问:
A special graph coloring problem

Let $G$ be an undirected graph. We define:

An Orientation $D$ from $G$: A subgraph $D$ with $V(D) = V(G)$ such that for every edge $\{u, v\} \in E(G)$, we take either $(u, v)$ or $(v, u)$ in $D$.

We call a function $c: V(D) \rightarrow \{1, ..., k\}$ a $k$-In-Out-Coloring if for every vertex $v \in V(D)$ the following holds:
$$\{c(u) | (u, v) \in E(D) \} \cap \{c(w) | (v, w) \in E(D) \} = \phi$$

In other words, no vertex in a $k$-in-out-colored, directed graph has an in-neighbor $u$ and an out-neighbor $w$, so that $c(u) = c(w)$. In this graph for example, the vertices $0$, and $2$ have to colored in different colors, so $c(0) \neq c(2)$. Note that we didn't impose restrictions on vertex $1$, it is allowed to be colored in $c(0)$ or $c(2)$.

For a function $f: A \rightarrow B$ and a subset $C \subseteq A$ we define a restriction from $f$ onto $C$ as $f_{\restriction C}(x):= f(x)$ if $x \in C$. So the function $f_{\restriction C}$ has the type $f_{\restriction C}: C \rightarrow B.$

Let $k \in \mathbb{N}$ be fixed, $G$ an undirected Graph (with no parallel edges or self-loops) with countably infinite vertices and let $c:V(G) \rightarrow \{1, ..., k\}$ be an $k$-in-out coloring function.

Prove: If every finite subgraph $G'$ from $G$ has an orientation which is $k$-in-out-colored by $c_{\restriction V}(G')$, then there must exist an orientation of $G$, which is $k$-in-out-colored by $c$.

There are too many layers in this proof and I don't know where to begin. A proof by contradiction seems suitable but even then, I don't see what follows.

If every subgraph $G'$ has an orientation which is $k$-in-out-colorable by $c_{\restriction V}(G')$, then the subgraph $H$ with $V(H) = V(G)$ and $E(H) = E(G)$ is $k$-in-out-colorable by $c_{\restriction V}(G')$. If $H$ is $k$-in-out-colorable by this restriction function, then it must be colorable by $c$, and so must be $G$, which is exactly the same graph as $H$. So I'm not really sure what the question is looking for. Any help would be much appreciated.

答:
As the vertex set is enumerable, so does the set of edges. Then let $\{e_{i}\}_{i \in \mathbb{N}}$ be an enumeration of $E(G)$. For each $i \in \mathbb{N}$, let $\{u_{i},v_{i}\} = \{e_{i}\} $ and define $G_{i} = (\bigcup_{j = 1}^i e_{i},  \{ e_{j} ~\colon j \in \{1,\ldots, i\}\})$. In other words, $G_{i}$ is the finite graph induced by the edges $\{e_{1},\ldots, e_{i}\}$. Let $G_{i}'$ be the orientation of $G_{i}$ respecting the property mentioned by OP.

We will define an orientation in $\{e_{i}\}_{i \in \mathbb{N}}$ by induction. First, we have two possibilities of orientation for $e_{1}$, $(u_{1},v_{1})$ or $(v_{1},u_{1})$. Moreover an orientation of the edge $e_{1}$ appears in an infinite number of graphs in the sequence $\{G_{i}'\}$(of course appears in all of them). Then, by the pigeonhole principle, at least one of theses orientations occurs in an infinite subsequence of $\{G_{i}'\}$. Then we define $\overrightarrow{e_{1}}$ to be the orientation of $e_{1}$ that occurs in all orientations of the subsequence $\{G_{i_{k}}\}_{k \in \mathbb{N}}$. Now notice that the edge $e_{2}$ occurs in an infinite number of orientations of $\{G_{i_{k}}\}$. Therefore, using the same argument one can find an infinite subsequence of $\{G_{i_{k}}\}$ where the orientation of $e_{2}$ is the same for all graph in this subsequence. And then define $\overrightarrow{e_{2}}$ to be this orientation. Notice that even in this news subsequence, all orientations of $e_{1}$ still are $\overrightarrow{e_{1}}$. By induction we can define the orientation of all edges like this. Than call by $A = \{\overrightarrow{e_{1}},\overrightarrow{e_{2}},\overrightarrow{e_{3}}, \ldots\}$.

Now we must prove that $A$ respects the property. Suppose otherwise and take two vertices $w_{1}$ and $w_{2}$ such that $c(w_{1}) = c(w_{2})$ and there is a vertex $z$ such that $(w_{1},z,w_{2})$ is a directed path in $(V(G),A)$. Let $i_{1}$ and $i_{2}$ be natural numbers such that $(w_{1}, z) = \overrightarrow{e_{i_1}}$ and $(z,w_{2}) = \overrightarrow{e_{i_{2}}}$. Without loss of generality suppose that $i_{2} > i_{1}$.

Let's remember the construction. When we are choosing the orientation of $e_{i_2}$, we are in a subsequence $\{G_{i_{k}}\}$ such that the graphs $\{G_{i_{k}}'\}$ contain all edges from $e_{1}$ to $e_{2} - 1$ and they are oriented in the same direction, in particular the direction of $e_{i_{1}}$ is the same as in $A$. Then we take a subsequence which by abuse of notation we will denote also by $\{G_{i_{k}}'\}$ where $e_{i_2}$ is oriented in the same direction as in $A$, $\overrightarrow{e_{2}}$. Let $G^*$ be an arbitrary graph in the  subsequence of orientations. As $(w_{1},z,w_{2})$ is a directed path in $G^*$ and the property holds by hypothesis for finite sets, we have that $c(w_{1}) \neq c(w_{2})$. An contradiction.

I know that it is far from being easy to understand this demonstration and maybe its because of my english mistakes, I am sorry in advance. If you don't understand feel free to ask.

Title: A special graph coloring problem


问:
Probability for normal distribution

The speed of cars is represented as normal distribution $N(87;144)$. What is the probability that a randomly selected car has a speed that is least $100$ km/h?

My solution is:

For $x = 100: Z = (100 - 87)/ 12 = 1.083$

$P(Z>1.083) = 1 - 0.85993 = 0.14$

So, the answer is that the probability of randomly selected car has speed $100$ km/h or higher is about $14$%.

And my question: is that a correct solution or did I do something wrong?

答:
Your steps and reasoning are correct but pay attention to the wording of your final answer. You mentioned:

$P(z>1.083) = 1 - 0.85993 = 0.14$ So, the answer is that the
probability of randomly selected car has speed $100$ km/h or higher is
about $14$%.

However that is not true, the probability is $0.14$ and not $14$% and that is what they asked for, the percentage is $14$%, remember that any probability must be between $0$ and $1$.

ContentLicense: CC BY-SA 4.0

Title: Probability for normal distribution


问:
Formula for the probabilities of guessing `x` correct pairings
Suppose:

we have a set A with n elements

we have a set B with n elements

every element in A has exactly one "friend" in B

every element in B has exactly one "friend" in A

someone blindly guesses x pairings correctly

p is the probability of guessing x pairings correctly

How might one calculate each p for x=0 to x=n?

Sorry in advance if there is a better way to word this problem; I am not a mathematician.

答:
Probability that at least one pairing is correct (probability that at least one letter is placed in the correct envelope) $=$ $$\sum_{k=1}^{k=n} \frac{(-1)^{k+1}}{k!}$$

Probability that exactly x pairings are correct (that exactly x letters are placed in the correct envelope) $=$
$$\frac{1}{x!}\sum_{k=0}^{k=n-x} \frac{(-1)^k}{k!}$$

These can be found by using inclusion-exclusion and derangement ideas.


问:
Trying to find the conditional expected value of X when X + Y = m, and X, Y are both independent binomial random variables w/ parameters n & p
If X and Y are independent binomial random variables with identical parameters n and p, calculate the conditional expected value of X given X+Y = m.

The conditional pmf turned out to be a hypergeometric pmf, but I'm a but unclear on how to relate that back into finding E[X|X+Y=m]

答:
$$E(X|X+Y)+E(Y|X+Y)=E(X+Y|X+Y)=X+Y$$

Given $X\perp \!\!\!\perp Y$ the are changeable too, thus $E(X|X+Y)=E(Y|X+Y)$

Concluding:

$$2E(X|X+Y)=X+Y$$

$$E(X|X+Y)=\frac{X+Y}{2}=\frac{m}{2}$$


问:
Volumes of n-balls: what is so special about n=5?

The volume of an $n$-dimensional ball of radius $1$ is given by the classical formula
$$V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}.$$
For small values of $n$, we have
$$V_1=2\qquad$$
$$V_2\approx 3.14$$
$$V_3\approx 4.18$$
$$V_4\approx 4.93$$
$$V_5\approx 5.26$$
$$V_6\approx 5.16$$
$$V_7\approx 4.72$$
It is not difficult to prove that $V_n$ assumes its maximal value when $n=5$.

Question. Is there any non-analytic (i.e. geometric, probabilistic, combinatorial...) demonstration of this fact? What is so special about $n=5$?

I also have a similar question concerning the $n$-dimensional volume $S_n$ ("surface area") of a unit $n$-sphere. Why is the maximum of $S_n$ attained at $n=7$ from a geometric point of view?

note: the question has also been asked on MathOverflow for those curious to other answers.

答:
At some point, the factorial must overtake the power function.  This happens at five dimensions for the sphere, and seven for the surface.   The actual race that is involved is between an alternation of $2$ and $\pi$, against $n/2$.  At $n=5$, $\frac{5}{2}>2$, but $\frac{6}{2}<\pi$.

That means that five dimensions is the last dimension that the sphere's volume stops increasing relative to the prism-product of the radius.

After 19 dimensions, the surface of the sphere is less than the prism-product of the radius.

ContentLicense: CC BY-SA 2.5

ContentLicense: CC BY-SA 3.0


问:
Prove if a discrete subset $E$ of $\mathbb{R}^n$ is compact, then it is finite.

Let $E\subseteq \mathbb{R}^n$ be a compact set and suppose that for
each $x\in E$, there is an $r_x \gt 0$ such that $E \cap B_{r_x}(x) = \{x\}.$

Prove that $E$ is a finite set.

I am a bit confused as to what "finite" means here. Do I have to prove that $E$ is closed and bounded? This is giving me memories of the Heine-Borel Theorem. Am I on the right track?

答: $E$ is covered by the balls $B_{r_x}(x)$. By compactness there is a finite subcover, say $E \subseteq \bigcup_{i=1}^{m} B_{r_{x_i}}(x_i)$. So $E \subseteq \{x_1,x_2,...,x_m\}$ proving that $E$ has finite number of elements.


问:
Does f(-x) also make the horizontal phase shift negative as well?
The $f(-x)$ is the mirror image of $f(x)$ with respect to the Y axis, so then for example if I have $f(x)=\sin(x+h)$ would $f(-x)=\sin(-x-h)$? Or just $\sin(-x+h)$?

Do I just make the $x$ negative or the horizontal shifts come with it as well?

答: You should only change the $x$, so $f(-x)=\sin(-x+h)$ would be the correct reflection. One easy way to verify is to check $f(0)=f(-0)=\sin(h)\neq\sin(-h)$


答:
The conditional pmf turned out to be a hypergeometric pmf...

Indeed it is, thanks to using Vandermonde's identity we see that...

$$\begin{align}\mathsf P(X=k\mid X+Y=m) &= \dfrac{\mathsf P(X=k,Y=m-k)}{\mathsf P(X+Y=m)}\\[1ex] &=\dfrac{\dbinom nk\dbinom n{m-k}p^m(1-p)^{2n-m}}{\displaystyle\sum_{h=0}^m \dbinom nh\dbinom n{m-h}p^m(1-p)^{2n-m}}\,\mathbf 1_{\langle k,m\rangle\in\Bbb N^2:0\leq k\leq m\leq n}\\[1ex]&=\dfrac{\dbinom nk\dbinom n{m-k}}{\dbinom{2n}{m}}\,\mathbf 1_{\langle k,m\rangle\in\Bbb N^2:0\leq k\leq m\leq n}\end{align}$$

$\therefore (X\mid X+Y=m)\sim\mathcal{Hypergeo}(2n,n,m)$

, but I'm a but unclear on how to relate that back into finding E[X|X+Y=m]

Well the expected value for a hypergeometric random variable where $\sim\mathcal{Hypergeo}(N,K,m)$ is $mK/N$, so...

Title: Trying to find the conditional expected value of X when X + Y = m, and X, Y are both independent binomial random variables w/ parameters n & p

Body:
If X and Y are independent binomial random variables with identical parameters n and p, calculate the conditional expected value of X given X+Y = m.

The conditional pmf turned out to be a hypergeometric pmf, but I'm a but unclear on how to relate that back into finding E[X|X+Y=m]


问:
System of linear equations having a real solution has also a rational solution.
I saw this question

Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose
  that the system of linear equations $Ax = b$ has a solution in
  $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?

and I thought I'd give an interesting, possibly wrong, approach to solving it. I'm not sure if such things can be done, if not maybe you can help me refine.

I considered the form of the equality as

$$ A^{(1)}x_1+\cdots+A^{(n)}x_n=b, $$

where $A^{(i)}$ is a column vector of $A$. I then noticed that for $x_i\in\mathbb{R\setminus Q}=\mathbb{T}$ then, and this is where I think I'm doing something forbidden, each $x$ has the represenation

$$ x_1  =  k_{11}\tau_1+\cdots+k_{1p}\tau_p \\
x_2= k_{21}\tau_1+\cdots+k_{2p}\tau_p \\
\vdots \hspace{4cm} \vdots \\
x_n=k_{n1}\tau_1+\cdots+k_{np}\tau_p,$$

where $\tau_i$ is a distinct irrational number, $k_{ij}\in\mathbb{R}$, and $p$ is the number of such distinct irrational numbers. I wound this out, but there may be a discrepancy with $p$ and $m$. I feel this method can lead me to the answer, but I'm not sure where to go from here.

EDIT$^1$:

I end up getting something like this, I believe, after substitution:

$$ A(k^{(1)}\tau_1+\cdots+k^{(p)}\tau_p)=b$$

Here, $k^{(i)}$ is the vector

$$\begin{pmatrix} k_{1i} \\ \vdots \\ k_{ni} \end{pmatrix}.$$

EDIT$^2$:

I think there is no discrepancy with $p$ and $m$ because $A\in M_{m\times n}(\mathbb{Q})$, $K\in M_{n\times p}(\mathbb{R})$, and $\tau\in M_{p\times 1}(\mathbb{T})$, so

$$ (m\times n)\cdot (n\times p)\cdot (p\times 1) = m\times 1. $$

答:
Yes, if the coefficients of $A$ and $b$ are rational and the system has at least one solution in reals, it also has at least one rational solution.

The elementary row operations performed during Gaussian elimination only use existing entries from the matrix and their inverses, thus keeping the matrix (and the vector $b$) rational all the time. Once the matrix is in reduced row-echelon form, some of the unknowns will be "free" to be set to any value (this happens if the rank of the matrix is smaller than the number of unknowns) and the others will be completely determined by them and the vector $b$. Setting the "free" ones to rational numbers yields completely rational solution to the original system.

More generally, if a system of linear equations over some field $F$ has a solution in its extension $E$, it also has a solution in $F$. Moreover, if we know there is at least one non-rational (or non-$F$ solution in the general version) solution, the cardinality of the set of solutions in $F$ will not be smaller than $|F|$, since there will be at least one "free" unknown.


问:
How to do power series expansion by long division?
So I have the question

$$\frac{2z+1}{z^2-z+0.5}$$

do I solve the problem of power series by long division like the following?

what I've done

I know the power series is infinite but I'm just a little confused on the terminology for the question I'm doing I'm supposed to go to the 4th coefficient but I'm not sure if I'm calculating correctly.

------------------------------------ EDIT

The exact wording of the question is as follows

A causal system is defined by transfer function

$$\frac{2z+1}{z^2-z+0.5}$$

Determine the first four non-zero coefficients of the system’s impulse response, ℎ(), using the power series expansion by long division method.

when I asked the TA who was teaching this section where I could find more info on this he said look at google so I don't know what to do because every example I get from google is talking about doing the inverse of a transfer function.

Body: Try to think of it as $$\frac{1}{z^4}\cdot\frac{2z^5+z^4}{z^2-z+0.5}=\frac{1}{z^4}(2z^3+3z^2+2z+\cdots)$$


问:
Integrating $\int \frac{dx}{\sqrt{x+1}+\sqrt[3]{x}}$ and $\int \frac{dx}{\sqrt[3]{1+x^2}}$
Good day folks, last couple of days I had been having a good time with indefinite integrals up until stumpled upon these two
$$\int \frac{dx}{\sqrt{x+1}+\sqrt[3]{x}}$$
and
$$\int \frac{dx}{\sqrt[3]{1+x^2}}$$
after what's been about 2 or 3 days I gave up and decided to use a calculator to help me out.
But I tried a bunch of them and they either dont compute or give a result in terms of non-elementary functions existence of which we (as current students) shouldn't even know about, let alone use for solving.

Are they really unsolvable (Well, unsolvable in terms of elementary functions, but those are synonyms at this point) ?

答:
For integral 1, making the substitution $u=x^{1/3},\frac{du}{dx}=\frac1{3u^2}$ as suggested in comments:
$$\int_0^y\frac1{\sqrt{x+1}+\sqrt[3]x}\,dx=\int_0^{y^{1/3}}\frac{3u^2}{u+\sqrt{u^3+1}}\,du$$
which is indeed a non-elementary integral (but can be expressed in terms of elliptic integrals – it gets really messy).

Integral 2 is explicitly covered in Byrd and Friedman (274.01), so is also non-elementary and elliptic:
$$\int_0^y\frac1{\sqrt[3]{x^2+1}}\,dx=\frac{3(1-\sqrt3)}{2\sqrt[4]3}F(\varphi,m)+3^{5/4}\int_0^{u_1}\frac1{1+\operatorname{cn}u}\,du$$
where $m$ is the the parameter and
$$\operatorname{cn}u_1=\cos\varphi=\frac{\sqrt3+1-\sqrt[3]{y^2+1}}{\sqrt3-1+\sqrt[3]{y^2+1}}\qquad m=\frac{2-\sqrt3}4$$
The integral with bounds $[0,u_1]$ above in turn evaluates by B&F 341.53 as
$$\int_0^{u_1}\frac1{1+\operatorname{cn}u}\,du=F(\varphi,m)-E(\varphi,m)+\frac{\operatorname{sn}u_1\operatorname{dn}u_1}{1+\operatorname{cn}u_1}$$


问:
What does a proof in an internal logic actually look like?
The nLab has a lot of nice things to say about how you can use the internal logic of various kinds of categories to prove interesting statements using more or less ordinary mathematical reasoning. However, I can't find a single example on the nLab of what such a proof actually looks like. (The nLab has a frustrating lack of examples in general.)

Can anyone supply me with some examples? I'd be particularly interested in the following kinds of examples:

I've heard in a topos one can internalize the real numbers and, in the topos $\text{Sh}(X)$ of sheaves on a topological space, this reproduces the sheaf of continuous real-valued functions $X \to \mathbb{R}$. Moreover, one can internalize "finitely generated projective $\mathbb{R}$-module" and in $\text{Sh}(X)$ this reproduces real vector bundles on $X$. What can you prove about vector bundles this way?

I'd also like to see examples of what you can do in the internal logic of Cartesian closed categories.

This MO question is related but it doesn't really satisfy my curiosity.

答:
There are a number of examples in "Sketches of an elephant" by Johnstone, but two that stand out are D1.3.13 and D1.3.14, dealing with regular categories and their internal logic.  The first shows that every cover is a coequalizer, and the second shows that the square formed by $A_1,A_2,A_1 \cap A_2$ and $A_1 \cup A_2$, along with their inclusions, is a pushout.  The great thing about these two examples is that they are also proven earlier using categorical methods, in A1.3.4 and A1.4.3.  This way you can see that the proofs with the internal language are much more familiar than the categorical way.

For vector bundles, the paper "A generalization of Swan's theorem" by Mulvey uses the internal language, if I remember correctly.

Also look at "Generic Galois theory of local rings" by Wraith, for examples related to the etale topos.


问:
prove the recurrence sequence is convergent

Consider the sequene $\{a_n\}$: $a_1>0$ and $\displaystyle a_{n+1}=\frac{2n}{2n+a_n}$ for $n=2,3,\cdots$. Prove that the sequence $\{a_n\}$ is convergent.

To prove $\{a_n\}$ is convergent, we have to to  show it is monotone and bounded. But I'm unable to show it. How can I do it? Does there any ther simplest way to show the convergence ?

Since $a_1>0$, from the recurrance relation, it is easily seen that $a_n>0$ for all $n\in \Bbb N$.

Now, $a_n-a_{n+1}=\frac{(a_n+n)^2-n(n+2)}{2n}$ from this I can nt say anything about the sign , so that it proves increasing/decreasing!

答:
Observe that
$$\alpha_{n+1}=\frac{2n}{2n+\alpha_n}\leq 1$$
for all $n$. This shows that $(\alpha_n)$ is bounded. Hence, $\alpha_n/2n\to 0$ as $n\to \infty$. Now,
\begin{align}
\alpha_{n+1}&=\frac{2n}{2n+\alpha_n}\\
&=\frac{2n}{2n(1+\frac{\alpha_n}{2n})}\\
&=\biggl(1+\frac{\alpha_n}{2n}\biggr)^{-1}\to 1
\end{align}
since $\alpha_n/2n\to 0$.


问:
If T and T' are two topologies on X and T ⊂ T', what does the connectedness of X in one topology imply about connectedness in the other topology?
Here is what I have: If (X,T) is disconnected, then  there exists a separation (U,V), where U and V are disjoint open non-empty subsets of (X,T) Since we are given T ⊂ T', then U and V are open non-empty subsets of (X,T'), and their union is X. Then, (U,V) is a separation of (X,T'), so (X,T') is disconnected. Conversely, if (X,T') is connected, then (X,T) is connected. Is this correct?

答: You're correct. Another way to see it:  the identity $\textrm{id}:(X,\tau') \to (X,\tau)$ is continuous and preserves connectedness.


问:
How do we know that the Sine function has no Non-Real Roots?
In this question and answers (How was Euler able to create an infinite product for sinc by using its roots?) we use the fact that the real roots of $f(x)=\sin x$ occur when $x$ is an integer multiple of $\pi$ to obtain an infinite product for $\sin x$ in terms of of its factors.

My question is, how do we know that $f(x)=\sin x$ has no non-real roots?

Thanks in advance.

答:
Since $\sin z = \frac{1}{2i} ( e^{iz} - e^{-iz} )$, we see that $\sin z = 0$ if and only if $e^{iz} = e^{-iz}$, or $e^{2iz}=1$. Taking the modulus of both sides, we see that
$$
1 = |1| = |e^{2iz}| = e^{\Re(2iz)} = e^{-2\mathop{\rm Im} z},
$$
which implies that $-2\mathop{\rm Im} z=0$ and thus $\mathop{\rm Im} z=0$.


问:
filter on a topological space- how to define filter
So $(X,\tau$) is a topological space. Let $\mathcal{C}$ be the collection of closed sets in $X$. Then a $\textit{filter}$, on $\mathcal{C}$, call it $\mathcal{F}$, is a sub collection of these closed sets such that

$(1)$ $\emptyset \notin \mathcal{F}$

$(2)$ closed under intersection, that is, if $C_i,C_j \in \mathcal{F}$, then $C_i \cap C_j \in \mathcal{F}$.

($3)$ closed under supersets?, that is if $C_i \subset C_j$ where $C_i \in \mathcal{F}$ and $C_j \in \mathcal{C}$, then $C_j \in \mathcal{F}$.

My problem says if every filter on $\mathcal{C}$ has non-empty intersection, prove that $(X,\tau)$ is compact.

I'd like how to rigorously define what it means for a filter to have nonempty intersection, so what intersection is nonempty? what is the subscript? where are the elements being pulled from?

so $\bigcap \mathcal{F} \neq \emptyset$ but I am so lost when it comes to the subscripts here, that is, the subscript of my $\bigcap$. Basically want to know how to formally define a filter on a top space. I have never heard of filters til now. Are they a way of sort of sifting out smaller sets and only including "bigger" sets? And can one put a filter on $\tau$ as well as the closed subsets? What is the purpose of a filter? Feel like once I get a hang on what a filter is I could possibly prove this statement, so I would like to know about this filter business not a solution to my problem I want to still attempt it myself I do not like spoon feeding.

答:
Brian already explained what the intersection means: the set of all points of $X$ that lie in all members of the filter $\mathcal{F}$.

If all (closed) filters have non-empty intersection. Then start with an arbitrary cover $\mathcal{U}$ of $X$. If it has a finite subcover you're done. If not, consider $\mathcal{G}=\{X\setminus U: U \in \mathcal{U}\}$ and consider the smallest filter that contains it...


问:
Given an orthonormal basis $\beta$, transformation $T$ and $T'$, show that $M_\beta(T')$ is the transpose of $M_\beta(T)$
Let $\beta=\{f_1,f_2,...,f_n\}$ be an orthonormal basis of an inner product space $V$. Given a linear operator $T:V \rightarrow V$, define $T':V \rightarrow V$ by
$$T'(v)=\sum_{i=1}^n \langle v,T(f_i) \rangle f_i$$
Show that $M_\beta(T')$ is the transpose of $M_\beta(T)$.

We know that for any $v \in V$, we can write $v= \sum_{i=1}^n \langle v,f_i \rangle f_i$. So in this case $T′(v)=∑_{i=1}^n⟨T′(v),f_i⟩f_i=∑_{i=1}^n⟨v,T(f_i)⟩f_i$. Sadly I do not really know how we can use this to our advantage. I have thought about using the fact that $⟨T′(v),f_i⟩ = ⟨v,T(f_i)⟩$, but I do not see how I can continue. Could anyone provide me with a hint?

答: But, isn't it true that $\langle Tv,w\rangle=\langle v,T^tw\rangle$?  This appears to make it rather trivial, since then $T'$ and $T^t$ agree on any basis.


问:
A problem regarding the stopped $\sigma$-field
Suppose $(\Omega,\mathcal F,\{\mathcal F_n\}_n,\mathbb P)$ is a filtered probability space. Let $\tau$ be a stopping time, and define a collection of sets $\mathcal F_\tau:=\{A\in\mathcal F: A\cap\{\tau=n\}\in\mathcal F_n\forall n<\infty\}$. We can easily show that $\mathcal F_\tau$ is a $\sigma$-algebra.

Let $M=\{M_n\}_n$ be a martingale. Show that $\mathbb E[M_n|\mathcal F_\tau]=M_{n\wedge\tau}, n\ge 0$.

Let $M=\{M_n\}_n$ be a uniformly integrable martingale with last element $M_\infty\in\mathcal F_\infty$. Show that $\mathbb E[M_\infty|\mathcal F_\tau]=M_\tau$.

My attempt:

I would use the definition of conditional expectation to show both of the identities.

For any $S\in\mathcal F_\tau$, we want to show that $\mathbb E[M_{n\wedge\tau};S]=\mathbb E[M_n;S]$. But then I got completely stuck for hours. Even after checking a lot of posts dealing with this kind of problem, I am still quite confused with the definition of $\mathcal F_\tau$. Can someone help me out? Thank you!

答:
First we have to show that $M_{n\wedge \tau}$ is $\mathcal F_{\tau}$ measurable. For this we have to show that $(M_{n\wedge \tau} \leq x)\cap (\tau=k) \in \mathcal F_k$. But  $(M_{n\wedge \tau} \leq x)\cap (\tau=k)=(M_{n\wedge k} \leq x)\cap (\tau=k)$ which is an intersection of two sets in $\mathcal F_k$.

Next we have to show that $\int_A M_{n\wedge \tau} dP=\int_A M_n dP$ for $A \in \mathcal F_{\tau}$. It is enough to show that $\int_{A\cap (\tau=k)} M_{n\wedge \tau} dP=\int_{A\cap (\tau=k)} M_n dP$ for each $k$. In other words we have  to show $\int_{A\cap (\tau=k)} M_{n\wedge k} dP=\int_{A\cap (\tau=k)} M_n dP$. This follows by martingale property if $n > k$ and trivially holds if $n \leq k$.

The second part is just obtained by taking limit as $n \to \infty$ since a uniformly integrable martingale converges almost surely and in $L^{1}$.


问:
Discrete analogy of partial integral formula.
Show that discrete analogy of partial integral formula as below.
$$
 \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)=x(n)y(n)-\sum\limits_{k=0}^{n-1}x(k+1) \Delta y(k) + c
$$

I have tried as below and I stuck.

\begin{align*}
  \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&=
  \sum\limits_{k=0}^{n-1} y(k)\left(x(k+1)-x(k)\right)\\
  &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k)y(k)\right)\\
  &=\sum\limits_{k=0}^{n-1} x(k+1)y(k)- \sum\limits_{k=0}^{n-1} x(k)y(k)\\
 \end{align*}

Now I can't expand it in form

$$=x(n)y(n)-\sum\limits_{k=0}^{n-1}x(k+1) \Delta y(k) + c.$$

Anyone can give me hint?

Body:
\begin{align*}
  \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&=
  \sum\limits_{k=0}^{n-1} y(k)\left(x(k+1)-x(k)\right)\\
  &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k)y(k)\right)\\
  &=\sum\limits_{k=0}^{n-1} \left(x(k+1)y(k)-x(k+1)y(k+1)+x(k+1)y(k+1)-x(k)y(k)\right)\\
  &=\sum\limits_{k=0}^{n-1} \left(-x(k+1)(y(k+1)-y(k))+x(k+1)y(k+1)-x(k)y(k)\right)\\
  &=-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k)+ \sum\limits_{k=0}^{n-1} \left(x(k+1)y(k+1)-x(k)y(k)\right) \\
  &=-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k)+  \left(x(n)y(n)-x(0)y(0)\right).
 \end{align*}
Let $c=-x(0)y(0)$, we have
\begin{align*}
 \sum\limits_{k=0}^{n-1} y(k)\Delta x(k)&=x(n)y(n)-\sum\limits_{k=0}^{n-1} x(k+1)\Delta y(k) +c.
\end{align*}


问:
How do I solve this differential equation $y'' + y^2 = 0$?
$y'' + y^2 = 0$

I tried in the following way:

$y' =u.$ Therefore $y'' = u'$

$u' + y^2 =0$

$\int du = -y^2 \int dx$

$u = -y^2x + c$

$y' = -y^2x +c$

I don't know how to proceed further.

答:
$$y''=-y^2$$
Multiply bt $2y'$ both sides to get
$$2y'y''=-2y^2y' \implies \frac{d}{dx}(y'^2)=-2y^2 \frac{dy}{dx}.$$
Integrating both sides w.r.t $x$, we get
$$y'^2=-\int 2y^2 dy +A \implies y'^2=\frac{2y^3}{3}+C.$$
$$\implies y'=\pm\sqrt{A+2y^3/3}$$
Integrating ww.r.t. $x$ we have
$$\int\frac{dy}{\sqrt{A+2y^3/2}}=\pm \int dx+B$$
$A,B$ are constant of integration to be found by further given conditions.


问:
Determine E(X) in terms of E(Y) and V(X) in terms of V(Y)
Let $X$ and $Y$ have respective cumulative distribution functions $F_x$ and
$F_y$, and suppose that for some constants $a \in \mathbb{R}$ and $b > 0$,

$F_x(x) = F_y((x − a)/b)$

(a) Determine $E(X)$ in terms of $E(Y)$ and $V(X)$ in terms of $V(Y)4.

I'm not sure about $V(X)$ but I think $E(X)$ should be $(E(y)-a)/b.$ Would this be right?

答:
For ease of notation, let's say $Y$'s CDF and density are $G(y)$ and $g(y)$ respectively. We have $f(x) = \frac{1}{b}g(\frac{x-a}{b})$, from which we have $EX = \int xf(x)dx = \int (bz+a)g(z)dz = bEY+a$.

From linearity of $E$ you can guess $Var(X) = b^2Var(Y)$. Also, it is not too difficult to go through the change of variables in the integration to evaluate $EX^2 = \int x^2f(x)dx$ as in $EX$.


问:
Find How Many Leaves a Ternary Tree Graph Has
I stumbled across a problem while programming and was wondering if it had a nice mathematical solution. This is my first post, hopefully I follow your rules well and my terminology is understandable. The problem is as follows:

Setup:

Given a list of numbers like {1,2,3,5,8,9,10}

You can move at most +3 from the number you are on to the next number

You must start on the lowest number and end on the highest number

Assume that the numbers allow for a solution

For the numbers {1,2,3,5,8,9,10} the tree would look like:

```<code>                     1
                  ┌──┴──┐
                  3     2
           ┌──────┘  ┌──┴──┐
           5         5     3
           │         │     └─────┐
           8         8           5
        ┌──┴──┐   ┌──┴──┐        │
        9     10  9     10       8
        │         │           ┌──┴──┐
        10        10          9     10
                              │
                              10
</code>```

It has 6 total leaves.
The problem could be solved by programmatically creating all possible paths then counting them, but that's boring and slow. Is there some way to count the leaves without having to do that?

答:
I believe the problem is to count the number of such paths.
This problem can be solved for general graphs, and in this case using a recurrence.

Let the numbers be $a_1<a_2<\ldots<a_n$.
Let $f(k)$ denote the number of paths from $a_1$ to $k$.

The recurrence is: $f(k)=f(k-1)g(k-1)+f(k-2)g(k-2)+f(k-3)g(k-3)$,
where $g(i)=1$ if $i$ is present in the sequence and $g(i)=0$ otherwise. The last step to $k$ could have been from one of $k-1,k-2,k-3$ which explains the recurrence.

The values $f(k)$ can then be computed in order for $k=a_1,a_2,\ldots,a_n$.


问:
Proving rigorously that $(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$ is an ascending chain of ideals of $A$?
I was reading that question here How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian? :

The question was:

Consider the subring $R=k[x,xy,xy^2,\ldots]$ of $k[x,y]$. I want to prove that $R$ is not noetherian.

An ascending chain of ideals is the following: $$(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$$ It is intuitively clearly to me that this is an ascending chain of ideals. But how do I prove it rigorously that $$xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$$ or that this chain of ideals can never stabillize?

And one of the answers was this:

Preliminary remark

Given a polynomial $f(x,y)\in k[x,y]$, write $[f]_{i,j }$ for its term $ax^iy^j$ in the monomial $x^iy^j$.

Then for a finite sum $f=\sum_lf_l$ of polynomials we have $[f]_{i,j }=\sum [f_l]_{i,j}$

Proof that $xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})$

Suppose that $xy^n=\sum _{l=0}^{n-1}g_lxy^l$ with $g_l\in R$.

From the preliminary remark we get $$  xy^n=[xy^n]_{1,n}=[\sum _{l=0}^{n-1}g_lxy^l]_{1,n}\stackrel{prel.rem.}{=} \sum _{l=0}^{n-1}[g_lxy^l]_{1,n}=\sum _{l=0}^{n-1}[g_l]_{0,n-l}xy^l$$  But all $[g_l]_{0,n-l}=0$ since $n-l\gt0$  and all non-constant terms of a polynomial in $R$ involve a positive power of $x$.

Contradiction.

1-My question is that I do not understand how this answer proves that $(x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots$ is an ascending chain of ideals of $R$? could anyone explain this to me please ? or is this was an answer to the OP second question "a proof that it can not stabilize"

2-Also, why we need in the finite sum $f$ of polynomials in the preliminary remark in this answer?

EDIT: I specifically do not understand what contradiction that we actually get? could anyone clarify this to me please?

答:
Regarding your question 1, you ask precisely one literal question:

But how do I prove it rigorously that
$xy^n \not\in (x,xy,xy^2,\dots,xy^{n−1})$
or that this chain of ideals can never stabillize?

The theorem you recite starts

Proof that $xy^n \not\in (x,xy,xy^2,\dots,xy^{n−1})$...

So evidently, the proof is precisely responding to your question.  Since you argue that this is sufficient, you should already know how the proof of this fact resolves your question.

Regarding your second question.  If one wishes to show that some object is not an element of generated set, one way is to take a fully generic generated element and shows that equating it to your object leads to a contradiction.  So a way to show that $xy^n$ is not an element of the ideal is to equate it to a generic element of the ideal and discover that this leads to a contradiction.

A generic element of this ideal is not the form that is written.  (I was briefly concerned that the proof was wrong for this reason.)  However, it is the generic element with $x$-degree $1$.  The function of the preliminary remark is this : in any sum of element of $k[x,y]$, terms of specific multidegree in the result are exactly the sum of the terms of that multidegree in the summands.  So the output term of $(x,y)$-multidegree $(1,n)$ is the sum of the $(1,n)$-multidegree terms in the summands.  This means we can restrict to terms with $x$-degree $1$ in the generic form.  (And so the generic form of an element is not an error in the proof.)

Since the product of any two (non-scalar, equiv. non-$k$, equiv. positive degree) elements of the ideal has $x$-degree greater than $1$, we need not consider products of terms in the generic form of an element of the ideal that is (for purpose of contradiction) set equal to $xy^n$.  Since $xy^n$ has no terms of multidegree other than $(1,n)$, all terms of other multidegrees in the sum must cancel.  We may therefore assume that all terms in the sum with nonzero coefficients have multidegree $(1,n)$.  But all the terms in the sum have multidegree $(1,\ell)$ with $\ell < n$, so the coefficients must have nonzero degree, which is a contradiction -- the coefficients are scalars (i.e., elements of $k$), having multidegree $(0,0)$.


问:
Concluding the proof of the Baire category theorem
Let $(M,d)$ be a complete metric space, $A_n\subseteq M$ be nowhere dense for $n\in\mathbb N_0$, $x_0\in M$ and $r_0>0$.

We can show that there is a $(x_n)_{n\in\mathbb N}\subseteq M$ and $(r_n)_{n\in\mathbb N}\subseteq(0,\infty)$ such that $$\forall n\in\mathbb N:r_n<\frac{r_0}{2^n}\text{ and }\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}\tag1.$$

In fact, since $A_1$ is nowhere dense, there is a $x_1\in B_{r_0}(x_0)\setminus\overline{A_1}\ne\emptyset$. And since $B_{r_0}(x_0)\setminus\overline{A_1}$ is open, there is a $r_1\in\left(0,\frac{r_0}2\right)$ with $$\overline B_{r_1}(x_1)\subseteq B_{r_0}(x_0)\setminus\overline{A_1}\tag2.$$ Continuing this procedure, we obtain $(x_n)_{n\in\mathbb N}$ and $(r_n)_{n\in\mathbb N}$ as claimed.

Now, by construciton, $$\forall n\in\mathbb N:\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\subseteq\overline B_{r_{n-1}}(x_{n-1})\tag3$$ and hence $$\forall n\ge m\ge0:x_n\in\overline B_{r_n}(x_n)\subseteq\overline B_{r_m}(x_m)\tag4.$$ Moreover, $$\forall n\in\mathbb N:\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\setminus\bigcup_{i=1}^n\overline{A_i}=B_{r_0}(x_0)\cap\bigcap_{i=1}^n\overline{A_i}^c\tag5$$ and hence $$\bigcap_{n\in\mathbb N}\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\cap\bigcap_{n\in\mathbb N}\overline{A_n}^c=B_{r_0}(x_0)\setminus\bigcup_{n\in\mathbb N}\overline{A_n}\tag6.$$ By $(4)$, $$d(x_m,x_n)\le r_{\min(m,\:n)}\xrightarrow{m,\:n\:\to\:\infty}0\tag7$$ and hence $(x_n)_{n\in\mathbb N_0}$ is Cauchy. Since $M$ is complete, there is a $x_\infty\in M$ with $$d(x_n,x_\infty)\xrightarrow{n\to\infty}0\tag8.$$ By $(4)$ and $(6)$, $$x_\infty\in\bigcap_{n\in\mathbb N_0}\overline B_{r_n}(x_n)\subseteq B_{r_0}(x_0)\setminus\bigcup_{n\in\mathbb N}\overline{A_n}\tag9.$$

The desired conclusion is that $M\ne\bigcup_{n\in\mathbb N}A_n$. Or, the even stronger conclusion, that $M\setminus\bigcup_{n\in\mathbb N}A_n$ is dense.

However, we should even have shown that $M\setminus\bigcup_{n\in\mathbb N}\overline{A_n}$ is dense, since we have shown that for arbitrary $x_0\in M$ and arbitrary small $r_0>0$, there is a $x_\infty\in M$ with $d(x_0,x_\infty)<r_0$.

From this it clearly follows that $M\setminus\bigcup_{n\in\mathbb N}A_n\supseteq M\setminus\bigcup_{n\in\mathbb N}\overline{A_n}$ is dense as well; which in turn clearly implies that $M\ne\bigcup_{n\in\mathbb N}A_n$.

However, my question is: If our goal is only to show the original claim (i.e. the denseness of $M\setminus\bigcup_{n\in\mathbb N}A_n$ wouldn't it be sufficient to choose the radii such that $B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}$ (instead of $\overline B_{r_n}(x_n)\subseteq B_{r_{n-1}}(x_{n-1})\setminus\overline{A_n}$) or would the given argument fail at some point?

答:
It’s not true in general.

Consider $\Bbb R$ with the usual metric, let $\Bbb Q=\{q_k:k\in\Bbb N\}$, and for $k\in\Bbb N$ let $A_k=\{q_k\}$; clearly each $A_k$ is nowhere dense in $\Bbb R$. Let $\alpha\in\Bbb R\setminus\Bbb Q$, and for $k\in\Bbb N$ let $x_k=2^{-k}\alpha$. Then $x_k\notin\bigcup_{k\in\Bbb N}\operatorname{cl}A_k=\Bbb Q$, but $\langle x_k:k\in\Bbb N\rangle$ converges to $0\in\Bbb Q$.


问:
Why does this formula about differential hold?
$f$ : $U(\subset \mathbb{R}^n) \to \mathbb{R}$, $x_0 \in U^i$

We fix $x \in U$ and we examine $f(x_0 + t(x-x_0))$ .

$\left (f(x_0 + t(x-x_0)) \text{ is a function of } "t". \right)$

Then, my textbook says that

\begin{align}
&\dfrac{d}{dt} f(x_0 +t(x-x_0))\\=
&\sum_{i=1}^n \dfrac{\partial f}{\partial x_i} (x_0+t(x-x_0))(x-x_0)_i.\\
&(x-x_0)_i \text{ means $i$-th component of } x-x_0.
\end{align}

I cannot understand why this holds.

I think that this is related to chain rule, but I cannot understand.(For example, why do I have to sum from $i=1$ to $i=n$ , and why $(x-x_0)_i$ appears.)

答:
Because when you write $f$ differential at $x_0$ using the partial derivatives you have

$$f^\prime(x_0).h =\sum_{i=1}^n \frac{\partial f}{\partial x_i}h_i$$

Where $h_i$ stands for the $i$-th coordinate of $h$. Then as you suspected, it is an application of the chain rule as the differential of $$t \mapsto x_0+t(x-x_0)$$ is $x-x_0$.


问:
Inverse Laplace Transform of $\frac{e^{a\sqrt{s}}}{s^2}$ where $a$ is a positive number?
I'm trying to invert the following equation back into the time domain

$$\frac{c}{s^{2}} \bigg(2-e^{-a\sqrt{s}}+e^{a\sqrt{s}}\bigg)$$ where ${a}$ and ${c}$ are positive constants.

I'm ok with solving the first two terms within the big brackets but I don't know how to deal with $\frac{e^{a\sqrt{s}}}{s^2}$. Are there any suggestions about how to tackle it?

答:
Hint: Note that the convolution of the functions $f$ and $g$ is given by $$f\ast g=\int_{0}^{t}f(\tau)g(t-\tau)d\tau$$
Also, we can see that $$\mathscr{L}^{-1}\left\{F(s)G(s)\right\}=f\ast g$$
Now, note that
$$\mathscr{L}^{-1}\left\{\frac{\exp(a\sqrt{s}}{s^{2}}\right\}=\mathscr{L}^{-1}\left\{\frac{1}{s^{2}}\cdot \exp(a\sqrt{s})\right\}=f(t)\ast g(t)$$
where $F(s)=\frac{1}{s^{2}}$ and $G(s)=\exp(a\sqrt{s})$.


问:
Relation between chords length and radius of circle
Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly</p>

答:
Draw a circle through c.  Let's suppose the chords are parallel to the x and y axis, and that the points a and b lie on some large circle, and the lines $x=a$ and $y=b$ cross at $A, B$

We then have the equation that $A^2+B^2=C^2$ for the crossing, which gives these lines, and thence $A^2 + b^2 = a^2 + B^2 = R^2$,  which leads to the equation $a^2+b^2+c^2 = 2R^2$.

Answer:   $R^2 = (a^2+b^2+c^2)/2$


问:
Cyclic inequality with fractions
Let $x,y,z\geq0$. Show that $x^4(y+z)+y^4(z+x)+z^4(x+y)\leq\frac{1}{12}(x+y+z)^5$

答:
Suppose $x \geqslant y \geqslant z$ and
$$f(x,y,z) = x^4(y+z)+y^4(z+x)+z^4(x+y).$$
We have
$$\begin{aligned}f(x,y,z)-f(x,y+z,0) & = y^4(z+x)+z^4(x+y)-x(y+z)^4\\ \\& = y^4(z+x)+z^4(x+y)-x(y+z)^4 \\ \\&=-yz\left[3x(y+z)^2+y^2(x-y)+z^2(x-z)\right] \leqslant 0.\end{aligned}.$$
Therefore $f(x,y,z) \leqslant f(x,y+z,0).$ Finally, we need to prove
$$f(x,y+z,0) = x^4(y+z)+x(y+z)^4\leqslant \frac{1}{12}(x+y+z)^5.$$
Setting $a = y +z,$ the inequality become
$$12ax(a^3+x^3) \leqslant (x+a)^5,$$
or
$$(a+x)(a^2-4ax+x^2)^2 \geqslant 0.$$
Which is true. The proof is completed.

Title: Cyclic inequality with fractions


问:
How to calculate this elliptic integral?
I need to calculate this integral
$\int \cos(t\theta)\sqrt{1-k^{2}cos(\theta)^{2}} \mathrm{d}\theta$

But according to Wolfram|Alpha there is no result in terms of standart mathematical functions.

How do we calculate this integral?

Edit: t is integer.

答:
$$I_n=\int_0^\varphi\cos n\theta\sqrt{1-k^2\cos^2\theta}\,d\theta=\int_0^\varphi T_n(\cos\theta)\sqrt{1-k^2\cos^2\theta}\,d\theta$$
where $T_n$ is a Chebyshev polynomial of the first kind. If $n$ is odd, $I_n$ is elementary because then $T_n(\cos\theta)=P_n(\sin^2\theta)\cos\theta$ with $P_n$ a polynomial; substituting $u=\sin\theta$ gives
$$I_n=\int_0^{\sin\varphi}P_n(u^2)\sqrt{1-k^2(1-u^2)}\,du$$
which is easily integrated. If $n$ is even, however, $I_n$ requires elliptic integrals: express $T_n(\cos\theta)=P_n(\sin^2\theta)$, then letting $\nu^2=\frac{k^2}{1-k^2}\ge0$,
$$I_n=\sqrt{1-k^2}\int_0^\varphi P_n(\sin^2\theta)\sqrt{1+\nu^2\sin^2\theta}\,d\theta=\int_0^\varphi\frac{P_n(\sin^2\theta)(1+\nu^2\sin^2\theta)}{\sqrt{1+\nu^2\sin^2\theta}}\,d\theta$$
This is a linear combination of integrals of the form
$$\int_0^\varphi\frac{\sin^{2j}\theta}{\sqrt{1+\nu^2\sin^2\theta}}\,d\theta=(1+\nu^2)^{-j-1/2}\int_0^{u_1}\operatorname{sd}^{2j}u\,du=(1-k^2)^{j+1/2}M_{2j}$$
The middle relation is Byrd and Friedman 282.04. The hidden parameter of the Jacobian elliptic function $\operatorname{sd}$ is $m=\frac{\nu^2}{1+\nu^2}=k^2$ and
$$\operatorname{sn}u_1=\sin\psi=\sqrt{\frac{(1+\nu^2)\sin^2\varphi}{1+\nu^2\sin^2\varphi}}$$
The integrals $M_{2j}$ are given by B&F 318.00, .02, .05:
$$M_0=F(\psi,m)\qquad M_2=\frac{E(\psi,m)-(1-m)u_1-m\operatorname{sn}u_1\operatorname{cd}u_1}{m(1-m)}$$
$$M_{2j+2}=\frac{2j(2m-1)M_{2j}+(2j-1)M_{2j-2}-\operatorname{cd}u_1\operatorname{nd}u_1\operatorname{sd}^{2j-1}u_1}{(2j+1)m(1-m)}$$


问:
Can subtraction be commutative
In primary and secondary school we all learnt that subtraction is not commutative, as only addition and multiplication is.

However, could subtraction be commutative for all real numbers that are subtracted from itself.

E.g 5-5=0
And 5-5=0

There are an infinite number of these pairs, so could subtraction be commutative when:
x=y
As in x-y?

Answers will be appreciated

答:
There is a sort of commutative property for subtraction. Observe that for all numbers $a, b, c$:  $$a-b-c = a-c-b.$$

But this is not usually what is meant by commutativity.

Title: Can subtraction be commutative


问:
Showing Completeness of a Space

Given $s\geq1$, $X$ is nonempty set and $d: X\times X \rightarrow [0,\infty)$ is $b_s$-pseudo metric on $X$ if for $x,y,z\in X$

i. $d(x,x)=0$

ii. $d(x,y)=d(y,x)$

iii. $d(x,y)\leq s[d(x,z)+d(z,y)]$

$F=\{d_n:n\in\mathbb{N}\}$ is a family of $b_s$-pseudo metric.

i. a sequence $\{x_n\}$ in $X$ is convergent-$F$ to $x\in X$ if for all positive $\varepsilon$ there exist $N_0\in\mathbb{N}$ such that $$d_k(x_n,x)<\varepsilon$$
for all $n\geq N_0$ and $d_k\in F$.

ii. a sequence $\{x_n\}$ in $X$ is Cauchy-$F$ if for all positive $\varepsilon$ there exist $N_0\in\mathbb{N}$ such that $$d_k(x_n,x_m)<\varepsilon$$
for all $n,m\geq N_0$ and $d_k\in F$.

iii. $X$ is complete-$F$ if each of its Cauchy-$F$ sequence is convergent-$F$.

I have $X =C([0,\infty),\mathbb{R})$ i.e. a set of all continuous function from $[0,\infty)$ to $\mathbb{R}$ and I also have $F=\{d_k:k\in\mathbb{N}\}$ where $$d_k(x,y)=\max_{t\in[0,k]}(x(t)-y(t))^2$$
for all $x,y\in X$

How can I show that $X$ is complete-$F$? I can't find $N_0$ such that its Cauchy-$F$ sequence is convergent-$F$.

答:
$d_k$ are indeed $b_s$-pseudo metrics on $X$. $(i),(ii)$ being obvious while for $(iii)$ we have
\begin{align}
d_k(x,y)&=\max_{t\in [0,k]}(x(t)-y(t))^2\\
&=\max_{t\in [0,k]}(x(t)-z(t)+z(t)-y(t))^2\\
&\leq 2\max_{t\in [0,k]}(x(t)-z(t))^2+2\max_{t\in [0,k]}(z(t)-y(t))^2\\
&=2d_k(x,z)+2d_k(z,y)
\end{align}
where in the third step we used the inequality $2\alpha \beta \leq \alpha^2+\beta^2$. Hence, $d_k$ are indeed $\beta_s$-pseudo metrics with $s=2$. Now, suppose we have an $F-$Cauchy sequence $(x_n)$ on $X$. Then, for $\epsilon>0$ we may find $N$ such that
$$\tag{1} d_k(x_n,x_m)=\max_{t\in [0,k]}(x_n(t)-x_m(t))^2<\epsilon$$
for every $n,m\geq N$. Fix some $t\geq 0$ and find $k\geq t$. Then, by $(1)$ we have
$$\tag{2}|x_n(t)-x_m(t)|\leq \sqrt{\epsilon}$$
for every $n,m\geq N$. This shows that $(x_n(t))$ is Cauchy sequence on $\mathbb{R}$. Hence, it converges to some limit denoted by $x(t)$. I.e. $\lim_{n\to \infty}x_n(t)=x(t)$ for every $t\geq 0$. Letting $n\to \infty$ in $(2)$ and squaring we obtain
$$|x(t)-x_m(t)|^2\leq \epsilon$$
for all $m\geq N$ and every $t\in [0,k]$. Taking $\sup$ to the above inequality we obtain
$$\tag{3} \sup_{t\in [0,k]}(x(t)-x_m(t))^2\leq \epsilon$$
for every $m\geq N$ and $k$. Now using $(3)$ we  can show that $f$ is continuous. Let $t_0\geq 0$ and $\epsilon>0$. By $(3)$ we may find $m$ such that
$$\tag{4} \sup_{t\in [0,k]}(x(t)-x_m(t))^2<\epsilon^2$$
for every $k$. By continuity of $x_m$ at $t_0$ there is and $\delta>0$ such that $|x_m(t)-x_m(t_0)|<\epsilon$ for every $|t-t_0|<\delta$. Pick $k>t_0+\delta$. Then, using $(4)$ and the continuity of $x_m$, for every $|t-t_0|<\delta$ we have
\begin{align}
|x(t)-x(t_0)|\leq |x(t)-x_m(t)|+|x_m(t)-x_m(t_0)|+|x_m(t_0)-x(t)|\leq 3\epsilon
\end{align}
Hence, $x$ is continuous. Now, returning to $(3)$ by the continuity of $x$ we can substitute $\sup$ with $\max$. Hence, rewriting $(3)$ we've shown that for $\epsilon>0$ there is an $N$ such that $d_k(x,x_m)\leq \epsilon$ for every $k$ and $m\geq N$. This shows that $x_m$ is convergent-$F$ to $x$.

Title: Showing Completeness of a Space


问:
Computing Homology group of three-dimensional manifolds
I'm preparing for an exam. I want to solve this past exam question:

Define three-dimensional torus by $T^3 = I^3/[(x,y,0)\sim (x,y,1), (x,0,z)\sim (x,1,z), (0,y,z)\sim (1,y,z)]$ and then other three-dimensional manifolds by $K^3 = I^3/[(x,y,0)\sim (1-x,y,1), (x,0,z)\sim (x,1,z), (0,y,z)\sim (1,y,z)]$.

I want to compute the homology groups of $T^3$ and $K^3$. For $T^3$, I think Hatcher does the the proof in page 142 and 143 of his book. Is the proof applicable for this question? More specifically, am I not supposed to use Mayers-Vitoris for this ? Also, on the same page 143, the homology groups are computed for $K\times S^1$, is this same as $K^3$? I think I may be confusing a bunch of stuff with each other.

I'd really appreciate if someone could write a solution for this especially for $K^3$

答:
Note that: $$T^{3} \cong S^1 \times S^1 \times S^1,$$ and $$K^3 \cong K \times S^1 $$ (as you had correctly noted.) To prove the second fact note that the image of the plane $\{y=0\}$ is a Klein bottle. Then, one may check that the map $K^3 \rightarrow K \times S^1$ given by $[(x,y,z)] \mapsto ([x,0,z],[y])$ is a diffeomorphism.

(but I think that in reality it is more instructive to draw a picture of a cube and see that the subsets $U_{K} = \{[(x,k,y)] : x,z \in [0,1] \}$ are Klein bottles, and indeed when the Klein-bottles are glued together there is no "twisting" in the $y$ direction i.e. we are just taking $K \times \mathbb{R}$ and just quotienting by the map $(p,t) \mapsto (p,t +1)$, so it is natural that we get a direct product with the circle).

Now the computation of the homology manifolds follows from the Kunneth Formula (i.e. the product formula for homology). The computations of the homology of the circle and the Klein bottle are easy to find.


问:
Strong characterization of $\mathbb C$ with respect to $\mathbb R$

$\mathbb R^2$ is not a field, but $2$-tuple arithmetic rules like $(a,b)(c,d)=(ac-bd,ad+bc)$ coupled with $\mathbb R^2$ make it a field, but are there other rules than $(a,b)(c,d)=(ac-db,ad+bc)$ combined with $\mathbb R^2$ that make it a field?

Is $\mathbb C$ the only field that contains $\mathbb R$ as a subfield or are there other fields that contain $\mathbb R$ as subfields?

Is $\mathbb C$ the only that can be made by coupling $\mathbb R^2$ and rules like $(a,b)(c,d)$ $=$ $(ac-db,ad+bc)$? Are there other rules that can be coupled with $\mathbb R^2$ to make it a field?

PS: Just trying to pin down some properties that make $\mathbb C$ special/unique from other fields that can/may be constructed using some specific algebraic structure of 2 tuples in $\mathbb R$ e.g. $(\sqrt 2,\pi)$. At this point this is just a wide net as I lack the terminology to make this question specific. But any hints on how to make this question specific to listing properties that make $\mathbb C$ unique or gives strong characterisation of relationship between  $\mathbb C$ and $\mathbb R$ are welcome.

答:
First off, ${\bf R}^2$ is not a field. Let me build up the idea you want. If $L/K$ is a field extension, then in particular $L$ is a vector space over $K$, and the degree of the extension $[L:K]$ is defined to be the dimension $\dim_KL$ of $L$ over $K$. So perhaps you want to ask

Question: What kinds of degree $2$ extensions of the reals are there?

Answer: The complex numbers are the only such extension.

This requires only the most basic field theory; if $K/{\bf R}$ is a degree two extension, then $K={\bf R}(a)$ for any $a\in K\setminus{\bf R}$, and such an $a$ has minimal polynomial of degree two over $\bf R$, the only irreducibles over $\bf R$ are of the form $(x+s)^2+b$ with $b>0$, so ${\bf R}(a)\cong {\bf R}(\sqrt{-b})={\bf R}(i)={\bf C}$.

Note: the notation $F(a)$, where $F$ is some field, means "the smallest field containing $F$ and $a$," where $a$ lives in some field containing $F$.

There are many fields containing $\bf R$ other than just $\bf C$. Field theory gives ways to construct field extensions of any field, in fact. Here is one way: if $F$ is a field that is not algebraically closed, then there is some polynomial $p(x)\in F[x]$ that does not have a root in $F$, and we can adjoin a root of the polynomial to $F$ via the quotient ring $F[x]/(p(x))$.

(However, it does make sense to adjoin different roots and obtain distinct extensions, though the extensions are isomorphic as fields, from the perspective of some overlying field $L/F$. For instance, adjoining the real cube root of $2$ to $\bf Q$ and adjoining the complex cube root of $2$ to $\bf Q$ yield two distinct number fields, finite field extensions of $\bf Q$.)

Another way is to adjoin a transcendental element. If $F$ is a field and we want $T$ to be a transcendental element over $F$, then $F(T)$ can be thought of as the field of all rational functions in the variable $T$ with coefficients from $F$. (In the case of positive characteristic, we should be quick to note that two rational functions, in fact two polynomials, can act as the same function even though they are different as abstract expressions.)

This makes sense: if there were two distinct rational functions $a(\cdot)/b(\cdot)$ and $c(\cdot)/d(\cdot)$ such that $a(T)/b(T)=c(T)/d(T)$ in $F(T)$, then $a(T)d(T)-b(T)c(T)=0$ would make $T$ algebraic, a contradiction. Thus $F(T)$ contains all rational functions in $T$ as distinct elements, and by definition of minimality this means $F(T)$ need contain no further elements.

There is an analytically inspired way of creating fields that I will also note. If a field $F$ is also a metric space, then the metric space completion of the field $F$ will also be a field. This can be constructed abstractly as the ring of all sequences of elements of $F$ that are Cauchy with respect to the metric, modulo the maximal ideal comprised of null sequences (those that converge to $0$ in the metric).

Question: What kinds of ways are there of constructing new fields from old ones?

Answer: Adjoining algebraic elements, adjoining transcendental elements, completing with respect to a metric, taking certain types of closures.

An example of a closure is an algebraic closure of a field. We say $\bar{F}$ is an algebraic closure of $F$ is it is (i) algebraically closed, (ii) contains $F$, and (iii) there is no field lying between $\bar{F}$ and $F$ that is also algebraically closed. In fact any two algebraic closures of a field $F$ will be isomorphic (a fact that requires some amount of choice.) We can restrict ourselves to other closures too, such as maximal abelian extension, maximal (totally/tamely) ramified extensions, etc.

There is more to say about uniqueness of field extensions and its relation to $\bf C$ than in the first section above. In fact with the axiom of choice,

Question How many algebraically closed fields of cardinal size $\kappa\ge{\frak c}:=|{\bf R}|$ are there?

Answer: There is always only one, up to isomorphism.

I don't actually know how to prove such a fact.


问:
Every definite positive matrix admits Cholesky decomposition: help with Demmel's proof
I am studying Cholesky decomposition by using the book "Applied Numerical Linear Algebra" of Demmel. In particular, I am trying to understand why a positive definite matrix $A\in\mathbb{R}^{n\times n}$ admits a Cholesky decomposition.

In the book, the proof is made by induction on the dimension of matrix $n$. I am stuck at the second equality below.

If $n=1$, choose  $ l_{11} = \sqrt{a_{11}}$ which exists since $a_{11} > 0$. As with Gaussian Elimination it suffices to understand the block 2-by-2 case. Write

$$  A = \begin{bmatrix} a_{11} & A_{12} \\ A_{21}^T & A_{22} \end{bmatrix} $$

$$ = \begin{bmatrix} \sqrt{a_{11}} & 0 \\ \frac{A_{21}^T}{a_{11}} & I \end{bmatrix}  \begin{bmatrix} 1 & 0 \\ 0 & \tilde{A}_{22} \end{bmatrix} \begin{bmatrix} \sqrt{a_{11}} & \frac{A_{12}}{a_{11}} \\ 0 & I \end{bmatrix}$$

$$ = \begin{bmatrix} a_{11} & A_{12} \\ A_{21}^T & \tilde{A_{22}} + \frac{A_{12}^TA_{12}}{a_{11}} \end{bmatrix} $$

Why can the matrix be written as the product of those three matrices? What is the point?

Could anyone please help me?

Thank you in advance!

答:
This decomposition tells us three things:

You can calculate $\widetilde{A}_{22}$ easily from $A$: we have $\widetilde{A}_{22}=A_{22}-\frac{A_{12}^TA_{12}}{a_{11}}$.

$\widetilde{A}_{22}$ is positive definite, because the positive definite matrix $A$ is congruent to the direct sum of $1$ and $A_{22}$. Therefore, by induction assumption, $\widetilde{A}_{22}$ admits a Cholesky decomposition $\widetilde{A}_{22}=\widetilde{L}\widetilde{L}^T$.

Now you get a Cholesky decomposition $LL^T$ of $A$ automatically, where
$$
L=\pmatrix{\sqrt{a_{11}}&0\\ \frac{A_{12}^T}{\sqrt{a_{11}}}&I}\pmatrix{1&0\\ 0&\widetilde{L}}
=\pmatrix{\sqrt{a_{11}}&0\\ \frac{A_{12}^T}{\sqrt{a_{11}}}&\widetilde{L}}.
$$


问:
What's the usage of knowing $p(\ln\sigma) = \text{const}$ in noninformative prior
In the PRML equation 2.239, for a non-informative prior of a scale parameter, the probability mass for interval $A \leq\sigma \leq B$ and $A/c\leq \sigma \leq B/c$ for any choice of all A and B should be the same. Hence $p(\sigma) \propto \frac{1}{\sigma} $.

In the following, it said it is convenient to consider prior distribution in its log form and have $p(\ln\sigma)=\text{const}$

But what's the usage of $p(\ln\sigma)=\text{const}$ ?

答:
The cited passage goes on to explicitly mention that the same probability mass falls in each decade: $[1,10]$, $[10,100]$, $[100,1000]$, ...  Equivalently, the same probability mass falls in each octave: $[1,2]$, $[2,4]$, $[4,8]$, ...  Equivalently, the same probability mass falls in each partition by integer powers of a positive constant different from $1$.

In the text between (2.236) and (2.237), we have the text

The parameter $\sigma$ is known as a scale parameter, and the density exhibits scale invariance.

Scale invariance (in dimension $1$) is a synonym for the previous sentences about power partitions.

Further, at the top of p. 118, we have

If we have a distribution $p(x\mid \lambda)$ governed by a parameter $\lambda$, we might be tempted to propose a prior distribution $p(\lambda) = \mathrm{const}$ as a suitable prior.

The text then explains that might be fine for a discrete parameter, but is (generally) incorrect for a continuous parameter.  Ending with

If, however, we are to choose a prior distribution that is constant, we must take care to use an appropriate representation for the parameters.

Taking these together in the context of (2.239), we should be looking for a (re-)parametrization leading to a constant probability density.  Since $\sigma > 0$ by assumption, one may parametrize by $ \lambda = \sigma$, $\lambda = \sigma +1$, $\lambda = \sigma^3$, $\lambda = \mathrm{e}^\sigma$, $\lambda = \sqrt{\sigma}$, or any of several other functions of $\sigma$.  There is one simple choice for which the probability density is constant for the new choice of parameter : $\lambda = \ln \sigma$.  (There are "silly" variations, like $\log_7 \sigma$ or $\ln(2\sigma)$, but these are just obscurations of the plain logarithmic relation.)


问:
Homeomorphic to a circle
Let $X:=[a,b]\times [c,d] \subset \mathbb{R}^2$ equipped with the subspace topology and let $U$ be an open connected set in $X$.
Show that the boundary of $U$ is homeomorphic to a circle.

Thanks in advance.

答: Let $U$ be an open annulus in $X$.  Then the statement is not true.</p>

Title: Homeomorphic to a circle


问:
Show that $\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$
I have to show that the left equation simplifies to $\tan\theta$:

Show that:
  $$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$

I do have prior knowledge that:
$$\tan \theta = \frac{\sin\theta}{\cos \theta}$$

But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.

答:
Here is a detailed answer.
Let's go!
$$
\require{cancel}
\begin{align}
\frac{1-\cos2\theta}{\sin2\theta}&=\frac{1-\left(\cos^2\theta-\sin^2\theta\right)}{2\sin\theta\cos\theta}\\
&=\frac{1-\cos^2\theta+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\sin^2\theta\cancel{+\cos^2\theta}\cancel{-\cos^2\theta}+\sin^2\theta}{2\sin\theta\cos\theta}\\
&=\frac{\bcancel2\cancelto{\sin\theta}{\sin^2\theta}}{\bcancel2\cancel{\sin\theta}\cos\theta}\\
&=\frac{\sin\theta}{\cos\theta}\\
&=\tan\theta
\end{align}
$$
I hope this helps.


问:
Vanishing Hessian on circle
I'm looking for examples of smooth functions $f: \mathbb{R}^2 \longrightarrow [0,\infty[$ such that the determinant of their Hessian matrix does only vanish on the unit circle, i.e.

$$\det \mathsf{H}_f(x,y)=f_{xx}(x,y)f_{yy}(x,y) -  f_{xy}^2(x,y)=0 \Longleftrightarrow x^2+y^2=1.$$

What can be said about the overall shape of such a function?

Title: Vanishing Hessian on circle

Body:
For $f(x,y)=g(x^2+y^2)$ it follows
$$\det \mathsf{H}_f(x,y)=4 g'(x^2 + y^2) (2 (x^2 + y^2) g''(x^2 + y^2) + g'(x^2 + y^2)).$$
Such that this vanishes on a circle iff
$$g'(1)=0 \ \ \text{or} \ \ 2g''(1) + g'(1)=0.$$


问:
Find all finitely generated $\Bbb Z/60\Bbb Z$-modules

Find all finitely generated $\Bbb Z/60\Bbb Z$ modules.

To do this, let $M$ be a finitely generated $\Bbb Z/60\Bbb Z$ module. Then $M$ is also $\Bbb Z$ module. Hence, we can write $M\simeq \Bbb Z^m\oplus\Bbb Z/p_1\Bbb Z\oplus\cdots\oplus\Bbb Z/p_n\Bbb Z$ where $m\geq 0$ and $p_1|p_2|\cdots|p_n$. Now, I quotient out this by $60\Bbb ZM$ so that it $\Bbb Z/60\Bbb Z$ module structure, but I don't know how $\Bbb Z/p_i\Bbb Z$ part would be like. Could you help?

答:
You don't need to quotient out by $60\Bbb Z$, as $M$ is already a $\Bbb Z/60\Bbb Z$-module, so $60M=\{0\}$.

In particular it implies that $m=0$ and each $p_i$ (which is by the way an unlucky notation) divides $60$.

And that's it.


问:
Total variation inequality liminf
I'm trying to understand the proof, given by t.b. here Space of Complex Measures is Banach (proof?)
.

In the course of the proof, the author uses the equality

$\left|\left| \mu - \mu_{m} \right|\right| \leq \liminf\limits_{n \to \infty} \left|\left|\mu_{n}-\mu_{m}\right|\right|$,

where $\left|\left| \cdot \right|\right|$ denotes the total variation norm and

$\mu(A):=\lim\limits_{n \to \infty} \mu_{n}\left(A\right)$

the limit of a sequence of measures $\mu_{n}$ with bounded total variation.

I would like to know, why the above inequality holds. I tried several thinks using the definition of the total variation and I came up with nothing.

Any help is most appreciated.

答:
We will use the notation from the proof that was mentioned.

By defintion, $| \mu_n(E) - \mu_m(E) | \le || \mu_n - \mu_m ||$, hence $\liminf_n | \mu_n(E) - \mu_m(E) | \le \liminf_n || \mu_n - \mu_m ||$.

But $\liminf_n | \mu_n(E) - \mu_m(E) | = \lim_n | \mu_n(E) - \mu_m(E) | = | \mu(E) - \mu_m(E) |$.

It follows that  $| \mu(E) - \mu_m(E) | \le \liminf_n || \mu_n - \mu_m ||$.

Taking supremum over E, we have
$|| \mu - \mu_m || \le \liminf_n || \mu_n - \mu_m ||$, q.e.d.

Title: Total variation inequality liminf


问:
Alternative proof that the sum of two convergent series is convergent
For my Real Analysis course I'm being asked to prove this basic concept:

If $\sum_{n=1}^\infty |a_n|$ and $\sum_{n=1}^\infty |b_n|$ converge absolutely then prove that $\sum_{n=1}^\infty (a_n+b_n)$ also converges absolutely.

Now this question is very basic. It can be proved by adapting the proof of the sum law for limits to series instead. I've done this but my professors says the answer is wrong. Instead, he wants to me solve this question without using epsilon-proofs. The only hint he provides is that I must consider the rearrangement of each series and apply the triangle inequality to prove convergence. This logic is incomprehensible to me but he refuses to help any further.

Is there another method to prove this question without using limit laws and epsilon-proofs?

答:
The series is bounded above, as $\sum|a_n+b_n|\le\sum |a_n|+\sum|b_n|$.

The series forms an increasing sequence, as every term $|a_n+b_n|$ is non-negative.


问:
$E(XY)=E(X)E(Y)\neq 0$ implies independance
I have a little question.
I know that $E(XY)=E(X)E(Y)$ does not imply independence. There are a lot of examples with $E(XY)=E(X)E(Y)=0$.

But if $E(XY)=E(X)E(Y)\neq 0$, is it implying that $X$ and $Y$ are independent ? If not, have you an example where $X$ and $Y$ are dependent ?

Thanks a lot !

答: Since $E(XY)-E(X)E(Y)$ doesn't change if you add a constant to one or both of $X,\,Y$, any dependent $X,\,Y$ with $E(XY)=E(X)E(Y)=0$ can be shifted (thereby preserving their dependence) to give $E(XY)=E(X)E(Y)\ne0$.


问:
$\binom{n - 1 + k}{k}$ as the basis for a polynomial vector space, and the use of $\frac{1}{(1-x)^n}$ to derive generating functions for $a_n = n^k$
I am trying to understand the following passage from my course notes:

Here is Lemma 3.1:

Discussion:

Starting from the beginning, I will try to explain in detail what I think I understand and ask questions where I am confused, and hopefully someone can set me straight.

(One) Where it says "generating funtions... of the form...", what is happening more precisely is that we are operating in some ring $R$, and the generating function $a_0 + a_1x + \dots$ is an element of the formal power series ring $R[[X]]$.

(Two) The "of the form $\frac{1}{(1-x)^n}$" part means that when following the standard multiplication rules for $R[[X]]$, we get that $(a_0 + a_1x + \dots)(1-x)^n = 1$. (I should also say that $(1 - x)^n$ is an element in $R[[X]]$ as well.)

(Three) Lemma 3.1 shows that $a_k = \binom{n - 1 + k}{k}$, so we know the explicit form of $a_0 + a_1x + \dots$. Algebraic manipulation of $\binom{n - 1 + k}{k}$ into $\frac{1}{(n - 1)!}(k + 1)^{(n - 1)}$ is straightforward.

(Four) In the above notes, we are considering $R = \mathbb R$ or $\mathbb C$, so that means $R$ is a field, and we can therefore consider the vector space $R[X]$ over $R$ where we define scalar multiplication in the form $R \times R[X] \to R[X], (\lambda, p) \to \lambda p$, meaning that the formal power series $b_0 + b_1x + \dots \in R[X]$ multiplied by $\lambda \in R$ becomes $\lambda b_0 + \lambda b_1 x + \dots$.

(Five) In our polynomial vector space $V := R[X]$, we are told that $B := \bigl\{ \frac{1}{(n - 1)!}(x + 1)^{(n - 1)}$ for $n = 1, 2, \dots \bigr\}$ are a basis, so we must have $\text{span}(B) = V$. I'm not sure how to show this. I suppose I can show linear independence by taking a finite subset of $B$, call them $v_1, \dots, v_m$, and noting that $\lambda_1 v_1 + \dots + \lambda_m v_m$ is a polynomial of degree equal to the highest degree of the vectors $v_1, \dots, v_m$. If this linear combination equals $0 \in R[X]$, then we must have all $\lambda_1, \dots, \lambda_m = 0 \in R$.

I need to show that $\text{span}(B) = V$ but I am not sure how to do this. I need to show that the intersection of all subspaces $U \subseteq V$ such that $B \subseteq U$ is equal to $V$ itself.

I guess I could suppose that $\text{span}(B) \neq V$, which would mean there were some element $v \in V$ that was not in the intersection of all such $U$. Therefore there is some $U^*$ in the intersection that does not contain $v$.

I want to show that this is a contradiction, meaning that I can create $v$ out of a linear combination of the basis elements in $B$, which will imply that $v \in U^*$, which will prove that $\text{span}(B) = V$.

If $v = c_0 + c_1x + \dots + c_r x^r$, then take the set $b := \bigl\{ \frac{1}{(n - 1)!}(x + 1)^{(n - 1)}$ for $n = 1, 2, \dots r \bigr\}$. This is a finite subset of the basis $B$, so we must have $b \subseteq U^*$.

If $b_i = \frac{1}{(i - 1)!}(x + 1)^{(i - 1)}$ for $i = 1, \dots, r$, then if I pick the right scalars $\lambda_1, \dots, \lambda_r$ I should be able to show that $\lambda_1 b_1 + \dots + \lambda_r b_r = v$. Figuring out the right scalars seems very messy, though. Is there any easy way to find the right scalars?

(Six) Assuming that $B$ is a basis for $V$, I think then notes are trying to explain how we can use this to find the generating function for a sequence defined by $a_n = n^k$ for $n = 0, 1, 2, \dots$ and fixed natural number $k$.

(Note: I think that strictly speaking the generating function for such a sequence will simply be $0^k + 1^k x + 2^k x + \dots$. What the notes really mean is that we can find the compact i.e. finite representation for this
generating function, much like how Lemma 3.1 shows that $\binom{0 + n - 1}{0} + \binom{1 + n - 1}{1}x + \dots = \frac{1}{(1 - x)^n}$.)

I don't understand the connection between $a_n = n^k, b_n = \binom{n - 1 + k}{k}$ and $\frac{1}{(1 - x)^k} = b_0 + b_1x + \dots$. Why $n^k$? I guess the idea is that we can write $0^k + 1^k x + 2^k x + \dots$ as a linear combination of terms $\binom{n - 1 + k}{k}$?

(Seven) Regarding the final paragraph concerning complex numbers, first of all I don't understand the substitution $y = cx$ when there no $y$ to be found anywhere. Are we supposed to be looking at $\binom{0 + n - 1}{0} + \binom{1 + n - 1}{1}y + \dots = \frac{1}{(1 - y)^n}$ and simply replace $y$ with $cx$ for $c \in \mathbb C$? Then we are supposed to be able to write $0^k + 1^k cx + 2^k c^2 x + \dots$ as a linear combination of terms $\binom{n - 1 + k}{k}c^n$? Since I can't understand the simpler instance in (Six) I don't think I can understand this part either.

Sorry about the length of this post. I appreciate any feedback.

Edit:

I think where I wrote $\binom{0 + n - 1}{0} + \binom{1 + n - 1}{1}y + \dots = \frac{1}{(1 - y)^n}$ in (Seven), I should have written instead $\binom{k + 0 - 1}{k} + \binom{k + 1 - 1}{k}y + \dots = \frac{1}{(1 - y)^n}$. In other words it is $k$ that is fixed here and $n$ changes for every term in the generating function.

答:
Your first four points are fine; I’ll take your last three points in reverse order.

The last paragraph could have been written more clearly. Yes, the idea is to start with

$$\frac1{(1-y)^k}=\sum_{n\ge 0}\binom{n-1+k}ky^n$$

and substitute $y=cx$, but one really might just as well work directly with $cx$:

$$\begin{align*}
\frac1{(1-cx)^k}&=\sum_{n\ge 0}\binom{n-1+k}k(cx)^n\\
&=\sum_{n\ge 0}\binom{n-1+k}kc^nx^n\,,
\end{align*}$$

so $\frac1{(1-cx)^k}$ is the generating function for the sequence $\left\langle\binom{n-1+k}kc^n:n\in\Bbb N\right\rangle$.

For $n\in\Bbb N$ let $p_n(x)=\frac1{n!}(x+1)^{\overline{n}}$, and assume that $\{p_n(x):n\in\Bbb N\}$ is a basis for $R[x]$. For each $k\in\Bbb N$ let $q_k(x)=x^k$. Since $\deg p_n(x)=n$, there are constants $c_{k,i}$ for $i=0,\ldots,k$ such that $q_k(x)=\sum_{i=0}^kc_{k,i}p_i(x)$.

$$\begin{align*}
\sum_{n\ge 0}n^kx^n&=\sum_{n\ge 0}\sum_{i=0}^kc_{k,i}p_i(n)x^n\\
&=\sum_{i=0}^kc_{k,i}\sum_{n\ge 0}p_i(n)x^n\\
&=\sum_{i=0}^kc_{k,i}\sum_{n\ge 0}\frac1{i!}(n+1)^{\overline{i}}x^n\\
&=\sum_{i=0}^kc_{k,i}\sum_{n\ge 0}\binom{n+i}nx^n\\
&=\sum_{i=0}^k\frac{c_{k,i}}{(1-x)^{i+1}}\,,
\end{align*}$$

so that $\sum_{i=0}^k\frac{c_{k,i}}{(1-x)^{i+1}}$ is the generating function for the sequence $\langle n^k:n\in\Bbb N\rangle$.

Probably the easiest way to see that $\{p_n(x):n\in\Bbb N\}$ is a basis for $R[x]$ is to observe that $\deg p_n(x)=n$ for each $n\in\Bbb N$; an easy induction on $k$ then shows that $q_k(x)=x^k\in\operatorname{span}\{p_n(x):n\in\Bbb N\}$ for each $k\in\Bbb N$. The actual coefficients in the linear combinations are a bit messy and involve Stirling numbers of the first kind.


问:
Find Minimum of Variance
Given the Variance $\sum_{n=1}^{\infty}\frac{c\Delta\mu_n-\Delta(\sigma_{n}^{²}+\mu_{n}^2)}{P_{n}}$ and $ \sum_{n=1}^{\infty}P_{n} = \mu$. I would like to minimize Var by choosing the best $P_{n}$.
Can somebody give me a hint how to get to the solution $P_{n}=const*\sqrt{c\Delta\mu_n-\Delta(\sigma_{n}^{²}+\mu_{n}^2)}$.

'const' stems from the constraint $ \sum_{n=1}^{\infty}P_{n} = \mu$.

I should probably use Lagrange and differentiate with respect to P. I tried it a couple of times but my Problem is I think to get the correct Lagrange function.

Thank you

答:
Using aLagrangian mulitplier, you minimize$$\sum_n \frac{v_n}{P_n}+\lambda\left(\sum_n Pn-\mu\right).$$

Then cancelling the gradient,
$$-\frac{v_n}{P_n^2}+\lambda=0$$

gives

$$P_n=\sqrt{\frac{v_n}{\lambda}}.$$

As you say,

$$c=\dfrac1{\sqrt\lambda}=\dfrac{\mu}{\displaystyle\sum_n\sqrt{v_n}}.$$

Title: Find Minimum of Variance


问:
Show that if  is uniformly distribution between 0 and 1 and  = (1 −  )^-1, then  has the same distribution function as .
Question is as titled.

So  was given by a previous question, the following probability density function which I've worked out, hopefully correctly.

\begin{array}ff_X(x)= \begin{cases} ^{-2} \ \ \ \geqslant \ 1, & \\0\ \ \ \ \ \ < \ 1. & \end{cases}\end{array}

I was asked to find the distribution function of X. So i gave it a go to work it out which I hope the answer below is correct, pdf-to-cdf transformation

\begin{array}\$\int_1^x    t^{-2} dt   =  \frac{\mathrm{1} }{\mathrm{1}} - \frac{\mathrm{1} }{\mathrm{x}} = \frac{\mathrm{x-1} }{\mathrm{x}}\end{array}
\begin{array}wwhere : x  \geq 1\end{array}

From what I gathered from the question,  is a uniform distribution that can be expressed like this:  ∼  (0, 1). This is a continious problem

What I can understand from the question is that the uniform distribution will have a probability of 1 between the values 0 and 1: \begin{array} 00\leq x \leq 1 \end{array}

My guess is the question wants me to find out whether: \begin{array} YY = (1 - W)^{-1}\end{array} equals the same as the above limit. I am not 100% sure. I did some research into probability transformations as i believe thats what my question is asking but didnt get to helping me solve this question.

The question was taken from my statistics coursework.

答:
First remember that, with a monotonic transformation $g$, you get

$$f_Y(y)=f_X[g^{-1}(y)]\Bigg|\frac{d}{dy}g^{-1}(y)\Bigg|$$

thus being

$$Y=\frac{1}{1-W}$$

you get $W=1-\frac{1}{Y}$ and obviously

$$f_Y(y)=1\cdot\Big|\frac{dw}{dy}\Big|=\frac{1}{y^2}\cdot\mathbb{1}_{[1;+\infty)}(y)$$

This because

$f_W(w)=1$ and thus also $f_W[g^{-1}(y)]=1$

$\Bigg|\frac{d}{dy}(1-\frac{1}{y})\Bigg|=\frac{1}{y^2}$

(same density and thus same distribution as $X$)


问:
How to interpret summation signs
I'm taking a course in statistics, and I really need to brush up my math to be able to follow the book at times.

I'm looking at formulas for sum of squares, and I am slightly confused about the capital sigma letter and how to interpret an equation with several signs like this:

$$SS_a = \sum^a_{i=1} \sum^b_{j=1} \sum^n_{k=1} (\bar{X}_i - \bar{X})^2 $$

I mean, it's only i which is used  the summation signs, but when I try it out (ignoring b and n) it seems like I get it wrong, and that it should be something like this:

$ b * n * \sum^a_{i=1}(\bar{X}_i -\bar{X})^2 $

Is this correct, and what is the rule behind it?

PS. I wasn't sure what tags really fits into this question, might it be basic algebra? DS

答: Note that $$ \sum^n_{k=1} (\bar{X}_i - \bar{X})^2\\=\underbrace{(\bar{X}_i - \bar{X})^2+(\bar{X}_i - \bar{X})^2+\dots+(\bar{X}_i - \bar{X})^2}_{\displaystyle n \,\text{times}}\\=n\cdot(\bar{X}_i - \bar{X})^2$$ Similarly $$\sum^b_{j=1}\sum^n_{k=1} (\bar{X}_i - \bar{X})^2\\=\sum^b_{j=1}n\cdot(\bar{X}_i - \bar{X})^2\\=n\cdot b\cdot (\bar{X}_i - \bar{X})^2$$Here the fact is $$\sum^n_{k=1} X_i=n\cdot X_i.$$That's all from me.</p>

Title: How to interpret summation signs


问:
Evaluating an indefinite integral $\int \frac{dx}{\sqrt{x^3+2x+3}}$

Evaluate the following integral
  \begin{equation}
J = \int \frac{dx}{\sqrt{x^3+2x+3}}
\end{equation}

I do not find suitable substitution to compute the above indefinite integral. Since $x^3+2x+3=(x+1)(x^2-x+3)$, substituting $z=\sqrt{x+1}$, we have $$J= 2\int \frac{dz}{\sqrt{z^4-3z^2+5}}.$$ I think this is not a good substitution. Any help is appreciated.

答:
Since the integral is just a cubic raised to the power $-\frac12$ and the cubic has no repeated roots, it is not elementary but may be expressed in terms of elliptic integrals.

Suppose we want to find the integral from the pole at $-1$ to some number $y$ greater than that – the indefinite integral from $0$ follows from this easily. The cubic's factorisation as given in the question is $x^3+2x+3=(x+1)(x^2-x+3)=(x+1)((x-1/2)^2+11/4)$, so Byrd and Friedman 239.00 gives for this
$$\int_{-1}^y\frac1{\sqrt{x^3+2x+3}}\,dx=5^{-1/4}F\left(\cos^{-1}\frac{\sqrt5-1-y}{\sqrt5+1+y},m=\frac{10+3\sqrt5}{20}\right)$$


Title: $E(XY)=E(X)E(Y)\neq 0$ implies independance

Body:
I have a little question.
I know that $E(XY)=E(X)E(Y)$ does not imply independence. There are a lot of examples with $E(XY)=E(X)E(Y)=0$.

But if $E(XY)=E(X)E(Y)\neq 0$, is it implying that $X$ and $Y$ are independent ? If not, have you an example where $X$ and $Y$ are dependent ?

Thanks a lot !

Body: Let $X \sim N(0,1)$. then $1+X^{2}$ and $1+X$ are not independent but $E(1+X^{2})(1+X)=E(1+X^{2})E(1+X)=2$.


问:
Give an example of $\phi$ and $\psi$ : $\forall x$ $\phi$ $\equiv$ $\forall$ $x$ $\psi$, but $\phi$ $\not\equiv$ $\psi$
Give an example of $\phi$ and $\psi$ :  $\forall x$$\phi$ $\equiv$$\forall$$\psi$, but  $\phi$$\not\equiv$$\psi$

I think it's about $\exists$ $x$ predicat, but i'm not sure.

答:
Here's an example in a language with two constant symbols $c$ and $d$: we set $$\phi(x): x=c\quad\mbox{and}\quad\psi(x): x=d.$$ Clearly $\phi\not\equiv\psi$ in general: in any structure where the symbols $c$ and $d$ are interpreted differently, $\phi$ and $\psi$ will in turn mean different things. However, the sentences "$\forall x\phi(x)$" and "$\forall x\psi(x)$" are much simpler: each sentence is true in a given structure $\mathcal{M}$ if and only if

 that structure has exactly one element.


问:
Algebraic Number theory, why is $\mathcal{O}_K/p \cong p^k/p^{k+1}$?
Assume that $\mathcal{O}_K$ is the integer ring of number field $K$, $p$ is its prime ideal, why $\mathcal{O}_K/p \cong p^k/p^{k+1}$, in which $k$ is a positive integer? Is $p/p^{k+1}$ a ring? Could it be considered as the ideal of $\mathcal{O}_K/p^{k+1}$?

答:
As Mummy pointed out, we are dealing with additive abelian groups morphisms (i.e modules over the integers).
In theorem (22.a), chapter 3 of Marcus' Number Fields there is the sketch of the proof of what you are saying. The point is that for unique factorization, for every $k \in \mathbb{Z}$, there is an $\alpha \in p^k \setminus p^{k+1}$. Then the mapping $ \varphi :\mathcal{O}_K \rightarrow p^k/p^{k+1} $ where $\varphi(x)= \alpha x+p^{k+1}$ (as a class) induces the isomoprhism that you are looking for. You should check that this mapping is onto and that the kernel is just $p$.


问:
Relation between chords length and radius of circle
Two chords of a circle, of lengths $2a$ and $2b$ are mutually perpendicular. If the distance of the point at which the chords intersect,from the centre of the circle is $c$($c<$radius of the circle),then find out the radius of the circle in terms of $a,b$ and $c$.Show some short-cut to do it quickly</p>

答: Another similar approach may be arisen as following: ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$


问:
Injection of Ext groups
Let $\mathcal A  \subset \mathcal B$ be an exact inclusion of a full abelian subcategory $\mathcal A$ into an abelian category $\mathcal B$. Assume, that both $\mathcal A$ and $\mathcal B$ have enough injectives (or enough projectives). Since the Ext functor classifies extensions, we have an inclusion $\mathrm{Ext}^i_{\mathcal A}(X,Y) \subset \mathrm{Ext}^i_{\mathcal B}(X,Y)$. The map can also be constructed via resolutions. Is there a way to show, that the map $\mathrm{Ext}^i_{\mathcal A}(X,Y) \to \mathrm{Ext}^i_{\mathcal B}(X,Y)$ is injective without any reference to extensions?

A standard example is the base restriction functor $K\text{-}\mathrm{Vect} \to K[x]\text{-}\mathrm{Mod}$ along $K[x] \to K, x \mapsto 0$, where $K$ is a field. Then $\mathrm{Ext}^1_{K}(K,K)=0$ and $\mathrm{Ext}^1_{K[x]}(K,K)=K$.

I don't want to assume, that the inclusion preserves injectives or projectives. In this case any resolution in $\mathcal A$ computes the derived functor in $\mathcal B$ and the map is bijective.

The claim is false, when the functor $\mathcal A \to \mathcal B$ is faithful, but not full.

This question is related: Exact functor and relationship between Ext functors

答:
I don't think the claim is correct: Consider e.g. a field $k$ and the full inclusion of $k[x]/(x^2)\text{-mod}$ into $k[x]\text{-mod}$. This cannot induce an injection on extension groups because $k[x]/(x^2)$ is of infinite global dimension, while $k[x]$ has global dimension $1$.

The problem with the attempt to reason in terms of Yoneda extensions is that $\text{Ext}^k(X,Y)$ are the connected components of the directed graph of length-$k$ extensions, and when considered in ${\mathscr B}$ there may be paths between extensions living in ${\mathscr A}$ which pass through intermediate extensions which are outside of ${\mathscr A}$. This cannot happen for $k=1$ because of the Five lemma. So, broadly speaking the problem is simply that $A\subset B$ doesn't imply $\pi_0(A)\subset\pi_0(B)$.

Title: Injection of Ext groups


问:
Small detail in proof of Borsuk-Ulam theorem.
I’m reading the following proof.

Corollary B13.2. There is no odd map $S^{m} \rightarrow S^{m-1}$.

Theorem $\mathbf{B 1 3 . 3}$ (Borsuk-Ulam Theorem). Given any continuous map $f: \mathbb{S}^{m} \rightarrow \mathbb{R}^{m}$, there exists a pair of antipodal points $\pm x \in \mathbb{S}^{m}$ with $f(x)=f(-x)$.

Proof. If there is no such $x$, we can set
$$
g(x):=\frac{f(x)-f(-x)}{\|f(x)-f(-x)\|}
$$
giving an odd map $g: \mathbb{S}^{m} \rightarrow \mathbb{S}^{m-1}$. This contradicts the previous corollary.

One small detail that I don’t get is how we can conclude that the image of $g$ is $\mathbb{S}^{m-1}$. Some elaboration would be appreciated.

答: The point is that $f$ is a map into $S^{m-1}$, as indicated by the fact that $||f(x)||=1$ for any $x$ and $f(x)\in\Bbb R^m$.


问:
limit value and LH rule
L=$\lim \limits_{x \to 0}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$ and L is finite. What is the value of a ?

After using LH rule,

we get L=$\lim \limits_{x \to 0}\frac{\frac{1}{\sqrt{a^2-x^2}}-\frac{1}{2}}{4x^2}$, since L is finite, we make the numerator zero i.e a=2. Which make L=0.

But when we apply a=2 directly to L=$\lim \limits_{x \to 0}\frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$ i.e

L=$\lim \limits_{x \to 0}\frac{2-\sqrt{4-x^2}-\frac{x^2}{4}}{x^4}$

L=$\lim \limits_{x \to 0}\frac{(2-\frac{x^2}{4})^2-(4-x^2)}{x^4(2-\frac{x^2}{4}+\sqrt{4-x^2})}=\lim \limits_{x \to 0}\frac{(4+\frac{x^4}{16}-x^2)-(4-x^2)}{x^4(2-0+2)}=\frac{x^4}{16}\frac{1}{x^4 4}$
, we get L as $\frac{1}{64}$

Which is the true value L=0 or L=$\frac{1}{64}$ ?

Thanks.

答: $a=2$ does not give you $L=0$. You just make the numerator $0$ at $x=0$ but when you take the limit of the ratio as $ x\to 0$ you still have an indeterminate form and you have to apply L'Hopital's Rule to get the limit. The correct limit is $\frac  1 {64}$.

Title: limit value and LH rule


问:
$n\log_2(n)\in\Omega\left(10^3n\right)$
I am struggling with a new topic in my discrete math class.
I must show that

$$n\log_2(n)\in\Omega\left(10^3n\right)$$

A detailed explanation will be awesome!

答:
First note that
$$f(x)\in\Omega(g(x))\ \mbox{as}\ x\to\infty$$
If
$$\limsup_{x\to\infty}\left\vert\frac{f(x)}{g(x)}\right\vert\gt 0$$
So now we have
$$\limsup_{n\to\infty}\left\vert\frac{n\log_2(n)}{10^3n}\right\vert$$
$$=\frac1{10^3}\limsup_{n\to\infty}\left\vert\log_2(n)\right\vert$$
I think you can take it from here. The basic idea is to show that the above limit is greater than zero.

Title: $n\log_2(n)\in\Omega\left(10^3n\right)$


问:
Integral of $\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$
I was trying to solve the following question:

Evaluate: $$\int{\frac{1}{\sqrt{x(1+x^2)}}dx}$$

This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.

But, I don't know how to start this. I want a hint to get started with this. Thanks!

答:
This is an unsolved question in my sample papers book and so I believe it should have an elementary primitive.

Well, there is no elementary primitive, because the roots of the cubic under the root are not repeated: $0,i,-i$. It requires an elliptic integral:
$$\int\frac1{\sqrt{x(1+x^2)}}\,dx=F\left(\cos^{-1}\frac{1-y}{1+y},m=\frac12\right)+C$$
On my side I prefer using the Byrd and Friedman tables rather than a CAS to work out elliptic integrals (the above formula is 239.00). Often the result is cleaner and the parameter $m$ ends up in the classic range of $[0,1]$.


问:
How to prove that the exponential function is primitive recursive?
I know that we can define the exponential by a function $f: \mathbb{N}^2 \rightarrow \mathbb{N}$ by letting:

$f(m,o) = 1$ and $f(m,n+1) = f_x(m^n,m)$ where $f_x$ is the multiplication function, which we know is recursive.

I would then let $g = s(z)$ and hence $g(n) = 1$ for each $n \in \mathbb{N}$ , and let $h = f_x$

However I am unsure where to go from here in order to complete a recursive definition for the exponential function. Could somebody please guide me into the approach to finish this proof?

答:
According to Martin Davis Computability, complexity, and languages, and the way he solves this kind of problem. with the following steps, we can prove Exponential function is primitive recursive

$f(x,y) = x.y\ which\ f \ is \ prim.rec.\\
h(x,y) = x^y\\
h(x,0) = 1\\
h(x,y+1) = x^{y+1} = g(y,h(x,y),x)\\
g(x_1,x_2,x_3) = f(\bigcup^3_2(x_1,x_2,x_3),\bigcup^3_3(x_1,x_2,x_3))\\
\bigcup^n_i(x_1,...,x_n) = x_i (1\leq i \leq n)
$

These steps show that we build the exponential function with initial functions so it's prim.rec.


问:
testing if a die is fair with hypothesis testing
Suppose a child plays with a die and the mean obtained after $25$ rolls is equal to $4.5$. Perform a statistical test to check if the die is fair at significance level $\alpha=0.05$.

Suppose we perform hypothesis testing with null hypothesis $\mu_0 = 4.5$. If we have a fair die the average number of points after throwing it $25$ times is just the same as after throwing it once so $E(\bar{X})=\frac{1+2+3+4+5+6}{6}=3.5$. The alternative I guess is that $\mu_1=3.5$ and we want to show  that the die is unfair. I am not sure if I am on the right track though? Or do we just convert this to probabilities, and say that under the null $p=\frac{1}{6}$. Thanks for any help.

答:
I managed to solve the exercise myself after the suggestion in the comments.
We suppose that $\mu_0=3.5$. Then we use the $z$-score:
$$\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}.$$

In this case $\bar{X}=4.5$, $\mu_0=3.5$ and the standard deviation can be calculated as the square root of the variance divided by the number of rolls
$\sigma / \sqrt{n}=\sqrt{0.1166}=0.34$

So
$$\frac{\bar{X}-\mu_0}{\sigma / \sqrt{n}}=\frac{1}{0.34}=2.94.$$
Then $p=P(Z\geq 2.94)=0.0033$ so we reject the null hypothesis.


问:
Would such a natural transformation exist?
So, I know the following definition of a Natural Transformation:

Let $F,G: \mathcal{C} \to \mathcal{D}$ be two functors. A natural transformation $\eta: F \implies G$ consists of morphisms $\eta_{x} \in \text{Hom}_{\mathcal{D}}(F(x), G(x))$ for every $x \in \text{Ob}(\mathcal{C})$ such that for every morphism $f: x \to y; y\in\mathcal{C}$ such that the diagram:

commutes.

Now, I wanted to know what happens when $\text{Hom}_{\mathcal{D}}(F(x), G(x)) = \emptyset$ for some $x$. If this happens for some $x$ would there exist some $\eta$ between $F$ and $G$ assuming that all $\eta_{x}$'s for $x$'s such that $\text{Hom}_{\mathcal{D}}(F(x), G(x)) \neq \emptyset$ follow the required criteria.

In other words, I wanted to know if $\text{Hom}_{\mathcal{D}}(F(x), G(x)) = \emptyset$ for some $x$ (which would mean $\eta_{x}$ for such $x$ would not exist) then it does not mean that the natural transformation $\eta$ could not exist, right??

答:
Yes, that is right, if $\operatorname{Hom}_\mathcal{D}(F(x), G(x)) = \emptyset$ then there cannot be a natural transformation $\eta: F \Rightarrow G$ because that would require $\eta_x \in \operatorname{Hom}_\mathcal{D}(F(x), G(x))$.

We could define a notion of partial natural transformation, much like you suggest: we only require that $\eta_x$ exists for some of the $x \in \mathcal{C}$ and we only require the relevant square to commute where it exists. It is not unthinkable that in some setting such a definition would be useful. However, I have never seen a definition like that, so I do not know if it is useful.


问:
Placing a circle in a plane without overlap
Given are $n$ non-overlapping unit circles in $\mathbb R^2$ and some number $r>1$. Let $C$ be one of the circles. I want to place a new circle of radius $r$ which is tangent to $C$ and additionally does not overlap any of the other circles.

How can I find a suitable center for this new circle efficiently, or at least decide whether this is even possible or not?

答:
Hint:

You can solve this using a trick: instead of placing circles of radius $r$ tangent to a given circle, you can inflate all circles by $r$ and place a point on the circumference of that circle.

This implies to compute the union of the inflated circles, a relatively complex geometric operation. When this is done, for every circle that still has a piece of circumference in this "union map", the answer is positive.

https://mathoverflow.net/questions/24859/finding-the-union-of-n-random-circles-arbitrarily-or-conspiratorially-placed-o


答:
In the picture below, suppose $C$ is the circle in black, while the other circles are in grey.  Surround $C$ with a concentric red circle radius $1+r$ and the other circles with pink circles radius $1+r$.

Any point on the red circle can be the centre of a new circle of radius $r$ which is tangent to $C$.  Any point not inside a pink circle can be the centre of a new circle of radius $r$ which does not overlap a grey circle.

So any point on the red circle not in a pink circle (reshaded blue) can be the centre of a new circle of radius $r$ which is tangent to $C$ and additionally does not overlap any of the other grey circles. If there are no such points (i.e. if the red circle is covered by the pink circles), then it is not possible.

Title: Placing a circle in a plane without overlap

Body:
Given are $n$ non-overlapping unit circles in $\mathbb R^2$ and some number $r>1$. Let $C$ be one of the circles. I want to place a new circle of radius $r$ which is tangent to $C$ and additionally does not overlap any of the other circles.

How can I find a suitable center for this new circle efficiently, or at least decide whether this is even possible or not?


问:
Why do we say limits which go to infinity are not defined?
I recently saw this article by oregonstate.edu about the differentiability of the function $y=\sqrt{X}$ at $x=0$ they claim that the derivative at $x=0$ does not exist.i understand the concept of one-sided differentiability and how the  $\lim_{h \to 0-} \frac{\sqrt{x+h}-\sqrt{x}}{h} $
does not exist but for values of greater than 0 and approaching from the right they say that the limit does not exist as it is positive infinity. my question is why isn't this limit defined surely we know it is approaching infinity so why can't we say it has an infinite slope

答:
This question gets into subtleties of how we define derivatives in terms of unidirecional limits. Let's consider the outcome in two different number systems.

Real analysis

The derivative exists iff two one-sided derivatives exist with equal value. The right-derivative$$\left.\lim_{h\to0^+}\frac{\sqrt{x+h}-\sqrt{x}}{h}\right|_{x=0}=\lim_{h\to0^+}\frac{\sqrt{h}}{h}=\color{blue}{+\infty}$$ is an extended real number; the left-derivative $\lim_{h\to0^-}\frac{\sqrt{h}}{h}$ does not exist.

Complex analysis

$f^\prime(0)$ exists with value $L$ iff $\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=L$ for all $\theta\in\Bbb R$. For $f(z)=z^{1/2}=e^{\tfrac12\ln z}$,$$\lim_{r\to0^+}\frac{f(re^{i\theta})-f(0)}{re^{i\theta}}=\lim_{r\to0^+}\frac{\sqrt{r}e^{i\theta/2}}{re^{i\theta}}=\color{red}{\infty}$$is an extended complex number. Do not confuse $\color{blue}{+\infty}$ with $\color{red}{\infty}$.

One more point: if we'd been differentiating $x^{1/3}$ at $x=0$, the two-sided derivative would again not exist in real analysis, but this time for a different reason: the left-derivative would exist, but would be $-\infty$ rather than $+\infty$.


问:
Prove if a discrete subset $E$ of $\mathbb{R}^n$ is compact, then it is finite.

Let $E\subseteq \mathbb{R}^n$ be a compact set and suppose that for
each $x\in E$, there is an $r_x \gt 0$ such that $E \cap B_{r_x}(x) = \{x\}.$

Prove that $E$ is a finite set.

I am a bit confused as to what "finite" means here. Do I have to prove that $E$ is closed and bounded? This is giving me memories of the Heine-Borel Theorem. Am I on the right track?

答:
Let $E\subseteq \mathbb{R}^n$ be a compact set and suppose that for each $x\in E$, there is an $r_x\gt0$ such that $E \cap B_{r_x}(x)$ = {x}.

Then all the points $x \in E$ are isolated points. If E is infinite set, E would have infinite isolated points, where that can not ne the case on the metric space $\mathbb{R}^n$. Because $\mathbb{R}^n$ is a separable compact metric space.

Take this problem as inlightness, and you can generlize it for the euclidian metric space:

Prove that every subset of $\mathbb{R}$ has countably many isolated points.

Let $A \subset \mathbb{R}$ and let $S$ the set of isolated points in $A$. Let $ x \in S$, so that there is an $ \epsilon >0 $ such that $(x-\epsilon, x+\epsilon) \cap A-\{x\}=\phi$. Now, let $a,b$ any two rational numbers such that $(x-\epsilon < a < x < b< x+\epsilon )$ or $x \in (a,b) \subset (x-\epsilon,x+\epsilon)$.

Define $\Phi: S \rightarrow \mathbb{Q}^2$ as $\Phi(x) = (a,b)$. Since there is  a correspondence between  the set $S$ and $\mathbb{Q}^2$. Thus $S$ is a countable set.


问:
Prove$ (2^n+3^n) ^{1/n}$ is a decreasing sequence
We want to prove this without using any calculus.</p>

答:
$$f(x)=(2^x+3^x)^{1/x}=3[1+(2/3)^x]^{1/x}$$
$$\implies \log f(x)= \log 3+\frac{1}{x}\log[1+(2/3)^x]$$
$$\implies f'(x)=-\frac{1}{x^2}\log[1+(2/3)^x]+\frac{1}{x}\frac{(2/3)^x \log(2/3)}{1+(2/3)^x}<0$$
Because the first term is negative, the second term is also negative because $\log(2/3)<0.$
Thus $f(x)$ or $f(n), n \in N$ is a decreasing function of $n$.


问:
What real number is exactly one less than its cube?
And does it have any of the special properties that the golden ratio (one less than its square) has?</p>

答:
The numbers that have the sorts of features that $\phi$ does, are to be found in the series of $J(n) = \frac{\sqrt{n+2}+\sqrt{n-2})}2$, where $J(3) = \phi$.

The exotic-looking formula reflects the use of isoseries, in the form that $t(a+1) = n t(a) - t(a-1)$.  Applied to n=3, we get alternating lucas numbers, eg 2 (1) 3 (4) 7 (11) 18 (29) ...

The value for $J(4)$ gives the heron triangles, and is used to hunt down fermat-like primes. It turns up in the dodecagon, for example.  The series 2, 4, 14, 52, 194, ..., are the middle side of a triangle, whose edges are n-1, n, n+1, and the area is integral.

The value for $J(6)$ is involved in the octagon, as well as the various approximations to the square-root of 2,  Also, the silver ratio is $1:J(6)$.

On the other hand $J(0)$ and $J(1)$ are complex numbers, which produce ultimately the gaussian and eisenstein integers.

All of the $J(n)$ eventually turn up in polygons: $J(5)$ can be constructed from the edges of a polygon of 21 sides, and $J(11)$ turns up as in the polygon of 13 sides.