问:
The condition for intersection of closed set to be empty
Let $(X,d)$ be a complete metric space and $\{A_n\}$ be the decreasing sequence of non empty closed subsets of $X$. Then Cantor's intersection theorem tells that if $d(A_n) \to 0$ then the set $A = \cap_{i} A_i$ consists a single point.

If any of the above conditions are dropped then $A$ would not be a singleton set. Suppose if we drop the condition of decreasing sequence then $A$ may not be bounded or if we drop closedness of $A_i$ then $A$ may be empty.

Now if we drop the condition $d(A_i) \to 0$ then I intuitively feel that $A$ should be non empty even though I'm not able to prove that or to find any counter examples. Does there exist a sequence like the diameter of them does not go to 0 but intersection is empty?

答: Let $A_n$ be the closed interval $[n,\infty)$.  They are decreasing but have empty intersection.


问:
integration increasing decreasing
Let f be a continuous strictly increasing function from $[0,\infty) \rightarrow [0,\infty)$ and $g=f^-1$ (i.e.$f(x)=y \iff g(y)=x$). Let $a,b>0$ and $a\neq b$.Then $ \int\limits_{0}^{a}f(x)dx$+$\int\limits_{0}^{b}g(y)dy$ is

(A) greater than or equal to ab

(B) less than ab

(C) always equal to ab

答:
Since you have only three choices, try some simple examples.

The very simplest is to take f(x)= x, a= b= 1.  Then g(x)= x and $\int_0^1 xdx+ \int_0^1 xdx= 2(x^2/2)_0^1= 1$.  That is equal to ab= 1 so we can drop "< ab" but not  "$\ge ab$".

To distinguish between those, try $f(x)= x^2$, a= b= 1.  Then $g(x)= x^{1/2}$ and $\int_0^1 x^2dx+ \int_0^1 x^{1/2}dx= \frac{1}{3}+ \frac{2}{3}= \frac{5}{3}$. That is strictly greatere that ab= 1 so  we must say "$\ge ab$".

ContentLicense: CC BY-SA 4.0

Title: integration increasing decreasing


问:
Show that $(\log n)^{\log n}\in\Omega (n)$
Show that $(\log n)^{\log n}\in\Omega (n)$

Proceeding with a common logarithm property, we get $$(\log n)^{\log n}=(n^{\log\log n})$$

How do I deduce that $$(n^{\log\log n})\in\Omega(n)$$

If I say that $n=b^{b}$ where $b$ is the base of the logarithm then it can be true, but doesn't it defeat the purpose of it being linear?

Here, $\Omega,\Theta,\mathcal{O}$ are the well known asymptotics, i.e. lower bound, upper bound and the big O.

This was found in MIT 6.006 Spring 2020, question number 6 although they assumed a base $2$ logarithm.

答: When $n$ is sufficiently large, $\log \log n \geq 1$.  Therefore $n^{\log \log n} \geq n^1 = n$ for $n$ sufficiently large.


问:
Having difficulty to figure out variables in correlation formula
From where does that "92110000" come from. For what I understood it is the summation of X squared.
But that would be 2300² = 453690000?

答:
Brackets matter:

Here $(\sum x_i)^2 =21300^2 = 453690000$

but $\sum (x_i^2) = 921100000$

and you should read the formula $n\sum x^2 - (\sum x)^2$ as $n\sum (x^2) - (\sum x)^2$


问:
upper bound on cross entropy or relative entropy
Am looking for upper bounds for relative-entropy or cross-entropy for non-gaussian case . All I am aware of is bounds for entropy based on determinant of covariance matrix.

What about for relative entropy?

答:
As the other commenter already stated, the KL-divergence aka relative entropy is not bounded.

As the cross entropy $H(p,q)$ can be written as
$$H(p,q) = H(p) + D_{KL}(p||q)$$

it is therefore also not bounded.


问:
Compute E(sin(X)|X+Y)
Okay, let $X,Y$ be independent random variables with the same exponential distribution. Compute $E(X|X+Y)$ and $E(\sin(X)|X+Y)$.

Solution:
$X+Y=E(X+Y|X+Y)=E(X|X+Y)+E(Y|X+Y)=2\cdot E(X|X+Y)$
Hence $E(X|X+Y)=\frac{X+Y}{2}$.

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

Hence
$E(\sin(X)|X+Y)=\sin(\frac{X+Y}{2})$.

Is that correct?

答:
Set $Z=X+Y$.
First we'll show $X|Z=z\sim \mathcal{U}(0,z)$ for $z>0$ fixed. For $x\in(0,z)$ we have

$$f_{X|Z=z}(x|z)=\frac{f_{XY}(x,z-x)\sqrt{2}}{\int_0^zf_{XY}(x,z-x)\sqrt{2}dx}=\frac{1}{z}$$ Here we're using the fact that $f_{XY}(x,y)=\lambda^2e^{-\lambda(x+y)}$ for $(x,y)\in (0,\infty)^2$ and $f_{XY}(x,y)=0$ elsewhere. Therefore, $$E(X|Z=z)=\int_0^zxf_{X|Z=z}(x|z)dx=\frac{z}{2}$$ Moreover, $$E(\sin(X)|Z=z)=\int_0^z\sin(x)f_{X|Z=z}(x|z)dx=\frac{2}{z}\sin^2\Big(\frac{z}{2}\Big)$$ So we get $E(X|X+Y)=\frac{X+Y}{2}$ while $E(\sin(X)|X+Y)=\frac{2}{X+Y}\sin^2\Big(\frac{X+Y}{2}\Big)$.

Note: The problem arises in your calculation when you claim that $$2E\Big(\sin(X/2)\cos(Y/2)\Big|X+Y=z\Big)=E(\sin(X)|X+Y=z)$$

Title: Compute E(sin(X)|X+Y)


问:
Application of Darboux's theorem:Prove that:there exists $\xi \in (0,1)$, s.t. $f''(\xi ) = f(\xi )$
In my textbook, there is an interesting problem in the section of Darboux's theorem:

$f(x)$ is second-order differentiable on $[0,1]$

$f(0)=2, f '(0)=0, f(1)=e+e^{-1}$

How to prove that:there exists  $\xi  \in (0,1)$, s.t. $$f''(\xi ) = f(\xi )$$

I noticed that it's required to construct a function and apply Darboux's theorem, just like the exercises of mean value theorem , and I found that maybe the solutions to $f''=f$:
$$y = {c_1}{e^x} + {c_2}{e^{ - x}}$$ is related to the problem since $f(1)=e+e^{-1}$.But I just fail to construct a nice function to do so.Can you give me some hint?  And are there some special techniques to find such functions? (Maybe solving some related differential equations is one of them) Thank you very much.

答:
Yes, you go over the solutions of this ODE in using them to straighten the problem. Consider $h(x)=f(x)\cosh(1-x)+f'(x)\sinh(1-x)$, where $\cosh(1-x)$ and $\sinh(1-x)$ are basis solution for the ODE $f''(x)=f(x)$.

Then one can compute

$h(0)=f(0)\cosh(1)+f'(0)\sinh(1)=2\cosh(1)$

$h(1)=f(1)\cosh(0)+f'(1)\sinh(0)=2\cosh(1)=e+e^{-1}$

It follows from the theorem of Rolle that there is a point $ξ\in(0,1)$ with
\begin{align}
0=h'(ξ)&=[f'(ξ)\cosh(1-ξ)-f(ξ)\sinh(1-ξ)]+[f''(ξ)\sinh(1-ξ)-f'(ξ)\cosh(1-ξ)]
\\
&=(f''(ξ)-f(ξ))\sinh(1-ξ).
\end{align}


问:
In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$.

In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$. If area of quadrilateral $BCED$ is $52$ $cm^2$ and $[\triangle ADE] = x$ $cm^2$, find $x$.

What I Tried: Here is a picture :-

The only thing I am not able to figure out is that how to use the area of the quadrilateral using the available information in the picture. Suppose I am able to find the value of $y$, but how will I use that to find $[\Delta ADE]$ too? I cannot find a way to do that either.

Can anyone help me? Thank You.

答:
You can use

$$ \dfrac{[ADE]}{[ABC]} = \dfrac{\tfrac{1}{2}\cdot AD \cdot AE \cdot \sin A}{\tfrac{1}{2}\cdot AB \cdot AC \cdot \sin A} $$

to know their ratios. And you know their difference.


问:
Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$
The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?

答:
Multiply by $\sin(x)\cos(x)$:
$$
\sin(x)\cos(x)=\sin(x)+\cos(x)\tag1
$$
Square and multiply by $4$:
$$
4\sin^2(x)\cos^2(x)=4+8\sin(x)\cos(x)\tag2
$$
Apply $\sin(2x)=2\sin(x)\cos(x)$:
$$
\sin^2(2x)-4\sin(2x)-4=0\tag3
$$
Solve:
$$
\sin(2x)=2-2\sqrt2\tag4
$$
Since $2+2\sqrt2\gt1$, we don't want that solution.

$(4)$ yields the $4$ possible solutions:
$$
\scriptsize x\in\left\{\color{#AAA}{\frac12\sin^{-1}\left(2-2\sqrt2\right),\frac\pi2-\frac12\sin^{-1}\left(2-2\sqrt2\right),}\pi+\frac12\sin^{-1}\left(2-2\sqrt2\right),\frac{3\pi}2-\frac12\sin^{-1}\left(2-2\sqrt2\right)\right\}\tag5
$$
Note that $x\mapsto\pi+x$ negates $\sin(x)+\cos(x)$ but leaves $\sin(x)\cos(x)$ unchanged. Thus, each solution will have a non-solution offset by $\pi$.

The first two solutions in $(5)$ give $\sin(x)\cos(x)=-\sin(x)-\cos(x)$, so we only want the last two solutions.


问:
Let $G$ be a group. Let $N\unlhd G$ with $G/N$ being cyclic. Let $H\leq G$. Show that $HN\unlhd G$.
Let $G$ be a group. Let $N\unlhd G$ with $G/N$ being cyclic. Let $H\leq G$. Show that $HN\unlhd G$.

I cannot figure out how to use the fact that $G/N$ is cyclic in this proof. Any help is much appreciated.

答: As $G/N$ is abelian, $HN/N$ is a normal subgroup of $G/N$. By the correspondence between subgroups of $G/N$ and subgroups of $G$ containing $N$ (which preserves normality), this implies that $HN$ is normal in $G$


问:
Inequality $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$

prove that for $a,b,c$ being positives and $a+b+c=1$:$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$$

This is a very interesting inequality which i came upon accidentally.We also see that the condition $a+b+c=1$ is not needed.I modified the inequality slightly using Nesbitt's inequality and hence it boils down to prove an even stronger inequality $$\sum_{cyc}\left(\frac{b}{a}-\frac{2c}{a+b}\right)\ge 0$$ Since this was a stronger version i checked WA which shows that it is valid.I have tried to get an SOS but failed.The problem is that even after complete expanding and cross-multiplying we get a cyclic inequality and hence Muirheads theorem fails.

答:
Using a computer algebra system, the inequality in the title turns out to be equivalent after expanding to the inequality:
$$
2\sum_{\text{cyclic}}a^4b^2
+
2\sum_{\text{cyclic}}a^3b^3
\ge
\sum_{\text{cyclic}}a^3bc(b+c)
+ 6a^2b^2c^2\ .
$$
Now we represent the monomial powers $(r,s,t)$ in the plane $r+s+t=6$, and in each "node" that appears in the inequality we place the corresponding coefficient. This makes it easier to find a domination scheme. The picture is as follows:

```<code>                b^6
                .
              .  .
            .   .  2
          2  -1  -1  2
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  .
a^6                        c^6
</code>```

A comment on the scheme. The "extremal vertices" are marked with $a^6$ alias $(6,0,0)$, $b^6$ alias $(0,6,0)$, and $c^6$ alias $(0,0,6)$.

Consider the "base line" joining the vertices for $b^6$ and $c^6$. Parallels to this line passing through the lattice points are lines with constant $a$-part in the monomial. So the parallel line "immediately after the line from $b^6$ to $c^6$" is the line from $ab^5$ to $ac^5$, and the lattice points on it are corresponding to  $ab^sc^t$ with $s+t=5$.

The inserted coefficients are the coefficients in the inequality to be shown. We can "dominate" with the positive coefficients on some "nodes" one (ans similarly more) negative coefficients on those nodes in the convex hull. For instance, a domination scheme can be applied by using from the positive positions marked with a bracket in

```<code>                b^6
                .
              .  .
            .   .  2
         [2][-1][-1][2]
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  .
a^6                        c^6
</code>```

and from each $[2]$ use only $[1]$ to dominate the $[-1]$ term.
Explicitly, we use:
$$
b^3(a^3-a^2c-ac^2+c^3)\ge 0\ .
$$
Use this pattern for all other $[-1]$ entries. Finally, the $-6$ in the middle is dominated by the remained positions. (Arithmetic mean is $\ge$ then geometric mean for instance.)

The claimed stronger inequality is false. Expanding, we would have to show equivalently an inequality corresponding to:

```<code>                b^6
                .
              .  .
            .  -1  1
          1   .   .  1
        1   .  -6   .  .
      .  -1   .   . -1  .
    .   .   .   1  1  .  .
a^6                        c^6
</code>```

But there is no chance to dominate the $-1$ entries. They escape from the convex hull of the positive entries. It is enough to consider the monomials in $a^4$. An inequality of the shape $a^4b^2-a^4bc\pm\dots\ge 0$ (where the dots cover $O(a^3)$) will never happen, just take $a=a(n)$ to be the polynomial $a(n)=n$, then $b,c$ constants with $c>b$, and pass with $n$ to infinity. Knowing how to produce the "bad case", makes it simple also to verify in the given inequality. The only terms having $a$ in numerator are in
$$
\frac ac-\frac{2a}{b+c}\ .
$$
Now use $a=a(n)=n$, $b=1$, $c=100$.

Body:
Using a computer algebra system, the inequality in the title turns out to be equivalent after expanding to the inequality:
$$
2\sum_{\text{cyclic}}a^4b^2
+
2\sum_{\text{cyclic}}a^3b^3
\ge
\sum_{\text{cyclic}}a^3bc(b+c)
+ 6a^2b^2c^2\ .
$$
Now we represent the monomial powers $(r,s,t)$ in the plane $r+s+t=6$, and in each "node" that appears in the inequality we place the corresponding coefficient. This makes it easier to find a domination scheme. The picture is as follows:

                b^6
                .
              .  .
            .   .  2
          2  -1  -1  2
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  .
a^6                        c^6

A comment on the scheme. The "extremal vertices" are marked with $a^6$ alias $(6,0,0)$, $b^6$ alias $(0,6,0)$, and $c^6$ alias $(0,0,6)$.

Consider the "base line" joining the vertices for $b^6$ and $c^6$. Parallels to this line passing through the lattice points are lines with constant $a$-part in the monomial. So the parallel line "immediately after the line from $b^6$ to $c^6$" is the line from $ab^5$ to $ac^5$, and the lattice points on it are corresponding to  $ab^sc^t$ with $s+t=5$.

The inserted coefficients are the coefficients in the inequality to be shown. We can "dominate" with the positive coefficients on some "nodes" one (ans similarly more) negative coefficients on those nodes in the convex hull. For instance, a domination scheme can be applied by using from the positive positions marked with a bracket in

                b^6
                .
              .  .
            .   .  2
         [2][-1][-1][2]
        2  -1  -6  -1  .
      .   .  -1  -1  .  .
    .   .   .   2  2  .  .
a^6                        c^6

and from each $[2]$ use only $[1]$ to dominate the $[-1]$ term.
Explicitly, we use:
$$
b^3(a^3-a^2c-ac^2+c^3)\ge 0\ .
$$
Use this pattern for all other $[-1]$ entries. Finally, the $-6$ in the middle is dominated by the remained positions. (Arithmetic mean is $\ge$ then geometric mean for instance.)

The claimed stronger inequality is false. Expanding, we would have to show equivalently an inequality corresponding to:

                b^6
                .
              .  .
            .  -1  1
          1   .   .  1
        1   .  -6   .  .
      .  -1   .   . -1  .
    .   .   .   1  1  .  .
a^6                        c^6

But there is no chance to dominate the $-1$ entries. They escape from the convex hull of the positive entries. It is enough to consider the monomials in $a^4$. An inequality of the shape $a^4b^2-a^4bc\pm\dots\ge 0$ (where the dots cover $O(a^3)$) will never happen, just take $a=a(n)$ to be the polynomial $a(n)=n$, then $b,c$ constants with $c>b$, and pass with $n$ to infinity. Knowing how to produce the "bad case", makes it simple also to verify in the given inequality. The only terms having $a$ in numerator are in
$$
\frac ac-\frac{2a}{b+c}\ .
$$
Now use $a=a(n)=n$, $b=1$, $c=100$.


问:
Why is projection matrix diagonalisable?
Suppose $π:V\to V$ is a projection matrix, does it follows that its eigenvalues are $0, 1$? Is $π$ diagonalisable?

One of the following answers is true and the other is false, but they both seem true to me, so can anyone clarify this?

1)Yes, because if π has an eigenvalue λ and an eigenvector v, it follows that $π^2(v)=π(λv)=λ^2v=π(v)=λv$, so $λ(λ-1)=0$ and $λ=0$ or $λ=1$

2)Yes, because $V=U \oplus W$, where $U=\operatorname{im}π, W=\ker π$. Then $π(v)=v$ and $π(w)=0$, for $v \in U, w \in W$. So $U=E(1,π), W=E(0,π)$ and  we then take a basis of $U$ and a basis of $W$ as a basis of eigenvectors of $V$.

答: The first proof is wrong. All that it proves is that if $\pi$ has eigenvalues, then they can only be $0$ or $1$. But it doesn't follow from this that there must be a basis of $V$ that consists only of eigenvectors of $\pi$ (that is, that $\pi$ is diagonalizable).


问:
Values of $a$ such that $x^5-x-a$ has quadratic factor
I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.

My Attempt

Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:

$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$

Hence:

$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$

and we have:

$$1=-bf-ce=b^4-3b^2c+c^2,$$

so that:

$$(2c-3b^2)^2=5b^4+4.$$

Question

How can I find all values of $n$ such that $5n^4+4$ is a perfect square?

My Attempt

If $m^2=5n^4+4$, then $m^2-5n^4=4$.

If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.

答:
Short version: in $w^2 - 5 v^2 = 4,$ the numbers $v$ are Fibonacci numbers, of which the largest perfect square is $144$

As you can see, my "v" numbers are alternate Fibonacci numbers, while "w" are Lucas. I will try to find a reference, it is known that the largest square Fibonacci number is 144.  Your largest $n$ is therefore $12,$ where your $m=322$

COHN 1963

Umm. Here is a Conway topograph for the quadratic form $x^2 - 5 y^2.$ This constitutes a proof that all solutions of $x^2 - 5 y^2 = 4$ are generated by  initial pairs
$$ (2,0) , (3,1) , ( 7,3), (18,8), (47,21), 123,55), (322,144), (843, 377)    $$
with recursions
$$ x_{n+6}  = 18 x_{n+3} - x_n $$
$$ y_{n+6}  = 18 y_{n+3} - y_n $$

These are from Cayley-Hamilton for
$$
\left(
\begin{array}{cc}
9&20 \\
4&9
\end{array}
\right)
$$

A little more work shows that we may interpolate, meaning
$$ x_{n+2}  = 3 x_{n+1} - x_n $$
$$ y_{n+2}  = 3 y_{n+1} - y_n $$

Let's see, the irrationals  in the Binet description of alternate Fibonacci numbers are
$$ \frac{3 \pm \sqrt 5}{2},  $$
while
$$ \left(\frac{3 \pm \sqrt 5}{2} \right)^3 = 9 \pm 4 \sqrt 5  $$
where $9 \pm 4 \sqrt 5$ are the Binet numbers from $\lambda^2 - 18 \lambda + 1 =0$

REsources on Conway's Topograph

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf   (Conway)

https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf    (Hatcher)

http://bookstore.ams.org/mbk-105/                  (Weissman)

http://www.springer.com/us/book/9780387955872             (Stillwell)

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How do we solve pell-like equations?

Diophantine equation $x^2 + xy − 3y^2 = 17$ <1,1,-3>

```<code>jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:
    9   20
    4   9
  Automorphism backwards:
    9   -20
    -4   9

  9^2 - 5 4^2 = 1

 w^2 - 5 v^2 = 4 =  2^2

Thu Nov 19 10:06:08 PST 2020

w:  2  v:  0  SEED   KEEP +-
w:  3  v:  1  SEED   KEEP +-
w:  7  v:  3  SEED   BACK ONE STEP  3 ,  -1
w:  18  v:  8  SEED   BACK ONE STEP  2 ,  0
w:  47  v:  21
w:  123  v:  55
w:  322  v:  144
w:  843  v:  377
w:  2207  v:  987
w:  5778  v:  2584
w:  15127  v:  6765
w:  39603  v:  17711
w:  103682  v:  46368
w:  271443  v:  121393
w:  710647  v:  317811
w:  1860498  v:  832040
w:  4870847  v:  2178309
w:  12752043  v:  5702887

Thu Nov 19 10:06:37 PST 2020

 w^2 - 5 v^2 = 4 =  2^2
</code>```

Body:
Short version: in $w^2 - 5 v^2 = 4,$ the numbers $v$ are Fibonacci numbers, of which the largest perfect square is $144$

As you can see, my "v" numbers are alternate Fibonacci numbers, while "w" are Lucas. I will try to find a reference, it is known that the largest square Fibonacci number is 144.  Your largest $n$ is therefore $12,$ where your $m=322$

COHN 1963

Umm. Here is a Conway topograph for the quadratic form $x^2 - 5 y^2.$ This constitutes a proof that all solutions of $x^2 - 5 y^2 = 4$ are generated by  initial pairs
$$ (2,0) , (3,1) , ( 7,3), (18,8), (47,21), 123,55), (322,144), (843, 377)    $$
with recursions
$$ x_{n+6}  = 18 x_{n+3} - x_n $$
$$ y_{n+6}  = 18 y_{n+3} - y_n $$

These are from Cayley-Hamilton for
$$
\left(
\begin{array}{cc}
9&20 \\
4&9
\end{array}
\right)
$$

A little more work shows that we may interpolate, meaning
$$ x_{n+2}  = 3 x_{n+1} - x_n $$
$$ y_{n+2}  = 3 y_{n+1} - y_n $$

Let's see, the irrationals  in the Binet description of alternate Fibonacci numbers are
$$ \frac{3 \pm \sqrt 5}{2},  $$
while
$$ \left(\frac{3 \pm \sqrt 5}{2} \right)^3 = 9 \pm 4 \sqrt 5  $$
where $9 \pm 4 \sqrt 5$ are the Binet numbers from $\lambda^2 - 18 \lambda + 1 =0$

REsources on Conway's Topograph

http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf   (Conway)

https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf    (Hatcher)

http://bookstore.ams.org/mbk-105/                  (Weissman)

http://www.springer.com/us/book/9780387955872             (Stillwell)

Generate solutions of Quadratic Diophantine Equation

diagrams

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http://math.stackexchange.com/questions/1737385/if-d1-is-a-squarefree-integer-show-that-x2-dy2-c-gives-some-bounds-i/1737824#1737824 "seeds"

Find all natural numbers $n$ such that $21n^2-20$ is a perfect square.

Is there a simple proof that if $(b-a)(b+a) = ab - 1$, then $a, b$ must be Fibonacci numbers? 1,1,-1; 1,11

To find all integral solutions of $3x^2 - 4y^2 = 11$

How do we solve pell-like equations?

Diophantine equation $x^2 + xy − 3y^2 = 17$ <1,1,-3>

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:
    9   20
    4   9
  Automorphism backwards:
    9   -20
    -4   9

  9^2 - 5 4^2 = 1

 w^2 - 5 v^2 = 4 =  2^2

Thu Nov 19 10:06:08 PST 2020

w:  2  v:  0  SEED   KEEP +-
w:  3  v:  1  SEED   KEEP +-
w:  7  v:  3  SEED   BACK ONE STEP  3 ,  -1
w:  18  v:  8  SEED   BACK ONE STEP  2 ,  0
w:  47  v:  21
w:  123  v:  55
w:  322  v:  144
w:  843  v:  377
w:  2207  v:  987
w:  5778  v:  2584
w:  15127  v:  6765
w:  39603  v:  17711
w:  103682  v:  46368
w:  271443  v:  121393
w:  710647  v:  317811
w:  1860498  v:  832040
w:  4870847  v:  2178309
w:  12752043  v:  5702887

Thu Nov 19 10:06:37 PST 2020

 w^2 - 5 v^2 = 4 =  2^2


答:
First inequality:
$$ LHS = \sum_{cyc} \dfrac{bc}{a(a+c)}  = \sum_{cyc} \dfrac{(bc)^2}{a^2bc+abc^2} \ge \dfrac{(ab+bc+ca)^2}{ 2abc(a+b+c)} \ge \dfrac{3}{2}$$

Title: Inequality $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$

Body:
prove that for $a,b,c$ being positives and $a+b+c=1$:$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$$

This is a very interesting inequality which i came upon accidentally.We also see that the condition $a+b+c=1$ is not needed.I modified the inequality slightly using Nesbitt's inequality and hence it boils down to prove an even stronger inequality $$\sum_{cyc}\left(\frac{b}{a}-\frac{2c}{a+b}\right)\ge 0$$ Since this was a stronger version i checked WA which shows that it is valid.I have tried to get an SOS but failed.The problem is that even after complete expanding and cross-multiplying we get a cyclic inequality and hence Muirheads theorem fails.


答:
You can connect $\overline{BE}$ and note that $$\frac{[\triangle ADE]}{[\triangle ABE]}=\frac{\overline{AD}}{\overline{AB}}=\frac{3y}{3y+y}=\frac{3}4$$
and also $$\frac{[\triangle ABE]}{[\triangle ABC]}=\frac{\overline{AE}}{\overline{AC}}=\frac{y}{y+3y}=\frac{1}4$$

Therefore you get $$\frac{[\triangle ADE]}{[\triangle ABC]}=\frac{3}{4}\cdot\frac{1}4=\frac{3}{16}=\frac{x}{x+52}\implies \color{red}{x=4}$$

Body:
In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$. If area of quadrilateral $BCED$ is $52$ $cm^2$ and $[\triangle ADE] = x$ $cm^2$, find $x$.

What I Tried: Here is a picture :-

The only thing I am not able to figure out is that how to use the area of the quadrilateral using the available information in the picture. Suppose I am able to find the value of $y$, but how will I use that to find $[\Delta ADE]$ too? I cannot find a way to do that either.

Can anyone help me? Thank You.


答:
Draw a parallel line to $BC$ from $D$ and it meets $AC$ at $F$.

Say $\triangle ABC = a$

$\triangle ADF = \frac{9}{16}a$ (both base and height ratio is $3:4$)

$\triangle EDF = \frac{2}{3} \triangle ADF = \frac{3a}{8}$ (same base but height is $2:3$, as $AE:EF = 2:3$)

So $52 = \triangle EDF + [BCDF] = \frac{3}{8}a + a - \frac{9}{16}a = \frac{13a}{16}$

So $a = 64$
$\triangle ADE = 64 - 52 = 12$

Title: In $\triangle ABC$ where $AB = AC$, $D$ and $E$ are points on $AB$ and $AC$ respectively, such that $AB = 4BD$ and $AC = 4AE$.


问:
Norm Convergence of Partial Sum Operator on Schwartz Space
Let's define the partial sum operator $S_R$ as
$$ (\widehat{S_R f}) = \chi_{B_R}\widehat{f}. $$

I need to prove that for $2 < p < \infty$ and $f \in \mathcal{S}(\mathbb{R})$, the Schwartz space, then $\|S_R f - f\|_p \rightarrow 0$ as $R \rightarrow \infty$.  Based on the line of questions, I think I basically need the two following facts:

$\|S_R f\| \leq C_p \|f\|_p$ for all $1 < p < \infty$,

Hausdorff-Young Inequality: If $f \in L^p$, $1 < p < 2$, then $\widehat{f} \in L^{p^\prime}$ and
$$ \|\widehat{f}\|_{p^\prime} \leq \|f\|_p .$$

My struggle is regarding how to use 2., which was given with the advice to use said inequality to estimate $\|S_Rf - f\|_p$ in terms of its Fourier transform.  That confuses me as it seems on the wrong side of the inequality to get my desired result in any case.  Any help and/or advice would be a great help.  Thank you.

答:
says that $\|g\|_{q} \leq \|\hat{g}\|_{q'}$ for $q > 2$, take $q=p,g=S_rf-f$

(from that $\hat{f}\in S$ we get that $|\hat{f}(x)| \le C/(1+ x^2)$ obtaining a bound for $\|\hat{g}\|_{q'}$)


答:
This is essentially an elliptic curve.

There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.

We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.

If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.

Now we use Sage to find all integer points on this curve. Paste the following codes into this site and press "Evaluate".

```<code>EllipticCurve([20, 0]).integral_points()
</code>```

The output:

```<code>[(0 : 0 : 1), (4 : 12 : 1), (5 : 15 : 1), (720 : 19320 : 1)]
</code>```

We see that the corresponding values of $(m, n)$ are $(2,0), (3,1), (322,12)$, respectively (negative values are not listed).

Title: Values of $a$ such that $x^5-x-a$ has quadratic factor

Body:
I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.

My Attempt

Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:

$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$

Hence:

$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$

and we have:

$$1=-bf-ce=b^4-3b^2c+c^2,$$

so that:

$$(2c-3b^2)^2=5b^4+4.$$

Question

How can I find all values of $n$ such that $5n^4+4$ is a perfect square?

My Attempt

If $m^2=5n^4+4$, then $m^2-5n^4=4$.

If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.

Body:
This is essentially an elliptic curve.

There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.

We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.

If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.

Now we use Sage to find all integer points on this curve. Paste the following codes into this site and press "Evaluate".

EllipticCurve([20, 0]).integral_points()

The output:

[(0 : 0 : 1), (4 : 12 : 1), (5 : 15 : 1), (720 : 19320 : 1)]

We see that the corresponding values of $(m, n)$ are $(2,0), (3,1), (322,12)$, respectively (negative values are not listed).


问:
derivate a function in 1
how are you? They ask me to derive the function in $1$, I have already done all the possible accounts and it gives me (depending on the account I do) $1$ or that it does not exist.
The functions is
$f:\left(0;+\infty\right)\rightarrow \mathbb{R}$

$$f(x)=\begin{cases}
\frac{\ln x}{1-e^{x-1}}&x\ne 1\\
-1 & x=1\\
\end{cases}
$$
So $f'(1)$ is....
i think that this function is not possible to derivative in $1$ , but it's strange because some times give me $1$.... I don't where start , if its beater Hlopital or what..
thanks

答:
Here is a computation using Taylor series.
The derivative at $x = 1$ (if it exists) is the limit
$$
\begin{align}
\lim_{h \to 0}\frac{f(1 + h) - f(1)}{h} &=
\lim_{h \to 0} \frac 1h \left[\frac{\ln(1 + h)}{1 - e^{h}} + 1\right]
\\ & = \lim_{h \to 0} \frac{\ln(1 + h) -(e^{h}-1)}{h(1 - e^{h})}
\\ & = \lim_{h \to 0} \frac{\frac{\ln(1 + h)}{h} -\frac{e^{h}-1}{h}}{1 - e^{h}}
\\ & = \lim_{h \to 0} \frac{1 - \frac 12 h + o(h) -(1 + \frac 12 h + o(h))}{-h+o(h)}
\\ & = \lim_{h \to 0} \frac{-h + o(h)}{-h+o(h)} = 1.
\end{align}
$$
So, the derivative at $x = 1$ exists and $f'(1) = 1$.

Title: derivate a function in 1


答:
Join $DC$

let $ar(\Delta ADE)=p,ar(\Delta) DEC=q$

$$\frac{p}{q}=\frac{y}{3y}=\frac{1}{3}$$
$$\frac{ar(\Delta ADC)}{ar ( \Delta CDB)}=\frac{q+p}{{52-q}}=\frac{3y}{y}=3$$
$p=?,q=?$


问:
Solving the given system of equations

$$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$

I got $x^3 + y^3 = 6xy = (x+y)^2$ which I simplified to $x^2 - xy + y^2 = x + y$ and I'm not sure what to do now. Please help.

答:
We are given that $$\frac{x^2}{y} + \frac{y^2}{x} = 6$$
$$\frac{x}{y} + \frac{y}{x} = 4$$
Consider the expression $(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)$:
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}\right)=x+y+\frac{x^2}{y}+\frac{y^2}{x}$$
so
$$(x+y)\left(\frac{x}{y}+\frac{y}{x}-1\right)=\frac{x^2}{y}+\frac{y^2}{x}$$
But using the initial conditions we see that
$$3(x+y)=6\implies x+y=2\implies x=y-2$$
which can be subsituted into your equations to provide solutions for $x$ and $y$.

This is a bit more of an experimental solution than the other answers; the expression considered intially is not necessarily obvious. Often though some fiddling around with expressions like that is useful.

I hope that helps :)

Title: Solving the given system of equations


问:
Expected size of a sample
A warehouse contains seven printing machines, three of which are defective.
A company worker has a task to send two of the defective machines for a
repair. In order to identify such a pair, he selects a printing machine at random and
inspects whether it is a defective or a non-defective one. Then he selects another
printing machine at random for the inspection and he repeats the process until he
identifies the second defective printing machine. What is the expected number of
printing machines he has to inspect?

I would really appreciate a hint on how to proceed with this problem. So far I have created a tree diagram, which made me believe the answer is 4. But I think there has to be another way to solve this.

答:
Assuming that non-defective machines aren't inspected again, the scenario is equivalent to shuffling the machines and having the worker inspect them in order. There are $\binom73=35$ possible orders and of those

$5$ have the second defective machine second-in-line ($1$ way to place the defective machine before it, $5$ ways to place the one after)

$8$ have the second defective machine third-in-line

$9$ have it fourth

$8$ have it fifth

$5$ have it sixth

Thus the expected number of machines inspected is
$$\frac{5×2+8×3+9×4+8×5+5×6}{35}=4$$

Title: Expected size of a sample


问:
Proof of Transitivity (Inference Rule) using Leibniz rule and Equanimity
In Tourlakis' Mathematical logic, we are given the following primitive rules of inference [p.40]:

Leibniz rule: (A↔B) / (C[p≔A]↔C[p≔B])

Equanimity: (A, A↔B) / B

Capture from the book

Then he goes to derive the following inference rule, using he previous two [p.47]:

Transitivity: (A↔B, B↔C) / (A↔C)

Capture from the book

I have trouble proving how to use the primitive rules to proof Transitivity.

Leibniz rule: I understand that the Leibniz rule replaces in C (which is well formed formulae) any occurrence of the Boolean variable (atomic) p with A to get C[p≔A] and then replaces p with B to get C[p≔B], which yields to C[p≔A]↔C[p≔B].

Equanimity: If we assume A to be t and A↔B to be t, (because that's our premise), we get that B must be also t, because that's required for A↔B to be t.

Now, I don't see how to use either of the primitive formulae to prove Transitivity, I don't quite get how to use the Leibniz rule in Tourlakis' way. I would even go further and to state Transitivity as another primitive rule of inference, but I think just two is a nicer approach.

So, how to prove Transitivity using Leibniz and Equanimity?

Thank you!

答:
Note: due to the fact that the example uses $A,B,C$ as schematic letters for formulas, I'll adopt a different letter for the formulation of Leibniz's rule:

from $A \equiv B$, infer $X[p := A] \equiv X[p := B]$.

Thus, to apply the rule with premise $B \equiv C$, we use as $X$ the formula $A \equiv p$.

The two substitutions are: $X[p := B] = (A \equiv p)[p := B] = A \equiv B$ and  $X[p := C] = (A \equiv p)[p := C] = A \equiv C$, respectively.

In conclusion:

$A \equiv B$

$B \equiv C$

$X[p := B] \equiv X[p := C] \text {  i.e.  } A \equiv B \equiv A \equiv C$ --- from 1) and 2) by Leibniz.

Regarding

in step "we use as $X$ the formula $A≡p$" we assume the formula $A$ to be atomic, so that, when replacing $p$ in $X$ for $A$ we get $A$,

the answer is: no. We assume as $X$ the "composite" formula $A \equiv p$, where $p$ is atomic.


问:
For an irreducible Markov chain on a finite state space, all states are positive recurrent
I'm reading Jeffrey's A first look at rigorous probability theory, in particular, Proposition 8.4.10. Here is a sketch of the proposition and its proof:

There are two places in the Proof I'm stuck on.

The first one is in the first paragraph. I'm confused about why a finite state space implies that $h_{ji}^{(m)} \geq \delta$ for all states $j$, and for some $m \in$ N and $\delta > 0$?
The second confusion follows immediately at the beginning of the second paragraph, that is, how does the author come up with the result $1 - h_{ii}^{(n)} \leq (1- \delta)^{\lfloor n/m \rfloor}$?

Any advice is greatly appreciated!!

答:
The first point follows from
$$
   \lim_{m \to \infty} h_{ji}^{(m)} = f_{ji} > 0.
$$
The idea is that for these things to be converging to something positive, eventually they all have to stay positive.

Let's make things concrete. For that limit above to hold, there has to be a value $k_j$ such that $h_{ji}^{(k_j)} \ge \frac12 f_{ji}$. (Otherwise, if $h_{ji}^{(m)}$ never reaches even $\frac12 f_{ji}$, how can it converge to $f_{ji}$?) Note that $h_{ji}^{(m)}$ is non-decreasing, so we have $h_{ji}^{(m)} \ge \frac12 f_{ji}$ for all $m \ge k_j$.

Now choose $\delta$ to be minimum of $\frac12 f_{ji}$ over all states $j$, and let $m$ be the maximum of $k_j$ over all states $j$. Then for each $j$, $h_{ji}^{(m)} \ge h_{ji}^{(k_j)} \ge \frac12 f_{ji} \ge \delta$.

For the second point: having $h_{ji}^{(m)} \ge \delta$ tells us that no matter where the random walk is at any time, with probability at least $\delta$ it will visit state $i$ in the next $m$ steps. We're looking at $1 - h_{ii}^{(n)}$: the probability that the random walk stays away from state $i$ for $n$ steps.

In the first $m$ steps, there's at least a $\delta$ probability of visiting $i$. If that doesn't work out, then in the next $m$ steps, there's at least a $\delta$ probability of visiting $i$. If that doesn't work out either, then in the next $m$ steps, there's again at least a $\delta$ chance of probability $i$, and so on.

So the probability of avoiding $i$ is at most $1-\delta$ after $m$ steps, at most $(1-\delta)^2$ after $2m$ steps, at most $(1-\delta)^3$ after $3m$ steps, and so on. It is at most $(1 - \delta)^{\lfloor n/m\rfloor}$ after $\lfloor n/m\rfloor m$ steps, by the same reasoning. Since $n \ge \lfloor n/m\rfloor m$, the probability is at most $(1 - \delta)^{\lfloor n/m\rfloor}$ after $n$ steps, too.


问:
Gluing diagrams for tori
I know that gluing opposite sides of a square gives a torus, and I think I've seen somewhere that gluing opposite sides of an octagon gives a 2-holed torus.  Is there a general way of constructing gluing diagrams for surfaces of genus $g$ / for $k$-holed-tori?

答:
Here's a depiction of the most common construction:

Gluing a 4k-gon by grouping the edges into groups of 4 like this produces a k-holed torus. Other gluings of 4k-gons also produce a k-holed torus, as long as they keep the surface orientable and ensure that all 4k vertices are glued together into a single point. This is a consequence of Euler's formula for CW complexes and the classification theorem for closed 2-manifolds. Gluing opposite sides together, as you suggest, works as well.

Title: Gluing diagrams for tori


答:
$4$ is indeed correct for sampling without replacement.

This is a negative hypergeometric distribution which in this particular formulation has mean $2\frac{7+1}{3+1}$

One approach is say: introduce another defective machine so you have $7+1=8$ machines of which $3+1=4$ are defective.  Place them in a circle at random.  Clearly the expected or average gap between successive defective machines is $\frac{7+1}{3+1}=2$, in effect counting from immediately after a defective machine through to the next defective machine.  Now break the circle at the machine which you had introduced, remove it and straighten the line. The expected count of machines to the second defective machine (i.e. two gaps) is then $2\frac{7+1}{3+1}=4$

Title: Expected size of a sample

Body:
A warehouse contains seven printing machines, three of which are defective.
A company worker has a task to send two of the defective machines for a
repair. In order to identify such a pair, he selects a printing machine at random and
inspects whether it is a defective or a non-defective one. Then he selects another
printing machine at random for the inspection and he repeats the process until he
identifies the second defective printing machine. What is the expected number of
printing machines he has to inspect?

I would really appreciate a hint on how to proceed with this problem. So far I have created a tree diagram, which made me believe the answer is 4. But I think there has to be another way to solve this.


问:
For two disjoint compact subsets $A$ and $B$ of a metric space $(X,d)$ show that $d(A,B)>0.$

I was thinking about the following problem: For two disjoint compact subsets $A$ and $B$ of a metric space $(X,d)$ show that $d(A,B)>0.$ I'm having doubt with my attemp. Please have a look and do comment:

Let $d(A,B)=\inf\{d(a,b):a\in A,b\in B\}=0.$ Then $\exists$ sequences $\{a_n\}\subset A,\{b_n\}\subset B$ such that $0\le d(a_n,b_n)<\dfrac{1}{n}~\forall~n.$ Since $A$ is compact, $\exists$ a convergent subsequene $\{a_{r_n}\}$ of $\{a_n\}$ converging to some $a\in A.$ Then $0\le d(a_{r_n},b_{r_n})<\dfrac{1}{r_n}~\forall~n.$ Similarly since $B$ is compact, $\exists$ a convergent subsequene $\{b_{r_{n_m}}\}$ of $\{b_{r_n}\}$ converging to some $b\in B.$ Then $0\le d(a_{r_{n_m}},b_{r_{n_m}})$$<\dfrac{1}{r_{n_m}}~\forall~m.$ We note that $r_{n_m}\ge n_m\ge m>0$$\implies0<\dfrac{1}{r_{n_m}}<\dfrac{1}{m}\to0$$\implies\dfrac{1}{r_{n_m}}\to0.$ Using the sqeezing lemma once again we can see that $\exists$ convergent sequences $\{a_n\}\subset A$ and $\{b_n\}\subset B$ such that $a_n\to a\in A,b_n\to b\in B.$ $\exists$ convergent sequences $\{a_n\}\subset A$ and $\{b_n\}\subset B$ such that $a_n\to a\in A,b_n\to b\in B.$ Then $d(a_n,b_n)\to d(a,b)$ Consequently, $d(a,b)=0\implies a=b,$ a contradiction to $A\cap B=\emptyset.$

答:
Your solution is correct. Here is another one.

Assume $d(A,B)=0$. Consider function $d(x, B)=\inf\{d(x,y):y\in B\}$ It is continuous, and defined on compact metric space $A$. So it attains its minimum, i.e. there exist $a\in A$ such that $d(a,B)=\inf\{d(x,B):x\in A\}=d(A,B)=0$. Since $d(a,B)=0$, then $a\in \operatorname{cl}_X(B)$. Since $B$ is compact it is closed, so $a\in \operatorname{cl}_X(B)=B$. Contradiction, since $A\cap B=\varnothing$.

ContentLicense: CC BY-SA 3.0


问:
Limit of $\frac{1 - F(x)}{f(x)}$
Suppose $f: [0, 1] \rightarrow \mathbb{R}+$ is a continuous density function. Let $F$ be its c.d.f. Suppose $f > 0$ over $(0, 1)$ and $$\frac{1 - F(x)}{f(x)}$$ is decreasing. Is it true that $$\lim_{x \rightarrow 1} \frac{1 - F(x)}{f(x)} = 0?$$

Trivially, if $f(1) > 0$, then it is true. But how about $f(1) = 0$?

答:
if the distribution is over $[0,1]$ for $f$ then it makes sense that the cumulative probability, $F(1)=1$ since this is the upper limit of pdf, meaning that:
$$\lim_{x\to 1}1-F(x)=0$$
However, if we do not know anything about the type of pdf $f$ is or its skewness how can be know the value of $f(x)$ at a given point, and therefore the value of $f(1)$? Without this I dont think it is possible to fully conclude the value of the limit given.

Title: Limit of $\frac{1 - F(x)}{f(x)}$


问:
Determinant of a permutation matrix plus identity

Let $A$ be a permutation matrix. Calculate $\det(I + A)$.

I tried with the definition of the determinant but couldn't find it. I also tried to use the decomposition of permutations as products of disjoint cycles.

答:
For the readers who want more, the question that follows is as the following :
Considering $T_n=\sum_{\sigma \in S_n}^{} det(I+P_{\sigma})$, prove that $T_{n+1}=2T_{n}+n(n-1)T_n$ to determine a simple formula for $T_n$


问:
correspondence between subspaces and left ideals in End$(V)$
Let $V$ be a right- n-dimensional vector space over a division ring $D$. Let $R=\text{End}(V_D)$. We know that for every subspace $W$ of $V$ the set $J_W = \{f \in R \mid f(W) = 0\}$ is a left ideal of $R$.

What I have to prove is that the mapping $W_D \mapsto J_W$ is a bijection between the the lattice of subspaces of $V$ and the lattice of all left ideals of $R$.

Injectivity is actually really clear. What about surjectivity? My attempt was defining, for each $J$ left ideal of $R$, the subspace $W_J = \{v \in V \mid f(v) = 0\, \forall f \in J\}$. Then we have to show that $W_J$ is mapped into $J$ by our correspondence. The fact that $J \subset J_{W_J}$ is trivial, but what about the other inclusion?

Any hint\solution is highly appreciated. Thanks!

答:
I am not sure if anything breaks down for division rings which are not fields, but this is how I would do it in a regular linear algebra setting, and then you can tell us if it works for division rings:

Take a basis $\mathcal{B}=\left\{e_1,\ldots,e_n\right\}$ of $V$, and consider the functionals $\phi_i\colon V\to D$, $\phi_i(e_j)=\delta_{ij}$. Also consider the co-functionals $\psi_i\colon D\to V$, $\psi_i(d)=e_id$.

By definition, $J_{W_J}=\left\{g\in \operatorname{End}(V):\bigcap_{f\in J}\ker f\subseteq\ker g\right\}$. Let $g\in J_{W_J}$.

Let us find $f_1,\ldots,f_N\in J$ such that $\bigcap_{j=1}^N\ker(f_j)=\bigcap_{f\in J}\ker(f)(=W_J)$.

Just choose any $f_1\in J$. If $\dim \bigcap_{f\in J}\ker(f)=\dim\ker(f_1)$, we are done;

If not, choose some $f_2\in J$ such that $\ker(f_2)\cap \ker(f_1)$ is strictly smaller than $\ker(f_1)$. If $\dim \bigcap_{f\in J}\ker(f)=\dim(\ker f_1\cap\ker f_2)$, we are done.

If not, keep "decreasing the dimension" of $\ker(f_1)\cap\ker(f_2)\cap\cdots$ until you get to the dimension of $\bigcap_{f\in J}\ker(f)$.

Then we have
$$\bigcap_{f\in J}\ker(f)=\bigcap_{j=1}^N\ker f_j=\bigcap_{j=1}^N\bigcap_{i=1}^n\ker(\phi_i f_j).$$
and, for each $k$,
$$\ker(g)\subseteq\ker(\phi_k g)$$
so $\bigcap_{j=1}^N\bigcap_{i=1}^n\ker(\phi_i f_j)\subseteq\ker(\phi_k g)$ for each $k$.

The following is standard (there should be a proof somewhere on this site as well):

Theorem (see Rudin, Functional Analysis, Lemma 3.9): If $p_1,\ldots,p_m,p$ are linear functionals on a vector space $V$, then $p$ is a linear combination of $p_1,\ldots,p_m$ iff $\bigcap_{j=1}^m p_j\subseteq\ker(p)$.

So for each $k$ we can find scalars $\alpha_{i,j,k}$ such that $\phi_k g=\sum_{i,j}\alpha_{i,j,k}(\phi_i  f_j)$ (if I am doing things right, spaces of operators between right $D$-modules should be left $D$-modules... right?)

Now remember the "cofunctionals" $\psi_k$ right from the beggining. We have $\operatorname{id}_V=\sum_k\psi_k\phi_k$. Then
$$g=\sum_k\psi_k\phi_kg=\sum_{i,j,k}\psi_k(\alpha_{i,j,k}(\phi_i f_j)),$$
and all $\psi_k(\alpha_{i,j,k}(\phi_i f_j))$ belong to $J$, as it is a left ideal by hypothesis, so $g$ does as well.


问:
Factorization of polynomial doubt
The question is to factorize x^4 + x to irreducible polynomials over Z2[x].
Here is my attempt but I have certain doubts:

Check roots first : p(0)= 0, so (x-0)=x is a factor;

p(1)=0, so (x-1) is a factor;

But when I check if they divide the polynomial or not, x divides it but (x-1) doesn't divide it.
So, my solution in this case is x^4 +x = x(x^3 +1).

Another approach: If p(x) is reducible to g(x) and h(x), then degree(p(x))= deg(g(x)) + deg(h(x)).
So the possibilities are 4= 1+ 3 or 2+2.
But Z2 only contains polynomials of degree 0 or 1. So, I am not considering the case (2+2). If I consider the case of (1+3), then the only irreducible polynomials of degree 1 in Z2 are x and x+1 and they both divide the polynomial. So my factorization then becomes:
x^4 +x = x(x^3 +1);

x^4 +x = (x+1)(x^3-x^2+x);

But my doubt here is that both (x^3+1) and (x^3-x^2+x) are of degree 3 and they don't belong to Z2.
So, my confusion is that it gives me one factorization using root method and 2 factorizations using the second approach but both of them have a 3rd degree polynomial. So, is is polynomial even reducible in Z2 or no or if I am doing something wrong here. Any suggestions or help will be appreciated.

答:
$\mathbb Z_2$ only contains polynomials of degree $0$ or $1$

False. There are irreducible polynomials of all degrees, even in $\mathbb Z_2$; $x^2+x+1$ is the unique degree-$2$ example. It can be shown irreducible by your second approach: if it were reducible it would have to have linear factors, which it does not.

Having found $x$ and $x-1$ as factors you have found the irreducible factors:
$$x^4+x=x(x+1)(x^2+x+1)$$
Remember that $1=-1$ in $\mathbb Z_2$!

Title: Factorization of polynomial doubt


问:
How do you isolate $\theta$ from $\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$?
This equation derives from physics force equations, and I have verified to make sure that the equation works. No actual physics pertains to this question. The part I need help on only requires math skills. Here is the equation:

$$\sin(\theta) - k\cos(\theta) = \frac{m_1}{m_2}$$

where $k$, $m_1$, and $m_2$ are variables, specifically the kinetic friction coefficient, mass one, and mass two respectively (irrelevant).

I have tried squaring both sides to then use double angle rules on the $2\sin\cos$ that yields, but all of my attempts have failed to simplify the equation further.

Could someone help me derive an equation for theta?

答:
Here's a general guide and explanation for problems of your type:

If we have an expression, $a\sin{x}+b\cos{x}$, let us assume it can be written in the form $R\sin(x+\alpha)$ Now to see if we can find values for $R$ and $\alpha$ in terms of $a$ and $b$. Using the compund angle formulae, also known as the addition formulae:
$$R\sin(x+\alpha)=R\sin{x}\cos{\alpha}+R\sin\alpha\cos x=a\sin{x}+b\cos{x}$$
So we have
$$R\cos\alpha=a,R\sin\alpha=b$$
So dividing the second equality by the first:
$$\tan\alpha=\frac{b}{a}$$
meaning we can find $\alpha$ in terms of $a$ and $b$, as we wanted. Now, to find $R$:
Squaring the $2$ equalities above we have
$$R^2\cos^2\alpha+R^2\sin^2\alpha=R^2(\cos^2\alpha+\sin^2\alpha)=R^2=a^2+b^2\implies R=\sqrt{a^2+b^2}$$
So, to finish off by recapping what we have learnt:
$$\tan\alpha=\frac{b}{a},R=\sqrt{a^2+b^2}$$
meaning
$$a\sin{x}+b\cos{x}\equiv\sqrt{a^2+b^2}\sin(\theta+\arctan{\frac{b}{a}})$$
Try applying that to your question. If you have any questions, please ask!

I hope that was helpful :)

Note: It's not always going to be the case that we take the positive square root for $R$. It depends on the sign of the quantities $a$ and $\cos\alpha$ ( I could've picked $b$ and $\sin\alpha$ as well; can you see why?).


答:
For every case where the second of three defective machines is not at position $k\ne 4$, there is an equally likely reflected case where the second of three is at position $8-k$. Hence we can pair all $k\ne 4$ cases such that the expected position of the second defective machine is $4$. Then that expected position is also $4$ overall.

Nitpicking mode. However, there are $5$ (out of $7\choose 3$) cases where our given knowledge that there are three defective machines saves us one check - namely when the last two remaining must be defective. Additionally, there is one case where we save yet another check - namely when the last three remaining machines are defective. So we might subtract $\frac 6{7\choose 3}$ from the simple answer above.

Title: Expected size of a sample


问:
Integrate: $ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $ without using complex analysis methods
Can this integral be solved without using any complex analysis methods: $$ \int_0^\infty \frac{\log(x)}{(1+x^2)^2} \, dx $$

Thanks.

答:
Integrate by parts
\begin{align}
& \int_0^\infty \frac{\log x}{(1+x^2)^2} dx
=\int_0^\infty \frac{\log x}{2x}d\left( \frac{x^2}{1+x^2} \right) \\
\overset{IBP}= &-\frac12 \int_0^\infty \frac{1}{1+x^2}dx+\frac12 \int_0^\infty \underset{x\to \frac1x} {\frac{\log x}{1+x^2}}dx
= -\frac12\cdot\frac\pi2+0=  -\frac{\pi}4
\end{align}


问:
Additive inverse notation confusion: $-2^2$ vs $(-2)^2$
My textbook says that the additive inverse of a number is represented by $-$ (minus sign).
For example the additive inverse of a number $-1$ is typed like this: $-(-1)$, and the additive inverse of a number $1$ is then typed like this; $-1$.

In the book there are also some example calculations and this is what confuses me:

$$-2^2 = -4\quad\text{and}\quad (-2)^2 = 4$$

I've always thought $-n^2$ simply like $(-n)\times(-n)$. But according to my textbook this is wrong and $-n^2$ actually means $-(n\times n)$

I tried to wrap my head around this and played with it for a while and found out that when i type $-2^2$ to the Microsoft Windows calculator, I'll get the (wrong?) answer of $4$. But when I type this to my calculator (Casio fx-991EX) I'll get the answer of $-4$.

I think many people will just write $-2^2$ when they actually mean $(-2)\times(-2)$ and depending how you'll look at the minus sign gives you two completely different answers.

So which one is right? Is the minus sign really a notation for additive inverse of the operation.

Which is the correct answer?

$$-2^2 = -4\quad\text{or}\quad-2^2 = 4$$

答: You're not alone in this confusion.  It's very common to see $-2^2$ and read it the way you would read the symbols left to right, i.e."negative two squared" as the order of operations.  However, remembering PEMDAS (parenthesis, exponents, multiplication, division, addition, subtraction) what $-2^2$ really means is "the negative of two squared" because the exponent needs to happen before multiplication by $(-1)$.  The first interpretation would make you think the answer is $4$ and the second would imply that the answer is $-4$.  However, the first is based on the order you read the symbols, and the second is based on the actual order of the mathematical operations involved, so $-2^2 = -4$ and $(-2)^2 = 4$ is correct.


问:
How to solve exponential equation without using logarithm , $10=(\frac{6}{5})^x$
Without using logarithms, how can I solve

$$10=\left(\frac{6}{5}\right)^x$$

答:
You could use the series expansion of the natural logarithm as follows:

$$
\frac{\sum\limits_{n=1}^{\infty}\frac{1}{n}\left(1-\frac{1}{10}\right)^{n}}{\sum\limits_{n=1}^{\infty}\frac{1}{n}\left(1-\frac{5}{6}\right)^{n}}
$$

Therefore, you don't use the logarithm directlry. But I am not sure, if this infinite sum is better for you. At least it will approach the correct result, the more terms you use.


问:
How to prove the converge of a sequence?
Let $t_{n} = \int_{1}^{n} (\ln(x)^p) \,dx$, where $p < 0$ is any real number.

How can I prove that this sequence converges?

My attempt: Clearly, $t_{n} \geq 0$, so by the linear property of Riemann integrals, $t_{n}$ is increasing. Then, if $t_{n}$ is bounded, it is also convergent. Can you find a bound?

答: This sequence does not converge. You are asking it $\int_1^{\infty}\ln(x)^pdx$ converges for any $p<0$, but for $p=-1$ we can use the comparison test with $\frac{1}{x}$ (i.e $\frac{1}{\log(x)}>\frac{1}{x}$) to see that the integral diverges since $\int_{1}^{\infty}\frac{1}{x}dx$ diverges.


问:
Graph problem with a deck of cards
I have a question that needs a response using graph theory but I can't think of a method how to use a graph on that.
Tha problem is ->

Two friends perform the following game card: the first choses at random six cards from a regular
deck (of 52 cards), looks at them, keeps one of them, and arrange the other five in order from
left to right, face up; the, the second correctly guesses the sixth card. Prove that the two friends
can agree on a system which always makes this possible, and find such a system.

I have a feeling that this is a graph matching problem but don't know where to go from here.

答:
Here's an easy way to do it:

First rank the cards from $1$ to $52$ however you like (standard would be $2\clubsuit=1$ to $A\spadesuit=52$ but it isn't important).

As there are $>4$ cards we must have some duplicated suit.  Choose the special card as one from the duplicate suits.  Doesn't matter if you have more than one duplicated suit, nor if you have more than two cards in that suit.  It will matter which card you choose to hide from the duplicated suit, we'll come to that later.

Place a card of the duplicate suit first. Now you know what suit the missing card has.  Note that there are only $12$ possible cards for the missing one.

There are $6$ permutations for the other three (viewing them as Low, Middle, High).  Thus the order in which they are placed determines a number from $1$ to $6$.  You can number the permutations however you like.

Viewing the duplicate suit cyclicly, add that number to the card on the far left.  Thus if you have a $6\diamondsuit$ on the far left and the permutation is number $4$ then the answer is $10\diamondsuit$.  If the far left card is $K\heartsuit$ then the answer is $4\heartsuit$ and so on).

The key point is that of the two cards in the same suit one of them must be $≤6$ away from the other, that tells you which one of the two to hide.

Title: Graph problem with a deck of cards


问:
Hypothesis Testing scenario steps

According to some folklore, if you examine n clover plants in a field and do not find any with four leaves, you may be reasonably confident that the fraction of four-leafed clovers is less than 3/n.  Justify the folklore.  You may assume n is reasonably large, but state your assumptions.

I assumed n is reasonably large, with n = 1000. Lets say the parameter is $\theta$, and we can assume  $X_{i}$ $\sim Ber(\theta)$. Thus we can also assume that because n is approximately large, we can invoke a binomial approx. to the normal.

Thus:

$H_{0}: \theta = 0.003$

$H_{a}: \theta < 0.003$

Y = $\sum X_{i}$ is binomially distributed with $u = np = 1000(0.003) = 3$

$\sigma = \sqrt{1000 *0.003*0.997} = 1.73$

$P(Y \leq 0.5) = \frac{0.5 - 3}{1.73} \sim N(0,1)$

$P(Z \leq -1.445) = .074$

For $\alpha$ level of $0.10$ you may reject the null hypothesis and reasonably conclude that $\theta < 0.003$

The continuity correction factor shifts the observed Y to $0.5$

Is this correct? Are my assumptions valid?

答:
The normal approximation to the binomial is not justified here as you are at the end of the tail of the binomial distribution.

A Poisson approximation to the binomial would be justifiable, or you could just say the probability of seeing zero four-leaf clovers is $(1-\theta)^n$ which for small $\theta$ and large $n$ is about $e^{-n/\theta}$.

If we want the probability of seeing seeing zero four-leaf clovers to be $5\%$ or more under the null hypothesis that the parameter is $\theta$, then we want $e^{-n/\theta} \ge 0.05$ so $\theta \le -\dfrac{\log_e(0.05)}{n}\approx \dfrac{2.996}{n}\approx \dfrac{3}{n}$.

This is known as the rule of three.

Title: Hypothesis Testing scenario steps


问:
integration increasing decreasing
Let f be a continuous strictly increasing function from $[0,\infty) \rightarrow [0,\infty)$ and $g=f^-1$ (i.e.$f(x)=y \iff g(y)=x$). Let $a,b>0$ and $a\neq b$.Then $ \int\limits_{0}^{a}f(x)dx$+$\int\limits_{0}^{b}g(y)dy$ is

(A) greater than or equal to ab

(B) less than ab

(C) always equal to ab

答:
Remember that the function $f^{-1}(x)$ is a reflection of the function $f(x)$ about the line $y=x$. If we look at the integral of the inverse first:
$$I_1=\int_0^bf^{-1}(x)dx$$
$x=f(y)$, $dx=f'(y)dy$ and so:
$$I_1=\int_0^{f(b)}yf'(y)dy$$
and since you said $f$ is strictly increasing on $(0,\infty)$ we can say $f'(y)>0$ and so clearly $I_1>0$. If we do IBP we get:
$$I_1=\left[yf(y)\right]_{y=0}^{f(b)}-\int_0^{f(b)} f(y)dy$$

Now if we look at $f$:
$$I_2=\int_0^af(x)dx$$
this is the best we can simplify for a general equation, and so:
$$I_1+I_2=\left[yf(y)\right]_{y=0}^{f(b)}+\int_{f(b)}^af(x)dx$$
$$=f(b)f(f(b))+\int_{f(b)}^a f(x)dx$$
Now what can we say about this statement, well clearly for $a>f(b)$ and if $f(x)>0$ for $(0,\infty)$ then:
$$I_1+I_2\ge f(b)f(f(b))+(a-f(b))f(b)$$

Title: integration increasing decreasing


问:
Theorem 4.25 - Rudin's Functional Analysis. Elementary observation about Quotient spaces
I'm studying the theorem in the title and the very first elementary observation is not really clear to me.
I think I can state it as follow:

Let $Y$ a locally convex space, $M_0$ a closed subspace of $Y$ and $\Sigma$ is the space of all continuous linear functionals on $Y$ that annihilate $M_0$ then
$$
\dim Y / M_0 \leq \dim \Sigma
$$

I'll quote the book from now on

Pick a positive integer $k \leq \dim Y / M_0$. Then there're vectirs $y_1,\ldots, y_k \in Y$ such that the vector space $M_i$ generated by $M_0$ and $y_1,\ldots y_i$ contains $M_{i-1}$ as proper subspace. By theorem 1.42 each $M_i$ is closed and by theorem 3.5 there're functionals $\Lambda_1, \ldots, \Lambda_k$ such that $\Lambda_i y_i = 1$ but $\Lambda_i y = 0$ for all $y \in M_{i-1}$. These functionals are linearly indipendent. The following conclusion is therefore reached.

I don't understand and I think the reason is beacause I'm missunderstanding the definition of $\Sigma$ which to me is simply $M_0^{\perp}$ and whose dimension is at least $k$ but it's not really helping me to reach the conclusion, so probably I don't understand the definition of $\Sigma$.

Can you clarify why the conclusion is true?

Update : Reference is Theorem 4.25 Rudin's Functional analysis. It's literally the first few lines of the proof of the theorem.

答:
The definition of $\Sigma$ is:
$$\Sigma= \{ f\in Y^*\mid f(y) =0\ \ \forall y\in M_0\}$$
this agrees with the definition of the annihilator $M_0^\perp$ that I know. Now $M_0$ is assumed to be closed and
$$M_i := M_{i-1}\oplus \Bbb C y_{i}$$
will inductively be closed since you are adding a one-dimensional space to a closed subspace (here we are assuming that $[y_1],..,[y_k]$ are linearly independent in $Y/M_0$ else you may have $M_i=M_{i-1}$).

Now the functional
$$\Lambda_i:M_i\to\Bbb C, \qquad (y+\lambda y_i)\mapsto \lambda$$
is well defined.

Further it is continuous: Suppose $y^{(\alpha)}+\lambda^{(\alpha)}y_i\to y+\lambda y_i$, by closedness of $M_{i-1}$ you get that $y^{(\alpha)}\to y$ and $\lambda^{(\alpha)}\to\lambda$, whence
$$\Lambda_i(y^{(\alpha)}+\lambda^{(\alpha)}y_i)= \lambda^{(\alpha)} \overset{\alpha\to\infty}\longrightarrow \lambda = \Lambda_i(y+\lambda y_i)$$
verifying continuity.

By Hahn-Banach you can extend the $\Lambda_i$ to be functionals on all of $Y$. Now note that $\Lambda_i\lvert_{M_0}=0$ for all $i$, hence they are also in $\Sigma$. Next see that for $i<j$ you have: $\Lambda_j(y_i)=0$ and $\Lambda_j(y_j)=1$, you can leverage these two equalities to find that the $\Lambda_i$ must all be independent. This gives $\dim(\Sigma)≥k$.


问:
Geometric quotient of $\mathbb{A}^2$ modulo $G$
I have the following exercise:
(This exercise deals with the action of a group $G$ on an affine closed.)

Let $G=\langle w \rangle$ be a cyclic group of order $n$ (in multiplicative notation).
Consider the action of $G$ on $\mathbb{A}^2$ defined by $$w(x,y):=(wx,w^{-1}y)$$
Prove that the geometric quotient $\mathbb{A}^2/ G$ is isomorphic to the closed affine set $Z(uv-w^n)\subset \mathbb{A}^3$

I thought about proceeding this way.

I know that a geometric quotient $Y=\mathbb{A}^2/ G$ is such that
$$K[Y]=K[\mathbb{A}^2]^G \hookrightarrow^{\psi^*}  K[\mathbb{A}^2]$$
where:

$K[Y]$ is the set of regular functions on $Y$,

$K[\mathbb{A}^2]^G=\{f\in K[\mathbb{A}^2]: f=f^g \; \forall g \in
   G\}$ is the set of regular functions on $\mathbb{A}^2$ which are
$G$-invariant

$\psi^*:  K[Y] \hookrightarrow  K[\mathbb{A}^2]$ is
such that for the corresponding $\psi: \mathbb{A}^2 \rightarrow Y$
the following property holds  $$\psi(p)=\psi(q) \iff \;\;  p \;
   \textit{and}\; q \;\textit{are in the same orbit of G}$$

So, I'd like to try to figure out who $Y$ is.

I have examined the orbits of $G$ studying the equivalence classes by means of the relation
$$p,q \in \mathbb{A}^2, \;\;\;\; p \sim q \iff \exists g \in G: q=p^g$$
in this case, because $G=\langle w\rangle$ of order $n$
$$(x,y),(a,b) \in \mathbb{A}^2 \;\;\; (x,y)\sim (a,b) \iff \exists t\le n : (a,b)=(w^tx,w^{-t}y)$$
so there are four possibilities

$a=x=b=y=0$ i.e the origin $\{0\}$

$a=x=0, b,y\neq0$ i.e the $x$ axis

$b=y=0, a,x\neq0$ i.e the $y$ axis

$a,x,b,y\neq0$ i.e $\mathbb{A}^2 \setminus \{0\}$

These are the orbits of G, aren't they?

Now, can I assert that Y is made up of these four classes?
I believe so because we are in a finite group and therefore the set quotient (and not geometric quotient) has a natural geometric structure and therefore $\psi$ coincides with the application of the set quotient.

Is everything said so far correct?

If everything is correct, then how can I find an isomorphism with $Z(uv-w^n)\subset \mathbb{A}^3$?
I can't picture this closed up in my mind. Could someone explain me better?

Why if $w$ generates $G$ of order $n$ we take $w ^ n$? Isn't it unity?
If it is the unit, then the origin $(0,0)$ and the points on the axes $(x, 0)$ and $(0, y)$ [which according to my idea are in the quotient $Y$] would not cancel the polynomial $uv-1$ and therefore they would not be in closed $Z(uv-w^n)$.
Where am I wrong?

答:
$$K[x,y]^G = \{ \sum_{a,b,\ n\ |\ a-b} c_{a,b} x^ay^b\}= K[x^n,y^n,xy]\cong K[u,v,w]/(w^n-uv)$$
The isomorphism is $G.(x,y)\to (x^n,y^n,xy)$


问:
Values of $a$ such that $x^5-x-a$ has quadratic factor
I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.

My Attempt

Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:

$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$

Hence:

$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$

and we have:

$$1=-bf-ce=b^4-3b^2c+c^2,$$

so that:

$$(2c-3b^2)^2=5b^4+4.$$

Question

How can I find all values of $n$ such that $5n^4+4$ is a perfect square?

My Attempt

If $m^2=5n^4+4$, then $m^2-5n^4=4$.

If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.

答:
Another way is to do a long division of $x ^ 5-x-a$ by the arbitrary trinomial $x ^ 2 + bx + c$ and set the remainder to zero. This gives the remainder
$$(c ^ 2-3b ^ 2c + b ^ 4-1) x + (cb ^ 3-2bc ^ 2-a) = 0$$ from where we have
$a = cb (b ^ 2-2c)$ and $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$.

This means that for every solution of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ we have a corresponding value $a = cb (b ^ 2-2c)$.

Some solutions of $c ^ 2-3b ^ 2c + b ^ 4-1 = 0$ are $(b,c)=(1,3),(0,1),(12,55),(12,377)$.

EXAMPLES.-$(b,c)=(1,3)$ gives $a=-15$ and we have
$$x^5-x+15=(x^2+x+3)(x^3-x^2-2x+5)$$
$(b,c)=(12,377)$ gives $a=-2759640$ and we have
$$x^5-x+2759640=(x^2+12x+377)(x^3-12x^2-233x+7320)$$


问:
Understanding the divergence of an asymptotic series
In class, we recently developed an asymptotic expansion for a perturbed Gaussian integral and I was wondering why it was divergent.

First, the $e^{-gx^4}$ term was Taylor expanded, then integral and summation swapped, and finally the formula of the $n$-th momentum of a Gaussian used to obtain:
$$
\int_{-\infty}^{+\infty} e^{-x^2-gx^4} dx \rightarrow \sum_{n=0}^\infty \frac{(-g)^n}{n!}
\int_{-\infty}^{+\infty} e^{-x^2} x^{4n}=\sqrt{\pi}\sum_{n=0}^{\infty}\frac{(-g)^n\cdot(4n-1)!!}{n!\cdot2^{2n}}
$$
By the Ratio Test this diverges for all $g\neq0$ because
$\frac{a_{n+1}}{a_n}=\frac{(-g)\cdot(4n+3)(4n-1)}{(n+1)\cdot2^2}\rightarrow\infty$.

So far I had only came across a bunch of asymptotic expansions and developed the (probably unjustified) intuition that the divergence of the asymptotics came from expanding a function beyond its radius of convergence:
$$
\int_0^\infty \frac{e^{-x}}{1+x}dx=\sum_{n=0}^\infty(-1)^n\int_0^\infty e^{-x}x^ndx=\textrm{divergent}
$$

because the expansion $\frac{1}{1+x}=\sum (-x)^n$ is not justified for the integration domain $\int_1^\infty$.

However, in the cause of the perturbed Gaussian, the Taylor expansion of $e^{-gx^4}$ is well justified for $\forall x$ and for $g\ge0$ the integral converges. I was expecting the series to converge for $g\ge 0$.

Even though I am aware that limits cannot arbitrarily be interchanged ($\int\sum^\infty \neq \sum^\infty\int$), I was wondering whether the divergence of the Gaussian series would admit some "intuitive" explanation appart from that.

答:
Not sure if anybody would be interested but just in case...

TL;DR
Interchanging $\int^\infty\lim_{N\to\infty}\sum^N \Rightarrow \lim_{N\to\infty}\int^\infty\sum^N$ adds an extra term (the bump) to the integrand caused by the "explosion" of the Taylor series at large $x$. Given that the area of the bump is divergent as $N\to\infty$ the series diverges.

I would blame the Taylor expansion $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ for being not uniformly convergent. Roughly speaking, if you decide to cut off the Taylor expansion of the exponential at some order $N$, the series $1+x+\frac{x^2}{2!}+\dots$ will only approximate $e^x$ on the finite range $(-\sqrt[N]{N!\ },+\sqrt[N]{N!\ })$. Beyond this point it 'explodes' and it would go like $\sim\frac{x^N}{N!}$.

In fact, cutting the Taylor expansion $e^{-x^2}e^{-gx^4}\rightarrow e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}$ at order $N$ results in the desired function in the range $(-\left(\frac{\sqrt[N]{N!\ }}{g}\right)^{1/4},+\left(\frac{\sqrt[N]{N!\ }}{g}\right)^{1/4})$ and an extra BIG bump starting at $x=\left(\frac{\root{N}\of{N!\ }}{g}\right)^{1/4}$:

Therefore, truncating the Taylor expansion inside the integral for large $N$ gives
$$\int_0^\infty e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}=\int_0^\infty e^{-x^2}e^{-gx^4}+\textrm{\{area of the bump\}}$$
where the area of the bump is roughly $\sim \int_0^\infty e^{-x^2} \cdot\frac{(-gx^4)^{N}}{N!} =
\frac{(-g)^N}{N!} \frac{(4N-1)!!}{2^{2N}}$ which $\to\infty$ for large $N$.

Consequently the divergence is caused by
$$\text{area of }e^{-x^2-gx^4} = \int_0^\infty e^{-x^2}\lim_{N\to\infty}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}$$
$$\neq \lim_{N\to\infty}\int_0^\infty e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}\ =\ \text{area of }e^{-x^2-gx^4}\ +\ \text{\{area of the bump\}}$$


问:
Find a generating function for unlabeled graphs
Let's take a graph(undirected) of $m$ edges and $3$ vertices. Now I want to derive a polynomial for which the coefficient of $x^m$ denotes the number of unlabeled graphs of $m$ edges.

How to derive the polynomial in this case? Also is deriving a polynomial means deriving a generating sequence?

答:
Here we require the cycle index of the action on the edges of the
symmetric group  $S_3$ on the vertices. We find:

 The identity: $a_1^3.$

 Three reflections: $3 a_1 a_2.$

 Two rotations: $2 a_3.$

We get for the cycle index

$$Z(G_3) = \frac{1}{6} ( a_1^3 + 3 a_1 a_2 + 2 a_3 ).$$

Evaluate at $1+z$ for the generating function:

$$Z(G_3; 1+z) = \frac{1}{6}
((1+z)^3 + 3 (1+z) (1+z^2) + 2 (1+ z^3))
\\ = \frac{1}{6}
(1+3z+3z^2+z^3 + 3+3z^2+ 3z + 3z^3 + 2+2z^3)
=  1 + z + z^2 + z^3.$$

This means there is one graph on three vertices taking into account
ismorphisms having  zero, one, two and three edges.

The general case is described and implemented at the following MSE
link.


问:
Bayesian average PHP question
I'm implementing this bayesian average in one of my sites:

```<code>$avg_num_reviews = 18;

// average number of reviews from all company's (collect all the reviews written the last 12 months and devide them by the amount of company's active on our platform)

$avg_rating = 3.7;

// average rating of all company's (calculate a new average based on reviews that are written in the last 12 months for each company, add them together and divide them by the ammount of company's active on our platform)

$this_num_reviews = 6;

// The number of reviews written for the specific company for the last 12 months

$this_rating = 4;

// The average rating of the specific company  based on reviews written the last 12 months

<span class="math-container">$bayesian_rating = ( ($</span>avg_num_reviews * <span class="math-container">$avg_rating) + ($</span>this_num_reviews * this_rating) ) / (<span class="math-container">$avg_num_reviews + $</span>this_num_reviews);

// The formula calculating the bayesian average
</code>```

The question

If I do it like this, read the comments. When I calculate the $avg_num_reviews and the $avg_rating should I also add the company's with 0 (zero) reviews to make this bayesian average fair? Or just the company's that have reviews?

I'm trying to make a fair sorting and ranking system, just as explained here:
https://fulmicoton.com/posts/bayesian_rating/

I anyone can assist me that would be great

答:
When I calculate the $avg_num_reviews and the $avg_rating should I also add the company's with 0 (zero) reviews to make this bayesian average fair?

It makes no difference. The avg_rating is computed by summing all the individual reviews, and dividing by the total number of such reviews. The companies with zero reviews do not sum anything, in the numerator or in the denominator.

Title: Bayesian average PHP question


问:
Non trivial Automorphism
Prove that every finite group having more than two elements has a nontrivial Automorphism.

It is from Topics in Algebra by Herstein. I am not able to solve.

答:
Suppose that $G$ is abelian with exponent greater than 2 (i.e. not every element has order 2). Then mapping $g\in G$ to $g^{-1}$ is an automorphism. It's clearly bijective, and since $G$ is abelian it is indeed a homomorphism.

If $G$ is abelian with exponent 2, then it is a vector space over $\mathbb Z/2$. Choose a basis for this vector space, and pick your favourite two basis vectors to interchange. This is a nontrivial automorphism.

If $G$ is not abelian, then the center of $G$, $Z(G)\neq G$. Now, let $G$ act on itself via conjugation. These are all automorphisms. Convince yourself that if $Z(G)\neq G$, there is a nontrivial one.

Title: Non trivial Automorphism


问:
Showing Every Polygonal function is Lipschitz
There is a problem in my (N.L. Carothers') analysis textbook that asks:

Prove every polygonal function is Lipschitz. Thus, the Lipschitz functions are dense in $C[a,b]$.

Is this question supposed to say: prove every polynomial function is Lipschitz? My reason for thinking this is because polygonal functions are just polynomials with degree $\leq 1$, and therefore,  would be constant and hence Lipschitz. On the other hand, if the author meant to ask this claim for polynomials in $C[a,b] : = \{f:[a,b] \subset \mathbb{R} \to \mathbb{R}: f \ is \ continuous\}$ we would only be studying polynomials on a compact domain in $\mathbb{R}$. Which, in turn would mean the polynomials we are considering are bounded and hence continuous.

I have definitely been staring at this problem for far too long, so any insight will be useful.

答:
No, "polygonal" isn't a typo. A polygonal function on $[a, b]$ is formed by splitting $[a, b]$ into a finite set of intervals and defining an affine function on each interval, such that the affine functions "join up" at the interval boundaries. In other words, a function that looks like this:

A polygonal function is definitely not just a polynomial of degree at most one, and certainly isn't necessarily constant.

A more formal definition could be this:

A function on $[a, b]$ is polygonal iff here exists a strictly increasing finite sequence $a=x_1<...<x_n=b$ such that for all $i$, the function is affine on $[x_i, x_{i+1}]$, and the function is continuous on $[a, b]$.

To prove such a function is Lipschitz, consider the following intuitions.

Think of the function $x(t)$ as representing the position of something over time. To say $x$ is $\lambda$-Lipschitz is precisely to say that the average speed of $x$ over any time interval is at most $\lambda$ (convince yourself of this).

To say $x$ is an affine function is precisely to say the particle travels at constant speed. Apply this to show an affine function is Lipschitz (and work out the Lipschitz constant).

To say $x$ is polygonal is precisely to say that the particle starts out traveling at speed $v_1$, then changes speed instantaneously to $v_2$, and so on a finite number of times. Think about why this makes the function Lipschitz (hint: there is a maximum speed $\max v_i$ the particle is ever travelling at).

In more detail, here is one way to do it.

Let $f$ be polygonal and let $x < y$ be two points in $[a, b]$. Let $\xi_1 < ... < \xi_n$ be the points in $[a, b]$ where $f$ "changes direction" that are between $x$ and $y$. Note that $f'$ is constant on each interval $[\xi_i,\xi_{i+1}]$, say $f'=\lambda_i$ on that interval. Similarly, $f'$ is constant on $[x, \xi_1]$ and on $[\xi_n, y]$, let's say equal to $\alpha$ and $\beta$ respectively. We have:

$$f(y)=f(x)+\alpha|\xi_1-x|+\sum_{i=1}^{n-1} \lambda_i |\xi_{i+1}-\xi_i|+\beta|y-\xi_n|$$

You can use this to obtain a bound on $|f(y)-f(x)|$ using the triangular inequality which will give you that $f$ is Lipschitz with constant $\max(\alpha, \beta, \lambda_1, ..., \lambda_{n-1})$.


问:
Is there a systematic way to calculate a correlated equilibrium? (in game theory)
Consider the following game:

The pure Nash eq. are $( x_1,y_2 )$ and $( y_1,x_2 )$.

Now, in a correlated equilibria situation, you want to assign a probably of zero to playing $( y_1,y_2 )$.

So effectively choosing $p_1$ and $p_2$ in $U = 7 \cdot p_1 + 7 \cdot p_2 + 8 \cdot (1-p_1-p_2) = 8 - p_1-p_2$ to maximise $U$ but such that they are strictly positive.

But it is a constrained optimisation, but how do I work out what the constraints are?

答:
Check out this paper on Correlated Equilibrium

Correlated-Q Learning

Amy Greenwald and Keith Hall

https://www.aaai.org/Papers/ICML/2003/ICML03-034.pdf


问:
Left (resp. right) adjoint functor fully faithful iff unit (resp. counit) isomorphism
Let $\mathcal{C}$ and $\mathcal{D}$ be categories and $\mathcal{F}:\mathcal{C}\longleftrightarrow\mathcal{D}:G$ functors such that $\mathcal{G}$ is right-adjoint to $\mathcal{F}$, ie. we have a natural isomorphism $\text{Hom}_{\mathcal{D}}(\mathcal{F}(X),Y)\cong \text{Hom}_{\mathcal{C}}(X,\mathcal{G}(Y))$ for all objects $X$ in $\mathcal{C}$ and $Y$ in $\mathcal{D}$. I have proven that right-adjoint functors are unique up to natural transformation and I have constructed the unit $\eta:\text{Id}_{\mathcal{C}}\longrightarrow G\circ F$ as well as the counit $\varepsilon:F\circ G\longrightarrow\text{Id}_{\mathcal{D}}$. But now I am stuck: How can I show that $\mathcal{F}$ (resp. $\mathcal{G}$) is fully faithful if and only if $\eta$ (resp. $\varepsilon$) is an isomorphism?

答:
For any pair of objects $X,X' \in \mathcal C$, we can consider the induced map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$. Via naturality, we know that this composite map is actually equal to $f \mapsto \eta_{X'} \circ f$. This is just $\mathrm{Hom}_{\mathcal C}(-,\eta_{X'})$. From the Yoneda lemma, we know that this is a bijection for all $X$ iff $\eta_{X'}$ is an isomorphism.

On the other hand, $\mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$ is always a bijection, since we have an adjunction, so that the composite map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X')) \to \mathrm{Hom}_{\mathcal C}(X,GF(X'))$ is a bijection iff the map $\mathrm{Hom}_{\mathcal C}(X,X') \to \mathrm{Hom}_{\mathcal D}(\mathcal F(X),\mathcal F(X'))$ is a bijection. This is precisely the condition that $\mathcal F$ is fully faithful.

The statement for $\varepsilon$ is dual to the one about $\eta$.


问:
Find all triples of positive integers (x, y, z)

I've no idea how to attempt such a question. I tried eliminating z from the equations to get a quadratic equation in terms of x, so I could use the fact that the discriminant would be a perfect square. Please help

答:
You have the $2$ equations

$$x^2 + y - z = 100 \tag{1}\label{eq1A}$$

$$x + y^2 - z = 124 \tag{2}\label{eq2A}$$

Next, \eqref{eq2A} minus \eqref{eq1A} gives

$$\begin{equation}\begin{aligned}
x - x^2 + y^2 - y & = 24 \\
-4x^2 + 4x + 4y^2 - 4y & = 96 \\
-(4x^2 - 4x + 1) + (4y^2 - 4y + 1) & = 96 \\
(2y - 1)^2 - (2x - 1)^2 & = 96 \\
(2y - 2x)(2y + 2x - 2) & = 96 \\
(y - x)(y + x - 1) & = 24
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Also, as Neat Math's comment states, it would have been simpler & easier to use $x - y + y^2 - x^2 = -(y - x) + (y - x)(y + x) = (y - x)(y + x - 1)$.

With $x$ and $y$ being positive integers means $y + x - 1 \gt 0$, so $y - x \gt 0$. Next, note the parity of $y - x$ and $y + x - 1$ are different, i.e., one is even and the other odd. The only odd & even set of divisors of $24$ are $(1, 24)$ and $(3, 8)$. Since $y + x - 1 \gt y - x$, this means we have first the possibility of

$$y - x = 1 \tag{4}\label{eq4A}$$

$$y + x - 1 = 24 \implies y + x = 25 \tag{5}\label{eq5A}$$

Adding \eqref{eq4A} and \eqref{eq5A} gives $2y = 26 \implies y = 13$, so $x = 12$, with \eqref{eq1A} and \eqref{eq2A} giving $z = 57$. The second possibility gives

$$y - x = 3 \tag{6}\label{eq6A}$$

$$y + x - 1 = 8 \implies y + x = 9 \tag{7}\label{eq7A}$$

Adding \eqref{eq4A} and \eqref{eq5A} gives $2y = 12 \implies y = 6$, so $x = 3$, with \eqref{eq1A} and \eqref{eq2A} giving $z = -85$. However, since this is not a positive integer, it's not allowed.

Thus, the only triple of positive integers solution is $(x, y, z) = (12, 13, 57)$.


问:
System of 4 linear equations with 4 variables and two parameters
I am trying to solve this system of equations with two parameters (through Gaussian elimination) and I have almost reached a solution but the numbers I get are terrible so I'd like to ask if I am on the right path to solving the problem. Unfortunately, my LaTex code in Word doesn't work if I paste it here, so I will share my solution through the photo below (please excuse me for the inconvenience:

答:
If $\mu\ne 1$ the system has one unique solution, for Cramer's theorem
$$\begin{cases}
x_1=\frac{\lambda  \mu +2 \lambda +5 \mu -8}{\mu -1}\\
x_2=\frac{9 \lambda  \mu +8 \lambda +38 \mu -55}{\mu -1}\\
x_3=\frac{\lambda-1 }{\mu -1}\\
x_4=-\frac{4 \lambda  \mu +4 \lambda +17 \mu -25}{\mu -1}\\
\end{cases}$$
if $\mu=1$ and $\lambda\ne 1$  the rank of the matrix of the coefficients is $3$ and the rank of the augmented matrix is $4$, therefore for Rouchè-Capelli theorem , as they are different, the system has no solutions.

If $\mu=1$ and $\lambda= 1$ the ranks are the same and are $3$ so we have infinite solutions depending on $4-3=1$ parameter.

$$
\begin{cases}
x_1= \frac{3}{8} \left(x_4+5\right)\\
x_2= \frac{1}{8} \left(19-17 x_4\right)\\
x_3= -\frac{1}{8} \left(x_4+21\right)\\
\end{cases}
$$


问:
Check the conditions of Lax-Milgram Lemma - proving coercivity.
Check the conditions of the Lax-Milgram Lemma for the problem
$$a(u,v) = \left< F,v \right>, v \in H_0^1(\Omega)$$
where $\Omega$ is bounded and
$$a(u,v) = \int_\Omega \left[ \sum_{i,j=1}^d (i+j)^2 \frac{\partial u}{\partial x_j} \frac{\partial v}{\partial x_i} + (1+x^2)uv \right] ~dx$$
$$\left< F,v \right> = \int_\Omega 2v ~dx$$
What I've managed to do:

$a$ is obviously bilinear.

Functional $F$ is bounded:
$$| \left< F,v \right>| \leq 2 \int_\Omega |v| dx \leq 2 |\Omega| \left( \int_\Omega |v|^2 ~dx \right)^{1/2} \leq C \| v \|_{H^1(\Omega)} $$

$a$ is continuous:
$$|a(u,v)| \leq \sum_{i,j=1}^d \int_\Omega (i+j)^2 \left| \frac{\partial u}{\partial x_j} \right| \left| \frac{\partial v}{\partial x_i} \right| + (1+x^2) |u| |v| ~dx \\ \leq d^2 n \cdot d^2 |u|_{H^1{\Omega}} |v|_{H^1(\Omega)} + (1+M^2) \| u \|_{L^2(\Omega)} \| v \|_{L^2(\Omega)} \\ \leq max\{ n d^4, 1+M^2 \} \left[ |u|_{H^1(\Omega)} |v|_{H^1(\Omega)} + \| u \|_{L^2} \| v \|_{L^2} \right] \\ \leq D \cdot \| u \|_{H^1} \| v \|_{H^1}$$
where $M$ is such that $1+x^2 \leq 1+M^2, \forall x \in \Omega.$

What I can't prove is coercivity of $a.$ I don't even know if it holds.

Any hints would be appreciated!

答:
Let $A$ be the leading matrix. For simplicity, we assume that $A$ is diagonal and $A_{11} < 0$ (this can be obtained after some rotation). Moreover, we assume that $0$ belongs to the interior of the domain $\Omega$.

Now, consider some $\psi \in C_c^\infty(\Omega)$, $\psi \ne 0$, such that the support of $\psi$ is a ball centered at $0$ (again, for simplicity).
We define
$$
\psi_n(x) := \psi(n \, x_1, x_2, \ldots, x_d).$$
Then, one can check
$$
\int_\Omega \frac{\partial^2}{\partial x_1^2} \psi_n(x) \, \mathrm dx
=
\int_\Omega n^2 \, \frac{\partial^2}{\partial x_1^2} \psi(n\,x_1,x_2,\ldots) \, \mathrm dx
=
n \, \int_\Omega \frac{\partial^2}{\partial x_1^2} \, \psi(x) \, \mathrm dx
=: n \, c_1.
$$
Similarly,
$$
\int_\Omega \frac{\partial^2}{\partial x_i^2} \psi_n(x) \, \mathrm dx
=
\frac1n \, \int_\Omega \frac{\partial^2}{\partial x_i^2} \psi(x) \, \mathrm dx
=: \frac1n \, c_i
$$
for $i > 1$.

Thus,
$$
\int_\Omega \nabla \psi_n^\top \, A \, \nabla\psi_n \, \mathrm dx
=
n \, c_1 \, A_{11} + \frac1n \, \sum_{i=2}^d c_i \, A_{ii}
$$
and this will be negative for $n$ large enough.


问:
injective immersions
if my function is an immersion and say it is defined on a path connected open of an euclidian space.

immersions are locally injective. if we add the path connectedness could I assert that my immersion is injective? If not I would be grateful to you for giving me a counterexample.

答: $\mathbb{R}$ is path connected, and $f:\mathbb{R}\rightarrow\mathbb{R}^2$ defined by $f(t)=(\cos(t),\sin(t))$ is an immersion because $$d_tf=[-\sin(t)\quad\cos(t)]$$ is never of rank 0, but it is not injective.</p>

Title: injective immersions


问:
Density of the image and closedness of the inverse of a bounded linear operator
Let $A \colon X \to X$ be a bounded linear operator, where $X$ is a Banach space.

$(Q1)$ Is it true that if $A$ is injective then the image of $A$ is dense in $X$?

$(Q2)$ Is it true that $A^{-1} \colon \text{rgA} \to X$ is closed?

My attempts of solution:

$(A1)$ I have no idea.

$(A2)$ I know that if $A$ is bounded then it is closed, i.e. if $$x_n \to x,\ \text{and}\ Ax_n \to y,\ \Longrightarrow x \in D(A)( = X)\ \text{and}\ Ax = y.$$
Let $y_n \to y$ and $A^{-1}y_n \to w$. $(*)$

By definition, there exist $\{x_n\}$ such that $Ax_n = y_n$ substituting this into $(*)$ we get $$Ax_n \to y,\ \text{and}\ A^{-1}Ax_n = x_n \to w.$$

By closedness of $A$ this implies that $w \in X$ (obviously) and that $Aw = y$, i.e. $w = A^{-1}y$.

Do you have any idea for $(Q1)$? is $(A2)$ correct?

Thank you very much for your time and help!!!

答:
(A2) is correct. Note that it assumes that $A$ be injective. And that $X$ be a normed vector space suffices.

For (Q1), a natural family of counterexamples is given by non surjective isometries on a Banach space, since their ranges are closed proper subspaces. For instance, the unilateral shift operator $S:\ell^2(\mathbb{N})\longrightarrow \ell^2(\mathbb{N})$
$$
S:(x_0,x_1,x_2,\ldots)\longmapsto (0,x_0,x_1,x_2,\ldots).
$$


问:
How to complete semi-lattice via Adjunctions?
I am thinking about free/complete lattice constructions from sets, posets and semilattices. For example, given a set $A$ we can construct (at least) two canonical posets from it as one being $(2^A,\subseteq)$ under inclusion order and the other being the trivial order (that is, $a\le b$ in $A$ iff $a=b$). I guess one can construct them as adjunctions to the forgetful functor $$\operatorname{Poset}\rightarrow\operatorname{Set}.$$ However I do not know how to obtain these left and adjoints. I have heard that one can write them as (co)limits, but never learn (or found a reference that explain in a way that I can understand) how to do it. So my first question is, can somebody explain me how to write explicit formulas for (possible) adjoints to this functor?

Next, I want to do the same thing for each forgetful functor in the chain
$$\operatorname{CompletelyDistributiveLattice}\rightarrow\operatorname{CompleteLattice}\rightarrow\operatorname{CompleteSemiLattice}\rightarrow\operatorname{SemiLattice}\rightarrow\operatorname{Poset}$$
Specially I was thinking about constructing complete semi-lattices from semi-lattices without hurting to existing (finite) joins. This is clearly related to the first question, since once I learn to obtain general left adjoints explicitly it become trivial.

答:
The left adjoint of the forgetful functor from posets to sets sends a set to the discrete poset on that underlying set, but there is no right adjoint; it's easy to check that if such a right adjoint $R$ existed then its value $R(S)$ on any set $S$ would have $s_1\le s_2$ and $s_2\le s_1$ for every $s_1,s_2$, so wouldn't be a poset at all.

For your problem of free complete semilattices, is a good idea to first notice that the left adjoint of the forgetful functor from complete semilattices to sets does send a set to its powerset, and a function to the associated direct image morphism of powersets. We can then readily solve the problem of constructing the free complete semilattice on a semilattice $S$, at least formally, by subjecting the powerset complete semilattice $\mathcal P(S)$ to the relations asserting that finitary joins in $\mathcal P(S)$ must be made to agree with their definition in $S$. That is, we quotient by the equivalence relation $\sim$ on $\mathcal P(S)$ that says a subset $\{a,b,\cdots\}$ is equivalent to the subset $\{a\vee b,\cdots\}$, with $\vee$ the supremum from $S$. Then complete semilattice maps out of $\mathcal P(S)/\sim$ are naturally identified with complete semilattice maps out of $\mathcal P(S)$ which respect $\sim$, or in other words which restrict to a semilattice homomorphism on $S$. Since complete semilattice maps out of $\mathcal P(S)$ in general are naturally identified with set functions out of $S$, this shows $\mathcal P(S)/\sim$ has the universal property of the free complete semilattice on $S$.

You can handle the free semilattice on a poset $T$ similarly: construct the free semilattice on $T$'s underlying set, namely the finite powerset of $T$, then quotient by the relations asserting that $\{a,b\}\sim b$ if $a\le b$ in $T$.

Interestingly, the forgetful functor from complete lattices to complete semilattices is a rare example of a functor between complete categories which preserves all limits but admits no left adjoint. The problem is that the free complete lattice on, say, the powerset of a three-element set $\{a,b,c\}$, forms a proper class. There is a construction of a proper class of distinct elements $p_\alpha$ given inductively by $p_{\alpha+1}=a\vee b\wedge c\vee a\wedge b\vee c\wedge p_\alpha$ (identifying $a$ with $\{a\}$ and associating, say, left to right) for successor ordinals and $p_\beta=\sup_{\alpha<\beta} p_\alpha$ for limit ordinals.

To summarize, sometimes showing a left adjoint exists is a bit easier than giving an explicit construction, but if it's an explicit construction you want then you need to rely on special aspects of your situation. There is a rather general answer for algebraic categories, like lattices, groups, rings, and complete lattices (defined by sets with total operations), which says that a left adjoint to a forgetful functor to sets takes $S$ to the object formed of all "words" constructed of elements of $S$ out of the operations of the algebraic category--at least if the class of such words is a set! And left adjoints between two such algebraic categories can be constructed out of the free construction on a set by imposing appropriate relations.


问:
The map $\frac1{1-z}$
I want to find the image of the unit circle under the map $\frac1{1-z}$. I first rewrote this as
$$
\frac{1+z}{1-z^2}
$$
But on the unit circle, $z^2 = 1$. So I get $\infty$ for the image of all points on the unit circle as the result. But I thought linear fractional always map circles to lines or circles. Can anyone point out what I did wrong?

EDIT: It turns out that I was confusing $z^2$ with $\bar{z}$, so not all points on the unit circle satisfy $z^2 = 1$. I should let $z = x + iy$ and plug that into the map instead.

答:
The easiest way to find the images of circles and lines (generlized cirlces) under linear fractional transformations (or Möbius transformations) is to simply look at three points on the preimage. Three points on the unit circle is 1, $i$ and -1. Now let

$$f(z) = \frac{1}{1-z}$$

and observe that

$$f(1) = \infty, \quad f(i) = \frac{1+i}{2}, \quad f(-1)=\frac{1}{2}.$$

This shows that the image of the unit circle under $f$ is the straight line through $\frac{1}{2}(1+i)$ and $\frac{1}{2}$, i.e. $\{ z \in \mathbb{C} \, | \, \Re(z) = \frac{1}{2} \}$.

Title: The map $\frac1{1-z}$


问:
Help to understand the proof about injective immersions
I'm currently working on the proof of the following theorem in the book of "Bröcker":

Let $M^m$ be a differential manifold and $f:M^m \to \Bbb R^n$ be a differentiable map and $2m < n$. Let $A \subset M$ be closed and let the restriction of $f$ to a neighbourhood $U$ of $A$ be an injective immersion. Then arbitrarily close of $f$ there exists an injective immersion $g: M^m \to \Bbb R^n$, s.t. $g \vert A = f \vert A$.

The proof is given as follow:

I don't understand here why, we have that $g_v$ are still immersions?

Also later, he used the Sard theorem to implies that the image of this map has measure zero. So we may choose $b_v$ not to be in this image.. I don't understand how he used here the theorem and how he can conclude that.

EDIT:

Here is the Lemma 7.6 but I don't see the relation.

Many thanks for some help.

答:
First, if $g_\nu$ has rank $m=\dim M$, then that means the map is an immersion.

Second, a set of measure $0$ has empty interior, and so the complement is dense. It certain follows that the complement is non-empty.


问:
Calculating Flux across a spherical surface in a vector field (Newtonian Gravity)
The question I am working on gives the Poisson Equation for Newtonian Gravity  $∇⃗ ∙ ⃗(⃗) = −4(⃗) $ with the mass density function $(⃗) =0/(1 + (⁄0)^2)$ and asks for the flux of the gravitational acceleration across a spherical surface with radius $r$.

Using the divergence theorem, so far I have:

$ \oint \oint  \overrightarrow{g} (\overrightarrow{r}) \cdot ds =  \iiint \overrightarrow{\nabla} \cdot \overrightarrow{g} (\overrightarrow{r}) dV$

$= -4\pi G \rho 0 \iiint 1/(1+(r/r0)^2) dV$

However I don't know how to solve the integral I have arrived at. I thought about changing it to spherical coordinates but am not sure if that would help?

答:
"Slicing" along the z-axis yields circles of radius $\sqrt{r^2-z^2}$, so parameterizing in terms of $r$ and $\theta$:
$$
\int_0^{\sqrt{r^2-z^2}}\int_0^{2\pi}\frac{r}{1+(\frac{r}{r_0})^2}d\theta dr
$$
So the entire integral should be:
$$
\iiint\nabla\cdot\vec{g}(\vec{r})dV=-4\pi G\rho_0\int_{-r}^r\int_0^{\sqrt{r^2-z^2}}\int_0^{2\pi}\frac{r}{1+(\frac{r}{r_0})^2}d\theta dr dz
$$
Which can be solved with integration by parts and partial fractions.


问:
How to prove the converge of a sequence?
Let $t_{n} = \int_{1}^{n} (\ln(x)^p) \,dx$, where $p < 0$ is any real number.

How can I prove that this sequence converges?

My attempt: Clearly, $t_{n} \geq 0$, so by the linear property of Riemann integrals, $t_{n}$ is increasing. Then, if $t_{n}$ is bounded, it is also convergent. Can you find a bound?

答:
An expansion on the discussion in the comments: For any $p<0$, we show that there exists some constant $c$ such that $\ln^p(x)>\frac{c}{x}$. We begin see that for any $N$ we have that

$$e^t=\sum_{n=1}^{\infty}\frac{t^n}{n!}>\frac{t^N}{N!}$$

changing variables $x=e^t$ we get that

$$x>\frac{\ln^N(x)}{N!}$$

Thus,

$$\ln^{-N}(x)>\frac{1}{N!x}$$

and so the choice $c=[-p]!^{-1}$ where $[\cdot]$ denotes the least integer above its argument is a satisfactory choice.


问:
Why is $x = c$ an inflection point only if the lowest-order (above the second) non-zero derivative is of odd order (third, fifth, etc.).
So according to wikipedia inflection point,

If the second derivative, $f''(x)$ exists at $x_0$, and $x_0$ is an inflection point for $f$, then $f''(x_0) = 0$, but this condition is not sufficient for having a point of inflection, even if derivatives of any order exist. In this case, one also needs the lowest-order (above the second) non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection, but an undulation point.

It is also mentioned that:

the condition that the first nonzero derivative has an odd order implies that the sign of $f'(x)$ is the same on either side of $x$ in a neighborhood of $x$

My question is why is it the case? For example, if $f^{3}(c)$ is nonzero, while $f''(c) = 0$.  How does the first condition contribute to the fact that $(x, f(x))$ is not an undulation point?.  I found this non-obvious but I couldn't find any explanation online. So I am seeking a proof (or any kind of intuition) for this. Thanks in advance.

答:
Assume that $f$ is $C^\infty$ in a neighborhood of $x=0$. Then the  the tangent to the graph of $f$ at $x=0$ is given by $y=t(x):=f(0)+f'(0) x$.
We now have to study the function $g(x):=f(x)-t(x)$ in the neighborhood of $x=0$. In any case $g(0)=g'(0)=0$.  When all derivative values $f^{(n)}(0)=0$ $\>(n\geq2)$ these derivative values cannot tell us anything about the behavior of $g$ near $x=0$, since $g$ has all derivatives $=0$ at $x=0$.

Now let us assume that there is some $n\geq2$ with
$$f^{(k)}(0)=0\quad (2\leq k<n),\qquad f^{(n)}(0)=:c\ne 0\ .$$
Then Taylor's theorem tells us that
$$g(x)=c{x^n\over n!}+o(x^n)=x^n\left({c\over n!}+o(1)\right)\qquad(x\to0)\ .$$
This says that $g(x)$ is of a single sign in a punctured neighborhood of $x=0$, when $n$ is even, and $g$ (having the $x$-axis as tangent at $x=0$) changes sign when $x$ passes through $0$, when $n$ is odd. The latter behavior you would describe as an inflection point of $f$ at $x=0$.