问:
Would someone explain the meaning of some notation relating to the definition of Weak Measurability?
I have been independently reading the text $Lebesgue \ Integration$ by Soo Bong Chae, but have gone down a bit of a rabbit hole trying to learn more about measurability beyond the scope of this book and have become lost in some of the notation and definitions. I have tried to distill many questions down to a few key ones which I hope will clarify a lot for me once answered, so I appreciate your indulgence.

So, I have the following definition:

For a real Banach space $V$, and a function $f: [0,T] → V$, $f$ is said to be weakly measurable if $t → \langle v,f(t)\rangle$ is measurable for all $v \, \epsilon \, V^*.$

First, what does the $\langle \,, \rangle$ notation denote? Typically I understand this to mean an inner product, but if $V$ is only assumed to be a Banach space, then I suppose this isn't the case here. It seems like I am assumed to know what it means, but I am afraid I do not, and have not been able to deduce its meaning from contextual clues.

I have also seen the more compact notation $f \, \epsilon$ $C([0,T];V)$, which I think I am right to believe is just $f$ as defined previously, only now with the addition that $f$ is a continuous such function.

However in trying to then understand subsequent examples given, I have encountered the collections $C^n([0,T];C^n([0,T]))$, and $C^n([0,T]^2)$, and I have no confidence that I am understanding these objects correctly. The first I believe is just, n$^{th}$-continuously differentiable functions on $[0,T]$ to the collection of n$^{th}$-continuously differentiable functions on $[0,T]$. But after much searching, I can only guess at the second. I have thought perhaps the Cartesian product, but am disinclined to think I am correct in this interpretation, because then I could not imagine any connection between the two.

In any case I am grateful for any light that can be shed on the subject, and hope these questions are not too ignorant.

答:
Some people use the notation $ \langle v, x \rangle$ for the value $v(x)$ of $v$ at $x$. $v:V \to F$ is a continuous linear map and $x \in V$ ($F$ being the scalar field).

Your guesses for $C([0,T];V), C^{n}([0,T])$ etc are all correct.

Incidentally,  $ \langle v, x \rangle$ becomes an actual inner product in the case of Hilbert spaces when the dual space is identified with the space itself using Riesz Theorem.


问:
Solving a system of equations with boolean variables in $\mathbb{Z}_3$
I have a problem that reduces to a system of $2n+1$ equalities on $4n$ variables all in $\mathbb{Z}_3$, where the variables are constrained to be in $\{0,1\}$, i.e.,

$$x_1 + x_2 + ... + x_k \equiv a\ (\mathrm{mod}\ 3)$$

$$x_2 + x_3 + ... + x_l \equiv b\ (\mathrm{mod}\ 3)$$

$$\cdots$$

$$x_m + ... + x_{4n} \equiv c\ (\mathrm{mod}\ 3)$$

The right hand side ($a$, $b$, $c$, $\ldots$) is known, but not necessarily in $\{0,1\}$ but in $\{0,1,2\}$.

I have found no better way of solving this
but a brute-force approach on the free variables, because doing Gaussian elimination, I can get a an upper triangular system, but that would only give me a solution if my variables were in $\{0,1,2\}$.

So my question is whether there is a more efficient, polynomial-time, way to solve this system that I may have missed? If not is there a complexity result that shows this problem to be hard in the general case ?

答:
If I understand correctly, you want to find a solution to the system :

$$Ax = b$$

Where $A$ is a binary matrix, $x = (x_1, x_2, ... , x_n)^T$ a vector of variables, and $b = (b_1, b_2, ..., b_m)$ a vector of right-hand side, with $b_i \in \{0, 1, 2\} \; \forall i \in \{1, ..., m\}$.

If so, you want to solve an Integer Program.

Unfortunately integer programs are NP-Complete in general, and your special case includes the set-partitioning problem, which is also NP-complete.

How to solve it

Because integer programs have a lot of real-life applications, specialized algorithms and software have been devised to solve them. In my experience, solving a problem with fewer than 1000 variables should take less than a second, but it really depends on the problem... The solvers offering the highest performances are probably cplex or gurobi (they are not free, but I think they offer a free student version). There are also open source solvers such as GLPK and lp_solve. If you have a relatively small problem (maybe less than 100 variables), you can even use the Excel solver, although it really not the best!

In any case, you do not want to code it yourself. Those software have decades of improvments into them and that is why they work well.

How the algorithm works

Every commercial solver solves the problem by branch-and-bound.

A branch-and-bound algorithm consists of a systematic enumeration of
candidate solutions by means of state space search: the set of
candidate solutions is thought of as forming a rooted tree with the
full set at the root. The algorithm explores branches of this tree,
which represent subsets of the solution set. Before enumerating the
candidate solutions of a branch, the branch is checked against upper
and lower estimated bounds on the optimal solution, and is discarded
if it cannot produce a better solution than the best one found so far
by the algorithm.

Your problem is a feasibility problem and not an optimization problem, so there is no need to compute bounds. However, the algorithm has to check wheather the current branch is infeasible. This is usually done by solving a linear relaxation of the problem (accepting solutions with fractional variables).

Commercial solvers also use many tricks to reduce to speed up the algorithm. Unfortunately, many of those tricks are secrets!

ContentLicense: CC BY-SA 4.0


问:
Proving $3\times3$ nilpotent matrix satisfies $A^3=0$
Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.

答:
Since you know that the eigenvalues of a nilpotent matrix are $0$, consider the Jordan normal form.

The Jordan normal form of this matrix will be strictly upper-triangular. Now, you can easily prove that for any strictly upper-triangular matrix of order $n \times n$, you have $A^n = O$, where $O$ is the zero matrix of the same order.

And your question is the case when $n=3$.


问:
Group structure associated to the equation $y^2 = x^2 + x + 1$
The elliptic curve group law allows us to add two solutions $(x_1, y_1), (x_2, y_2)$ of the equation $y^2 = x^3 + ax + b$ to obtain a new solution $(x_3, y_3)$. Is there an analogous group law associated to the equation $y^2 = x^2 + x + 1$?

答:
If $char(k)\ne 2$ then

$y^2=x^2+x+1$ is isomorphic through $x=z-1/2$ to $y^2=z^2+3/4$

Which is isomorphic through $y = \frac{u+v}{2},z=\frac{u-v}{2}$ to $uv=3/4$.

If $char(k)\ne 3$ then it is isomorphic to $uw=1$ through $v=\frac43w$

Which is the multiplicative group $GL_1(k)$ whose group law is $(u_1,w_1).(u_2,w_2)=(u_1u_2,w_1w_2)$

If $char(k)=2$ then through $y=t-x$ the curve is isomorphic to $\Bbb{A}^1_k$ which is a group. If $char(k)=3$ then the curve is isomorphic to $y^2=z^2$ which is not irreducible, removing $(0,0)$ (the only singular point) it becomes $GL_1(k)\cup GL_1(k)= GL_1(k) \times \pm 1$ which does have a group law.

OwnerDisplayName: Gautam


答:
There are a few ways to do that. If you know about minimal polynomials, then you can use that the minimal polynomial of $A$ has degree at most $3$ and divides $A^n$.

A less advanced way: Note that for each $k$, the range of $A^k$ is a subspace of ${\mathbf R}^3$, and $range(A^3) \subset range(A^2)\subset range(A) \subset {\mathbf R}^3$. The dimensions must correspondingly be nonincreasing as you go down the list. If $range(A^3)$ weren't of dimension zero, then you'd have $range(A^k) = range(A^{k+1}) \neq \emptyset$ for some $k \leq 2$. Then by induction $range(A^l) = range(A^k)$ for all $l \geq k$. This would include $l = n$, a contradiction.

Title: Proving $3\times3$ nilpotent matrix satisfies $A^3=0$

Body: Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.


问:
Determine the minimum of $\frac{\int_0^1{x^2\left( f'\left( x \right) \right) ^2 dx}}{\int_0^1{x^2\left( f\left( x \right) \right) ^2dx}}$
For all non-zero continuously differentiable function $f:[0,1]\to\mathbb R,f(1)=0$, determine the minimum of
$$\dfrac{\displaystyle \int_0^1{x^2\left( f'\left( x \right) \right) ^2\mathrm dx}}{\displaystyle \int_0^1{x^2\left( f\left( x \right) \right) ^2\mathrm dx}}$$

I totally have no idea how to start. Can anyone help?

答:
Let $g = xf$, notice

$$\begin{align}|g'|^2 &= (xf' + f)^2 = (xf')^2 + 2xff' + f^2 =
(xf')^2 + x(f^2)' + x'f^2\\
&= (xf')^2 + (xf^2)'\end{align}$$
Intergate both sides over $[0,1]$, we obtain

$$\require{cancel}
\int_0^1 |g'|^2 dx = \int_0^1 (xf')^2 dx + \color{red}{\cancelto{0}{\color{gray}{\left[xf^2\right]_0^1}}} = \int_0^1 (xf')^2dx$$
because $f(1) = 0 \implies \left[xf^2\right]_0^1 = 0$. As a result,

$$\mathcal{I(f)} \stackrel{def}{=} \frac{\int_0^1 (xf')^2 dx}{\int_0^1 (xf)^2 dx} = \frac{\int_0^1 |g'|^2 dx}{\int_0^1 g^2 dx}$$
Notice $f \in C^1 \implies g = xf \in C^1$ and $g(0) = g(1) = 0$.

By Wirtinger's inequality, we have

$$\int_0^1 |g'|^2 dx \ge \pi^2\int_0^1 |g|^2 dx$$
This implies $\mathcal{I}(f)$ is bounded from below by $\pi^2$.
It is easy to see this lower bound is attained by
$$g(x) = \sin(\pi x)\quad\iff\quad f(x) = \frac{\sin\pi x}{x}$$
The minimum we seek equals to $\pi^2$.


问:
How to integrate $\int e^{2x} \sec{3x} \,dx$
$\int e^{2x} \sec{3x} \,dx$

I have tried every elementary (e.g. methods taught in first course of integration) way to integrate this function . But couldn't make it.
Please someone solve it and mention the concepts you have used to solve it.

答:
As said in comments, there is no closed form solution and the only possible result is a gaussian hypergeometric function ... which corresponds to an infinite summation.

So, as @ person suggested, use
$$\sec(t)=\sum_{n=0}^\infty  \frac{\left|E_{2 n}\right|}{(2 n)!}t^{2n}$$ making
$$I=\int e^{2x} \sec(3x) \,dx=\sum_{n=0}^\infty 9^n \int \frac{\left|E_{2 n}\right|}{(2 n)!}x^{2n}\,e^{2x}\,dx$$ that is to say
$$I=\sum_{n=0}^\infty \frac{ 9^n \left|E_{2 n}\right| }{2^{2 n+1}\,(2 n)!}\,\Gamma (2 n+1,-2 x)$$
$$J=\int_0^t e^{2x} \sec(3x) \,dx=\sum_{n=0}^\infty \frac{ 9^n \left|E_{2 n}\right| }{2^{2 n+1}\,(2 n)!}\,\Big[\Gamma (2 n+1,-2 t)-\Gamma (2 n+1)\Big]$$

Expanded as a Taylor series
$$J=\sum_{n=1}^\infty \frac{a_n}{n!} t^n$$ coefficients $a_n$ make the sequence
$$\{1,2,13,62,637,4802,70993,742142,14537497,195123842,\cdots\}$$ which is not found in $OEIS$.


问:
Derivative of integral in one direction
Suppose I have a 3D scalar field $\phi(x,y,z)$, where $x,y,z$ are cartesian coordinates in $\mathbb{R}^3$. I can define another 3D field, which is the integral of the field with respect to one dimension

$$ \Phi(x,y,z) = \int_{z}^{\eta} \phi(x,y,z^\prime) \,d z^\prime$$

This has the interpretation of integrating from a depth $z$ to a surface $\eta$, where $\eta(x,y)$ is known and fixed for every $x$ and $y$. If $\phi$ has the interpretation of a density*gravity, this is something like the weight of a material above a point in space.  I am interested in 3 questions:

What is the partial derivative $\frac{\partial}{\partial z} \Phi = \frac{\partial}{\partial z} \int_{z}^{\eta} \phi(x,y,z^\prime) \,d z^\prime$ ?

What is the total derivative $\frac{d}{dz} \Phi = \frac{d}{dz} \int_{z}^{\eta} \phi(x,y,z^\prime) \,d z^\prime$ ? Is it the same as the partial derivative in this case?

How could I write the line integral $\Phi(z)$ (from a depth $z$ upwards, that is from $(x_0,y_0,z)\to(x_0,y_0,\eta)$) as a volume integral over all 3 spatial coordinates? Something like $\Phi(z) = \int_{z}^{\eta}\int_{A} \phi(x,y,z^\prime) D(x,y,z)\,dA \,dz^\prime$ where $D(x,y,z)$ is something like a delta function. I ask because I need to approximate the integral over predefined 3D blocks where I know 3D quadrature rules.

Any insight for any of the 3 questions would be appreciated!

edit: changed the order of the questions to ask partial derivative first

答:
1)) If for fixed $x$ and $y$, $f(z)=\phi(x,y,z)$ is a continuous function of $z$ then for any primitive $F(z)$ of $f(z)$ we have $$\Phi(x,y,z)=\int_{z}^{\eta} \phi(x,y,z^\prime) \,d z^\prime=F(\eta)-F(z)$$ by Newton-Leibnitz' formula. Then
$$\frac{\partial}{\partial z} \Phi(x,y,z)=(F(\eta)-F(z))’=-f(z)=-\phi(x,y,z).$$

Title: Derivative of integral in one direction


问:
Why do these trig functions "overpower" each other?
For example, $\sin(x)\cos(x)$ can be written as $\sin(2x)/2$, the limit as $x$ approaches $0$ of $\sin(x)\cos(x)$ is $0$, and the limit as x approaches $\pi/2$ is $0$. I don't see a reason why sine always increases/decreases faster than cosine. Why does sine overpower cosine near x=0, and why does cosine overpower sine near x=pi/2, I feel that if they did increase the same the limit should be somewhere in the middle, such as 1/2 for both of these values. If you try to take the derivative of each with respect to some other to see why one increases faster, you end up with the same problem because the derivatives just keep giving you negative versions of the original problem. </p>

答:
the limit of $\sin(x)\cdot \cos(x)$ as $x\rightarrow \pi /2$ is $0$.
So in both cases you have $\sin(x)\cdot \cos(x)=0$.
Infact they do increase/decrease with the same rate. Infact $\sin(\pi/2-x)=\cos(x)$ if you vary $x$ they have the same values so they increase/decrease with the same rate.</p>

ContentLicense: CC BY-SA 3.0


问:
Divisibility of Number Theory
Can anyone explain this? I'm new to this topic.
Prove that for all integers a, b, c , if a | c and b | c , then ab | c² .</p>

答: That implies $c=ak$ and $c=bq$ so multiplying the EQ we get $c^2=ab(Kq)$ therefore $ab|c^2$

Title: Divisibility of Number Theory


问:
Solve a simple equation with log in it
I'm stuck with solving this equation,

$$2 \log x = \log 9 $$

This is how far I made it:

\begin{align}
\log x &= \log 4,5 \\
x &= ?
\end{align}

I'm a beginner at logarithms so I appreciate ways to solve it and not just an answer.

答:
If we start with $$2 \log x = \log 9,$$ the first step is to move the 2, but you can't divide it over like that.  The rule is that $a \log b = \log b^a$, so we get $2 \log x = \log x^2$.

Now our equations is $$\log x^2 = \log 9.$$  The next step is to use the fact that $\log A = \log B$ means $A = B$.  In our case, that means $$x^2 = 9,$$ and you can solve for $x$.  Remember to check that the answer makes sense.  For instance $\sqrt{9} = \pm 3$, but you can't take the log of a negative, so $x \neq -3$.

Title: Solve a simple equation with log in it


答:
Since $M_{n\times n}$, the space of $n\times n$ matrices, is an $n^2$ dimensional vector space, there must be a linear dependence between the matrices $A^{0}, A^1, A^2,\ldots, A^{n^2}$ for any $A$.  Therefore, $A$ satisfies some polynomial of degree at most $n^2$.  However, if $p(A)=0$ and $q(A)=0$, then for any other polynomials $a, b$, we have $a(A)p(A)+b(A)q(A)=0$, so by the Bezout lemma, $\gcd(p,q)(A)=0$.  On the one hand, this establishes that $A$ satisfies a unique monic minimal degree polynomial, but on the other hand establishes that if $A$ satisfies that if $p(A)=0$, then the minimal polynomial of $A$ will be some divisor of $p$.

By the Cayley-Hamilton theorem, $A$ satisfies its characteristic polynomial, which shows that the minimal polynomial of $A$ is actually of degree at most $n$.

Putting this all together, in your particular case, the minimal polynomial divides $x^n$ and has degree at most $3$.  In particular, it divides $x^3$.

However, if we do not have the Cayley-Hamilton theorem, then we must make use of something else to get our bound of $3$.  One good way is the following lemma:

If $\ker A^k=\ker A^{k+1}$, then $\ker A^k=\ker A^i$ for all $i>k$.

Once you show this, we note that we have a chain of subspaces
$$\ker A^0\subset\ker A \subset \ker A^{2} \subset \ker A^3 \subset \cdots.$$
Since $A$ is acting on a 3 dimensional vector space, and since we have a proper inclusion only when two subspace have different dimensions, the dimension of the kernel must increase by at least $1$ every time until it stabilizes (at 3, since it is eventually the whole space).  In particular, if $\dim \ker A^n=3$ for some $n$, then $\dim A^3=3$.


Title: How to solve $x^3=-1$?

Body:
How to solve $x^3=-1$? I got following:
$x^3=-1$

$x=(-1)^{\frac{1}{3}}$

$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...

Body:
$$x^3+1=0$$

$$(x+1)(x^2-x+1)=0$$

$$x=-1,\frac{1\pm \sqrt{3}i}{2}$$

$$x=e^{i\pi},e^{i2\pi/3},e^{-i2\pi/3}$$


问:
Is it possible to plot a function with a vertical tangent line while the plot of the function has no vertical line segment?
Is it possible to plot a function with a vertical tangent line while the plot of the function has no vertical line segment?</p>

答: Yes, eg- x^(1/3) (real to real only not complex) and has a vertical tangent at x=0</p>


答:
Let $v$ be a non-zero vector and consider the sequence

$$
v, Av, A^2v, ..., A^kv, 0, 0, ...
$$

where $k < n$ is the largest integer such that $A^kv \ne 0$. We will see that the non-zero elements of the sequence are linearly independent. Therefore, in a three-dimensional space, at most three, i.e. $v, Av, A^2v$, can be non-zero. Thus, $A^3v = 0$ for every $v$.

To prove linear independence, note first that the last non-zero element of the sequence, i.e. $A^kv$ lies in the kernel of $A$, but the other vectors, i.e. $v, Av, A^2v, ..., A^{k-1}v$ do not. Therefore none of the vectors
$v, Av, A^2v, ..., A^{k-1}v$ can be a multiple of $A^kv$. Similarly, the last two non-zero vectors of the sequence, i.e. $A^{k-1}v, A^kv$ lie in the kernel of $A^2$, but the preceding vectors do not. Therefore none of the preceding vectors is a linear combination of $A^{k-1}v, A^kv$.

Continuing this reasoning we see that $A^iv, ..., A^kv$ lie in the kernel of $A^{k-i+1}$ while none of $v, Av, ..., A^{i-1}v$ do. Therefore, none of $v, Av, ..., A^{i-1}v$ is a linear combination of $A^iv, ..., A^kv$. Therefore, the non-zero vectors in the sequence $v, Av, A^2v, ..., A^kv, ...$ are linearly independent.

Therefore there cannot be more than three of them, so $A^3 = 0$.


问:
What's more general than category theory?
First there was arithmetic with numerical calculations (i.e., one unknown on one side of an equation). Then algebra with manipulations of variables (many unknowns anywhere in an equation). Then systems are studied that differ from ordinary arithmetic but share some of the same properties (equations where the unknowns represent all sorts of things - even functional equations) and then these properties are abstracted in abstract algebra and whole classes are studied such as groups and rings. Then category theory studies maps between structures (functorial equations), then n-category theory, then ...

Where do we go now? Is category theory the end of the road for the foreseeable future?
  Is the only way forward to go backwards and generalize in a different direction (like "generalized equations" of optimization or something)?

答:
I don't really see a coherent logical progression in the branches of mathematics you're putting forward.  Mathematics isn't just about abstraction and generalizing, making things more and more general.  It's most often about solving particular problems.  Category theory was born out of algebraic topology, in many ways as a notational convenience, to make sense of the tremendous and difficult-to-follow messes algebraic topologists were producing.

If you've ever programmed in a language like "C" you know the concept of a "macro".  This is an idea that is re-usable in many different contexts.  You plug in different objects and the macro continues to make sense.  That's much of the point of category theory, as there are so many ideas that are duplicated over and over again in mathematics that it's confusing to give them special names.  So we call these ideas by generic names that make sense in a wide-array of contexts, like "the co-product (or whatever) in the category C (name your category)", etc.  It saves time and energy.   Moreover, once you've reduced the "bulk" of your notation sufficiently, there is a phenomena where the concepts are lighter and easier to play with.  So by using category theory you sometimes "lighten the load" a little, making other discoveries perhaps a little easier (if you're lucky).

In that regard what gets called "category theory" I think of as more of an attempt to find the natural language for certain types of ideas.  The general idea being that certain types of problems become easy when using appropriate notation. Not all, but some.  Some problems are just hard -- like the Poincare conjecture, or the Schoenflies problem, the classification of finite-simple groups, or existence and uniqueness of solutions to Navier-Stokes (and if you look at the work that's been done on these problems you will see almost no category theory at all, just a tiny little bit on the high-dimensional Poincare and Schoenflies problems). In programming category theory might be analogous to the study of data types, and how one structures memory efficiently.

ContentLicense: CC BY-SA 2.5


问:
How to work out a Mahalanobis distance
Currently going through my Lecturer's notes and came across the Mahalanobis distance but nothing in the notes on how to work this out and googling hasn't really turned up anything either. My maths isn't great but I've been at it for 20 minutes now trying to come up with the same answer and can't.

The Mahalanobis norm is defined by $D_M^2(x) = (x-\mu)^T C^{-1}(x-\mu)$, and for my calculation, I have $(x-\mu)^T = [-90, -100]$ with covariance matrix

$C=\begin{pmatrix}
0.00025  &  -0.00015 \\
-0.00015 &  0.00025
\end{pmatrix}$

Thanks in advance for your help.

答: Plugging in, we have $C^{-1}=\begin{pmatrix} 6250 & 3750 \\ 3750 & 6250 \end{pmatrix}$, and so $D_M^2= \begin{pmatrix}-90 & -100 \end{pmatrix}\begin{pmatrix} 6250 & 3750 \\ 3750 & 6250 \end{pmatrix} \begin{pmatrix}-90 \\ -100\end{pmatrix}=1.80625*10^8.$  Taking square roots, we see that the norm of $x$ is approximately $13439.7$.</p>

Title: How to work out a Mahalanobis distance


问:
A combination problem of picking m out of n
I am thinking of a problem of putting 4 identical beans onto 4 grids of a 3x3 chessboard. The possible number of selection is
$$
  C(9, 4) = \frac{9!}{4!5!} = 126
$$

Now assuming we need to put at least 1 bean on the first column and at least 1 bean on the second column, I want to find out the possible configuration. To solve it, I need to pick 1 spot on the first column and 1 spot on the second column, there will be 3x3 = 9 ways to do it. I then pick another 2 spots out of the remaining 9-2=7 spots for all my 4 beans. The total combination will be
$$
  3 \times 3 \times C(7,2) = 189
$$

I do not understand how come this combination is even more than the total combination (126). Any idea what I missed in the calculation? Thanks.

答:
If there are more than one bean in the first or second columns, that configuration is double-counted (or triple-counted).

To solve this problem, one should consider the complement: the cases when columns 1 or 2 do not contain any beans. There are $C(6,4) \times 2$ such cases (they cannot be both empty!) Hence the number of cases where at least 1 bean is on the first column and at least 1 bean is on the second column is $C(9,4) - 2 \times C(6,4) = 96$.


答:
That is because $\, 3×3 × {7 \choose 2}$ has duplicates.

For example, take case of choosing cell $(1,1) \,$ first column first row as part of $3$ selections from column $1$.

You again choose them as part of ${7 \choose 2}$ when you select cell $(1,2)$ from the first column.

The easiest way to do this will be to find number of ways to keep column $1$ empty = ${6 \choose 2}$. Similarly for column $2$. As both column $1$ and $2$ cannot be empty at the same time, your answer based on principle of inclusion exclusion is simply

$ = 126 - 2 \times {6 \choose 2} = 96$.

Title: A combination problem of picking m out of n

Body:
I am thinking of a problem of putting 4 identical beans onto 4 grids of a 3x3 chessboard. The possible number of selection is
$$
  C(9, 4) = \frac{9!}{4!5!} = 126
$$

Now assuming we need to put at least 1 bean on the first column and at least 1 bean on the second column, I want to find out the possible configuration. To solve it, I need to pick 1 spot on the first column and 1 spot on the second column, there will be 3x3 = 9 ways to do it. I then pick another 2 spots out of the remaining 9-2=7 spots for all my 4 beans. The total combination will be
$$
  3 \times 3 \times C(7,2) = 189
$$

I do not understand how come this combination is even more than the total combination (126). Any idea what I missed in the calculation? Thanks.


问:
asymptotic normality of z-estimator
I'm working on Problem 5.4.1 in Bickel and Docksum's Mathematical Statistics
Let $X_1, \dots, X_n$ be i.i.d. random variables distributed according to $P\in\mathcal{P}$. Suppose $\psi:\mathbb{R}\to\mathbb{R}$:

(i) is monotone nondecreasing

(ii)$\psi(-\infty)<0<\psi(\infty)$

(iii)$|\psi|(x)\le M<\infty$ for all $x$

And suppose for all $P\in\mathcal{P}$, $\theta(P)$ is the unique solution of $\mathbb{E}_P\psi(X_1-\theta) = 0$. Let $\hat{\theta}_n = \theta(\hat{P})$, where $\hat{P}$ is the empirical distribution of $X_1,\dots,X_n$. Now set $\lambda(\theta) = \mathbb{E}_P\psi(X_1-\theta)$ and $\tau^2(\theta) = \text{var}\psi(X_1-\theta)$. Assume $\lambda'(\theta)<0$ exists and that
$$\frac{1}{\sqrt{n}\tau(\theta)}\sum_{i=1}^n\left[\psi(X_i-\theta_n)-\lambda(\theta_n)\right] \xrightarrow{\mathcal{L}} N(0,1)$$
for every sequence $\{\theta_n\}$ with $\theta_n = \theta = t/\sqrt{n}$ for $t\in\mathbb{R}$.\

$\textbf{The problem is to show that}$
$$\sqrt{n}(\hat{\theta}_n-\theta) \xrightarrow{\mathcal{L}} N(0,\frac{\tau^2(\theta)}{[\lambda'(\theta)]^2})$$
Hint: $P(\sqrt{n}(\hat{\theta}_n-\theta)<t) = P(\hat{\theta}_n<\theta_n) = P(-\sum_{i=1}^n\psi(X_i-\theta_n)<0)$.

$\textbf{My question}$ is that I don't understand the last equality in the hint, and also how this could lead to the conclusion. Any help will be appreciated! (I manage to prove the consistency of this z-estimator $\hat{\theta}_n$, but not sure whether this could help this problem or not)

答:
Here's how to understand the last equality. The $z$-estimator $\hat\theta_n$ by definition solves the equation $$\mathbb{E}_{\hat P}\psi(X-\hat\theta_n)=0.\tag1$$ The empirical distribution $\hat P$ assigns mass $\frac1n$ to each of $X_1,\ldots,X_n$, so the above translates to
$$\frac1n\sum_{i=1}^n \psi(X_i-\hat\theta_n)=0.\tag2$$
The mapping $t\mapsto \sum_{i=1}^n\psi(X_i-t)$ is decreasing, since the function $\psi$ is increasing. Therefore the statement $$\hat\theta_n<\theta_n\tag3$$ is equivalent to the statement $$\sum \psi(X_i-\hat\theta_n)>\sum \psi(X_i-\theta_n)\tag4$$ (all sums run from $i=1$ to $n$). By (2), the LHS of (4) equals zero, so (4) is equivalent to
$$\sum \psi(X_i-\theta_n)<0.\tag5$$ (The negative sign in your version is a typo?) To compute the probability of (5), the natural manipulation is to write the algebraically equivalent statement
$$\frac1{\sqrt n\tau(\theta)}\sum\left[\psi(X_i-\theta_n)-\lambda(\theta_n)\right]<-\sqrt n\frac{\lambda(\theta_n)}{\tau(\theta)}\tag6
$$
since by assumption, the LHS of (5) converges in distribution to a standard normal as $n\to\infty$. What about the RHS? Recall that $\theta_n:=\theta + \frac t{\sqrt n}$ where $\theta$ solves $\lambda(\theta)=0$. Hence we can rearrange the RHS of (6) to the form
$$-\sqrt n\frac{\lambda(\theta_n)}{\tau(\theta)}=-\frac t{\tau(\theta)}\left(\frac{\lambda\left(\theta+\frac t{\sqrt n}\right)-\lambda(\theta)}{\frac t{\sqrt n}}\right)\tag7
$$
which by the definition of derivative converges to $\displaystyle-\frac t{\tau(\theta)}\lambda'(\theta)$ as $n\to\infty$.

Title: asymptotic normality of z-estimator


问:
How can one show that $f(x) = 3 + 5^{\sin(x)}$ is continuous on $[0, 2\pi]$ but also differentiable on $(0, 2\pi)$?
How can one show that $f(x) = 3 + 5^{\sin(x)}$ is continuous on $[0, 2\pi]$ but also differentiable on $(0, 2\pi)$? This has to do with verifying the usage of the Mean Value Theorem.

答:
Now we can use composite function properties so

$f(x)=3+x , g(x)=5^x, h(x)=sinx$

All of these functions are continuous and differentiable on R therefore

f°g°h(x)=3+5^sinx is also continuous and differentiable on R


问:
How to notate a concatenation of a finite number of sequences?
Say you have a time series sequence $A=(X_n)_{n \in I}$. There can be many subsequences of $A$, defined as follows:

$B = (X_n)_{n \in J}$ where $J \subseteq I$.

In other words, subsequence $B$ can be any of a finite number of subsequences within $A$, which we obtain by using possible subsets of the index set $I$, denoted with $J \subseteq I$.

How would one mathematically notate a new sequence that is a concatenation of all those possible subsequences that $B$ can be? This is easy enough to do with code but I don't know how to notate this correctly in mathematics.

答:
There's no standard notation for this, and you could basically define any notation you want as long as you say (with words) that this object is what you want it to be.

That said, here is one idea:

$$\prod_{J \subseteq I} A \upharpoonright J$$

Here $A \upharpoonright J$ is relatively standard notation for "the sequence you get when you restrict $I$ to $J$".

It is common to view sequences as a free monoid, and in this context concatenation is multiplication. This is why we use the symbol $\prod$.

One thing to keep in mind is that order matters, and the subsets of $I$ don't come equipped with a natural total ordering. This notation says that you're concatenating all the subsequences together, but it doesn't say what order you're concatenating them in.

So if you're going to use this, you'll want to specify how you're planning to order the subsets of $I$. If it doesn't matter for your purposes, you just want some sequence that contains every subsequence of $A$, then you can say that as well.

I hope this helps ^_^


问:
Current carrying wires
A current carrying wire takes the form of a plane circular loop of radius a. If the current in the loop is I, find the magnetic firld strength B at a point on the axis of the circle, distance b from the centre.

I think I'm supposed to use the Biot-Savart Law, but I don't know how to integrate everything in the form I have it in.

答:
The Biot-Savart law tells us that the magnetic field is given by
$$\mathbf{B} = \frac{\mu_0}{4\pi}\int_O\frac{I\mathrm{d}\mathbf{l}\times\mathbf{r}}{|\mathbf{r}|^3},$$
where $O$ is a circle of radius $a$.

Let's parametrise the circle in the usual way: $\mathbf{r}=(a\sin(\theta),a\cos(\theta),b)$ for $\theta\in[0,2\pi)$. Then the line element becomes $\mathrm{d}\mathbf{l} = (a\cos(\theta),-a\sin(\theta),0)\mathrm{d}\theta$ and
$$\mathrm{d}\mathbf{l}\times\mathbf{r}= (-ab\sin(\theta),-ab\cos(\theta),a^2)\mathrm{d}\theta,$$ thus

$$\mathbf{B} = \frac{\mu_0}{4\pi}\int_0^{2\pi}\frac{I}{\sqrt{a^2+b^2}^3}(-ab\sin(\theta),-ab\cos(\theta),a^2)\mathrm{d}\theta.$$
We may evaluate this componentwise to get
$$\frac{\mu_0 I}{4\pi\sqrt{a^2+b^2}^3}(ab\cos(\theta),-ab\sin(\theta),a^2\theta)|_0^{2\pi} = \frac{a^2\mu_0 I}{2\sqrt{a^2+b^2}^3}(0,0,1).$$

Title: Current carrying wires


问:
Is a metric space with unique balls infinite/uncountable?
Let $M$ be a non-empty metric space with the following property: if $B(x,r) = B(y,s)$, then $x=y$ and $r=s$, where $B(x,r)$ is the closed ball with center $x$ and radius $r$. Must $M$ be infinite? In fact, must $M$ be uncountable?

答:
Uncountable, no. The space of rational numbers with the usual metric has your property but is countable.

Infinite, yes. If $B(x,1),B(x,2),B(x,3),\dots$ are all different then $M$ must be infinite, since for each $n\in\mathbb N$ there is a point $y_n$ such that $d(x,y_n)\in(n,n+1]$.


问:
how can I show that for projection matrix $P^2 = P$ without applying trigonometry
I am following MIT 18.06 lectures on linear algebra and would like to understand how one can show that $P^2 = P$ for a projection matrix.

I understand that for 2-d case:

$P = \frac{a a^t}{a^t a}$

$P^2 = \frac{a a^t}{a^t a} \frac{a a^t}{a^t a}$

I assume I should easily show it by using some properties of transpose matrices/vectors, but I couldn't quite figure out how to derive it.
(I am familiar with the proof that relies on trigonometric formulas)

答:
What you are asking is indeed just some matrix calculation:

First note that $a^ta = |a|^2$ and that matrix multiplication is associative both with respect to scalars and the matrices themselves and that scalars can be put in front:

$$P^2 = \frac{a a^t}{a^t a} \frac{a a^t}{a^t a}=\frac 1{|a|^4}(a a^t)(a a^t)$$$$=\frac 1{|a|^4}(a \underbrace{(a^ta)}_{=|a|^2} a^t)=\frac{1}{|a|^2}aa^t=P$$


问:
Determining the Taylor series of $\sqrt{x}$ on $x_0 = 1$
I've attempted to determine the Taylor series of $\sqrt{x}$, centered on $x_0 = 1$. Here's what I have thus far:

$f(1) = 1$

$f'(1) = \frac{1}{2\sqrt{x}} = \frac{1}{2}$

$f''(1) = -\frac{1}{4x^{3/2}} = -\frac{1}{4}$

$f^{(3)}(1) = \frac{3}{8x^{5/2}} = \frac{3}{8}$

$\vdots$

Trivially, the Taylor series is of the form:

$$1 + \frac{1}{2(1!)}(x-1) - \frac{1}{4(2!)}(x-1)^2 + \frac{3}{8(3!)}(x-1)^3 + \cdots$$

Notice that the denominators are of the form $2^n(n!)$. We also have an alternating term, $(-1)^{n+1}$

$$\to \sum_{k=0}^{\infty} = \frac{(-1)^{k+1}(x-1)^n}{2^n(n!)}$$

When I graph this, however, it doesn't seem to be a particularly nice approximation of $\sqrt{x}$ anywhere.

Body:
Let $f(x)=\sqrt x=x^{1/2}$. Then:

$f'(x)=\frac12x^{-1/2}$;

$f''(x)=-\frac1{2^2}x^{-3/2}$;

$f^{(3)}(x)=\frac{3}{2^3}x^{-5/2}$;

$f^{(4)}(x)=-\frac{3\times5}{2^4}x^{-5/2}$

and so on… So,$$n>1\implies f^{(n)}(1)=\frac{(-1)^{n-1}(2n-3)!!}{2^n}.$$


答:
The easiest way to see this is to think about what a projection does. A projection map $P = P_{U}$ maps every vector $x$ into the subspace $U$. In particular, for any vector $u \in U$, $P(u) = u$. Then since you know that $P(x) \in U$ for all $x$, we have that for all $x$,
$$P^{2}(x) = P(P(x)) = P(x).$$

Title: how can I show that for projection matrix $P^2 = P$ without applying trigonometry

Body:
I am following MIT 18.06 lectures on linear algebra and would like to understand how one can show that $P^2 = P$ for a projection matrix.

I understand that for 2-d case:

$P = \frac{a a^t}{a^t a}$

$P^2 = \frac{a a^t}{a^t a} \frac{a a^t}{a^t a}$

I assume I should easily show it by using some properties of transpose matrices/vectors, but I couldn't quite figure out how to derive it.
(I am familiar with the proof that relies on trigonometric formulas)


问:
Show that $\lim_n \int_A f(nx)dx =\lambda(A) \frac1T \int_0^T f(t) dt$ for a measurable, bounded, T-periodic function.
I found this exercice :

Let $f : \mathbb{R} \to \mathbb{R}$ a measurable bounded and T-periodic function. Show that for every bounded borel set A, $$\lim_n \int_A f(nx)dx =\lambda(A) \frac1T \int_0^T f(t) dt$$

I can prove it when A is an bounded interval of $\mathbb{R}$. But, I can't prove the general result.

My first idea was to use that the borel set is generated by open intervals. I proved that the null set satifies the property, if $A$ and $B$ satifies the property then $A\setminus B$ also, if $(A_n)$ satifies the property and $A_n\subset A_{n+1}$ then $\cup A_n$ also. But, I wasn't able to prove that is $A_1, ..., A_N$ satisfies the property then $\cap_{i=1}^N A_n$ also.

My second idea was to use the definition of the Lebesgue measure as the infimum of lenght of covering by open intervals. But I obtained only one inequality.

答: Hint: Given $\epsilon >0$ there exists a set $B$ which is  a finite disjoint union of intervals of the type $(a,b]$ with $\int |\chi_a-\chi_b| <\epsilon$. Ref: Theorem D, p. 56 of Measure Theory by P R Halmos.


问:
Show y is odd in the equation $y^3 =x^2 +2$
Found a question which asks you to find all integer solutions $y^3 =x^2 +2$ .    In order to do so, it first asks you to show that y is odd.  How would one go about showing this?

Many Thanks

答:
If $y$ is even, $y^3\equiv0\pmod 8\equiv0\pmod 4$

So, $x^2=y^3-2\equiv-2\pmod 4\equiv2$

Now, $0^2\equiv0\pmod 4,(\pm1)^2\equiv1, (\pm2)^2\equiv0$


问:
Necessary and sufficient condition for weak convergence of gamma distribution
Let $(X_n)_n$ be a sequence of random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x),$ where $\alpha_n>0,\lambda_n>0.$

Suppose that $\alpha_n=1,\forall n \in \mathbb{N}.$ Find a necessary and sufficient condition on $(\lambda_n)_n$ such that $(X_n)_n$ converges in distribution.

More generally, find a necessary and sufficient condition on $\alpha_n,\lambda_n$ so that $(X_n)_n$ converges in distribution.

The first part is easy, it converges in distribution if and only if $0<\liminf_n\lambda_n=\limsup_n\lambda_n$.

Concerning part 2), is it true that a condition of weak convergence is the convergence of $(\alpha_n)$ and $(\lambda_n)$?

答:
Let $\{\mu_n\overset{\text{d}}{=}\Gamma(\alpha_n,\lambda_n), n\ge 1\}$ be a sequence of Gamma-distributions.

The necessary and sufficient conditions for weak convergence of Gamma-distribution
could be expressed as follows:

The necessary and sufficient conditions of $ \mu_n\Rightarrow\delta_0 $ (where $\delta_0$ is a single point probability distribution concentrated at $0$) are
\begin{equation*}
   \lim_{n\to\infty}\alpha_n\log\Big(1+\frac1{\lambda_n}\Big)= 0. \tag{1}
\end{equation*}

The necessary and sufficient conditions of $ \mu_n\Rightarrow\delta_c$ (where $\delta_c$ is a single point probability distribution concentrated at $c>0$) are

\begin{equation*}
 \lim_{n\to\infty}\lambda_n=\infty, \qquad
 \lim_{n\to\infty}\frac{\alpha_n}{\lambda_n}= c.  \tag{2}
\end{equation*}

The necessary and sufficient conditions of $ \mu_n\Rightarrow\mu\ne\delta_c(c\ge 0)$ are
\begin{equation*}
 \lim_{n\to\infty}\alpha_n=\alpha\in(0,\infty),\qquad
\lim_{n\to\infty}\lambda_n=\lambda\in(0,\infty).  \tag{3}
\end{equation*}

If $\mu$ is a probibilty distribution on $\mathbb{R}_+=[0,\infty)$, then define the Laplace
transform of $ \mu $ as follows:
$$ L(u)=\int_0^\infty e^{-ux}\mu(dx),\qquad u>0,
$$
and
$$ \psi(u) =-\log L(u).
$$
(For the Laplace transform of distribuions please refer to the
W. Feller, An introduction to Probability Theory and Its Applications(2nd Ed.),
Vol.II John Wiley & Sons, Inc.(1971). Ch.13.)
If $\mu=\Gamma(\alpha,\lambda)$ is a $\Gamma$-distribution, then it is a infinitively divisible distribution on $\mathbb{R}_+$, and its  Laplace transform is
$$ L(u)=\Big(1+\frac{u}{\lambda}\Big)^{-\alpha}.
$$
and
\begin{gather*}
 \psi(u)=-\log L(u)=\alpha\log\Big(1+\frac{u}{\lambda}\Big). \tag{4}\\
 \alpha=\frac{\psi(u)}{\log\Big(1+\dfrac{u}{\lambda}\Big)}\qquad
 \lambda=\frac{u}{e^{\psi(u)/\alpha}-1}.  \tag{5}
\end{gather*}
In particular,
$$ \psi_{\delta_0}(u)=0,\qquad \psi_{\delta_c}(u)=cu.
$$
For $\mu_n\overset{w}{\to}\mu^\ast$, iff the following limit is true(c.f. Feller's book),
\begin{gather*}
 L_n(u)\to L^\ast(u),\qquad \forall u\ge0,\\
 \begin{aligned}
  \psi_n(u)&=-\log L_n(u)=\alpha_n\log\Big(1+\frac{u}{\lambda_n}\Big)\\
  &\to\psi^\ast(u)=\log L^\ast(u),\qquad
  \forall u\ge0.
 \end{aligned} \tag{6}
\end{gather*}
Using the similar method for characteristic functions we could also deduce the convergence in (6) is uniform at $ u=0 $, that is
\begin{equation*}
 \lim_{n\to\infty}\psi_n(u_n)=0.\qquad \forall u_n\to0. \tag{7}
\end{equation*}
Firstly, to prove the sufficiency of (1), it is suffice to use following fact:
\begin{align*}
  0<c_1(u)&\le \inf_{\lambda>0}\frac{\log\Big(1+\dfrac{u}{\lambda}\Big)}{\log\Big(1+\dfrac1{\lambda}\Big)}\\
&\le \sup_{\lambda>0}\frac{\log\Big(1+\dfrac{u}{\lambda}\Big)}{\log\Big(1+\dfrac1{\lambda}\Big)}
\le C_2(u)<\infty,\qquad \forall u>0.
\end{align*}

The proofs of sufficiency of (2),(3)  are direct.

Next the necessity discussed only.
In the following we always assume $\mu_n\overset{w}{\to}\mu^\ast$ and (6) holds.

A. If $ \psi^\ast(u)\ne0 $, then
\begin{equation*}
 \varliminf_{n\to\infty} \alpha_n>0,\qquad \varliminf_{n\to\infty} \lambda_n>0. \tag{8}
\end{equation*}

\textit{Proof}:
Prove (8) using contradictory. If (8) is not true, then there exists a subsequence
$ \alpha_{n_k}(\text{or }\lambda_{n_k})\to 0 $. From (6),
\begin{gather*}
 \lambda_{n_k}=\frac{\psi_{n_k}(u)}{e^{\psi_{n_k}(u)/\alpha_{n_k}}-1}\to0.\\
 \Big(\text{or } \alpha_{n_k}=\frac{\psi_{n_k}(u)}{\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)}\to0.\Big)
\end{gather*}
Hence both $\alpha_{n_k}\to 0$, $\lambda_{n_k}\to 0$. If $\psi^\ast(a)\ne 0$,
\begin{equation*}
  \lim_{k\to\infty} \alpha_{n_k}\log\Big(1+\frac{a}{\lambda_{n_k}}\Big)
  =\lim_{k\to\infty} \psi_{n_k}(a)=\psi^{\ast}(a)\ne0.    \tag{9}
\end{equation*}

Now take $ u_{n_k}=\sqrt{\lambda_{n_k}^2+a\lambda_{n_k}}-\lambda_{n_k} $,
then $ u_{n_k}\to0 $ and
\begin{align*}
 \psi_{n_k}(u_{n_k})&=\alpha_{n_k}\log\Big(1+\frac{u_{n_k}}{\lambda_{n_k}}\Big)\\
 &=\frac12\alpha_{n_k}\log\Big(1+\frac{a}{\lambda_{n_k}}\Big)
 \to \frac12\psi^\ast(a)\ne0 \tag{10}
\end{align*}

(10) contradicts with (7). Hence (8) holds.

B. If $ \psi^\ast(u)\ne0 $ and there exists a subsequence  $\alpha_{n_k}\to\infty$(or $ \lambda_{n_k}\to\infty $ ), then
\begin{gather*}
 \lim_{n\to\infty}\alpha_n=\lim_{n\to\infty}\lambda_n=\infty,\\
 \lim_{n\to\infty}\frac{\alpha_n}{\lambda_n}=c, \qquad \psi^\ast(u)=cu.
\end{gather*}

\textit{Proof}: From (6), if $\alpha_{n_k}\to\infty$(or $ \lambda_{n_k}\to\infty $ ), then
\begin{equation*}
 \lambda_{n_k}=\frac{\psi_{n_k}(u)}{e^{\psi_{n_k}(u)/\alpha_{n_k}}-1}\to\infty,\quad
 \Big(\text{or } \alpha_{n_k}=\frac{\psi_{n_k}(u)}{\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)} \to\infty\Big)
\end{equation*}
and
\begin{align*}
 \lim_{k\to\infty}\frac{\alpha_{n_k}}{\lambda_{n_k}}
 &=\lim_{k\to\infty}\frac{\psi_{n_k}(u)}{\lambda_{n_k}\log\Big(1+\dfrac{u}{\lambda_{n_k}}\Big)}\\
 &=\frac{\psi^\ast(u)}{u}\ne0,
\end{align*}
$$ \psi^\ast(u)=cu.$$
In this case, if there exists a convergent subsequence $ \{\alpha_{n_k},k\ge 1\} $ (or $\{\lambda_{n_k},k\ge 1\}$), but not converge to $ \infty $, then the  $\psi_{n_k}(u)\to\psi^{\ast\ast}(u)\ne cu=\psi^\ast(u) $, so doesn't exist any subsequence $ \{\alpha_{n_k},k\ge 1\} $ (nor $\{\lambda_{n_k},k\ge 1\} $) which converge to a finite limit. B holds.

C. If $ \psi^\ast(u)\ne cu(c\ge 0) $, then
\begin{gather*}
  \lim_{n\to\infty}\alpha_n=\alpha^\ast\in(0,\infty),\\
\lim_{n\to\infty}\lambda_n=\lambda^\ast\in(0,\infty).\\
\psi^\ast(u)=\alpha^\ast\log\Big(1+\frac{u}{\lambda^\ast}\Big).
\end{gather*}

\textit{Proof}: From {\bfseries B, C},  if $ \psi^\ast(u)\ne cu(c\ge 0) $, then
\begin{gather*}
 0<\varliminf_{n\to\infty} \alpha_n\le \varlimsup_{n\to\infty} \alpha_n<\infty,\\
 0<\varliminf_{n\to\infty} \lambda_n\le \varlimsup_{n\to\infty} \lambda_n<\infty.
\end{gather*}
This means that $S=\{(\alpha_n,\lambda_n), n\ge 1\} $ is a relative compact set in  $(0,+\infty)\times(0,+\infty)$(The closure $\overline{S}\subset(0,+\infty)\times(0,+\infty)$ and $\overline{S}$ is compact). Now we prove that $ A $ has unique limit point.
Suppose that  $\{(\alpha_{n_k},\lambda_{n_k}), k\ge 1\} $ is a convergent subsequence
in $ S $ and
\begin{align*}
 \alpha^\ast=\lim_{k\to\infty}\alpha_{n_k},\qquad \lambda^\ast=\lim_{k\to\infty}\lambda_{n_k}
\end{align*}
then
$\mu^*\overset{\text{d}}{=}\lim\limits_{k\to\infty}\mu_{n_k}$ is a $ \Gamma(\alpha^\ast,
\lambda^\ast) $-distribution and
\begin{equation*}
 \alpha^\ast=\frac{(\mathsf{E}_{\mu^\ast}[X])^2}{\mathsf{V}_{\mu^\ast}[X]},\quad
 \lambda^\ast=\frac{\mathsf{E}_{\mu^\ast}[X]}{\mathsf{V}_{\mu^\ast}[X]}.
\end{equation*}
This also means that for $ S $ there exits unique limits point $(\alpha^\ast,\lambda^\ast)$ and $ \mu^\ast\overset{\text{d}}{=}\Gamma(\alpha^\ast,\lambda^\ast) $.  C holds.

From above facts, it is easy to get the necessary conditions for weak convergence of Gamma-distribution.


问:
Proving $3\times3$ nilpotent matrix satisfies $A^3=0$
Let $A$ be a $3$ $\times$ $3$ matrix. Prove that if there exists $n$ $>$ $3$ such that $A^n$= $0$, then $A^3$ = $0$. Can someone help explain how to prove this? I know that since A is nilpotent, its only eigenvalue ($\lambda$) is $0$. So, I was thinking that I could argue the following: $A^3$$\dot{\vec{v}}$=$\lambda^3$$\dot{\vec{v}}$=$\dot{\vec{0}}$, which implies that $A^3$=$0$. I don't think this proof is fully correct, however, because it doesn't use the fact that $n>3$. I could have made the same argument to suggest that $A^2=0$, which is not necessarily true when a $3$ $\times$ $3$ matrix A is nilpotent.

答:
Pick an arbitrary nonzero vector $v$. Let $f$ be the monic polynomial of the lowest degree such that $f(A)v=0$. Such $f$ must exist because $A^nv=0$. Also, $k:=\deg(f)\le3$ because $v,Av,A^2v,A^3v$, being four vectors in a three-dimensional vector space, must be linearly dependent.

However, $f$ must divide $x^n$, otherwise $g=\operatorname{gcd}(f,x^n)$ will satisfy $g(A)v=0$ (because $g=af+bx^n$ for some polynomials $a$ and $b$) but $\deg(g)<\deg(f)$, which contradicts minimality of $\deg(f)$. Therefore $f(x)=x^k$ for some $k\le 3$. Hence $A^3v=A^{3-k}f(A)v=0$ and we conclude that $A^3=0$ because $v$ is arbitrary.


答: What do you worry about ? The four first terms that you computed do match the square root fairly closely.

Title: Determining the Taylor series of $\sqrt{x}$ on $x_0 = 1$

Body:
I've attempted to determine the Taylor series of $\sqrt{x}$, centered on $x_0 = 1$. Here's what I have thus far:

$f(1) = 1$

$f'(1) = \frac{1}{2\sqrt{x}} = \frac{1}{2}$

$f''(1) = -\frac{1}{4x^{3/2}} = -\frac{1}{4}$

$f^{(3)}(1) = \frac{3}{8x^{5/2}} = \frac{3}{8}$

$\vdots$

Trivially, the Taylor series is of the form:

$$1 + \frac{1}{2(1!)}(x-1) - \frac{1}{4(2!)}(x-1)^2 + \frac{3}{8(3!)}(x-1)^3 + \cdots$$

Notice that the denominators are of the form $2^n(n!)$. We also have an alternating term, $(-1)^{n+1}$

$$\to \sum_{k=0}^{\infty} = \frac{(-1)^{k+1}(x-1)^n}{2^n(n!)}$$

When I graph this, however, it doesn't seem to be a particularly nice approximation of $\sqrt{x}$ anywhere.


问:
if $|f''(x)|\le |f'(x)|,\forall x\in R$ show that either $f'(x)>0$ or $f'(x)<0,\forall x\in R$
Edit:let $f:R\to R$ be twice differenriable on $R$, such that
$$|f''(x)|\le |f'(x)|,\forall  x\in R$$
show that either:
$f'(x)>0$
or $f'(x)<0,\forall x\in R$

since
$$0\ge (f''(x))^2-(f'(x))^2=(f''(x)-f'(x))(f''(x)+f'(x))$$
so
$$f"(x)+f'(x)\ge 0,\mbox{and} f''(x)-f'(x)\le 0$$
or
$$f"(x)+f'(x)\le 0,\mbox{and} f''(x)-f'(x)\ge 0$$

答: It is just an application of the Gronwall's lemma. Since $\left|f^{\prime}\left(x\right)\right|>\left|f^{\prime\prime}\left(x\right)\right|$ then $\left|f^{\prime}\left(x\right)\right|>f^{\prime\prime}\left(x\right)$. Now, assume there are $a,b\in\mathbb{R}$ such that $f^{\prime}\left(a\right)=0$ and $f^{\prime}$ is positive in $\left(a,b\right)$ Hence, for every $x\in\left[a,b\right)$ we have $f^{\prime}\left(x\right)>f^{\prime\prime}\left(x\right)$ and so, by the Gronwall's lemma, $$f^{\prime}\left(x\right)\leq f^{\prime}\left(a\right)e^{x}=0,\,\forall x\in\left[a,b\right)$$ which is a contradiction. The other cases are similar.


问:
Algebraic expression depending if variable is even or odd
I'd like to find an algebraic expression for $f(p)$ that meets the following:

$$
f(p) = \begin{cases}
1 \, \, \text{if $p$ is even} \\
0 \, \, \text{if $p$ is odd}
\end{cases}
$$

答:
As suggested by user Kavi Rama Murthy, a possible solution is

$\frac{1+(-1)^p}{2}$.


问:
Interesting real integrals from $ I=\oint\limits_{C}\sin(\frac{1}{z})\,\mathrm{dz} = 2\pi i $
I have separated and equated Real and Imaginary Parts of the following equation by plugging $\mathrm{z}=e^{it}$    :

$$I=\oint\limits_{C}\sin\frac{1}{z}\,\mathrm{dz} = 2\pi i,$$

where $C$ is the boundary of unit circle centered at origin.

which gives $$ K=\int_{-\pi}^{+\pi}e^{it}\sin(e^{-it})\mathrm{dt} = 2\pi$$

simplifying gives the following:-

$\int_{0}^{\frac{\pi}{2}}\cos(t)\sin(\cos(t))\cosh(\sin(t))+\sin(t)\cos(\cos(t))\sinh(\sin(t))\mathrm{dt} = \frac{\pi}{2}$

the first and second part of the above integral translates to $I_1$ and $I_2$ respectively under a simple substitution for each.

$$ I_1 = \int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\mathrm{dx}\\ I_2 = \int_{0}^1 \cos(x)\sinh(\sqrt{1-x^2})\mathrm{dx}$$

which implies, $I_1 + I_2 = \frac{\Large\pi}{2}$

Interestingly, I have found that $I_1 $ and $ I_2$ are equal (using a calculator), but stuck proving them.
Any ideas how they are equal or finding another method to calculate $I_1 $ or $ I_2$?

EXTRA: some integrals that show up but eventually cancel out( as they are odd functions, $f(x)=f(-x)$ ) in calculating Imaginary part of equation $K$:-

$$ I_3 = \int_{0}^1 \sin(x)\cosh(\sqrt{1-x^2})\mathrm{dx}\\ I_4 = \int_{0}^1 \sinh(x)\cos(\sqrt{1-x^2})\mathrm{dx}$$

however $I_3 $ and $ I_4$ are not equal and their numerical values are approx. 0.584 and 0.418 , can $I_3 $ and $ I_4$ be calculated in closed form?

答:
The value of the integral $$I_{3} = \int_{0}^{1} \sin(x) \cosh \left( \sqrt{1-x^{2}} \right) \, \mathrm dx $$ can  be expressed in terms of the cosine integral $\operatorname{Ci}(x)$ and the hyperbolic cosine integral $\operatorname{Chi}(x)$.

$$ \begin{align} I_{3} &= \int_{0}^{\pi/2} \sin(\sin t) \cosh(\cos t) \cos(t) \, \mathrm dt \\ &= \Im \int_{0}^{\pi/2} \sinh(e^{it}) \cos (t) \, \mathrm dt \\ &= \Im \int_{C} \sinh(z) \,  \frac{z+\frac{1}{z}}{2} \frac{dz}{iz} \\ &= -\frac{1}{2} \, \Re \int_{C} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz, \end{align}$$
where $C$ is the portion of the unit circle in the first quadrant of the complex plane.

But since the integrand is analytic in the first quadrant, we have$$ \begin{align} I_{3} &= - \frac{1}{2} \, \Re \int_{1}^{i} \left( \sinh(z) + \frac{\sinh (z)}{z^{2}} \right) \, \mathrm dz \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} \Bigg|^{i}_{1} + \int_{1}^{i} \frac{\cosh (z)}{z} \, \mathrm dz \right) \\ &= - \frac{1}{2} \Re \left(\cosh (z) - \frac{\sinh (z)}{z} + \operatorname{Chi}(z) \Bigg|_{1}^{i} \right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \Re \left(\operatorname{Chi}(i)  \right)- \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= - \frac{1}{2} \left(\cos(1)- \sin(1) + \operatorname{Ci}(1) - \cosh(1) + \sinh(1) - \operatorname{Chi}(1)\right) \\ &= \frac{1}{2} \left(\sin(1) - \cos(1) - \operatorname{Ci}(1) + \frac{1}{e} + \operatorname{Chi}(1) \right) \\ & \approx 0.58475. \end{align}$$

The evaluation of  $$I_{4} = \int_{0}^{\pi/2} \sinh(\sin t) \cos(\cos t) \cos (t) \, \mathrm dt =  \Im \int_{0}^{\pi/2} \sin(e^{it}) \cos(t) \, \mathrm dt$$ should be similar.


问:
Integrating a second-order partial differential equation
I haven't dealt with solving many partial differential equations before. How would I go about solving:

$$
A = \frac{\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} v}{\partial y^{2}}
$$

where A is a constant.

Ideally, I could get an equation of  v(x,y)  but I am not sure if that's possible. If there isn't a "pretty" solution to this. How could I go about approximating the solution?

Thanks

答:
$$
\frac{\partial^{2} v}{\partial x^{2}} + \frac{\partial^{2} v}{\partial y^{2}}=A
$$Without boundary condition one cannot recommend a specific convenient method.

An obvious form of solution is for example :
$$v(x,y)=\frac{A}{2}x^2+f(x+iy)+g(x-iy)$$
$f$ and $g$ are arbitrary functions.

$\frac{\partial^{2} v}{\partial x^{2}}=A+f''(x+iy)+g''(x-iy)$

$f''(X)=\frac{d^2f(X)}{dX^2}$

$\frac{\partial^{2} v}{\partial y^{2}}=(i^2)f''(x+iy)+(-i)^2g''(x-iy)=-f''(x+iy)-g''(x-iy)$

$\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}=\left(A+f''(x+iy)+g''(x-iy)\right)+\left(-f''(x+iy)-g''(x-iy)\right)=A\quad$ which satisfies the PDE.


问:
The law of large numbers - limits of $\max$ vs $\max$ of a limit.
Assume that $X_{1,1}, \dots , X_{1,n}, X_{2,1},\dots, X_{2,n}, \dots ,X_{n,1}, \dots , X_{n,n}$ are i.i.d. random variables, and that $\mathbb EX_{i,j}$ exists and is finite. From the strong law of large numbers we have
$$\max_i\left\{\lim_{n\to\infty}\left\{\frac{1}{n}\sum_{j=1}^{n}X_{i,j}\right\}\right\} \overset{a.s.}{=} \mathbb EX_{i,j} ,$$
because $\max_i$ is redundant. However, do we have
$$\lim_{n\to\infty}\left\{\max_i\left\{\frac{1}{n}\sum_{j=1}^{n}X_{i,j}\right\}\right\}  \overset{a.s.}{=} \mathbb EX_{i,j} ? $$
I.e., can  we change the order of $\max$ and $\lim$ without affecting the result?</p>

答:
IF the all moments exist, then the answer is as follows.

Without loss of generality (and by doing and affine transformation $X\rightarrow aX+b$),  I assume
$$\mathbb{E}[X]=0; \hspace{3mm}\mathbb{E}[X^2]=1$$
Then, the moment generating function
$$g(s)\equiv \log\mathbb{E}[e^{sX}]$$
would satisfy
$$g(0)=0; \hspace{3mm}g^\prime(0)=0; \hspace{3mm}g^{\prime\prime}(0)=1$$
For each $i$, the variate $Y_i\equiv\frac1n\sum_jX_{ij}$ has
$$f(s)\equiv\log\mathbb{E}[e^{sY}]=ng(s/n)=\frac{s^2}{2n}+\mathcal{O}(\frac{1}{n^2})$$
Therefore the density for $Y$ is
$$p(y)=\sqrt{\frac{n}{2\pi}}e^{-\frac{ny^2}{2}}+\frac{e_n(y)}{n^2}$$
For some finite function $e_n(y)$. Now the question is: What is the asymptotic behavior of
$$Z\equiv\max_{i\in[n]}Y_i$$
Let's start by the cumulative distribution function
$$F_Z(z)=F_Y^n(z)=\Big(F_G(z\sqrt n)+\frac{e_n(z)}{n^2}\Big)^n$$
$$= F_G^n(z\sqrt n)+\frac{e_n(z)}{n}+\mathcal{O}(\frac{1}{n^2})$$
where $F_G$ is the cumulative distribution for a standard gaussian. This approximation holds for all real $z$ unless $|z|\ll \frac{1}{\sqrt{n}}$.

Next, let us try to calculate the moments for $Z$ using known results about the maximum of a set of independent gaussian variables.
$$\mathbb{E}[Z]=\int z F_Z^\prime(z)dz=\frac{C\log n}{\sqrt{n}}+\mathcal{O}(\frac{1}{n})$$
$$\mathbb{E}[Z^2]=\int z^2 F_Z^\prime(z)dz=\frac{C\log^2 n}{n}+\mathcal{O}(\frac{1}{n})$$
These mean
$$\boxed{Z\overset{a.s.}{=}0}$$
There are (small) gaps in my solution but the bounty is due, I had to hurry!


问:
Let $ABCD$ be a quadrilateral in a circle with diameter $AC$, and let $DE \perp AB$ with $E$ on $AB$. If $AD = DC$, $[ABCD] = 24$ cm$^2$.

Let $ABCD$ be a quadrilateral in a circle with diameter $AC$, and let $DE \perp AB$ with $E$ on $AB$. If $AD = DC$, $[ABCD] = 24$ cm$^2$. Find $(DE)^2$.

What I Tried: Here is a picture :-

This problem has been eating my brain for quite some time now, and I tried it in many ways but in the end I succeeded doing it. The picture shows it.

First, we have that $BC$ $||$ $ED$ , which is obvious from the information. Next is that since $AD = DC$ , $BD$ is the angle bisector of $\angle BAC$ which gives $BED$ an isosceles triangle and $DE = BE$. Now draw a perpendicular from $C$ to $DE$ and let it intersect at $F$ . We have that $\Delta AED \cong \Delta DFC$ from $RHS$ , and this gives all the necessary piece of info in the picture, also noting that $CFEB$ is a rectangle. This total areas of the $2$ triangles and the rectangle gives the area of the quadrilateral, which looks really good.

From here :-
$$[ABCD] = 2[\Delta AED] + [CFEB]$$
$$\rightarrow 24 = x(x + y) + y(x + y)$$
$$\rightarrow 24 = (x + y)^2$$
$$\rightarrow (x + y) = DE = \sqrt{24}$$

Hence $DE^2 = 24$ .

This solution was so elegant that I couldn't resist posting it in Stack Exchange, maybe others will be helpful with this.

As a note, I will be waiting for some other solutions too :) .

答:
Here is an alternative solution.

First, we note that $ABC$ and $ADC$ are right triangles, and angles $A$ and $C$ in triangle $ADC$ are $45^{\circ}$.

We may also note that $\widehat{DBA}$ faces the same arc as $\widehat{DCA}$ , therefore it should be $45^{\circ}$ too. As a result, in the right triangle $DEB$ we have $DE = EB$  (This interesting information is not part of my solution but I just added it for aesthetic reasons).

Let's add $YXA$ , a rotated copy of the triangle $ABC$ to the shape, as shown in the figure below.

$XA$ is on the same line as $BA$ . We have:
$$\widehat{BCA}=\widehat{XAY}$$
$$\widehat{BCA} + \widehat{BAC} = 90^{\circ}$$
It follows that
$$\widehat{XAY} + \widehat{BAC} = 90^{\circ}$$
$$\widehat{CAY} = 180 - (\widehat{XAY} + \widehat{BAC}) = 90^{\circ}$$
$$\widehat{DAY} = \widehat{CAY} - \widehat{DAC} = 45^{\circ}$$
The triangles $ADC$ and $ADY$ have an equal angle ($\widehat{DAC} = \widehat{DAY} = 45^{\circ}$) with equal sides ($AC=AY$ and they share $AD$) , therefore they are equal and:
$$DY=DC$$
$$\widehat{ADY} = \widehat{ADC} = 90^{\circ}$$
This means that $Y$ , $D$ and $C$ are on the same line. Moreover, this line is divided into equal segments by the three parallel lines $CB$ , $DE$ and $YX$ . In other words, in the trapezoid $CBXY$ , line segment $DE$ , which is parallel to the bases, bisects the legs $CY$ and $BX$ . It follows that
$$DE = \frac{1}{2}(CB + YX)$$
And recalling that $YX = AB$ , we have:
$$DE = \frac{1}{2}(CB + AB)$$

Now we can move on to the calculation of area.
$$S_{ABCD} = S_{ABC} + S_{ADC}$$
$$= \frac{1}{2}(AB.CB + AD^2)$$
$$= \frac{1}{2}(AB.CB + \frac{1}{2}AC^2)$$
$$= \frac{1}{2}(AB.CB + \frac{1}{2}(AB^2 + CB^2))$$
$$= \frac{1}{4}(AB + CB)^2$$
$$= DE^2$$
Voila: $S_{ABCD} = DE^2$ ! (This is an exclamation mark, not a factorial)


问:
Central Limit Theorem and Lightbulbs
"Suppose I have lightbulbs that have a lifetime which is exponentially distributed with an average lifespan of 5 years. Each time a bulb dies, it is replaced with an another of the same type.

What is the probability that in 10 years, at least three bulbs break?"

So far I have that $\mu=5$, E$[X]=\frac{1}{5}=0.2$, Var$[X]=\frac{1}{\lambda^2}=\frac{1}{25}=0.04$.

My next step was going to be approaching the problem by plugging in these values into the formula for the central limit theorem, namely:

$\chi=\frac{N-0.2}{0.04}$

But this just returns $245$ when I plug in $N=10$ years which doesn't seem to be right. If anyone could explain where my understanding of the topic is failing and where I'm going wrong that would be great.

答:
Multiple Exponential Waiting Time as Poisson or Gamma

Poisson (as in Comments by @MatthewPilling and me): The ten-year mean is $\lambda_{10} = 10(1/5) = 2.$
Let $X \sim \mathsf{Pois}(\lambda_{10}=2),$ then you
seek $P(X \ge 3) = 1 - P(X \le 2) = 0.3233.$ In R, where
ppois is a Poisson CDF:

```<code>1 - ppois(2, 2)
[1] 0.3233236
</code>```

Gamma (equivalently): The waiting time for three successive
exponential failures (mean=5, rate=1/5) is
$W\sim\mathsf{Gamma}(\mathrm{shape}=3,\mathrm{rate}=1/5).$
You seek $P(W \le 10) = 0.3233.$
In R, where pgamma is a gamma PDF:

```<code>pgamma(10, 3, 1/5)
[1] 0.3233236
</code>```

Simulation:

```<code>set.seed(1112)
w = replicate(10^6, sum(rexp(3,.2)))
mean(w <= 10)
[1] 0.323841

hist(w, prob=T, br=60, col="skyblue2",
     main="Wait for 3")
 curve(dgamma(x, 3, .2), add=T, lwd=2, col="orange")
 abline(v=10, lty="dotted", lwd=3, col="darkgreen")
</code>```

Title: Central Limit Theorem and Lightbulbs

Body:
Multiple Exponential Waiting Time as Poisson or Gamma

Poisson (as in Comments by @MatthewPilling and me): The ten-year mean is $\lambda_{10} = 10(1/5) = 2.$
Let $X \sim \mathsf{Pois}(\lambda_{10}=2),$ then you
seek $P(X \ge 3) = 1 - P(X \le 2) = 0.3233.$ In R, where
ppois is a Poisson CDF:

1 - ppois(2, 2)
[1] 0.3233236

Gamma (equivalently): The waiting time for three successive
exponential failures (mean=5, rate=1/5) is
$W\sim\mathsf{Gamma}(\mathrm{shape}=3,\mathrm{rate}=1/5).$
You seek $P(W \le 10) = 0.3233.$
In R, where pgamma is a gamma PDF:

pgamma(10, 3, 1/5)
[1] 0.3233236

Simulation:

set.seed(1112)
w = replicate(10^6, sum(rexp(3,.2)))
mean(w <= 10)
[1] 0.323841

hist(w, prob=T, br=60, col="skyblue2",
     main="Wait for 3")
 curve(dgamma(x, 3, .2), add=T, lwd=2, col="orange")
 abline(v=10, lty="dotted", lwd=3, col="darkgreen")


问:
Show eigenvalues of a positive definite matrix are larger than eigenvalues of another positive definite matrix
Let $A,B \in \mathbb{R}^{n \times n}$ be two symmetric matrices with $A$ being positive semidefinite. Matrix $B$ may not positive semidefinite but all its diagonal entries are positive.

Let $\{\mu_i\}_{i=1}^{n}$ be eigenvalues of matrix $A + B$ and $\{\nu_i\}_{i=1}^{n}$ be eigenvalues of matrix $A - B$; both are sorted in decreasing order. I can prove both matrices $A+B$ and $A-B$ are positive semi-definite, and my simulation shows that it is always the case that $\mu_i \geq \nu_i \geq 0$ for any $1 \leq i \leq n$.

I would like to formally prove that $\mu_i \geq \nu_i$ for each $1 \leq i \leq n$. So I just wanted to ask for any advice that, in general, what are the sufficient conditions that people commonly use to prove similar things? Any suggestions are very welcome!

答:
This is not true. Random counterexample: $A=I$ and
$$
B=\pmatrix{1&2&-2\\ 2&1&-2\\ -2&-2&1}.
$$
The sorted eigenvalues of $A+B$ are $6,0,0$ and the sorted eigenvalues of $A-B$ are $2,2,-4$.


问:
Total variation distance between pairs when one variable is independent of the rest
Let $\Delta(\cdot, \cdot)$ denote total variation distance. Let $X, Y, U, V$ be random variables, where $U$ is independent of $(X, Y, V)$. Does this imply $\Delta((X, Y), (U, Y)) \leq \Delta((X, Y), (U, V))$? For simplicity, assume all the random variables are discrete.

Surely there is a simple proof or a simple counterexample, but so far I can't find either. If we drop the independence assumption, it's easy to find counterexamples, e.g., sample $A \in \{0, 1\}$ uniformly at random and let $X = Y = A$ and $U = V = 1 - A$.

答:
As mentioned in the comments, to remove some notational ambiguity I am rephrasing your question as

Let $p$ be a probability distribution on some discrete space $\mathcal{X}\times\mathcal{Y}$, with marginals $p_X$ and $p_Y$; and let $u,v$ be two probability distributions on $\mathcal{X}$ and $\mathcal{Y}$ respectively. Is it always the case that $\Delta(p, u\otimes p_Y) \leq\Delta(p, u\otimes v)$
?

The answer is no. Take for instance $\mathcal{X}=\mathcal{Y}=\{0,1\}$, and $p,u,v$ defined as follows:
$$
p(0,0) = p(1,1)=0, \quad p(0,1)=p(1,0)=\frac{1}{2}
$$
and
$$
u(0)=\frac{1}{10}, \;u(1)=\frac{9}{10} \qquad\text{and}\qquad v(0)=\frac{9}{10}, \;v(1)=\frac{1}{10}
$$
Note that the marginal $p_Y$ of $p$ is uniform on $\{0,1\}$.

Then, we can compute
$$
2\Delta(p, u\otimes p_Y) = |\frac{1}{10}\cdot\frac{1}{2}-0|+|\frac{1}{10}\cdot\frac{1}{2}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{2}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{2}-0|=1
$$
while
$$
2\Delta(p, u\otimes v) = |\frac{1}{10}\cdot\frac{9}{10}-0|+|\frac{1}{10}\cdot\frac{1}{10}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{9}{10}-\frac{1}{2}|+|\frac{9}{10}\cdot\frac{1}{10}-0|=\frac{98}{100}
$$
so $\boxed{\Delta(p, u\otimes p_Y) > \Delta(p, u\otimes v)}$.


问:
Show a series is analytic
Let
$$
f(z) = \sum_n z^{n!}
$$

I am trying to show that $f(z)$ is analytic on the open unit disk ${|z| <1}$.

My thought is to find its radius of convergence $R$, but this series does not have a form of a power series (which I learned how to find $R$ from the various tests) so I am not sure if I am doing it correctly.

答: You can express you serie as : $\sum_n b_n z^n$ with $b_n = n$ if $n = k!$ for some $k$ and $b_n = 0$ for the other cases. Then, according to the Hadamard theorem (https://en.wikipedia.org/wiki/Cauchy%E2%80%93Hadamard_theorem) the radius of convergence $R$ satistfies $\frac1R = \limsup |b_n|^{\frac1n} = 1 $, thus $R = 1$.

Title: Show a series is analytic


问:
Determine the minimum of $\frac{\int_0^1{x^2\left( f'\left( x \right) \right) ^2 dx}}{\int_0^1{x^2\left( f\left( x \right) \right) ^2dx}}$
For all non-zero continuously differentiable function $f:[0,1]\to\mathbb R,f(1)=0$, determine the minimum of
$$\dfrac{\displaystyle \int_0^1{x^2\left( f'\left( x \right) \right) ^2\mathrm dx}}{\displaystyle \int_0^1{x^2\left( f\left( x \right) \right) ^2\mathrm dx}}$$

I totally have no idea how to start. Can anyone help?

答:
A Calculus of Variations approach:

Since multiplying $f$ by a constant does not change the minimum or the boundary condition, the denominator can be scaled to any constant, to give a constrained minimization problem $$\textrm{Minimize }\int_0^1x^2f'(x)^2\,dx\textrm{ subject to }f(1)=0, \int_0^1x^2f(x)^2\,dx=c$$
Let the Lagrangian function be $L=x^2(f')^2-\lambda x^2f^2$. Then the Euler-Lagrange equation $\frac{d}{dx}\frac{\partial L}{\partial f'}=\frac{\partial L}{\partial f}$ becomes $$f''+\frac{2}{x}f'+\lambda f=0$$ This has the solution $f(x)=A\frac{\sin\sqrt{\lambda}x}{x}$ (the second solution is divergent at $x=0$).

Substituting $f(1)=0$, gives $\sqrt{\lambda}=\pi$, hence the minimum is $$\frac{\int_0^1x^2f'(x)^2\,dx}{\int_0^1x^2f(x)^2\,dx}=\frac{\pi^2/2}{1/2}=\pi^2$$

Edit: Strictly speaking $\sqrt{\lambda}=n\pi$ but the cost function is then $n^2\pi^2$, so the minimum occurs for $n=1$. Also, substituting $f=g/x$ simplifies the differential equation.


问:
Entropy contribution from variable length segment of a sequence.

If I have a sequence which is comprised of one of $10$ prefixes, one of $5$ suffixes and a variable length middle, how do I compute the entropy of the sequence?

Using Shannon-Entropy $$H= -\sum_{i=1}^{m} p_i \ln(p_i)$$

I can compute the entropy contributions from the $10$ starting and $5$ ending sequences using this sum, but I am unsure how to computer the variable length section.
If the distribution of length is:

$0=0.5$

$1=0.25$

$2=0.2$

$3=0.05$

And each character can be A-F say, for $16$ possibility with equal probability how do I calculate the number of bits of entropy? If I simply use the sum above, I am underestimating the contribution from the longer sequences, which have more possibilities. Do I also need to ignore the $p_0$ term, as $0.5\ln(0.5)$ isn't actually adding any additional characters to the sequence.

答:
If all $16$ possibilities has an equal probability, that entropy must $\ln 16\approx 2.77$
its; $H\sim \max= \sum[ \frac1n \ln(\frac1n) ]$. Adding new information with "equal" probability is not add extra information into entropy.</p>


问:
$f(f(x))=x \,\forall x\in [0,1], f(0)=1$and $f$ is differentiable.Find $f$.
$f(f(x))=x \,\forall x\in [0,1], f(0)=1$and $f$ is differentiable.Find $f$.

I can guess two such

$f(x)=1-x, f(x)=\frac {1-x}{ax+1}$ where $a > -1$ is some real.

I got one general solution for $f(f(x))=x$.

Suppose $h(x)$ is one such function and g be any invertible  function
then $f(x)=g(h(g^{-1}(x)))$

答:
Hinnt:

$f(x)=\frac{ax+b}{cx-a}, x\ne c/a$ follows $f(f(x))=x$

$f(x)=(1-x^k)^{1/k}$ follows $f(f(x))=x$.


问:
Prove $\mu_{\ast}(A\cap E)=0$ and $\mu^{\ast}(A\cap E)=\mu(E)$
Show that there exists a non-measurable set $A\subseteq \mathbb{R}$ such that:

$\mu_{\ast}(A\cap E)=0$ and $\mu^{\ast}(A\cap E)=\mu(E)$

For each measurable set E with $\mu(E)>0$.

So my attempt was trying to get a Vitali set (which is known to be unmeasurable) and constructing a set E that can have inner measure 0 and outer measure equals its Lebesgue measure, but I'm stuck in constructin a good set E that can satisfy these conditions.

答:
I'm sure there's a simpler answer to this question, but this was the best I could come up with.

One can modify the Vitali set construction in $[0,1]$ such that the resulting nonmeasurable set $V$ has outer measure $1$ and inner measure $0$. This problem is discussed at length here. Then put
$$A = \bigcup_{n\in\mathbb{Z}} (V+n),$$
which is clearly nonmeasurable.

We claim the inner measure $0$ property is retained. Indeed, given $F$ compact with $F\subset A$, we have $(F\cap [n,n+1])-n\subset V$, so we find $\mu(F\cap[n,n+1]) = 0$ and thus $\mu(F)=0$. This proves $\mu_*(A\cap E)\leq\mu_*(A) = 0$ for any measurable set $E$.

To prove the outer measure property, we first assume the measurable set $E$ is bounded, so there is positive integer $N$ large enough that $E\subset [-N,N]$. Also put $A_N: = A\cap [-N,N]$, which is easily seen to have outer measure $2N$. Applying Carathéodory criterion gives
$$\mu^*(A_N) = \mu^*(E\cap A_N)+\mu^*(E^c\cap A_N).$$
Since $E^c\cap A_N\subset [-N,N]\setminus E$, we have
$$\mu^*(E^c\cap A_N)\leq \mu^*([-N,N]\setminus E) = 2N-\mu(E).$$
Thus our inequality becomes $2N\leq \mu^*(E\cap A_N)+2N-\mu(E)$. Rearranging,
$$\mu(E) \leq \mu^*(E\cap A_N)\leq \mu^*(E) =\mu(E),$$
so $\mu(E)= \mu^*(E\cap A_N) = \mu^*(E\cap A)$ in this case.

In the general case, we use inner regularity of measure and continuity of outer measure from below. Measurability of $E$ implies $E=F\cup Z$, where $F$ is $F_\sigma$ and $Z$ is measure $0$. It is sufficient to prove $\mu^*(F\cap A) = \mu(E)$ (exercise). Now, there is a sequence of compact sets $F_n\nearrow F$ such that $\mu(F_n)\nearrow\mu(F)=\mu(E)$ as $n\to\infty$. Each $F_n$ satisfies $\mu(F_n) = \mu^*(F_n\cap A)$. Yet $F_n\cap A\nearrow F\cap A$, so
$$\mu^*(F\cap A) = \lim \mu^*(F_n\cap A) = \lim \mu(F_n) = \mu(E).$$
This completes the proof.


问:
Infinite dimensional dual space vs finite dimension
I just have a question about why some of the theorems involving dual space fails for infinite dimension.
A standard problem is below:

"Suppose V is finite-dimensional and v is an element of V with v does not equal 0. Prove that there exists an element in the dual space of V such that phi(v)=1"

The answer is to extend v to a basis of V, then we have a corresponding basis for V'(V' is the dual space of V) and by definition, phi1(v)=1 and 0 for all the other basis of V. So why does the question requires V to be finite dimension? Could we not do the same if V was infinite dimensional?

答:
The argument works fine in infinite dimensions except that you need the axiom of choice to guarantee that infinite-dimensional vector spaces have bases, and in fact "every vector space has a basis" is equivalent to the axiom of choice; this is due to Blass.

In the finite-dimensional case this is true in ZF and it's not necessary to get into the set-theoretic weeds like this.

Here is a nice relatively explicit example: $\mathbb{R}$ is an infinite-dimensional vector space over $\mathbb{Q}$. Can you write down any nonzero linear functional $\mathbb{R} \to \mathbb{Q}$ whatsoever? In fact it's consistent with ZF that there don't exist any, and any such linear functional must be non-measurable.


问:
Partition 100 people, 4 from each country into 4 groups with conditions
This is a problem from the $2005$ All-Russian Olympiad. Problem is as follows:

$100$ people from $25$ countries, four from each country, sit in a circle.
Prove that one may partition them onto $4$ groups in such way that no
two countrymen, nor two neighboring people in the circle, are in the
same group.

My approach was to try and find an algorithm that can monotonically reduce the number of bad states. However, I haven't been able to figure out a way to do this for every possibility. Does anyone have any hints on how to approach this problem?

I've tried to start with an arbitrary partition where each partition only has 1 person from each country, and preserve that with each move.

I also claim that there is always a country for which not all of its representatives' neighbors are in a single partition. Proof idea: Suppose not. WLOG, say that country $1$ has all of its representatives' neighbors in the first partition. So if $1_i$ is the representative from country $1$ in partition $i$, we have $1_1$ and its neighbor, WLOG from country $2$, is also in group $1$. So we have $1_1, 2_1$. Then $2_1$'s other neighbor must also be in partition $1$ or we are done. Continuing in this way, we find that all the representatives are in partition $1$, contradiction.

Using this, I try to decrease the number of "bad" pairs (i.e. neighbors or same country) in each partition monotonically. However, I haven't been able to figure out how to do so.

In particular I'm stuck on cases like the following: country $1$ has its representatives in partitions $1-4$ s/t the neighbors of $1_2,1_3,1_4$ are all in the first partition. $1_1$'s neighbors are in partition $1$ and another partition, say $2$. Trying to swap the placements of country $1$ representatives seems to only increase number of "bad" pairs, and trying to swap the neighbors may also increase the number of bad pairs.

Is there an observation that I'm missing here, or is this approach not the right way to go about it?

答:
Ah yes, this is such a "russian" problem, if you know what I mean.

$\text{Partial Solution}$

Anyway, the solution is based on induction. Suppose that if we place $4n$ people at a table from $n$ countries, $4$ people from each country, we can separate them in $4$ groups such that no $2$ people from the same country are in the same group and no $2$ people sitting next to each other are from the same group.

Now, lets consider $4n+4$ people on our circle. There are $2$ cases:

$\text{Case 1:}$

There are $2$ people from the same country that are sitting next to each other. Suppose that country is $c_i$. Take all $4$ people from country $c_i$ and ignore them. For the other $4n$ people on the circle, distribute them in $4$ groups such that the conditions are satisfied (this can be done, from the induction hypothesis).

If we prove that the $4$ people from country $c_i$ can be distributed to the $4$ groups such that the conditions in the statement still hold, then the induction is complete. Here are the $3$ possible cases of the positions of the people from country $c_i$ on the circle (here, the $X_i$s represent people from other countries and $A_i$s people from country $c_i$)

$(X_1-A_1-A_2-X_2)$, $(X_3-A_3-X_4)$, $(X_5-A_4-X_6)$

$(X_1-A_1-A_2-A_3-X_2)$, $(X_3-A_4-X_5)$

$(X_1-A_1-A_2-A_3-A_4-X_2)$

Notice that because we "ignored" the $A_i$s when distributing the $4n$ people to the $4$ groups, in each of the cases above $X_1$ and $X_2$ are in different groups; $X_3$ and $X_4$ are in different groups and $X_5$ and $X_6$ are from different groups.

It is easy to observe that, no matter the groups the $X_i$s are from, in each of the above cases the $A_i$s can be distributed in groups to still satisfy the conditions. Example: (instead of $X_i$s I just put $1$, $2$, $3$ or $4$ for the group they are in)

$(1-A_1-A_2-2)$, $(1-A_3-3)$, $(3-A_4-4)$. Take $A_3=2$, $A_4=1$, $A_1=3$, $A_2=4$ (I will not prove all the above cases, but it doesn't take much and it simply is case-work)

So in this case, the induction holds.

$\text{Case 2:}$

There aren't $2$ people from the same country which sit next to each other. In this case we do not need induction, we will just prove that if we have $4n$ people on the circle such that there aren't $2$ people from the same country which sit next to each other we can distribute them to the $4$ groups like it was specified in the statement.

However, I haven't yet found what the algorithm is in this case. I will come back when I find it.


问:
Has this optimization problem already been solved?
Here's a simplified version of the problem I'm thinking of.

Say there are three levers I can pull. Lever 1 takes 1s to pull, provides me \$1, and has a cooldown of 2s (i.e. after it's been pulled, I cannot pull it again for 2 seconds). Lever 2 takes 3s to pull, provides \$10, and has a cooldown of 10s. Lever 3 takes 2s to pull, provides \$4, and has a cooldown of 3s. What is the fastest way for me to reach \$100?

I'm not so much thinking about the solution to this problem, but rather if an algorithm to find the optimal solution is already known. I genuinely have no idea. It vaguely sounds like the travelling salesperson problem, though. If this problem is also NP-hard and has no efficient solution, then that is an answer I will accept.

答:
My understanding of the problem: I believe that if the target amount is large, a periodic pattern will emerge. Every single lever has a known yield, i.e. the number of \$ it rewards in the time it takes before you can repeat it (that includes the cooldown). Then two levers pulled in turn also have a yield, which is better than that of the two individual levers, because you don't have to wait for whole cooldowns. This generalizes to longer sequences.

By expanding the tree of lever choices, you get the yield for various patterns, and some of them will be "cool". A cool pattern is one that you can replay immediately, because the first lever has already cooled down.

My intuition is that the optimal solution for large amounts is made by repetition of the cool pattern with the best yield (in case of ties, take the shortest). I also believe that the exact target value to be reached will break the periodicity to some distance of the end. I would try a greedy approach based on the optimal patterns and do exhaustive search for the last, incomplete period.

This reasoning should be validated more formally.


问:
Simple zero $a$ is equal to $\frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz$ for $f$ holomorphic within $C$?
I'm struggling with this question:\

Show that the root $a$ which is a simple zero of $f(z)$ is given by $$a = \frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz, $$
where $f$ is holomorphic function in a simple closed contour $C$ and $f(z) \neq 0$ and $a$ is in $C.$ Also, $f(z) \neq 0$ for $z \in \{\mathbb{C}-a \} \ \text{in} \ C.$

I just know that since $a$ is a simple pole, then $f(a) =  0$ and $f'(a) \neq 0.$ Maybe I will use Residue theorem... but I don't know where to start.

答:
Let $a \neq 0$. $\frac {zf'(a)} {f(z)}$ has a simple pole at $a$ and the residue is $\lim_{z \to a} \frac {z(z-a)f'(a)} {f(z)}=\lim_{z \to a} \frac {z(z-a)f'(a)} {f(z)-f(a)}$. By definition of the derivative $f'(a)$ this limit is simply $a$. [Divide numerator and denominator by $z-a$ and note that $\frac {f(z)-f(a)} {z-a} \to f'(a)$ as $z \to a$].  Hence, Residue Theorem gives the value of the integral divided by $2\pi i$ as $a$.

As pointed out in the comment below the result follows by Cauchy's Theorem when $a=0$ since there is no pole in that case.


答: By the residue theorem,$$\frac1{2\pi i}\int_C\frac{zf'(z)}{f(z)}\,\mathrm dz=\operatorname{res}_{z=a}\left(\frac{zf'(z)}{f(z)}\right).$$Since $a$ is either a removable singularity or a simple pole of $f$,\begin{align}\operatorname{res}_{z=a}\left(\frac{zf'(z)}{f(z)}\right)&=\lim_{z\to a}(z-a)\frac{zf'(z)}{f(z)}\\&=\lim_{z\to a}\frac{zf'(z)}{\frac{f(z)-f(a)}{z-a}}\\&=\frac{af'(a)}{f'(a)}\\&=a.\end{align}

Title: Simple zero $a$ is equal to $\frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz$ for $f$ holomorphic within $C$?

Body:
I'm struggling with this question:\

Show that the root $a$ which is a simple zero of $f(z)$ is given by $$a = \frac{1}{2\pi i}\int_C \frac{zf'(a)}{f(z)} dz, $$
where $f$ is holomorphic function in a simple closed contour $C$ and $f(z) \neq 0$ and $a$ is in $C.$ Also, $f(z) \neq 0$ for $z \in \{\mathbb{C}-a \} \ \text{in} \ C.$

I just know that since $a$ is a simple pole, then $f(a) =  0$ and $f'(a) \neq 0.$ Maybe I will use Residue theorem... but I don't know where to start.


问:
NP problems understanding on specific example
Let's say we have a game where we should discover a number of length n, generated by RNG. If someone says us a solution it is easy to verify - we enter the number into machine and it compares it to the generated number (O(n)). On the other hand, it should be obvious, that to find the solution in general case takes exponential time - there is no way to discover a random number in polynomial time without any information. So, it should be a NP problem that is not in P

What I miss here?

答:
It all hinges on what you mean by "RNG." You haven't been very clear on whether you mean a pseudorandom number generator, and in your second sentence it's not clear at which point you're talking about a seed for the PRNG and at which point you're talking about a value generated from a seed.

But if you clean up the argument a bit, this is roughly a version of the argument that the existence of one-way functions (functions easy to compute but hard to invert on random outputs) would imply $P \neq NP$. Naturally, this means that the existence of one-way functions is an open problem; in other words, nobody knows if it's possible to write down an RNG which is "random enough" to make this argument work. So the part where you say "it should be obvious" - it's not at all obvious! ("Obvious" and "clearly" are very dangerous words to allow yourself to use in proofs!)


问:
How is this step reached in the sum?
I have the following equation from this paper:

https://dspace.mit.edu/bitstream/handle/1721.1/106948/Roy_Aggresive%20flight.pdf?sequence=1&isAllowed=y

In this paper the author supposedly uses 104 to reach from 107 to 108. What I don't understand is how the external sum that has bounds from 0 to 2N was reached at. How was the external sum reached at?

答:
I guess you could somehow fine-tune the limits of the summation more if you wanted, but the point is that you do not have to. And you end up with a nicer looking formula if you enlarge the limits slightly.

More concretely: already in (107), you can start the summation at $n=0$ since for $n<r$ there is a zero in the product $\prod_{m=0}^{r-1}(n-m)$, so you are just summing up zeros.

Then of course, by squaring the polynomial, the degree of the polynomial doubles. So instead of summing up to $N$, you have to sum up to $2N$. If you were even more lazy, you could consider all the summations from zero to infinity. If you adopt the convention that $p_i=0$ for every $i<0$, then everything would work fine again.

Title: How is this step reached in the sum?


问:
Proving that the exponential function is continious at $0$
Given $b \in \mathbb R,b>0$, I have the exponential function $f(x)=b^x$ and I want to show that it is continious at $x=0$. From epsilon-delta definition, given an $\epsilon>0$, I need to find $\delta$ s.t $$|x-0|<\delta \implies |b^x-b^0|<\epsilon \text{, or}$$
$$|x|<\delta \implies |b^x-1|<\epsilon.(1)$$

From $(1)$, I have $$1-\epsilon<b^x<1+\epsilon$$ and by taking the logarithm of both sides in base b, I get

$$log_b(1-\epsilon)<x<log_b(1+\epsilon).$$

So I have chosen $\delta=min\text{{$log_b(1+\epsilon),-log_b(1-\epsilon)$}}.$

Is this proof valid? If not, can you give me a valid proof using the epsilon-delta definition or the sequential characterization of continiuty?

答:
The function is convex

$\lambda f(a) + (1-\lambda) f(b) \ge f(\lambda a + (1-\lambda)b)$  with $0\le\lambda\le 1$

Or more relevant to this problem.  $\lambda f(1) + (1-\lambda) f(0) \ge f(\lambda)$

But, it seems to me we need two cases for when $b>1$ and when $0<b\le 1$

Let $\delta = \min(1,\frac {\epsilon}{|f(1)|},\frac {\epsilon}{|f(-1)|})$


问:
Can rationals be completed to reals?
It's a well-known theorem that any metric space can be completed. Its proof essentially uses the completeness of the reals.
The rationals $\mathbb{Q}$ can be treated as an uncomplete metric space with the metrics $d:(x,y)\rightarrow |x-y|$.
The question arises: can the metric space $(\mathbb{Q},d)$  be completed without using the reals?
If I correctly remember it, I saw that construction in a course of mathematical analysis lectured at Moscow University
in 70s of the 20-th century. A proof and/or reference are welcome.

答:
First of all, completions don't need the real numbers.

Look in Kelley's book for the chapter on uniform spaces, where it is hinted that they can be described by a set of axioms not so different from the axiom of a topology. You can find the system in Bourbaki, but it's not difficult to derive it. The axiom system is basically a description of entourages, special subsets of $X\times X$ containing the diagonal: an entourage for a metric space $(X,d)$ would be a subset of $X\times X$ containing a set of the form $\{(x,y):d(x,y)<r\}$, for some $r\in\mathbb{R}$. However, the axiom system never mentions the real numbers.

It is possible to define the concept of convergence and of Cauchy sequence only using entourages and to show that every uniform space has a completion using the set of Cauchy sequences modulo an equivalence relation, which set can be endowed with a uniform space structure and every Cauchy sequence in it converges.

Metric spaces are uniform spaces having a countable basis of entourages (basis in the sense of filter basis).

About getting the reals from the rational numbers, here's a construction similar to the one described above, which is a formalization of Cantor’s idea.

A C-sequence in the rational numbers is a sequence $(x_n)$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for every $m,n>k$, $|x_m-x_n|<\varepsilon$.

Real numbers are not used, as you see.

Now it's not difficult to show that the set $\mathscr{C}$ of C-sequence can be made into a ring, with pointwise addition and multiplication: the sum and product of C-sequences are C-sequences.

A Z-sequence in the rational numbers is a sequence $(x_n)$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for every $n>k$, $|x_n|<\varepsilon$.

It's easy to see that every Z-sequence is a C-sequence and that the set $\mathscr{Z}$ of Z-sequences is an ideal of $\mathscr{C}$.

Now the quotient ring $\mathscr{R}=\mathscr{C}/\mathscr{Z}$ can be shown to be complete in the sense that every C-sequence in it converges. And, guess what? It is a field that's isomorphic to the real numbers.

Basic steps.

$\mathscr{Z}$ is a maximal ideal in $\mathscr{C}$, so $\mathscr{R}$ is a field.

Define an order relation on $\mathscr{C}/\mathscr{Z}$ by declaring an element of the form $[(x_n)]=(x_n)+\mathscr{Z}$, where $(x_n)$ is a C-sequence, to be positive if there exist $\varepsilon\in\mathbb{Q}$ and $k$ such that $\varepsilon>0$ and $x_n>\varepsilon$, for every $n>k$. Prove that $\mathscr{R}$ becomes an ordered field.

Embed $\mathbb{Q}$ into $\mathscr{R}$ in the unique possible way.

Define a C-sequence in $\mathscr{R}$ as at the beginning (only rational numbers).

Define a sequence $(x_n)$ to be convergent in $\mathscr{R}$ if there exists $x\in\mathscr{R}$ such that, for every $\varepsilon\in\mathbb{Q}$, $\varepsilon>0$, there exists $k$ with the property that, for all $n>k$, $|x_n-x|<\varepsilon$.

Prove that all C-sequences converge in $\mathscr{R}$.

What's the idea in the last step? Take a sequence $(x_n)$ in $\mathscr{R}$. Each $x_n$ is the residue class of a sequence $(x_n^{(1)},x_n^{(2)},\dotsc)$ in $\mathbb{Q}$. Show that the sequence $y$ defined by $y_n=x_n^{(n)}$ is a C-sequence in $\mathbb{Q}$ and that the given C-sequence converges to $[y]$.

Title: Can rationals be completed to reals?


问:
Continuity in the Strong(Whitney) Topology
Let $P,M$ and $N$ are smooth manifolds and let $F:P\times M \to N$ be a smooth map. We know its associated map $\tilde F:P \to C^\infty(M,N)$ given by $p \mapsto F_p(m)$ is continuous if and only if $F$ is continuous whenever we equip $C^\infty(M,N)$ with the weak topology. However this is not true in genreal if $C^\infty$ comes with the the strong (Whitney) topology; I feel the counterexample should not be that hard but I am unable to find it. Does anybody know such a function? Thanks in advance.</p>

答:
$M$ must be non-compact or else the topologies are the same, so the answer should exploit non-compactness in some way. Let $F: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ send $(t, x)$ to $tx$. We wish to show that

\begin{align*}
  \tilde{F}: \mathbb{R} & \to C^\infty(\mathbb{R}, \mathbb{R}) \\
t & \mapsto F(t, -) \\
\end{align*}

is not continuous in the Whitney topology. Let $U = \{ (x, y) \in \mathbb{R}^2 \mid |y| < 1\}$. Then $M(U) = \{ f\in C^\infty(\mathbb{R}, \mathbb{R}) \mid \text{graph}(F(t,-)) \subset U\}$ is a basic open subset of $C^\infty(\mathbb{R}, \mathbb{R})$ in the Whitney $C^0$ topology and hence sub-basic in the Whitney $C^\infty$ topology. (Here I am using notation and definitions from Golubitsky and Guillemin). This is because $j^0 f (X)$ of $f: X \to Y$ is simply the graph of $f$ in $X\times Y$. But notice that
$$\tilde{F}^{-1}(M(U)) = \{ t \in \mathbb{R} \mid \quad |tx| < 1 \quad \forall x\} = \{ 0\}$$

so $\tilde{F}$ is not continuous. This is because for any $t \neq 0$ there exists an $x$ such that $|xy| \geq 1$.


答: Speaking about the squential characterization: take a sequence $x_n$ that converges to zero. I want to prove that  $b^{x_n}\rightarrow 1$. Since $x_n$ is infinitesimal, for every $\varepsilon >0$ you can say that from a certain $n$ on, you have $|x_n|<\varepsilon$, in other words $-\varepsilon<x_n<\varepsilon$ now you take the exponential and you are done (just observe that if $0<b<1$ you need to switch  the $b^\varepsilon$ and $b^{-\varepsilon}.$)

Title: Proving that the exponential function is continious at $0$

Body:
Given $b \in \mathbb R,b>0$, I have the exponential function $f(x)=b^x$ and I want to show that it is continious at $x=0$. From epsilon-delta definition, given an $\epsilon>0$, I need to find $\delta$ s.t $$|x-0|<\delta \implies |b^x-b^0|<\epsilon \text{, or}$$
$$|x|<\delta \implies |b^x-1|<\epsilon.(1)$$

From $(1)$, I have $$1-\epsilon<b^x<1+\epsilon$$ and by taking the logarithm of both sides in base b, I get

$$log_b(1-\epsilon)<x<log_b(1+\epsilon).$$

So I have chosen $\delta=min\text{{$log_b(1+\epsilon),-log_b(1-\epsilon)$}}.$

Is this proof valid? If not, can you give me a valid proof using the epsilon-delta definition or the sequential characterization of continiuty?


问:
Find the limits of integration of $f(x,y)$ using polar coordinates over the following Domain $D=\{(x,y):4x\leq x^{2}+y^{2}\leq 5x\}$.
So, it's given on this question the domain $D=\{(x,y):4x\leq x^{2}+y^{2}\leq 5x\}$. It also said that we use polar coordinates of the form $$x=R\cos t$$ and $$y=R\sin t$$ then we have to find the limits of integration of the double integral over this domain. I'm kind of confused because I thought the limits would be $2$ and $5/2$ for the outsider integral relative to $dr$, and $0$ to $2\pi$ on the insider integral relative do dt. But both of those answers are wrong and I'm having trouble finding why. Visual approaches still let me confuse with double integrals over domains.

答:
You have:
$$x^2+y^2-5x\le0$$
$$(x-5/2)^2-25/4+y^2\le0$$
$$(x-5/2)^2+y^2\le25/4$$
and you have:
$$x^2+y^2-4x\ge 0$$
$$(x-2)^2-4+y^2\ge0$$
$$(x-2)^2+y^2\ge 4$$
So your region can be represented as the inside of a circle centre $(5/2,0)$ and radius $5/2$ and the outside of a circle centre $(2,0)$ and radius $2$. If you sketch these graphs you will notice that one is inside the other, so your region looks something like an uneven ring donut. Hope this helps :)


问:
$R$ Noetherian, $M$ finitely generated module. If $P$ is minimal over $\operatorname{ann}M$, how is $M' = \ker (M \to M_P)$ a $P$-primary submodule?
Let $R$ be Noetherian and $M$ a finitely generated $R$-module.

We know a theorem which says: $M$ is $P$-coprimary iff $P$ is minimal over $\operatorname{ann}M$ and $M$ injects into $M_P$.

Now the claim is:

If $M$ is any finitely generated $R$-module and $P$ is minimal over the annihilator of $M$, then the submodule $M' \subset M$ defined by $M' = \ker (M \to M_P)$ is $P$-primary because $M/M'$ injects into $(M/M')_P \cong M_P$.

Why is this true? Is $P$ minimal over $\operatorname{ann}(M/M')$?

答:
First note that, if $P$ is minimal over $ann M$, then $Ass(M_P)=\{P\}$.

Now $M/M'\hookrightarrow M_P$, hence $Ass(M/M')\subset Ass(M_P)=\{P\}$. But $M/M'$ is a non-zero module, so $Ass(M/M')=\{P\}$. Thus $M'$ is $P$-primary.


问:
Diameter and Radius and Well Known Inequality, Which one is preferable?
An un-directed graph $G$ is given.

The diameter of a graph is the maximum of shortest paths between two vertices of the graph.

$L(S)$ is maximum length of shortest paths from $S$ to other vertices.

The radius of a graph is minimum value of $L(S)$ among all vertices.

If $d$ and $r$ be diameter and radius of graph two following inequalities always hold:

$A)$ $r \leq d$

$B)$ $r \geq \frac{d}{2}$

But my note tells me that $B$ is better than of $A$  (maybe by 'better' word, it means that it always holds or any condition that be logical... I don't know for example one of them works with negative weights or anything else in Graph)

I see both of them in all of graph books but anyone can distinguish between these two case, is there any difference here that we can prefer $B$ to $A$?

答:
Let me begin by saying that your question almost makes no sense. As @MorganRodgers very explicitly said, when comparing bounds, you need $2$ lower ones or $2$ higher ones, to find which one is better.

A an example, here are $2$ upper bounds for $n!$

$$n!<n^n\text{ and } n!<\bigg(\frac{n+1}{2}\bigg)^n$$

The first one comes from the fact that $1,2,...,n-1<n$ and the second
one comes from the $Am-Gm$ inequality. You can clearly tell that the
second higher bound is better (or tighter), because it is smaller.

You cannot prefer a lower bound over a higher bound or viceversa. They do different things It's like saying you prefer to draw with a spoon over eating soup with a pencil.

However, I think what you want to prove is the fact that $d-r>r-\frac{d}{2}$, in other words, the lower bound is closer to the actual value than the higher bound.

Lets begin by clarifying the notions. Consider a graph $G$ and let the set of the vertices of $G$ be $V(G)$.

For any pair of vertices $v,u\in V(G)$, the distance between $u$ and
$v$ is denoted with $d(u,v)$ and is equal to the length of the
shortest path that connects $u$ and $v$. If $u$ and $v$ are not
connected this is equal to $\infty$ and if $u=v$ then the distance is
$0$

For any $v\in V(G)$, the eccentricity of $v$ is denoted by
$\varepsilon(v)$ and satisfies the following relation
$$\varepsilon(v)=\max_{u\in V(G),u\neq v}d(u,v)$$

The radius of a graph $G$ is $$r=\min_{v\in V(G)}\varepsilon(v)$$

The diameter of a graph $G$ is $$d=\max_{v\in V(G)}\varepsilon(v)$$

Using what we have just found out, we can see that sometimes the higher bound is closer to $r$ than the lower bound and sometimes the lower bound is closer to $r$ than the higher bound.

For example take a complete graph $K_n$. In this case, $d=r=1$ so $d-r=0<\frac{1}{2}=r-\frac{d}{2}$ so in this case we could say that the higher bound is "better".

But when we take a complete bipartition $K_{n,n}$, $d=2$ and $r=1$ so in this case $d-r=1>0=r-\frac{d}{2}$

So to conclude the answer, sometimes the $\frac{d}{2}$ is closer to $r$ and sometimes $d$ is closer to $r$. You cannot prefer a bound over another.


问:
Solving exponential equation with negative base
I have to solve the equation: $(x-2)^{x^2-6x+1}\leq 1 $.

If $x-2>1 $ i.e. $x> 3 $ we have $3< x \leq 3+2\sqrt{2}$.

If $0< x-2<1 $ i.e. $ 2< x<3$ we get no solutions.

If   $x -2=1$ we get solution  $x=3$ and if $x -2= -1$ we have  $x=1$.

If $x=2$ we have $0^{-7} $ which is undefined.

Question:  I wonder how this equation could  be solved if $x<2$. I considered special case $x= 1$,but how  could I solve it in general? Wolfram Alpha gives results $x = 3- \sqrt{n+8} , n \in \mathbb{Z}, n \geq -6$, but how can these solutions be obtained without program?

I found in some book that $a^x, a < 0$ is defined for $x = \frac{p}{q}, p \in \mathbb{Z}, q \in  \mathbb{N}$, $q$ is odd. Can we apply that in some way and how. Please help,I would be greatly thankful.

答:
From the top answer in this question, it seems that only the second interpretation is useful here: the first one does not allow negative exponents for a negative base, and the third one yields complex results which cannot be compared with 1 in general. It seems that you are also referring to the second interpretation.

Hence in the case that the base is negative, $x^2-6x+1\in\mathbb Q$. Either from the rational roots theorem or from the fact that $\mathbb Z$ is integrally closed domain, $x^2-6x+1$ must be an integer.

Say $x^2-6x+1=n\in\mathbb Z$. Then $(x-3)^2=n+8$, which eventually yields the results given by Wolfram Alpha. Then use the inequality to bound $n$, and at last check what you get are in fact solutions.


答: Let's repeat @HenryLee's reasoning in polar coordinates. We have $4R\cos t\le R^2\le5R\cos t$, i.e. $4\cos t\le R\le5\cos t$. At least one $R$ satisfies this iff $\cos t\ge0$, so it's convenient to give $t$ a range on which this condition is satisfied contiguously. In particular we take $t\in[-\pi,\,\pi]$, so in $D$ this is limited to $t\in[-\pi/2,\,\pi/2]$. For a function $f$, $\iint_Dfdxdy=\int_{-\pi/2}^{\pi/2}dt\int_{4\cos t}^{5\cos t}RfdR$.

Title: Find the limits of integration of $f(x,y)$ using polar coordinates over the following Domain $D=\{(x,y):4x\leq x^{2}+y^{2}\leq 5x\}$.

Body: So, it's given on this question the domain $D=\{(x,y):4x\leq x^{2}+y^{2}\leq 5x\}$. It also said that we use polar coordinates of the form $$x=R\cos t$$ and $$y=R\sin t$$ then we have to find the limits of integration of the double integral over this domain. I'm kind of confused because I thought the limits would be $2$ and $5/2$ for the outsider integral relative to $dr$, and $0$ to $2\pi$ on the insider integral relative do dt. But both of those answers are wrong and I'm having trouble finding why. Visual approaches still let me confuse with double integrals over domains.


答:
Lets look at this region. $4x\le x^2 + y^2 \le 5x$

Start with $x^2+y^2 = 5x$

$(x^2 - 5x) + y^2 = 0\\
(x^2 - 5x + (\frac {5}{2})^2) + y^2 = (\frac {5}{2})^2\\
(x^2 - \frac {5}{2})^2 + y^2 = (\frac {5}{2})^2$

It is a circle centered $\frac {5}{2}$ units from the orgin, of radius $\frac 52$ units.  The origin lies on the circle.

And $x^2 + y^2 = 4x$ is a circle centered $2$ units from the origin, of radius $2$ units.

The region we are interested in is a little crescent between the two circles.  The reason to do this analysis and have a picture of what we are working with is that it tells us that the entire region is to the right of the y axis.

Now for an approach that is raw algebra.

$x = r\cos\theta\\
y = r\sin\theta\\
r^2 = x^2 + y^2$

Now state your region in terms of $r$ and $\theta.$

$4r\cos\theta \le r^2\le 5r\cos\theta$

Since $r\ge 0$ we can safely divide through by $r.$

$4\cos\theta \le r\le 5\cos\theta$

And we have our limits for $r.$

$\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \int_{4\cos\theta}^{5\cos\theta}  f(r,\theta) r\ dr\ d\theta$