问:
Inverse image sheaf of an $\mathcal{O}_Y$ Module
I am trying to understand the inverse image sheaf since I am not comfortable with sheafification and taking direct limits. So I tried to prove the following to help my understanding:

Let $f:X \rightarrow Y$ be a map of (locally) ringed topological spaces. Let $\mathcal{G}$ be an $\mathcal{O}_Y$ module. Show that $f^{-1}\mathcal{G}$ is an $f^{-1}\mathcal{O}_Y$ module.

My attempt:

Given any $U \in X$ open, we know that the inverse image presheaf $f^{-1,pre}\mathcal{G}(U)$ is given by lim$_{V \supseteq f(U)}\mathcal{G}(V)$. So any element in the sheafification of this will be given by a function $s:U\rightarrow \bigcup_{p \in U}f^{-1,pre}\mathcal{G}_p$.

Now given that stalks on the inverse image presheaf $f^{-1,pre}\mathcal G_p$ are isomorphic to stalks on $\mathcal G_{f(p)}$, we get that any element of the sheafification is given by a function $s: U \rightarrow \bigcup _{p \in U}\mathcal{G}_{f(p)}$. Similarly, any element in $f^{-1}\mathcal{O}_Y(U)$ is given by a function $r:U \rightarrow \bigcup_{p \in U} \mathcal{O}_{Y,f(p)}$.

So we can define the module structure as $(r \cdot s)(p) = r(p)\cdot s(p)$ where the multiplication on the right is given by the module structure of $\mathcal{G}_{f(p)}$ as an $\mathcal{O}_{Y,f(p)}$ module induced by the $\mathcal{O}_Y$-module structure on $\mathcal G$.

Since we are using the induced module structure on stalks, this does indeed define a module and we only need to check that the module structure defined respects restrictions. But $(rs)|_{V}(p) = r(p)s(p)|_V = r|_V(p) = s|_V(p)$ for any $V \subseteq U$ open. And we are done.

My questions:

Is the module structure and reasoning for why it defines a module structure correct?

I'm a bit unsure about how to explicitly write out the restriction maps and so would appreciate any explanation on how the restriction map explicitly works (viewed as restrictions of the functions r,s - if my understanding till there was fine)

Is there anything else that needs to be checked aside from the module structure and restriction maps that I am missing.

Thanks

答:
If $f:X \rightarrow Y$ is a map of topological spaces and if $A$ is a sheaf of rings on $Y$ you define the presheaf on $X$:

$$F_A(U):=lim_{f(U) \subseteq V}A(V).$$

Since $A(V)$ is a ring for all $V$ and since $F_A(U)$ is a direct limit of rings it follows $F_A(U)$ is a presheaf of rings on $X$. It follows $F_A^+:=f^{-1}(A)$ is a sheaf of rings on $X$. Hence the inverse image of a sheaf of rings is a sheaf of rings. If $f: X \rightarrow Y$ is a map of schemes there is a canonical map of sheaves of rings

$$ f^{\#}: f^{-1}(\mathcal{O}_Y) \rightarrow \mathcal{O}_X.$$

If $E$ is a sheaf of $\mathcal{O}_Y$-modules it follows $f^{-1}(E)$ is a sheaf of $f^{-1}(\mathcal{O}_Y)$-modules:

Since $E(V)$ is an $\mathcal{O}_Y(V)$-module for all $V$ it follows
$lim_{f(U) \subseteq V} E(V)$ is a $lim_{f(U)\subseteq V}\mathcal{O}_Y(V)$-module for all $U$,
hence $f^{-1}(E)(U)$  is an $f^{-1}(\mathcal{O}_Y)(U)$-module for all $U$.
It follows that $f^{-1}(E)$ is an $f^{-1}(\mathcal{O}_Y)$-module.
You define the following presheaf

$$ F_{f^*E}(U):= \mathcal{O}_X(U)\otimes_{f^{-1}(\mathcal{O}_Y(U)}f^{-1}(E)(U)$$

and define

$$ f^*E:=F^+_{f^*E}:=\mathcal{O}_X \otimes_{f^{-1}(\mathcal{O}_Y)} f^{-1}(E)$$

which is canonically a sheaf of $\mathcal{O}_X$-modules.

Question:   "1. Is the module structure and reasoning for why it defines a module structure correct? 2. I'm a bit unsure about how to explicitly write out the restriction maps and so would appreciate any explanation on how the restriction map explicitly works (viewed as restrictions of the functions r,s - if my understanding till there was fine)
3. Is there anything else that needs to be checked aside from the module structure and restriction maps that I am missing."

Answer: Instead of working with stalks it is better to work with direct limits and tensor products. Then the module structure is "canonical".


问:
$p$ is prime and $a\in \mathbb Z$ Prove that this set is a complete residue system $\bmod p$
if $p$ is prime and $a\in \mathbb Z$ such that $p\nmid a$, Show that the set
$$ R=\{0,a,2a,3a,...,(p-1)a\}$$

Is a complete residue system modulo $p$.

This problem is from my textbook.

My attempt:

I think that the best way to prove it it's by contradiction, so let $x,y \in R$ such that :
$$x\ne y, x\equiv y \pmod p$$

and we gonna prove that it's impossible that $x,y\in R$ and $x\equiv y \pmod p$... But I don't know how to do it.

ContentLicense: CC BY-SA 4.0

Body:
Yes, $a$ is missing. Let $x=k_1a,y=k_2a$ where $k_i\in\{0,1,2,...,p-1\}$. Then $k_1-k_2\in\{-(p-1),...,-1,0,1,2,...,p-1\}$.

$p|(x-y)\implies p|(k_1-k_2)a$ and since $p\nmid a$, it must be that $p|(k_1-k_2)$. So $k_1-k_2=0$. Thus $x=y$.


问:
Uniform Distributions Ratio
Let $\xi$ is $U(0, 1)$, $\nu$ is $U(0, 1)$. What type of distribution is $g = \frac{\xi}{\nu}$? I have build logarithm of $g$, here is the plot but it seems like it is not normal distribution:
Distribution Hist

答:
Assming independence...

Consider the following system

$$\begin{cases}
g=\frac{\xi}{\nu} \\
u=\xi
\end{cases}$$

The Jacobian is $|J|=\frac{\xi}{g^2}$

thus your density is

$$f_G(g)=\frac{1}{2}\cdot\mathbb{1}_{[0;1)}(g)+\frac{1}{2g^2}\cdot\mathbb{1}_{[1;+\infty)}(g)$$

Title: Uniform Distributions Ratio


问:
How to show that $f(x)=x^2$ is continuous at $x=1$?
How to show that $f(x)=x^2$ is continuous at $x=1$?</p>

答: In his 1821 text "Cours d'Analyse", Cauchy defined continuity of $y=f(x)$ by requiring that an infinitesimal $x$-increment should necessarily produce an infinitesimal change in $y$.  According to this definition, if $f(x)=x^2$, then for $\alpha$ infinitesimal, the change in $y$ is precisely $f(x+\alpha)-f(x)=(x+\alpha)^2-x^2=(x+\alpha+x)(x+\alpha-x)=\alpha(2x+\alpha)$.  Since $2x+\alpha$ is finite, the product $\alpha(2x+\alpha)$ is infinitesimal. Therefore $f(x)=x^2$ is continuous by definition.</p>

ContentLicense: CC BY-SA 3.0


问:
Help with the proof of Zorn's Lemma: if $A$ and $B$ are conforming subsets, why must it be that $A\subseteq B$ or $B\subseteq A$?
Here we can find a simple proof of Zorn's Lemma, but I find something that I can't really understand. There is an statement that say: If A and B are conforming subsets of $X$ and $A\neq B$, then one of these two sets is an initial segment of the other.

If I take, by example, the set $X=\{0,1,2,3,4,5,6\}$, I can create two conforming subsets which satisfy the definition but aren't initial segment of each other, like $B=\{0,2,4,5\}$ and $C=\{0,2,3,4\}$.

This proof is similar to another used by Gorodentsev in his book Algebra I, so there is a thing that I am missing here. Sorry for my bad English and thanks in advantage.

Definition

We shall say that a subset $A$ of $X$ is conforming if the following two conditions hold:

The order $\le$ is a well order of the set A.

For every element $x \in A$, we have $x = f(P(A, x))$, where $f$ maps the initial segment to its strict upper bound x.

Definition

If $C$ is a chain in $X$ and $x\in C$, then we define
$P(C, x) = \{ y \in C: y < x\}$. A subset of a chain $C$ that has the form $P(C, x)$ is called an initial segment in $C$.

答:
Probably the error is here:

If I take, by example, the set $X=\{0,1,2,3,4,5,6\}$, I can create two conforming subsets which satisfy the definition but aren't initial segment of each other, like $B=\{0,2,4,5\}$ and $C=\{0,2,3,4\}$.

You need to define the function $f$ first, which assigns a strict upper bound to a set, and then show your set satisfies $x = f(P(B,x))$ for all $x\in B$ to say $B$ is conforming. And similarly for $C$.

That is,
$$\begin{align}
&\text{for } B & &\qquad \text{for } C \\
& f(\{\}) & = 0 & \qquad \text{same} \\
& f(\{0\}) & = 2 & \qquad \text{same} \\
& \color{red}{f(\{0,2\})} & = \color{red}4 & \qquad \color{red}{f(\{0,2\})} & = \color{red}3 \\
& f(\{0,2,4\}) & = 5 & \qquad f(\{0,2,3\}) & = 4\\
\end{align}$$

The third row shows at least one of $B, C$ can not be 'conforming'.


问:
Accumulation points of $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$
I'm having some trouble finding the accumulation points o set $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$. I know that it only has one accumulation point (which is $e$), but I don't know the proof.

If anyone could help me, it would be great.

Thank you very much.

答: Using the definition of an accumulation point, you can see that in any neighbourhood of an accumulation point there is an infinite number of elements of the set. In this case, this means that there are subsequences of $(1+1/n)^n$ converging to any accumulation point of this set. Since the limit exists, all subsequences converge to the same number ($e$) and therefore...


问:
Accessing earlier values in odeint
I'm having some trouble using odeint function from scipy. I'm translating a discrete system into a continuous one, but some equation in the discrete model requires that I access to the previous value of a variable that I'm currently integrating. How could I translate this behaviour?

```<code>import numpy as np
days_of_prediction = 15
N = 100
discrete_S0 = np.zeros((days_of_prediction, 1))
discrete_I0 = np.zeros((days_of_prediction, 1))
discrete_Q0 = np.zeros((days_of_prediction, 1))
discrete_H0 = np.zeros((days_of_prediction, 1))
discrete_D0 = np.zeros((days_of_prediction, 1))
discrete_S0[0] = 99
discrete_I0[0] = 1
discrete_Q0[0] = 0
discrete_H0[0] = 0
discrete_D0[0] = 0
v=0.1
alpha = 0.3
gamma = 1/21
psi = 0.2
k_h=0.1
k_q=0.1
eta_h=0.3
eta_q=0.3
 for t in range(days_of_prediction - 1):
 discrete_S0[t + 1] = discrete_S0[t] - v * discrete_S0[t] * discrete_I0[t] / (N - discrete_Q0[t] - discrete_H0[t] - discrete_D0[t])
discrete_I0[t + 1] = discrete_I0[t] + v * discrete_S0[t] * discrete_I0[t] / (N - discrete_Q0[t] - discrete_H0[t] - discrete_D0[t]) - gamma * discrete_I0[t] - alpha * discrete_I0[t] - psi * discrete_I0[t]
discrete_Q0[t + 1] = discrete_Q0[t] + alpha * discrete_I0[t] - eta_q * discrete_Q0[t] - k_h *discrete_Q0[t] + k_q * discrete_H0[t]
discrete_H0[t + 1] = discrete_H0[t] + psi * discrete_I0[t] - eta_h * discrete_H0[t] + k_h * discrete_Q0[t] - k_q * discrete_H0[t] - zeta * discrete_H0[t]
discrete_R0[t + 1] = discrete_R0[t] + eta_q * discrete_Q0[t] + eta_h * discrete_H0[t]
</code>```

I've posted a snippet of the code, the problem is with the denominator of the first two equations. Thanks in advance.

答:
From the interaction terms you can directly read off the ODE system
\begin{align}
\dot S &= \nu\frac{S·I}{N-H-Q-D}
\\
\dot I &=-\nu·\frac{S·I}{N-H-Q-D} - \gamma·I - \alpha·I - \psi·I
\\
\dot Q &= \alpha·I - \eta_q·Q - \kappa_h·Q + \kappa_q·H
\\
\dot H &=  \psi·I - \eta_h·H + \kappa_h·Q -  \kappa_q·H - \zeta·H
\\
\dot R &=  \eta_q·Q + \eta_h·H
\end{align}
If you now solve that with the Euler method with step size $1$, you get the discrete method back. However, if you simualate that with any other method, that is, get closer to the exact ODE solution, the result might be distinctly different. Compare Question about Euler's Method and the SIR epidemic model using a spreadsheet for a similar question (only backwards, from ODE to large-step Euler) on the basic Zombie Apocalypse model.

Title: Accessing earlier values in odeint


问:
How to use axioms for a proof in axiomatic set theory?
I am learning set theory and find it confusing sometimes to understand how to use an axiom in order to solve an exercise. My main problem is beginning the proof and understanding how to proceed.

Example:
Let X be a set. Show that { {y} : y ∈ x} is a set.
The way I saw how to solve this question was first to use the power set axiom, and say that {y} (singleton set) is a subset of X. I do not think this is the right way to go about the question and was wondering if anyone could explain/ give advice on how to solve these types of exercises with the axioms in set theory. Also, someone said that using the axiom of separation is actually another way to solve this question?

Thanks!

答:
The axioms tell us the allowed ways to "make"/define new sets from given sets, and here we're just given $x$. This has elements and a set like $\{y\}$ is a subset of $x$ so it is logical to use the power set axiom, and then filter out the singletons. We can use the formula

$$\phi(y):= (\exists z: z \in y) \land \left(\forall z: \forall z': ((z \in y) \land (z' \in y)) \to (z=z')\right)$$

to state that the set $y$ has exactly one element.

Then we know $\mathscr{P}(x)$ is a set and we can justify the existence of your set by the axiom of separation applied to the power set and $\phi$ from above:

$$\{\{y\}: y \in x\} = \{y \in \mathscr{P}(x) \mid \phi(y)\}$$

Alternatively for each set $y$ we can define $\{y\}$ by the pairing axiom applied to $y$ and $y$... and we know from the axiom of extentionality that $\{y\}$ is uniquely defined from $y$. So then we can apply the axiom of replacement with the "class function" $y \to \{y\}$ with domain the given set $x$ to justify that indeed
$\{\{y\}: y \in x\}$ is a set as well.

Afterthought: we could also have used $$\psi(y,x) = \exists z: (z \in x) \land ( y=\{z\})$$

and separation (and also pairing, really) in the first approach. Many ways lead to Rome.


问:
Prove that $\exists a,g,h\in G\colon h=aga^{-1}, g\neq h ,gh=hg$ in a finite non-abelian group $G$.
Let $G$ be a finite and non-abelian group.  How do I prove the following statement? $$\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.$$

Thanks in advance.

答:
Here is a slight variation of the other two answers.

The idea is still to use the observation that we can assume all proper subgroups are abelian.

Note now that it is sufficient to find a proper normal subgroup, which is not central (since it is proper, it will be abelian, so we just need two distinct conjugate element in it, which we get since it is normal and not central).

To show that any minimal non-abelian group (minimal non-abelian means precisely that all proper subgroups are abelian and the group itself is not abelian) has such a subgroup, let us note the following:

Any proper subgroup of any quotient of a minimal non-abelian group is abelian (easy exercise).

A minimal non-abelian group is not simple (I could not find a really elementary proof of this, but it follows easily from observing that any non-normal Sylow subgroup will be central in its normalizer and applying Burnside's transfer theorem).

Finally, since the center is never maximal, we can find our proper normal non-central subgroup by pulling back a proper non-trivial normal subgroup from the quotient by the center.


答: Since $\log 1/\nu$ and $\log 1/\xi$ are exponentially distributed, your plot, $\log g$, follows the Difference Between Exponential Distributions.

Title: Uniform Distributions Ratio

Body:
Let $\xi$ is $U(0, 1)$, $\nu$ is $U(0, 1)$. What type of distribution is $g = \frac{\xi}{\nu}$? I have build logarithm of $g$, here is the plot but it seems like it is not normal distribution:
Distribution Hist


问:
Golden ratio rectangles
I'm designing a layout and I would like to use four golden ratio rectangles. The total width of the layout is 960px. How do I find the height (x)? Below is a diagram of the layout.

答:
golden ratio is

(a+b)/a = a/b

so assuming that a is width, and b is height, you have:
$$
(960 + h)/960 = 960/h \\
960^2 = 960h + h^2
$$
a simple quadratic equation to solve, and the result is
$$
h = 480*(\sqrt 5-1) \approx 593.3
$$

Title: Golden ratio rectangles


问:
Evaluating a long product using rules of product (Big PI)
$$T= \prod_{i=2}^{N} \left ( i^3 - 1 \right )  ~~ \text{Find T in terms of N}$$

This seems like a super easy task but somehow I cannot figure it out. My try was to convert it to a sum of logs, denote

$$ \prod_{i=2}^{N} \left ( i^3 - 1 \right ) = T \\ \lg \left (\prod_{i=2}^{N} \left ( i^3 - 1 \right ) \right) = \lg (T) \\ \sum_{i=2}^{N} \lg(i^3 -1) = \lg(T)$$

But here I am stuck as $i^3 -1$ does not factor nicely, so it is a dead-end.. (I am aware that $i^2 -1$ would word nice actually, but here it does not work).

I know that:

$$\prod_{i=2}^{N}   i^3 = 2^3 \cdot 3^3 \dots \cdot N^3 = \left (N! \right)^3$$

And so I just need to "add $+1$" to each term and then find the value, from a property of multiplication:

$$ (A-1)\cdot B = A \cdot B - B$$

And so, we add $+1$ to the $2^3 -1$ term, then subtract the rest:

$$ T = 2^3 \cdot (3^3 -1) \dots \cdot (N^3-1)  \\ ~~ - (3^3 -1) \dots (N^3-1) $$

We do the same for $3^3-1$ and so on... and we get this 'triangular' shape:

$$ T = 2^3 \cdot 3^3 \dots \cdot N^3  \\ ~~ - (3^3 -1) \dots (N^3-1) \\ ~~~ - 2^3(4^3-1)\dots(N^3-1) \\ ~~~~ \vdots$$

But again, this seems like a dead end, and it is hard to evaluate by itself!

Any help would be appreciated! Thank you.

答:
This is sequence $A158620$ in $OEIS$. There is an approximation  Vaclav Kotesovec gave in $2015$ which is
$$T_n\sim \frac{2\sqrt{2 \pi }}{3}   \cosh \left(\frac{\sqrt{3\pi}  }{2} \right)\,n^{3 n+\frac{3}{2}}\, e^{-3 n} $$

Generating the values for $2 \leq n \leq 100$ (remember that $T_{100}\sim 6.58\times 10^{472}$ !), I made a quick and dirty linear regression
$$\log(T_n)=a+b\, n+c \log(n)+d\, n \log(n)$$

The results are reported below
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 a & +2.77075 & 0.00562 & \{+2.75958,+2.78191\} \\
 b & -2.98305 & 0.00105 & \{-2.98513,-2.98097\} \\
 c & +1.37264 & 0.00499 & \{+1.36274,+1.38254\} \\
 d & +2.99710 & 0.00019 & \{+2.99672,+2.99747\} \\
\end{array}$$

Forcing $d=3$ now shows that $b$ is extremely close to $-3$. So using $b=-3$ and $d=3$ leads to
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 a & 2.64594 & 0.00727 & \{2.63151,2.66037\} \\
 c & 1.47615 & 0.00193 & \{1.47232,1.47997\}
\end{array}$$ Now $c\sim \frac 32$ and this leads closer and closer to Vaclav Kotesovec's approximation.

Edit

Using a CAS
$$T_n=\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{3 \pi }\Gamma (n) \Gamma \left(n+\frac{3+i \sqrt{3}}{2}\right) \Gamma
   \left(n+\frac{3-i \sqrt{3}}{2}\right)$$ where appears, as natural, the product of three gamma functions.

Using now Stirling  approximation (any other could be possible) of the gamma function, we have
$$T_n=\frac{2\sqrt{2 \pi }}{3}   \cosh \left(\frac{\sqrt{3\pi}  }{2} \right)\,n^{3 n+\frac{3}{2}}\, e^{-3 n+f(n)} $$ where $$f(n)=\frac{1}{4 n}+\frac{1}{2 n^2}-\frac{61}{120 n^3}+\frac{1}{4 n^4}+\frac{43}{420
   n^5}-\frac{1}{3 n^6}+\frac{139}{560
   n^7}+O\left(\frac{1}{n^8}\right)$$

This truncated expression gives $T_2=6.98683$ (instead of $7$) and $T_3=181.979$ (instead of $182$).

Now, since
$$U_n=\prod_{i=2}^n i^3=\Big[\Gamma(n+1)\Big]^3$$
$$\prod_{i=2}^{n} \left ( 1 - \frac{1}{i^3} \right )=\frac {T_n}{U_n}\sim\frac{\cosh \left(\frac{\sqrt{3} \pi }{2}\right)}{3 \pi } \exp\Bigg[\frac{1}{2 n^2}-\frac{1}{2 n^3}+\frac{1}{4 n^4}+O\left(\frac{1}{n^5}\right) \Bigg]$$


Title: Accumulation points of $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$

Body:
I'm having some trouble finding the accumulation points o set $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$. I know that it only has one accumulation point (which is $e$), but I don't know the proof.

If anyone could help me, it would be great.

Thank you very much.

Body:
$(1+1/n)^n
\lt e
\lt (1+1/n)^{n+1}
$
so
$0 \lt e-(1+1/n)^n
\lt (1+1/n)^{n+1}-(1+1/n)^n
= (1+1/n)^n((1+1/n)-1)
= (1+1/n)^n/n
\lt e/n
$.


问:
Functions in two variables satisfying $f_{xy} f = f_x f_y$
What is the family of analytic functions $f : \mathbb{R}^2 \to \mathbb{R}$ which satisfy the condition$f_{xy} f = f_{x} f_{y}$? Is there any interesting significance to this condition?

This holds for any   function of the form $f(x,y) = g(x)h(y)$. Are there any other possibilities?

The condition is equivalent to saying that $\frac{\mathrm{d}}{\mathrm{d}x} \log (f) = \frac{\mathrm{d}}{\mathrm{d}x} \log (f_y)$.

答:
Per your last equation, it's also equivalent to $$\frac{d}{dx}[\log(f_y)-\log(f)]=\frac{d}{dx}\log(f_y/f) =0. $$ Thus $\log(f_y/f)$ is independent of $x$, i.e., $f_y/f=\frac{\partial}{\partial y}\log f$ is a function of $y$ alone. This implies that $\log f(x,y)=G(x)+H(y)$ for functions $G,H$, which is equivalent to $f(x,y)$ being separable.

A more symmetric presentation: Recalling the quotient rule, the relation $f_x f_y = f_{xy}f$ implies

$$(\log f)_{xy}=(f_x/f)_y=\frac{f_{xy}f-f_x f_y}{f^2}=0$$ which again implies $\log f(x,y)=G(x)+H(y)$.


问:
Proving $[A:B]=[A^f:B^f][A_f:B_f]$
I'm having a hard time showing an equality of indices holds. This is exercise 1.44 from Lang's Algebra.

Suppose $f\colon A\to A'$ is a homomorphism of abelian groups, and $B\leq A$. Denote by $A^f$ and $A_f$ the image and kernel of $f$ in $A$ respectively, and similarly for $B^f$ and $B_f$. Show that $[A:B]=[A^f:B^f][A_f:B_f]$ in the sense that if two of these three indices are finite, so is the third, and the stated equality holds.

My initial thought was to compose a bijection from $A/B$ to $A^f/B^f\times A_f/B_f$. I tried to map a coset $aB$ to $(f(a)B^f,\underline{\hspace{.75cm}})$ but couldn't think of a nice candidate for the right coordinate.

Does anyone have suggestions on the right approach here? Thanks for your input.

答:
Let $P$ be a representant system for cosets of $B^f$ in $A^f$. Let $Q$ be a representant system for cosets of $B_f$ in $A_f$. For every $p\in P$ choose an element $a_p\in A$ such that $f(a_p)=p$. We will prove that the indexed family $(a_p+q)_{(p,q)\in P\times Q}$ is a representant system for cosets of $B$ in $A$.

For $x\in A$ then $f(x)\in A^f$, so there is a $p\in P$ such that $f(x)\in p+B^f$, so $f(x)-f(a_p)\in B^f$, so there is a $b\in B$ such that $f(x)-f(a_p)=f(b)$, so $x-a_p-b\in A_f$, so there is a $q\in Q$ such that $x-a_p-b\in q+B$, so $x\in a_p+q+B$.

Also, if $a_p+q+B=a_{p'}+q'+B$ then $a_p+q-a_{p'}-q'\in B$, so $f(a_p)+f(q)-f(a_{p'})-f(q')\in B^f$, so $p-p'\in B^f$, so $p=p'$, so $q-q'\in B$, but $q-q'\in A_f$, so $q-q'\in B_f$, so $q=q'$.

Title: Proving $[A:B]=[A^f:B^f][A_f:B_f]$


问:
Is the Cardinality of all Lines in the Extended Euclidean Plane $\mathfrak{c}$?
Consider the surjection $[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\rightarrow L$ (such that $L$ is the set of all lines determined by the binary operation $*$ between a point and an angle). Therefore, $\left|L\right|\leq\left|[0,2\pi)*\left(\mathbb{R}^2\cup\ell_\infty\right)\right|=\mathfrak{c}*(\mathfrak{c}^2+\mathfrak{c})=\mathfrak{c}*\mathfrak{c}$. Is $\mathfrak{c}*\mathfrak{c}=\mathfrak{c}$? How would I show this?

Similar Questions:

Cardinality of the set of all straight lines in $\mathbb R^2$

Is the set of points or the set of lines on a plane "larger"?

What is the Cardinality of the Extended Euclidean Plane?

答:
I will answer here to your comment question. Again, there are general results that allow us to directly proof that $2^\mathfrak{c}-\mathfrak{c}$ is $2^\mathfrak{c}$, but their proof is not easy at all. So I've been thinking for a long time for a constructive proof of this particular case and I came to this, but there might be some failure on my proof. In that case let me know.

To prove this it suffices to prove that there is a bijection between the sets $2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}$ and $2^\mathbb{R}$. Proving that $|2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}|\leq|2^\mathbb{R}|$ is direct, so let's focus just on the other one.

Consider the following map $f:2^\mathbb{R}\longrightarrow2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}$: $$S\longmapsto f(S):=\begin{cases}S &\text{ if } S \text{ is infinite} \\ S\cup \{2·s_k\} &\text{ if } S=\{s_1,\ldots,s_k\} \text{ and } s_1<\ldots <s_k\end{cases}$$

Our claim is that $f$ is injective.
To check injectivity let $S_1,S_2\in 2^\mathbb{R}$, $S_1\neq S_2$:

If $S_1,S_2$ are infinite then $f(S_1)=S_1\neq S_2=f(S_2)$.

If $S_1,S_2$ have a distinct number of elements then $f(S_1)$ and $f(S_2)$ also do, so $f(S_1)\neq f(S_2)$

If $S_1,S_2$ have both $k$ elements, let $S_1=\{a_1,\ldots,a_k\}$ and $S_2=\{b_1,\ldots,b_k\}$ with $a_1<\ldots < a_k$ and $b_1<\ldots < b_k$. Because $S_1\neq S_2$ we can choose $a_i\in S_1\setminus S_2$. If $a_i=a_k$ then $b_k\neq a_k$. If $b_k<a_k$, we have that $\max f(S_1)=2\cdot a_k>2\cdot b_k=\max f(S_2)$, hence $f(S_1)\neq f(S_2)$. If $b_k>a_k$ we also have that $f(S_1)\neq f(S_2)$ with the same argument. In other case, $a_i\neq a_k$. If $a_i= 2· b_k$, then $2·a_k\geq a_k>a_i=2·b_k$ and again $\max f(S_1)>\max f(S_2)$. If $a_i\neq 2· b_k$ then $a_i\in S_1\setminus(S_2\cup\{2 · b_k\})=S_1\setminus f(S_2)\subset f(S_1)\setminus f(S_2)$, so $f(S_1)\neq f(S_2)$.

In any of the cases, $f(S_1)\neq f(S_2)$, so $f$ is injective, which proves that $|2^\mathbb{R}|\leq|2^\mathbb{R}\setminus\{\{x\}:x\in\mathbb{R}\}|$, and so $2^\mathfrak{c}-\mathfrak{c}=2^\mathfrak{c}$.


问:
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.

Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator

答: Here is a geometric way to show that from one of my friends. I don't know if it counts as a proof but it looks beautiful.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: The fact that the sum of the first n odd numbers is n squared. Also, not just this fact, but the nice visual “wrapping” proof of it (as opposed to the induction proof).</p>

LastEditorDisplayName: anon


答:
From the elementary theory of polynomials we know that the cubic polynomial $ax^3+bx^2+cx+d$ can be factored as

$$ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3),$$

where $x_k$ ($k=1,2,3$) are the roots of the general cubic equation

$$ax^3+bx^2+cx+d=0.$$

If we use the substitution $x=t+\frac{13}{3}$, the given equation
$$
\begin{equation*}
x^{3}-13x^{2}+32x+20=0\tag{1}
\end{equation*}
$$
is transformed into the reduced cubic equation
$$
\begin{equation*}
t^{3}+pt+q=0,\qquad p=-\frac{73}{3},q=-\frac{110}{27},\tag{2}
\end{equation*}
$$
a solution of which is$^1$
$$
\begin{eqnarray*}
t_{1} &=&\left( \frac{-q+\sqrt{\Delta }}{2}\right) ^{1/3}+\left( \frac{-q-
\sqrt{\Delta }}{2}\right) ^{1/3}, \qquad \Delta  &=&q^{2}+\frac{4p^{3}}{27}.\tag{3}
\end{eqnarray*}
$$
When the discriminant $\Delta <0$ the three solutions of the original cubic $(1)$ are real. By using complex numbers we can write them in the form

$$
\begin{equation*}
x_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2}
\sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\pi (k-1)}{3}\right) -\frac{b}{3}
,
\end{equation*}\tag{4}
$$
where $k=1,2,3$, and $b=-13$ is the coefficient of $x^{2}$ in $(1)$. Since $$\Delta =-\dfrac{57\,184}{27}<0,$$ we have:
$$
\begin{eqnarray*}
x_{1} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) \right) +\frac{13}{3} \\
&\approx &9.347\,9, \\
&& \\
x_{2} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{2\pi }{3}\right) +\frac{13}{3} \\
&\approx &-0.513\,6, \\
&& \\
x_{3} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{4\pi }{3}\right) +\frac{13}{3} \\
&\approx &4.165\,7.
\end{eqnarray*}
$$

Therefore the factorization of $(1)$ is
$$
\begin{equation*}
x^{3}-13x^{2}+32x+20=(x-x_1)(x-x_2)(x-x_3).\tag{5}
\end{equation*}
$$

--

$^1$ A deduction of $(3)$ and $(4)$ can be found in this blog post of mine, in Portuguese.

Title: factorise, $x^3-13x^2+32x+20$

Body:
factorise, $x^3-13x^2+32x+20$

Let, $f(x)=x^3-13x^2+32x+20$

$f(x)=x(x^2-13x+30)+2x+20$

$f(x)=x(x-3)(x-10)+2x+20$

$f(-1)\lt 0$, $f(0)\gt 0$, which shows there is a root between $x=-1$ and $x=0$

$f(4)\gt 0$, $f(5)\lt 0$, which shows there is a root between $x=4$ and $x=5$

$f(9)\lt 0$, $f(10)\gt 0$, which shows there is a root between $x=9$ and $x=10$


问:
Solving the diophantine equation $y^{2}=x^{3}-2$
It is known that the diophantine equation $y^{2}=x^{3}-2$ has only one positive integer solution $(x,y)=(3,5)$. The proof of it can be read from the book "About Indeterminate Equation" (in Chinese, by Ke Zhao and Sun Qi). But the method I have known of solving this problem is algebraic number theory.

My question:

Does there exists an elementary method for solving the diophantine equation $y^{2}=x^{3}-2$?

I have gotten an elementary method by the assumption that "$x$" is a prime. That is, we can get a conclusion from the equation that $3\mid x$. Since $x$ is a prime we get the only solution $(x,y)=(3,5)$. But if $x$ is not a prime, the answer seems to be difficult.

Thank you for your help.

答:
A completely elementary proof can be found on page 561 of the Nov 2012 edition of The Mathematical Gazette, where a descent mechanism first used by Stan Dolan in the March 2012 edition is adapted (as per his challenge to the “interested reader”) to solve both of Fermat’s “elliptic curve” theorems. The method uses math which was clearly available in Fermat’s time, and in particular to Fermat himself.

Lemma.
Let $a$ and $b$ be coprime integers, and let $m$ and $n$ be positive integers such that $a^2+2b^2=mn$. Then there are coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$. Furthermore, for any such choice of $r$ and $s$, there are coprime integers $t$ and $u$ such that $a=rt-2su$, $b=ru+st$, and $n=t^2+2u^2$ divides $bt-au$.

Proof.
Assume the theorem is false, and let $m$ be a minimal counterexample. Evidently $m > 1$ since the theorem is trivially true for $m=1$.

Note that $b$ is coprime to $m$. Let $A$ be an integer such that $Ab \equiv a\!\pmod{m}$, chosen so that $\tfrac{-m}{2} < A \le \tfrac{m}{2}$. Then $A^2+2 = lm$ for some positive integer $l < m$. Clearly $l$ cannot be a smaller counterexample than $m$, and so there exist coprime integers $r$ and $s$ such that $m=r^2+2s^2$ divides $br-as$.

Let $t = \tfrac{ar+2bs}{m}$ and $u=\tfrac{br-as}{m}$. Direct calculation confirms the equations for $a$, $b$, and $n$. From $n=t^2+2u^2$, we deduce that $t$ is an integer because $u$ is an integer, and $t$ and $u$ are coprime because $\gcd(t,u)$ divides both $a$ and $b$. Finally, note that $n$ divides $bt-au=sn$.

Hence $m$ is not a counterexample, contradicting the original assumption. $\blacksquare$

Corollary.
Let $a$ and $b$ be coprime integers with $m$ and integer such that $m^3=a^2+2b^2$. Then there are coprime integers $r$ and $s$ such that $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$.

Proof.
Evidently $m$ is odd since $a^2+2b^2$ is at most singly even. And $a$ and $m$ must be coprime. Using the theorem, we have $m=r^2+2s^2$ and $m^2=t^2+2u^2$. Then $m$ divides $a(ur-ts)=t(br-as)-r(bt-au)$, and therefore $m \mid (ur-ts)$. The theorem can then be reapplied with $a$ and $b$ replaced by $t$ and $u$. Repeating the process, we eventually obtain integers $p$ and $q$ such that $p^2+2q^2=1$. The only solution is $q=0$ and $p=\pm1$. Ascending the path back to $a$ and $b$ (reversing signs along the way, if necessary) yields $a=r(r^2-6s^2)$ and $b=s(3r^2-2s^2)$, as claimed. $\blacksquare$

Theorem.
The Diophantine equation $X^3 = Y^2+2$ has only one integer solution, namely $(x,y) = (3, \pm 5)$.

Proof.
Evidently $y$ and $2$ are coprime. By the corollary, we must have $b=1=s(3r^2-2s^2)$ for integers $r$ and $s$. The only solutions are $(r,s)=(1,\pm 1)$. Hence $a=y=r(r^2-6s^2)=\pm 5$, so $(x,y)=(3,\pm 5)$. $\blacksquare$

I personally believe this finally puts to rest any questions of whether Fermat could have had a proof of these two claims.


问:
Calculate $\sum_{i=0}^\infty i\binom{i+10}{i}(\frac{1}{2})^i$
I'm not great with summation and its various techniques, so go easy on me. If someone can point me in the right direction I would be very grateful.

This is the sum in display mode: $$\sum_{i=0}^\infty i\binom{i+10}{i}\frac{1}{2^i}$$

By using the symmetry rule for binomial coefficients I end up with this:
$$\frac{1}{10!}\sum_{i=0}^\infty \frac{i}{2^i}\prod_{k=1}^{10}(k+i)$$ which doesn't seem to help.
I tried working something else on the binomial to get rid of i and I got:
$$11\sum_{i=0}^\infty \frac{1}{2^i}\binom{i+10}{i-1}$$
but yet again, I'm stuck.

I have a feeling that I have to somehow simplify the sum and then use the recursive method?

答:
Hint:

You may note that
$$
\sum_{i\ge0}i\binom{10+i}i x^i=
x\frac{d}{dx}\sum_{i\ge0}\binom{10+i}i x^i,
$$
reducing the problem to known series.

$$=x\frac{d}{dx}\sum_{i\ge0}\binom{-11}i (-x)^i=x\frac{d}{dx}\frac1{(1-x)^{11}}=\left.\frac{11x}{(1-x)^{12}}\right|_{x=\frac12}=22528.$$


问:
Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?

Determine if the set $A= \{(x,y) \mid |y|\leqslant|x|, x \ne 0\}$ is closed?

I know that $A$ is closed if $A^c$ is open. Now $A^c$ is just $A^c = \{(x,y) \mid |y| >|x| \}$. To show that $A^c$ is open I would have to show that for every $x \in A^c$ there is a disk $B(x,r) \subset A^c$, but I'm not sure how to approach this?

答: For any $x\neq 0$, we have $(x,0) \in A$, thus, $B =\mathbb R \times \{0\}\cap A = \mathbb R\setminus \{0\}\times \{0\}$, and $\overline B = \mathbb R \times \{0\}$ so $B$ is not closed. But, $\mathbb R \times \{0\}$ is closed, so $A$ must not be closed.


问:
Weiner Process Markovian property

Wiener process definition:

$W(t) \sim \mathcal{N}(0, \sigma\sqrt{t})$

$W(t)$ is Markovian

$W(t+t')-W(t)$ and $W(t)$ are independent.

I don't know of a case where $2$ is true but $3$ is not & vice versa. From Markovian property definition, the future at $t+t'$ should depend on $t$. So the definition itself should mean $3$ is true. What am I missing here? Can you give examples for both cases?

答:
3 does not have to hold in order for 2 to hold. What the Markov property says is that $W(t+t')-W(t)$, conditional on the value of $W(t)$, does not depend on $W(s)$ for any $s<t$. It can still depend on $W(t)$ itself, but in the specific case of the Wiener process it doesn't. Basically for the Wiener process, you move the same way regardless of where you start; it is spatially homogeneous. So knowing $W(t)=x$ just translates the distribution of $W(t+t')$ to be centered at $x$ instead of $0$.

For an example where 2 holds and 3 does not, consider basically any SDE with coefficients depending on position, for instance $dX_t=X_t dW_t$.

I am less confident about whether 3 implies 2. What does imply 2 is $W(t_3)-W(t_2)$ being independent of $W(t_2)-W(t_1)$ whenever $t_1<t_2<t_3$. Maybe you can massage 3 into this but I would have to play with it.

Title: Weiner Process Markovian property


问:
Where to learn algebraic analysis
I have been studying categories, sheaf cohomology and complex analysis (the basics since I know just a little). Then recently I tried to find out more about algebraic analysis and these microlocal stuffs, but I couldn't find any introductory material.

I know about the books and papers by Kashiwara and they are a little advanced for me. (However, I have never seen Foundations of algebraic analysis since this book is not so easily found.) Any material, lecture notes, etc are welcome.

答: [KKK] "Foundations of Algebraic analysis" is not easy for beginners, I highly recommend to read A. Kaneko's "Introductions to Hyperfunctions", and Morimoto "An introduction to Sato's Hyperfunctions". And there is another one "Fundamentals of Algebraic microlocal analysis" by Struppa and Kato, also very good. [SKK]/LNM 287 is another classical book, but very hard to read. Please try to search M. Sato, M. Kashiwara, P. Schapira, and else.</p>

Title: Where to learn algebraic analysis


问:
Proving a set X is dense in [0,1] equivalence relation
Let the relation in $\mathbb{R}: x \equiv y \ \mbox{mod} \ \mathbb{Z}$, when $x-y \in \mathbb{Z}$. For each $n \in \mathbb{N}$, let $x_n \in [0,1)$ such that $x_n \equiv \sqrt{n} \ \mbox{mod} \ \mathbb{Z}$. Prove $X = \{x_n : n \in \mathbb{N} \}$ is dense in $[0,1]$. I don't know where to start, but I have doubts about using the greatest integer.

答:
Suppose $I$ is an  interval  with $I\subseteq (0,1),$ and with  length $l(I)=(\sup I)-(\inf I)>0.$ Suppose $S$ is a finite subset of $[0,1]$ with $0\in S$ and $1\in S,$ such that if $s,s'$ are two $consecutive$ members of $S$ then $|s-s'|<l(I).$ Then we have $I\cap S\ne\emptyset.$

So take $k\in \Bbb N$ with $1/2k<l(I).$ For integer $j$ with $0\le j\le 2k+1,$ let $\sqrt {k^2+j}=k+D(j).$ Let $S=\{D(j):0\le j\le 2k+1\}.$


问:
A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$
We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.

We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$

Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{  This is what we need to prove}$$
$$
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing  and is bounded sufficient?

答:
Your proof of growth is presented in a weird backward way. You could write

$$c_{n}\ge c_{n-1}\iff\sqrt{c_{n}+2}\ge\sqrt{c_{n-1}+2}\iff c_{n+1}\ge c_n.$$ You need a base case, which is missing. We indeed have $\sqrt{2+\sqrt2}\ge\sqrt2$.

You can also condense in a single, non-inductive proof of

$$0\le c_n\le2\implies c_n\le c_{n+1}\le2$$

or equivalently

$$0\le x\le2\implies x\le\sqrt{x+2}\le\sqrt{2+2}$$
because
$$x^2-x-2\le0$$ for $x\in[-1,2]$.


问:
Pairing Between Alternating Forms and Antisymmetric Product.
Let $V$ be a vector space and $V^*$ be the dual of $V.$ For $\omega \in \wedge^k V^* \subseteq V^{* \otimes k} \cong \left (V^{\otimes k} \right )^*$ and $(v_1 \otimes v_2 \otimes \cdots \otimes v_n ) \in V^{\otimes k}$ we denote $\omega (v_1 \otimes v_2 \otimes \cdots \otimes v_k)$ by $\left \langle \omega, v_1 \otimes v_2 \otimes \cdots \otimes v_k \right \rangle.$

Proposition $:$ Let $\omega \in \wedge ^k V^*$ and $\eta \in \wedge^l V^*.$ Then for all $v_1,  v_2, \cdots,  v_{k+l}$ we have $$\left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle = \frac {k!\ l!} {(k+l)!} \sum_{\sigma \in S_{k,l}} \varepsilon (\sigma) \left \langle \omega, v_{\sigma (1)} \otimes v_{\sigma (2)} \otimes \cdots \otimes v_{\sigma (k)} \right \rangle \times \left \langle \eta, v_{\sigma (k+1)} \otimes v_{\sigma {(k+2)}} \otimes \cdots \otimes v_{\sigma {(k+l)}} \right \rangle$$ where $S_{k,l} : = \left \{\sigma \in S_{k+l}\ \bigg |\ \sigma (1) \lt \sigma (2) \lt \cdots \lt \sigma (k), \sigma (k+1) \lt \sigma (k+2) \lt \cdots \lt \sigma (k+l) \right \}$ is the set of all $(k,l)$-shuffles and $\varepsilon (\sigma) = \text {sign} (\sigma),$ $\sigma \in S_{k+l}.$

Proof $:$ For each $\sigma \in S_{k+l}$ let $\tilde {\sigma} \in S_{k,l}$ be the element such that $\left \{\sigma (1), \sigma (2), \cdots , \sigma (k) \right \} = \left \{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \right \}.$ Consequently $\left \{\sigma (k+1), \sigma (k+2), \cdots , \sigma (k+l) \right \} = \left \{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \right \}.$ Let $\sigma_1' \in S_{k+l}$ be such that $\sigma_1' (\tilde {\sigma} (j)) = \sigma (j),\ 1 \leq j \leq k$ and $\sigma_1' (\tilde {\sigma} (j)) = \tilde {\sigma} (j),\ k+1 \leq j \leq k+l$ and $\sigma_2' \in S_{k+l}$ be such that $\sigma_2' (\tilde {\sigma} (j)) = \tilde {\sigma} (j),\ 1 \leq j \leq k$ and $\sigma_2' (\tilde {\sigma} (j)) = \sigma (j),\ k+1 \leq j \leq k+l.$ Then $\sigma = \sigma_1' \sigma_2' \tilde {\sigma}.$ Hence $\varepsilon (\sigma) = \varepsilon (\sigma_1') \varepsilon (\sigma_2') \varepsilon (\tilde {\sigma}).$

Now $$\begin{align*} \left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle & = \color {red} {\frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle  \omega \wedge \eta, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma (k+l)} \right \rangle} \\ & =  \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle  \omega, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma(k)} \right \rangle \times \left \langle  \eta, v_{\sigma (k+1)} \otimes v_{\sigma(k+2)} \otimes \cdots \otimes v_{\sigma(k+l)} \right \rangle \\ & = \color{red} {\frac {1} {(k+l)!} \sum\limits_{\substack {\tilde {\sigma} \in S_{k,l}} \\ \ {\sigma_1', \sigma_2'}} \varepsilon (\tilde {\sigma})\ \varepsilon (\sigma_1') \left \langle  \omega, v_{\sigma_1' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_1' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_1' \left (\tilde {\sigma} (k) \right )} \right \rangle \times \varepsilon (\sigma_2') \left \langle  \eta, v_{\sigma_2' \left (\tilde {\sigma} (k+1) \right )} \otimes v_{\sigma_2' \left (\tilde {\sigma} (k+2) \right )} \otimes \cdots \otimes v_{\sigma_2' \left ( \tilde {\sigma} (k+l) \right )} \right \rangle} \\ & = \color{red} {\frac {k!\ l!} {(k+l)!} \sum\limits_{\sigma \in S_{k,l}} \varepsilon (\sigma) \left \langle  \omega, v_{\sigma (1)} \otimes v_{\sigma(2)} \otimes \cdots \otimes v_{\sigma(k)} \right \rangle \times \left \langle  \eta, v_{\sigma (k+1)} \otimes v_{\sigma(k+2)} \otimes \cdots \otimes v_{\sigma(k+l)} \right \rangle}   \end{align*}$$

This completes the proof.

Can anybody please explain the equalities in red? I am having hard time understanding these equalities. Thanks for your time.

EDIT $:$ What I know is that $$\omega \wedge \eta = \text {Alt}_{k+l} (\omega \otimes \eta) = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \rho (\sigma) (\omega \otimes \eta)$$ where $\rho(\sigma) : V^{*\otimes {(k+l)}} \longrightarrow V^{*\otimes {(k+l)}}$ is a linear isomorphism given by $$\rho (\sigma) (v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l}) : = v_{\sigma^{-1} (1)} \otimes v_{\sigma^{-1} (2)} \otimes \cdots \otimes v_{\sigma^{-1} (k+l)}.$$ So if $\omega = \omega_1 \otimes \omega_2 \otimes \cdots \otimes \omega_k$ and $\eta = \omega_{k+1} \otimes \omega_{k+2} \otimes \cdots \otimes \omega_{k+l}$ then it follows that $$\begin{align*} \left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \prod\limits_{j=1}^{k+l} \omega_{\sigma^{-1} (j)} (v_j) \\ & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \prod\limits_{j=1}^{k+l} \omega_{j} (v_{\sigma (j)}) \\ & = \frac {1} {(k+l)!} \sum\limits_{\sigma \in S_{k+l}} \varepsilon (\sigma) \left \langle \omega \otimes \eta, v_{\sigma (1)} \otimes v_{\sigma (2)} \otimes \cdots \otimes v_{\sigma (k+l)} \right \rangle  \end{align*}$$ which is not precisely the first equality. Also I have no idea how to get the third and the fourth equalities.

答:
Let $\omega := \omega_1 \otimes \omega_2 \otimes \cdots \otimes \omega_k \in \wedge^k V^*$ and $\eta := \omega_{k+1} \otimes \omega_{k+2} \otimes \cdots \otimes \omega_{k+l} \in \wedge^l V^*.$ Since $\text {Alt}$ is an idempotent linear operator it follows that $\text {Alt}_k (\omega) = \omega$ and $\text {Alt}_l (\eta) = \eta.$ This shows that $\text {Alt}_k (\omega) = \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k)$ and $\text {Alt}_l (\eta) = \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l}).$ Before going to prove the equalities we note that the third equality can easily be obtained from the second equality just by decomposing $\sigma$ as $\sigma_1' \sigma_2' \tilde {\sigma},$ where $\tilde {\sigma} \in S_{k,l},$ $\sigma_1'$ is a permutation of $\{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \}$ and $\sigma_2'$ is a permutation on $\{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \}.$ So for a given $\tilde {\sigma},$ the number of such $\sigma_1'$ and $\sigma_2'$ are respectively $k!$ and $l!.$ Another thing to note here is that the multi-linear map $\left ( \text {Alt} \circ \otimes \right )$ is alternating linear. So to obtain the fourth equality from the third one let us first analyze $F_{\sigma_1}: = \left \langle \omega, v_{\sigma_1' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_1' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_1' \left (\tilde {\sigma} (k) \right )} \right \rangle.$

Clearly $$F_{\sigma_1} = \left ( \left (\left (\text {Alt}_k \circ \otimes_k \right ) \circ \mu \left ({\tilde {\sigma}}^{-1} \sigma_1' \tilde {\sigma}, \cdot \right ) \right ) (\omega_1,\omega_2, \cdots, \omega_k) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right ) = \varepsilon \left ({\tilde {\sigma}}^{-1} \sigma_1' \tilde {\sigma} \right ) \left ( \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1,\omega_2, \cdots, \omega_k ) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right )$$ where for a given $\tau \in S_k$ we have $\mu (\tau,\cdot) (v_1,v_2, \cdots, v_k) = \left (v_{\tau^{-1} (1)}, v_{\tau^{-1} (2)}, \cdots, v_{\tau^{-1} (k)} \right ),$ $(v_1,v_2, \cdots, v_k) \in V^{\times k}.$ Hence for any $\tilde {\sigma} \in S_{k,l}$ and for any permutation $\sigma_1'$ of the set $\{\tilde {\sigma} (1), \tilde {\sigma} (2), \cdots, \tilde {\sigma} (k) \}$ we have $$\varepsilon \left (\sigma_1'  \right ) F_{\sigma_1} = \left ( \left ( \text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k ) \right ) \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right ).$$ Similarly if we define $G_{\sigma_2} := \left \langle \eta, v_{\sigma_2' \left (\tilde {\sigma} (1) \right )} \otimes v_{\sigma_2' \left (\tilde {\sigma} (2) \right )} \otimes \cdots \otimes v_{\sigma_2' \left (\tilde {\sigma} (k) \right )} \right \rangle$ then for any $\tilde {\sigma} \in S_{k,l}$ and for any permutation $\sigma_2'$ of the set $\{\tilde {\sigma} (k+1), \tilde {\sigma} (k+2), \cdots, \tilde {\sigma} (k+l) \}$ we have $$\varepsilon \left (\sigma_2' \right ) G_{\sigma_2} = \left ( \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l} ) \right )  \left (v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right ).$$ Having obtained all those things the third equality simplifies to $$\begin{align*} \frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left ( \left (\text {Alt}_k \circ \otimes_k \right ) (\omega_1, \omega_2, \cdots, \omega_k ) \right )  \left (v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (l)} \right ) \times \left ( \left (\text {Alt}_l \circ \otimes_l \right ) (\omega_{k+1}, \omega_{k+2}, \cdots, \omega_{k+l} ) \right ) \left (v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right ) \end{align*}$$ which further simplifies to $$\frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k)} \right \rangle \times \left \langle \eta, v_{\tilde {\sigma} (k+1)} \otimes v_{\tilde {\sigma} (k+2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle$$ which again simplifies to $$\frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega \otimes \eta, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle.$$ Hence $$\left \langle \omega \wedge \eta, v_1 \otimes v_2 \otimes \cdots \otimes v_{k+l} \right \rangle = \frac {k!\ l!} {(k+l)!} \sum\limits_{\tilde {\sigma} \in S_{k,l}} \varepsilon \left (\tilde {\sigma} \right ) \left \langle \omega \otimes \eta, v_{\tilde {\sigma} (1)} \otimes v_{\tilde {\sigma} (2)} \otimes \cdots \otimes v_{\tilde {\sigma} (k+l)} \right \rangle.$$


问:
intersection of analytic set : codimension?
Hello. Let $A$ and $B$ be analytic sets of pure dimension in an analytic set X of $\mathbb{C }^{n}$. Suppose $A$ is a local complete intersection (i.e. for every point $a$ of $A$, there exists $\Omega$ an open neighbourhood of $a$ such that  $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorphs distincts and $p = codim_{X}(A)$).
Then, if $A \cap B \neq \emptyset$ why do we have $codim_{X}(A \cap B) \leq codim_{X}(A) + codim_{X}(B)$?

I tried to solve this by induction on $dim(A)$; here is the proof I first made (I realised afterwards the step $*$ is false). It's trivial if $dim(A) = 1$. Then, if $dim(A) > 1$, I choose $a \in A \cap B$ and I wright $A \cap \Omega = \{x \in \Omega | f_{1}(x) = ... = f_{p}(x) = 0\}$ where $f_{i}$ are holomorph distincts and $p = codim_{X}(A)$ and $B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p'}(x) = 0\}$ where $g_{i}$ are holomorph and distincts. Since the problem is located in $a$, we will wright $dim_{a} = dim$ and the same for $codim$. Notice that I have $p' \leq codim_{X}(B) $. We can wright $A \cap B \cap \Omega = \{x \in \Omega | g_{1}(x) = ... = g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\}$ where $p'' \leq p'$ is such that there is no repetition between $g_{j}$ and $f_{i}$. Now let's have a look at : $A' := \{x \in \Omega | g_{p''}(x) = f_{1}(x) = ... = f_{p}(x) = 0\} \subset A$. We wright $B' := \{x \in \Omega | g_{1}(x) = ... = g_{p''-1}(x) = 0\}$. By looking$*$ the local irreductible components of $A$( which are all of dimensions $dim(A)$ since $A$ is of pure dimension) in $a$ we have $A'$ is of dimension $dim(A) - 1$ so of codimension $codim_{X}(A) + 1$. Thus, $A'$ is a local complete intersection and so by induction $codim_{X}(A \cap B) = codim_{X}(A' \cap B') \leq codim_{X}(A') + codim_{X}(B') = codim_{X}(A) + 1 + codim_{X}(B')$. But $1 + codim_{X}(B')$ doesn't necessary have to be $\leq codim_{X}(B)$ (imagine $B$ is an irreductible component of $B'$ in $a$).

$*$Here I use the (6.2) Theorem page 106 of this curse https://www-fourier.ujf-grenoble.fr/~demailly/manuscripts/agbook.pdf . But it dosen't work since $g_{p''}$ can be constant on some irreductible components of $A$ in $a$.

Thank you for your help.

I wish you a good day.

答:
I add this : my problem is the exercise 11.11. page 128 of this book : https://www-fourier.ujf-grenoble.fr/%7Edemailly/manuscripts/agbook.pdf . The dimension is defined page 98 of this same curse and the codimension in $X$ of $A$ at a point $a \in A$ is define by $codim_{X, a}(A) := dim_{a}X - dim_{a}A$. For my background, I readed the two first chapters of the book I link (and I made the most exercises of those two chapters except the one I'm asking the question for). I'm reading at complex geometry and I also have the agregation.

I hope it enhances the question (I tried to respect the meta read). I'm french so sorry for the eventuals mistakes.


问:
Integrability of Thomae function and value of its integral
I am trying to solve the following exercise (from Axler's Measure, Integration and Real Analysis) and I would like to have some help in finishing my proof.

"Define $f:[0,1]\to\mathbb{R}$ as follows:
$
  f(a) =
  \begin{cases}
                                   0 & \text{if $a$ is irrational} \\
                                   \frac{1}{n} & \text{if $a\in\mathbb{Q}$ and $n$ is the smallest positive integer such that $a=\frac{m}{n}$ for some integer $m$}
  \end{cases}
$.

Show that $f$ is Riemann integrable and compute $\int_{0}^{1}f$."

My solution: (NOTE: THIS SOLUTION IS WRONG, I HAVE WRITTEN A CORRECT ONE IN THE ANSWER TO THIS QUESTION BELOW)

We immediately note that for any partition $P$ of $[0,1]$ we have $L(f,P,[0,1])=\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f=\sum_{j=1}^{n}(x_j-x_{j-1})\cdot 0=0$ since in any sub-interval of $[0,1]$ there will be an irrational number.

Now we let $\varepsilon>0$ and consider the size of the set $D_{\varepsilon}:=\{x\in [0,1]:f(x)\geq \varepsilon\}\subset\{x\in [0,1]:f(x)\geq\frac{1}{n}\}=:D_{1/n}$, where $\varepsilon\geq\frac{1}{n},\ n>1+\frac{2}{\varepsilon}$ fixed (such an $n$ must exist, since $\frac{1}{n}\overset{n\to\infty}{\to}0$).

We note that if $\varepsilon>1$ we have $D_{\varepsilon}=\emptyset\Rightarrow \#(D_{\varepsilon})=0$,

if $\varepsilon = 1$ we have  $D_{\varepsilon}=\{0,1\}\Rightarrow \#(D_{\varepsilon})=2$,

if $\frac{1}{2}\leq\varepsilon\leq 1$ we have $D_{\varepsilon}\subset D_{1/2}=\{0,1,\frac{1}{2}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/2})=3$,

if $\frac{1}{3}\leq\varepsilon\leq \frac{1}{2}$ we have $D_{\varepsilon}\subset D_{1/3}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/3})=5$,

if $\frac{1}{4}\leq\varepsilon\leq \frac{1}{3}$ we have $D_{\varepsilon}\subset D_{1/4}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4}\}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/4})=7$,

if $\frac{1}{5}\leq\varepsilon\leq \frac{1}{4}$ we have $D_{\varepsilon}\subset D_{1/5}=\{0,1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5}\}\Rightarrow \#(D_{\varepsilon})\leq \#(D_{1/5})=11$,

$\dots$,

so we see that in each case $D_{1/N},\ (N\geq 2)$ contains $0,1$ and the $k-1$ elements $\frac{1}{k},...,\frac{k-1}{k}$ for $2\leq k\leq N$ (this is an overestimate for the size of $D_{1/N}$ since some of the elements in this set might be present also in $D_{\frac{1}{N-1}}$ - see for example $D_{1/4}$) hence an upper bound for the size of $D_{1/N}$ is $$2+\sum_{k=2}^{N}(k-1)=2+\sum_{k=1}^{N}k-\sum_{k=1}^{N}1=1+\frac{N(N+1)}{2}-(N-1)=\frac{N^2-N+4}{2}$$
which in turn implies that $\#(D_{\varepsilon})\leq \frac{N^2-N+4}{2}$.

So, we consider again now the aformentioned $n$ such that $\frac{1}{n}\leq\varepsilon$, set $M:=\frac{n^2-n+4}{2}$ and consider the partition $P_{\varepsilon}$ of $[0,1]$ such that
$$P_{\varepsilon}:=\{0,n_1-\frac{\varepsilon}{4M}, n_1+\frac{\varepsilon}{4M},\dots, n_M-\frac{\varepsilon}{4M}, n_M+\frac{\varepsilon}{4M},1\}.$$

We have that $$U(f,P_{\varepsilon},[0,1])\leq (n_1-\frac{\varepsilon}{4M})\cdot 1+\frac{\varepsilon}{2M}\cdot 1+(n_2-n_1-\frac{\varepsilon}{2M})\cdot\frac{1}{n}+\frac{\varepsilon}{2M}\cdot 1+(n_3-n_2-\frac{\varepsilon}{2M})\frac{1}{n} +\dots +\frac{\varepsilon}{2M}\cdot 1+ (1-n_M-\frac{\varepsilon}{4M})\cdot 1= M\cdot\frac{\varepsilon}{2M}+(1-n_M-(\frac{\varepsilon}{4M}+\frac{\varepsilon}{4M}+M\frac{\varepsilon}{2M}))\frac{1}{n}=\frac{\varepsilon}{2}+(1-n_M-\frac{\varepsilon}{2M}+\frac{\varepsilon}{2})\frac{1}{n}<\frac{\varepsilon}{2}+(1+\frac{\varepsilon}{2})\frac{1}{n}<\varepsilon$$ so $U(f,P_{\varepsilon},[0,1])-L(f,P_{\varepsilon},[0,1])=U(f,P_{\varepsilon},[0,1])<\varepsilon$.

We have thus shown that for each $\varepsilon>0$ there exists a partition of $[0,1]$ such that $U(f,P_{\varepsilon},[0,1])-L(f,P_{\varepsilon},[0,1])<\varepsilon$ so $f$ is Riemann integrable on $[0,1]$, and $\int_{0}^{1}f=U(f,[0,1])=L(f,[0,1])=0$.

答:
There are two mistakes here, in my view:

The first is a mistake at the end:
You implicitly assume that in the intercal $[0,n_1  - \frac{\epsilon}{4M}]$ the supremum of $f$ is equal to zero, which is not true (and similarly for a lot of other intervals).
At most you can estimate this supremum with $\frac{1}{n+1}$, so you get an extra term of at most $\frac{M+1}{n+1}(1 - \frac{\epsilon}{2M})$, which does not go to zero if $n \rightarrow \infty$ (or am I missing something?).

Second (and third) you should define your $n_i$, ($1 \leq i \leq M$), and you assumed that there exist exactly $M$ such points (but it is an upper bound as you stated), of course there is no harm in picking exactly $\tilde{M}_n := \#(D_{\frac{1}{n}})$ and then filling up the remainin $M - M_n$ with random ones, but you should state this explicitely.

All in all, a very nice idea of a proof, maybe you can make it work with a few changes (to add something positive to all the criticism, but then again you literally asked for it ;))


答: Assuming the nested radical is infinite we have $x=\sqrt{2+\sqrt{2+...+\sqrt{2}}}\Rightarrow x^2-2=x \iff x^2-x-2=0\Rightarrow x=2$

Title: A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$

Body:
We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.

We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$

Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{  This is what we need to prove}$$
$$
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing  and is bounded sufficient?


答:
Hint: A well-known series (see here, e.g.) is:

$$\frac{1}{(1-x)^{11}} = \sum_{k=0}^\infty \binom{10+k}{k}x^k$$

Its derivative is close to what you want:

$$\frac{d}{dx} \frac{1}{(1-x)^{11}} = \sum_{k=1}^\infty k\binom{10+k}{k}x^{k-1}$$ At the end evaluate at $x=\frac12$ of course.

Title: Calculate $\sum_{i=0}^\infty i\binom{i+10}{i}(\frac{1}{2})^i$

Body:
I'm not great with summation and its various techniques, so go easy on me. If someone can point me in the right direction I would be very grateful.

This is the sum in display mode: $$\sum_{i=0}^\infty i\binom{i+10}{i}\frac{1}{2^i}$$

By using the symmetry rule for binomial coefficients I end up with this:
$$\frac{1}{10!}\sum_{i=0}^\infty \frac{i}{2^i}\prod_{k=1}^{10}(k+i)$$ which doesn't seem to help.
I tried working something else on the binomial to get rid of i and I got:
$$11\sum_{i=0}^\infty \frac{1}{2^i}\binom{i+10}{i-1}$$
but yet again, I'm stuck.

I have a feeling that I have to somehow simplify the sum and then use the recursive method?


问:
Physics books recommendations
I’m a maths PhD student looking for recommendations for physics books.

I haven’t really done much physics, and I’m wanting to learn the main topics, mechanics (Euler Lagrange etc), electromagnetism and quantum mechanics. Everything that is covered in the Feynman lectures but having looked through them, they don’t seem to go as far as I would like and he seems to develop the maths needed (in a handwavy way) to cover the material.

I imagine I might need multiple references ideally assuming the reader already knows the necessary maths or the maths is at the start and I can just skip it!

Thanks in advance.

答:
Series' of books

The theoretical minimum by Landau: ten physics-packed volumes over a variety of topics that Landau considered the 'theoretical minimum' for any physicist to know: from classical mechanics to condensed matter and field theory. They assume a level of mathematical competence on the reader's part, and are very focused on the physics, so might be what you're looking for. On the other hand, each volume has a lot of detail that you might not be interested in.

A more modern version of this idea is Greiner's Theoretical physics. In a way, this series is less 'complete' than 'The theoretical minimum': the statistical physics volume is introductory only, and there is no treatment of kinetic theory or fluid dynamics. However, the books are focused on introducing modern theories, e.g. gauge theory, and are really well explained.

A very 'student-friendly' version of this idea is the books by Griffiths, which I do like; but I cannot recommend to you, because they are very hand-wavy, more so than The Feynman lectures.

Individual books

I was reminded in the comments about the excellent Physics from symmetry by Jakob Schwichtenberg: a very readable introduction to field theory, and really emphasizes the 'big picture' stuff, while still actually doing (some) calculations. I was hesitant to recommend this at first, but you should probably at least give it a chance because of its conceptual clarity.

Here are the 'canonical' textbooks$^\dagger$ in the topics you mentioned:

For electromagnetism: Classical electrodynamics by Jackson, infamous amongst grad students. Since you will probably be mostly interested in the first few chapters, which are covered well in Greiner's book: Classical electrodynamics, I can't really recommend Jackson to you.

For classical mechanics: Goldstein is standard. You might prefer the more geometrically oriented Classical dynamics: A contemporary approach by Jose and Saletan.

For quantum mechanics: Probably the book by Sakurai. However, I think you might enjoy Quantum mechanics: A modern development by Ballentine. It has the very least hand-waving in a QM book that I've seen, and is clearly intended for those of a mathematical inclination. This book justifies things at a level that is simply not seen in other introductory texts.

$\dagger$ Obviously this is subjective. I'm basing these on "likely to be set in an introductory grad school course".

Title: Physics books recommendations


问:
What is the name for a fraction whose expansion extends forever?
Irrespective of whether it is a rational number like $\frac{1}{3}$, an irrational number, like $2^{1/2}$ or a transcendental number, like $\pi$, is there a word for all decimal fractions which continue without end?

答:
Every real number can be written as a nonterminating decimal expansion using trickery like $.999999... = 1$, so the question should really be, which numbers are forced to be written as a nonterminating expansion.

First consider a slightly easier question, which numbers are forced to be written as a nonrepeating expansion? The answer to this question is independent of the base (so long as it is integral >= 2) chosen for the expansion, and it is the irrational numbers. The proof of this goes by long division and noting that division by an integer gives only finitely many possible choices for remainder, so repetition will eventually occur. Thus, in any base any irrational number must be represented in a nonterminating expansion.

Then the question remains, of the rational numbers, which amongst them can be written in a nonterminating expansion. E.g. $1/10 = .1$ but $1/3=.33333333...$ must repeat base 10.
This question's answer obviously does depend on the base, so lets fix a base $b$.
To say that $x$ has a terminating expansion in base $b$ means that
$x = \sum_{i=-k}^na_i b^{-i}$ for a finite $n,k$ and some $a_{-k}, \ldots, a_0, a_1, \ldots, a_n \in \{0,1,\ldots, b-1\}$. This is possible iff multiplying by a large power of $b$ gives an integer, so we conclude that $x$ has a finite representation in base $b$ if and only if it is a rational number which can be written as $a/b^n$ for $a, n $ integers with $n \geq 0$.

Such numbers are called $b$-adic numbers, most commonly considered only for primes $p$, and referred to as $p$-adic numbers. https://en.wikipedia.org/wiki/Dyadic_rational

Thus the answer to your question is that $x$ must have a nonterminating expansion in base $b$ iff it $x$ is not $b$-adic (this includes all irrationals).


问:
How to evaluate seemingly simple sum to infinity?
The sum I am tying to evaluate is the following:

$$\sum\limits_{r=0}^{\infty}{\dfrac{3^r+4^r}{5^r}}$$

This sum seems to tend towards $0$ as the numerator is increasing slower than the denominator, yet I am not seeing a pattern between the terms to exploit.

All attempts to find a common ratio between terms were futile.

Any ideas are appreciated!


问:
Unbounded operators, Hilbert spaces and Borel sets
Let $H$ be a Hilbert space and $\langle , \rangle$ its inner product. Let $X : $ be a self-adjoint linear operator and let $F_X : H \to \mathbb{R}$ be defined as $F_X(a) = \langle a , Xa \rangle$. Let $U \subseteq \mathbb{R}$ be an open set. Is $F_X^{-1}(U)$ a Borel set (in the Borel algebra of the topology induced by the norm of the Hilbert space)?

From what I understand, if $X$ is bounded then it is continuous, therefore $F_X$ is continuous, therefore $F_X^{-1}(U)$ is an open set and therefore a Borel set.

What can I say in the unbounded case?

答:
May we assume that $X$ is a linear operator? If so then one has boundedness if and only if continuity as you describe. The missing step then would be show that that your self-adjoint operator is bounded using the Uniform Boundedness Principle. The rest follows by using the characterisation of a continuous map as one where the pre-image of open sets are open.

Let me elaborate on the UBP part:
Our goal is to show that $X$ is bounded as this implies the boundedness of $F_X$ and we're done. That is to say, we wish to show that $\sup_{\vert\ x \vert \leq 1}\vert\vert Xx \vert\vert < \infty$. To do this we start by defining the following family of linear operators $T_x:H\rightarrow H$ defined $T_x(y) := <Xx,y>_H$. One may show that for any choice of $x$ that these are bounded, in particular we have:
$\sup_{\vert y \vert \leq 1}\vert\vert T_x y \vert\vert = \sup_{\vert y \vert \leq 1}<Xx,y> = \vert\vert Xx\vert\vert$. Similarly, one may show that $\sup_{\vert x \vert \leq 1}\vert\vert T_x y \vert\vert = \sup_{\vert x \vert \leq 1}<Xx,y> = \sup_{\vert x \vert \leq 1}<x,Xy> = \vert\vert Xy\vert\vert$. Having satisfied the conditions of the UBP we conclude that $\vert X \vert := \sup_{\vert x \vert \leq 1}\vert Xx \vert = \sup_{\vert x\vert\leq 1, \vert y \vert \leq 1}\vert T_xy\vert < \infty$


问:
Showing that $n\sin(x/n)$ does not converge uniformly on $\mathbb{R}$
I know the version of this question with closed interval has been asked to death in this site, but how about if the function sequence $f_n(x) = n\sin(x/n)$ is considered over $\mathbb{R}$? At first I thought that the sequence would not converge to the limiting function $f(x) = x$, since at the points $x_n = n\pi$, $f(x_n) = n\pi$, but $f_n(x_n) = 0$. But is this sufficient to show that $\lim_{n \to \infty}\sup\{\left|f(x) - f_n(x)\right|\} \neq 0$?

答:
Hint:
In fact
$$
\forall n \in \mathbb N:\sup_{x\in \mathbb R}\{|f(x) - f_n(x)|\} = \sup_{x\in \mathbb R}\{|x-n\sin(x/n)|\} =\ "\infty"
$$
i.e. there is no supremum for any $n \in \mathbb N$ and so the limit
$$
\lim_{n \to \infty}\sup\{\left|f(x) - f_n(x)\right|\}
$$
is meaningless


问:
Find the function $f(x)$.
Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?

I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.

答:
Suppose that $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots$ where $n$ is the degree of the polynomial. What is the order of the polynomial $f(x+2)-f(x)$ in this case?
You should then be able to use the equality to ascertain $n$. At this point your task is much easier - using the fact that $f(0)=5$ and the equality you should be able to completely determine $f$.

Title: Find the function $f(x)$.


问:
$A\cong A^{op}$ means $A$ is commutative
Sorry for my bad English.

Let $A$ be (not necessarily commutative) ring.

By here, if $A\cong A^{op}$ as ring,  $A$ is commutative, but I cannot proof.

Is this trivial? Please help me.

答:
If $A$ is commutative then trivially $A\cong A^{op}$.

I think that the converse is not true, and I think that the following is a counter example:

Let $\mathbb H$ be the Quaternion algebra. Define
$$
f(i)=i \\
f(j)=k \\
f(k)=j  \\
$$
Then $f$ extends to an isomorphism $f: \mathbb H \to \mathbb H^{opp}$.


问:
Free models of infinitary essentially algebraic theories
Let $\mathbb{T}$ be an essentially algebraic theory in which the operations are allowed to have infinite arity. It is known that not every such theory has a free/initial model; for example, there is no free complete Boolean algebra on countably many generators. However, there are infinitary essentially algebraic theories that do have initial models, for example the theory of sheaves on a small site.

My question is: is it known whether there are any conditions on an infinitary essentially algebraic theory $\mathbb{T}$ that will guarantee that it has a free/initial model? For example, will $\mathbb{T}$ have an initial model if the arities of all the operations are bounded by some cardinal?

答: As @varkor pointed out, if $\mathbb{T}$ is a $\lambda$-ary essentially algebraic theory for some regular cardinal $\lambda$, then $\mathbb{T}$ does have an initial model. This follows (e.g.) by (the proof of) Theorem 3.36 in the book Locally Presentable and Accessible Categories by Adamek and Rosicky, in which it is shown that if $\mathbb{T}$ is a $\lambda$-ary essentially algebraic theory over a $\lambda$-ary signature $\Sigma$, then its category of (set-based) models is a reflective subcategory of the category of partial $\Sigma$-structures. An easy computation then shows that the reflection of the initial/empty partial $\Sigma$-structure is the initial model of $\mathbb{T}$.


问:
Numerical method to solve PDE system $\mathbf{A}\dot{\mathbf{U}} + \mathbf{B}\mathbf{U}^{'} + F(\mathbf{U}) = 0$
I would like to numerically solve a system of PDEs of the form

$$\mathbf{A}\dot{\mathbf{U}} + \mathbf{B}\mathbf{U}^{'} + F(\mathbf{U}) = 0$$

Where, $\dot{\mathbf{U}}(s, t)$ represents the time derivative, $\mathbf{U}^{'}(s, t)$ represents the space derivative, $\mathbf{U}$ is the vector of unknowns, and $F(\mathbf{U})$ is a nonlinear function of $\mathbf{U}$.

Both $\mathbf{A}$, and $\mathbf{B}$ are constant singular matrices.

So, the given system is a coupled system of PDEs and ODEs.

How to numerically solve such system of PDEs given initial and boundary conditions?. Which numerical method will be suitable?

答:
Have you looked at the Keller Box scheme? It's described in this paper http://dx.doi.org/10.1016/j.amc.2009.07.054. It's relatively straight forward in principle and suitable for PDE's and non-linear systems. Depending on the size of your system though it can be a bit of a hassle to work out.

I'm not sure what your problem is but I've used it for solving one dimensional Stefan problems, where I have time and space derivatives and non-linear terms. It essentially applies the finite difference operators defined in the paper (pg 1614) and then you can linearise the problem using Newton's method.


答:
We have two conditions, as the result is $(8x-2)$.

To be solvable we need to use a quadratic with two parameters
$$f(x)=ax^2+bx+5$$
so we have
$$f(x+2)-f(x)=4 a x+(4 a+2 b)\equiv 8x-2$$
setting
$$
\begin{cases}
4a=8\\
4a+2b=-2\\
\end{cases}
$$
we have $a=2,b=-5$ so finally
$$f(x)=2 x^2-5 x+5$$

Title: Find the function $f(x)$.

Body:
Let $f(x)$ be a polynomial function. If $f(x+2) - f(x) = 8x - 2$ and $f(0) = 5$, then what is $f(x)$?

I tried to replace $x$ with $0,2,4,\ldots $ for discovering some regular pattern but I have no idea after doing that.


问:
Are there non-continuable functions that become continuable when raised to some power?
Let $X$ be a complex algebraic variety (integral, separated scheme of finite type over $\mathbb C$) and $U\hookrightarrow X$ an open subvariety. I will say that $f\in\mathscr O_X(U)$ is continuable if there exists some $g\in \mathscr O_X(X)$ with $g|_U=f$. My question is simply: Can it happen that $f$ is not continuable, but there exists a $k\in\mathbb N^+$ with $f^k$ continuable?

I was debating with a friend who thinks it is possible but could not give an example, while I thought that it is impossible but wasn't able to give a satisfying proof. So, I'd be really glad if you could provide one of the two (that is, a simple example of such a phenomenon or a proof that it can not happen).

答:
On the other hand, for smooth curves it is true: If $K$ is the function field, then $f \in K$ is continuable iff $f \in \mathcal{O}_X(X)$ iff $\mathrm{ord}_P(f) \geq 0$ for all closed points $P \in X$, where $\mathrm{ord}_P$ denotes the valuation at $P$. Since $\mathrm{ord}_P(f^k)=k \cdot \mathrm{ord}_P(f)$, it follows that $f$ is continuable iff $f^k$ is continuable.

What about smooth varieties in higher dimensions?


问:
About extended of continuous function
Assume we have an open interval $I\subset\Bbb R$ and a set $D\subset I.$ Let $f\colon D\to \Bbb R$ be a continuous such that $f$ can not be extended to a continuous on $I.$ What we can conclude from that ?

答:
$\mathbf{ Claim:}$ The set $ D $ is not close in $ I $.

To see this we proceed by contradiction, in particular we will show that if $ D $ is close then there exists an extension. Suppose $ D $ close and non-empty. Let $ a = \min\{ x \in D \} $ and $ d = \max \{ x \in D \} $. We extend $ f $ in $ \mathbb{R} \setminus [a,b]  $ by setting $ \tilde{f}(x) = f(a) $ for $ x < a $ and  $ \tilde{f}(x) = f(b) $ for $ x > b $. Let $ x \in (a,b) $, we define $ y_-(x) = \sup\{ y \in D $ such that $ y < x \} $ and $ y_+(x) = \inf\{ y \in D $ such that $ y > x \} $. Define
$$ \tilde{f}(x) = f(y_{-}(x)) + \frac{x-y_{-}(x)}{y_{+}(x)-y_{-}(x)}(f(y_{+}(x))-f(y_{-}(x))).$$
The function $ \tilde{f} $ is a continuous extension of $ f $.

$ \textbf{Consequences:} $ if $ f$ cannot be extended to $ I $ then $ f $ cannot be extended to the closure of $ D $ that we denoted $ \bar{D} $.

One can wonder why one cannot extend $ f $ to $ \bar{D} $ and this happens if and only if there exists $ x \in \bar{D} \setminus D $ and a sequence $ x_{n} $ that converge to $ x $ such that $ f(x_{n}) $ does not converge.

$ \textbf{ Some examples:} $

$ f(x_{n}) $ diverge to $ \infty $ (imagine 1/x around zero)

$ f(x_{n}) $ does not converge (imagine a function that oscillate a lot like $sin(1/x)$ close to zero)

$ f(x_{n}) $ converge from the ''left and the right''  but the limit is different (imagine sing(x) around zero)

Title: About extended of continuous function


答:
Let's try to determine the degree of the polynomial $f(x).$

If $\ f(x)\ $ has degree $0$, then $f(x) = a_0\ $ which implies $\ f(x+2)-f(x) = 0 \neq 8x-2.$

If $\ f(x)\ $ has degree $1$, then $f(x) = a_0 + a_1x\ $ which implies $\ f(x+2)-f(x) = 2a_1 \neq 8x-2.$

If $\ f(x)\ $ has degree $2$, then $f(x) = a_0 + a_1x + a_2x^2\ $ which, after some cancelling, implies $\ f(x+2)-f(x) = 4a_2x + (2a_1+4a_2).$ For this to equal $8x-2,\ $ we must have $a_2 = 2\ $ and $a_1 = -5.$ So $f(x) = a_0 - 5x + 2x^2.\ $ The last condition, $f(0) = 5\ $ implies that $a_0 = 5.$ So $f(x) = 5 - 5x + 2x^2\ $ is certainly a solution - the only solution of degree 2, since we got there by deduction.

Are there any other solutions (of degree $\geq 3$)?

Well if the degree of $f(x) \geq 3,\ $ then $$f(x+2)-f(x) = a_0 + a_1(x+2) + a_2(x+2)^2 + ... + a_n(x+2)^n - (a_0 + a_1x+a_2x^2+...+a_nx^n),\ $$ where $n\geq3.\ $ From the binomial expansion, the $x^{n-1}\ $ coefficient of $f(x+2)-f(x)\ $ is equal to $a_{n-1} + 2na_n - a_{n-1} = 2na_n.\ a_n \neq 0\ $ and $n \neq 0,\ $ therefore the $x^{n-1}\ $ term in $f(x+2)-f(x)\ $ has at degree of at least $2\ $, and so $f(x+2)-f(x)\neq 8x-2.$ So no, there are no other solutions.

Title: Find the function $f(x)$.


答:
Indeed, every division algebra that has order two in the Brauer group of its center (such as the quaternions) is a counterexample.

But surely the simplest to grasp counterexample is the matrix ring $R=M_{2\times 2}(\Bbb{R})$. It is not commutative. And taking the transpose, i.e. the mapping
$$A\mapsto A^T$$
is an isomorphism from $R$ to $R^{opp}$ by the known rule $(AB)^T=B^TA^T$.

Title: $A\cong A^{op}$ means $A$ is commutative

Body:
Sorry for my bad English.

Let $A$ be (not necessarily commutative) ring.

By here, if $A\cong A^{op}$ as ring,  $A$ is commutative, but I cannot proof.

Is this trivial? Please help me.


问:
Herbrand Quotient
I am trying to solve exercises from Lang's Algebra, and I am stuck on a problem about Herbrand quotients.

Let $G$ be a finite cyclic group of order $n$ generated by an element $\sigma$. Assume that $G$ operates on an abelian group $A$, and let $f, g : A \rightarrow A$ be the endomorphisms of $A$ given by $$f(x) = \sigma x - x$$ and $$g(x) = x + \sigma x + \ldots + \sigma ^ {n -1} x.$$

Define the Herbrand quotient by the expression $q(A) = (A_f : A ^ g) / (A_g : A ^ f)$, provided both indices are finite. Assume now that $B$ is a subgroup of $A$ such that $G B \subseteq B$.

(a) Define in a natural way an operation of $G$ on $A / B$.

(b) Prove that $$q(A) = q(B) q(A / B)$$
in the sense that if two of these quotients are finite, so is the third, and the stated equality holds.

(c) If $A$ is finite, show that $q(A) = 1$.

The $A_f, A ^ f$ in the question denotes the image and kernel of $f$ as a map from $A$ to $A$. Similarly for $B_g$ and $B ^ g$.

I am currently stuck on $(b)$ and also I do not understand why $f, g$ are endomorphisms of groups to start with. If $G$ is acting on $A$, it is not necessary that each $g \in G$ induces an automorphism on $A$ right? Thanks in advance!

答:
I will use the result in this page. Let $\pi:A\rightarrow A/B$ be the canonical surjection.

It is sufficient to prove:

$$[A_f:A^g][B_g:B^f][(A/B)_g:(A/B)^f]=[A_g:A^f][B_f:B^g][(A/B)_f:(A/B)^g].$$

But:

$$[A_f:A^g]=[(A_f)^\pi:(A^g)^\pi][(A_f)_\pi:(A^g)_\pi]$$

$$[B_g:B^f]=[(A_\pi)_g:(A_\pi)^f]=[(A_g)_\pi:(A^f)_\pi][(A^f)_\pi:(A_\pi)^f]$$

$$[(A/B)_g:(A/B)^f]=[(A^\pi)_g:(A^\pi)^f]=[(A^\pi)_g:(A_g)^\pi][(A_g)^\pi:(A^f)^\pi]$$

Analogously:

$$[A_g:A^f]=[(A_g)^\pi:(A^f)^\pi][(A_g)_\pi:(A^f)_\pi]$$

$$[B_f:B^g]=[(A_f)_\pi:(A^g)_\pi][(A^g)_\pi:(A_\pi)^g]$$

$$[(A/B)_f:(A/B)^g]=[(A^\pi)_f:(A_f)^\pi][(A_f)^\pi:(A^g)^\pi]$$

So the problem is equivalent to prove that:

$$[(A_f)^\pi:(A^g)^\pi][(A_f)_\pi:(A^g)_\pi][(A_g)_\pi:(A^f)_\pi][(A^f)_\pi:(A_\pi)^f][(A^\pi)_g:(A_g)^\pi][(A_g)^\pi:(A^f)^\pi]$$

is equal to:

$$[(A_g)^\pi:(A^f)^\pi][(A_g)_\pi:(A^f)_\pi][(A_f)_\pi:(A^g)_\pi][(A^g)_\pi:(A_\pi)^g][(A^\pi)_f:(A_f)^\pi][(A_f)^\pi:(A^g)^\pi].$$

So it suffices to prove that:

$$[(A^f)_\pi:(A_\pi)^f][(A^\pi)_g:(A_g)^\pi]=[(A^g)_\pi:(A_\pi)^g][(A^\pi)_f:(A_f)^\pi].$$

We will prove that $[(A^f)_\pi:(A_\pi)^f]=[(A^\pi)_f:(A_f)^\pi]$. Let $P$ be a representant system of cosets of $(A_\pi)^f$ in $(A^f)_\pi$. For $p\in P$ choose an element $a_p\in A$ such that $f(a_p)=p$. We will prove that the family $(\pi(a_p))_{p\in P}$ is a representant system of cosets of $(A_f)^\pi$ in $(A^\pi)_f$.

For $y\in (A^\pi)_f$, then there is $x\in A$ such that $y=\pi(x)$, so $0=f(y)=f(\pi(x))=\pi(f(x))$, so $f(x)\in(A^f)_\pi$, so there is $p\in P$ such that $f(x)\in p+(A_\pi)^f$, so there is $b\in A_\pi$ such that $f(x)=f(a_p)+f(b)$, so $x-a_p-b\in A_f$, so $\pi(x)-\pi(a_p)-\pi(b)\in (A_f)^\pi$, so $y\in\pi(a_p)+(A_f)^\pi$.

If $\pi(a_p)+(A_f)^\pi=\pi(a_q)+(A_f)^\pi$, then $\pi(a_p)-\pi(a_q)\in(A_f)^\pi$, so there is $x\in A_f$ such that $\pi(a_p)-\pi(a_q)=\pi(x)$, so $a_p-a_q-x\in A_\pi$, so $f(a_p)-f(a_q)-f(x)\in(A_\pi)^f$, so $p-q\in(A_\pi)^f$, so $p+(A_\pi)^f=q+(A_\pi)^f$, so $p=q$.

Title: Herbrand Quotient


问:
Calculating the probability of drawing two cards of the same suit from a deck of dual-suit cards
I'm making a tabletop roleplaying game which uses a handmade deck of cards as a chance generator. All we need to know is that there are 16 cards and four suits, but:

Each card is of two suits;

Six cards in the deck are of two different suits;

And one card in the deck is of the same suit twice.

Thus, our cards look like this (with Hearts highlighted as an example):

Card
Suits

1
Hearts and Clubs

2
Hearts and Spades

3
Hearts and Diamonds

4
Hearts and Hearts

5
Clubs and Hearts

6
Clubs and Spades

7
Clubs and Diamonds

8
Clubs and Clubs

9
Spades and Hearts

10
Spades and Spades

11
Spades and Diamonds

12
Spades and Clubs

13
Diamonds and Spades

14
Diamonds and Diamonds

15
Diamonds and Hearts

16
Diamonds and Clubs

If I choose Hearts as my suit, and turn over a Hearts card, that's called a success. If I turn over the card with both suits as Hearts, that's two successes.

I need to calculate the probability of getting at least n successes when I draw x number of cards without replacement. For instance, it is quite common to need two successes and to draw three cards. The following sets of cards would all qualify:

One Hearts card, one Hearts card, and one other card (2 successes)

One double Hearts card and two other cards (2 successes)

One double Hearts card, one Hearts card, and one other card (3 successes)

Three Hearts cards (3 or 4 successes)

My knowledge of probability maths is basic at best, and I've set up what feels like quite a challenging system, so I'm getting nowhere calculating it. I'd be grateful for the specific probability of getting 2 successes on 3 cards, but a general formula for me to calculate successes would be even better!

答:
In your specific case, it is probably easier to find the probability of drawing one or zero hearts.

So

zero hearts: $\dfrac{9 \choose 3}{16 \choose 3}=0.15$

one heart: $\dfrac{{6 \choose 1}{9 \choose 2}}{16 \choose 3}\approx0.386$

and then subtract both from $1$ to give $\frac{13}{28}\approx 0.464$


问:
What equation to use when finding 6 variables?
I'm programming an application which will need to calculate and invert 6 variables.

e.g.

[4][20][15][16][0][1] -> (perform calculation) -> 450

450 -> (perform calculation) -> [4][20][15][16][0][1]

What would be the best direction to go to calculate and invert the calculation for these variables or nth variables if I wish to expand later on? I was looking at Matrices, linear equations, etc, but not sure the best and easiest method for programmatically making this a bidirectional equation. Edit: I’m looking to get the smallest number possible, and the range is 0 to 55 for my variables.

答:
Imagine that the variables you have had the range $0$-$9$, i.e. single digits. In that case you could obviously append them to form a single 6-digit number.

To do that in a computer program, appending a digit $d$ to a number $t$ is done by calculating $10t+d$. The multiplication by $10$ shifts it to the left by appending a zero, and then adding $d$ sets the last digit to $d$. So you can simply do this six times in a loop, appending the six variables to zero.

You'll also want to extract the digits one by one. You can get the last digit of a number $t$ by evaluating $t \bmod 10$, which is $t\%10$ in some computer languages. Once you have that digit $d$, you can calculate $\frac{t-d}{10}$, removing the digit and shifting it to the right, so that you can then access the next digit.

In your case you have variables in the range $0$ to $55$ instead of $0$ to $9$. Therefore you should use $56$ at every spot I used $10$ in the above explanation. It may be easier to implement it base 10 first, so that it is easy to see that it works the way you want to before changing it to base 56.


问:
Accumulation points of $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$
I'm having some trouble finding the accumulation points o set $A=\left[(1+\frac{1}{n})^n : n=1,2,...\right]$. I know that it only has one accumulation point (which is $e$), but I don't know the proof.

If anyone could help me, it would be great.

Thank you very much.

答:
Claim 1: The ordered sequence $a_n = (1+\frac 1n)^n$ is monotonically increasing.

Claim 2: $a_n$ is bounded above.

Now any monotonically increase sequence $A= \{a_n\}$ the following will hold

$\sup A$ exists

(all bounded above sets in the reals have a least upper bound) and that

$\sup A \not \in A$.

If $\sup A = a$ for some $a\in A$ then $\sup A =a_n$ for some $n$ and $\sup A=a_n < a_{n+1} \in A$, a contradiction.

$\sup A$ is a limit point:

Because for any $r> 0$ then $\sup A-r$ is not an upper bound so there is an $a\in A;a\ne \sup A$  so that $\sup A -r < a < \sup A$ so $a \in B_r(\sup A)$.

$\sup A$ is the only limit point.

If $x < \sup A$ then there $a\in A; a>x$. And by well-ordering principal there is a least $m$ so that $a_m > x$.  If $x \le a_1$ there are no $b\in A; b < x$ and if we let $r = a_m -x$ then $B_r(x)$ will have no  elements of $A$ (other than $x$ if $x=a+1$).  And if $a_1 < x$ then there will be a maximum $a_k < a$.  Let $r =\min(x-a_k, a_m -x)$ and then $B_r(x)$ will have no elements of $A$ (other than maybe $x$). So $x$ is not a limit point.

And if $x > \sup A$ then let $r = x -\sup A$ and $B_r(x)$ has no points of $A$.

....

So that's that.

$\sup A = \lim_{n\to \infty} (1+\frac 1n)^n = e$ is the only limit point.

....

I'll leave it to you to prove (or look up.... or take as a given) claims 1 and 2.


问:
Need help to solve ODE $y'(x) = -\tanh(y(x))$
I want to solve the ODE

$$y'(x) = -\tanh(y(x))$$

It is separable, so diving by $\tanh(y(x))$ and integrating wrt $x$ we get $$\int \frac{y'(x)}{\tanh(y(x))} \,dx = \int -1\,dx$$

I am having trouble finding the first integral. How do I proceed?

答:
Separate $y'(x) = -\tanh(y(x))$ as

$$\frac{dy}{\tanh y} =-dx$$

Then, integrate both sides

$$\ln (\sinh y )=-x+C$$


问:
Multistep ODE Solvers
Write both a fourth order Adams Bashforth and Adams Moulton procedure to solve

$$x'(t) = x(t)-y(t)-\exp(t);$$

$$y'(t) = x(t)+y(t)+2\exp(t)$$

with initial values $x(0) = -1, y(0) =- 1$ on the interval $0 \le t \le 4.$

Use the analytical solution to find the staring values.
The analytical solution is:

$$x(t) = \exp(t)\cos(t)-2\exp(t);$$

$$y(t) = \exp(t)\sin(t)-\exp(t)$$

I need to plot the solution (by PYTHON, if possible?) and verify the convergence rate.

答:
Adams-Bashfourth Four-Step Method

The Adams-Bashfourth Four-Step Method is given by:

$w_0 = a = 0, w_1 = \alpha_1, w_2 = \alpha_2, w_3 = \alpha_3$

$\displaystyle w_{i+1} = w_i + \frac{h}{24}\left[55 f(t_i, w_i) - 59 f(t_{i-1}, w_{i-1}) + 37 f(t_{i-2}, w_{i-2}) - 9 f(t_{i-3},w_{i-3})\right]$

where $i = 3, 4, \ldots N-1$.

The local truncation error is $\displaystyle \tau_{i+1}(h) = \frac{251}{720}y^{(5)}\left(\mu_i\right)h^4, t_{i-3} \lt \mu_i \lt t_{i+1}$.

Adams-Moulton Four-Step Method

The Adams-Moulton Four-Step Method is given by:

$w_0 = a = 0, w_1 = \alpha_1, w_2 = \alpha_2, w_3 = \alpha_3$

$\displaystyle w_{i+1} = w_i + \frac{h}{720}\left[251 f(t_{i+1}, w_{i+1}) + 646 f(t_i, w_i) - 264 f(t_{i_1}, w_{i-1}) + 106 f(t_{i-2}, w_{i-2}) - 19 f(t_{i-3}, w_{i-3})\right]$

where $i = 3, 4, \ldots N-1$.

The local truncation error is $\displaystyle \tau_{i+1}(h) = -\frac{3}{160}y^{(6)}\left(\mu_i\right)h^5, t_{i-3} \lt \mu_i \lt t_{i+1}$.

These methods are typically used in concert as a Predictor (Adams- Bashfourth) - Corrector (Adams - Moulton) approach. We calculate the predictor step and immediately follow that calculation by a corrector step (in other words, we calculate an $x_{i,P}, y_{i,P}$ and follow that up with calculating $x_{i,C}, y_{i,C}$).

For our system, we have two equations, so just do two sets of calculations at each step.

To calculate the four starting values, we use the given analytic solution since these values are needed for each method.

We will use $N = 10$ (change this if you were asked for a different value).

$h = \dfrac{b-a}{N} = \dfrac{4-0}{10} = \dfrac{2}{5}$

Using, $t = 0.0, 0.4, 0.8, 1.2$, this yields:

$x(t) = e^t \cos t - 2 e^t \rightarrow \{-1, -1.60959, -2.90053, -5.43716\}$

$y(t) = e^t \sin t - e^t  \rightarrow \{-1, -0.910881, -0.629036, -0.225638\}$

These are our values for $w_0 = a = 0, w_1, w_2, ~~ \text{and} ~~~ w_3$.

Now, lets use all of this and do a calculation for the next set of values using all of the methods, that is, for $i = 3, t = 1.6$, we have:

Adams-Bashfourth (Predictor Steps)

$\displaystyle x_4 = x_3 + \frac{0.4}{24} \left[ 55 f_3 - 59 f_2 + 37 f_1 - 9 f_0 \right] = -5.43716 +(0.4/24)(55 (-8.53164) -59(-4.49704) +37(-2.19053) -9(-1)) = -10.0365675$

$\displaystyle y_4 = y_3 + \frac{0.4}{24}\left[55 g_3 - 59 g_2 + 37 g_1 - 9 g_0\right] = -0.225638 +(0.4/24)\left[55 (0.977436) -59(0.921522) +37(0.463178) -9(0)\right] = 0.04980813333$

$f(x,y,t) = x - y -e^t \rightarrow f_4 = f(x_4, y_4, t_4) = -10.0365675 - 0.04980813333 -e^{1.6} = -15.0394080577251148$

$g(x,y,t) = x + y +2e^t \rightarrow g_4 = g(x_4, y_4, t_4) = -10.0365675 + 0.04980813333 +2e^{1.6} = -0.080694517879770392691427287$

Adams-Moulton (Corrector Steps)

$\displaystyle x_4 = x_3 + \frac{0.4}{720}\left[251 f_4 + 646 f_3 - 264 f_2 + 106 f_1 - 19 f_0\right] = -5.43716 +(0.4/720)\left[251 (-15.0394080577251148) + 646 (-8.53164) -264 (-4.49704) + 106 (-2.19053) -19 (-1)\right] = -10.05510915693833545266666666$

$\displaystyle y_4 = y_3 + \frac{0.4}{720}\left[251 g_4 + 646 g_3 - 264 g_2 + 106 g_1 - 19 f_0\right] = -0.225638 +(0.4/720)\left[251 (-0.080694517879770392691427287) + 646 (0.977436) -264 (0.921522) + 106 (0.463178) -19 (0)\right] = 0.006019995562320906352473194979444444444$

The previous two values for $x_i, y_i$ are the values we enter into the table as corrected to use for the next calculations.

Lets create a table of these values for you to fill out as follows (I will let you fill in the rest of the table so you can get your hands around what is going on).

$$\begin{array}{|c|c|c|c|c|c|}
\hline i  & t_i & x_i & y_i & f_i & g_i  \\ \hline
00 & 0.0 &  -1 & -1  & -1  & 0    \\ \hline
01 & 0.4 & -1.60959 & -0.910881&-2.19053 & 0.463178 \\ \hline
02 & 0.8 & -2.90053  & -0.629036   & -4.49704     & 0.921522 \\ \hline
03 & 1.2 & -5.43716  & -0.225638   & -8.53164     & 0.977436 \\ \hline
04 & 1.6 &  -10.0551091 & 0.00601999&-15.0394080 & -0.0806945 \\ \hline
05 & 2.0 & & & &  \\ \hline
06 & 2.4 & & & &  \\ \hline
07 & 2.8 & & & &  \\ \hline
08 & 3.2 & & & &  \\ \hline
09 & 3.6 & & & &  \\ \hline
10 & 4.0 & & & &  \\ \hline
\end{array}$$

Note: if you didn't want to use the two methods in concert, you are still generating all of the same data and can add more columns to the table for the AB method, AM method, the error for each method, plus exact results for comparison purposes.

For the second parts of your question:

Plot the values of tables for $x_i, y_i$ for calculated versus analytic using whatever you like (why Python, unless being used in class).

In other words, here is the analytic plot for (you can plot it as a parametric plot also) $x(t), y(t)$, over the range $0 \le t \le 4$:

For the convergence rates, I provided the truncation error and you have the analytic solution, so can use those to perform that calculation and verification.

Title: Multistep ODE Solvers


答:
You stated, that $f$ is a polynomial function, hence we can write $f$ as $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ plugging that into your given equation we get

$$f(x+2)-f(x)=a_n(x+2)^n+a_{n-1}(x+2)^{n-1}+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$

using the binomial formula, we can extend this to

$$a_n(x^n+2nx^{n-1}+...+2^{n-1}nx+2^n)+a_{n-1}(x^{n-1}+2(n-1)x^{n-2}+...+2^{n-2}(n-1)x+2^{n-1})+...+a_1(x+2)+a_0-(a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0)$$

We see, that the $a_nx^n$ cancel each other out; if we look at the coefficient of $x^{n-1}$ we have $2na_n+a_{n-1}-a_{n-1}=2na_n$. This is nonzero, as $a_n$ as the leading coefficient can't be zero.

Thus your polynomial is of degree $2$. That means $f(x)=a_2x^2+a_1x+a_0$. Since $f(0)=5$, we know that $a_0=5$. Plugging what we know so far back into our original equation we get:

$$f(x+2)-f(x)=a_2(x+2)^2+a_1(x+2)+5-a_2x^2-a_1x-5=4a_2x+4a_2+2a_1=8x-2$$

which yields $a_2=2$ and thus $a_1=-5$ giving us the complete polynomial $f(x)=2x^2-5x+5$.

We can plug that back in to verify our solution.

Title: Find the function $f(x)$.


答: Start with:$$\int \frac{dy}{\tanh(y)} = \int\frac{\cosh(y)}{\sinh(y)}\,dy.$$  Let $u = \sinh(y), du = \cosh(y)\,dy$.  This gives us $$\int\frac{\cosh(y)}{\sinh(y)}\,dy = \int \frac{du}{u} = \ln|u| + C$$ and so $$\int \frac{dy}{\tanh(y)}  = \ln|\sinh(y)| + C.$$

Title: Need help to solve ODE $y'(x) = -\tanh(y(x))$

Body:
I want to solve the ODE

$$y'(x) = -\tanh(y(x))$$

It is separable, so diving by $\tanh(y(x))$ and integrating wrt $x$ we get $$\int \frac{y'(x)}{\tanh(y(x))} \,dx = \int -1\,dx$$

I am having trouble finding the first integral. How do I proceed?


答: $$\int \coth y dy =-x+c \implies \ln \sinh y=(-x+c)$$ $$\implies y=\sinh^{-1}e^{c-x}.$$


问:
calculation expect pension payout given survival probabilities
Just wondering. How is the expected pension payout calculated, given an agreed fixed annual payment of a made up figure of 1000, whilst linking it to survival probabilities coming from life tables (I deliberately ignore inflation and other factors). Is it simply a question of multiplying each amount by the pensioner's survival probability depending on age as indicated in this table:

Any feedback would be very much appreciated. Thanks!

PS:

This above survival probabilities come (I think) from UK life tables. These tables contain information like the following depending on age and gender:

mx

is the central rate of mortality, defined as the number of deaths at age x last birthday in the three year period to which the National Life Table

relates divided by the average population at that age over the same period.

qx

is the mortality rate between age x and (x +1), that is the probability that a person aged x exact will die before reaching age (x +1).

lx

is the number of survivors to exact age x of 100,000 live births of the same sex who are assumed to be subject throughout their lives to the

mortality rates experienced in the three year period to which the National Life Table relates.

dx

is the number dying between exact age x and (x +1) described similarly to lx, that is dx=lx-lx+1.

ex

is the average period expectation of life at exact age x, that is the average number of years that those aged x exact will live thereafter

based on the mortality rates experienced in the three year period to which the National Life Table relates.

I would think, that 1 - qx represents the survival probability but I am not 100% sure.

答:
With a fixed annual payment of 1000 per year, we can treat the pension plan as a life annuity. The actuarial present value (APV) of a life annuity paying 1 per year is given by $$APV=a_x=\sum _{t=1}^\infty v^t{_tp_x}$$
where $v^t=(1+i)^{-1}$ is the discount factor and $_tp_x$ is the probability of a life aged $x$ surviving $t$ years. In your case of $i=0$ (no interest rate), we have $v^t=1$, and the above expression becomes $$APV=a_x=\sum_{t=1}^\infty{_tp_x}$$
So the expected payout of your annuity is just $$1000\sum_{t=1}^\infty {_tp_x}$$
Where $$_tp_x=\prod _{k=0}^{t-1}{p_{x+k}}$$


问:
Bayes theorem tree diagram

A bag I contain $4$ white and $6$ black balls while another Bag II contains $4$ white and $3$ black balls. One ball is drawn at random from one of the bags, and it is found to be black. Find the probability that it was drawn from Bag $I$.

By using NLP, I can detect spam e-mails in my inbox. Assume that the word ‘offer’ occurs in $80$% of the spam messages in my account. Also, let’s assume ‘offer’ occurs in $10$% of my desired e-mails. If $30$% of the received e-mails are considered as a scam, and I will receive a new message which contains ‘offer’, what is the probability that it is spam?

I have drawn tree digram for both and it works for the first question but not for the second question

Can you explain it?

Thank you!

答:
Question 1:
$$P(\text{bag I | black ball})\\= \frac{P(\{b\})}{P(\{b,d\})}\\= \frac{\frac12\times\frac6{10}}{\frac12\times\frac6{10}+\frac12\times\frac37}\\= \frac7{12}$$

Question 2:
$$P(\text{spam | contains ‘offer’}) \\= \frac{P(\{a\})}{P(\{a,c\})}\\= \frac{0.3\times0.8}{0.3\times0.8+(1-0.3)\times0.1}\\= \frac{24}{31}$$

Notice that:

the outcomes (the rightmost column) of an experiment constitute the sample space;

each event is a subset of the sample space: e.g., the event of
obtaining a black ball $=\{b,d\};$

the probability of a given outcome, say, $d,$ is obtained by multiplying the
values on all the branches leading to it;

the values on the branches of a probability tree are conditional
probabilities;

point (4) justifies why point (3) applies even when the trials (the underlined columns) are dependent, as in the two examples above;

abusing notation, $P(\{b,d\})$ and $P(\{HH,HT\})$ are typically simply written as $P(b,d)$ and $P(HH,HT)$.

Title: Bayes theorem tree diagram


问:
Braid invariants resource
What are some braid invariants (analogous to the idea of knot invariants) or a resource where I can find them?</p>

答:
For a solid reference on this topic, I would suggest either

J. S. Birman Braids, Links and Mapping Class Groups, Annals of Mathematical Studies 82, Princeton University Press, 1975.

for a classical approach to a solution for the word problem of the braid groups (contained in the first two chapters), or

V. L. Hansen, Braids and Coverings, London Mathematical Society Student Texts 18, Cambridge University Press, 1989.

for a more refined and modern approach, closely based on Birman's text.

To directly answer your question, piecewise linear (non-backtracking) braids on $n$ strands can be entirely classified up to ambient isotopy by 'solving the word problem' in the braid group $B_n$. This means that, given two words in the standard braid generators, we can tell whether these words represent ambient isotopic braids or not in an algorithmic way, as outlined in either of the above references (this amounts to writing braids in a 'standard form' which is shown to exist, and be unique, for every braid).

There are of course many other ways to classify braids. You may want to classify them up to conjugacy, because conjugate braids in $B_n$ will give the same link, up to isotopy, when taking the braid closure (although the converse does not hold. Non-conjugate braids can close to give isotopic links).

You may want to consider classifying braids up to 'homotopy'. This is where you consider two braids to be the same if you allow standard ambient isotopies to get from one to the other, but you also allow single braid strings to 'pass through themselves' (but not each other). A good reference for this concept of homotopic braids is

K. Murasugi & B. Kurpita, A Study of Braids, Kluwer Academic Publishers, 1999.

Title: Braid invariants resource