a field of characteristic of $0$ is separable.

I found the proof was actually not too hard.
Suppose $f(x) \in F[x]$ is irreducible of degree $n$, then the only constant factors of $f(x)$ are $1$ and $f(x)$ up to constant, then $Df(x)$ and $f(x)$ shares no common factors, hence they can't have any common zeros. Therefore, $f(x)$ must have distinct zeros.

But my questions is, it seems like I did not use the assumption that the characteristic of $F$ is $0$ in my proof or maybe I used this assumption implicitly but I did not notice.

Can anyone tells me where did I used the assumption that $F$ has to have a characteristic $0$?

答: If the characteristic is not zero, $Df$ can be zero ( consider  $Df$ where $f=X^p-a$, $p$ is the characteristic), and every element is a root of $Df$.


问:
Proving Wu's Formula for Steenrod Squares (Milnor & Stasheff, Problem 8-A)
I am stuck at the following problem from chapter 8 of Characteristic Classes by Milnor & Stasheff.

Problem 8-A. It follows from 7.1 that the cohomology classes $\operatorname{Sq}^kw_m(\xi)$ can be expressed as a polynomial in $w_1(\xi), \dots, w_{m+k}(\xi)$. Prove Wu's explicit formula

$$\operatorname{Sq}^k(w_m) = w_kw_m + \binom{k-m}{1}w_{k-1}w_{m+1} + \cdots + \binom{k-m}{k}w_0w_{m+k},$$

where $\binom{x}{i} = x(x-1)\cdots(x-i+1)/i!$, as follows. If the formula is true for $\xi$, show that it is true for $\chi\times\gamma^1$. Thus by induction it is true for $\gamma^1\times\cdots\gamma^1$, and hence for all $\xi$.

Using the hint given, I have started to work out the following.

First, we prove the result for the line bundle $\gamma^1$ over $P^\infty$ as follows. Let $w_m \in H^m(P^\infty) \implies w_m=a^m$ where $a\in H^1(P^\infty)$. Claim:

$$\operatorname{Sq}^k(w_m) = \sum_{i = 0}^{k} {k-m \choose i} a^{k-i}a^{m+i}=\sum_{i = 0}^{k} {k-m \choose i}a^{k+m}$$ but I cannot proceed further. Using the total Steenrod square I was able to prove that $\operatorname{Sq}^k(w_m) =  {m \choose k} a^{k+m}$ but in vain.

Now suppose I have proved the above claim and it's true for the bundle $\xi$ then if we show that the result is true for $\xi \times\gamma^1$ then by induction argument we can say that it will be true for $({\gamma^1})^n$. Why is this enough to give the result for any n-plane bundle, I mean how exactly are we using the universality of $\gamma^n$.

Proof of second claim

\begin{align*}
&\ \operatorname{Sq}^k(w_m(\xi\times\gamma^1))\\
=&\ \operatorname{Sq}^k(w_m(\xi)\times 1+ w_{m-1}(\xi)\times w_1(\gamma^1))\\
=&\ \operatorname{Sq}^k(w_m(\xi)\times 1) + \operatorname{Sq}^k (w_{m-1}(\xi)\times w_1(\gamma^1))\\
=&\ \sum_{i +j=k}\operatorname{Sq}^i(w_m(\xi))\times \operatorname{Sq}^j(1)+ \operatorname{Sq}^i(w_{m-1}(\xi))\times \operatorname{Sq}^j(w_1(\gamma^1))\\
=&\ \operatorname{Sq}^k(w_m(\xi))\times 1 + \operatorname{Sq}^{k-1}(w_{m-1}(\xi))\times (w_1(\gamma^1))^2 + \operatorname{Sq}^{k}(w_{m-1}(\xi))\times w_1(\gamma^1)\\
=&\ \sum_{i = 0}^{k} {k-m \choose i} w_{k-i}(\xi)w_{m+i}(\xi)\times 1+\sum_{i = 0}^{k-1} {k-m \choose i} w_{k-i-1}(\xi)w_{m+i-1}(\xi)\times (w_1(\gamma^1))^2+\sum_{i = 0}^{k} {k-m+1 \choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).
\end{align*}

That's where I'm stuck as I'm not sure how to manipulate the third summand coefficients to get the desired thing as I need to have all the coefficients to be ${k-m \choose i}$. How can I get it?

Thanks & regards.

Edit 1: I think I figured out the use of Universality-
Since $(\gamma^1)^n$ is an n-plane bundle over ${(P^\infty)}^n$ so we have a map $f: {(P^\infty)}^n\to G_n(R^\infty) $covered by a bundle map $\gamma^1\to\gamma^n$ inducing an algebra monomorphism on the cohomology. Hence if the result is true for $(\gamma^1)^n$ it will be for $\gamma^n$ as we have that $$f^*(\operatorname{Sq}^k(w_m(\gamma^n)) = \operatorname{Sq}^k(w_m(f^*\gamma^n)) = \operatorname{Sq}^k(w_m(\gamma^1)).$$ Therefore we can again invoke the universality to say that the formula is true for any $n$-plane bundle. Any suggestions are welcome.

Edit 2: Here's how I manipulated the coefficients to complete the proof.
We can split the third summand of R.H.S using Pascal's identity to get

$$\sum_{i = 0}^{k} {k-m \choose i-1} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1)+\sum_{i = 0}^{k} {k-m\choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).$$

Now we can just replace $i-1$ by $i$ (as for $i=0$ in the former case it's just zero) in the first summand here to get

$$\sum_{i = 0}^{k} {k-m \choose i} w_{k-i-1}(\xi)w_{m+i}(\xi)\times w_1(\gamma^1)+\sum_{i = 0}^{k} {k-m\choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).$$

Now substituting this above we get

\begin{align*}
&\ \operatorname{Sq}^k(w_m(\xi\times\gamma^1))\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}[ w_{k-i}(\xi)w_{m+i}(\xi)\times 1 + w_{k-i-1}(\xi)w_{m+i-1}(\xi)\times (w_1(\gamma^1))^2\\
&\qquad\qquad\qquad + w_{k-i-1}(\xi)w_{m+i}(\xi)\times w_1(\gamma^1)+ w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1)]\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}(w_{k-i}(\xi)\times1+w_{k-i-1}(\xi)\times w_1(\gamma^1))\cup(w_{m+i}(\xi)\times1+w_{m+i-1}(\xi)\times w_1(\gamma^1))\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}w_{k-i}(\xi\times\gamma^1)\cup w_{m+i}(\xi\times\gamma^1)
\end{align*}

concluding the proof.

Kindly help me with the first doubt.

答:
The two expressions for $\operatorname{Sq}^k(w_m(\gamma^1))$ are equal.

First note that

\begin{align*}
\binom{k - m}{i} &= \frac{(k-m)(k-m-1)\cdots(k - m - i + 1)}{i!}\\
&= (-1)^k\frac{(m-k)(m-k+1)\cdots(m - k + i -1)}{i!}\\
&= (-1)^k\binom{m - k + i - 1}{i}.
\end{align*}

In particular, $\displaystyle\binom{k-m}{i} = \binom{m-k+i-1}{i}$ modulo $2$.

Now note that by repeated application of Pascal's rule we have

\begin{align*}
\binom{m}{k} =& \binom{m-1}{k} + \binom{m-1}{k-1}\\
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \binom{m-2}{k-2}\\
 \vdots & \\
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \dots + \binom{m-k}{1} + \binom{m-k}{0}\\
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \dots + \binom{m-k}{1} + \binom{m-k-1}{0}\\
=& \sum_{i = 0}^k\binom{m-i-1}{k-i}\\
=& \sum_{i = 0}^k\binom{m-k+i-1}{i}.
\end{align*}

Therefore

$$\sum_{i=0}^k\binom{k-m}{i}a^{k+m} = \sum_{i=0}^k\binom{m-k+i-1}{i}a^{k+m} = \binom{m}{k}a^{k+m}.$$

Due to the modulo $2$ equality of binomial coefficients mentioned above, Wu's formula is more commonly written as

$$\operatorname{Sq}^k(w_m) = \sum_{i=0}^k\binom{m-k+i-1}{i}w_{k-i}w_{m+i}.$$


问:
What is the domain in the First Order Logic expression?
I think I'm misunderstanding how to apply the domain for these expressions.

The question is:

Let $P(x), Q(x), R(x),$ and $S(x)$ be the statements “$x$ is a duck,” “$x$ is one of my poultry,” “$x$ is an officer,” and “$x$ is willing to waltz,” respectively. Express each of these statements using quantifiers; logical connectives; and $P(x), Q(x), R(x),$ and $S(x)$.

One of my answers.

No ducks are willing to waltz.

$∃x(S(x) \to ¬P(x))$

The answer that pops up when I google.

$∀x(S(x) \to ¬P(x))$

The way that I interpret this, the domain is everyone and my answer reads "If somebody is a duck, then they are not willing to waltz".

I interpret the seemingly correct answer as, "Everyone is a duck, so they are not willing to waltz".

答:
Your issue isn't with domains, it's with how to parse the logical expressions syntactically.

Look at the parentheses, and be careful to translate quantifiers very literally. An expression of the form "$\exists x(stuff)$" means "There is some $x$ such that $stuff$." Similarly, an expression of the form "$\forall x(stuff)$" means "For every $x$, $stuff$." So your proposed answer reads, in somewhat-natural language, as follows:

There is some $x$ such that if $x$ is willing to waltz then $x$ is not a duck,

or a bit more cleanly, as:

There is something which, if it is willing to waltz, is not a duck.

This is very obviously not the same as "No ducks are willing to waltz" - it merely indicates that there is at least one thing which is not a willing-to-waltz duck.

By contrast, the correct answer $\forall x(P(x)\rightarrow\neg S(x))$ translates to

For every $x$, if $x$ is a duck then $x$ isn't willing to waltz

which is indeed the same as

No duck is willing to waltz

if a bit more indirect sounding.

(As an aside, your incorrect reading of the correct answer as "Everyone is a duck, so they are not willing to waltz" also suggests another issue, namely that you're trying to read $\rightarrow$ as indicating causality somehow - per the word "so" in your reading.)


答: The total discount is just what you said given that the customers know $n$.  Each customer should guess the smallest number not yet handed out.  If $i$ numbers have been handed out and the rest are randomly ordered, the chance this number wins is $\frac 1{n-i}$.  The discount you award is then $\sum_{i=1}^n \frac 1{n-i}=\sum_{i=1}^n\frac 1i=d$ because you are adding the same terms in the reverse order.  The point is that each customer knows that his/her number is still in play.  If the number is less than the highest given out, it is in play.  If you have given out $1$ through $k$ and no more, $k+1$ must be in play because you have not closed up.  The customer does not know the chance they will win because they don't know $n$, but they know this number is as good as they can do.

Title: Expected discount shop owner will have to shell out in a guessing game with customers

Body:
This question is a bit of a corollary to this one: Generate a random permutation of the first $n$ numbers on the fly

I have some customers coming to my shop. I'll accept the first $n$ and then close the shop. I'll assign each customer an ID between $1$ and $n$. The customers know they're going to get an integer ID, but don't know what $n$ is. Before they're assigned their ID, they will be given a chance to guess it. If they get it right, they'll be given a $1\$$ discount on any purchase.

The customers are very smart and will maximize their chances of getting the discount. They also talk to each other and so know the previous ID's. I want to minimize the discount I end up shelling out.

It seems intuitive that the best strategy for assigning ID's is to do it in a random order. So, take the numbers $1$ to $n$ and put them in an array. Randomly permute that array using the Fisher Yates shuffle and assign the IDs based on successive indices of the array. If the customers knew what $n$ was, the first customer would guess anything from $1$ to $n$, the second one would rule out the ID of the first customer and so on, making the total discount I'd end up giving becoming on average:

$$d = \sum\limits_{i=1}^n\frac{1}{i}$$

But, what will $d$ become when they don't know $n$?


问:
A doubt about what is straightedge and compass construction
The doubt comes from a try. To bisect a segment, I wonder whether I can first choose an arbitrary degree of opening for the compass, then use the compass to measure the segment until I can measure it with two times of use of the compass. It seems that we get the bisection now. Is this foolish method a rigorous straightedge and compass construction? And why?

I tend to believe it is wrong. Because once you allow such 'try method', I think then those impossibility in straightedge and compass construction will become possible, that is horrible. But I actually don't know why it's wrong.

答:
The issue here is twofold: a compass and straightedge construction must occur in finitely many steps and it must constitute an actual geometric proof.

You haven't shown that your procedure of "try a radius then try again" will terminate in finitely many steps. Perhaps you can do some sort of iterative method so that your sequence of guesses for the radius converges to half the length of your line segment, but this is no longer a compass and straightedge proof since those must be done in finitely many steps.

Suppose by some happenstance your method does, in finitely many steps, get you precisely the radius you want. How do you verify that this is actually correct? You arrived at this radius by guessing but even if you randomly get the right answer, you need to prove that it is so. You could say that you take your compass with the correct radius $r$ and draw one circle starting at $a$ and another at the intersection of the line segment $\overline{ab}$ with this circle of radius $r$. Since we're assuming you did get the right radius, it will be the case that this second circle actually intersects $b$, but you cannot actually verify this. What if you are actually some tiny length like $10^{-TREE(3)}$ away from $b$? the exit condition of your procedure is "when I get the right radius" but since your procedure does not constitute a proof, you can only ever guess at when the radius looks right. The beauty of these constructions is that they are utterly ambiguous. A valid compass and straightedge construction is 100% precise without worrying about imprecise guesses. That's why the straightedge is not a ruler with markings! We don't allow measurements that are not provable from our idealized tools.

To explain this second point some more, let me give you a correct construction of the bisection of a line segment $\overline{ab}$. Put one leg of your compass on $a$ and another on $b$, the rotate your compass around $a$ to form a circle $C_1$. This is one of the allowed moves of this game, since the points $a, b$ are assumed to be known to us. Then, put your compass in the same position but rotate around $b$ to get another circle $C_2$. These circles intersect in precisely two points, say $c, d$. This is not from us eyeballing - it is provably so! Now, we know of the points $c, d$ so draw the line segment $\overline{cd}$. Again, drawing a line between known points is one of our legal moves. The intersection of $\overline{cd}$ and $\overline{ab}$ will precisely bisect $\overline{ab}$. If my dictation wasn't clear, look at this link. You can press z and x to go back and forth.

Now I have not actually given a formal geometric proof of this construction, but my point is that the I could translate it to one should I choose to do so. I could prove rigorously that my circles $C_1$ and $C_2$ intersect in precisely two points and I could prove that $\overline{cd} \cap \overline{ab}$ forms the bisection point. I did all of this in finitely many steps and I never had to guess to say "well this looooks close enough to the bisection." I can precisely prove using the formal definitions of compasses and straightedges (which come from Euclid's postulates) that the point I arrive at is the bisection point.

EDIT: I'll answer your comment in here, since it's too long to fit in a response.

That's a fair objection, and I think the only ways to counter it are through formalism and philosophy. For the formal viewpoint, compass and straightedge constructions are visual representations of proofs from Euclid's postulates, which I linked in my answer. The steps I give like "put a leg on $a$" are representatives of postulates like the existence of a circle with a certain center and radius. Then by a "precise" construction I mean one following from these postulates, which yours does not. Also, a proof is necessary a finite sequence of deductions so any classical construction must therefore also be finite. This is the formal difference between our approaches, but it's not very motivating beyond "well Euclid says so." You can ask more deeply why we choose to take these postulates as "precise." I think this inevitably begs questions of history and philosophy. I would also like to point out that I am much more familiar with the formalism than the philosophy, but I'll try my best to get across some ideas and, more importantly, bring up some relevant terms.

I think it is best to try to start by asking why the Greek's were so interested in compasses and straightedges. After all, the story goes that at the door of Plato's academy there was an inscription reading "Let no one ignorant of geometry enter." Whether or not this is apocryphal[1] is not really the point; it is merely illustrative of the philosophical importance the Greeks placed upon geometry. A large part of this was due to the theory of forms[2]. Roughly (which is how well I understand it...), forms are immaterial, perfect, ideal objects. For instance, a circle I draw with an actual physical compass is just a shadow of the ideal form of a circle[0]. In the Republic[3], Plato said that

[geometers] make use of the visible forms and talk about them, though they
are not thinking of them but of those things of which they are
a likeness, pursuing their inquiry for the sake of the square as
such and the diagonal as such, and not for the sake of the image
of it which they draw

So to the geometers at the time of Plato, a compass and straightedge construction was not really just the physical sequences of moves (I put this leg here and rotate about this, I draw a line from here to here, etc). These were a way to grapple with the ideal geometry forms by using "the image of it which they draw." If you're familiar with Plato's allegory of the cave this may ring a bell.

What I mean to say by all of this is that as I understand it, these compass and straightedge constructions arise from an attempt to understand these precise objects like circles and lines and points via their imprecise drawings and likenesses in the real world. When I say that your "guess until your touch $b$" method is imprecise, I suppose I am saying that that is an instance of reasoning with the "shadows on the wall" of the ideal forms we're trying to understand. When I invoke one of Euclid's postulates, I am making a statement about the ideal objects, and when I draw using an actual compass or straightedge, I am merely "[making] use of the visible forms" while "not thinking of them but of those things of which they are a likeness." Your approach of "guess until you touch $b$" is a different type of thought to what the Greek geometers were trying to do. It is working directly with your imperfect drawings rather than using the drawings as a proxy for this higher, immaterial truth the Greeks were after.

This is all very highfalutin, and I am far from knowledgeable in this realm, but I hope my summary and references will help you understand a bit more why your construction is not classically valid. I suppose to sum it up, when you make an imprecise statement like "put your compass leg on $a$" it is because you are using images to deal with these immaterial forms. However, when you have to guess you are only able to do this for the image itself. The idealized forms of these constructions are perfectly precise. The construction I describe is unquestionably the bisection. What I draw by hand or by computer is an imprecise mental model of that perfect precision, whereas the imprecision of guessing is reflective only of the shadows, not the forms themselves.

I hope my rambling helped. Please feel free to discuss this with me further, and here are some of the articles I referred to:

https://topologicalmusings.wordpress.com/2008/01/18/platonic-idealism-and-geometry/ [0]

https://www.dialogues-de-platon.org/faq/faq009.htm [1]

https://en.wikipedia.org/wiki/Theory_of_forms [2]

https://link.springer.com/book/10.1007/978-1-4612-0803-7 [3]


问:
Show spectrum of a bilateral shift is contained in a donut
Let $W$ be a bilateral shift with weight sequence $(w_n) \in \ell^{\infty}(\mathbb{Z})$, for $w_n \geq 0$ for each $n \in \mathbb{Z}$.

i.e. we have that $We_n = we_{n+1}$ for $n\in \mathbb{Z}$, where $(e_n)$ is an orthonormal basis.

Suppose that there exists $\delta >0$ such that $\delta \leq w_n \leq 1$.

Show that $\sigma(W) \subset \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.

Things I've done:

I showed that if $\alpha \in \sigma(W)$, then we have $\{z\in \mathbb{C}: |z|=\alpha \} \subset \sigma(W)$.

But I want to show that $\alpha \in \{ z\in \mathbb{C}: \delta \leq |z| \leq 1\}$.

Any help is appreciated!

答:
I guess you mean $We_n = w_ne_{n+1}$ and that your space is $\ell^2(\mathbb Z)$. Then $W = SD_w$, where $S$ is the usual bilateral shift and $D_w$ is the diagonal matrix with the $w_n$ on the diagonal. So, $\|Wx\| = \|SD_wx\| = \|D_wx\|$, which implies
$$
\delta\|x\|\le\|Wx\|\le\|x\|.
$$
Similarly, $\|W^*x\|\le\|x\|$ and $\|W^*x\| = \|D_wS^*x\|\ge\delta\|x\|$.

From here you should be able to see that the spectrum of $W$ is in that annulus.


答:
It may prove helpful to understand the general conceptual background that underlies this definition. Congruence is an equivalence relation (generalized equality) so it partitions the integers into equivalences classes, here $\,[a] = a + n\Bbb Z\,$ is the class of all integers $\equiv a\pmod{n}.\,$

Conceptually, modular arithmetic is actually working with these classes, e.g. the congruence sum rule $\,a'\equiv a,\,b'\equiv b\Rightarrow a'+b'\equiv a+b\,$ is equivalent to saying that $\,[a]+[b] = [a+b]\,$ is a well-defined addition on the classes. Ditto for multiplication, so these classes inherit the addition and multiplication from the integers. Further they inherit the ring laws from $\,\Bbb Z,\,$ i.e. associative, commutative, and distributive laws. Thus the classes enjoy the same algebraic (ring) structure as the integers $\Bbb Z,\,$ so we can perform arithmetic on them just as we do with  integers.

But when calculating it is often more convenient to work with numbers than sets (classes), so we often choose convenient representatives ("normal forms") for each class.  A complete system of residues is precisely that. The most common choice here is the least natural in $[a],$ which is simply its remainder ("residue") $= a\bmod n,\,$ so this complete system of rep's  $\bmod n\,$  is $\,\{0,1,\ldots, n-1\}.\,$ Another common choice are reps of least absolute value, e.g. $\{0,\pm1,\pm2\}\bmod 5$.

It may help to consider the analogy with fractions. Here the congruence relation for fraction equivalence is $\,a/b \equiv c/d\iff ad = bc.\,$ and the common normal form reps are again the least terms reps, i.e. fractions with coprime numerator and denominator. The above sum rule is usually never proved at elementary level - which should make you ponder why you accepted its truth without proof. This will be rectified if you study fraction fields (or localizations) in abstract algebra (which also generalizes modular arithmetic in the study of general quotient (residue) rings, and congruences, and their relationship).

Remark $ $ The choice of the remainder as a normal form rep also works in polynomial rings over a field (e.g. this yields the complex numbers $\,\Bbb C \cong \Bbb R[x]/(x^2+1)\,$ as the complete system of reps of real polynomials mod $\,x^2+1),\,$ or in any domain with a (computable) Euclidean algorithm. But in general there may not be any (computable) natural  way to choose such normal forms, so we may have no choice but to work with the classes themselves.

Title: Struggling to understand basics of complete residue system

Body:
I'm really struggling to understand the literal arithmetic being applied to find a complete residue system of modulo $n$. Below is the definition my textbook provides along with an example.

Let $k$ and $n$ be natural numbers. A set $\{a_1,a_2,...,a_k\}$ is called a canonical complete residue system modulo $n$ if every integer is congruent modulo $n$ to exactly one element of the set

I'm struggling to understand how to interpret this definition. Two integers, $a$ and $b$, are "congruent modulo $n$" if they have the same remainder when divided by $n$. So the set $\{a_1,a_2,...,a_k\}$ would be all integers that share a quotient with $b$ divided by $n$?

After I understand the definition, this is a simple example provided by my textbook

Find three residue systems modulo $4$: the canonical complete residue system, one containing negative numbers, and one containing no two consecutive numbers

My first point of confusion is "modulo $4$". $a{\space}mod{\space}n$ is the remainder of Euclidean division of $a$ by $n$. So what is meant by simply "modulo $4$"? What literal arithmetic do I perform to find a complete residue system using "modulo $4$"?


问:
Non-negativity in Fatou's Lemma and Lebesgue Dominated Convergence Theorem
Im currently busy with Measure Theory and noticed that the main theorems that I want to use, require the non-negativity condition.

Fatou's Lemma has the condition that we need a sequence $\{f_n\}$ of non-negative measurable functions.

and Lebesgue Dominated Convergence Theorem a sequence $\{f_n\}$ of measurable functions such that $|f_n| < g$ which will imply $g - |f_n|$ is non-negative.

i) Is it correct that in order to continue with any proof, non-negativity is critical?

ii) According to me, if a sequence is integrable or measurable, it does not satisfy this condition. Is this correct?

iii) I only have $f_n < g$ but it does not mean $g - f_n$ is non-negative.

Apologies, I do not want to state too much as this is part of an assignment.
I know how to answer my questions depending on the non-negativity condition.

What might I be missing that could also imply a function/sequence is non-negative?

Thank you!

答:
Some remarks:

Non-negativity is important in part because many authors first define measurability for positive functions (e.g. $L^+$), and then define general measurability using the decomposition
$f=f^+-f^-$ of $f$ into positive and negative parts:
$$f^+(x)=\max(f(x),0)\quad \text{and} \quad f^-(x)=-\min(f(x),0).$$
Here, we note that $f^+$ and $f^-$ are both positive functions, and we say that $f$ is measurable if both $f^+$ and $f^-$ are (there are issues of indeterminate forms we must avoid, but let's sweep this under the rug).  In this sense, non-negativity is important because this is how measures are commonly introduced.

If $f_n <g$, then $g-f_n>0$ (as functions).  That is, $g-f_n$ is non-negative.  I'm not sure what the confusion is here.  I will make clear that the inequalities here are all point-wise, i.e. $g(x)-f_n(x)>0$ for all $x$ in our domain.


答:
Indeed the radius of convergence of the function $f$ is zero, which means that the series

$$ \sum_{k=0}^{\infty} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k $$

diverges for all $z \neq 0$. Although this can be directly verified by using Stirling's approximation, a more fundamental reason for this is that $0$ is a branch point of $f(z)$. Consequently, there is no way $f(z)$ extends to an analytic function about $0$. For instance, the closed-form

$$ f(z) = \sqrt{\frac{3}{2\pi z}} \, e^{3/4z} K_{1/4}(3/4z) \tag{*} $$

in Claude Leibovici's answer has the branch cut $(-\infty, 0]$, as seen from the domain coloring plot of $\text{(*)}$ on the rectangle with the corners $\pm 5 \pm 5i$:

That said, your best hope for an asymptotic expansion of polynomial form is

$$ f(z) = \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k + \mathcal{O}(z^N) \qquad \text{as } z \to 0 \tag{1} $$

within Stolz angle for each fixed $N \geq 0$.

Addendum. Here is a proof that $\text{(1)}$ holds at least within the region $\operatorname{Re}(z) \geq 0$, although I suspect that this holds within larger sectors.

Note that for each fixed $N$, Taylor's theorem tells that for $\operatorname{Re}(z) \geq 0$,

\begin{align*}
&f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl[ \exp\left(-\frac{zy^4}{24}\right) - \sum_{k=0}^{N-1} \frac{1}{k!} \left( -\frac{zy^4}{24} \right)^k \Biggr] e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{z^N (1-t)^{N-1}}{(N-1)!} \left(-\frac{y^4}{24}\right)^N \exp\left(-\frac{zty^4}{24}\right) \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y,
\end{align*}

where the last line follows from the Lagrange form of the remainder. Taking absolute value,

\begin{align*}
&\left| f(z) - \sum_{k=0}^{N-1} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k \right| \\
&\leq \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \Biggl( \int_{0}^{1} \frac{\left| z \right|^N (1-t)^{N-1}}{(N-1)!} \frac{y^{4N}}{24^N} \, \mathrm{d}t \Biggr) e^{-\frac{y^2}{2}} \, \mathrm{d}y \\
&= \frac{(4N-1)!!}{N!24^N} \left|z\right|^N.
\end{align*}

This proves that the error term in $\text{(*)}$ is indeed $\mathcal{O}(z^N)$, at least within the region $\operatorname{Re}(z) \geq 0$.

Title: Taylor expansion of a gaussian integral

Body:
I tried Taylor expanding the following function for small $x \ll 1$:
\begin{align*}
f(x) = \frac{1}{\sqrt{2\pi}} \int\limits^{+\infty}_{-\infty} e^{\frac{-xy^4}{24}-\frac{y^2}{2}}\,\mathrm{d}y.
\end{align*}
I already know that $f$ takes on finite values for $x \geq0$. My goal is to get a series expression for $f$ of the following form:
\begin{align*}
f_N(x) = \sum^N_{n = 0}a_n x^n.
\end{align*}

So after using the formula (for $a = 0$)
\begin{align*}
f(x) = \sum^\infty_{n = 0} \frac{f^{(n)}(a)}{n!}(x-a)^n,
\end{align*}
I get up to the $N$-th term:
\begin{align*}
f_N(x) = \sum^N_{n = 0} \frac{1}{n!}\frac{(-x)^n}{24^n}(4n-1)!!
\end{align*}
However, when plugging in a small value for $x$ (for example $x = 0.1$) in Wolfram Alpha, I get that till $N = 25$ the approximation is very good, but after around $N = 30$ the series diverges away from the true value ($\approx 0.988306$ ). So for $N \rightarrow + \infty$, the series seems to diverge.

My questions are:

How can this be ? I thought that for larger and larger $N$ the Taylor approximation would be more and more better ? (well, for small $x$ atleast)

Does this mean that the radius of convergence is equal to $0$ and that the interval of convergence is just the point $x = 0$ ? If not, then what is the radius of convergence ? If yes, doesn't this contradict the fact that $f$ is finite for $x \geq1$ ?


问:
In quadratic interpolation of 3 points - is the minimum guaranteed to be within the left and right points?
I have 3 ordered points, $(x_1,y_1),(x_2,y_2),(x_3,y_3),$ with $x_1<x_2<x_3$, of which I calculate the interpolationg quadratic function. In my case, $(x_2,y_2)$ is a local minimum in or maximum in this list (i.e. ($y_2>y_1$ and $y_2>y_3$) or ($y_2<y_1$ and $y_2<y_3$). My question: is the parabola's minimum or maximum guaranteed to be within the support of these points, so that $x_1<x_\text{min}<x_3$?

I thought about just working out the formula for $x_\text{min}$, but surely there is a more easy argument?

答:
I suppose something like this works?

Assume $x_\text{min} < x_1$. Then if $x_1<x_2<x_3$, $y_1,y_2,y_3$ is a monotonic sequence because the parabola is monotonic after $x_\text{min}$. Similarly for $x_\text{min}>x_3$. Hence, $x_1<x_\text{min}<x_3$.


问:
Probabilities of no ace and exactly not one ace in throwing 10 dies
The probabilities i am looking for are used for a problem in Feller's introduction to probability that asks for the probability $p$ of finding $2$ or more aces in throwing $10$ dies. For this problem i considered that:

$p = 1-x-y$

Where $x$ is the probability of finding no ace at all, and $y$ being the probability of finding exactly one ace. Of course $x = \frac{5^{10}}{6^{10}}$... i think. But i am a little more confused about finding $y$, more because the answer to this exercise is that: $p = 1- 10\times\frac{5^9}{6^{10}-5^{10}} $ but i don't really understand where this result can come from, more considering the result i got for $x$

答:
The book is wrong. And your factorization is slightly off. (You need a plus sign inside the parenthesis).$1-\frac{5^{10}}{6^{10}}-10*\frac{5^9}{6^{10}}=1-\frac{5^9}{6^{10}}\left(5+10\right)\approx 0.5154833$ whereas the answer in the book is $1-10*\frac{5^9}{6^{10}-5^{10}}\approx 0.6147724$.

This can be computed also without using complementary events as ${10\choose2}(1/6)^2(5/6)^8+...+{10\choose 10}(1/6)^{10}(5/6)^0$. Using code, this is

```<code>> pbinom(1, 10, 1/6, lower.tail=FALSE)
[1] 0.5154833
</code>```

Body:
The book is wrong. And your factorization is slightly off. (You need a plus sign inside the parenthesis).$1-\frac{5^{10}}{6^{10}}-10*\frac{5^9}{6^{10}}=1-\frac{5^9}{6^{10}}\left(5+10\right)\approx 0.5154833$ whereas the answer in the book is $1-10*\frac{5^9}{6^{10}-5^{10}}\approx 0.6147724$.

This can be computed also without using complementary events as ${10\choose2}(1/6)^2(5/6)^8+...+{10\choose 10}(1/6)^{10}(5/6)^0$. Using code, this is

> pbinom(1, 10, 1/6, lower.tail=FALSE)
[1] 0.5154833


问:
Notation - Two adjacent vectors?
I'm studying multivariable calculus at the moment and have come across equations involving two bolded variables placed side by side, like so:

$$ \nabla \mathbf{f}=\frac{\partial {{f}_{j}}}{\partial {{x}_{i}}}{{\mathbf{e}}_{i}}{{\mathbf{e}}_{j}} $$

Is this meant to be a dot product? Or something else?

答:
Since the result of $\nabla\mathbf{f}$ gives us a second-rank tensor (it can be represented as a two-by-two matrix with components $\frac{\partial f^j}{\partial x^i}$), the usual vector basis $\mathbf{e}_i=(0,\cdots,1,0,\cdots,0)$ (where the one is in the $i^{th}$ place) doesn't quite cut it.  However, we can generalize them.  If we define the matrices $\mathbf{e}_i\mathbf{e}_j$ to have all zeroes except for a one in the $(i,j)$ spot, then the sum
$$\sum_{i,j=1}^n \frac{\partial f^j}{\partial x^i} \mathbf{e}_i\mathbf{e}_j$$
gives us the desired matrix, just as the sum $\sum_{i=1}^n \frac{\partial g}{\partial x^i} \mathbf{e}_i$ gives us $\nabla g$.

Properly, we should be writing $\mathbf{e}_i\bigotimes\mathbf{e}_j$ as we're actually dealing with the Kronecker product of $\mathbf{e}_i$ and $\mathbf{e}_j$, though in many contexts it seems people drop the direct sum symbol.

Title: Notation - Two adjacent vectors?


问:
A bag contains 4 red, 5 white and 6 black balls. 3 balls are selected simultaneously. Find the probability that one of colour is missing.
I know the correct method to do this ie. making cases such as RRW, BBW, WWB etc.

But why can’t complementary method be used?
$$P=1-\frac{4\cdot 5 \cdot 6}{\binom {15}{3}}$$

Which implies the probability that at least one of the colour can repeat

答:
Your approach is correct, but you might have misinterpreted the question. The “correct” answer you gave, $\frac{301}{455},$ counts ones which are missing exactly one color. You answer includes when you are missing $2$ colors, that is, when you select three balls of the same color.

The correct answer under this interpretation is:

$$1-\frac{6\cdot5\cdot 4+\binom63 +\binom53+\binom43}{\binom{15}3}$$

This is still easier than the other approach.


问:
Different definition of uniform integrability
I'm reading Tao's An introduction to measure theory.And the definition of uniform integrability in this book is

$(X,{\mathfrak  {M}},\mu)$ is a measure space(not necessarily finite). A sequence $f_n:X \rightarrow \mathbb{C}$ of absolutely Integrable functions is said to be uniformly Integrable if the following three statements hold

1.(Uniform bound on $L^1$ normal) One has $\sup_n\|f_n\|_{L^1(\mu)}=\sup_n\int_{X}|f_n|d\mu <+\infty$.

2.(No escape to vertical infinity) One has $\sup_n\int_{|f_n|\ge M}|f_n|d\mu\xrightarrow{M\rightarrow\infty} 0$.

3.(No escape to width infinity) One has $\sup_n\int_{|f_n|\le\delta}|f_n|d\mu\xrightarrow{\delta\rightarrow0} 0$.

However, the definition of uniform integrability in Wikipedia is

Let $(X,{\mathfrak  {M}},\mu ) $ be a positive measure space. A set ${\displaystyle \Phi \subset L^{1}(\mu )}$ is called uniformly integrable if to each ${\displaystyle \varepsilon >0} $  there corresponds a ${\displaystyle \delta >0}$  such that
${\displaystyle \int _{E}|f|\,d\mu <\varepsilon }
$ whenever ${\displaystyle f\in \Phi }$ and ${\displaystyle \mu (E)<\delta .}$

Tao leaves the equivalence of this two definitions when $(X,\mathfrak{M},\mu)$ is finite(namely $\mu(X)<+\infty$) as an exercise, and I've prove it. But I can't prove the equivalence of them when $\mu(X)=+\infty$. Can anyone prove or disprove it?

Any help is appreciated.

EDIT:I've also prove that the first definition implies the second one(which is also an exercise in Tao's book). So the problem becomes whether the second one implies the first one.

答: The condition in Wikipedia is to weak to imply 1), 2) and 3). Consider $\mathbb  N$ with power set and the counting measure. Any family of functions satisfies the Wikipedia  condition since $\mu (E) <1$ implies that $E$ is the empty set. You can easily write down a family of functions each of which satsifies the condition $\|f||_1=\sum |f(n) |<\infty$ but the norms are not bounded when$f$ varies over the family. [I will write down an explicit example if you ask for it].


问:
$A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements.
$(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$

I know I have to start with $A = ...$ to get to $B$. I incorporated $A = A \cap (A \cup C)$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $(A \cup C) = (B \cup C)$? I've looked at the set identities, and think that I have to use absorption law again to show that $A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$. Can you guys please help me solve this problem, I'm stuck.

答:
Assuming $(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$. Prove $A=B$.

Proof by Contradiction:

Assume $x \in A$ and $x \notin B$.

Then $x \in B \cup C \to x \in$ C.

But this leads to a contradiction. Because $x \notin B \cap C$ but $x \in  A \cap C$.

Therefore, if $x \in A$ then $x \in B$.

And the same argument works for $x \in B$ but $x \notin A$.

Therefore, if something is in $A$ then it is in $B$ and vice versa.


问:
Finding eigenvalues of transition matrix of a given Markov process (solution verification)
Consider sampling the uniform measure $\mathbb{Z}_L$ by the Markov chain $X^{(k)}$ with
$$P_i[X^{(i)}=(i+1)\mod L]=1$$
with initial condition $X^{(0)}=L-1$. Here $P_i$ means conditioning on $X_0=i$.
$$P_i[X^{(k)}=j]=\begin{cases}1, &\text{if }j=(i+k)\mod L\\0, &\text{otherwise}\end{cases}$$
First, I compute the transition matrix. The state space is $\{0,1,\ldots,L-1\}$, and the Markov process permutes one element to the right each time, i.e.
$$[0,1,2,\ldots,L-1]\rightarrow[1,2,\ldots,L-1,0]\rightarrow[2,3,\ldots,L-1,0,1]\rightarrow\ldots$$
so
$$T=\begin{pmatrix}0&1&0&0&\ldots&0&0\\0&0&1&0&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\cdots&\vdots&\vdots\\0&0&0&0&\ldots&0&1\\1&0&0&0&\ldots&0&0\end{pmatrix}_{L\times L}$$
I am interested in finding the eigenvalues of $T$, $T^k$ for any $k$, and $F=\frac{1}{L}\sum_{k=1}^{L}T^k$.

For $T$, the eigenvalues are solution of $\lambda^L-1=0$, which are $L$-roots of unity $\lambda=e^{2\pi i\ell/L}$, where $\ell=0,1,\ldots, L-1$.

For $T^k$, the eigenvalues are the same as the first case, since the effect is just permuting the columns of $T$ cyclically, so the determinant of characteristic polynomial remains unchanged.

For $F$, first consider $J=\sum_{k=1}^{L}T^k$. Each column of $J$ consists of sum of all standard unit vectors $e_i$ and all the entries are $1$, so $J$ is $L\times L$ matrix whose entries are all $1$. So the rank of $J$ is just $1$, with eigenvalue $\lambda=0$ with multiplicity $L-1$ and $\lambda=L$, which is the result of $\pi T=\pi$, where $\pi$ is the stationary distribution, and we are viewing it as an eigenvector. Hence, the eigenvalues of $F=J/L$ is just either $0$ (with multiplicity $L-1$) or $1$.

I'd like to know whether my solutions are correct. If there are any flaws, can anyone point that out? Thank you.

Edit:

The second part of my original solution is wrong. Use the fact that if $\lambda_{\ell}$ are eigenvalues of $T$, then $\lambda_{\ell}^k$ are eigenvalues of $T^k$, we have the result.

答: If $p$ is a polynomial, then the eigenvalues of $p(T)$ are $p(\lambda)$, where $\lambda$ is an eigenvalue of $T$.


答:
If the element is in $A$ then either it's in $C$ or it isn't.

If it is in $C$ then it is in their intersection, which implies that it is in the intersection of $B$ and $C$ which implies that it is in $B$.

If it is not in $C$, it is still in the union of $A$ and $C$, which implies that it is in the union of $B$ and $C$, which implies that it is in $B$ (since it is assumed here that it is not in $C$).

By a symmetrical argument, $B$ is also a subset of $A$.

Title: $A, B,$ and $C$ are all sets that lie in a common universal set. Prove $A = B$ given the following statements.

Body:
$(A \cup C) = (B \cup C)$ and $(A \cap C) = (B \cap C)$

I know I have to start with $A = ...$ to get to $B$. I incorporated $A = A \cap (A \cup C)$ as the first step using absorption law, but I don't know where to go from here. How do I show that it equals $(A \cup C) = (B \cup C)$? I've looked at the set identities, and think that I have to use absorption law again to show that $A \cap (A \cup C) = (A \cap (A \cup C)) \cap ((A \cap (A \cup C)) \cup (B \cup C))$. Can you guys please help me solve this problem, I'm stuck.


问:
Proof Leibniz pi series using Wallis product or viceversa
Hi I am trying to get a smart way to connect this two well known expressions for $\pi$:

Leibniz serie:

$$\frac{\pi}{4}=\sum_{n=1}^{\infty}{\frac{(-1)^{n+1}}{2n-1}}$$

Wallis product:

$$\frac{\pi}{2}=\prod_{n=1}^{\infty}{\frac{(2n)^2}{(2n-1)(2n+1)}}$$

and only saw this method through Brouckner's continued fraction:

https://link.springer.com/article/10.1007/BF00348349

It seems a little bit too tricky to be the simplest way to stablish such a connection. Any advice will be valuable.

答:
A few ideas (may be)

A small typo in the Wallis product
$$\frac{\pi}{\color{red}{2}}=\prod_{n=1}^{\infty}{\frac{(2n)^2}{(2n-1)(2n+1)}}$$

Now, we could think from
$$\frac{(2n)^2}{(2n-1)(2n+1)}=1-\frac{1}{2 (2 n+1)}+\frac{1}{2 (2 n-1)}$$ and consider the logarithm and then some Taylor series after some value of $n$


问:
How to interpret minimal sufficient statistics?
I know the definition of minimal sufficient statistics is:

A sufficient statistic T(X) is called a minimal sufficient statistic if, for any other sufficient statistic T'(X), T(X) is a function of T'(X).

I want to know how to interpret this definition from function view. T is smaller than T', if T is a function of T'. What does that mean?

答:
If $T'\le T$, then there exists a surjective function from T to T'. For this you will need to recall what the definition of a function is. The right is not a surjective function, but the left is.

As an example, consider $\mathscr {Unif}(-\theta, \theta)$ with $\theta>0$. A sample has likelihood function of $f(x|\theta)=\frac{1}{2^n\theta^n}I(X_{(1)}\ge-\theta)I(X_{(n)}\le\theta)$ and so by the factorization theorem $T=\{X_{(1)}, X_{(n)}\}$ is a sufficient statistic for $\theta$. It is not minimal sufficient though, because $f(x|\theta)\propto I(-X_{(1)}\le\theta)I(X_{(n)}\le\theta)=I(\max\{-X_{(1)},X_{(n)}\}\le\theta)$. So $T'=\max\{-X_{(1)},X_{(n)}\}$ is a minimal sufficient statistic. There exists a surjective function from T to T', namely take the max of the components of T. But there is no surjective function from T' to T. In this sense you can construct T' from T but not the other way around. For this example, if you know the largest and smallest value, you know the largest absolute value. But if you know the largest absolute value, you don't necessarily know the smallest or largest value. (It could be either the largest or smallest value, and then the other will be unknown.)


问:
Find the maximum and minimum values in a range
I'm trying to understand how to find the minimum & maximum values of this function:

$$ f(x,y) = xy-y^2 $$

In the following range D:

$$ D = \{(x,y) \in R^2 : 0 \leq x \leq 1, |y| \leq x^2 \} $$

Obviously I tried to use Lagranage multipliers, but I was a little confused about the absolute value.
Should I divide it to two different equastion systems, one for positive y and one for negative?

Thanks in advance.

答:
The region specification $ \ |y| \leq x^2 \ $ is a "shorthand":  it could as easily have been written as $ \ -x^2 \ \leq \ y \ \leq \ x^2 \ $ .  So the region lies between two parabolas and between the origin and the vertical line $ \ x = 1 \ \ , $ as seen in the graph below.

For the interior region, you would use Lagrange multipliers as you proposed; the result shown by Dr. Sonnhard Graubner tells us that

$$ f_x \ = \ y \ = \ 0 \ \ , \ \ f_y \ = \ x - 2y \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 2y \ \ . $$

So there is a single critical point in the interior of the region at $ \ (0,0) \ . $  Since the function is $ \ f(x,y) \ = \  xy \ - \ y^2 \ = \ y · (x-y) \ , $ we have $ \  \ f(x,0) \ = \  0 \  $ and $ \  \ f(x,x) \ = \  0 \  $ .  So this critical point is a little peculiar in that it lies at the intersection of two lines (the $ \ x-$ axis and $ \ y \ = \ x  $ ) for which the function is zero.  The index $ \ f_{xx}f_{yy} \ - \ (f_{xy})^2 \ = \ 0 · (-2) \ - \ 1^2 \ = \ -1 \ < \ 0 \ $ is constant for the function, suggesting that $ \ (0,0) \ $ is a saddle point. [Indeed, it is at the junction of four regions of alternating sign as one circulates around the origin.]  Since for our region, $ \ x \ > \ y \ $ , $ \ f(x,y) \ > \  0 \ $ for $ \ y \ > \ 0 \ $ and $ \ f(x,y) \ < \  0 \ $ for $ \ y \ < \ 0 \ . $

It is when we check the boundary of the region for extremal points that we need to work with the two individual parabolas (and the vertical line).

On $ \ x \ = \ 1 \ $ , the function becomes $ \  \ f(1,y) \ = \  1·y \  - \ y^2 \  , $ so we have
$$ \ \frac{df}{dy} |_{x=1} \ = \ 1 - 2y \ = \ 0  \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ \ , $$

giving us an extremal point (local maximum) at $ \ \left(1 \ , \ \frac{1}{2} \right) \ $ with $ \  \ f \left(1 \ , \ \frac{1}{2} \right) \ = \  \frac{1}{2} \  - \ \left( \frac{1}{2} \right)^2 \ = \ \frac{1}{4} \ \ . $  [This is where we might have expected to find such a point from our earlier result of $ \ f_y  \ = \ 0 \ $ on the line $ \ x \ = \ 2y \ . $ ]

Along the "upper" parabolic boundary,
$$ \  \ f(x,x^2) \ = \  x·x^2 \  - \ (x^2)^2 \ \ = \ \ x^3 \ - \ x^4 \ \ \Rightarrow \  \ \frac{df}{dy} |_{x^2} \ = \ x^2 · (3 - 4x) \ = \ 0  \ \ , $$

so there is a local maximum at $ \ \left(\frac{3}{4} \ , \ \frac{9}{16} \right) \ $ with $ \  \ f \left(\frac{3}{4} \ , \ \frac{9}{16} \right) \ = \  \frac{3}{4} · \frac{9}{16} \  - \ \left( \frac{9}{16} \right)^2 \ = \ \frac{27}{256} \ \approx \ 0.105 \ \ . $

For the  "lower" parabolic boundary,
$$ \  \ f(x,-x^2) \ = \  x·(-x^2) \  - \ (-x^2)^2 \ \ = \ \ -x^3 \ - \ x^4 \ \ \Rightarrow \  \ \frac{df}{dy} |_{-x^2} \ = \ -x^2 · (3 + 4x) \ = \ 0  \ \ , $$

for which the solution is outside of our region.

It remains to test the "vertex"  $ \ (1  ,  -1) \ $ , as we already know that   $  \ f (0  ,  0) \ = \ f(1  ,  1) \ = \ 0 \ $ :

$ \  \ f (1  ,  -1 ) \ = \  1 · (-1) \  - \  ( -1  )^2 \ = \ -2 \ \ . $

The absolute maximum in the region is thus $ \  \ f \left(1 \ , \ \frac{1}{2} \right) \ = \   \frac{1}{4} \ \   $  and the absolute minimum is $ \  \ f (1  ,  -1 ) \ =  \ -2 \ \ . $


问:
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$.
Tried this one a couple of times but can't seem to figure it out.

I am trying to simplify the expression:

$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$

So my attempt at this is:

$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$

$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$

$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$

$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$

$$=\dfrac{6x^3-2x^2}{9x^3-x}$$

This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$

Edit, missed a difference of squares:

$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$

Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$

答: HINT: Pull out everything that you can: $6x^3-2x^2=2x^2(3x-1)$, and $9x^3-x=x(9x^2-1)$. Then notice that $9x^2=(3x)^2$, so that $9x^2-1=(3x)^2-1^2=(3x-1)(3x+1)$. Finally, do the cancellations that are now apparent.</p>


问:
Struggling to understand basics of complete residue system
I'm really struggling to understand the literal arithmetic being applied to find a complete residue system of modulo $n$. Below is the definition my textbook provides along with an example.

Let $k$ and $n$ be natural numbers. A set $\{a_1,a_2,...,a_k\}$ is called a canonical complete residue system modulo $n$ if every integer is congruent modulo $n$ to exactly one element of the set

I'm struggling to understand how to interpret this definition. Two integers, $a$ and $b$, are "congruent modulo $n$" if they have the same remainder when divided by $n$. So the set $\{a_1,a_2,...,a_k\}$ would be all integers that share a quotient with $b$ divided by $n$?

After I understand the definition, this is a simple example provided by my textbook

Find three residue systems modulo $4$: the canonical complete residue system, one containing negative numbers, and one containing no two consecutive numbers

My first point of confusion is "modulo $4$". $a{\space}mod{\space}n$ is the remainder of Euclidean division of $a$ by $n$. So what is meant by simply "modulo $4$"? What literal arithmetic do I perform to find a complete residue system using "modulo $4$"?

答:
You can draw a picture for modulo 6. Get a piece of paper and place it with the longer dimension horizontal.  With a dark pen and a ruler, draw seven horizontal lines completely across the page. In about the middle of the page write the integers from 0 to 5 ONE PER REGION, starting at the bottom.

Enter all the other integers (or at least a lot of them) as follows:
In each region, repeatedly add 6 to the number on the left until you run out of paper.

And then return to the middle and repeatedly subtract 6 from every number on the right.

(A complete residue system will be 6 integers, no two being from the same region.)


问:
Function in $L^2(\Omega)$ is in $H^1(\Omega)$
I want to show the equivalence between the two statements: ($\Omega$ is a bounded open subset of $\mathbb{R}^n$)

$v\in L^2(\Omega)$ is in $H^1(\Omega)$,

There is a constant $C$ s.t. $$ |\int_\Omega v\nabla \varphi dx| \leq C||\varphi||_{L^2(\Omega)},\ \forall \varphi\in C_c^\infty(\Omega).$$

$1\to2$ is clearly, but when I want to prove $2\to1$, I could find $\{\varphi_k\}_{k=1}^\infty\in C_c^\infty(\Omega)$ s.t. $\varphi_k\to v$ in $L^2$ norm, then from limits and Green's formula,
$$
\lim_{k\to\infty}|\int_\Omega \nabla\varphi_k\varphi dx|\leq C||\varphi||_{L^2(\Omega)},\ \forall \varphi\in C_c^\infty(\Omega).
$$
Then I have no idea for the inequality( if we let $\varphi$ take $\varphi_k$, then we would have $\lim_{k\to\infty} ||\nabla\varphi_k||_{L^2(\Omega)}\leq C$, but this couldn't prove the convergence in $L^2(\Omega)$. I want to ask the complete proof or some ideas aobut it.

答: Define $f(\varphi):=\int_\Omega v\nabla\varphi\,dx$. Then the condition implies that $f$ extends to a bounded linear functional on $L^2$. By Hilbert's representation theorem, we have $f(\varphi) = \langle \varphi,\psi\rangle = \int_\Omega\varphi\overline\psi\,dx$ with some $\psi\in L^2$. Can you go on?


问:
Linear Algebra -- Non invertible Change of Coordinate Matrix
Correct me if anything is wrong what I say, I just start to study linear algebra

The question that I encounter is as following:

Suppose that S, the subspace of $\mathbb{R}^{2 \times 2}$ formed by symetric matrices of 2nd order ($A \in \mathbb{R}^{2 \times 2}: A^T = A)$. We also know that $\mathcal{B} = \{A_1,A_2,A_3\}$. $A_1 = \begin{pmatrix} 1& -2 \\ -2 & 1\end{pmatrix}, A_2 = \begin{pmatrix} 1& 3 \\ 3 & 6\end{pmatrix}, A_3 = \begin{pmatrix} -1& 1 \\ 1 & -3 \end{pmatrix}$ forms a basis of $S$. Suppose we have $x = \begin{pmatrix} a & b \\ b & c\end{pmatrix} \in S$, find the the coordinate relative to the basis, $[x]_{\mathcal{B}}$

We know that a linear transformation like following $T: \mathbb{R}^{2 \times 2} \to \mathbb{R}^4$ should be isomorhpism which means:

$$\displaystyle T(\begin{pmatrix} a & b \\ c& d\end{pmatrix}) = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}$$

From the textbook I have learnt that, if we have a basis $\mathcal{B} = \{b_1,b_2,\dots,b_n\}$, then the vector equation $x = c_1b_1 + c_2b_2 + \dots + c_nb_n = P_{\mathcal{B}}[x]_{\mathcal{B}}$ and $[x]_{\mathcal{B}}$ is the coordinate of x rleative to the basis.

Also, from the textbook, I learned that since $b_1,b_2,\dots,b_n$ form a basis, then they are inverivle and so is $P_{\mathcal{B}}$. As a result, we have $[x]_{\mathcal{B}} = P_{\mathcal{B}}^{-1}x$

This is what I plan to do with the question I have encountered. However, if I try to write $A_1,A_2,A_3$ into column form and combine them, I would have a matrix of $4 \times 3$ which is not invertible since it is not a square matrix.

I am not sure where I did wrong

答: $S$ is a $3$-dimensional space, so the change of coordinate matrix is $3\times3$.  $S$ is isomorphic to $\Bbb R^3$, for instance via $\begin{pmatrix}a&b\\b&c\end{pmatrix}\mapsto\begin{pmatrix}a\\b\\c\end{pmatrix}$.  Then the change of basis matrix would be $\begin{pmatrix}1&1&-1\\-2&3&1\\1&6&-3\end{pmatrix}$.


问:
Best definition for distributive property of multiplication
I am teaching math for kids in Perú, however, I find so many different definitions for the distributive property of multiplication. Do you have a solid definition in words with a book reference?

Thanks.

答:
Teach them FOIL (first outside inside last). $(a+b)(c+d)=\underbrace{ac}_{first}+\underbrace{ad}_{outer}+\underbrace{bc}_{inner}+\underbrace{bd}_{last}$. But then tell them that b is 0 here. So $a(c+d)=(a+0)(c+d)=ac+ad+0c+0d=ac+ad$. This is actually a high school Algebra II technique so it might be overkill.

Here is my definition.

$a\times (b+c)=a\times b+a\times c$

or from the right,

$(a+b)\times c=a\times c+b\times c$.

That's about it, you just need to remember to teach them both sides.


问:
Find all the points where a function is not differentiable
Let $f\left( x \right)$ be a differentiable function $\mathbb{R}\to \mathbb{R}$. Find all the points where the function $\left| f\left( x \right) \right|$ is not differentiable.
Of course, I understand that $\left| f\left( x \right) \right|$ is not differentiable where its graph has a corner because of the modulus, e.g., ${{\delta }_{+}}f\left( a \right)+{{\delta }_{-}}f\left( a \right)=0$. Obviously, it is only possible where $f\left( x \right)$ changes its sign since $f\left( x \right)$ is be a differentiable function.
Is there any way to write this informal idea in a more formal manner?

答: If $f(x) >0$ then $f>0$ in some open interval $(x-r,x+r)$ (by continuity) so $|f|=f$ in this inetrval and so $|f|$ is differentiable at $x$. Similarly, $f(x) <0$ the $|f|$ is differentiable at $x$. But if $f(x)=0$ we cannot say whether $f$ is differentiable or not at $x$: If $f=0$ the $|f|$ is differentiable at all points but if $f(x)=x$ then $|f|$ is not differentiable at $0$. If $f(x)=0$ then $f$ is differentiable at $x$ iff $f(y)=o(y-x)$ as $ y \to x$.


答:
Let me try a direct approach. For sets $A$ and $C$, we have one of the following two possibilities:

$A \subset C$, this includes the case where both sets are equal;

$A \cap C$ is some proper subset of $A$, which may be empty.

For the first case, we know that $A \cup C = C = B \cup C$, hence $B \subset C$. By the second equality given in the question, we have $A = A \cap C = B \cap C = B$.

For the second case, let $A \cap C = D = B \cap C$. By the first equality in the question, $(A \cup C) \backslash C = (B \cup C) \backslash C \implies A \backslash D = B \backslash D$. Taking union with $D$ on both sides gives us the desired result.


问:
normality of unilateral forward shift
Let $H$ be an infinite dimensional, separable complex Hilbert space with orthonormal basis $(e_n)_n$.

Let $T$ be the unilateral forward shift with $Te_n= e_{n+1}$.

Let $\mathcal{K}(H)$ denote the set of compact operators acting on $H$.

Show $T$ is essentially normal.

Show that there doesn't exist a normal $N \in B(H), K\in \mathcal{K}(H)$ such that $T= N+K$.

Things I've done:

To show that $T$ is essentially normal, we want to show that $\pi(T)\pi(T^*) = \pi(T^*)\pi(T) $, where $\pi$ is the canonical quotient map $\pi: B(H) \rightarrow \mathcal{L}(H)$, and $\mathcal{L}(H)$ is the Calkin algebra.

From here I'm not sure how to continue.

We will proceed by contradiction.

Suppose there exists a normal $N \in B(H), K\in \mathcal{K}(H)$ such that $T= N+K$.

Then we have $ e_{n+1}=T(e_n)= N(e_n)+ K(e_n)$. From here, I don't know how to continue to have a contradiction.

I also know that if $N$ is normal and $K$ is compact, then $N+K$ is essentially normal.

Any help or ideas will be appreciated!

Thank you!

答: $T$ is Fredholm with index $-1$. So, if $T = N+K$, then $N = T-K$ would be Fredholm with index $-1$. But a normal Fredholm operator always has index zero (because $H = \ker N\oplus \operatorname{ran}N$). Contradiction!

Title: normality of unilateral forward shift


问:
How do I parse P(A | B, C) and P(A, B | C)?
Is there a universally accepted way of parsing the statement $P(A | B, C)$ where there is conditional and joint probability in the same expression? Should I read it as:

a conditional probability $P(A | (B,C))$ ? or

a joint probability $P( (A|B), C)$ ?

Likewise, how do I parse: $P(A, B|C)$ ?

Are there general rules? And if so, do they extend to 3+ variables?

答:
It is interpretation 1.

For 3+ variables $P(A_1 \mid A_2, A_3, \ldots, A_n)$ is read as "the probability of event $A_1$ occurring, given that events $A_2$ through $A_n$ are occurring."


答: Usually, $P(A|B,C)=P\left(A|(B,C)\right)$. And $P(A,B|C)=P\left((A,B)|C\right)$. So the general rule would be $P(A_1,...,A_n|B_1,...,B_m)=P\left((A_1,...,A_n)|(B_1,...,B_m)\right)$. Treat the things on the left and right hand sides of the vertical bar as their own groups with the commas meaning "and" or "intersection" ($\cap$). Unless you have a good reason to not parse the conditional probability that way, this is almost universally the way things are intended. I guess, the reason is because if you are given something, you're not going to be given it for some things and not other things. $P((A|B),C)$ implies that you have knowledge of B for A but not for C. But since they're part of the same sample space, you'd know that B happened for C too.

Title: How do I parse P(A | B, C) and P(A, B | C)?

Body:
Is there a universally accepted way of parsing the statement $P(A | B, C)$ where there is conditional and joint probability in the same expression? Should I read it as:

a conditional probability $P(A | (B,C))$ ? or

a joint probability $P( (A|B), C)$ ?

Likewise, how do I parse: $P(A, B|C)$ ?

Are there general rules? And if so, do they extend to 3+ variables?


答:
The set of points at which $|f(x)|$ is not differentiable is

$$\left\{a\in\mathbb{R}\left|\lim_{x\to a}\frac{f(x)}{x-a} \text{ exists and is nonzero}\right.\right\}$$

Title: Find all the points where a function is not differentiable

Body:
Let $f\left( x \right)$ be a differentiable function $\mathbb{R}\to \mathbb{R}$. Find all the points where the function $\left| f\left( x \right) \right|$ is not differentiable.
Of course, I understand that $\left| f\left( x \right) \right|$ is not differentiable where its graph has a corner because of the modulus, e.g., ${{\delta }_{+}}f\left( a \right)+{{\delta }_{-}}f\left( a \right)=0$. Obviously, it is only possible where $f\left( x \right)$ changes its sign since $f\left( x \right)$ is be a differentiable function.
Is there any way to write this informal idea in a more formal manner?


答:
You can factor a $2x^2$ out of the top to get $6x^3-2x^2=2 x^2 (3x-1 )$.  After this, you can factor an $x$ out of the bottom to get $(9x^2-1)$, which then factors into $(3x-1) (3x+1)$ as it is a difference of squares.  So, putting that all together,
$$\frac{\left(6x^3-2x^2\right)}{\left(9x^3-x\right)}=\frac{2 x^2 (3x-1 )}{x (3x-1) (3x+1)}=\hspace{2pt}\ldots\hspace{2pt}\Large{?}$$</p>

Title: Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$.

Body:
Tried this one a couple of times but can't seem to figure it out.

I am trying to simplify the expression:

$$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$

So my attempt at this is:

$$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\div\bigg(\dfrac{9x^2}{x^2}-\dfrac{1}{x^2}\bigg)$$

$$=\bigg(\dfrac{6x-2}{x}\bigg)\div\bigg(\dfrac{9x^2-1}{x^2}\bigg)$$

$$=\dfrac{6x-2}{x}\cdot\dfrac{x^2}{9x^2-1}$$

$$=\dfrac{(6x-2)(x^2)}{(x)(9x^2-1)}$$

$$=\dfrac{6x^3-2x^2}{9x^3-x}$$

This is the part that I get stuck at. I can't decide what to factor out:
$$=\dfrac{x(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{(6x^2-2x)}{(9x^2-1)}$$

Edit, missed a difference of squares:

$$=\dfrac{2x^2(6x^3-2x^2)}{x(9x^3-x)}$$

$$=\dfrac{2x^2(3x-1)}{x(3x-1)(3x+1)}$$

Giving a final answer of:
$$=\boxed{\dfrac{2x}{3x+1}}$$