问:
If we place $1$ to $n^2$ in an $n\times n$ table, what is the smallest $s$ where $s$ is the max of $a+b$ where $a,b$ are numbers in adjacent cells?
Primary question : Let $n$ be an integer greater than 1.  Numbers $1,2,...,n^2$ are placed in an $n\times n$ table, one number per cell.  For each such configuration, let $s$ be the maximum value of $a+b$ where $a$ and $b$ are numbers in adjacent cells (sharing sides).  What is the minimum possible value of $s$?

Write $S=S_n$ for the minimum value of $s$.  Because $n>1$ we obviously have $S\ge n^2+2$ because $n^2$ is adjacent to at least two cells.  It is clear also that when $n>3$, $S>n^2+2$ because either $n^2-1$ is adjacent to $\ge 3$ cells or both $n^2$ and $n^2-1$ are in some corners, so one of the $4$ neighbours of $n^2$ and $n^2-1$ is at least $4$.  Hence $S\ge n^2+3$ for $n>3$.  I know $S_2=6$, $S_3=11$, $S_4=19$.  I am not seeing a pattern here.

I was asked for $S_{10}$, so an answer for $n=10$ is already good for me.  But it'd be great if $S_n$ is known for an arbitrary $n$.

If you know where this problem came from, plz let me know.  The same friend from another question asked me this question but didnt know if this problem is a contest problem from somewhere.

Modified questions :

(1) what is the smallest value of $s$ if we define 'adjacent' to mean 'sharing at least one vertex'?  If $S'_n$ is the smallest number of the modified question, it is easy to see that $S'_n\ge n^2+5$ for $n>3$.  We have $S'_2=5$, $S_3'=12$.  I'd like to know the answer to this question too.

(2) The same question as the primary question but we are filling $1,2,...,mn$ in an $m\times n$ table.

(3) The same question as the modified question (1) but we are filling $1,2,...,mn$ in an $m\times n$ table.

答:
Indeed, a minimum of $n^2 + \lfloor \frac n 2\rfloor + 1$ can be arranged for any $n \geq 3$, as follows, where $m = \lfloor \frac n 2 \rfloor$:
\begin{matrix}
n^2, & 1, & n^2 - 1, & 2, & n^2 - 2, & \dots \\
m + 1, & n^2 - n + m, & m + 2, & n^2 - n + m + 1, & m + 3, &\dots \\
n^2 - n, & n + 1, & n^2 - n - 1, & n + 2, & n^2 - n - 2, & \dots \\
n + m + 1, & n^2 - 2n + m, & n + m + 2, & n^2 - 2n + m + 1, & n + m + 3, & \dots \\
\dots & \dots & \dots & \dots & \dots & \dots
\end{matrix}

As for the optimality, it seems that the following is true:

On a $2k \times 2k$ grid, if we mark $k^2$ cells such that no two are adjacent to each other, then the number of cells that are adjacent to at least one marked cell is at least $k^2 + k$.

If this statement is true, then the above arrangement for $n = 2k$ is optimal: we consider the largest $k^2$ numbers $4k^2, 4k^2 - 1, \dots, 3k^2 + 1$. If two of them are adjacent, then the sum is at least $6k^2 + 3$; otherwise, by the above statement, there are at least $k^2 + k$ numbers adjacent to them, among which the largest one must be at least $k^2 + k$. Since this number is adjacent to a number at least $3k^2 + 1$, their sum is at least $4k^2 + k + 1$.

However I'm not able to prove the above statement, although it seems very likely to be true.


问:
Find the parametrization of the intersection of two surfaces
I need to find the parametrization of the intersection for the two surfaces defined by these equations:

$x^2+y^2=25$ and $z^2+y^2=25$

I'm not quite sure how to do it. What is the best way?

答:
You see the union of two circles $x=z, x^2+y^2=25$ and $x=-z, x^2+y^2=25$

A good parametrization is for instance for the first circle, $x= 5 \cos(u), y=5 \sin(u), z= 5 \cos(u)$ and $x= 5 \cos(u), y=5 \sin(u), z= -5 \cos(u)$ for the second.

ContentLicense: CC BY-SA 4.0


问:
Find the number of 4 digital numbers whose greatest common divisor with 18 is 3
So I found this question shared by another student in my Maths doubt group.As the question says,we have to find the total number of 4-digit numbers whose greatest common divisor with 18 is 3.

Here's what I tried:-

Find total number of 4 digit numbers divisible by 3.
The numbers are :- 1002,1005,1008........9999

So by the nth term of an AP formula,we get:-

1002+(n-1)3 = 9999

So n comes out to be 3000.After this I tried to filter out the numbers.

I now divided the entire series by 3.We get:-
334,335,336......3333.

Now the condition to have common divisor with 18 equal to 3 is that the number in the above series shouldn't be divisible by 3 or by 2 I think.

So first of all the even numbers in the series are disqualified.Now we are left with :-
335,337,339......3333,i.e 1500 numbers still remain. Now the numbers divisible by 3 in the above series are:-

339,341...3333.

Therefore number of digits divisible by 3 between 339 and 3333 are 999(using the same nth term of an AP formula I used earlier).

4)In the original series I wrote in step 3,there were 1498 numbers between 339 and 3333, both included.Of this 999 are divisible by 3.Hence we are left with 499.Now we add 2 to this(335,337),and the total final answer comes to be 501.

But the answer key says the answer is 1000. Can someone please tell me what mistakes I have made?

Thanks!

答: You were OK until the last step.  When you count multiples of 3 in the sequence $335, 337, 339, 341, 343, ... , 3333$ you must assume they are six units apart because you dropped out all the even numbers.  The smallest multiple of 3 in this list is $339$, the largest is $3333$.  Then $[(3333-339)/6]+1=500$ which, subtracted from the $1500$ numbers surviving the previous steps, properly leaves $1000$.


问:
If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$
I have come across the following comment in the lecture:

If for a prime $q$ holds $q \equiv 5 \pmod 8$, then $q-2$ has a prime factor $p$ with $p \equiv 3 \pmod 8$ or $p \equiv 5 \pmod 8$.

I do not see why this is true. I attempted to prove it by rewriting $q-2$ as $q-2 = 8k+3$ for some integer $k$ and tried to see what happens when I divide it by an arbitrary prime. However, this did not help. Could you please give me a hint?

答:
An odd number must have odd prime factors.

An odd number $K \equiv 1,3,5, -1 \pmod 8$.

Now if you have product of odd numbers $K_i$ and each $K_i \equiv \pm 1 \pmod 8$ then $\prod_i K_i \equiv \prod_i (\pm 1)\equiv \pm 1\pmod 8$

So if an odd number $K$ has only prime factors that are $p_i \equiv \pm 1 \pmod 8$ then $K \equiv \pm 1 \pmod 8$.

So if it is NOT the case, if $K \equiv \pm 3,5 \pmod 8$ then at least one of the prime factors $p_i \not \equiv \pm 1 \pmod 8$.

But if $p_i$ is odd, and $p_i \not \equiv \pm 1 \pmod 8$. then $p_i \equiv 3, 5 \pmod 8$ as those are the only options left.

So if $K =q-2 \equiv 3 \pmod 8$ then $q-2$ is odd and at least one of its prime factors is $\not \equiv \pm 1 \pmod 8$ so at least one of its prive factors is $3,5 \pmod 8$.


问:
Convergence of bisection method to a zero of $f: [a,b] \to\mathbb{R}$

Let $f: [a,b] \to\mathbb{R}$ be a continuous function such that $f(a) < 0 \leq f(b)$.

Show that the bisection method converges to a zero of $f$. That is, show that
$\bigcap\limits_{i=0}^{\infty} I_{i} = \{x_0\}$ for some $x_0\in [a,b]$ and $f(x_0) = 0$.

I really don't know where to go from here, I've tried using the iterating method for the bisection method, namely:

Iterating the intervals $I_i = [a_i,b_i]$ such that $I_0 = [a,b]$;

And then for every iteration $I_{i+1} = [a_{i+1},b_{i+1}] =
 \begin{cases}
      [a_i, \frac{a_i + b_i}{2}], & \text{if } f(\frac{a_i + b_i}{2}) >0, \\
      [\frac{a_i + b_i}{2}, b_i], & \text{otherwise}.\\
   \end{cases}$

After doing this though I have no clue what to do next.

If anyone had advice I'd gladly take it, thanks.

答:
You want the Nested Interval Theorem. This tells you that if you have a sequence of nested closed and bounded intervals, where each interval is a subset of the previous one (like in your case), the infinite intersection is non-emtpy, so the intersection you have defined will contain some points.

Now you have to ask yourself, in your case, what points can it contain? Suppose a point $x$ is in the infinite intersection. Either $f(x)$ is positive, it is negative, or it's zero. If if is positive, you should be able to show that eventually $x$ will be left out of one of the intervals, and therefore it cannot be on the intersection, yielding a contradiction. If it's negative, you argue in the same way. The only possibility left is that $x$ is such that $f(x)=0$. You can check that specific $x$ will be inside all of your intervals, and therefore also in the intersection.

Note you might also what to think about what happens if $f$ has multiple zeros on the interval $[a,b]$.

Edited for clarity.


问:
Question on proof of Abel's limit theorem
Let be $\sum\limits_{k=0}^{\infty}a_kx^k$ a power series with $a_k\in\mathbb{R}$ , radius of convergence $0<R<\infty$ and assume that $\sum\limits_{k=0}^{\infty}a_kR^k$ exists.

Then, Abel's limit theorem states:

$\lim\limits_{x\to R-}\sum\limits_{k=0}^{\infty}a_kx^k=\sum\limits_{k=0}^{\infty}a_kR^k$ (continuity from the left).

I have worked through the proof and got it so far - except that our professor begins with the simplification "without loss of generality we assume $R=1$". Due to this assumption all the subsequent manipulations become a lot easier. But why is he allowed to assume that?

答:
Suppose the theorem is true for $R=1$.  Now let $R>0$, and for $|x|<R$ write $x=Ry$, with $|y|<1$.

We have $\lim_{x\to R^-}\sum a_kx^k=\lim_{y\to1^-}\sum a_kR^ky^k=\sum a_kR^k$ because $a_kR^k$ is real and  $\sum a_kR^k$ converges.


问:
How can we prove that f(a)*g(a)=(f*g)(a) for any two polynomials f,g?
I have an idea how to prove it, but I'm not sure if it's right. So a verification would be nice, and if it's wrong, a nudge in the right direction would be appreciated.

This is my solution:

Let $deg(f)=k; deg(g)=m;$ and let $n$ be the degree of $f*g$.

Then $$(f*g)(a)= \sum_n C_n*a^n $$ where $$C_n = \sum_i f_i * k_{n-i}$$

Now we take $f(a)*g(a)$

$$ f(a)*g(a) = (f_k*a^k+f_{k-1}*a^{k-1}+...+f_0)*(g_m*a^m+g_{m-1}*a^{m-1}+...+g_0) $$

If we expand this, we'll see:

$$ f(a)*g(a) = (f_0 * g_0)*a^0 + (f_0 * g_1 + f_1 * g_0)*a^1 + ... + (f_0 * g_{k+m} + ... + f_{k+m} * g_0)*a^{k+m} $$

We can shorten this expression by displaying each coefficient as a sum, $C$ such that

$$C_{k+m} = \sum_i f_i * k_{k+m-i}$$

Finally, we can conclude that

$$f(a)*g(a) = \sum_{k+m} C_{k+m}*a^{k+m} $$

Now what's left to prove is that $n=m+k$ which is already a defined property of polynomials:

$$deg(f*g)=deg(f)+deg(g)=m+k$$

答:
Part of this answer is to clarify some confusion which arose in the comment section. To begin with, you presented everything one needs to consider to show what I assume you intended to ask.

The important part here might be to understand the difference between polynomials as maps and polynomials as somewhat algebraic objects. For simplicity I will only refer to $\mathbb{R}$, but feel free to always replace that with any algebra you have in mind.

A polynomial map of degree $n\in\mathbb{N}$ is a map $f:\mathbb{R}\rightarrow\mathbb{R}$ that takes a value $x$ and associates a value $f(x):=\sum_{k=0}^na_kx^k$, where $x^k$ is the $k$-th power of $x$ and $a_k\in\mathbb{R}$, $a_n\neq 0$. Since we now how to multiply functions with values in $\mathbb{R}$, that is the value of $f\times g$ at at point $x$ is defined to be $(f\times g)(x):=f(x)\times g(x)$, we can show that for two polynomial maps $f,g:\mathbb{R}\rightarrow\mathbb{R}$, $g(x):= \sum_{k=0}^mb_kx^k$ the product again is a polynomial map given by $h(x):=(f\times g)(x)=\sum_{k=0}^{n+m}c_kx^k$, where $c_k:=\sum_{l=0}^ka_lb_{k-l}$ .

Since this product formula for polynomials seems to be true in the usual algebras that come to mind quickly, one might want to abstract the object behind the map. This can be done algebraically by introducing an abstract variable $t$ and its powers $t^k$, $k\in\mathbb{N}$, which have no other meaning than being different from each other. Having these we can introduce the set
\begin{align*}
\mathbb{R}[t]:=\{\sum_{k=0}^na_kt^k:\ a_k\in\mathbb{R}, n\in\mathbb{N}\}
\end{align*}
which can be equipped with a scalar multiplication $\lambda\times f:=\sum_{k=0}^n(\lambda a_k)t^k$, $\lambda\in\mathbb{R}$ and an addition $f+g:=\sum_{k=0}^n(a_k+b_k)t^k$ with $f:=\sum_{k=0}^na_kt^k$ and $g:=\sum_{k=0}^nb_kt^k$ elements in $\mathbb{R}[t]$. Hence this set is a real vector space with basis $(1,t,t^2,\ldots)$. On top of that we can equip $\mathbb{R}[t]$ with a multiplication, making it an algebra, by $f\times g:=\sum_{k=0}^{n+m}c_kt^k$, where $c_k:=\sum_{l=0}^ka_lb_{k-l}$. This definition is designed to match the multiplication of polynomial maps!

Having developed this notion of algebraic polynomials the link between these and polynomial maps is given by the so called evaluation maps. For each $a\in\mathbb{R}$ there is a map $E_a:\mathbb{R}[t]\rightarrow \mathbb{R}$, taking a polynomial $f$ and assigning to it the real number $\sum_{k=0}^na_ka^k$, or to be more precise it sends the coefficient $a_k$ of the base element $t^k$ to $a_k\times a^k$ and sums over all these real numbers afterwards. Then OP's question might read as 'How to prove that $E_a$ is an algebra homomorphism, especially that $E_a$ is multiplicative, i.e.
\begin{align}
E_a(f)\times_\mathbb{R} E_a(g)=E_a(f\times_{\mathbb{R}[t]} g).
\end{align}
Since one doesn't naturally care between the difference of algebraic polynomials and polynomial maps, one often writes $f(a)=E_a(f)$, meaning both at once or whatever is needed in that moment (regarding $f$!).

Returning to your question: Your calculation and expansion revalidates the knowledge of the coefficients of the product of two polynomial maps, which is needed here! To the abstract polynomial $f$ we consider the polynomial map $\tilde{f}:\mathbb{R}\rightarrow\mathbb{R}$ (or just $f$ if we don't care) with the exact same coefficients. Then we have that $E_a(f)=\tilde{f}(a)$ and hence $E_a(f)\times E_a(g)=\tilde{f}(a)\times\tilde{g}(a)=(\tilde{f}\times \tilde{g})(a)$, where the last equation is the definition of multiplication of maps! Now we have $(\tilde{f}\times \tilde{g})(a)=E_a(f\times g)$ by the definition of multiplication in $\mathbb{R}[t]$ which was designed to fit in that way.

I hope that clarifies a bit of the questions that occured.


问:
Evaluating the surface integral of a paraboloid $z = x^2 + y^2$ for $0\le z \le 4$
Evaluate $\iint_S x^2 dS$ where S is the paraboloid $z=x^2 + y^2$ for $0\le z \le 4$.

So far, I have parametrized my function as:
$r(x,y) = \langle x, y, x^2 + y^2\rangle$ which gives me the partial sums:
$$ r_x = \langle 1, 0, 2x\rangle$$
$$ r_y = \langle 0, 1, 2y\rangle$$
which gives me $ r_x \times r_y = \langle-2x, -2y, 1\rangle $.
I'm a bit confused which orientation should I use and what my surface integral will be.

答:
Rather than keeping the variables $x$ and $y$, consider parameterizing $S$ in cylindrical coordinates as

$$\vec s(r,\theta)=\langle r\cos\theta,r\sin\theta,r^2\rangle$$

which follows from the change of coordinates,

$$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=x^2+y^2\end{cases}$$

where $0\le r\le2$ and $0\le\theta\le2\pi$. (Keeping the parameterization in terms of $x$ and $y$ would make the limits of integration more annoying to deal with.)

Now let the normal vector (again, direction doesn't matter) be

$$\vec n=\vec s_\theta \times \vec s_r=\langle 2r^2\cos t,2r^2 \sin t,-r\rangle$$

which makes the surface element

$$\mathrm dS=\|\vec n\|\,\mathrm dr\,\mathrm d\theta=r\sqrt{4r^2+1}\,\mathrm dr\,\mathrm d\theta$$

So the surface integral is

$$\begin{align}
\iint_S x^2\,\mathrm dS&=\int_0^{2\pi}\int_0^2 (r\cos\theta)^2 r\sqrt{4r^2+1}\,\mathrm dr\,\mathrm d\theta\\[1ex]
&=\left(\int_0^{2\pi}\cos^2\theta\,\mathrm d\theta\right)\left(\int_0^2 \, r^3 \sqrt{4r^2+1}\,\mathrm dr\right)
\end{align}$$

which is easy to compute.

If you want to hold on to your original parameterization, you would have

$$\vec r(x,y)=\langle x,y,x^2+y^2\rangle$$

with $-2\le x\le2$ and $-\sqrt{4-x^2}\le y\le\sqrt{4-x^2}$. Then the normal vector would be as you found,

$$\vec n=\vec r_x\times\vec r_y=\langle-2x,-2y,1\rangle$$

so that the surface element is

$$\mathrm dS=\|\vec n\|\,\mathrm dx\,\mathrm dy=\sqrt{4x^2+4y^2+1}\,\mathrm dx\,\mathrm dy$$

and the integral would be

$$\iint_S x^2\,\mathrm dS=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}x^2\sqrt{4x^2+4y^2+1}\,\mathrm dy\,\mathrm dx$$


答:
If we can show this for a radius of $1$, then we can show
$$
\begin{align}
\lim_{x\to R^-}\sum_{k=0}^\infty a_kx^k
&=\lim_{x\to R^-}\sum_{k=0}^\infty\left(a_kR^k\right)\frac{x^k}{R^k}\tag1\\
&=\lim_{u\to 1^-}\sum_{k=0}^\infty\left(a_kR^k\right)u^k\tag2\\
&=\sum_{k=0}^\infty\left(a_kR^k\right)\tag3
\end{align}
$$
Explanation:

$(1)$: $1=\frac{R^k}{R^k}$

$(2)$: substitute $u=\frac xR$

$(3)$: apply the theorem for a radius of $1$

Title: Question on proof of Abel's limit theorem

Body:
Let be $\sum\limits_{k=0}^{\infty}a_kx^k$ a power series with $a_k\in\mathbb{R}$ , radius of convergence $0<R<\infty$ and assume that $\sum\limits_{k=0}^{\infty}a_kR^k$ exists.

Then, Abel's limit theorem states:

$\lim\limits_{x\to R-}\sum\limits_{k=0}^{\infty}a_kx^k=\sum\limits_{k=0}^{\infty}a_kR^k$ (continuity from the left).

I have worked through the proof and got it so far - except that our professor begins with the simplification "without loss of generality we assume $R=1$". Due to this assumption all the subsequent manipulations become a lot easier. But why is he allowed to assume that?


答:
Your integrand is a scalar. As the surface is defined by $z = x^2 + y^2$, you can apply the surface area formula directly to map $dS$ to $dA$ which is projection in $XY$ plane.

$dS = \sqrt{1 + \big(\frac{\partial z}{\partial x} \big)^2 + \big(\frac{\partial z}{\partial y} \big)^2} \ dA$

Now, $\sqrt{1 + \big(\frac{\partial z}{\partial x} \big)^2 + \big(\frac{\partial z}{\partial y} \big)^2} = \sqrt{1 + 4x^2+4y^2}$

Converting it into polar coordinates, $x = r \cos \theta, y = r \sin \theta$,

$0 \leq z \leq 4 \implies 0 \leq r \leq 2$

So the integral becomes,

$\displaystyle \int_0^{2\pi} \int_0^2 r^3 \cos^2\theta \ \sqrt{1 + 4r^2} \ dr \ d\theta$

Title: Evaluating the surface integral of a paraboloid $z = x^2 + y^2$ for $0\le z \le 4$

Body:
Evaluate $\iint_S x^2 dS$ where S is the paraboloid $z=x^2 + y^2$ for $0\le z \le 4$.

So far, I have parametrized my function as:
$r(x,y) = \langle x, y, x^2 + y^2\rangle$ which gives me the partial sums:
$$ r_x = \langle 1, 0, 2x\rangle$$
$$ r_y = \langle 0, 1, 2y\rangle$$
which gives me $ r_x \times r_y = \langle-2x, -2y, 1\rangle $.
I'm a bit confused which orientation should I use and what my surface integral will be.


问:
A simple method to calculate minimal area enclosed between a tangent to $f(x)$ and coordinate axes
Given a function $y=f(x)$, take a tangent to the function at some point. We are to find the minimum area between this line and the coordinate axes. An example question is the coordinate axes have minimum area.

I faced two different algorithms to find the solution. The first one is straight-forward:

Pick a point on the function: $(x_1, y_1) = (x_1, f(x_1))$.

Find the derivative of the function at $x_1$ to calculate the slope of the line. ($m=f'(x_1)$)

Derive the tangent-line formula. $y-y_1=m(x-x_1)$

Find the formulations of intersections with coordinate axes $(0, y_0)$, $(x_0, 0)$.

Calculate the formulation of the area of triangle as $A = x_0 y_0/2$.

$A$ shall be a function of $x_1$, minimize that to calculate minimum area.

Second algorithm is very short compared to that:

Take the function $g(x) = 2 x f(x)$.

Minimize that function to calculate the result.

I cannot figure out how the second algorithm works, or when it works. I checked both against the following family of functions, and both algorithms give the same result:

$f(x) = ax+b$

$f(x) = k/x$

$f(x) = 3b^2 - a^2 x^2$

$f(x) = \frac{b}{a} \sqrt{a^2-x^2}$

Question is: Can we prove/disprove the second algorithm? If disproved, under what conditions does the second algorithm work?

答:
Alternative proof with Lagrange Multipliers.

If $y=a(x)$, then, equivalently, we have a constraint function in x and y, $f(x,y)=y-a(x)=0$. So if $y=4-x^2$, $f(x,y)=y-4x^2=0$.

The slope of the tangent line to an arbitrary $f(x,y)$ is $m=\frac{-\delta f / \delta x}{\delta f / \delta y}$.

From above $\delta f/ \delta y =1$, and $\delta f / \delta x = -da/dx$.

Also: $\frac{\delta^2 f}{\delta x \delta y}=\frac{\delta ^2 f}{\delta y^2}=0. $

Area is:

$$A = \frac{(y+x\frac{\delta f / \delta x}{\delta f / \delta y})^2}{2}\frac{\delta f / \delta y}{\delta f / \delta x}$$

Cleaning up:

$$A = \frac{ (y\frac{\delta f}{\delta y}+x\frac{\delta f }{\delta x})^2  }{2\frac{\delta f}{\delta x}\frac{\delta f }{\delta y}}=\frac{(y+x \frac{\delta f}{\delta x})^2}{2 \frac{\delta f}{\delta x}}$$

$$A_x = \frac{2 \frac{\delta f}{\delta x}\cdot 2\cdot(y+x \frac{\delta f}{\delta x})(\frac{\delta f}{\delta x}+x\frac{\delta ^2f}{\delta x^2})-2\cdot \frac{\delta ^2 f}{\delta x^2}(y+x\frac{\delta f}{\delta x})^2 }{4(\frac{\delta f}{\delta x})^2}=y+x\frac{\delta f}{\delta x}-\frac{y^2 \frac{\delta ^2 f}{\delta x^2}}{2 (\frac{\delta f}{\delta x})^2}+\frac{x^2\frac{\delta ^2 f}{\delta x^2}}{2}$$

$$A_y =\frac{4\frac{\delta f}{\delta x}(y+x\frac{\delta f}{\delta x})(1+x\cdot \frac{\delta ^2 f}{\delta x \delta y})-2\frac{\delta ^2f}{\delta x \delta y}(y\frac{\delta f}{\delta y}+x\frac{\delta f}{\delta x})}{4(\frac{\delta f}{\delta x})^2} =\frac{y+x\frac{\delta f}{\delta x}}{ \frac{\delta f}{\delta x}}$$

By Lagrange Multipliers, $\lambda f_x=A_x$ and $\lambda f_y = A_y$

$f_y=1$, so $\lambda=\frac{y+x \frac{\delta f}{\delta x}}{\delta f/ \delta x} $, and $\lambda f_x=y+x\frac{\delta f}{\delta x}=y+x\frac{\delta f}{\delta x}-\frac{y^2 \frac{\delta ^2 f}{\delta x^2}}{2 (\frac{\delta f}{\delta x})^2}+\frac{x^2\frac{\delta ^2 f}{\delta x^2}}{2}$

So:

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

And:
$$(x\frac{\delta f}{\delta x}-y)(x\frac{\delta f}{\delta x}+y)=0$$

If the second term is zero, we have zero area, so set the first term to zero.

But, $\frac{\delta f}{\delta x}=-\frac{dy}{dx}$, so $(x\frac{\delta f}{\delta x}-y)=0 \implies \frac{d(xy)}{dx}=0$

Since $y=x\frac{\delta f}{\delta x}$, $A=2x^2\frac{\delta f}{\delta x}=2xy$


问:
New x coordinate of a rotated line
I need help finding the equation to find $x$

I work in GIS and I'm working on a script that uses the new x coordinate of a rotated line. I havent work with trigonometry in a long long time so I would much appreciate help with this equation.

I calculated the distance variable myself so it might be wrong $ \sqrt{(y_2 - y_1)^2 +(x_2-x_1)^2}$.

答:
firstly the distance you have find between two known points,is wrong:
$$\sqrt{(36000-35500)^2+(653000-653500)^2}\implies\sqrt{(500)^2+(-500)^2}\implies707.1\, approx.$$
then find its slope $m_2=\dfrac {653000-653500}{36000-35500}\implies m_2=-1$

suppose slope of 2nd line is $m_1$ then angle between two lines :

$$\tan \theta=|\dfrac{m_1-m_2}{1+m_1m_2}|$$
$$\tan 45^\circ=|\dfrac{m_1-m_2}{1+m_1m_2}|$$
$$1=|\dfrac{m_1+1}{1-m_1}|\implies m_1=0$$ it means this line is parallel to X-axis.So the coordinate $(x,y)$ will be $(x,653000)$.
Now use distance formula (I assume that your given value $538.5m$ is right.)

$$\sqrt{(x-36000)^2}=538.5$$
$$x=36538.5m$$

ContentLicense: CC BY-SA 3.0

Title: New x coordinate of a rotated line


问:
If $f,g\in L^2(\mathbb{R})$ then $\lim_{n\to \infty}\int_{\mathbb{R}}f(x)g(x+n)=0$
I don't know if this is true or false. I tried counterexamples but couldn't find any so then I tried to prove this with some theorems regarding convolutions because it looks a little bit like a convolution but couldn't find any that I could use. Can someone give me a hint?</p>

答:
Let $f,g$ be square integrable functions (not equivalent classes
of functions). Let $S_{f}=\{x\in\mathbb{R}\mid f(x)\neq0\}$ and $S_{g}=\{x\in\mathbb{R}\mid g(x)\neq0\}$
be the supports of $f$ and $g$ respectively.

Case 1. $S_{f},S_{g}$ are bounded subsets of $\mathbb{R}$: Note
that for sufficiently large $n$, $f(x)g(x+n)=0$ for all $x$, so
$\int f(x)g(x+n)dx=0$.

Case 2. Let $f_{n}(x)=\begin{cases}
f(x), & \mbox{if }x\in[-n,n]\\
0, & \mbox{otherwise}
\end{cases}$ and $g_{n}=\begin{cases}
g(x), & \mbox{if }x\in[-n,n]\\
0, & \mbox{otherwise}
\end{cases},$ then $f_{n},g_{n}\in L^{2}$ with bounded supports. Note that $|f_{n}-f|^{2}\leq|f|^{2}$
and $|f_{n}-f|^{2}\rightarrow0$, so by Dominated Convergence Theorem,
$||f_{n}-f||_{2}\rightarrow0$. Similarly, $||g_{n}-g||_{2}\rightarrow0$.
Choose $M>0$ such that $||f||_{2}\leq M$ and $||g||_{2}\leq M$.
Let $\varepsilon>0$ be given. Choose $N$ such that $||f_{N}-f||_{2}<\frac{\varepsilon}{M}$
and $||g_{N}-g||_{2}<\frac{\varepsilon}{M}$. By Case 1, choose $K$
such that $\int f_{N}(x)g_{N}(x+k)dx=0$ whenever $k\geq K$. For
any $k\geq K$, we have that
\begin{eqnarray*}
 &  & \int f(x)g(x+k)dx\\
 & = & \int\left[f(x)g(x+k)-f_{N}(x)g(x+k)\right]dx+\int\left[f_{N}(x)g(x+k)-f_{N}(x)g_{N}(x+k)\right]dx.
\end{eqnarray*}
By Cauchy-Schwarz inequality,
\begin{eqnarray*}
 &  & \left|\int\left[f(x)g(x+k)-f_{N}(x)g(x+k)\right]dx\right|\\
 & \leq & ||f-f_{N}||_{2}||g(\cdot+k)||_{2}\\
 & = & ||f-f_{N}||_{2}||g||_{2}\\
 & \leq & \varepsilon.
\end{eqnarray*}
By Cauchy-Schwarz inequality again,
\begin{eqnarray*}
 &  & \left|\int\left[f_{N}(x)g(x+k)-f_{N}(x)g_{N}(x+k)\right]dx\right|\\
 & \leq & ||f_{N}||_{2}\cdot||g(\cdot+k)-g_{N}(\cdot+k)||_{2}\\
 & \leq & ||f||_{2}\cdot||g-g_{N}||_{2}\\
 & \leq & \varepsilon.
\end{eqnarray*}
This shows that $\lim_{k\rightarrow\infty}\int f(x)g(x+k)dx=0.$


问:
Vanishing partial trace
Given a square matrix $A$ (could be chosen to be unitary, and possibly also involutive if it helps), then how do we find a matrix $M$ such that
$$\text{pt}(MA) = 0?$$

On a pure tensor, the partial trace above acts as $\text{pt}(A\otimes B) = \text{Tr}(A) B$ and can then be extended to general $n^2 \times n^2$ matrices by linear combinations.

In the case of a full trace, we of course can say for example that if $MA = BC - CB$ for some $C$ and $B$, then the trace vanishes, and there are also other possiblities. But what about the case of a partial trace?

答:
Express $A$ in the form
$$
A = \sum_{j,k = 1}^n E_{jk} \otimes A_{jk}
$$
where $E_{jk}$ denotes the matrix with $1$ as its $j,k$ entry and zeros elsewhere. Equivalently, each $A_{jk}$ is an $n \times n$ "block-entry" of $A$. Partition $M$ in the same fashion.

We can write the $p,q$ entry of $\operatorname{pt}(MA)$ as
$$
\operatorname{tr}\left(\sum_{j=1}^n M_{pj} A_{jq}\right) = \operatorname{tr}(M C_{pq}^T),
$$
where $C_{pq}$ denotes the matrix whose block-entries are zero except for the $p$-th row, and the block-entries of the $p$th row are $A_{1q}^T,\dots,A_{nq}^T$. Now, in terms of the vectorization operator, we have
$$
\operatorname{tr}(MC_{pq}^T) = \operatorname{vec}(C_{pq})^T \operatorname{vec}(M).
$$
With that, we see that $\operatorname{vec}(M)$ must be a solution to the equation $C x = 0$, where $C$ is the matrix whose rows are $\operatorname{vec}(C_{pq})^T$ for all $1 \leq p,q \leq n$.

To simplify the above a bit: we can write
$$
C_{pq} = \sum_{j=1}^n E_{pj} \otimes A_{jq}^T,
$$
so that
$$
\operatorname{vec}(C_{pq}) = \sum_{j=1}^n \operatorname{vec}(E_{pj} \otimes A_{jq}^T) = \sum_{j=1}^n \sum_{k=1}^n e_j \otimes e_k \otimes e_p \otimes A^T_{jp}[k],
$$
where $M[k]$ denotes the $k$th column of the matrix $M$.

Title: Vanishing partial trace


问:
Finding the area of this figure inside a circle?
Let's say I have a unit circle and I draw a triangular sector with an angle $\pi/2$. Next to it, I draw a sector with an angle $\pi/8$, next $\pi/18$, so that the $n^{\text{th}}$ sector has an angle of $\pi/(2n^2)$. Now I am only looking at the area of the triangle inside each sector. What is the area of the triangles if I were to draw infinitely many of them, as shown in the figure below?

I was only able to find an approximation for this. I got $f(1)/2 + \int_1^\infty f(x)dx$, where $f(x)=(1/2)\sin(\pi/(2x^2)) \approx 1.95$. How would you find the exact area of this?

答:
As you seem to know, the area of an isosceles triangle with common side length 1 and angle between them $\theta$ is $\sin(\theta)/2$.  Thus, your area is

$$\sum _{i=1}^{\infty } \frac{1}{2} \sin \left(\frac{\pi }{2 i^2}\right),$$

which is just a bit bigger than 1.  I don't see any easy reason to think that this is expressible in closed form; Mathematica, at least, can't seem to do so.


问:
Conversion of minimization to maximization objective under Frobenius norm
I want to understand how Schnass $^\dagger$ arrived at the maximising objective as described below (on page 5)

\begin{aligned} \displaystyle \min_{D,A} \Vert X-DA \Vert_F^2 &= \min_D \sum_n \min_{|I|\leq S} \Vert x_n -D_ID_I^{\dagger}x_n \Vert_2^2 \\ &= \Vert X \Vert_F -   \max_D \sum_n \max_{|I|\leq S} \Vert D_ID_I^{\dagger}x_n \Vert_2^2 \end{aligned}

where $\dagger$ is the pseudo-inverse, $S$ denotes the cardinality of column vectors $a_n$ of matrix $A$.

$^\dagger$ K. Schnass, On the Identifiability of Overcomplete Dictionaries via the Minimisation Principle Underlying K-SVD, Applied and Computational Harmonic Analysis, 37(3):464--491, 2014.

答:
First, decompose

$$
\| x_n - A x_n \|_2^2 = \| x_n \|_2^2 + \| A x_n \|_2^2 - 2 \langle x_n, A x_n \rangle
$$

In your case, $A = D_I D_I^{\dagger}$. By the properties of the Moore-Penrose pseudoinverse, we know that $D_I D_I^{\dagger} = D_I D_I^{\dagger} D_I D_I^{\dagger}$. Therefore,

$$
\| x_n - D_I D_I^{\dagger} x_n \|_2^2 = \| x_n \|_2^2 + \| D_I D_I^{\dagger} x_n \|_2^2 - 2 \langle x_n, D_I D_I^{\dagger} D_I D_I^{\dagger} x_n \rangle
= \| x_n \|_2^2 - \| D_I D_I^{\dagger} x_n \|_2^2.
$$

The rest follows from the fact that $\sum_{n} \|x_n \|_2^2 = \| X \|_F^2$ and the property $\min_{z} x - f(z) = x - \max_{z} f(z)$.

Another proof of the first decomposition: $D_I D_I^{\dagger}$ is a projection, so by the Pythagorean theorem you have

$$
\| x_n \|_2^2 = \| D_I D_I^{\dagger} x_n \|_2^2 + \| (I - D_I D_I^{\dagger}) x_n \|_2^2 \Rightarrow
\| x_n - D_I D_I^{\dagger} x_n \|_2^2 = \|x_n\|_2^2 - \|D_I D_I^{\dagger} x_n \|_2^2.
$$


问:
How to setup a function that has a defined second derivative and have two maxima but no minima?
For no minima on the endpoint, I am thinking a using a polynomial with odd degree so that the both ends will go upward to positive and negative infinity respectively.

However, I have no idea how to have two maxima but no minima in between.

Please leave a hint, and much appreciated!

答:
If a function of one variable has two local maxima, it must have a local minimum in between.

A higher dimensional function could have two local maxima and no other critical points. Take for a example
$$f(x,y)=-(x^2-1)^2-(x^2y-x-1)^2.$$
This function has two local maxima at $(-1,0)$ and at $(1,2)$. You can write out the partial derivatives with respect to $x$ and $y$ and see that the only place they’re both zero is at the two local maxima.

Here’s a plot of the function:


问:
Let $f'\ge0, f''\ge0 \ \forall x \in \mathbf{R}$. If there is $c$ such that $f'(c)>0$, then $\lim_{x\to\infty}f(x)=\infty$

Let $f$ be a twice differentiable function. Suppose $f'(x)\ge0, f''(x)\ge0 \ \forall x \in \mathbf{R}$. If there exists $c \ \in R$ such that $f'(c)>0$, show that $\lim_{x\to\infty}f(x)=\infty$. My attempt is to prove the statement by absurd. So, I would like to have a feedback on my proof, please. Setting myself that $f(c_x)>0$ in my proof seems to me a little bit strange..

Suppose by absurd that $f$ converges to $L$: $\lim_{x\to\infty}f(x)=L$. As $f''\ge0$, then $f$ is convex and we can deduce that $f'$ is increasing.

Consider the interval $[x,x+a]$ with $a>0$.  As $f$ is differentiable on $\mathbf{R}$, it is differentiable on $[x,x+a]$ and so continuous on $[x,x+a]$. By mean value theorem there exists $c_x\in ]x,x+a[$  such that: $f'(c_x)=\frac{f(x+a)-f(x)}{a}$. As $x$ is "arbitrary" in the interval, we can suppose that $f'(c_x)=\frac{f(x+a)-f(x)}{a}>0$. But,

$f'(c_x)=\frac{f(x+a)-f(x)}{a} \iff \underbrace{\lim_{x\to\infty}f'(c_x)}_{(*)}=\lim_{x\to\infty}\frac{f(x+a)-f(x)}{a}=\frac{L-L}{a}=0\ngtr 0$. So we got a contradiction. Thus, $f$ diverges.

(*) As $f'$ is increasing, as $x$ goest to infinity the inequality might hold and so $\lim_{x\to\infty}f'(c_x)>0$ might hold.

答:
I think your proof is fine, but the idea behind the proof is somewhat obscured. To me, it seems the key point is to prove there is a line of positive slope that bounds $f$ from below. Indeed, if $f'(c) > 0$, then $f(x) - f(c) = f'(c_x)(x-c) \geq f'(c)(x-c)$. The right side clearly tends to $\infty$ as $x \to \infty$, hence so must $f$.

Edit, I should also point out that this proof also works for convex $f$ (not assumed to be twice differentiable), as convex implies $f$ is above any tangent line, which is all we really used.


问:
Let $f:\mathbb R \to \mathbb R$ be continuous. Then which of the following statements implies that $f(0)=0$?

Let $f:\mathbb R \to \mathbb R$ be continuous. Then which of the
  following statements implies that $f(0)=0$?

(A)$\lim_{n \to \infty}\int_{0}^{1}f(x)^n dx=0.$

(B)$\lim_{n \to \infty}\int_{0}^{1}f(x/n) dx=0.$

(C)$\lim_{n \to \infty}\int_{0}^{1}f(nx) dx=0.$

(D) None of the above.

(A)Suppose $f(x)=1-x, \int_{0}^{1}(1-x)^n dx=-\frac{1}{n+1},$ which converges to zero. I tried to find the counter example for the (B) and (C). I couldn't find. please help me.

答:
B ) $0=\lim_{n\to \infty} \int_0^1f(\frac{x}{n})\, dx\\=\lim_{n\to \infty} f(\frac{c}{n})(1-0)  for \, some \, c\in (0,1)\\= \lim_{n\to \infty}f(\frac{c}{n})\\= f(0) \, since \, f \, is \, continuous .
$


问:
Finding a vector subspace
Considering the exponential map $\exp : M_{n*n}(\mathbb{C}) → GL(n, \mathbb{C})$ we know that it is a local diffeomorphism at $0$.
How can we find the appropriate vector subspaces $V( U(n))$ such that the restriction:

$$ \exp: V(U(n)) \longrightarrow U(n)$$

is a local diffeomorphism at $0$?

My attempt:

Choosing $A \in V(U(n))$ and writing it in its eigenbasis it will only have elements, whose norm is $1$, in the main diagonal so its easy to see that:

$$ \exp(-A) = [\exp(A)]^{-1}=[\exp(A)]^{\dagger}=\exp(A^{\dagger})$$

Applying the exponential map to:

$$ A^{\dagger}+A=0 \longrightarrow \exp(A^{\dagger}+A)= \exp(A^{\dagger}) \exp(A) = [\exp(A)]^{\dagger}\exp(A)=\exp(-A)\exp(A) = \exp(A-A)= I$$

then, the restriction that we must apply is:

$$ A^{\dagger}+A=0$$

Is my attempt correct?

答:
Hint: For $A\in V(U(n))$, you want
$$ \exp(-A)=\exp(A)^{-1}=\exp(A)^H=\exp(A^H)$$

Title: Finding a vector subspace


问:
Conjecture: $-\partial_b F(b-a,b-a;b;z)$ is increasing in $z$ (proof of special case provided in answers)
See answers below for a proof of a special case to the following conjecture.

I am looking to prove the following conjecture:

Conjecture:
Let $0<b<a<b+1$. Then,
$$
f(z):=-\frac{\mathrm d}{\mathrm db}{_2F_1}\left({b-a,b-a\atop b};z\right)
$$
is positive and strictly increasing on $z\in(0,1)$.

I have successfully shown for all $a,b>0$ and $z\in(0,1)$ that $\operatorname{sign}(f(z))=\operatorname{sign}(b-a)$, which gives the necessary result for the first part of the conjecture. However, I have been unable to establish monotonicity in $z$ for the full domain of $a$. So is this conjecture indeed true? The proof below shows that $f>0$ for $b<a<b+1$ and I believe it contains clues for establishing that $f$ is also increasing in $z$ (it does...see answer below for special case).

Proof that $f>0$ for $0<b<a<b+1$:

Using DLMF 15.8.1 we write
$$
\begin{aligned}
f(z)
&=-\partial_b F(b-a,b-a;b;z)\\
&=-\partial_b (1-z)^{2a-b}F(a,a;b;z)\\
&=(1-z)^{2a-b}(Q_2-Q_1),
\end{aligned}
$$
where $Q_1=-\log(1-z)F(a,a;b;z)$ and $Q_2=-\partial_b F(a,a;b;z)$. Upon inspection of their power series expansions in $z$, both $Q_1$ and $Q_2$ have nonnegative coefficients and thus are positive on $z\in(0,1)$. Therefore, all we need to show is is $Q_1<Q_2$. Using the series expansion for the logarithm and hypergeometric function we find for the product
$$
Q_1=\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k}\sum_{\ell=0}^{k-1}\frac{(a+k)_{-\ell-1}^2}{(b+k)_{-\ell-1}(k-\ell-1)!}\frac{1}{\ell+1}z^k,
$$
where $(a)_n:=\Gamma(a+n)/\Gamma(a)$ is the Pochhammer symbol (rising factorial). The series coefficients are positive and $(x+k)_{-\ell-1}$ is a positive decreasing function in $x$ on $x\in\Bbb R^+$. Since $0<b<a$ it follows that $0<(a+k)_{-\ell-1}<(b+k)_{-\ell-1}$ and
$$
\begin{aligned}
Q_1%
&<\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k}\sum_{\ell=0}^{k-1}\frac{(b+k)_{-\ell-1}}{(k-\ell-1)!}\frac{1}{\ell+1}z^k\\
&=\sum_{k=1}^\infty\frac{(a)_k^2}{(b)_k\,k!}(\psi(b+k)-\psi(b))z^k\\
&=Q_2,
\end{aligned}
$$
with $\psi(z):=\partial_z\log\Gamma(z)$ is the digamma function. The proof is now complete. $\quad\square$

答:
$\newcommand{\pFq}[5]{{_{#1}F_{#2}}\left({#3\atop #4};#5\right)}$
I was able to find a proof that holds for the special case: $b+\color{red}{1/2}\leq a<b+1$. As you will see in the proof, there is only one step that introduces the restriction $b+1/2\leq a$ and my hope is someone can improve on this step to prove the conjecture in the affirmative. I will point out this step in the proof by highlighting it in red. Intuition tells me this may come in handy for a proof of the general case of the conjecture. We need two lemmas and then the proof will follow.

Lemma 1: For $x>0$, $k\in\Bbb N$, and $\ell=0,\dots,k-1$:
$$
(x+k+1/2)_{-\ell-1}^2\leq(x+k)_{-\ell-1}(x+k+1)_{-\ell-1}.
$$

Proof: The proof is conducted by minimizing $g(a)=(x+k+a)_{-\ell-1}(x+k+1-a)_{-\ell-1}$ on $a\in[0,1]$. One can easily show that $g^\prime(1/2)=0$ and $g^{\prime\prime}(1/2)>0$. $\quad\square$

Lemma 2: For $x>0$ and $k\in\Bbb N$:
$$
\sum_{\ell=0}^{k-1}\frac{(x+k)_{-\ell-1}}{(k-\ell-1)! (\ell+1)}=\frac{\psi(x+k)-\psi(x)}{k!},
$$
where $\psi(z):=\partial_z\log\Gamma(z)$ is the digamma function.

Theorem: Let $b+\color{red}{1/2}\leq a<b+1$. Then $\partial_z f>0$ on $z\in(0,1)$.

Proof: Using the definition of $f(z)$ and the hypergeometric transformation of DLMF 15.8.1 we have
$$
\begin{aligned}
\partial_z f(z)%
&=-\partial_b\partial_z\pFq{}{}{b-a,b-a}{b}{z}\\
&=-\partial_b\frac{(b-a)^2}{b}\pFq{}{}{b+1-a,b+1-a}{b+1}{z}\\
&=-\partial_b\frac{(b-a)^2}{b}(1-z)^{2a-b-1}\pFq{}{}{a,a}{b+1}{z}\\
&=\cdots\\
&=\frac{a-b}{b}(1-z)^{2a-b-1}\left(\left(1+\frac{a}{b}\right)\pFq{}{}{a,a}{b+1}{z}+(a-b)(Q_2-Q_1)\right),
\end{aligned}
$$
where $Q_1=-\log(1-z)F(a,a;b+1;z)$ and $Q_2=-\partial_b F(a,a;b+1;z)$. Using the hypergeometric series, logarithmic series, and Cauchy product of power series we have after some algebraic manipulation
$$
Q_1=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\sum_{\ell=0}^{k-1}\frac{(a+k)_{-\ell-1}^2}{(b+1+k)_{-\ell-1}(k-\ell-1)!(\ell+1)}z^k
$$
and
$$
Q_2=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\frac{\psi(b+1+k)-\psi(b+1)}{k!}z^k.
$$
Both $Q_1$ and $Q_2$ have positive coefficients so that $Q_1,Q_2>0$ on $z\in(0,1)$. Furthermore, noting that $(x+k)_{-\ell-1}$ is positive and decreasing in $x$ one can show that $Q_2-Q_1<0$. Thus, our goal will be to bound $\partial_z f$ from below by bounding $Q_2-Q_1$ from below and then showing that this lower bound for $\partial_z f$ is positive.

$\color{red}{\text{If we restrict $b+1/2\leq a<b+1$}}$ then we may use the monotony of $(x+k)_{-\ell-1}$ and Lemma 1 to write
$$
\begin{aligned}
Q_1%
&<\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k}\sum_{\ell=0}^{k-1}\frac{(b+k)_{-\ell-1}(b+k+1)_{-\ell-1}}{(b+1+k)_{-\ell-1}(k-\ell-1)!(\ell+1)}z^k\\
&=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_k k!}(\psi(b+k)-\psi(b))z^k\\
&=:Q_1^\ast.
\end{aligned}
$$
It follows using the recurrence relation for the digamma function that
$$
\begin{aligned}
0 &>Q_2-Q_1\\
&>Q_2-Q_1^\ast\\
&=\sum_{k=1}^\infty\frac{(a)_k^2}{(b+1)_kk!}\left(\frac{1}{b+k}-\frac{1}{b}\right)z^k\\
&=-\frac{a^2 z}{b(b+1)^2}\sum_{k=0}^\infty\frac{(a+1)_k^2}{(b+2)_kk!}\frac{b+1}{b+k+1}z^k\\
&=-\frac{a^2 z}{b(b+1)^2}\pFq{3}{2}{b+1,a+1,a+1}{b+2,b+2}{z}.
\end{aligned}
$$

Now using the integral representation in DLMF 16.5.2 we write
$$
\begin{aligned}
Q_2-Q_1^\ast%
&=-\frac{1}{b}\int_0^1 t^b \frac{a^2 z}{b+1}\pFq{}{}{a+1,a+1}{b+2}{zt}\,\mathrm dt\\
&=-\frac{1}{b}\int_0^1 t^b \partial_t\pFq{}{}{a,a}{b+1}{zt}\,\mathrm dt\\
&=\text{IBP}\cdots\\
&=-\frac{1}{b}\pFq{}{}{a,a}{b+1}{z}+\int_0^1 t^{b-1} \pFq{}{}{a,a}{b+1}{zt}\,\mathrm dt\\
&=-\frac{1}{b}\pFq{}{}{a,a}{b+1}{z}+\frac{1}{b}\pFq{3}{2}{b,a,a}{b+1,b+1}{z},
\end{aligned}
$$
Substituting this result into the expression for $\partial_z f$ and simplifying we find
$$
\begin{aligned}
\partial_z f%
&>\frac{a-b}{b}(1-z)^{2a-b-1}\left(\left(1+\frac{a}{b}\right)\pFq{}{}{a,a}{b+1}{z}+(a-b)(Q_2-Q_1^\ast)\right)\\
&=\frac{a-b}{b}(1-z)^{2a-b-1}\left(2\pFq{}{}{a,a}{b+1}{z}+\frac{a-b}{b}\pFq{3}{2}{b,a,a}{b+1,b+1}{z}\right)\\
&>0,
\end{aligned}
$$
since all terms are positive. $\quad\square$

All that is needed is to correct the one step that introduces the restriction on $a$.


问:
reducing a feasible solution to a basic feasible solution
I have a doubt in Switching the solutions in linear programming. Here, we are reducing a feasible solution to a basic feasible solution. I know the result that in an LP, the optimal solution always occurs at extreme points (i.e. basic feasible solutions). However, what I am unable to understand is:

Why is there a need to reduce a feasible solution to a basic feasible solution? Also, a feasible point can always be written as a convex combination of the extreme points of a feasible region, so why reduce it?

I just want to confirm that my thought is true (it is likely very trivial, I apologize): We can reduce a feasible point to a basic feasible solution in multiple ways by taking different non-zero variable $0.$

答:
Some algorithms, typically the simplex algorithm, work by moving around along the corner points (because of the guaranty that the optimal solution is among them). This is nice, but to start this you have to find one first point to begin with. So the global algorithm is : Find any feasible point (you can solve this as a simpler LP problem), then use it to find an corner point, and then only move between the corner points of the problem.

Yes, "reducing" isn't a very well defined operation. But it can be expressed as... a simpler linear programming problem. Which can be solved in multiple ways.


问:
How to prove that set A is Lebeg measurable?
I'm wondering how to solve this.
We have an infinite countable set A ⊂ R is such that λ * (A) =0,
. I would appreciate any help. Thanks.</p>

答: Any countable set is measurable since it is a countable union of singletons, which are measurable.</p>


问:
strong law of large number with Covariance
I'm supposed to prove that $ \frac{1}{n} \sum_{j=1}^n (X_j-\overline X_n)(Y_j-\overline Y_n)$ converges almost surely to $Cov(X,Y)$ assuming that $ (X_i,Y_i)$ are iid with the same distribution as $(X,Y)$ for $i=1,\dots,n$ .  Clearly that is the same as proving that $ \frac{1}{n} \sum_{j=1}^n X_j Y_j $ converges almost surely to $ E[XY]$ because $ \frac{1}{n} \sum_{j=1}^n (X_j-\overline X_n)(Y_j-\overline Y_n) =  \frac{1}{n} \sum_{j=1}^n X_j Y_j - \overline X_n \overline Y_n$ as $ \overline X_n \overline Y_n$ converges almost surely to $ E[X]E[Y]$ and we have the formula $ Cov(X,Y)=E[XY]-E[X]E[Y]$

However I'm not sure how to prove that because it'll require to prove that the $X_1Y_1,\dots,X_nY_n$ are i.i.d. (if we use the strong law of large number) and I don't know how to prove that...any hints ?
Thans you

Note : $ \overline X_n = \frac{1}{n}\sum_{j=1}^n X_j $ (same for $\overline{Y_n}$)

答:
For any  $t\in\mathbb R$, let $D_t=\{(x,y): xy \leqslant t\}$. This is Borel set in $\mathbb R^2$. Then
$$F_{X_1Y_1}(t)=\mathbb P(X_1Y_1 \leqslant t) = \mathbb P((X_1,Y_1) \in D_t)=\mathbb P((X_2,Y_2) \in D_t)=\mathbb P(X_2Y_2 \leqslant t)=F_{X_2Y_2}(t).$$
The central equality is valid since $(X_1,Y_1)$ has the same distribution as $(X_2,Y_2)$.

So we obtain that $X_1Y_1,\ldots, X_nY_n$ are identically distributed.

Next, for any $t,s$
$$
\mathbb P(X_1Y_1 \leqslant t,\, X_2Y_2 \leqslant s) = \mathbb P((X_1,Y_1) \in D_t,\, (X_2,Y_2) \in D_s)
$$
(use independence of $(X_1,Y_1)$ and $(X_2,Y_2)$)
$$
= \mathbb P((X_1,Y_1) \in D_t)\cdot \mathbb P( (X_2,Y_2) \in D_s) = \mathbb P(X_1Y_1 \leqslant t) \cdot \mathbb P(X_2Y_2 \leqslant s).
$$

So we get independence of  $X_1Y_1, X_2Y_2$. The same way independence of  $X_1Y_1,\ldots, X_nY_n$ can be proved.


问:
From geometric sequence to function
I have this question:

Find the functions which equal the sums:
$$
x + x^3 + x^5 + ..
$$

Now, I can see in my result list, that its supposed to give

$$
\frac{x}{(1-x^2)}
$$

I can see why the numerator should be x, but I fail to see why the denominator should be
$$
(1-x^2)
$$
Can someone please explain this to me?

David

答: Rewrite it as $x\left(1+(x^2)^1+(x^2)^2+\dots+(x^2)^k+\dots\right)$, and use what you know about the series $\sum_{k=0}^\infty \alpha^k$ for $|\alpha| < 1$.</p>

Title: From geometric sequence to function


问:
Replace linear return with cosine return
Below you will find the position profile in time for a servo-device I am building.  Currently the motion follows the orange profile moving away from zero to a maximum value and then quickly returning to zero to start the cycle all over again.  At $2$ Hz the period is $0.5$ seconds and the maximum amplitude is $1563$ mm. The linear "return to zero" line (orange) has sharp corners where excessively large accelerations are required to follow the profile.  Instead I'd like to define a portion of a cosine function (shown in blue) running from something less than zero to something beyond $\pi$ as the graphical solution shows in blue.  Clearly the majority of the cosine function will be something like $y = 1563\cdot((\cos(\theta)/2)+0.5)$ for $\theta = 0$ to PI.  But I don't know how to figure out how far to extend the function beyond those endpoints to have it's slope exactly match the slope of the $2$ orange lines shown on the image.  The cosine function including the extensions must also fit in the $0.1483$ seconds window allocated for "return to zero".  I believe the solution can be found by differentiating the position equation and setting it equal to the specified velocities (slopes) but I don't know how to do that.

答:
You are looking to replace the "linear return" with a function of the form
$$
f(t) = A \cos(Bt + C) + D
$$
for a suitable choice of coefficients $A,B,C$. the conditions that must hold are as follows:

The maximum of $f$ is $1563 \implies A + D = 1563$

$f'(.2151) = 2228 \implies -AB \sin(.2151\cdot B + C) + D = 2228$

At $t = .3634$ the slope is $7897 \implies -AB \sin(.2151\cdot B + C) + D = 7897$

You do not say this explicitly, but it seems that the cosine function is supposed to connect to the data at the start and end points. So, we will have two more equations of the form

$f(.2151) = y_1 \implies A \cos(.2151 \cdot B + C) + D = y_1$

$f(.3634) = y_2 \implies A \cos(.3634 \cdot B + C) + D = y_2$.

What you're looking for is a solution for $A,B,C,D$ to all $5$ of these equations.
Because there are $5$ equations for only $4$ variables, this is a system of equations that is likely overdetermined; for that reason, you should probably ignore the first equation and solve the rest. I do not believe that this equation is analytically solvable, but it should be possible to obtain a numerical solution with any equation solver (e.g. using SymPy).

Note: We might be able to simplify the solution process here by rewriting $f$ as
$$
f(t) = A\cos(Ct) + B\sin(Ct) + D
$$
and then using trigonometric identities to rewrite the function in the desired form.

A more standard approach towards the effect that you are trying to achieve is to connect the two data points with a natural cubic spline. The computational process would be more straightforward.

In particular, you would be looking to replace the linear return with a function of the form
$$
f(t) = At^3 + Bt^2 + Ct + D.
$$
Equations 2-5 would define a linear system:
$$
f'(.2151) = 2228 \implies (3 \cdot (.2151)^2)A + (2 \cdot .2151)B + C = 2228\\
f'(.3634) = 7897 \implies (3 \cdot (.3634)^2)A + (2 \cdot .3634)B + C = 7897\\
f(.2151) = y_1 \implies (.2151)^3 A + (.2151)^2 B + (.2151) C + D = y_1\\
f(.3634) = y_1 \implies (.3634)^3 A + (.3634)^2 B + (.3634) C + D = y_2.\\
$$
In matrix form, the system of equations is
$$
\pmatrix{3 \cdot (.2151)^2 & 2 \cdot .2151 & 1 & 0\\
3 \cdot (.3634)^2 & 2 \cdot .3634 & 1 & 0\\
(.2151)^3 & (.2151)^2 & .2151 & 1\\
(.3634)^3 & (.3634)^2 & .3634 & 1} \pmatrix{A\\B\\C\\D} = \pmatrix{1563\\2228\\y_1\\y_2}.
$$


问:
What are $\vec{x}$ and $\vec{b}$ in $A \vec{x} = \vec{b}$?
The condition of a matrix gives a bound on how inaccurate the solution $\vec{x}$ will be after approximation. So given $A \vec{x} = \vec{b}$, how much does $\vec{x}$ change when there is a change in $\vec{b}$. Here is an example:

$$\pmatrix{1&1\\1&1.001} \pmatrix{x\\y} = \pmatrix{2\\2} \implies x=2, y=0$$
$$\pmatrix{1&1\\1&1.001} \pmatrix{x\\y} = \pmatrix{2\\2.001} \implies x=1, y=1$$

This is an ill-conditioned matrix because a small change in $\vec{b}$ lead to a big change in $\vec{x}$. But here's where I am confused. The matrix $A$ can be considered as observed values or input data. However, I am not sure what $\vec{x}$ and $\vec{b}$ are in this case.

答:
This question isn't really about condition number at all, but rather "how do I interpret a system of linear equations".  In general, $A$ is a matrix of coefficients, $\vec{x}$ is a vector of unknowns, and $\vec{b}$ is a vector of constants.  The single matrix equation you have above relates to two linear equations in two unknowns.  Specifically:

$$x + y = 2$$

and

$$x + 1.001y = 2$$ (or 2.001, depending on which version you are using)

So one (geometric) interpretation of this matrix equation is to find the intersection points between the two lines $y = -x + 2$ and $y = \frac{-1}{1.001}x + 2$.  In this sense, the vector $\vec{b}$ represents the y intercepts.  Of course, this assumes that your quantities are in $\mathbb{R}^2$ and that you are able to use this interpretation for useful results.

Otherwise, "what do these quantities represent" is answered by "whatever they need to for your application".


问:
Reference for surface area (volume) formula
I am having a hard time finding a reference book for the following theorem.  It seems to be always presented in special cases in elementary books, and then the advanced (graduate) books skip it get bogged down in complicated technology.

Theorem: Suppose $S$ is a smooth surface in $\mathbb R^n$.  In other words, there is a function $f : V \to \mathbb R^n$, where $U \subseteq V$ are bounded open subsets of $\mathbb R^m$ ($m \leq n$) such that the closure $\bar U \subseteq V$, $f$ is injective and continuously differentiable on $V$, and $S = f[U]$.  Then the volume of $S$ is: $\operatorname{vol}_f(S)=$ [insert formula involving Jacobian of $f$].  And if $g : U’ \to \mathbb R^n$ is any other smooth parameterization of $S$, then $\operatorname{vol}_g(S) = \operatorname{vol}_f(S)$.

Much thanks.

答: Mathematical Analysis II by Vladimir Zorich.</p>


答: $f(x)=-x^{-2}$ has two maxima at $\pm \infty $ and no minimum.

Title: How to setup a function that has a defined second derivative and have two maxima but no minima?

Body:
For no minima on the endpoint, I am thinking a using a polynomial with odd degree so that the both ends will go upward to positive and negative infinity respectively.

However, I have no idea how to have two maxima but no minima in between.

Please leave a hint, and much appreciated!


答:
Let your series be $s(x)$ with $s(x)=xp(x)$

Then note that $x^2p(x)=p(x)-1$ which reduces to $p(x)=\cfrac 1{1-x^2}$

Title: From geometric sequence to function

Body:
I have this question:

Find the functions which equal the sums:
$$
x + x^3 + x^5 + ..
$$

Now, I can see in my result list, that its supposed to give

$$
\frac{x}{(1-x^2)}
$$

I can see why the numerator should be x, but I fail to see why the denominator should be
$$
(1-x^2)
$$
Can someone please explain this to me?

David


问:
Example of $\mathbb{E}[X_t|\mathcal{F}_0]=X_0$ but $X_t$ is not a martingale?
Edit: Some great examples in the comments and in the answer below by @NullUser: it would be great to see even more examples, if anyone can come up with more please. Happy to award a small bounty on this if incentive needed. Trying to collect as many different examples as possible.

I am looking for examples of processes that satisfy the condition that:

$$\mathbb{E}\left[X_t|\mathcal{F}_{t_0}\right]=X_{t_0}$$

But where $X_t$ is not a martingale, i.e. where for some $0<s\leq t$:

$$\mathbb{E}\left[X_t|\mathcal{F}_s\right]\neq X_s$$

I can't think of any such process at the moment, although I distinctly remember seeing examples of such processes in the past. As many interesting examples as possible would be very much appreciated here,

Thank you so much.

答:
If $\mathcal{F}_0$ is trivial, then the condition you are asking for is simply that $E[X_t] = X_0$ but $X_t$ is not a martingale.

Take any integrable process $Y_t$ with $Y_0 = 0$, and a uniform $\pm1$ valued $U$ independent of $Y_t$.
Then the process $X_t:=UY_t$ has this property, and is usually not a martingale.

Indeed, $E[UY_t] = \frac12 E[Y_t] + \frac12 E[-Y_t] = 0$ for all $t$.


答:
Starting from the geometric series we have
$$\frac{1}{1-x}=1+x+x^2+x^3+...$$
When we fill in $x^2$ instead of x we get
$$\frac{1}{1-x^2}=1+x^2+x^4+x^6$$
Multiplying with x
$$\frac{x}{1-x^2}=x+x^3+x^5+x^7+...$$
On het other hand is a function uniquely determined by his Taylor expension so that the only possible answer to this question is $\frac{x}{1-x^2}$.</p>

Title: From geometric sequence to function


问:
$Ind_H^G(\pi\oplus \mu) \simeq Ind_H^G(\pi)\oplus Ind_H^G(\mu)$
Let $\pi$ and $\mu$ be representations of a subgorup $H\leq G$. ($G$ a finite group.) My question is whether or not it is true that
$$
Ind_H^G(\pi\oplus \mu) \simeq Ind_H^G(\pi)\oplus Ind_H^G(\mu)
$$
That is, is the induction of the direct sum of two representations equal to the direct sum of the induced representations?

I have been working with inducing irreducible representations, but I want to understand how how induce representations that are not irreducible (so are direct sums of irreducibles). I was hoping the above would hold.

答:
Yes, this is true. One way to get it for free is to note that both sides have the same universal property:

A map of $G$-modules out of $\text{Ind}_H^G (\pi \oplus \mu)$ into some  $G$-representation $\rho$ is, by definition of $\text{Ind}$, a map of $\pi \oplus \mu$ in the category of $H$-representations to  $\rho$ (viewed as an $H$ representation.)
By the universal property of the direct sum, this is a pair of maps of $H$-representations, one from $\pi$ to $\rho$ and one from $\mu$ to $\rho$. By the universal property of induction again, this is a pair of map from $G$-representations to $\rho$, which, by the universal property of sums again, is a map from the right hand side to $\rho$.


问:
Weyl group of a compact Lie group vsWeyl group of a root system
Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$ and $T$ be a maximal torus with Lie algebra $\mathfrak{t}$. I read that the Weyl group $W$ of $G$ is "the group of automorphisms of $T$ which are restrictions of inner automorphism of $G$": $N(T)/T$ where $N(T)$ is the normaliser of $T$. Now I have trouble seeing why/how this is related to the symmetries of the roots of the adjoint action of $\mathfrak{t}$ on the complexidied of $\mathfrak{g}$. Can someone help me here? Or is it a too general question?
Thank you

答:
The nLab article is correct but, of course, is very much incomplete (and it is not meant to be complete). The definition you should get out of their article is that
$$
W≃N_G(T)/T,
$$
as an abstract group. In fact, there is more to this formula, as it also gives you the standard linear representation of $W$: You let $N_G(T)$ act on the Lie algebra ${\mathfrak t}$ of $T$ via the adjoint representation (the linearization of the action of $N_G(T)$ on $T$ by conjugation). The kernel of this linear representation   of $N_G(T)$ is exactly $T$ and, hence, you obtain the standard linear representation of $W$.

Connecting this to root systems is a very long story and many books are written on it. It is too long to be reproduced here. In brief: There is a canonical $W$-invariant (positive-definite) inner product on ${\mathfrak t}$, obtained by restricting the Killing form on ${\mathfrak g}$. One then studies codimension 1 subspaces (called "walls") in ${\mathfrak t}$ which are fixed by nontrivial elements of $W$. (In terms of the normalizer $N_G(T)$, the walls correspond to the codimension 1 subgroups of $T$ which have larger centralizer than the entire $T$.) These elements are involutions (since they are orthogonal transformations with codimension 1 fixed-point sets). The walls divide  ${\mathfrak t}$ into connected components, called "chambers" and one proves that $W$ acts simply-transitively on the set of chambers. Then one picks one of the chambers $\Delta$ (called a positive chamber) and proves that $W$ is generated by reflections fixing the walls passing  the codimension 1 faces of $\Delta$. These are called simple reflections in $W$. Roots of the root system of $G$ (or or ${\mathfrak g}$) are certain (carefully chosen) linear functionals on ${\mathfrak t}$ whose kernels are the walls in ${\mathfrak t}$. Then one verifies that these linear functionals satisfy a certain set of axioms, which then one uses to define an abstract root system. Later on, one relates this to the root system of the complex Lie algebra ${\mathfrak g}^{\mathbb C}$, the complexification of ${\mathfrak g}$. One can easily spend the entire semester getting to this point.

As for references, you can use the ones from the nLab or from this Wikipedia article.

This is a freely available set of notes on the subject which I found simply by googling.


问:
Are these particular cases of the universal property of the coproduct and product?
I was talking to a friend about the results presented here and here.

He told me that these are particular cases of the universal property of the coproduct and of the product (in a category). I have learnt a bit about universal properties, but I can't see why what he says is true. For instance, those universal properties (that of a direct product of groups, for instance) don't mention anything about having isomoprhisms in some cases. So, could someone enlighthen me a bit?

答:
The point is that the Cartesian product $A\times B$ of two groups $A,B$ with the projection maps $\pi_A,\,\pi_B$ satisfies the universal property:

whenever we have a similar diagram, i e. homomorphisms $f:C\to A,\ g:C\to B$, then there's a unique homomorphism $u:C\to A\times B$ which makes
$$f=\pi_A\circ u\ \text{ and }\ g=\pi_B\circ u\,.$$
Namely, this condition forces $u(c)=(f(c),\,g(c))$, and this also works for defining $u$.

One more important thing is that whenever an object $C$ with maps $f:C\to A,\ g:C\to B$ also satisfies the universal property, then $u$ above turns out to be an isomorphism, hence $C\cong A\times B$.


问:
$\int \phi \chi_{_{[0,1/n]}}dx\to \phi(0)$ for $\phi\in C_0(\mathbb{R})$.
Let $\phi\in C_0(\mathbb{R})$ which is the space of functions that vanish at infinity (the completion of $C_c(\mathbb{R})$. I would like to show that $\int \phi\cdot n\chi_{_{[0,1/n]}}dx\to \phi(0)$. This seems intuitive since we put more and more weight on the point $x=0$ but I don't know where to start to show it properly. Also,it doesn't really seem important that $\phi$ is vanishing at infinity, is it?

答:
Hint Let $\epsilon >0$. Since $\varphi$ is continuous at $0$ there exists some $\delta >0$ such that, for all $x \in (-\delta, \delta)$ you have
$$
|\varphi(x)- \varphi(0) | <\epsilon \,.
$$

Now, for all $n$ such that $\frac{1}{n} < \delta$ you have
$$
\left|\int_\mathbb{R}\phi(x)\cdot n\chi_{[0,1/n]}\,\mathrm{d}(x) -\varphi(0) \right|=\left|\int_{0}^{\frac{1}{n}}(\phi(x)\cdot n -\varphi(0) \cdot n)\mathrm{d}(x)\right| \leq
\int_{0}^{\frac{1}{n}}\left| \phi(x)\cdot n -\varphi(0) \cdot n\right|\mathrm{d}(x)
$$


问:
Find all solutions for $dy/dx=y^2$
Started to learn ODE, and the second example was of the equation $y^\prime=y^2$

We showed that the group of solutions is $\{y(x)=0 \}\cup \{ y(x)=\frac{1}{c-x} | c \in \mathbb{R} \}$ and mentioned that there are no other solutions.

I would like to prove that fact.

I tried to use a method like the one we used to prove that for the equation $$dy/dx=y$$

the solutions are only $\{y=Ce^x|C>0\}$: what we did there was to define $g = ye^{-x}$ and by looking at the derivative of $g$ we showed that $g$ is constant. But I can't figure how to do it in my case. (I guess I cant define $h(x) = y\cdot (c-x)$ because it is dependent on $c$, right?)

答:
You can do the same thing here to show that the solution is (at least locally) unique.

So let's fix $x_0\in\Bbb R$ such that $y(x_0) \neq 0$ and an open neighborhood $U$ of $x_0$. By shrinking $U$, we may assume that $y(x) \neq 0$ for any $x \in U$.

We now define $g = x + 1/y$ and proceed to calculate $g'$:
$$g' = 1 - \frac 1{y^2} \cdot y' = 0$$ which shows that $g$ is constant on $U$.

To obtain the global result, we may "glue" all the local pieces together.

The above argument already shows that, whenever $U$ is an open set on which $y$ is nonzero, there exists $c$ such that $y(x) = \frac 1{c - x}$ on $U$.

Let $E$ be the set $\{x \in \Bbb R: y(x) = 0\}$. We show that $E$ is either empty or the whole $\Bbb R$.

Assume the contrary. Then there is an open set $U \subset \Bbb R \backslash E$ such that $E \cap \bar U \neq \emptyset$, where $\bar U$ denotes the closure of $U$.

Thus we have $y(x) = \frac 1{c - x}$ on $U$, which extends by continuity to $\bar U$. But this leads to a contradiction for any $x \in E \cap \bar U$.

Therefore either $y = 0$ on the whole $\Bbb R$, or $y$ is never zero on $\Bbb R$, in which case we have $y = \frac 1 {c - x}$ for some $c$.

Title: Find all solutions for $dy/dx=y^2$


问:
Why, when we impose restrictions such as the ones of our physical world, new properties arise?
Unfortunately I am not and expert mathematician nor a philosopher so I don't have the right words to phrase the concept, but the following example should be able to make the question clear:

Take the complex numbers. We can't impose an order relationship between them (>, <). But as we impose a new constrain by saying that $\sqrt{-1}$ doesn't exist, we find that we can now tell which number is bigger than another.

How can we rigorously explain what's going on?

EDIT: Second example concerning Achilles and the tortoise removed since multiple users pointed out that it contained numerous errors. Minor corrections.

答:
First, regarding your first point, I agree with comments and answers pointing out that the Paradox of Achilles and the Tortoise has virtually nothing to do with uncountable infinities.

Regarding your second point, I would like to add something which I think is not stressed enough in other answers: Namely that inclusion of structures $A \subset B$ says virtually nothing about how "complicated" or "interesting" the structures $A$ and $B$ are relative to each other. That is because one could be interested in different structural properties of them.

As a graphical example, the Menger sponge (approximation on the right) is contained in the cube (on the left):

(Image from Wikipedia https://en.wikipedia.org/wiki/Menger_sponge#/media/File:Menger_sponge_(Level_0-3).jpg)

Now which of these objects do you find more interesting, which has "more" or "less" properties? It very much depends on what you want. With the cube, you can compute a volume, study its symmetry group, do many non-trivial things ... But still the sponge on the right seems to have other, and more intricate, properties.

I like to compare that with the inclusion $\mathbb R\supset \mathbb Z$ (the real numbers being a bit like the cube).

If you have, say, a series $\sum a_n$ with the $a_n \in \mathbb R$, there's quite an intricate theory about whether it absolutely converges or conditionally converges or does not converge at all etc.

But if that series has terms $a_n$ taken just from $\mathbb Z$, this is absolutely boring: It converges if and anly if all but finitely many $a_n$ are $=0$, and then it'S just a finite sum.

So in that sense (namely, calculus / topology), the structure $\mathbb R$ is far more interesting than $\mathbb Z$.

But:

For $n \ge 3$, does the equation $x^{n} + y^{n} =z^{n}$ have solutions with all $x,y,z \neq 0$ in $\mathbb R$? Well, trivially yes, infinitely many, you can graph them nicely and compute them to great accuracy with a pocket calculator.

Does the equation $x^{n} + y^{n} =z^{n}$ have solutions with all $x,y,z \neq 0$ in $\mathbb Z$? No. But it took mathematicians hundreds of years and very intricate theories to figure that out.

So in that sense (namely, solving equations / arithmetic), the structure $\mathbb Z$ is much more interesting than $\mathbb R$.


问:
How to setup a function that has a defined second derivative and have two maxima but no minima?
For no minima on the endpoint, I am thinking a using a polynomial with odd degree so that the both ends will go upward to positive and negative infinity respectively.

However, I have no idea how to have two maxima but no minima in between.

Please leave a hint, and much appreciated!

答:
The function $f(x)=x^2-x^4$ has its maximum value of $f(c)=1/4$ attained at the two points $c=\pm \sqrt{1/2}.$ It has a local minimum at $x=0$ and no other local minima.

This function can be adjusted  by restricting its domain to all reals except those in the closed interval $[-1/2,1/2].$ Calling the resulting restricted function $g(x)$ we now have an example of a twice differentiable function having two local maxes and no minimum. [Its graph does approach two non-attained low points at $(\pm 1/2, 3/8),$ but these are not technically local minima (not on the graph)]


问:
Find the number of pairs $(x,y)$ of integer numbers (positive or negative) such that $2020x+2022y+1=xy$.
I was struggling for ages on this question, and even after careful considering I cannot make sense of this with my knowledge of algebra or set theory.

How can I solve this?

Body:
$$xy-2020x-2022y=1$$

$$(x-2022)(y-2020)=1+2020*2022$$

$$2020*2022+1=2021^2=(43*47)^2$$

$$x-2022=\pm(1,43,47,43^2,2021,47^2,2021^2)$$

Then you can just solve for x and y algebraically


问:
Existence and unicity ODE
I've been reading "A text book on ordinary differential equations" of Sahir Ahmad & Ambrossi. On page 37 we have the following excercise:

"Explain why $x'+\dfrac{\sin(t)}{e^t+1}x=0$ cannot has solution $x(t)$ such that $x(1)=1$ and $x(2)=-1$."

As far I know, $f(t,x)$ is continuos everywhere since $e^t\ne -1$ for every $t$, and $\frac{\partial f}{\partial x}$ doesn't depends on $x$, so its constant and therefore continous, then the existence and unicity theorem must apply on neighborhood of those points. Where am I wrong? Can anyone explain  me? Im very stuck af this assignament.

答:
Yes, the conclusion follows directly from the uniqueness part of the Picard–Lindelöf theorem for initial value problems.

The differential equation is of the form $ x'(t) = f(t, x(t))$
where $f(t, x) = -\frac{\sin(t)}{e^t+1} x$ is continuous, and uniformly Lipschitz continuous in $x$.

$\tilde x(t) = 0$ is a solution of that differential equation on the interval $I = [1, 2]$.

If $x$ is any solution of the differential equation on $I$ with $x(t_0) = 0$ then $x$ and $\tilde x$ are both solutions of the same initial value problem $x'(t) = f(t, x(t))$, $x(t_0) = 0$, and therefore $x = \tilde x$.

It follows that any solution of the differential equation on $I$ is either identically zero, or has no zeros at all.

In particular, there can be no solution which takes both positive and negative values.

Title: Existence and unicity ODE


问:
$|\int f \ d\alpha| \leq \int |f| \ d\alpha$ (Rudin 6.13)
I'm reading Rudin's proof of theorem 6.13,

$$\left|\int_a^b f\ d\alpha \right| \le \int_a^b |f| \ d\alpha$$

and realizing there's a step I can't seem to understand.  We let $c=\pm 1$ such that

$$\left|\int f \right| = c\int f = \int c f \leq \int |f|$$

The last step is justified by $cf\leq |f|$.  But it seems to me that it could be possible that both $c = 1$ and also $f(x) < 0$ for some values of $x$.  I assume we must be saying here that, at every $x\in[a,b]$ we have $cf(x) \leq |f(x)|$, but this doesn't seem true.  I'm sure I'm misinterpretting or misunderstanding something, but can't find what it is.  Because if we're saying that $cf(x)\leq |f(x)|$ holds for all $x$ then that seems false.  And if we say that it holds for some $x$ then it seems like we can't infer $\int cf \leq \int |f|$.

My thought about how to approach this problem would have been to break $f$ into regions where it is positive and negative, and integrate on each region.  However, that would assume $f$ is piece-wise continuous, and we'd like a more general result.  So it seems to me that this style of proof would be inadequate.

答: The number $c$ is either $1$ or $-1$, and therefore $cf(x)$ is either $f(x)$ or $-f(x)$. And both $f(x)$ and $-f(x)$ are smaller than or equal to $|f(x)|$. So, yes, $(\forall x\in[a,b]):cf(x)\leqslant|f(x)|$. And it follows from this that, indeed$$\int_a^bcf(x)\,\mathrm dx\leqslant\int_a^b|f(x)|\,\mathrm dx.$$


问:
Ergodic process and random walk
Is a gaussian random walk process an ergodic process? If Yes, does someone knows the proof?

Thanks in advance

答: No.  The standard random walk has no invariant distribution.  The solution is N(0,t) for $t<\infty$.  The OU process (with drift towards 0) is ergodic. </p>

Title: Ergodic process and random walk


问:
If two random variables are defined on the same probability space do they have the same distribution?
Let's say we have two random variables $X$ and $Y$ both defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$. So $X$ and $Y$ are measurable functions from $\Omega$ to $\mathbb{R}$.

Do $X$ and $Y$ necessarily have the same distribution? Can we for example have $X\sim Bern(p)$ and $Y\sim N(0,1)$? Or would we then say that $X$ and $Y$ have for example the measure $\mathbb{P}_X$ and $\mathbb{P}_Y$ respectively?

答: No, they do not. For example, the random variable on the interval with the uniform measure which maps everything to $1$ has a delta function distiribution, while the one that maps $x$ to $x$ has uniform distribution, while the one that maps $x$ to $x^2$ has some non-uniform distribution.


答:
I don't see how the matrix $A$ constitutes either observed values or input data in an example like this. A matrix in a context like this is a relationship between two sets of values

Often $A$ is a model of how we think two things are related. These two things could be the input to some process and the output of the process.
So if you are able to observe the output of the process but not the input, and you want to estimate what the input might be,
you could set the values of $\vec b$ to the observed output values
and let $\vec x$ be the unknown input values,
and then you would want to solve for the unknown $\vec x$
in the system
$$ A \vec x = \vec b.$$

Anyway, if you're looking for two distinct things called "observed values" and "input data" in the equation $A \vec x = \vec b$, my first guess would be that $\vec b$ is the "observed values" and $\vec x$ is the "input data."

Title: What are $\vec{x}$ and $\vec{b}$ in $A \vec{x} = \vec{b}$?

Body:
The condition of a matrix gives a bound on how inaccurate the solution $\vec{x}$ will be after approximation. So given $A \vec{x} = \vec{b}$, how much does $\vec{x}$ change when there is a change in $\vec{b}$. Here is an example:

$$\pmatrix{1&1\\1&1.001} \pmatrix{x\\y} = \pmatrix{2\\2} \implies x=2, y=0$$
$$\pmatrix{1&1\\1&1.001} \pmatrix{x\\y} = \pmatrix{2\\2.001} \implies x=1, y=1$$

This is an ill-conditioned matrix because a small change in $\vec{b}$ lead to a big change in $\vec{x}$. But here's where I am confused. The matrix $A$ can be considered as observed values or input data. However, I am not sure what $\vec{x}$ and $\vec{b}$ are in this case.


问:
Use the definition of convergence to prove that the sequence $\left\{\frac{2n}{3n+2}\right\}$ converges.
The definition for convergence: The sequence $\{x_n\}$ converges to $L$ where $L\in \mathbb{R}$ provided that for every $\epsilon > 0$ there exists a corresponding integer $N\in \mathbb{N}$ such that $n \geq N$ $\Rightarrow$ $|x_n-L| < \epsilon$.

The sequence given is $$\left\{\frac{2n}{3n+2}\right\}.$$

So far for my proof I have:

Let $\epsilon >0$ be given to us. We must show there is a number N such that $n \geq N\Rightarrow |\frac{2n}{3n+2} - \frac{2}{3}| < \epsilon$.

Then for scratch work I have
$|2n/(3n+2) - 2/3| = |6n-6n-4/(9n+6)| = \\|(-4/(9n+6)| = 4/9n+6 \\< 4/9n = 4/9 * 1/n < 1/n <\epsilon $

答: The limit is $\frac23$, not $\frac7{10}$. Note that$$\frac{2n}{3n+2}-\frac23=-\frac4{9n+6}$$and that therefore$$\left|\frac{2n}{3n+2}-\frac23\right|=\frac4{9n+6.}$$So, given $\varepsilon>0$, take $N\in\Bbb N$ such that $N>\frac4{9\varepsilon}$ and then, if $n\geqslant N$,\begin{align}\frac4{9n+6}&<\frac4{9n}\\&\leqslant\frac4{9N}\\&<\varepsilon.\end{align}


答: Perhaps if you write the nth term of the sequence as $\frac{(2/3)(3n+2)}{3n+2}-\frac{4/3}{3n+2}=\frac{2}{3}-\frac{4/3}{3n+2} $ the limit should become clear and then you just need to find $n$ large enough so that $\frac{4/3}{3n+2}<\epsilon $.

Title: Use the definition of convergence to prove that the sequence $\left\{\frac{2n}{3n+2}\right\}$ converges.

Body:
The definition for convergence: The sequence $\{x_n\}$ converges to $L$ where $L\in \mathbb{R}$ provided that for every $\epsilon > 0$ there exists a corresponding integer $N\in \mathbb{N}$ such that $n \geq N$ $\Rightarrow$ $|x_n-L| < \epsilon$.

The sequence given is $$\left\{\frac{2n}{3n+2}\right\}.$$

So far for my proof I have:

Let $\epsilon >0$ be given to us. We must show there is a number N such that $n \geq N\Rightarrow |\frac{2n}{3n+2} - \frac{2}{3}| < \epsilon$.

Then for scratch work I have
$|2n/(3n+2) - 2/3| = |6n-6n-4/(9n+6)| = \\|(-4/(9n+6)| = 4/9n+6 \\< 4/9n = 4/9 * 1/n < 1/n <\epsilon $


问:
Some doubts about Trace Theorem (for Sobolev functions).
Let $\Omega\subset\mathbb{R}^N$ be a bounded regular domain and consider the Sobolec space $W^{1,p}(\Omega)$ for $p\in (1,\infty)$. I have some doubts with th trace theorem.

Roughly speaking, the Trace theorem states that there is a linear functional $$T:W^{1,p}(\Omega)\to L^p(\partial \Omega)$$

such that $Tu=u_{|\partial\Omega}$ for every $u\in C^\infty(\overline{\Omega})$.

My questions is:

The only functions for which $Tu=u_{|\partial\Omega}$ are the $C^\infty(\overline{\Omega})$ functions, or can we find, for instance, functions not continuous in $W^{1,p}(\Omega)$ such that the same equality holds?

Thank you

答:
The reason that the equality $Tu=u_{|\partial \Omega}$ is stated for  functions in $C(\overline{\Omega})$ is that for other functions it is not clear what $u_{|\partial \Omega}$ means. Of course, we can take any function $u\in W^{1,p}(\Omega)$ and extend its domain to $\overline{\Omega}$ by letting $u$ be equal to $Tu$ on the boundary. This will be an instance of $Tu=u_{|\partial \Omega}$  being true despite $u$ not necessarily being in $C(\overline{\Omega})$. But this is merely a tautology that does not teach us anything new.

But there are useful and nontrivial results along the lines of your question: starting  with $\phi\in L^p(\partial \Omega)$, one can extend $\phi$ to a function in $\Omega$ (by solving the Dirichlet problem for some elliptic operator) and then recover $\phi$ from $u$ via nontangential limits. This does not require $\phi$ to be continuous. For a simple example, take $\Omega=\{z\in\mathbb C: |z|<1\}$ and $u=\log|z-1|$. Then $u\in W^{1,p}(\Omega)$ for $p<2$. The trace operator agrees with the boundary values of $u$ understood in the sense of nontangential limits.


问:
Proof Verification- proving that a sequence is a Cauchy sequence
Prove that \begin{array}{l}\left\{\frac{2^n-1}{2^n}\right\},\ n\in\mathbb{N}\\
\end{array} is a Cauchy sequence.
This is what I have so far:
Let $\epsilon>0$ and $m,n >k$.
Then,$$\begin{gather}\frac{1}{2^m}+\frac{1}{2^n}<\frac{1}{2^k}+\frac{1}{2^k}\\
\therefore \frac{1}{2^m}+\frac{1}{2^n}<\frac{2}{2^k}\end{gather}$$
Then I was thinking of assuming $\frac{2}{2^k}<\epsilon$.
Am I on the right track at all?

答:
Why not just observe that
$$\lim_n \frac{2^n-1}{2^n}= \lim_n (1-2^{-n}) = 1$$
and convergent sequences are Cauchy sequences?


问:
Does exchangeability imply conditional independence?
Let $A,B,C$ be a sequence of exchangeable random variables, i.e. $P(A=a,B=b,C=c) = P(B=b,A=a,C=c)=P(C=c,B=b,A=a)$

and so on for any permutation of $\{A=a,B=b,C=c\}$.

Is $P(A=a,B=b|C=c) = P(A=a|C=c)P(B=b|C=c)$?

I'm working through a proof that assumes this, but I can't seem to prove this simple result (and even suspect it isn't true).

I know we want to show the following equality
$\frac{P(A=a,B=b,C=c)}{P(C=c)} = \frac{P(A=a,C=c)}{P(C=c)} \times \frac{P(B=b,C=c)}{P(C=c)}$. But I'm not sure how.

答: It is not true that $A$ and $B$ are conditionally independent given $C$. For example, shuffle a deck of three cards, two of which are black and one of which is red, and let $A$ (resp. $B,C$) equal $1$ when the first (resp. second, third) card is red, and $0$ otherwise. Then $A,B,C$ are exchangeable, but neither two are independent given the third.


问:
Slope of tangent in $(x,y)$ on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with initial conditions
I came across an exercise with a sample solution that I unfortunately don't fully understand given that it's shortened.

Let $(x,y)$ be a point on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with $y\neq0$. Show that the slope of the tangent on $K$ in $(x,y)$ is $-\frac{x}{y}$ and find two solutions for the differential equation $y'=-\frac{x}{y}$ that adhere to the initial conditions $y(1)=1$ or $y(-1)=2$.

The sample solution is as follows:
$$x^2+y^2=r^2$$
$$2x+2y\frac{dy}{dx}=0$$
$$\Rightarrow\frac{dy}{dx}=-\frac{x}{y}$$
$$\int y\;dy=\int -x\;dx$$
$$y^2=c-x^2$$
$$c=2\;or\;c=5$$
$$\Rightarrow x^2+y^2=2\;or\;x^2+y^2=5$$

First of all I'm interested in line 2.

From what I could find and understand, we come to this point through implicit differentiation of line 1. I assume the $r^2$ disappeared because it is treated as a constant but why exactly is that the case?

Furthermore, why is the constant of integration omitted on one side of the equation and how exactly do we find the solutions $2$ and $5$ for it?

Thank you very much in advance!

答:
$r$ is indeed constant (otherwise $K$ wouldn't be a circle)

It doesn't matter if separate constants of integration are included on both sides. (Two arbitrary constants can simply be combined: $c_1 + c_2$ is still just a constant).

$c=2$ and $c=5$ come from the initial values: $y(1) = 1$ and $y(-1) = 2$, respectively. In the first case $x=1$, $y=1$ which gives $c=2$. Similarly for the second case.


问:
Metric spaces "$\iff$" proof
Let $(A, d_1)$ and be a metric. Suppose $d_2: A \times A \rightarrow \mathbb{R}$ be defined in this way:

For $a, b \in A$, $d_2(a, b) = \frac{d_1(a, b)}{1 + d_1(a, b)}$.

Show that $E \subset A$ is open with respect to $d_1 \iff E$ is open with respect to $d_2$.

My friend and I's attempt:

Assume $E$ is open inside $(A, d_2)$. Let $x \in E$. We need $B_{\rho}^{d_1} \subset E$ for some $\rho > 0$. We know that there is some $r > 0$ so that $B_{r}^{d_2} \subset E$.

Choose $\rho = ?$ so that $B_{\rho}^{d_1} \subset B_{r}^{d_2}$. (Not sure what $\rho$ to choose).

In essence, we need $d_1(x, a) < \rho \implies d_2(x, a) < r$.

Going the other way, assume $E$ is open with respect to $(A, d_1)$. Then $d_2(a, b) = \frac{d_1(a, b)}{1 + d_1(a, b)} \leq d_1(a, b) < r$.

Not sure how to proceed from here.

QED.

Is this proof correct so far? We are new to metric spaces and tried to connect the dots as much as possible. Can somebody fill in the holes in the proof and explain it? Any assistance much appreciated.

答:
I have replaced $a, b$ with $x, y$*

Proof.

Assume $E$ is open inside $(A, d_2)$. Then there is a $r > 0$ s.t. the open ball $B_2(x, r) = \{y \in A: d_2(x, y) < r\}$ is contained in $E$.

We need to show that for any $x \in E$, there is an $\rho > 0$ so that the open ball $B_1(x, \rho)  = \{y \in A: d_1(x, y) < \rho\}$ is contained in $E$.

Let $r < 1$ and set $\rho = \frac{r}{1 - r}$ then for $y \in B_1(x, \rho)$ we have

$$d_2(x, y) = \frac{d_1(x, y)}{1 + d_1(x, y)} < \frac{\rho}{1 + \rho} = r$$

So $y \in B_2(x, r)$ and so $B_1(x, \rho) \subset E$. Hence $E$ is open inside $(A, d_1)$.

The other way.

Assume $E$ is open inside $(A, d_1)$. Then there is a $\rho > 0$ s.t. the open ball $B_1(x, \rho) = \{y \in A: d_1(x, y) < \rho\}$ is contained in $E$.

We need to show that for any $x \in E$, there is an $r > 0$ so that the open ball $B_2(x, r)  = \{y \in A: d_2(x, y) < r\}$ is contained in $E$.

Set $r = \rho$ then for $y \in B_2(x, r)$ we have:

$$d_2(x, y) = \frac{d_1(x, y)}{1 + d_1(x, y)} \leq d_1(x, y) < \rho = r$$

So $y \in B_1(x, \rho)$ and $B_2(x, r) \subset E$. We may conclude $E$ is open inside $(A, d_2)$.

QED.

Title: Metric spaces "$\iff$" proof


问:
Natural isomorphisms between finite-dimensional vector spaces and their duals
There are several questions on this already, and my question arose after reading the explanation given in the following:

In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?

We know a natural isomorphism $\eta:F \to G$ is defined as a natural transformation whose components are all isomorphisms in $\mathbf{D}$, where $F,G: \mathbf{C} \to \mathbf{D}$ are functors. What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

Also, in the answer to the quoted question, the author states that there is a natural isomorphism between a Euclidean vector space and its dual, and gives it as $V \mapsto V^*, v \mapsto \langle v,- \rangle$. What does this notation mean? And also, if there are natural isomorphisms between some finite-dimensional vector spaces, how can we "prove" that there are none in general, as the OP wants? It seems like the best we could do is a counterexample? Or am I missing a key distinction between Euclidean spaces and finite-dimensional spaces?

答:
A Euclidean space is not just a finite-dimensional vector space. It is a finite-dimensional vector space over the reals together with an inner product space. So it is an object of a very different category than mere vector spaces (over the reals). The extra structure gives in this case a particular choice for a mapping $V\to V^*$, namely using the inner product. In the absence of inner products, that is if all you have is a finite-dimensional vector space over the reals, then there is no canonical choice for a mapping $V\to V^*$ even though the two spaces have the same dimension. So, they are isomorphic but there is no canonical choice for any particular isomorphism.

Further, the construction $V\mapsto V^*$ is contravariant, so it does not even make sense to ask for a natural transformation between the identity functor on $\mathrm {Vect}$ and the (contravariant!) functor $-^*$. The double dual is covariant and so the question becomes meaningful and there is then a natural isomorphism between the functors.

You could account for the reversal of arrows and try to compare the identity functor and the contravariant dualisation. Once this is given precise meaning it can be shown that any assumed such transformation leads to a contradiction, so it does not exist.


问:
Number of Bijections such that $f(x+y)=f(x)+f(y)$
Find the number of bijective functions $f:Z \to Z$ such that $f(x+y)=f(x)+f(y)$, $\forall x, y \in Z$

I substituted $x=y=0$ we get $f(0)=0$

Now Keep $x=-y$, we get
$$f(0)=f(x)+f(-x)$$
$\implies$
$$f(-x)=-f(x)$$
Now let $f(1)=m$ for some $m \in Z$

Then $f(-1)=-m$

So once the mapping of any integer is done, the mapping of its negative is simultaneously done.

But how to count number of bijections?

答: There are only two: $f(x)=x$ and $f(x)=-x$. If $m=0$, then $f(1)=f(0)$ and then $f$ is not injective. And if $m>1$ or $m<-1$, then, for each $n\in\Bbb N$, $f(n)$ is a multiple of $m$, and therefore $f$ is not surjective.


问:
Use average value theorem to find c for indefinite integral (piecewise)
Given a piecewise function

$$   f(x)=\left\{
\begin{array}{ll}
      x & 0 \leq x \leq 1 \\
      3-x & 1 < x \leq 3\\
\end{array}
\right. $$

and it is integrable in $[0,3]$. Then Can I apply the average value theorem ($\frac{1}{b-a} \int_{a}^{b}f(x)dx$)? I think so because it is integrable, so it is continuous and thus I can use it. But the next question is: is there multiple $c$, for which $f(c)= \frac{1}{3-0} \int_{3}^{0}f(x)dx$? Notice that $a=0$ and $b = 3$ since this is the interval I said is integrable. For this question I think I should divide the integral into 2 parts, $\int_{0}^{1}f(x)dx$ and $\int_{1}^{3}f(x)dx$ because of the piecewise function. I did that and still can't see if there is more than one $c$ value.

答:
By definition, a function is continuous at some point if

$\lim_{x \rightarrow a}f(x) = f(a)$

You can see that it is not continuous if you apply the $\epsilon, \delta$ definition for example.

And then by definition of the Mean value theorem $f$ should be continuous, but it is not, therefore you need to partition the integral into two parts.


问:
Can LP with bounded feasible region be converted to LP with unbounded feasible region after adding artificial variables?
In Big-M method we add artificial variables to a linear programming problem(call it $p$) to convert it to a secondary problem(call it $p^\prime$) so we can find basic feasible solution to start Simplex algorithm.

Does there exist a linear programming problem with bounded feasible region which will have unbounded feasible region after adding artificial variables?

答:
This holds true for all feasible bounded problems. A constraint $a^Tx \geq b$ with $b > 0$ becomes $a^Tx-s_i+ a_i = b$ with $s_i$ the slack variable and $a_i$ the excess variable. If $(x,s,a)$ is feasible, then $(x,s+\alpha,a+\alpha)$ is also feasible for any $\alpha>0$, so the new feasible region is unbounded.

Based on the comment it seems like you are interested in whether the projection of the feasible region of $p'$ onto the variables in $p$ is unbounded. That can also be the case, but not necessarily. As a simple example, consider the set $S_1=\{x\in\mathbb{R} : 1 \leq x \leq 2\}$. After adding slacks and an artificial variable, this becomes the set $S_2=\{(x,a,s_1,s_2) \in\mathbb{R}^4 : x-s_1+a=1, x+s_2=2\}$. The constraint $x \geq 1$ can be violated arbitrarily by increasing the value of $a$ (for example, $(0,1,0,2) \in S_2$), so the projection of $S_2$ onto $x$ is $x \leq 2$.


答:
What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

I assume that the most puzzling aspect of this is the fact that the term "isomorphism" is used for two different kinds of objects; one is a map between vector spaces, and the other is a family of such maps. In the most general sense, "isomorphism" can refer to any reversible structure preserving map. An invertible linear map between vector spaces is called an isomorphism because it preserves the "linear" structure of a vector space. In particular, the essential thing that makes an object a vector space is the ability to take linear combinations of elements, i.e. to ("nicely" and "meaningfully") refer to an object $ax + by$ for scalars $a,b$ and vectors $x,y$. $f:V \to W$ is an isomorphism, then this essential structure is preserved in the sense that
$$
f(ax + by) = af(x) + bf(y).
$$
In other words, taking a linear combination in the first space and then applying an isomorphism yields the same result as applying the isomorphism and then taking (the same) linear combination in the second space.

On the other hand, the essential thing that makes a map $F:\mathbf C \to \mathbf D$ a functor is the fact that $F(f \circ g) = F(f) \circ F(g)$ holds for all compatible morphisms $f,g \in \mathbf C$. A natural isomorphism $\eta:F \to G$ preserves this structure in the sense that for any morphism $f:X \to Y$ in $\mathbf C$, we have
$$
\eta_Y \circ F(f) = G(f) \circ \eta_X.
$$
In other words, applying the first functor to a morphism and then applying the natural isomorphism yields the same result as applying the natural isomorphism and then applying the second functor to that same morphism.

Title: Natural isomorphisms between finite-dimensional vector spaces and their duals

Body:
There are several questions on this already, and my question arose after reading the explanation given in the following:

In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?

We know a natural isomorphism $\eta:F \to G$ is defined as a natural transformation whose components are all isomorphisms in $\mathbf{D}$, where $F,G: \mathbf{C} \to \mathbf{D}$ are functors. What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

Also, in the answer to the quoted question, the author states that there is a natural isomorphism between a Euclidean vector space and its dual, and gives it as $V \mapsto V^*, v \mapsto \langle v,- \rangle$. What does this notation mean? And also, if there are natural isomorphisms between some finite-dimensional vector spaces, how can we "prove" that there are none in general, as the OP wants? It seems like the best we could do is a counterexample? Or am I missing a key distinction between Euclidean spaces and finite-dimensional spaces?


问:
Reference on Fixed Point Theory
Could you please recommend me some references for Fixed Point Theory? I am after references containing the main fixed point theorems, their variants, and sufficient (neccessary?) conditions for uniqueness of the fixed points. Mainly I've been looking at Brouwer's and Kakutani's, but any others like Banach's are welcome. I searched here in stack exchange but I don't think I saw much covering these aspects in a comprehensive manner.</p>

答:
Granas-Dugundji's book on fixed point theory is a good starting point for an encyclopedic survey of the classical fixed point theory of continuous mappings including some of the modern extensions as well.

Fixed Point Theory

Authors: Granas, Andrzej, Dugundji, James

ISBN 978-0-387-21593-8

Title: Reference on Fixed Point Theory


答:
$r$ is indeed a constant. Think about it this way - the distance from the origin, which would be $x^2 + y^2$ remains constant. Geometrically, that's a circle. In another sense, if $r$ weren't a constant, the implication is that it depends on $x$. A circle's radius shouldn't depend on anything.

The constant of integration is omitted on one side because the sample solution "skipped a step." Really, both integrals should end up with a constant $c$. However, since both constants are arbitrary, we can just lump them into one constant and put it on one side of the equation.

You can solve for $c$ by plugging in the initial values for $x$ and $y$ into your solution equation.

Title: Slope of tangent in $(x,y)$ on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with initial conditions

Body:
I came across an exercise with a sample solution that I unfortunately don't fully understand given that it's shortened.

Let $(x,y)$ be a point on a circle $K=\{(x,y)\in\mathbb{R}^2|x^2+y^2=r^2\}$ with $y\neq0$. Show that the slope of the tangent on $K$ in $(x,y)$ is $-\frac{x}{y}$ and find two solutions for the differential equation $y'=-\frac{x}{y}$ that adhere to the initial conditions $y(1)=1$ or $y(-1)=2$.

The sample solution is as follows:
$$x^2+y^2=r^2$$
$$2x+2y\frac{dy}{dx}=0$$
$$\Rightarrow\frac{dy}{dx}=-\frac{x}{y}$$
$$\int y\;dy=\int -x\;dx$$
$$y^2=c-x^2$$
$$c=2\;or\;c=5$$
$$\Rightarrow x^2+y^2=2\;or\;x^2+y^2=5$$

First of all I'm interested in line 2.

From what I could find and understand, we come to this point through implicit differentiation of line 1. I assume the $r^2$ disappeared because it is treated as a constant but why exactly is that the case?

Furthermore, why is the constant of integration omitted on one side of the equation and how exactly do we find the solutions $2$ and $5$ for it?

Thank you very much in advance!


问:
Growth of finite simple groups
Background.  There is an important group-theoretic notion of growth rate, defined for  finitely-generated groups $G$ equipped fixed finite generating sets $S$. The growth rate is (the equivalence class of) the function $\gamma_{G,S}(n)$ "counting" the number of vertices in the ball of radius $n$ in the Cayley graph of $G$ with respect to the generating set $S$. For instance, one of the foundational results in this direction is Gromov's theorem on groups of polynomial growth.

Much is known about  functions $\gamma_{G,S}(n)$ but much remains unknown.

Question:

What is known about growth rates of sequences of finite groups?

A bit more precisely:

What is known  about asymptotics of the sequences $\gamma_{G_i,S_i}(n_i)$, for various classes of  sequences $(G_i)$ of finite groups satisfying $|G_i|\to\infty$?

I am especially interested in the setting when all finite groups $G_i$ are simple. What is known about growth rates in this case?

答:
To expand my comment:

In order to form a meaningful question along these lines, one has to consider an infinite sequence of finite groups $G_i$ with generating sets $S_i$, such that $|G_i|\to\infty$. Then one would define an asymptotic growth functions of this family $\gamma(n)$ and study its asymptotics (rate of growth) in relation to group-theoretic properties of the groups $G_i$. One example of this approach is the article

Black, Sarah, Asymptotic growth of finite groups, J. Algebra 209, No. 2, 402-426 (1998). ZBL0922.20036.

Her setup is to fix a uniform bound on cardinalities of the generating sets $S_i$ and define
$$
\gamma(n):= \sup_i \gamma_{G_i,S_i}(n).
$$

Then she gets some interesting results which imply, for instance, that in the case of finite simple groups growth cannot be (sub)polynomial.   She conjectures that for sequences of finite nonabelian simple groups, growth is (totally) exponential, i.e. exponential regardless of the choices of bounded generating sets and proves  some interesting results towards this conjecture.

Title: Growth of finite simple groups


问:
Show that $\dim V = \dim W$
$V$ and $W$ are subspaces of $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. $m, n$ be positive integers. Let $A$ be a $(n,m)$ matrix and $B$ be a $(m,n)$ matrix assume that $\lambda \neq 0$

$V= \{x\in \mathbb{C}^m |$ for a positive integer $k$ , $(BA-\lambda I_m)^k x=0$ holds}

$W= \{y\in \mathbb{C}^n |$ for a positive integer $l$ , $(AB-\lambda I_n)^l y=0$ holds}

Show that $\dim V= \dim W$

I have proved $BA$ and $AB$ have the same eigenvalue $\lambda$ but then how do we involve these statements, any hints would be appreciated.

答:
As mathmath has already posted a handmade solution, I feel free to write a more sophisticated proof.

Recall that if $T$ is an endomorphism and $\sigma$ is an eigenvalue, its geometric multiplicity is the dimension of $\ker (T-\sigma I)$, while its algebraic multiplicity is the multiplicity of $\sigma$ a a root of the characteristic polynomial of $T$, $p_T(x)$ and you know that the first one is less than or equal to the second one.

When studying the Jordan normal form, one needs to consider the generalised eigenspaces $\ker (T-\sigma)^k$ and they form a tower that stops at some point $$\ker (T-\sigma I)^1 \subsetneq \ker (T-\sigma I)^2 \subsetneq \ldots \subsetneq \ker(T-\sigma I)^N = \ker(T-\sigma I)^{N+1}=\ldots$$And it turns out that the algebraic dimension of $\sigma$ is the dimension of this top space $\ker (T-\sigma I)^N$.

Now, turning back to your problem, and using your notation, we have $$V= \ker (BA-\lambda I) \oplus \ker(BA-\lambda I)^2 \oplus \ldots \qquad W= \ker (AB-\lambda I) \oplus \ker(AB-\lambda I)^2 \oplus \ldots$$Therefore, $\dim V$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $BA$ and $\dim W$ is the algebraic multiplicity of $\lambda$ as an eigenvalue of $AB$.

It is well known that if $A$ and $B$ are square matrices of the same dimension, then the characteristic polynomials of $AB$ and $BA$ are the same (I'll let you think about this or look it by yourself). What is not so well known is that a similar result holds when $A$ and $B$ are not square, as in your case. If $m<n$, for example, add zeroes at the matrices $A$ and $B$ to get two square $n \times n$ matrices $A'$, $B'$, and use that $p_{A'B'}(x) = p_{B'A'}(x)$ to obtain $p_{AB}(t) = t^{n-m}p_{BA}(t)$ and from there and all what I have commented, the result follows. Note the importance that $\lambda \neq 0$

Title: Show that $\dim V = \dim W$


问:
Random Walk decided by a coin
It's a discrete math question, so hope it's okay to post here. I have this lattice graph, where you start at Z and have to end in one of the S. You can only go down (so all the lines are actually down arrows), and every walk has to end in one of the S. The way you go (left or right) is decided by a coin. I have to calculate the probability of the chance that a walk ends in the different S's. I have figured out that there is only one way to get to S0 and S6, so the probability would be:

$(1/2)^6 = 0,0156$

So my question is, how do I calculate the probability of the other S's? (I have also a normal picture of the lattice graph if needed)

```<code>             Z
            / \
           Y0 Y1
          / \ / \
         X0  X1 X2
        / \ / \ / \
       W0  W1  W2 W3
      / \ / \ / \ / \
     V0  V1  V2  V3 V4
    / \ / \ / \ / \ / \
   U0  U1  U2  U3  U4  U5
  / \ / \ / \ / \ / \ / \
 S0 S1  S2  S3  S4  S5  S6
</code>```

答:
Let's start with the X row.  The number of ways to get to X0, X1 and X2 are 1,2,1.

For the W row it's 1,3,3,1.  Try and find them.

For each row it's the numbers in Pascals triangle (look this up if you aren't used to it).
So for the S row it's 1,6,15,20,15,6,1.  Since they add up to 64 the probabilities are
$1/64, 6/64, 15/64 $, etc...

Title: Random Walk decided by a coin