答:
The topology you have in mind - more commonly called the cofinite topology - looks a lot like the discrete topology in a few ways. One of these ways is the following:

Every self-bijection is continuous with respect to the cofinite topology.

The proof is basically immediate: the preimage of a cofinite set under an injective map must be cofinite.

Now if $X$ is any set with more than one element, there is a permutation of $X$ with no fixed points (exercise$^1$), so in fact no space with the cofinite topology which has more than one point has the fixed point property.

$^1$Amusingly, this actually requires a small amount of the axiom of choice: it is consistent with $\mathsf{ZF}$ (= set theory without choice) that there is a set with more than one element such that every self-bijection of the set has a fixed point. In fact it is consistent with $\mathsf{ZF}$ that there is an infinite set $A$ such that every self-bijection of $A$ fixes all but finitely many points! This is very difficult to prove however (as all $\mathsf{ZF}$-consistency results are), but Cohen's original model of $\mathsf{ZF+\neg AC}$ provides an example. In such a model we do indeed have infinite sets which, when equipped with the cofinite topology, have the fixed point property. But these are extremely pathological models.

Title: Does a Space with the Finite Closed Topology have Fixed Point Property?

Body: I wanted to know if a space $(X, \tau)$ with $\tau$ the finite-closed topology, have the Fixed Point Property, that is if every continuous mapping $f$ of $(X, \tau)$ into itself has a fixed point.


问:
Half order derivative of $ {1 \over 1-x }$
I'm new to this "fractional derivative" concept and try, using wikipedia, to solve a problem with the half-derivative of the zeta at zero, in this instance with the help of the zeta's Laurent-expansion.

Part of this fiddling is now to find the half-derivative  $$ {d^{1/2}\over dx^{1/2}}{1 \over 1-x}$$

First I would like to understand, whether there is a short/closed form for this ot whether I have to express the fraction as a power series first and then to differentiate termwise.

Next I would like to know the value at $x=0$.

答:
See Reference  for a complete solution of the problem of differentiation and integration of real order of rational polynomials.

Formal approaches to fractional derivative: There are several definitions for Fractional derivative. The most widely known one is the Riemann-Liouville fractional derivative

$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)} \frac{d^k}{dx^k} \int_{a}^{x}\,
(x-t)^{k-q-1}\,f(t)\,dt\>, \quad (k-1 < q < k )\,, $$

and according to this definition you will have the following answer

$$ f^{\left(\frac{1}{2}\right)}(x)={\frac {{_2F_1\left(1,1;\,\frac{1}{2};\,x\right)}}{\sqrt {x}\sqrt {\pi }}}, $$

where $ _2F_1 $ is the hypergeometric function. The limit of the above function as $x\to 0^+$ goes to infinity. However, there is another definition known as the Caputo definition and is defined as

$$ f^{(q)}(x) = \frac{1}{\Gamma(k-q)}  \int_{a}^{x}\,
(x-t)^{k-q-1}\,\frac{d^k}{dt^k}f(t)\,dt\>, \quad (k-1 < q < k )\,, $$

and this definition will give you a different answer, namely

$$ f^{\left(\frac{1}{2}\right)}(x) = {\frac {1}{\sqrt {\pi }  \left( x-1
 \right) ^{3/2}}} \left( {\it \rm arctanh} \left( {\frac {\sqrt {x}}{
\sqrt {x-1}}} \right) -\sqrt {x}\sqrt {x-1}\right), $$

and the limit in this case goes to $0$.

Note: The simplest approach to your problem is summarized in the steps

1) Compute the Taylor series of the function,

2) Use the Formula

$$ \frac{d^q}{dx^q} x^m = \frac{\Gamma(m+1)}{\Gamma(m-q+1 )} x^{m-q}\,, $$

which corresponds to the Riemann-Liouville definition of the function $x^m$, or the formula (2.29), page 15 which corresponds to Caputo definition.

See here to see the power series technique.

ContentLicense: CC BY-SA 3.0


问:
Show that a set is not a complex vector space
Let vector space $W$ be $\{p(z)\in P_{3}(\mathbb{C}):p(a)=0 \Rightarrow a\in \mathbb{R}\}$ under usual addition and usual scalar multiplication.

How to show that W is not a vector space over $\mathbb{C}$?

It should violate one or more of the 10 axioms of complex vector space, but I am stuck here, I tried and it all seems that 10 axioms holds, how can W not be an complex vector space?

答: Jose already picked out an answer, but it’d also suffice to observe that the zero polynomial isn’t in $W$.

ContentLicense: CC BY-SA 4.0


问:
Equivalence of the persistence landscape diagram and the barcode?
I am studying persistent homology for the first time. I was reading Peter Bubenik's paper "Statistical Topological Data Analysis using Persistence Landscapes" from 2015 introducing persistent landscapes. I am quite confused on the approach on finding the values of the persistence landscape function using a barcode/persistence diagram. I feel like I have a naive misunderstanding of this topic as I shall attempt to explain.

Suppose $X$ is a finite set of points in Euclidean space. From my understanding, if we consider the (finite length) persistence vector space given by the simplicial complex homology for a fixed dimension $l$, $\{H_l(X_k)\}_{k=1}^n$ with maps $\{\delta_{k,k'}\}$, for $1\leq k,k'\leq n$, we have
$$\{H_l(X_k)\}_{k=1}^n\cong\oplus_{i=1}^m \mathbb{I}(b_i, d_i)$$
(Theorem 4.10 of this paper) for some multiset $\{(b_i, d_i)\}_{i=1}^m$, where $\mathbb{I}(b,d)$ gives the persistence vector space of length $n$,
$$0\rightarrow ... \rightarrow 0\rightarrow \mathbb{R} \rightarrow \mathbb{R}\rightarrow \mathbb{R}\rightarrow... \rightarrow 0 ... \rightarrow 0$$
with non-zero vector spaces at values of the specified interval.

This multiset corresponds to the persistence diagram/barcode so that the $k$-th Betti number can be identified by finding the number of lines of the barcode that intersect the line $x=k$ in $\mathbb{R}^2$.

Now Bubenik defines the Betti number of the persistence vector space for an interval $[a,b]$ by
$\beta^{a,b}=\dim(\text{im}(\delta_{a,b}))$, and the persistence landscape functions $\lambda_k:\mathbb{R}\to [-\infty,\infty]$, for $k\in\mathbb{N}$ by
$$\lambda_k(t) = \sup \{m\geq 0 \mid \beta^{t-m, t+m}\geq k\}$$

Shouldn't $\beta^{t-m, t+m}$ then correspond to the number of lines on the barcode that contain the interval $[t-m, t+m]$, so that $\lambda_k(t)$ is the largest value of $m$ that has at least $k$ lines of the barcode intersecting $[t-m,t+m]$?

I am confused on how the triangle construction is equivalent to the persistence landscape function instead. Any help would be much appreciated!

答:
To read off the value of the landscape function $\lambda_k(t)$ from the barcode for fixed $k$, imagine expanding an interval centered at $t$.  A small interval might be contained in many bars of the barcode, but as the interval grows, fewer and fewer bars of the barcode can contain all of it.  Continue expanding the interval until just before it is contained in only $k - 1$ bars.  The radius $m$ of that interval is equal to $\lambda_k(t)$.

Of course, I prefer to obtain landscapes from persistence diagrams.  To do this, drop two lines down and right from each point on the diagram to the diagonal and rotate the entire picture $45^\circ$ clockwise.  Imagine the "diagonal" (which is now horizontal) as the $t$-axis and the vertical distance as the $m$-axis.  What you see formed from the lines you drew is the union of the graphs of the functions $\lambda_k(t)$ for $k \geq 1$.  The topmost ridge formed from the lines is the graph of $\lambda_1$, the second topmost ridge formed is the graph of $\lambda_2$, and so on.


问:
For primes, $p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$
I am looking for primes that would satisfy this equation:

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

I started by proving that an odd amount of these primes (two or four) cannot be equal to $2$ due to the parity of the left hand side and right hand side.

Then I assumed that exactly three of these primes would be equal to  $2$ and easily found a solution of $(2,2,2,3)$ and its permutations.

My problem now is, I cannot say that there isn't a fitting set of primes, out of which exactly one would be equal to $2$, or for all primes being odd. That leaves me with

$p_1+p_2+p_3=2p_1 p_2 p_3 - 17$

and

$p_1 + p_2 +p_3+p_4 = p_1 p_2 p_3 p_4 - 15$

respectively, and I am not sure how to go about those.

Is there a way to solve this assuming one of the primes is equal to $2$ and none of them is equal to  $2$?

Body:
The first thing I think is that $p_1p_2p_3p_4$, which is a multiple of many $p_1$s,$p_2$s, $p_3$s, and $p_4$s, is certainly going to be a lot bigger than $p_1+p_2+p_3+p_4$.

Indeed.  $p_1p_2p_3p_4 = (p_1p_2p_3 -1)p_4 + p_4=$

$(p_1p_2p_3-1)(p_4-1) + p_1p_2p_3 -1 + p+4 =$

$(p_1p_2p_3-1)(p_4-1) + (p_1p_2 -1)p_3 -1 + p_3 + p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) +p_1p_2 -2 +p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)p_2 -2 + p_2+p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) + p_1 -3 p_2+p_3+p_4=$

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) -3+ (p_1 + p_2+p_3+p_4)\ge $

$(8-1)(2-1) + (4-1)(2-1)+(2-1)(2-1)-3+ (p_1 + p_2+p_3+p_4)=$

$11+ (p_1 + p_2+p_3+p_4)$.

Now we want $p_1+p_2+p_3+p_4 = p_1p_2p_3p_4 -15 =(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1) -3+ (p_1 + p_2+p_3+p_4)-15$

or in other words

$(p_1p_2p_3-1)(p_4-1)+(p_1p_2-1)(p_3-1) + (p_1-1)(p_2-1)=18$.

Note: $p_1p_2p_3-1 \ge 7$ so $p_4-1 \le 2$.  And $p_4-1 \ge 1$ so $p_1p_2p_3-1< 17$.

The tells us that $p_4 = 2,3$ and at least two of $p_1,p_2,p_3=2$.

If $p_4 = 3$ the only option is $p_1=p_2=p_3=2$.

If $p_4 =2$ we can have $p_1p_2p_3-1=7, 11$.

We can brute force this out, but as it was arbitrary which terms we factored out first.  We can assume without loss  of generality that $p_4\ge p_3,p_2,p_1$.  So we have either $p_3=3$ and $p_1=p_2=p_3=2$.

Or that $p_3=2$ and all the primes are $2$.  Butbut... $2+2+2+2\ne 2*2*2*2-15$ so $(2,2,2,3)$ is the only solution.

==== old answer below ====

If we assume $p_1 \le p_2 \le p_3\le p_4$ then

$p_1+p_2+ p_3 + p_4 = p_1p_2p_3p_4-15$

$15 = p_1p_2p_3p_4 - p_1-p_2-p_3-p_4=$

$p_1(p_2p_3p_4 -1) - p_2 - p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1) + (p_2p_3p_4 -1)- p_2 - p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_3p_4-1)p_2 -1-p_3-p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + p_3p_4 - 2 - p_3 - p_4=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_4-1)p_3 -p_4 -2=$

$(p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1) -3$

So

$12 = (p_1-1)(p_2p_3p_4-1)+ (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1)$

Now $p_2p_3p_4 - 1\ge 7$ so $p_1-1 < 2$ so $p_1-1 =1$ and $p_1 = 2$.

And $p_2p_3p_4-1 \ge 7$ so

$5 \ge (p_2-1)(p_3p_4-1) + (p_3-1)(p_4-1)$ and $p_3p_4-1 \ge 3$ so $p_2-1 < 2$ and $p_2 = 2$.

And as $p_3p_4-1 \ge 3$ we have

$2\ge (p_3-1)(p_4-1)$.

If $p_3-1\ge 2$ then $p_4-1 \ge 2$ and that'd imply $2\ge 4$ which is a contradiction, so $p_3-1=1$ and $p_3 = 2$

And with that we have

$2+2+2+p_4 = 8p_4 -15$

$21 = 7p_4$

$p_4 =3$.

So that's the only (up to order) solution.  $p_1,p_2,p_3,p_4 = 2,2,2,3$


问:
Best approximation to reflexive subspace - converse
Let $X$ be a normed vector space.

I know if $X$ is reflexive, then for every normed space $Y$ that $X$ is a subspace of $Y$, $X$ has a best approximation to every $y \in Y$ (which is to say there exists $x\in X$ that $d(y, X) = \|x-y\|$).

I need to prove the converse of the above fact. To be more precise, if for every normed space $Y$ that $X \subset Y$, $X$ has a best approximation to every $y \in Y$, is it true that $X$ is reflexive?

答: A proof can be found here (page 399)</p>


问:
Is this transformation linear?
The input is $v=(v_1,v_2)$

$T(v)=v_1v_2$

Is this linear transformation?

答:
Hint: Take:

$$v = (1,0)$$

and

$$u = (0,1)$$

Then check whether:

$$T(u+v) = T(u) + T(v)$$

Title: Is this transformation linear?


问:
Closed curves in 3D
Is $R^3$ a union of disjoint closed curves? (Obviously $R^3$ minus a line is). Is this a classical problem?</p>

答: This is probably not what you wanted, but I can do it trivially: For every $r \in \mathbb{R}^3$, let $$C_{r}:I \to \mathbb{R}^3$$ be the map given by $C_{r}(x) = r$. That is, we let each curve be a trivial one point curve. These are all closed. Then we have that $$\mathbb{R}^3 = \bigcup_{r \in \mathbb{R}^3} C_r$$ and the union is disjoint.</p>

Title: Closed curves in 3D


问:
Elementary algebraic system with integer and positive rational unknowns
The problem is to determine the solutions of the system $x+y+z=3$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=5$, given that $x$ is integer and $y,z$ are positive rationals.

From the hypotheses I have obtained that $y$ and $z$ are rational solutions of the quadratic equation
$$(5x-1)t^2-(5x-1)(3-x)t+(3-x)x=0$$
with integer coefficients, therefore its discriminant $\left( 5\,x-1 \right)  \left( x-3 \right)  \left( 5\,{x}^{2}-12\,x+3
 \right)$ has to be a perfect square.

But I cannot see a clear continuation from here.

答:
(There was an old problem which looks similar.)

Solution:
First note that x<3, so if $x>0$ we can easily find $x=1$ is not good; and $x=2$ yields $(y,z)=(\frac 13, \frac 23)$ or $(\frac 23, \frac 13)$.

Now suppose $x<0, w=-x, w \in \mathbb N$. Your discriminant becomes $$(5w+1)(w+3)(5w^2+12w+3)=(5w^2+16w+3)(5w^2+12w+3)\\=(5w^2+14w+3)^2-(2w)^2$$

So we have a Pythagorean triple. There are only two possibilities:

Case 1: $5w^2+14w+3=k(m^2+n^2), 2w=2kmn,m>n, (m,n)=1, $ and not both odd.

But
$$k(m^2+n^2)>5w^2+14w=5k^2m^2n^2+14kmn>5k^2m^2+14kn^2>km^2+kn^2, \Rightarrow \Leftarrow.$$

Case 2: $5w^2+14w+3=k(m^2+n^2), 2w=k(m^2-n^2),m>n, (m,n)=1, $ and not both odd.

Again
$$k(m^2+n^2) > 5w^2 = 5 \frac{k^2}{4} (m+n)^2(m-n)^2 > k^2(m+n)^2 > k(m^2+n^2), \Rightarrow \Leftarrow.$$

Therefore the only solutions are $(2,\frac 13, \frac 23), (2, \frac 23, \frac 13).$


问:
Proof that a well-ordering is $\omega$-saturated only if it is finite
I wanna prove that an infinite well-ordering is not $\omega$-saturated. How can I do this? Does the proof involve Lowenheim-Skolem theorem?

答:
As indicated in Noah's and Atticus's answers, and my comment, it's probably best to understand this fact as an application of the theorem that $\omega$-saturated models realize all consistent types in $\omega$-many free variables, so an $\omega$-saturated infinite linear order realizes the type of a countable descending sequence. But in this answer, I'll try to explain how you could answer the question without using the more general theorem.

Let $(M;<)$ be an infinite $\omega$-saturated linear order. We would like to show that $M$ is not a well-order. To do that, we would like to build an infinite descending chain in $M$. Let's try to do this by induction. We start by picking an element $a_0$. Then, having chosen $a_n$, we pick some $a_{n+1}<a_n$.

There's an obvious problem with this strategy: What ensures that there is some $a_{n+1}<a_n$? That is, if we accidentally pick $a_n$ to be the least element of the order, we get stuck. Saturation is no help: if $a_n$ is the least element of $M$, then $M\models \forall x\, \lnot (x < a_n)$, so the formula $x<a_n$ is just inconsistent.

Ok, so instead let's pick our sequence by induction, making sure to never pick the least element. But now what ensures that there is some $a_{n+1}<a_n$ which is not the least element? If $a_n$ has only one element below it in the order, we again get stuck. Clearly, we're in trouble if we ever pick an element $a_n$ with only finitely many elements below it in the order.

New idea: Make sure that each element of the sequence has infinitely many elements below it. So we start by picking an element $a_0$ with infinitely many elements below it. Then, having chosen $a_n$ with infinitely many elements below it, we pick some $a_{n+1}<a_n$ such that $a_{n+1}$ has infinitely many elements below it. But what ensures that there is some $a_{n+1}<a_n$ with infinitely many elements below it? Now saturation is comes into play: This condition amounts to realizing a certain type. Details after the break.

For each natural number $n>0$, let $\varphi_n(x)$ be the formula $$\exists y_1\dots\exists y_n\,(y_1<y_2<\dots<y_n < x).$$ Let $\Phi(x)$ be the partial type $\{\varphi_n(x)\mid n\in \mathbb{N}\}.$ This partial type says "$x$ has infinitely many elements below it".

We construct a descending sequence $(a_n)_{n\in \mathbb{N}}$ from $M$ by induction, ensuring that each $a_n$ realizes $\Phi(x)$.

In the base case, we show that $\Phi(x)$ is consistent by compactness: Given any finite subset $\{\varphi_n\mid 1\leq n\leq N\}$, we can pick any $N+1$ elements of $M$, listed in increasing order $b_1<\dots<b_{N+1}$ (here we use the fact that $M$ is infinite). Then taking $x = b_{N+1}$, the elements $b_i$ serve as witnesses for the variables $y_i$ in the formulas $\varphi_n$ with $n\leq N$. Since $\Phi(x)$ is consistent and $M$ is $\omega$-saturated, we can find $a_0\in M$ realizing $\Phi(x)$.

Now suppose by induction that we have picked $a_n$ realizing $\Phi(x)$. We show that the partial type $\Phi(x)\cup \{x<a_n\}$ is consistent by compactness: Given any finite subset $\{\varphi_n\mid 1\leq n\leq N\}$, recall that $a_n$ satisfies $\varphi_{N+1}$. So we can find $b_1<\dots<b_{N+1}<a_n$ in $M$. Then taking $x = b_{N+1}$, we have $x < a_n$, and the elements $b_i$ serve as witnesses for the variables $y_i$ in the formulas $\varphi_n$ with $n\leq N$. Since $\Phi(x)\cup \{x<a_n\}$ is consistent and $M$ is $\omega$-saturated, we can find $a_{n+1}\in M$ realizing $\Phi(x)\cup \{x<a_n\}$.


问:
Does $\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$ converge, absolutely converge?
$\sum_{k=1}^{\infty} \frac{1}{k}\cdot(\frac{1+i}{1-i})^k$, $a_k=\frac{1}{k}\cdot(\frac{1+i}{1-i})^k \in\mathbb{C}$

What I figured out so far:

$a_k=\left\{\begin{array}{0} \frac{1}{k}, & k\equiv0 \text{ (mod } 4) \\
         \frac{1}{k}i, & k\equiv1 \text{ (mod } 4)\\
-\frac{1}{k}, & k\equiv2 \text{ (mod } 4) \\
-\frac{1}{k}i, & k\equiv3 \text{ (mod } 4)\end{array}\right. $

I know that the series converges if the sequence of partial sums $s_n:=\sum_{k=1}^{n} a_k$ converges.

So $s_1 = -1, s_2=-\frac{3}{2}, s_3=\frac{5}{6}, s_2=\frac{13}{12}, s_2=\frac{53}{60},s_2=-\frac{7}{60}$.

Therefore I assume that $\lim_{n\to\infty} s_n$ exists.

I don't know how to handle the different cases for $a_k$ in my $s_n$.

I appreciate any hints.

答: Consider the function $f(z)=\frac 1 {1+z}$. Observe that, for $|z|<1$, we have $f(z)=\sum\limits_{n=0}^{\infty} {(-z)^n}$, integrating from $0$ to $w$ (with $|w|\leq 1$, we obtain $\log(w+1)=\int_0^w f = \sum_{n=0}^\infty \frac{-(-z)^{n+1}}{n+1} \Bigr|_0^w$. What happens when you take $w=-i$?


问:
Ideals in $\mathbb{C}[x,y]$ which are not the product of another ideal
I feel like I understand the theory of ideals (at least at the basic level required for this), but struggle with actual computations. I am trying to figure out an example of two ideals $I, A$ in $\mathbb{C}[x,y]$, $I\subseteq A$, such that there does not exist another ideal $J$ such that $I=AJ$.

I proved that in a commutative ring with identity, if an ideal $I$ is contained in a principal ideal $(a)$, then there exists an ideal $J$ such that $I=(a)J$.

So in $\mathbb{C}[x,y]$, I thus need to find a non-principal ideal and look at ideals contained within that.

My first instinct is to go with $A=(x,y)$, and then consider $I=(x)$ or $I=(x-y)$, but in computing I am not creating any contradictions taking $J$ to be some finitely generated ideal.

Is there some higher-level theorem I could use to point myself in a better direction?

Edit: I imagine it may have something to do with polynomials of minimal degree perhaps being impossible in the product of $A$ and $J$, however that is only a hunch.

答: $I=(x)$ and $A=(x,y)$ work. In fact, let $J$ be such that $JA=I$. We have $$JA=xJ+yJ=(x)$$ and therefore $x\mid yf$ for all $f\in J$. This means that $x\mid f$ for all $f\in J$ and, therefore, that $J\subseteq (x)$. If $J\subsetneq (x)$, then $JA\subseteq J\subsetneq (x)$: hence, $J=(x)$ is necessary. But $IA=(x^2,xy)\ne (x)$, because, for instance, if $f\in (x^2,xy)$ then either $f=0$ or the least total degree of a monomial of $f$ is at least $2$.


问:
Does $\lim\limits_{n\to+\infty}|\sin{2^n}|^{\frac1n}$ exist?
More generally, for what real number x does $\displaystyle\lim_{n\to+\infty}|\sin2^nx|^\frac1n$ exist?

Actually it's already known that if you replace the $2^n$ with $n$, then the answer is 1. That is, $$\lim\limits_{n\to+\infty}|\sin n|^\frac1n=1$$.

But when it seems that I could not get the same information about how $\frac{2^n} m$ goes to $\pi$.

To answer this question, one has to consider two things:

$\liminf\limits_{n\to+\infty}|2^n-m\pi|$ $=$ ?

If the answer to 1) is $0$, then how fast does the infimum actually come to $0$?

答:
Just a few elementary facts that you didn't mention.

Let $f(t)= t-\lfloor t +1/2\rfloor$. Note that $f(2t) = f(2f(t))$, thus for $f(t)\in (-1/4,1/4)$ we have $f(2t)=2f(t)$.

If $|\sin(2^n \pi x)|^{1/n}\to 0$ then $f(2^nx)\to 0$.

Thus for all $n\ge N$ large enough we have $f(2^n x)\in (-1/4,1/4)$ so that $f(2^{n+1} x)=2 f(2^n x)$ and hence $f(2^n x) = 2^{n-N} f(2^N x)$ which implies that $f(2^N x)=0$ ie.  $x\in \Bbb{Z}[2^{-1}]$.

Next assume that $|\sin(2^n \pi x)|^{1/n}\to r\in (0,1)$, thus $|\sin(\pi f(2^n  x))|= |\sin(2^n \pi x)| \sim r^n, \pi f(2^n  x) \sim r^n$ so that $\pi f(2^{n+1}  x) \sim 2 r^n$ and $r^{n+1} \sim 2 r^n$ is a contradiction.

Whence for $x\not\in \Bbb{Z}[2^{-1}]$, if $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ exists then the limit is $1$.

If $x\in \Bbb{Q}-\Bbb{Z}[2^{-1}]$ then the binary expansion of $x$ is infinite periodic so that $f(2^n x)$ is periodic and $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}=1$.

With things like $x=\sum_{m\ge 1} 2^{-m!}$ then $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ doesn't exist.

For $x\not \in\Bbb{Z}[2^{-1}]$ let $c_l$ be the indice of the first $\underbrace{0\ldots 0}_l$ or $\underbrace{1\ldots 1}_l$ motif found in the binary expansion of $x$ (if it doesn't exist set $c_l=\infty$). Then $\lim_{n\to \infty}|\sin(2^n \pi x)|^{1/n}$ exists (thus is $=1$) iff $c_l/l\to \infty$.


问:
Differentiate $\log_{10}x$
My attempt:

$\eqalign{
  & \log_{10}x = {{\ln x} \over {\ln 10}}  \cr
  & u = \ln x  \cr
  & v = \ln 10  \cr
  & {{du} \over {dx}} = {1 \over x}  \cr
  & {{dv} \over {dx}} = 0  \cr
  & {v^2} = {(\ln10)^2}  \cr
  & {{dy} \over {dx}} = {{\left( {{{\ln 10} \over x}} \right)} \over {2\ln 10}} = {{\ln10} \over x} \times {1 \over {2\ln 10}} = {1 \over {2x}} \cr} $

The right answer is: ${{dy} \over {dx}} = {1 \over {x\ln 10}}$ , where did I go wrong?

Thanks!

答: You made one small mistake when you looked at $v^2$. Note that: $$(\ln {v})^2\ne 2\ln(v)$$That would only be true if the expression was: $$\ln\bigl(v^2\bigr)$$</p>

Title: Differentiate $\log_{10}x$


问:
how to determine if a vector filed is a gradient field or a rotation field
Which of the following is neither a gradient field nor a rotation field?

$$\mathbf{f}(\mathbf{r})= ||\mathbf{r}||^2\mathbf{r}$$

or $$\mathbf{f}(\mathbf{r})= \mathbf{r} \times \mathbf{i}$$

I know that if the curl is $0$ then it is a gradient field however I don't understand how to work that out.

答:
Hint:

$f(r) = ||r||^2 \,\vec{r} = ||r||^2 (x\hat{i}+y\hat{j}+z\hat{k})$

$curl F = \nabla \times F = ||r||^2 \,\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z\end {vmatrix} = ?$

In second case,

$f(r) = \vec{r} \times \hat{i} = \begin{vmatrix} i & j & k \\ x & y & z \\ 1 & 0 & 0\end {vmatrix} = ?
$

Once you get $f(r)$, find its curl just like in the first case.


问:
What is the Fourier transform of $e^{2 \pi i / x}$?
The Fourier transform of $e^{2 \pi i / x}$ makes sense as a distribution, I believe. Does it have a nice expression in terms of functions and well-known distributions (e.g. Dirac delta)? </p>

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\expo{2\pi\ic/x}
\expo{2\pi\nu x\ic}\,\dd x} =
\int_{0}^{\infty}\cos\pars{2\pi\root{\nu}\bracks{{1 \over \root{\nu}x} + \root{\nu}x}}
\end{align}
With $\ds{x = {\expo{\theta} \over \root{\nu}}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{-\infty}^{\infty}\expo{2\pi\ic/x}
\expo{2\pi\nu x\ic}\,\dd x} =
\int_{-\infty}^{\infty}
\cos\pars{4\pi\root{\nu}\cosh\pars{\theta}}
{\expo{\theta} \over \root{\nu}}\,\dd\theta
\\[5mm] = &\
{1 \over \root{\nu}}\int_{0}^{\infty}
\cos\pars{4\pi\root{\nu}\cosh\pars{\theta}}
\cosh{\theta}\,\dd\theta
\\[5mm] = &\
\left.{1 \over \root{\nu}}\,\partiald{}{\alpha}\
\underbrace{\int_{0}^{\infty}
\sin\pars{\alpha\cosh\pars{\theta}}
\,\dd\theta}_{\ds{{\pi \over 2}\on{J}_{0}\pars{\alpha}}}\,\right\vert_{\,\alpha\ =\ 4\pi\root{\nu}}\label{1}\tag{1}
\\[5mm] = &\
\left.{1 \over \root{\nu}}\,{\pi \over 2}\,
\partiald{\on{J}_{0}\pars{\alpha}}{\alpha}\
\,\right\vert_{\,\alpha\ =\ 4\pi\root{\nu}}  =
\bbx{-\,{\pi \over 2\root{\nu}}\,\on{J}_{1}\pars{4\pi\root{\nu}}}
\label{2}\tag{2} \\ &
\end{align}

(\ref{1}): See a $\ds{\on{J}_{0}}$ Integral Representation.

(\ref{2}): See the $\ds{\on{J}_{0}}$ Recurrence.

(\ref{2}) as a function of $\ds{\nu}$:


答: Elaborating a bit on @Clement's answer, $\left(\frac{\ln x}{\ln 10}\right)'=\frac{1}{\ln 10}(\ln x)'$, because $\frac{1}{\ln 10}$ is a constant.</p>

Title: Differentiate $\log_{10}x$

Body:
My attempt:

$\eqalign{
  & \log_{10}x = {{\ln x} \over {\ln 10}}  \cr
  & u = \ln x  \cr
  & v = \ln 10  \cr
  & {{du} \over {dx}} = {1 \over x}  \cr
  & {{dv} \over {dx}} = 0  \cr
  & {v^2} = {(\ln10)^2}  \cr
  & {{dy} \over {dx}} = {{\left( {{{\ln 10} \over x}} \right)} \over {2\ln 10}} = {{\ln10} \over x} \times {1 \over {2\ln 10}} = {1 \over {2x}} \cr} $

The right answer is: ${{dy} \over {dx}} = {1 \over {x\ln 10}}$ , where did I go wrong?

Thanks!


问:
Finding the largest of $S_1,S_2,S_3,\cdots,S_{13}$
Let $S_n$ be the sum of the first $n$ terms of a arithmetic sequence $\{a_n\}.$ If $a_3 = 12,  S_{12} > 0$ and $S_{13} <0,$ then find the largest out of $S_1, S_2, S_3, \ldots, S_{13}.$

I first let $a$ be the first term and $d$ be the common difference. Then, I got that
\begin{align*}
S_1 &= a \\
S_2 &= 2a + d \\
S_3 &= 3a + 3d \\
S_4 &= 4a + 6d \\
S_5 &= 5a + 10d \\
S_6 &= 6a + 15d \\
S_7 &= 7a + 21d \\
S_8 &= 8a + 28d \\
S_9 &= 9a + 36d \\
S_{10} &= 10a + 45d \\
S_{11} &= 11a + 55d \\
S_{12} &= 12a + 66d \\
S_{13} &= 13a + 78d.
\end{align*}
However, I am unsure how to proceed from here. Can someone help please?

答:
From the given relations, it may be shown that

$$18 < a < \frac{132}{7}$$ and $$-\frac{24}{7} < d < -3$$

Observe that if $a_n = a+d(n-1) < 0$, then $S_{n-1} > S_n$

However, $S_{13} = 13a + 78d < 0 \implies a + 6d < 0 \implies a_7 < 0 \implies S_6 > S_7$

Since $d < 0$, this means $S_6 > S_7 > \ldots > S_{13}$

Can we show that $S_1 < \ldots < S_6$?

In fact, it is enough to show that $S_5 < S_6 \equiv a_6 = a + 5d > 0$

Given: $S_{12} = 12a + 66d > 0$

$\implies 2a + 11d > 0$, it follows that $2(a + 5d) > 0$ since $d < 0$


问:
Using an open and compact set to find another compact set within the original compact set.
I am trying to show that if U a subset of R^n is open, and K a subset of U is compact, then there is a compact set such that K is a subset of D^o and D is a subset of U.

I know that since K is compact and K is a subset of U, we can use the boundary of K (represented as delta(K) = K - K^o). I think I need to find a delta that can be used to find the biggest epsilon ball at any point on K and that would be the bound for D. I'm just stuck on how I find the value of delta. Can anyone help?

答: Hint: Let $r=d(K,\mathbb R^{n}\setminus U)$. Since $K$ is compact and $\mathbb R^{n}\setminus U$ is closed we see that $r>0$. Let $D=\{x: d(x, K) \leq r/2\}$. Then $D$ is closed and bounded, hence compact. Also $K \subseteq D$. Now $K \subseteq ${x: d(x, K) < r/2}$ and $$D=\{x: d(x, K) < r/2\}$ is an open set conatined in $D$ It follows that $K$ is contained in the interior of $D$.


Title: Finding the largest of $S_1,S_2,S_3,\cdots,S_{13}$

Body:
Let $S_n$ be the sum of the first $n$ terms of a arithmetic sequence $\{a_n\}.$ If $a_3 = 12,  S_{12} > 0$ and $S_{13} <0,$ then find the largest out of $S_1, S_2, S_3, \ldots, S_{13}.$

I first let $a$ be the first term and $d$ be the common difference. Then, I got that
\begin{align*}
S_1 &= a \\
S_2 &= 2a + d \\
S_3 &= 3a + 3d \\
S_4 &= 4a + 6d \\
S_5 &= 5a + 10d \\
S_6 &= 6a + 15d \\
S_7 &= 7a + 21d \\
S_8 &= 8a + 28d \\
S_9 &= 9a + 36d \\
S_{10} &= 10a + 45d \\
S_{11} &= 11a + 55d \\
S_{12} &= 12a + 66d \\
S_{13} &= 13a + 78d.
\end{align*}
However, I am unsure how to proceed from here. Can someone help please?

Body:
If $a_3=12$ this means that $a+2d=12\to a = 12-2d$

Knowing that $S_{12}>0$ tells us that $12a+66d>0\to 42 d+144>0 $

and if $S_{13}<0$ we know that $13a+78d<0\to  156 + 52 d<0$

Thus we have the limitation $-\frac{24}{7}<d<-3$

Anyway the largest sum is $63$ substituting $d=-3$ in $S_6,S_7$

$$\{12-2 d,24-3 d,36-3 d,48-2 d,60,3 d+72,7 d+84,12 d+96,\\18 d+108,25 d+120,33 d+132,42 d+144,52 d+156\}$$

$$\{18,33,45,54,60,63,63,60,54,45,33,18,0\}$$


问:
Dornhoff definition of R-module in "Group representation theory"
I have a question regarding the definition given by L. Dornhoff in his book: "Group representation theory", part A.

He states that "and additive abelian group M is called a module over a ring A (left A-module) if we have a multiplication $am \in M$, for $a\in A$, $m \in M$ such that $(a+a')m=am+a'm$, $a(a'm)=(aa')m$, $1m=m$".

However, the definition of A-module that I know also requires also (as an axiom) that $a(m+m')=am+am'$ and, as far as I tried, I couldn't derive this last one from the others.

My questions are: did he forget to include this last axiom? Or it is possible to derive it from the others? Or maybe it is a weaker valid definition suitable for his purposes?

If anyone knows, it would help a lot

答:
Whether or not it follows from the other axioms (which I strongly doubt, though I'm not interested enough to concoct a formal counter-example... since it wouldn't really mean anything), surely that property is desired/intended/required of modules over rings.

In fact, I'd think it was/is so strongly intended that the writer lost interest in spelling out "the usual details" before quite finishing. :)


问:
Prove that the set of a product space is closed.
Given that $(r_n) \rightarrow 0 $ is a null sequence of irrational numbers in $\underline R$.

Also $(s_{n,m}) \rightarrow r_n$ is a sequence of rational numbers in $\underline R$.

Prove that: $$ A = \left\{ \left(n + \frac 1 m,\ s_{n,m} \right) \ | \ n \in N, m \in N, m \ge 2 \right\} $$
is a closed in $\underline R \times \underline Q$.

I guess we have to show that $A$ is sequentially-closed or that any convergent sequence from it converges to the element of $A$. I'm not sure how to do that in the case of product spaces.

答:
The set $A$ is discrete, meaning for any point you can find an open ball around it which contains only the point. This is because the set $\frac{1}{m}$ is discrete, and so is $s_{n,m}$ for any fixed $n$ since the only limit point of a converging sequence is it's limit, in this case the limit is irrational hence not in the product topology.

It is important to notice the set $A$ is not closed in the product $\mathbb{R}$, since the points $(n,r_n)$ would be limit points of $A$, but not in $A$.

To be a bit more detailed, for any $(n_0+\frac{1}{m_0},s_{n_0,m_0})\in A$ consider the ball of radius $\frac{1}{m_0+1}$ around $n_0+\frac{1}{m_0}$ and the ball of radius inf$_{\forall m}|s_{n_0,m_0}-s_{n_0,m}|$ which exists since $r_n$ is irrational, and converging sequences, have exactly one limit point.


问:
prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$
How to prove that any positive integer-valued random variable with memoryless property has the geometric distribution for some $p$.

By memoryless property,

$$P(X=i+s | X>i)=P(X=s)$$

How to get distribution for X from the above ?

答:
This is essentially a reworking of Did's answer. Fix $i=1$, $q = P\{X > 1\}$
and note that
$$P\{X = s+1\mid X > 1\} = \frac{P\{X = s+1, X > 1\}}{P\{X > 1\}}
  = \frac{P\{X = s+1\}}{q} = P\{X=s\}$$
which gives
$$\begin{align}
P\{X=2\} &= qP\{X=1\}\\
P\{X=3\} &=q P\{X=2\} = q^2P\{X=1\}\\
P\{X=4\} &=q P\{X=3\} = q^3P\{X=1\}\\
\vdots\qquad &\vdots \qquad \vdots\\
P\{X = n\} &= qP\{X = n-1\} = q^nP\{X=1\}\\
\vdots\qquad &\vdots \qquad \vdots
\end{align}$$
so the probability masses decrease as a geometric series.</p>


问:
If $f(x+1/n)$ converges to a continuous function $f(x)$ uniformly on $\mathbb{R}$, is $f(x)$ necessarily uniformly continuous?
Suppose $f:\mathbb{R}\mapsto\mathbb{R}$ is a continuous function. Define $f_n(x)=f(x+1/n)$ for all $n\in\mathbb{N}^+$. If $f_n(x)\to f(x)$ uniformly on $\mathbb{R}$, can we conclude that $f$ is actually uniformly continuous? If not, can you give a counterexample of $f$ to be not uniformly continuous but satisfies all above conditions?

I know that if $f_n(x)$'s are all uniformly continuous and $f_n(x)\to f(x)$ uniformly, then $f(x)$ must be uniformly continuous. However, here we do not assume $f(x)$ to be uniformly continuous, and $f_n(x)$ has specific structure related to $f(x)$, so at least you cannot directly conclude they are uniformly continuous. I guess the statement above is wrong but it's hard for me to find a counter-example. Can anyone help me?

答:
Updated answer:

No, $f$ need not be uniformly continuous, as shown by the example below. The example is tailored to the sequence $\{1/n\}_{n \in \mathbb{N}}$, and does not shed light on the general question of whether the condition that $f(x + \epsilon_n) \to f(x)$ uniformly implies that $f$ is uniformly continuous, for a given sequence $\{\varepsilon_n\}$ converging to zero.

First, define a sequence recursively by $a_1 = 3$ and $a_{n+1} = (a_n)!$ for $n \geq 1$. Let $h$ be the periodic function with period $1$ such that $h(x) = 2x$ for $x \in [0, 1/2]$ and $h(x) = 2-2x$ for $x \in [1/2, 1]$. In particular, $h(k) = 0$ for each integer $k$, and $h$ is continuous and Lipschitz with constant $2$. Now using the sequence, for each $n$ we define the function $g_n$ by
$$g_n(x) = \begin{cases} 0 & x < n \\ \frac{1}{n} h(a_n x) & x \geq n\end{cases}$$
so $g_n$ is continuous and Lipschitz with constant $2a_n/n$. Finally, define $f(x) = \sum_{n=1}^\infty g_n(x)$, where $f$ is continuous since each $g_n$ is continuous, and all but finitely many $g_n$ vanish on $(-\infty, b)$ for each $b$.

We will now show that $f_m \to f$ uniformly, where $f_m(x) = f(x + 1/m)$. Fix $m$, and let $k$ be the smallest index such that $a_k \geq m$. In particular, this means that $m$ divides $a_n$ for all $n \geq k+1$, and thus for each such $n$, $1/m$ is a multiple of $1/a_n$, so $g_n(x + 1/m) = g_n(x)$ for all $x$ not in $(n-1, n)$. Therefore for any $x$, there is at most one $n \geq k+1$ for which $g_n(x + 1/m) \neq g_n(x)$, and this $n$ necessarily satisfies $|g_n(x + 1/m) - g_n(x)| \leq \frac{1}{n} \leq \frac{1}{k}$. It follows that for sufficiently large $m$ we have

\begin{align*}
|f(x + 1/m) - f(x)|
&= \left|\sum_{n=1}^\infty g_n(x + 1/m) - g_n(x)\right| \\
&\leq \frac{1}{k} + \sum_{n=1}^k |g_n(x + 1/m) - g_n(x)| \\
&\leq \frac{1}{k} + \frac{1}{k} + \frac{1}{k-1} + \sum_{n=1}^{k-2} |g_n(x + 1/m) - g_n(x)| \\
&\leq \frac{4}{k} + \sum_{n=1}^{k-2} \frac{2a_n}{n} \cdot \frac{1}{m} \\
&\leq \frac{4}{k} + \frac{2ka_{k-2}}{m} \\
&\leq \frac{5}{k}
\end{align*}
where the last inequality follows from $m \geq a_{k-1} \geq 2k^2 a_{k-2}$, which clearly holds for sufficiently large $k$. Then if we define $k_m$ to be the smallest $k$ for which $a_k \geq m$, we see that $k_m \to \infty$ as $m \to \infty$, so since $|f(x + 1/m) - f(x)| \leq 5/k_m$ uniformly for sufficiently large $m$, it follows that $f(x + 1/m)$ converges to $f(x)$ uniformly.

However, $f$ is not uniformly continuous. Note that $\int_0^1 h(a_nx) \,dx = 1/2$, so for any positive integer $k$, $\int_0^1 f(k + x) \,dx = \frac{1}{2}\sum_{n=1}^k \frac{1}{n}$, and thus there is some $x_k \in (0, 1)$ with $f(k + x_k) \geq \frac{1}{2}\sum_{n=1}^k \frac{1}{n}$. On the other hand $f(k) = 0$ always. But if $f$ were uniformly continuous, there would be some $M$ for which $|f(x) - f(y)| \leq M$ whenever $|x - y| \leq 1$, which clearly does not hold, so $f$ cannot be uniformly continuous.

Original answer:

Too long for a comment:

Whatever the answer is, it might depend on the properties of the sequence $\{1/n\}_{n \in \mathbb{N}}$. Below is an example which shows that if we instead define $f_n(x) = f(x + 3^{-n})$, then the condition that $f_n \to f$ uniformly on $\mathbb{R}$ does not imply that $f$ is uniformly continuous. I was unable to construct a similar example for the sequence $\{1/n\}_{n \in \mathbb{N}}$.

Let $h(x)$ be the "triangle wave" function, such that $h(x)$ is periodic with period $1$ and has $h(x) = 2x$ for $x \in [0, 1/2]$ and $h(x) = 2 - 2x$ for $x \in [1/2, 1]$. In particular, $h$ is continuous and has $h(0) = h(1) = 0$, $h(1/2) = 1$, and $|h(x) - h(y)| \leq 2|x - y|$ for all $x, y$. Next, for each $n$, define $g_n(x)$ so that $g_n(x) = 0$ for $x < n$ and $g_n(x) = \frac{1}{n} h(3^n x)$ for $x \geq n$. With this definition, $g_n$ is continuous, and satisfies $|g_n(x) - g_n(y)| \leq \frac{2}{n} \cdot 3^n|x - y|$ for all $x, y$. Finally, define $f(x) = \sum_{n=1}^\infty g_n(x)$. Note $f$ is continuous since each $g_n$ is continuous, and all but finitely many $g_n$ vanish on $(-\infty, a)$ for each $a$.

Now we will show that $f_m \to f$ uniformly, where $f_m(x) = f(x + 3^{-m})$. For $n \geq m$, $g_n(x + 3^{-m}) = g_n(x)$ so long as $x \not \in (n - 3^{-m}, n)$, meaning $g_n(x + 3^{-m}) \neq g_n(x)$ only in $(n-1, n)$. Thus for any $x$, there is at most one $n$ with $n \geq m$ and $g_n(x + 3^{-m}) \neq g_n(x)$, and this $n$ necessarily satisfies $|g_n(x + 3^{-m}) - g_n(x)| \leq \frac{1}{n} \leq \frac{1}{m}$. Thus we have
\begin{align*}
|f(x + 3^{-m}) - f(x)|
&= \left| \sum_{n=1}^\infty (g_n(x + 3^{-m}) - g_n(x)) \right| \\
&\leq \frac{1}{m} + \sum_{n=1}^{m-1} |g_n(x + 3^{-m}) - g_n(x)| \\
&\leq \frac{1}{m} + \sum_{n=1}^{m-1} \frac{2}{n} \cdot 3^{-(m - n)} \\
&\leq \frac{C}{m}
\end{align*}
for some sufficiently large $C$ not depending on $m$. Thus $|f_m - f| \leq C/m$, so $f_m \to f$ uniformly.

On the other hand, $f$ is not uniformly continuous. Consider $x$ of the form $k + \frac{1}{2}$ for $k$ a positive integer. For $n \leq k$ we have $g_n(x) = \frac{1}{n}$ since $3^n x = 1/2$ modulo $1$, so $f(k + \frac{1}{2}) = \sum_{n=1}^k \frac{1}{n}$. In particular, letting $k \to \infty$ we see $f(k + \frac{1}{2}) \to \infty$, while $f(k) = 0$ always, so $f$ is not uniformly continuous.


问:
Doubt about sets open in metric subspaces need not be open in the respective metric space.
In the example given the set [0,a),(a,1] is open on set S=[0,1] but closed on R. As there are no examples provided to illustrate and directly proceeded to other theorems, i am trying to decipher. Cant we choose radius 'r' finitely large such that we can find an open ball for which d(x,y)<r . or does it happen that the given set will have points which are not interior to the set. What exactly is happening on R which is not happening on S, that the set is not open on R. Hope my doubt is understandable. Thank You.

[Theorem][1]

[1]: https://i.stack.imgur.com/c928H.png**strong text**

答:
A topological space is a pair $(X, T_X)$ where $X$ is any set and $T_X$ is a collection (a set) of some or all of the subsets of $X,$ subject to:

$\phi$ (the empty set) and $X$ belong to $T_X.$

If $C$ is any set of members of $T_X$ then the common union $\cup C=\cup_{t\in C}\,t $ belongs to $T_X.$

If $D$ is any $finite$ set of members of $X$ then the common intersection $\cap D=\cap_{d\in D}\,d$ belongs to $T_X.$

$T_X$ is called a topology on $X.$ The members of $T_X$ are called the open sets, but this is meaningful only when talking about a particular $T_X.$ And the complements, in $X,$ of the members of $T_X$, are called the closed sets. Again, this makes sense only when talking about a particular $T_X.$ It is common to say "the space $X$", especially when it is understood from the context (that is, from other things that were already said), which $T_X$ is being considered.

When $(X,T_X)$ is a topological space and $S\subseteq X$ we can define a topological space $(S,T_S)$ where $T_S=\{t\cap S:t\in T_X\}.$ This $T_S$ is called the subspace topology on $S$ with respect to $T_X.$ It is common to say "the subspace $Y$".

If $T_{\Bbb R}$ is the "usual" or " standard" topology on $\Bbb R,$  then any $C\subset \Bbb R$ belongs to $T_{\Bbb R}$ iff (if & only if) for every $x\in C$ there exists $r_x>0$ such that $(-r_x+x,r_x+x)\subseteq C.$ With respect to this topology, if $S=[0,1]$ then $[0,a)\cup (a,1]$ belongs to the subspace topology on $S$ because the set $t=(-1,a)\cup (a,2)$ belongs to $T_{\Bbb R}$ and $[0,a)\cup (a,1]=t\cap S.$


问:
Functions: domain, codomain, range and graph
I have a question in my course.

Let $A = \{0, 1\}$, $B = \{4, 5, 6, 7\}$ and $C = \mathbb{N}$ (natural numbers). Consider the following functions:

$f:A\to B$ s.t. $f(x) = x+5$

$g:B\to C$ s.t. $g(x) = 2x−2$

$h:C \to A$ s.t. $h(x) = \begin{cases}
                                   0 & \text{if $x < 10$} \\
                                   1 & \text{if $x \ge 10$} \\
  \end{cases}$

Which of the following compositions are defined?
If they are defined, compute their domain, codomain, range and graph.

(a) $f \circ g$

(b) $g \circ f$

(c) $h \circ g$

(d) $g\circ h$

My answer:

•   Domain of $f$ is $A = \{0, 1\}$; domain of $g$ is $B = \{4,5,6,7\}$; domain of $h$ is $C = \mathbb{N}$;

•   Codomain of $f$ is $B = \{4,5,6,7\}$; codomain of $g$ is $C = \mathbb{N}$; codomain of $h$ is $A = \{0,1\}$.

•   Range of $f$ is $\{5, 6\}$; range of $g$ is $\{4,8,10,12\}$; range of $h$ is $\{0,1\}$.

a) $f \circ g = f(g(x)) = 2x + 5 -2 = 2x + 3$

b) $g \circ f = g(f(x)) = 2(x + 5) – 2 = 2x + 8$

c) $h \circ g = -2 \text{ if } x < 10;\ h \circ g = 0 \text{ if } x \ge 10 $

d) $g \circ h = g(h(x)) = -2\text{ if } x < 10; g \circ h = 0 \text{ if } x \ge 10$

Could I please get a review and how to calculate the graph?

答:
The composite function $f \circ g$ is only defined if the codomain of $g$ is contained in the domain of $f$.  Depending on the definition you are using, you may need the stronger condition that the codomain of $g$ equals the domain of $f$.

In this case, the codomain of $g$ is $\mathbb{N}$ and the domain of $A = \{0, 1\}$.  Moreover, as noted in the comments, since the range of $g$ is $\{6, 8, 10, 12\}$, the elements in the range of $g$ are not contained in the domain of $f$.  Thus, the composite function $f \circ g$ is undefined.

Since $f: A \to B$ and $g: B \to C$, the composite function $g \circ f$ is defined since the codomain of $f$ equals the domain of $g$.  As you correctly concluded, the function $g \circ f: A \to C$ is defined by
$$(g \circ f)(x) = g(f(x)) = g(x + 5) = 2(x + 5) - 2 = 2x + 8$$
The graph of the composite function $g \circ f$ is the set of all ordered pairs $(a, c) \in A \times C$ such that $(g \circ f)(a) = c$ for some $a \in A$.  Since $(g \circ f)(0) = 8$ and $(g \circ f)(1) = 10$, the graph of the function $g \circ f$ is the set
$$G_{g \circ f} = \{(0, 8), (1, 10)\}$$

Since $g: B \to C$ and $h: C \to A$, the composite function $h \circ g$ is defined since the codomain of $g$ equals the domain of $h$.  Since $h: C \to A$ is defined by
$$h(x) = \begin{cases}
         0 & \text{if $x < 10$}\\
         1 & \text{if $x \geq 10$}
         \end{cases}
$$
and $g: B \to C$ is defined by $g(x) = 2x - 2$, observe that if $x \in B$, then
$$(g \circ h)(x) = \begin{cases}
                    0 & \text{if $x < 6$}\\
                    1 & \text{if $x \geq 6$}
                   \end{cases}
$$
To see this, observe that if $x < 6$, $g(x) = 2x - 2 < 2 \cdot 6 - 2 = 10$, and that if $x \geq 6$, $g(x) = 2x - 2 \geq 2 \cdot 6 - 2 = 10$.  Since
\begin{align*}
(h \circ g)(4) & = h(g(4)) = h(6) = 0\\
(h \circ g)(5) & = h(g(5)) = h(8) = 0\\
(h \circ g)(6) & = h(g(6)) = h(10) = 1\\
(h \circ g)(7) & = h(g(7)) = h(12) = 1
\end{align*}
the graph of $h \circ g$ is
$$G_{h \circ g} = \{(4, 0), (5, 0), (6, 1), (7, 1)\}$$

Observe that $h: C \to A$ and $g: B \to C$, so the codomain of $h$ is not contained in the domain of $g$ since $A$ is not a subset of $B$.  Thus, the composite function $g \circ h$ is not defined.


问:
Prove $\{(x,y,z)\in \mathbb{R^3}:x^2+y^2+z^2=3\}$ is compact set
Prove $\{(x,y,z)\in \mathbb{R^3}:x^2+y^2+z^2=3\}$ is compact set

答: Not necessarily.  The unit sphere, for example, is the image of $[0,1]$ under a continuous mapping.  You can construct it using a space-filling curve.


问:
Prove that the existence of a step function implies the existence of a continuous function in L2
The title of this question isn't exactly a complete description, but couldn't come up with a better one-liner

Given the following two theorems, I am trying to prove that Theorem (1) implies Theorem (2):

Theorem (1): Let $f \in L^2(-\pi, \pi).$ Given $\epsilon > 0$, there exists a step function $\phi$ on $[-\pi,\pi]$ such that $||f(x)- \phi(x)||_2 < \epsilon$

Theorem (2): Let $f \in L^2(-\pi, \pi).$ Given $\epsilon > 0$, there exists a continuous function $\psi$ on $[-\pi,\pi]$ such that $||f(x)- \psi(x)||_2 < \epsilon$

答:
Given an interval $I_k=[a_k, b_k) \subset [-\pi, \pi]$  $(a_k<b_k)$, we define $k(x):=d(x,I_k)=\inf\{d(x,y):y \in I_k \}$. For each $n \in \mathbb{N}$ we define $$g_{k,n}(x):=\frac{1}{1+nk(x)} .$$ Notice that for each $n \in \mathbb{N}$ we have $g_{k,n}$ is a continuous function, and $g_{k,n}(x)=1$ whenever $x \in \bar{I_k}=[a_k, b_k]$. If $x \notin \bar{I_k}$, then $k(x)>0$ and so $\lim_{n \to \infty} g_{k,n}(x)=0$. In other words, $g_{k,n} \to \chi_{\bar{I_k}}$ pointwise. Since $(\chi_{I_k}-g_{k,n} )^2 \leq \chi_{I_k}^2$ (the characteristic function of $I_k$), we have that $\lim_{n \to \infty} \| \chi_{I_k}-g_{k,n}\|_2=0$.

Now notice that the step function $\phi$ is a finite linear combination of such characteristic functions. Moreover, a finite linear combination of continuous functions is a continuous function.


问:
Statistics Workshop for High School Students
We are going to hold an introductory workshop about the statistics. The participants will be students who have just finished their 8th or 9th grade. The workshop consists of 10 two-hour sessions. The students have a very limited background about the statistics: They can compute arithmetic average of a set of data, create pie charts, etc.

So we’re looking for some good educational material to introduce the students to some of the main statistical ideas, methods and concepts as well as to illustrate the importance of statistics in daily life.

Your suggestions would be of great help.

Thanks.

答:
I posted this first in a comment, but it is really worth status as an answer.:

A middle school teacher gifted me an old book: "How to Lie with Statistics" by Darrell Huff:

"Darrell Huff runs the gamut of every popularly used type of statistic, probes such things as the sample study, the tabulation method, the interview technique, or the way results are derived from the figures, and points up the countless number of dodges which are used to fool rather than to inform."

I was captivated while reading it, and remember feeling so empowered with the knowledge I gained after having read it! It is very highly relevant to life in the real world, as much today as decades ago, perhaps more so, given the information saturation of the current day and age.

When googling for a link to a preview of the book, I see that there are many powerpoints and pdf's available which are similarly titled: addressing the ways in which statistics can be used, and abused, both of which budding teen-agers will find engaging.


问:
Triangle inequality for split-complex numbers
After much research and head-scratching to no avail, I was hoping someone here could shed some light on certain properties of the split-complex numbers, namely:

1) Is there any meaningful analog of the triangle inequality for the split-complex numbers using their appropriate definition of modulus?

2) If not, what would be a suitable means by which to check whether (or not) an infinite series of split-complex numbers has a bounded modulus?

The issue I'm having is with the fact that the quadradic form which defines the modulus on the split-complex numbers is not positive-definite. Is it the case that every algebra with a positive-indefinite modulus has no meaningful analog of the triangle inequality?

Any insight into how one might approach this would be greatly appreciated. Reccomendations for reading material would also be more than welcome.

Thanks!

答:
In case of split-complex numbers it is easier to get the reverse triangle inequality than the usual one!

Let us restrict our attention to split-complex numbers $z = a+bj$ with $a^2 \ge b^2$. For them we can define the modulus to be $|z| = \sqrt{a^2-b^2} \in \mathbb{R}^+$. The set of such split-complex numbers form the following blue region on the split-complex plane:

Note that this region is non-convex. So even if both $z_1$ and $z_2$ belong to the blue region, their sum may fall out of this region, and then the modulus $|z_1+z_2|$ will be ill-defined. To fix this issue, we further restrict our attention to one of its convex subregions: either with $a \le 0$ or $a \ge 0$.

Working within one of these subregions (red or green), one can see that for any $z_1 = a_1 + b_1 j$ and $z_2 = a_2 + b_2 j$ the following inequalities hold:

$$
a_1 a_2 \ge b_1 b_2, \quad \quad (a_1+a_2)^2 \ge (b_1+b_2)^2. \tag{*}
$$

Now we start with the obviously true inequality and go step by step:

$$
0 \le (a_1b_2 - a_2 b_1)^2,
$$
$$
-(a_1b_2)^2-(a_2b_1)^2 \le -2a_1a_2b_1b_2,
$$
$$
\color{gray}{ (a_1a_2)^2+(b_1b_2)^2}-(a_1b_2)^2-(a_2b_1)^2 \le \color{gray}{ (a_1a_2)^2+(b_1b_2)^2}-2a_1a_2b_1b_2,
$$
$$
(a_1^2-b_1^2)(a_2^2-b_2^2) \le (a_1a_2-b_1b_2)^2
$$
$$
\Downarrow \text{ using (*)}
$$
$$
\sqrt{(a_1^2-b_1^2)(a_2^2-b_2^2)} \le (a_1a_2-b_1b_2),
$$
$$
2\sqrt{(a_1^2-b_1^2)(a_2^2-b_2^2)} \le 2 a_1a_2-2b_1b_2,
$$
$$
\color{gray}{a_1^2-b_1^2+a_2^2-b_2^2}+  2\sqrt{(a_1^2-b_1^2)(a_2^2-b_2^2)} \le \color{gray}{a_1^2-b_1^2+a_2^2-b_2^2} + 2 a_1a_2-2b_1b_2,
$$
$$
\left(\sqrt{a_1^2-b_1^2}+\sqrt{a_2^2-b_2^2}\right)^2 \le (a_1+a_2)^2-(b_1+b_2)^2
$$
$$
\Downarrow \text{ using (*)}
$$
$$
\sqrt{a_1^2-b_1^2}+\sqrt{a_2^2-b_2^2} \le \sqrt{(a_1+a_2)^2-(b_1+b_2)^2}.
$$
So we get the reverse triangle inequality:
$$
|z_1|+|z_2| \le |z_1+z_2|.
$$

It is easy to show by induction that if all $\{z_n\}$ belong to the same convex region (red or green), then the reverse triangle inequality still holds:

$$
\sum|z_n| \le \left| \sum z_n \right|.
$$

In contrast to that, the usual triangle inequality $|z_1| + |z_2| > |z_1 + z_2|$ may hold only if $z_1$ and $z_2$ belong to different regions (e.g., $z_1 \in$ red, $z_2 \in$ green). Note that such an arrangement cannot be continued by induction, so it seems that no simple conditions for $\sum|z_n| > \left| \sum z_n \right|$ can be found. I believe that (in case of split-complex numbers with $|z| \in \mathbb{R}^+$) it is more natural to work with the reverse triangle inequality than with the usual one.


问:
Self-Adjoint of Volterra Operator
I have the Volterra operator defined as following:

$$
A : L_{2}([0,1]) \to L_{2}([0,1])
$$

$$
A f(x) = \int_{0}^{x} f(t) dt
$$

I was asked to show that:

$$
A^{*}A f(x) = \int_{0}^{1} (1- \max(x,t)) f(t) dt
$$

This is my attempt:

Using the Fubini's theorem and changes of variables:
I can show that:

$$
<f, Ag> = \int_{0}^{1} f(x) \int_{0}^{x} g(y) dy dx = \int_{0}^{1} g(y) \int_{y}^{1} f(x) dx dy =
$$

So, I get the adjoint operator $ A^{*} f(x) = \int_{x}^{1} f(t) dt$. But I do not see how to get the term $\max(x,t)$. From this, I only see that:

$$
A^{*}A f(x) = \int_{x}^{1} \int_{0}^{y} f(z) dz dy
$$

How can I show that $A^{*}A f(x)$ as in the question that asks me to do?

答:
$ \langle A^{*}f,g \rangle= \langle f, Ag \rangle=\int_0^{1} g(y)\int_y^{1} f(x)dxdy$. We can write this as $ \langle A^{*}f,g \rangle= \langle h, g \rangle$ where $h(y)=\int_y^{1} f(x)dx$. Since this is true for all $g$ it follows that $A^{*}f=h$.

Now $A^{*}Af(x)=\int_x^{1} Af(t)dt=\int_x^{1} \int_0^{t} f(s)ds dt$. Switch the order of integration to finish.

Title: Self-Adjoint of Volterra Operator


问:
How to prove that two sequence formulas are the same
I derived the formula for the nth term of a sequence to be $$\frac{1-\displaystyle\sum_{i=1}^{n-1} a_i}{2},$$ but upon inspection this appeared to be the same as $$\left(\frac{1}{2}\right)^n.$$ I was wondering how to prove that these are the same more rigorously?

答:
Hint: You can use induction. For the inductive step, you'll need the formula
$$x+x^2+\dots+x^n=x(1+x+\dots+x^{n-1})=x\,\frac{1-x^n}{1-x}$$
with $x=\frac12$.


问:
Find the value of $A/B$ for this definite integration.
$$\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x$$

Given that the integral above is equal to $\frac{3\pi^A}{B}$ for rational numbers $A$ and $B$, find the value of $A/B$.

I thought of taking $x^{2/3}$ as t but that didn't help. Any other suggestion?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{-\root{3}
x^{2/3}}\sin\pars{x^{2/3}}\,\dd x}
\,\,\,\stackrel{x^{2/3}\,\,\, \mapsto\ x}{=}\,\,\,
{3 \over 2}\int_{0}^{\infty}x^{1/2}\,\expo{-\root{3}
x}\sin\pars{x}\,\dd x
\\[5mm] = &\
{3 \over 2}\,\Im\int_{0}^{\infty}x^{1/2}\,
\expo{-\pars{\root{3} - \ic}x}\,\,\dd x
\\[5mm] = &\
{3 \over 2}\,\Im\int_{0}^{\infty}x^{\color{red}{3/2} - 1}\,
\bracks{\sum_{k = 0}^{\infty}\color{red}{\pars{\root{3} - \ic}^{k}}\,
{\pars{-x}^{k} \over k!}}\,\,\dd x
\end{align}
With Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{-\root{3}
x^{2/3}}\sin\pars{x^{2/3}}\,\dd x} =
{3 \over 2}\,\Im\bracks{\Gamma\pars{3 \over 2}\pars{\root{3} - \ic}^{-3/2}}
\\[5mm] = &\
{3\root{\pi} \over 4}\,\Im\bracks{2\expo{-\ic\pi/6}}^{-3/2} =
{3\root{\pi} \over 4}\,2^{-3/2}\,\sin\pars{\pi \over 4}
\\[5mm] = &\
\bbx{3\root{\pi} \over 16} \approx 0.3323 \\ &
\end{align}


问:
Inter-arrival time distribution of a Poisson process
I'm actually looking at a rigorous way to prove that if $\{N_t\}_{t\geq 0}$is a Poisson process of rate $\lambda >0$, then the inter-arrival times $(H_n)_{n\in \mathbb{N}}$ are independently and identically distributed exponential random variable with rate parameter $\lambda$.

Note that for this I use the two definitions of Poisson process that does not use the above statement as a definition that is (the equivalence between the two definition is already proved):

Def 1:
A Poisson process $\{N_t\}_{t\geq 0}$ of rate $\lambda >0$ is a stochastic process with values in $\mathbb{N}$ satisfying:

$N_0=0$ a.s

The increments are independent, that is given $n\in \mathbb{N}$ and $0\leq t_0<\dots<t_n$ $N_{t_0}, N_{t_1}-N_{t_0},\dots,N_{t_{n-1}}-N_{t_n}$ are independent.

$P(N_t-N_s=k)=P(N_{t-s}=k), t\geq s$

$P(N_{t+\delta}-N_t=1)=\lambda\delta+o(\delta)$

$P(N_{t+\delta}-N_t \geq 2)=o(\delta)$

Def 2:
A Poisson process $\{N_t\}_{t\geq 0}$ of rate $\lambda >0$ is a stochastic process with values in $\mathbb{N}$ satisfying:

$N_0=0$ a.s

The increments are independent, that is given $n\in \mathbb{N}$ and $0\leq t_0<\dots<t_n$ $N_{t_0}, N_{t_1}-N_{t_0},\dots,N_{t_{n-1}}-N_{t_n}$ are independent.

$P(N_t-N_s=k)=P(N_{t-s}=k), t\geq s$

$N_t \sim Poi(\lambda t)$

What I want to avoid is what seems to be the classic proof that consider object like $P(H_2>t|H_1=t_1)$ which has no sense for me as $H_1$ is continuous ( namely exponential) and $H_2$ could be (at this step) any type of random variable. (and hence $P(H_1=t)=0 \forall t$)

Thank's by advance to the one that will propose me a good proof for this.

答:
We will assume that $N$ has right-continuous sample paths. Define the firth arrival time $T_1$ by

$$ T_1 = \inf \{ t \geq 0 : N_t \geq 1 \}. $$

By the right-continuity, this implies that $N_{T_1} = 1$. Then we will prove the following claim, which is a special case of the strong Markov property:

Claim. Define a new stochastic process $\tilde{N} = ( \tilde{N}_t )_{t\geq 0}$ by
$$ \tilde{N}_t = N_{T_1 + t} - N_{T_1} = N_{T_1 + t} - 1. $$
Then we have the following:

$T_1$ and $\tilde{N}$ are independent.

$\tilde{N}$ is a Poisson process of rate $\lambda$.

$T_1$ is an exponential random variable of rate $\lambda$.

Proof. Fix $0 < t_0 < t_1 < \cdots < t_n$. Also, let $\delta > 0$ be sufficiently small so that $0 < t_0 - \delta$ and $t_{i-1}+\delta < t_i-\delta$ holds for all $i = 1, \dots, n$.

Now, for any parameters $s, s_1, \dots, s_n \geq 0$, we consider

$$ Y = sT_1 + \sum_{i=1}^{n} s_i (\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}}). $$

Then, conditioned on $\{T_1 \in ((k-1)\delta, k\delta] \}$, we have

$$ Y \geq s (k-1)\delta + \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}}) $$

and so,

$$ \mathbb{E}[e^{-Y}]
\leq \sum_{k=1}^{\infty} \mathbb{E}\left[ \exp\left\{ - s (k-1)\delta - \sum_{i=1}^{n} s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}}) \right\} \mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta]  \}} \right]. \tag{1} $$

By noting that $\{T_1 \in ((k-1)\delta, k\delta] \} = \{ N_{(k-1)\delta} = 0 \text{ and } N_{k\delta} \geq 1 \} $, all of

$$ \mathbf{1}_{\{T_1 \in ((k-1)\delta, k\delta]  \}}, \quad N_{(k-1)\delta + t_1} - N_{k\delta + t_{0}}, \quad \dots, \quad N_{(k-1)\delta + t_n} - N_{k\delta + t_{n-1}} $$

are independent, and hence the bound $\text{(1)}$ reduces to

\begin{align*}
\mathbb{E}[e^{-Y}]
&\leq \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} \mathbb{E}\left[ e^{-s_i (N_{(k-1)\delta + t_i} - N_{k\delta + t_{i-1}})} \right] \right) \mathbb{P} \left( T_1 \in ((k-1)\delta, k\delta] \right) \\
&= \sum_{k=1}^{\infty} e^{-s(k-1)\delta} \left( \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) } \right) e^{-\lambda (k-1)\delta} (1 - e^{-\lambda \delta}) \\
&= \frac{1 - e^{-\lambda \delta}}{1 - e^{-(s+\lambda)\delta}} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1} - \delta) (1 - e^{-s_i}) }. \tag{2}
\end{align*}

As $\delta \to 0^+$, the bound $\text{(2)}$ converges to

$$ \mathbb{E}[e^{-Y}] \leq \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) }. \tag{3} $$

By a similar computation applied to

$$ Y \leq s k\delta + \sum_{i=1}^{n} s_i (N_{k\delta + t_i} - N_{(k-1)\delta + t_{i-1}}), $$

we can also obtain the reverse direction of $\text{(3)}$. So it follows that

$$ \mathbb{E}[e^{-Y}] = \frac{\lambda}{s+\lambda} \prod_{i=1}^{n} e^{-\lambda(t_i - t_{i-1}) (1 - e^{-s_i}) } \tag{4} $$

holds for any choices of parameters $s, s_1, \dots, s_n \geq 0$. However, since the multidimensional Laplace transform uniquely determines the law of a given random vector, this proves that:

$ T_1$, $\tilde{N}_{t_1} - \tilde{N}_{t_{0}}$, $\ldots$, $\tilde{N}_{t_n} - \tilde{N}_{t_{n-1}} $ are mutually independent,

$\tilde{N}_{t_i} - \tilde{N}_{t_{i-1}} \sim \operatorname{Poisson}(\lambda(t_i - t_{i-1}))$ for each $i = 1, \dots, n$, and

$T_1 \sim \operatorname{Exp}(\lambda)$.

Therefore the desired claim follows. $\square$

Now repeatedly applying this claim shows that inter-arrival times are independent exponential random variables of rate $\lambda$ as desired.


问:
Kleiner' s Lemma and Slice Theorem
Let $μ : G×M→M$ be an isometric action and let $γ : [0,1]→M$ be a
geodesic segment between G(γ(0)) and G(γ(1)) that realizes the distance between
these orbits. There exists a subgroup H of G such that $G_{γ(t)} = H$ for $t \in (0,1)$ and
$H$ is a subgroup of $G_{γ (0)}$ and $G_{γ (1)}$.

Suggestion of prove:

If $μ : G×M→M$ is an isometric action, então $\mu^g: M \longrightarrow M $ is a isometry, where $\mu^g(x)$ is the action of $g$ in $x$,i.e, $\mu^g(x)=\mu(g,x)$

Let

$H :=\{ g ∈ G : μ(g,γ (t)) = γ (t)), \space \forall  t \in [0,1]\}$. Suppose there exists
some $t_0 ∈ (0,1)$ and $g \in G_{γ(t_0)}$ such that $g \notin H$. In particular,
$$d(μ^g)_{γ (t_0)}(γ'(t_0))\neq γ'(t_0)$$.

Define a piecewise smooth path

$γ :[0,1]→M $ by setting $γ|_{[0,t_0]} := γ[0,t_0]$ and $γ|_{[t_0,1]} := μ(g,γ[t_0,1])$. Then

$\gamma$ joins $G(γ(0))$ to $G(γ(1))$ and has the same length as $γ$, contradicting the fact that minimizing geodesic segments are smooth.

my question:
I) how to justify the part of, in particular $$d(μ^g)_{γ (t_0)}(γ'(t_0))\neq γ'(t_0),$$ using Slice's Theorem?

II) And how to justify the application of the first variation formula in the final part.

答:
To answer I, I think you have a typo.  Namely, for $g\in G_{gamma(t_0)}$, I think $g\in H$ iff $d(\mu^g)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$.  This is a consequence of the fact that geodesics are determined by a single point and single tangent vector.

Thus, $\mu^g(\exp(t \gamma'(t_0))) = \exp(t (d(g^\mu)_{\gamma(t_0)} \gamma'(t_0)))$ since they agree to first order at $t=0$.  Moreover, $d(g^\mu)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$ iff $\exp(d(g^\mu)_{\gamma(t_0)} \gamma'(t_0)) = \exp( t\gamma'(t_0))$ for all $t$ for which either side is defined.   Putting these together, we see that $d(g^\mu)_{\gamma(t_0)} \gamma'(t_0) = \gamma'(t_0)$ iff $g\in H$.

To answer II, note that we just need to show that minimizing curves must be smooth.  This follows from the existence of totally normal neighborhoods (that is, an open set for which any two points are connected by a unique length minimizing geodesic).

Specifically, suppose $\alpha$ has a corner at $t=0$.  Pick a totally normal neighborhood $U$ of $\alpha(0)$.  Pick $\epsilon$ small enough so that $\alpha(\pm \epsilon)\in U$.  Because we are in a totally normal neighborhood, there is a unique minimizing geodesic $\beta$ connecting $\alpha(-\epsilon)$ to $\alpha(\epsilon)$.  Geodesics are $C^\infty$, so $\beta$ is $C^\infty$.  On the other hand $\alpha$ is not $C^\infty$ so $\alpha\neq \beta$.  Since $\beta$ is the unique length minimizer between $\alpha(-\epsilon)$ and $\alpha(\epsilon)$, it follows that $\alpha$ does not minimize distance.

Title: Kleiner' s Lemma and Slice Theorem


问:
Intersection of two $L^p$ spaces is complete.
Given $1 \leq p < q < \infty$, I wish to show that $L^p \cap L^q$ is a Banach space given the norm $\|f\| = \|f\|_p + \|f\|_q$. I have shown that this space is a vector space and that the norm is indeed a norm, all I have left is showing that this space is complete and am have some trouble. I was able to deduce that if a sequence is Cauchy in $L^p \cap L^q$ then it must be Cauchy in $L^p$ and $L^q$, but I am not sure where to go from here. Any advice would be appreciated.

答: If $(f_n)$ is Cauchy it converges in $L^{p}$ to some $g$ and in $L^{q}$ to some $h$. There is a subsequence $(f_{n_i})$ which converges to $g$ almost everywhere and there is a further subsequence which converges to $h$ almost everywhere. It follows that $g=h$ almost everywhere. It is now obvious that $f_n \to g$ in the given norm.


问:
Differentiate $\log_{10}x$
My attempt:

$\eqalign{
  & \log_{10}x = {{\ln x} \over {\ln 10}}  \cr
  & u = \ln x  \cr
  & v = \ln 10  \cr
  & {{du} \over {dx}} = {1 \over x}  \cr
  & {{dv} \over {dx}} = 0  \cr
  & {v^2} = {(\ln10)^2}  \cr
  & {{dy} \over {dx}} = {{\left( {{{\ln 10} \over x}} \right)} \over {2\ln 10}} = {{\ln10} \over x} \times {1 \over {2\ln 10}} = {1 \over {2x}} \cr} $

The right answer is: ${{dy} \over {dx}} = {1 \over {x\ln 10}}$ , where did I go wrong?

Thanks!

答:
$${\rm{lo}}{{\rm{g}}_{10}}x = {{\ln x} \over {\ln 10}} = \dfrac{1}{\ln(10)}\ln x$$

No need for the chain rule, in fact, that would lead you to your mistakes, since $\dfrac 1 {\ln(10)}$ is a constant.

So we differentiate only the term that's a function of $x$: $$\dfrac{1}{\ln(10)}\frac d{dx}(\ln x)= \dfrac 1{x\ln(10)}$$

Title: Differentiate $\log_{10}x$


问:
Distribution of 5 points on a plane where the lines connecting the points are either parallel and/or perpendicular to another
The problem is a basic geometry textbook problem (undergraduate level) but has obsessed me for weeks. We are in R^2, with 5 points on the plane. You must find the relative locations of these points so that you achieve that every line connecting two points is either parallel and/or perpendicular to another line. An additional condition is that the points cannot be co-linear and at least one perpendicular relationship exists.
I put an example. You draw a square. Let's say we take coordinates so that A is located at (0,0), B at (1,0), point C at (1,1) and finally point D at (0,1). You add a final fifth point E at (0.5, 1.5). Of all lines connecting the points (10), only 8 of them meet the conditions. Segments AE and BE are not parallel not perpendicular to any other segment (or "line"), but we have 2 sets of parallel lines and BD perpendicular to AC and in the other hand AC and DE are parallel each other and BD is parallel to CE, so only 8 lines meeting the criteria. We must find a distribution which has ALL possible lines parallel and perpendicular (so the maximum will be 10 lines meeting criteria) and demonstrate a) if such a distribution exists,  b) and if so, if it is unique or there are a bunch (maybe infinite) of them, and list the properties that they must exhibit.

答:
Hint: Angle subtended by a diameter is $90^\circ$.

Next hint:

 And again, angle subtended by a diameter is $90^\circ$.

One family of infinitely many non-isomorphic configurations:

 Let $AB, CD$ be perpendicular diameters of a circle (IE a square). (which is why the hint appears twice.)

 Where can we place $E$ in accordance with the hint?

I cannot yet prove that this configuration is unique.

If we don't require that a perpendicular exists, then the regular pentagon is another example.


问:
Probability Joint PDF and Joint CDF
Let R be the bounded region between y = x and y = x². A point (X, Y) is chosen uniformly at random from the region R. Please find the joint PDF and the joint CDF of X and Y.

I believed that the joint PDF formula is 1 / area(R) if (X, Y) is inside the region R. However, since area(R) is 1/6, then the PDF becomes 6(?) It confuses me.

Can someone help me please...

答:
You are correct.   The function (including its support) is: $$f_{\small X,Y}(x,y)= 6\cdot\mathbf 1_{0\leqslant x^2\leqslant y\leqslant x\leqslant 1}$$

There is no issue with the magnitude.   Since it is a probability density, rather than mass, function it may be greater than one.   You just need to ensure the total probability mass equals one.

$$\int_0^1\int^x_{x^2} 6 ~\mathrm d y~\mathrm d x= 6{\bigl[\tfrac 12x^2-\tfrac 13x^3\bigr]}_{x=0}^{x=1}=1$$

Title: Probability Joint PDF and Joint CDF


问:
Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$
I need help with this exercise.

Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$

a) Let $g(x)=f(x)e^{-x}$. Find $g'(x)$.

So this is what I did

$$g'(x)=f'(x)\cdot e^{-x}+e^{-x}\cdot -1 \cdot f(x)$$

$$g'(x)=f'(x)\cdot e^{-x}-e^{-x} \cdot f(x)$$

$$g'(x)=e^{-x}[f'(x)-f(x)]$$

$$g'(x)=e^{-x}[f(x)+1-f(x)]$$

$$g'(x)=e^{-x}[1]$$

$$g'(x)=e^{-x}$$

b) Find all the differentiable functions $f$ satisfying $f'(x)=f(x)+1$

I'm stuck with this one. Please help!

答:
The second part just needs you to solve the differential equation. If you let $f(x) =y$, you can write $\frac {dy} {dx} = y+1$.

You can separate and integrate to get: $\int \frac 1{y+1}dy = \int 1 dx$.

You then get: $\ln|y+1| = x + c$, where $c$ is an arbitrary constant.

Taking the exponent of both sides:

$|y+1| = e^xe^c = Ae^x$ (we're just assigning another constant $A>0$ such that $A = e^c$).

Now, since the right hand side is always positive, we can remove the absolute value sign and rearrange to get $y = \pm(Ae^x) - 1$, which you can write as: $f(x) =Be^x - 1, \ B \ne 0$.

You now note that when $B =0$, you get $f(x) = - 1$ which trivially satisfies the differential equation.

Hence the final solution is $f(x) = Be^x - 1$, without restriction on $B$. If you're working in the reals, $B$ can be any real number.


问:
How is $ B_n = 1- \sum_{k=0}^{n-1} \binom{n}{k} \frac{B_k}{n-k+1} $ Where $B_n$ are the Bernoulli Numbers with $B_1 = \frac{1}{2}$
So I was browing Wikipedia just looking at Bernoulli Number identities and I stumbled across this
$$ B_m^+ = 1- \sum_{k=0}^{m-1} \binom{m}{k} \frac{B_k^+}{m-k+1} $$
The Wikipedia page said that this identity was derived from the identity :
$$
\sum_{k=0}^{m}\left(\begin{array}{c}
m+1 \\
k
\end{array}\right) B_{k}^{+}=m+1
$$
My question is, how are these identities derived? Wikipedia gives the reference for these theorems to be from the page on Bernoulli Numbers on WolframMathWorld but I can't find the proof of any of these identities there. Can anyone guide me to or state a proof of either identity? (From the definition of Bernoulli Numbers stated below)
$$
\frac{t}{1-e^{-t}}=\frac{t}{2}\left(\operatorname{coth} \frac{t}{2}+1\right) \quad=\sum_{m=0}^{\infty} \frac{B_{m}^{+} t^{m}}{m !}
$$

答:
The definition of $B_k$ is $$\frac{z}{e^z-1}=\sum_{k=0}^\infty \frac{B_k}{k!} z^k$$
it gives
$$\sum_{m=0}^\infty \frac{z^{m+1}}{m!}=(e^z-1)\frac{-z}{e^{-z}-1}=\sum_{m=0}^\infty z^{m+1} \sum_{k=0}^m\frac{(-1)^k B_k}{k! (m-k+1)!} $$

Conclude by equating the coefficients and using $B_k^+ = (-1)^k B_k$.


答:
Proof without words:

Image source: Wikipedia's page about geometric sequence

Title: How to prove that two sequence formulas are the same

Body: I derived the formula for the nth term of a sequence to be $$\frac{1-\displaystyle\sum_{i=1}^{n-1} a_i}{2},$$ but upon inspection this appeared to be the same as $$\left(\frac{1}{2}\right)^n.$$ I was wondering how to prove that these are the same more rigorously?


问:
Is there a theorem in real analysis that relates the interval of continuity of a function to the interval of continuity of its derivative?
Is there a theorem in real analysis that relates the interval of continuity of a function to the interval of continuity of its derivative? So for example, in complex analysis, we are given a theorem that if a complex function is analytic in a domain, it is infinitely differentiable and analytic in the same domain. Does real analysis have a similar theorem or lemma?</p>

答:
No, there is not. For instance, the Weierstrass function is a concrete example of a continuous everywhere but differentiable nowhere. By integrating it however many times you like, you get a function which is $n$-fold differentiable everywhere but nowhere $(n+1)$-fold differentiable.

Note that there's some mismatch between the question asked and the result from complex analysis cited. We can whip up a complex version of the Weierstrass function to demonstrate that in complex analysis there is also no way to deduce differentiability from continuity. However, what we can't do is the more "iterative" construction above: holomorphicity/analyticity in complex analysis is a vastly more powerful property than differentiability in real analysis.


问:
How is the graph $K_m^n$ defined?
I'm reading Diestel's Graph Theory but I can't find a definition for $K_m^n$. How should I draw graphs like these?

答: On Pg $17$, Section $1.6$, the definition of $K^r_s$ is given. I found this by looking at the symbol index in the back. I'm using the fifth edition. Lots of well written books should have some version of a symbol index.

Title: How is the graph $K_m^n$ defined?


问:
Laurent's series for the function $\frac{1}{z^2+z}$ on the annular region $1<|z|<R$
I am studying alone complex variables cause I'm not having class this year due to this pandemic. Now I started to study Laurent's series and I would love to know the answer of this question just to take a start point to move on with my exercises. If somebody could help me I'll be extremely grateful.</p>

Body: This can be done for any $R\gt1$, by writing $1/(z^2+z)=1/z\cdot1/(z+1)=1/z-1/(z+1)=1/z-1/z\cdot1/(1-(-1/z))=1/z-1/z\cdot\sum_{n\ge0}(-1/z)^n$.


问:
Find all possible values of the determinant of matrix $A$
I am given that $A$ is an $m$ by $m$ matrix and that $A$ and $A^{-1}$ are both filled with integer values. How can I find all possible values of the determinant of $A$?

I tried solving this by stating $\det(AA^{-1})=1$. Then $\det(A)\det(A^{-1})=1$ and $\det(A)= \det(A^{-1})^{-1}$. However, when simplified, this just gives me $\det(A)= \det(A)$.

Any guidance is much appreciated!

答: From the definition of determinant (either by the Leibniz formula or Laplace expansion), it's easy to see that $\det(A^{-1})$ is an integer.  From your question, its multiplicative inverse $\det(A^{-1})^{-1}$ also has to be an integer, so only $1$ and $-1$ can satisfy this.


问:
Heron's sequence
Given: $x_{n+1}$=$\frac{1}{2}\Bigl(x_{n}+ \dfrac{c}{x_{n}}\Bigr)$

I'm trying to prove the following:
$\;\Bigl[\dfrac{c}{x_{n+1}},x_{n+1}\Bigr]\subset\Bigl[\dfrac{c}{x_{n}},x_{n}\Bigr]$.

I have no idea how to prove it. I would be very grateful if someone could give me a hint and help me further.

答:
This is false if $c>0$ and $\sqrt c >x_n>0.$ E.g. if $c=4$ and $x_n=1.$

If $c>0$ and $x_n>\sqrt c,$ observe that if $x_n>x_{n+1}> \sqrt c \,$ then $c/x_n<c/x_{n+1}<\sqrt c<x_{n+1}<x_n.$

So prove that if $x^2-c>0 $ then $0<((x+c/x)/2)^2-c <x^2-c.$

Title: Heron's sequence


问:
Express addition of natural numbers ($+$) in structure $(N;=;S;*)$ where $S(n)$ is n + 1
Express addition of natural numbers ($+$) in structure $(N;=;S;\cdot)$ where $S(n)$ $\rightarrow$ $n + 1$

I need to exress operationd of addition in this structure. I tried to: $c = a + b = a * S(0) + b * S(0) $ but stucked. Also tried $a = S(a-1) = S(S(a-2))$, $S(a-1) * S(b-1) = ab $

Please help me a have to send the solution of this problem today.

答:
This is my interpretation of your question, as Alex stated in his comments; let me know if you mean something different.

Let $\mathcal{L}=\{S,\times\}$, and consider $\mathbb{N}$ as an $\mathcal{L}$-structure. Show that addition is definable in $\mathbb{N}$ – ie, find an $\mathcal{L}$-formula $\varphi(v_1,v_2,w)$ such that $\mathbb{N}\models\varphi(a_1,a_2,b)$ iff $b=a_1+a_2$, for any $a_1,a_2,b\in\mathbb{N}$.

To this end, first define $\psi(v_1,v_2,w)$ to be the following $\mathcal{L}$-formula: $$S(v_1\times w)S(v_2\times w)=S(w\times w\times S(v_1\times v_2)).$$ Expanding the left hand side of $\psi$ out in usual arithmetic notation gives $$(v_1w+1)(v_2w+1)=w^2v_1v_2+w(v_1+v_2)+1,$$ and doing the same for the right hand side gives $$w^2(v_1v_2+1)+1=w^2v_1v_2+w^2+1.$$ Thus we will have $\mathbb{N}\models\psi(a_1,a_2,b)$ if and only if $b^2=b(a_1+a_2)$. If $b$ is non-zero this reduces to $b=a_1+a_2$, as desired, but if $b=0$ it will hold for any $a_1,a_2\in\mathbb{N}$, so we have to handle this case differently.

Note for this that $0$ is definable, eg by the formula $\theta(u):=\neg\exists v(u=S(v))$. So, putting this together with $\psi$, we can define our desired $\varphi(v_1,v_2,w)$ by $$(\theta(v_1)\wedge\theta(v_2)\wedge\theta(w))\vee(\neg\theta(w)\wedge\psi(v_1,v_2,w)).$$ This formula is due to Julia Robinson, whose other work on definability you may be interested in if you enjoyed this problem. See here for instance for a nice reference, which gives an exposition of her remarkable result that $\mathbb{Z}$ is definable in $(\mathbb{Q},+,\times,0)$.


问:
Is it true that any continuous injective map from a circle to circle is surjective?
Here is my idea:

Suppose $f : \mathbb S^1 \to \mathbb S^1$ is continuous and injective.

If $f$ is not surjective, its image is homeomorphic to a subset of $\mathbb S^1 \setminus \{ c\} \cong \mathbb R$ (by considering the projection).

Then, the restriction of $f$'s codomain $g : \mathbb S^1 \to f(\mathbb S^1)$ is injective and surjective, so it is bijective.

Also, since $\mathbb S^1$ is compact and connected, its image under the continuous map is also compact and connected, so $f(\mathbb S^1)$ is homeomorphic to some closed interval $[a, b]$, which is simply connected.

Noting $f(\mathbb S^1)$ is Hausdorff and $\mathbb S^1$ is compact, by the closed map lemma, it follows $g$ is a homeomorphism. This contradicts that $\mathbb S^1$ is not simply connected, since $[a, b]$ is simply connected. Thus, $f$ must be surjective.

Does it work?

答: Your argument is completely correct, you should maybe add an argument why any compact connected subset of the reals is a closed interval (which is not difficult). Btw I don't see why the other proof presented here as an answer is a "little bit shorter", it is exactly the same argument and just explains less</p>


问:
Frobenius on affine space is a bijection
Similar questions have been asked in various different settings, but I am not satisfied with the array of answers which have been received. If something truly is a duplicate on the nose, I will be happy to be referred to this question.

Let $q=p^s, A=\mathbb{F}_q [x_1,\dots ,x_n],$ and  $\phi: A\to A$ be the map which takes $x_i\mapsto x_{i}^p$. This is a ring map and induces a map of schemes $f: \text{Spec } A\to \text{Spec } A$ given by taking preimage of $\phi$.

I am told that $\phi$ is a bijection. How is this clear? I don't see why injectivity or surjectivity is obvious.

Certainly this map is clearly not an isomorphism; I have no issue regarding this fact. But the bijectivity seems to require something which I do not see.

答:
Notice that $\phi^s$ is just the Frobenius map $x\mapsto x^q$. Thus we have

$$f^s(\mathfrak p)=(\phi^s)^{-1}(\mathfrak p)=\{x\in A\mid \phi^s(x)\in\mathfrak p\}=\{x\in A\mid x^q\in\mathfrak p\}=\mathfrak p.$$

where on the last equality I've used the fact that $\mathfrak p$ is prime. So $f^s$ is the identity map on the level of sets, in particular a bijection, which implies $f$ is a bijection.


Body:
I need help with this exercise.

Let $f(x)$ be a differentiable function satisfying $f'(x)=f(x)+1$

a) Let $g(x)=f(x)e^{-x}$. Find $g'(x)$.

So this is what I did

$$g'(x)=f'(x)\cdot e^{-x}+e^{-x}\cdot -1 \cdot f(x)$$

$$g'(x)=f'(x)\cdot e^{-x}-e^{-x} \cdot f(x)$$

$$g'(x)=e^{-x}[f'(x)-f(x)]$$

$$g'(x)=e^{-x}[f(x)+1-f(x)]$$

$$g'(x)=e^{-x}[1]$$

$$g'(x)=e^{-x}$$

b) Find all the differentiable functions $f$ satisfying $f'(x)=f(x)+1$

I'm stuck with this one. Please help!

Body:
$$g'(x)=e^{-x}$$

$$g(x)=-e^{-x}+C$$

$$f(x)=g(x)e^x=-1+Ce^x$$


问:
Suppose you invest \$10 at 10.2% per annum compounded annually. How many years would it take for your investment to grow to \$15 000?
I'am solving a simlar equation to this and just trying to figure out how they did it?

the only part I don't understand is how they got the number.... 1.102

15000 = 10(1.102)n ¬1 mark
1500 = 1.102n

答:
The 1.102 comes from two places.

First, "%" means hundredths. If you see "10.2%", that really means $\frac{10.2}{100}$, which is 0.102.

Second, if your money is earning 10.2% interest annually, that means that at the end of the year, the bank takes your balance, multiplies it by 10.2%—that is, by 0.102—and hands you that much more in interest.

Your new balance after a year is equal to your old balance, plus the interest.  Or, in symbols:

new balance = old balance + (old balance × 0.102)

or written another way:

new balance = (old balance × 1) + (old balance × 0.102)

which adds up to:

new balance = old balance × (1 + 0.102)

And that $1 + 0.102$ is there the 1.102 came from.

Okay?


问:
Possible application of Rouche's theorem or Hurwitz theorem.
Let $p(z,w)=\sum_{j=0}^n\sum_{k=0}^mp_{jk}z^jw^k\in\mathbb{C}[z,w]$. Suppose $p(z,w)\neq 0$ for $z, w\in\mathbb{D}$. Then for each $w_0\in\partial\mathbb{D}$ either $w-w_0$ divides $p(z,w)$ or $p(z,w_0)\neq 0$ for $z\in\mathbb{D}$.Here $\mathbb{D}$ is a unit disk.

Can I apply Rouche's theorem or Hurwitz theorem to see this? Actually I have very little idea about this problem. Any idea or help is greatly appreciated!

答:
Take $q_{w_0}(z)=p(z,w_0)=\sum_{j=0}^n\sum_{k=0}^mp_{jk}z^jw_0^k$ and note that $w-w_0$ divides $p(z,w)-q_{w_0}(z)$ so we need to prove that either $q_{w_0}=0$ or $q_{w_0}(z)$ doesn't vanish inside the unit disc. So wlog assume $q_{w_0}(z)$ is not identically zero, but there is $|z_0|<1, p(z_0,w_0)=0,|w_0|=1$

But now picking $w_n \to w_0, |w_n|<1$ we have that $p(z,w_n) \to p(z,w_0)$ normally (actually uniformly on any large disc since they are polynomials of fixed degree) so by Hurwitz, there are $z_n \to z_0, n \ge N, p(z_n,w_n)=0$ which is a contradiction since $|z_n| <1, n \ge N$ as $|z_0|<1$


问:
Working out Vector Variances using index notation
I was hoping someone could provide some intuition to some proofs of vector variances. The expectation case seems rather simple but I get confused when trying to work it out for the variance.

In the case of $$Var_x(a^T x) = \int \sum_i a_i (x_i - \bar x_i) \sum_j a_j(x_j - \bar x_j)p(x_i, x_j) d(\bar x)$$

$$= \sum_i \sum_j a_i a_j\int (x_i - \bar x_i)(x_j - \bar x_j)p(x_i, x_j) d(\bar x)$$

The result is $$a^TV_x(x)a$$ butI do not quite see how this works?

Similarly for the matrix for $$Var_x(Ax)$$ I don't know how to get to the result $$A \Sigma A^T$$

I would really appreciate if someone could explain how to see the transposes, specifically for index case as that is the method we are following in our class.

Thanks

答: You can write $Ax = \sum_i a_i x_i$, where $a_i$ are columns of $A$ and $x_i$ is the $i^{th}$ component of $x$. This means $Var(Ax) = Var(\sum_i a_i x_i) = \sum_i \sum_j a_ia_jcov(x_i, x_j) = A\Sigma A^T$, where $\Sigma$ is the variance-covariance matrix of $x$.


问:
$9$-card-long (or longer) straight flush in a $13$-card hand
The other day at an online bridge site I got dealt a $13$-card hand containing $3,4,5,6,7,8,9,10,J$ of spades, i.e. a $9$-card-long straight-flush (s-f).  This is so rare that I decided to count how many $13$-card hands contain a s-f of length $9$ or longer.

For the purpose of this question:

A s-f is a consecutive sequence in the same suit, and aces always count high (since I was playing bridge!) so $A,2,3,4,5$ of spades is not a $5$-card s-f for this purpose.

Let us just count s-f in spades.  If we want to count all suits we can simply multiply by $4$ since a hand cannot contain two s-f's of different suits each of length $\ge 9$.

I was trying to use inclusion-exclusion but got very confused.  Finally I decided to count explicitly as follows.  My questions:

(1) Can someone verify my answer below?

(2) Can someone show how this can be counted via inclusion-exclusion, or explain why that's not a good idea?

My attempt:

$13$-long s-f: No. of hands $=1$.

exactly $12$-long s-f: There are two such ($2-K$ or $3-A$), and each must be completed with an extra card (to make a $13$-card hand) but the extra card must not complete a $13$-long s-f.  No. of hands $= 2 \times {39 \choose 1}$.

exactly $11$-long s-f: Three such; the $2-Q$ hand must not include the $K$ (leaving $40$ valid cards to choose $2$ to fill out the hand), the $3-K$ hand must not include $2$ or $A$ (leaving $39$ valid cards), and the $4-A$ hand must not include the $3$.  No. of hands $= {40 \choose 2} + {39 \choose 2} + {40 \choose 2}$.

exactly $10$-long s-f: Similar to above, no. of hands $= {41 \choose 3} + {40 \choose 3} + {40 \choose 3} + {41 \choose 3}$.

exactly $9$-long s-f: Similar to above, no. of hands $= 2 \times {42 \choose 4} + 3 \times {41 \choose 4}$.

According to wolfram alpha total $= 571130$.

Body:
Your count is correct, and here is the inclusion-exclusion approach, where the five properties correspond to the five straight flushes:
\begin{align}
&\binom{5}{1}\binom{52-9}{13-9} \\
&-\left[4\binom{52-10}{13-10}+3\binom{52-11}{13-11}+2\binom{52-12}{13-12}+1\binom{52-13}{13-13}\right] \\
&+\left[3\binom{52-11}{13-11}+4\binom{52-12}{13-12}+3\binom{52-13}{13-13}\right] \\
&-\left[2\binom{52-12}{13-12}+3\binom{52-13}{13-13}\right] \\
&+\binom{5}{5}\binom{52-13}{13-13} \\
&=
617050
-48461
+2623
-83
+1 \\
&=571130
\end{align}


问:
Simple description of double coset
Let $G=SL_2(\mathbb{F}_p)$, two by two matrices over $\mathbb{F}_p$ whose determinant is one. Let $H$ be the subgroup of $G$ that is upper triangular matrices.
Is there any simple way to describe ${H}\backslash^{G}/_{H}$?

Context is that in applying Mackey's irreducibility creterion, I am trying to find an element in ${H}\backslash^{G}/_{H}$ that is not identical to identity, but I have trouble understanding this double coset.

答: $G=\mathrm{SL}_2\mathbb{F}_p$ acts on $\mathbb{F}_p\mathbb{P}^1$ by Mobius transformations. The stabilizer $B$ of $\infty$ is comprised of upper triangular matrices, which are precisely the affine transformations $[\begin{smallmatrix} a & b \\ 0 & a^{-1}\end{smallmatrix}]x=a^2x+b$. The coset space $G/H$ may be identified with $\mathbb{F}_p\mathbb{P}^1$ (orbit-stabilizer theorem), and the double cosets $H\backslash G/H$ may be identified with the $H$-orbits of $\mathbb{F}_p\mathbb{P}^1$, of which there are clearly two: $\{\infty\}$ and $\mathbb{F}_p$ (on which translations act transitively).

Title: Simple description of double coset


问:
Derivation of bivariate Gaussian copula density
The multivariate Gaussian copula density, derived here, is

$$c(u_1,\ldots,u_n;\Sigma)=|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}(\Sigma^{-1}-I)x\right)$$
where $\Sigma$ is the covariance matrix, and  $x=[\Phi^{-1}(u_1),\ldots,\Phi^{-1}(u_n)]^{\top}$.

The bivariate Gaussian copula density, based on the pair-wise correlation coefficient $\rho$, is
$$
c\left(u_{1}, u_{2} ; \rho\right)=\frac{1}{\sqrt{1-\rho^{2}}} \exp \left\{-\frac{\rho^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho x_{1} x_{2}}{2\left(1-\rho^{2}\right)}\right\}
$$

What is the derivation of the second formula from the first?

答:
Note that with standard normal marginals

$$\Sigma=\left[\begin{array}{cc} 1 & \rho  \\ \rho  & 1 \end{array}\right],\,\, |\Sigma| = 1 - \rho^2$$

and

$$\Sigma^{-1}= \frac{1}{1- \rho^2}\left[\begin{array}{cc} 1 & -\rho \\ -\rho  & 1 \end{array}\right], \,\, \Sigma^{-1}-I= \frac{1}{1- \rho^2}\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho  & \rho^2 \end{array}\right]$$

Hence,

$$- \frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x} = \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho  & \rho^2 \end{array}\right]\left[\begin{array}{cc} x_1 \\ x_2  \end{array}\right] \\= \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2x_1  -\rho x_2 \\ -\rho x_1  + \rho^2 x_2 \end{array}\right] \\= -\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)},$$

and, thus,

$$|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x}\right) = \frac{1}{\sqrt{1- \rho^2}} \exp \left\{-\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)} \right\}$$


问:
Show that a matrix is invertible and that its inverse has integral entries
Let $A$ and $B$ be 2 × 2 matrices with integer entries such that $A$, $A + B$, $A + 2B$, $A + 3B$, and $A + 4B$ are all invertible with integral entries. I want to prove that $A + tB$ is invertible and that its inverse has integral entries, assuming that $t ∈ Z$.

I started by proving that det$(A) = \pm 1$. Let matrix $A = [a_{ij}]$. Since $A$ and $A^{-1}$ both have integral values, det$(A)$ and det$(A^{-1})$ must both be integers. Also, $\det(AA^{-1})=1$. Therefore,  $\det(A)\det(A^{-1})=1$. We already established that det$(A^{-1})$ had to be an integer. Thus, the only possible solutions for $\det(A)$ are $\pm 1$.

Now det$(A + tB) = a + tc + t^2b$ where $a =$ det$(A)$, $c = (a_{11}b_{22} + a_{22}b{11}) - (a_{12}b_{21} - a_{21}b_{12})$, and $b =$ det$(B)$. From the problem statement, since $A$, $A + B$, $A + 2B$, $A + 3B$, and $A + 4B$ all have integral entries, it follows that: $$a = \pm 1$$ $$ a + c + b = \pm1$$ $$ a + 2c + 4b = \pm1$$ $$ a + 3c + 9b = \pm1$$ $$ a + 4c + 16b = \pm1$$

I'm not quite sure how to proceed to complete the proof. Any guidance is much appreciated!

Edit:
Solving the system of equations yields: $$a = \pm 1, b = 0, c = 0$$

Thus, det$(A + tB) = a + tc + t^2b = a =$ det$(A)$ for any integer $t$ and $A + tB$ is invertible and has integer entries for all integers $t$.

答:
(With the assumption that $ (A + tB)^{-1}$ must all have integer entries for $ t = 0, 1, 2, 3, 4$.)

(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Note: You forgot $ a = \pm 1$, so we have 5 equations. This is crucial.

Hint: Using the Pigeonhole principle, conclude that $ b = c = 0$.

 Further hint: By PP, 3 of the 5 equations have the same value. Thus, the quadratic is a constant function.

Corollary: Hence, $det(A + tB) = \pm 1$, and so we are done.


问:
Inspiration behind substitution of main variable in Cardano's solution
Suppose the solutions to a general cubic equation $ax^3+bx^2+cx+d=0$ are to be found. Then according to Cardano's method, First a variable substitution must be carried on to convert the general cubic to depressed cubic.
$$ax^3+bx^2+cx+d=0\rightarrow t^3+pt+q=0\ \text{where $t=x-\frac{b}{3a}$}$$
This seems pretty clear but what happens next is not so obvious for me.

Now we take $t=u+v$ to get $u^3+v^3+(3uv+p)(u+v)+q=0$. Assuming $3uv+p=0$, we get the system
$$u^3+v^3=-q$$ $$uv=\frac{-p}{3}$$. Using Vieta's formula, the quadratic equation with $u^3$ and $v^3$ as roots is $$x^2+qx-\frac{p^3}{27}=0$$
Now the roots for quadratic can be found by analysing the discriminant.

My doubt is that what was the inspiration behind assuming $3uv+p=0$, except helping to form the quadratic equation.

Please help

THANKS

答:
One source of inspiration may have been looking at the binomial expansion of $(u+v)^3$ and mapping it to the cubic $t^3 + pt +q = 0$.

$$\begin{align*}(u+v)^3 &= u^3 +3 u^2v +3uv^2+v^3 \\
\\
&= u^3+v^3+3uv(u+v) \\
\\
(u+v)^3 - 3uv(u+v)-(u^3+v^3)&= 0\\
\end{align*}$$

Substituting $t =u+v$ yields

$$t^3-3uvt-(u^3+v^3) =0$$

Which forces the equalities
$$ p = -3uv$$
$$q= -(u^3+v^3)$$
to complete the mapping of the depressed cubic equation to the equation for the cube of a binomial.

Is this what really happened historically?  I don't know.  Probably not.


问:
Global sections of integral proper $k$-scheme is finite field extension of $k$
I am trying to show the following:
if $X$ is an integral proper $k$-scheme, $k$ a field, then $O_X(X)$ is a finite field extension of $k$.

I have succeeded to show that $O_X(X)$ is a field but I don't see why it must be a finite field extension.

(To show it is a field I used that a global section s corresponds to a morphism $X \to \operatorname{Spec} k[x]$, one can show the image is a closed point, so if $s \neq 0$ there is an irreducible polynomial  $g \in k[x]$ such that $g(s)=0$, so it is invertible.)

I would like to avoid using cohomology/Grothendieck's finiteness result for proper morphisms. a similar question was asked here but I am not assuming $X$ is geometrically integral.

答: The solution has been worked out in the comments. As restriction maps for the structure sheaf on an integral scheme are injective, $O_X(X)$ embeds in to $O_X(\operatorname{Spec} A)=A$ for $\operatorname{Spec} A$ any affine open subscheme of $X$. Let $\mathfrak{m}$ be a maximal ideal of $A$. Then $O_X(X)\subset A$ does not intersect $\mathfrak{m}$, so the map from $O_X(X)$ to its image in $A/\mathfrak{m}$ is an injection. Thus $O_X(X)$ embeds in to a residue field of a finite type scheme over $k$, and all such residue fields are finite extensions of $k$ by Zariski's lemma.


问:
Flux across a cone
Compute the flux of $F=(x,y,z^4)$ across the cone $z=\sqrt{x^2+y^2}, z \in [0,1]$ in the downward direction. (Answer is $\pi/3$.)

For this question I tried to use the divergence theorem: $\int\int_SF = \int\int\int_V\nabla F$

I got $\nabla F = 2+4z^3$ and used cylindrical coordinates: $\int_0^{2\pi}\int_0^1\int_r^1(2+4z^3)rdzdrd\theta$ but the answer I got was $4\pi/3$. Is this correct or did I do something wrong?

答:
Same solution as "1 Answer" except:

The first display with a union, not an intersection, and the restriction $x^2+y^2\le 1$ is specified. Also what is called surface $S$, the intersection of $z=1$ and the solid cone $z \le \sqrt{x^2+y^2}$, is parameterized by

$$r(a,t)=\langle a \cos(t),a \sin(t),1\rangle, \; 0 \le a \le1,\; 0 \le t \le 1.$$

Title: Flux across a cone


问:
Let $G$ a connected graph. If every vertex of $G$ has even degree
Let $G$ a connected graph. If every vertex of $G$ has even degree, then for any vertex $v\in V(G)$, $$\omega(G-v)\le\dfrac{\delta(v)}{2}$$ Where, $\delta(v)$ denote degree of $v$ and $\omega(G-v)$ is the number of connected components of $G-v$.

I am looking for a way to obtain that inequality, but I still have no results.

I consider two cases:

$v$ is not a cut vertex. In this case the result is satisfied.

$v$ is a cut vertex. In this case I have no idea

答:
Let $v \in V(G)$. Since every vertex of $G$ has even degree, $G$ is Eulerian, i.e. $G$ has an Euler cycle $C$. We can say $C$ starts (and thus ends) at $v$. Starting at $v$, we travel a sequence of vertices, say $A_1$, along $C$, and must eventually reach $v$ once more. This occurs $\delta(v)/2$ many times (why?). This gives us $A_1,A_2,\dots, A_{\delta(v)/2}$, where $(G-v)[A_i]$ is a connected component of $G-v$. If the $A_i$ are pairwise disjoint, then we achieve equality, i.e. $\omega(G-v) = \frac{\delta(v)}{2}$. An example of this is a number of cycles joined at the same vertex such as the Friendship Graph. Otherwise, the inequality is strict (why?).

I've left out some detail, but this is the gist of it. Note by $G[A]$ we mean the subgraph of $G$ induced by the vertex subset $A$.


问:
Inverse Laplace transform of the function: $F(s)=e^{-a\sqrt{s(s+r)}}$
I would like to find inverse Laplace transform of the function: $$F(s)=e^{-a\sqrt{s(s+b)}}$$
which $a$ and $b$ are positive real numbers and $s$ is a complex variable. It would be appreciated if someone can help me.</p>

答:
See my solution to this problem.  The analysis is identical, save for the denominator.

$$\oint_C ds \,e^{-a \sqrt{s (s+b)}}\, e^{s t}$$

where $C$ is the following contour pictured below:

This one's a bit odd because we are removing the branch point singularities by subtracting the inner contour $C_i$ from the outer contour $C_o$, i.e. $C=C_o+C_i$.

The integral about $C$ consists of the integral about vertical path (that ultimately becomes the ILT) and the integral about the outer arc.  We expect that this latter integral vanishes as the radius of the arc gets larger; this only happens when $a \lt t$ as shown below, by letting $s=R e^{i \phi}$; this integral over this arc is, for large $R$:

$$\left| \int_{\pi/2}^{3 \pi/2} d\phi \,e^{R (t-a) \cos{\phi}} \right|$$

Note that $\cos{\phi} \lt 0$ over the integration region, so that the integral only vanishes for $a \lt t$ when the contour opens to the left.

This leaves the integral over $C_i$ which consists of integrals above and below the real axis, between the branch points, and integrals about small circular arcs about the branch points.  These latter integrals vanish in the limit of the radii of the arcs going to zero. Thus, we are left with

$$\int_{c-i \infty}^{c+i \infty} ds e^{-a \sqrt{s (s+b)}}\, e^{s t} - \underbrace{ \int_0^{b} dx\, e^{i a \sqrt{x (b-x)}}\, e^{-x t}}_{s=e^{-i \pi} x} + \underbrace{ \int_0^{b} dx\, e^{-i a \sqrt{x (b-x)}}\, e^{-x t}}_{s=e^{i \pi} x} = 0$$

We set the sum of the integrals to zero because of Cauchy's theorem, i.e., there are no poles inside the contour $C$.  The second integral is the integral over the line below the real axis; the third is the integral over the line above the real axis.  Note that the phase transition was continuous from $-\pi$ to $\pi$ because we did not cross the branch cut; we did not see this explicitly because that information was included in the integrals about the branch points which vanished.

We do a little algebra and we have a real-valued integral for the ILT:

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-a \sqrt{s (s+b)}} e^{s t} = \frac{1}{\pi} \int_0^b dx \, \sin{\left [ a \sqrt{x (b-x)}\right]} e^{-x t}$$

We make similar substitutions as with the other solution and we find the RHS is equal to

$$\begin{align}\frac{b}{2 \pi} e^{- b t/2} \int_0^{\pi} d\phi \, \sin{\phi} \, \sin{\left[\frac12 a b \sin{\phi}\right]} \, e^{b t/2 \cos{\phi}}\\ = -e^{-b t/2} \frac{1}{\pi} \frac{\partial}{\partial a} \int_0^{\pi} d\phi \,  \cos{\left[\frac12 a b \sin{\phi}\right]} \, e^{b t/2 \cos{\phi}}\\ = -e^{-b t/2} \frac{\partial}{\partial a} I_0\left( \frac{b}{2} \sqrt{t^2-a^2}\right)\end{align}$$

Note again that the contour is only closed to the left when $t>a$; when $t<a$ it is closed to the right.  Therefore the ILT of $e^{-a  \sqrt{s (s+b)}}$ is

$$\frac{a b}{2} e^{-b t/2} \left(t^2-a^2\right)^{-1/2} \, I_1\left(\frac{b}{2}\sqrt{t^2-a^2}\right) H(t-a)$$


问:
A function is given by $f(x)=x^4−2x^3−2x^2+6x−3$. What is the solution set for $f(x)\geqslant 0$?
A function is given by  $f(x)=x^4−2x^3−2x^2+6x−3$. What is the solution set for  $f(x)\geqslant 0$?

答:
$f(1)=0$.  Trivially obvious.   Looking at the derivative $4x^3-6x^2-4x+6\ge 0$ for $x\ge 0$, none other.

Note that $f'(1)=0$, making that point a double root.

See my comment below - I missed other root.