问:
A caboodle of Pell's equation in one? $x^2+y^2-5xy+5=0$
I saw this twitter post that reads:

Find all the pairs of positive integers $(x,y)$ satisfying $$  x^2 + y^2 - 5xy + 5 = 0 . $$

I don't know how to tackle this and I ended up summoning WolframAlpha which shows that there are infinitely many solutions. What's interesting is that there are (at least?) 12 general forms that looks like they all resemble the solutions to Pell's equation.

For example, the first general solution presented reads:

$$ \begin{array} { r c l }
 x &=&
\dfrac1{42} \Big [ 21 \left(55 - 12\sqrt{21} \right)^n
- \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 21 \left(55 + 12\sqrt{21} \right)^n
+ \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big ] \\ \phantom0 \\
 y &=&
\dfrac1{42} \Big [ 63 \left(55 - 12\sqrt{21} \right)^n
- 13 \sqrt{21} \left(55 - 12\sqrt{21} \right)^n
+ 63 \left(55 + 12\sqrt{21} \right)^n
+ 13 \sqrt{21} \left(55 + 12\sqrt{21} \right)^n \Big] \end{array} $$ for $n = 0,1,2,3\ldots $.

Plugging $n=0,1,2,3$ gives the first few solutions

$$(x,y) = (1,3), (67,321), (7369,35307), (810523,3883449)
$$

I tried converting these solutions to the form of $X^2 - DY^2 = A$ or something similar to Pell's equation but I got nothing.

Is it a coincidence that this innocuous-looking quadratic Diophantine equation is actually a caboodle of Pell's equation in disguise? If so, how can we derive them all?

答:
The equation is symmetric and
it is easy to see solutions if, for example, we solve for $y$.

$$x^2 + y^2 - 5xy + 5 = 0 \implies\quad
y =  \frac{5 x \pm \sqrt{21 x^2 - 20}}{2}\qquad |x|\ge 1$$

Note that the absolute value of $x$ must be at least $1$ for the radical to be non-negative and therfore, for $y$ to be real.

Given this equation, we can also see that there are $2$ $y$-values for every valid $x$. There are $28$ solutions for
$\space -50000\le x \le 50000.\quad$ Here is that "sample" of $\quad (x,y_1,y_2.)\quad$

$$
(-7369,-35307,-1538)\quad
(-4729,-22658,-987)\quad
(-1538,-7369,-321)\\
(-987,-4729,-206)\qquad
(-321,-1538,-67)\qquad
(-206,-987,-43)\\
(-67,-321,-14)\qquad
(-43,-206,-9)\qquad
(-14,-67,-3)\\
(-9,-43,-2)\quad
(-3,-14,-1)\quad
(-2,-9,-1)\quad
(-1,-3,-2)\\
(1,2,3)\qquad
(2,1,9)\qquad
(3,1,14)\qquad
(9,2,43)\qquad
(14,3,67)\\
(43,9,206)\qquad
(67,14,321)\qquad
(206,43,987)\qquad
(321,67,1538)\\
(987,206,4729)\qquad
(1538,321,7369)\qquad
(4729,987,22658)\\
(7369,1538,35307)\quad
(22658,4729,108561)\quad
(35307,7369,169166)$$

Note that all the negative $x$-values have positive counterparts and that we have a counterpart solution for $x$.

$$x =  \frac{5 y \pm \sqrt{21 y^2 - 20}}{2}$$
so the $x,y$ values should be interchangeable.


问:
Constant speed parametrization
I am studying the book "Riemannian Manifolds: An Introduction to Curvature" by Lee. The following statement is from that book.

Let $S^n$ be the sphere of radius $1.$ The Geodesics of $S^n$ are precisely the great circles with constant speed parametrization.

I want to know what he means by ``constant speed parametrization" ad how to obtain such a parametrization when one is already given with a parametrization.

答:
If $\gamma:I\to R^n$ is a curve, then its derivative $\gamma':I\to R^n$ is the velocity vector and its modulus is the so-called speed. So the parameterization $\gamma$ has constant speed if $|\gamma'(t)|=v$ for some positive constant $v>0$.

Any regular curve can be reparameterized to have unit speed. It does not make sense to write down the details here as this is something that you can find in the first chapter of every book on classical differential geometry. For example, in Do Carmo book.

ContentLicense: CC BY-SA 4.0

Title: Constant speed parametrization


问:
Double refinement of Nesbitt's inequality
It's a double refinement of Nesbitt's inequality :

Claim :Prove or disprove this statement.

Let $a,b,c>0$ such that $1\leq c\leq b\leq 2$ then we have :

$$\frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+  \frac{(a-b)^2}{ab+bc+ca}} $$

As in my refinement here I have tried to use convexity and chords and it seems to works .Next again as in my link I use repeatedly Jensen's inequality . So i can prove it but it's not a nice method !

So have you better ?

Source :
Stronger than Nesbitt's inequality $\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$

Stronger than Nesbitt inequality

Conjecture about :$\frac{a}{\sqrt[4]{8(b^4+c^4)}}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac94+\frac32 \cdot \frac{(a-b)^2}{ab+bc+ca}}$

答:
Define a (real valued) function $F(a,b,c)$ and re-define the inequality as follows:
$$
F(a,b,c) = \frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b} - \sqrt{\frac94+\frac{(a-b)^2}{ab+bc+ca}} \geq 0
$$
Define new variables $x = a/c$ and $y = b/c$.
Due to the restrictions for $a,b,c$, and scaling properties of $F$, the problem can be rewritten as a two-dimensional one:
$$
f(x,y) = \frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \geq 0
$$
With $1 \leq y \leq 2$ and $x \gt 0$.
So the problem boils down to an investigation of function behaviour in 2-D.

To get a first impression, we start with a pictorial exercise: contouring of the function.
After some attempts, it seems that the following window may be interesting:

```
  xmin := 0; xmax := 2;
  ymin := 1; ymax := 2;
```

Maximum and minimum vales of $f(x,y)$ are determined first and the place of the minimum is estimated numerically:
```
 1.39061000309967E-0007 < f < 4.38447187191170E-0001
f( 1.00100100100100E+0000, 1.00200803212851E+0000) =  1.39061000309967E-0007
```
There are $75$ (vertically) equidistant isolines between the minimum and the maximum. The picture is colored blue for funcion values
$f(x,y) < 0.001$.
It is suggested - not formally proved - that maybe a global minimum equal to zero is found for $x$ and $y$ equal to one: $f(1,1)=0$.
Let's check this. A computer algerba system (MAPLE) has been employed when calculations became tedious (and essentially uninteresting as well):
```
> f(x,y) := x/((4*(y^3+1))^(1/3))+y/(x+1)+1/(x+y)-sqrt(9/4+(x-y)^2/(x*y+x+y));
> simplify(subs(x=1,y=1,f(x,y)));
                                       0
```
Meaning that the exact value of $f(1,1)$ is zero, as expected.

Next calculate the partial derivatives $\;\large\frac{\partial f}{\partial x}\small (1,1)\;$ and
$\;\large\frac{\partial f}{\partial y}\small (1,1)\;$. Verify that they are zero:
```
> simplify(subs(x=1,y=1,diff(f(x,y),x)));
                                               0
> simplify(subs(x=1,y=1,diff(f(x,y),x)));
                                               0
```
Calculate the second order partial derivatives $\;\large\frac{\partial^2 f}{\partial x^2}\small(1,1)\;$ ,
$\;\large\frac{\partial^2 f}{\partial y^2}\small(1,1)\;$ , $\;\large\frac{\partial^2 f}{\partial x \partial y}\small(1,1)\;$:
```
> simplify(subs(x=1,y=1,diff(diff(f(x,y),x),x)));
                                                      5/18
> simplify(subs(x=1,y=1,diff(diff(f(x,y),y),y)));
                                                      1/36
> simplify(subs(x=1,y=1,diff(diff(f(x,y),x),y)));
> simplify(subs(x=1,y=1,diff(diff(f(x,y),y),x)));
                                                    - 1/36
```
So the Hessian matrix of our function at $(1,1)$ is:
$$
\begin{bmatrix} +5/18 & -1/36 \\ -1/36 & +1/36 \end{bmatrix}
$$
The Hessian is positive definite here, because of its eigenvalues $\lambda_1,\lambda_2$:
$$
\left.\begin{matrix} \mbox{Trace} = \lambda_1 + \lambda_2 &=& 5/18 + 1/36 = 11/36 > 0 \\
\mbox{Determinant} = \lambda_1 . \lambda_2 &=& 5/18.1/36 - (1/36)^2 = 9/36^2 > 0 \end{matrix}\right\}
\Longrightarrow \lambda_1 > 0 \; , \; \lambda_2 > 0
$$
This ensures that we have found indeed a (local) minimum.

Inspection at the left side of the picture - i.e. for $\;x \downarrow 0\;$ - shows that there is no reason to assume that
function values $f(x,y)$ become somehow close to zero there. But it's not forbidden to investigate e.g. the borderline case:
$$
\lim_{x\to 0} \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\} \\
= y + \frac{1}{y} - \sqrt{\frac{9}{4} + y}
$$
In view of this, define the function $\color{red}{f(x) = x + 1/x - \sqrt{9/4 + x}}$.

There is no such borderline case at the right side of the picture, because the $x$-coordinate has no upper bound. So there might
be uncertainty for larger values of $x$, formally $\; x \to \infty\;$. To gain more insight, we calculate the following limit,
because it is suspected that the first term is dominant:
$$
\lim_{x\to\infty} \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\}\large / x \\
= \lim_{x\to\infty} \left\{
\frac{1}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x(x+1)}+\frac{1}{x(x+y)}-\sqrt{\frac{9}{4x^2}+\frac{(1-y/x)^2}{xy+x+y}} \right\} \\
= \frac{1}{\left[(4(y^3+1)\right]^{1/3}}
$$
In view of this, define the function $\color{blue}{g(x) = 1/\left[(4(x^3+1)\right]^{1/3}}$.

Both functions are displayed in the picture below, for $0 < x < 2$ and $0 < y < 5$, to ensure that both are positive valued
(which by the way is trivial for the blue one). It is noticed that the red function has a minimum above zero.
Of course $\color{red}{f(x)}$ and $\color{blue}{g(x)}$ can be investigated analytically as well, but without a computer algebra
system it quickly becomes complicated and hence error prone. What's worse: there is hardly any additional value.
That's why I prefer the graphics:

For $x\to\infty$ it is concluded that there are no values near zero in that far away region.
But where does the far away region start? To settle this issue, consider the expression:
$$
\delta(x,y) = 1 - \frac{\left[(4(y^3+1)\right]^{1/3}}{x} \times \\ \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\}
$$
Or, alternatively (approximately the same for large $x$-values) :
$$
\delta(x,y) = \ln\left(\frac{x}{\left[(4(y^3+1)\right]^{1/3}}\right) - \\
\ln\left(\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}}\right)
$$
with the interpretation attached that $\delta$ is a relative error of the limiting value $x/\left[(4(y^3+1)\right]^{1/3}$.
Then perform numerical experiments for $1 \leq y \leq 2$ (as usual) and for $x = 10$, giving: $\delta \lt 90\,\%\;$. Ongoing from there,
$\delta$ is continuously decreasing with increasing $x$, as expected. There is enough margin in the blue $\color{blue}{g(x)}$
to ensure that any $\delta(x,y)$ for $x \gt 10$ do not make it close to zero. So we only have to investigate function behaviour for
say $0 \lt x \leq 10$, a limited range of $(x,y)$ values.

Having done all this, we must conclude that the Double refinement of Nesbitt's inequality, as proposed by the OP, is valid.

Body:
Define a (real valued) function $F(a,b,c)$ and re-define the inequality as follows:
$$
F(a,b,c) = \frac{a}{\sqrt[3]{4(b^3+c^3)}}+\frac{b}{a+c}+\frac{c}{a+b} - \sqrt{\frac94+\frac{(a-b)^2}{ab+bc+ca}} \geq 0
$$
Define new variables $x = a/c$ and $y = b/c$.
Due to the restrictions for $a,b,c$, and scaling properties of $F$, the problem can be rewritten as a two-dimensional one:
$$
f(x,y) = \frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \geq 0
$$
With $1 \leq y \leq 2$ and $x \gt 0$.
So the problem boils down to an investigation of function behaviour in 2-D.

To get a first impression, we start with a pictorial exercise: contouring of the function.
After some attempts, it seems that the following window may be interesting:

  xmin := 0; xmax := 2;
  ymin := 1; ymax := 2;

Maximum and minimum vales of $f(x,y)$ are determined first and the place of the minimum is estimated numerically:

 1.39061000309967E-0007 < f < 4.38447187191170E-0001
f( 1.00100100100100E+0000, 1.00200803212851E+0000) =  1.39061000309967E-0007

There are $75$ (vertically) equidistant isolines between the minimum and the maximum. The picture is colored blue for funcion values
$f(x,y) < 0.001$.
It is suggested - not formally proved - that maybe a global minimum equal to zero is found for $x$ and $y$ equal to one: $f(1,1)=0$.
Let's check this. A computer algerba system (MAPLE) has been employed when calculations became tedious (and essentially uninteresting as well):

> f(x,y) := x/((4*(y^3+1))^(1/3))+y/(x+1)+1/(x+y)-sqrt(9/4+(x-y)^2/(x*y+x+y));
> simplify(subs(x=1,y=1,f(x,y)));
                                       0

Meaning that the exact value of $f(1,1)$ is zero, as expected.

Next calculate the partial derivatives $\;\large\frac{\partial f}{\partial x}\small (1,1)\;$ and
$\;\large\frac{\partial f}{\partial y}\small (1,1)\;$. Verify that they are zero:

> simplify(subs(x=1,y=1,diff(f(x,y),x)));
                                               0
> simplify(subs(x=1,y=1,diff(f(x,y),x)));
                                               0

Calculate the second order partial derivatives $\;\large\frac{\partial^2 f}{\partial x^2}\small(1,1)\;$ ,
$\;\large\frac{\partial^2 f}{\partial y^2}\small(1,1)\;$ , $\;\large\frac{\partial^2 f}{\partial x \partial y}\small(1,1)\;$:

> simplify(subs(x=1,y=1,diff(diff(f(x,y),x),x)));
                                                      5/18
> simplify(subs(x=1,y=1,diff(diff(f(x,y),y),y)));
                                                      1/36
> simplify(subs(x=1,y=1,diff(diff(f(x,y),x),y)));
> simplify(subs(x=1,y=1,diff(diff(f(x,y),y),x)));
                                                    - 1/36

So the Hessian matrix of our function at $(1,1)$ is:
$$
\begin{bmatrix} +5/18 & -1/36 \\ -1/36 & +1/36 \end{bmatrix}
$$
The Hessian is positive definite here, because of its eigenvalues $\lambda_1,\lambda_2$:
$$
\left.\begin{matrix} \mbox{Trace} = \lambda_1 + \lambda_2 &=& 5/18 + 1/36 = 11/36 > 0 \\
\mbox{Determinant} = \lambda_1 . \lambda_2 &=& 5/18.1/36 - (1/36)^2 = 9/36^2 > 0 \end{matrix}\right\}
\Longrightarrow \lambda_1 > 0 \; , \; \lambda_2 > 0
$$
This ensures that we have found indeed a (local) minimum.

Inspection at the left side of the picture - i.e. for $\;x \downarrow 0\;$ - shows that there is no reason to assume that
function values $f(x,y)$ become somehow close to zero there. But it's not forbidden to investigate e.g. the borderline case:
$$
\lim_{x\to 0} \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\} \\
= y + \frac{1}{y} - \sqrt{\frac{9}{4} + y}
$$
In view of this, define the function $\color{red}{f(x) = x + 1/x - \sqrt{9/4 + x}}$.

There is no such borderline case at the right side of the picture, because the $x$-coordinate has no upper bound. So there might
be uncertainty for larger values of $x$, formally $\; x \to \infty\;$. To gain more insight, we calculate the following limit,
because it is suspected that the first term is dominant:
$$
\lim_{x\to\infty} \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\}\large / x \\
= \lim_{x\to\infty} \left\{
\frac{1}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x(x+1)}+\frac{1}{x(x+y)}-\sqrt{\frac{9}{4x^2}+\frac{(1-y/x)^2}{xy+x+y}} \right\} \\
= \frac{1}{\left[(4(y^3+1)\right]^{1/3}}
$$
In view of this, define the function $\color{blue}{g(x) = 1/\left[(4(x^3+1)\right]^{1/3}}$.

Both functions are displayed in the picture below, for $0 < x < 2$ and $0 < y < 5$, to ensure that both are positive valued
(which by the way is trivial for the blue one). It is noticed that the red function has a minimum above zero.
Of course $\color{red}{f(x)}$ and $\color{blue}{g(x)}$ can be investigated analytically as well, but without a computer algebra
system it quickly becomes complicated and hence error prone. What's worse: there is hardly any additional value.
That's why I prefer the graphics:

For $x\to\infty$ it is concluded that there are no values near zero in that far away region.
But where does the far away region start? To settle this issue, consider the expression:
$$
\delta(x,y) = 1 - \frac{\left[(4(y^3+1)\right]^{1/3}}{x} \times \\ \left\{
\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}} \right\}
$$
Or, alternatively (approximately the same for large $x$-values) :
$$
\delta(x,y) = \ln\left(\frac{x}{\left[(4(y^3+1)\right]^{1/3}}\right) - \\
\ln\left(\frac{x}{\left[(4(y^3+1)\right]^{1/3}}+\frac{y}{x+1}+\frac{1}{x+y}-\sqrt{\frac{9}{4}+\frac{(x-y)^2}{xy+x+y}}\right)
$$
with the interpretation attached that $\delta$ is a relative error of the limiting value $x/\left[(4(y^3+1)\right]^{1/3}$.
Then perform numerical experiments for $1 \leq y \leq 2$ (as usual) and for $x = 10$, giving: $\delta \lt 90\,\%\;$. Ongoing from there,
$\delta$ is continuously decreasing with increasing $x$, as expected. There is enough margin in the blue $\color{blue}{g(x)}$
to ensure that any $\delta(x,y)$ for $x \gt 10$ do not make it close to zero. So we only have to investigate function behaviour for
say $0 \lt x \leq 10$, a limited range of $(x,y)$ values.

Having done all this, we must conclude that the Double refinement of Nesbitt's inequality, as proposed by the OP, is valid.


问:
Is there a meaning behind the dot product divided by the length of the vector?
I am reading a proof in which instead of using the common dot product they define the inner product as the standard dot product divided by the dimension of the vectors. For example:

$\langle x, y\rangle = \frac{1}{n} (x \cdot y)$

Does this have a common interpretation or could this be just be convenient for the proof? I am looking for a more general understanding, but if it helps I have written the specific proof below:

Here is the proof, it is from my notes in class.

Let $G$ be an $r$-regular graph with eigenvalues $\lambda_{0} \geq \lambda_{1}, \ldots \geq \lambda_{n-1}$. Show that the stability number (largest independent set) $\lvert S \rvert$ satisfies:

$\lvert S \rvert \leq \frac{-n \lambda _{n-1}}{\lambda _{n-1}}$

Let $\langle x,y \rangle$ be the inner product such that $\langle x,y \rangle = \frac{1}{n}(x \cdot y)$. Let $\left \| x \right \| = \sqrt{\langle x,x\rangle}$. Let $A$ be the adjacency matrix of the graph $G$ and let $\left \{ v_{0}, v_{1}, \ldots, v_{n-1} \right \}$ be an orthonormal basis of eigenvectors such that $Av_{i} = \lambda_{i} v_{i}$. As $G$ is $r$-regular $\lambda_{0}=r$ and $v_{0}$ is the all-ones vector $\vec{1}$.

Let $S$ be a set of vertices forming an independent set (there are no edges between them). Let $u$ be the characteristic vector of $S$ and let $\alpha = \frac{\lvert S\rvert}{n}$

Note that

$u^{T}Au = \sum_{x,y \in S}^{} a_{x,y} = 0$

as the entries $a_{x,y}$ are all 0 in $A$ (by definition of independent set).

As all $v_{i}$ form an orthonormal basis we can express $u$ as a linear combination of $v_{i}$ with coefficients $\left \{ a_{0}, a_{1}, \ldots, a_{n-1} \right \}$.

$u = \sum_{i=0}^{n-1}a_{i}v_{i}$

As $v_{i}$ is an orthonormal basis we know that $\langle u, v_{i}\rangle = a_{i}$. Hence $\langle u, v_{0}\rangle = \langle u, \vec{1} \rangle = \frac{\lvert S \rvert}{n} = \alpha =  a_{0}$

Using Parseval's identity:

$a_{0}^{2} + a_{1}^{2} + \ldots + a_{n-1}^{2} = \left \| u \right \|^{2} = \alpha$

Hence as $a_{0} = \alpha^{2}$ we can rearrange such that:

$\alpha - \alpha^{2} = \sum_{i=1}^{n-1}a_{i}^{2}$

Now we will put everything together an create the bound:

Note that:

$\langle u, Au \rangle = \frac{1}{n}u^{T}Au = 0$
( $u^{T}Au = 0$ as we saw above)

Hence we have:

0 = $\langle u, Au \rangle = \sum_{i=0}^{n-1}\langle u, Aa_{i}v_{i} \rangle = \sum_{i=0}^{n-1}\langle u, \lambda_{i}a_{i}v_{i} \rangle = \sum_{i=0}^{n-1}\lambda_{i}a_{i} \langle u, v_{i} \rangle = \sum_{i=0}^{n-1}\lambda_{i}a_{i}^{2} = \lambda_{0}a_{0}^{2} + \sum_{i=1}^{n-1}\lambda_{i}a_{i}^{2}$

Now we establish an inequality by setting every eigenvalue in $\sum_{i=1}^{n-1}$ to the smallest eigenvalue $\lambda_{n-1}$

$0 = \sum_{i=0}^{n-1}\lambda_{i}a_{i}^{2} \geq \lambda_{0}a_{0}^{2} + \lambda_{n-1}\sum_{i=1}^{n-1}a_{i}^{2}$

Remembering that we found that $\sum_{i=1}^{n-1}a_{i}^{2} = \alpha - \alpha^{2}$ ,that $\lambda_{0} = r$ and that $a_{0}^{2} = \alpha^{2}$ we have

$0 = \geq r\alpha^{2} + \lambda_{n-1}(\alpha - \alpha^{2})$

Rearranging and using the fact that $\frac{\lvert S \rvert}{n} = \alpha$ we find:

$\lvert S \rvert \leq  \frac{n\lambda_{n-1}}{r-\lambda_{n-1}}$

答:
It just saves us manually dividing out powers of $n$ in certain proofs. A vector whose entries are all $1$ in some privileged basis is often denoted $1$ or $e$; let's go with $e$. On the given definition, $\langle e,\,e\rangle=1$, so by Cauchy-Schwarz $|\langle e,\,x\rangle|\le\sqrt{\langle x,\,x\rangle}$, which is just the QM-AM inequality (in fact, a generalization thereof where $x$ can have components $\le0$), i.e. the statement that the entries in $x$ have variance $\ge0$.

Addendum, just before I post this: I seemy example is used early in the proof you've edited in, albeit with an eye to combinatorial inequalities such as in adjacency matrices.


问:
what does mean by $\max_i$ in metric space
what does mean by :
$$d(x,y) = \max_i|x_i - y_i|, \quad\text{where} \hspace{0.25cm}x_i,y_i \in \mathbb{R}^n $$
Can some one explain me what does it mean by "$\max_i$"?

(note: I am studying metric space in physics, maybe my question looks trivial)

答:
Let $x = (x_1, ... , x_n)$ and $y = (y_1, ... , y_n)$ be elements of $\mathbb R^n$.  Then

$$\operatorname{max}_i |x_i - y_i|$$

means the largest of the numbers $|x_1 - y_1|, |x_2-y_2|, ... , |x_n - y_n|$.


问:
Local connection of closure of open connected subspaces in some metric spaces.
Let $X$ be a metric, connected, locally connected and locally compact space, and let $U$ be an open connected subspace of $X$. Is $\overline{U}$ always locally connected?

I´m writing a thesis about Peano spaces and this would be useful if it were true, but I haven´t been able to prove it or find any counterexamples.

答:
No. Let $X=\mathbb{R}^2$ and define $U$ as follows:

$$U_n=\bigg(\frac{1}{n}-\epsilon_n,\frac{1}{n}+\epsilon_n\bigg)\times\mathbb{R}$$

We choose each $\epsilon_n>0$ so that closures of those subsets are pairwise disjoint, something like $\frac{1}{2n(n+1)}$.

Then we put

$$V=\mathbb{R}\times(-1,1)$$
$$U=V\cup\bigcup_{n=1}^\infty U_n$$

The role of $V$ is only to ensure that $U$ is connected. $U$ is also open but for $W=\{0\}\times\mathbb{R}$ we get

$$\overline{U}=\overline{V}\cup\bigcup_{n=1}^\infty \overline{U_n}\cup W$$

which is not locally connected (points at $W$ outside of $\overline{V}$ don't have small enough connected neighbourhoods).

This example can easily be modified to exclude $X$ compact as well. Simply consider $X=[-x,x]^2$ square for $x>1$ and then our open subset is $U\cap X$.


问:
Image and ker of a matrix
Why is it true that given $A \in M_n(\Bbb R)$, if $X\in \operatorname{Ker}(A^T)$, then $X \in (\operatorname{Im}A)^\perp$ ?

答: Hint: the $i$-th entry of the vector $A^TX$ is the dot product of the $i$-th row of $A^T$ with $X$.

Title: Image and ker of a matrix


问:
Showing continuity - epsilon-delta situation
Let $f$ be continuous on $[a,b]\times [c,d]\subset\mathbb{R}^2$ and $h(x,y):= \int_a^x f(z,y)dz$. Then, $h$ is continuous on $[a,b]\times [c,d]$.

For $(x,y),(x',y')\in [a,b]\times [c,d]$ we have

$$
|h(x,y)-h(x',y')|
= |\int_a^{x'}f(z,y)dz + \int_{x'}^{x}f(z,y)dz - \int_a^{x'}f(z,y')dz| \\
\leq \int_a^{x'}|f(z,y)-f(z,y')|dz + |\int_{x'}^x f(z,y)dz|$$

Now, $f$ is bounded and equicontinuous on $[a,b]\times [c,d]$. It is intuitively clear to me that the integrals on the right side get arbitrarily small, if just $(x,y)$ and $(x\prime,y\prime)$ get close enough. However, I fail to express this in terms of epsilon-delta.

答:
Since $f$ is bounded there exists $M > 0$ such that $|f(x,y)| \leqslant M$ for all $(x,y) \in [a,b]\times [c,d]$.  Since $f$ is uniformly continuous, given $\epsilon > 0$ there exists $\delta(\epsilon) > 0$ such that $|f(x_1,y_1) - f(x_2,y_2)| < \epsilon$ when $\sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2} < \delta(\epsilon)$.

Suppose $\sqrt{(x-x')^2 + (y - y')^2} < \min \left[\frac{\epsilon}{2M},\delta\left(\frac{\epsilon}{2(b-a)}\right)\right]$.

It follows that $|x-x'| \leqslant \sqrt{(x-x')^2 + (y - y')^2} < \frac{\epsilon}{2M}$.

We also have   $\sqrt{(z-z)^2 + (y - y')^2} = |y-y'| < \delta\left(\frac{\epsilon}{2(b-a)}\right)$, and by uniform continuity that $|f(z,y) - f(z,y')| < \frac{\epsilon}{2(b-a)}$ for all $z \in [a,x']$.

Thus,

$$|h(x,y)-h(x',y')| \leqslant \int_a^{x'}|f(z,y)-f(z,y')|\,dz + \int_{x'}^x |f(z,y)|\,dz\\ < \frac{\epsilon}{2(b-a)}|x'-a|+M|x-x'|< \frac{\epsilon}{2(b-a)}(b-a)+M\frac{\epsilon}{2M}= \epsilon$$.


问:
Is it true that for all $U$, $W$ non-empty sets in $\mathbb{R}^{n}$ if $U \cup W=\mathbb{R}^{n}$ then $U=\mathbb{R}^n$ or $W=\mathbb{R}^{n}$
I've been thinking about this one for a while, at first I thought it only to be valid if $U,W$ were open (the union of open sets in $\mathbb{R}^{n}$ is open and $\mathbb{R}^{n}$ is an open set), then I remembered that $\mathbb{R}^{n}$ is also a closed set and that the union of closed sets is closed.

I thought about proving this by contradiction, however I am confused in how to proceed.

答: No, of course not. Consider $n = 1$ and, say, $U = \mathbb{Q}$ and $W = \mathbb{R} - \mathbb{Q}$.


问:
if $f(0)=1$ and f is analytic, there can't be $f(z_0)=0$ with $|z_0|<M$
Let be $f$ analytic  in the disc with radius $$1 from the beginning, and continuous on the boundary of the disc. let $f(0)=1$ and also let $z$ on the boundary satisfies $1<|f(z)|<M$.
Then there can't be $|z_{0}|<\frac{1}{M}$ satisfying  $f(z_0)=0$ .
I thought about using Cauchy integral formula but it doesn't seem to be useful.

Edit: I thought about that idea: $M$ is the maximum value on the disc. So for every sub disc the maximum value must be $M$. Let be $z_0$ a zero on the sub-disc, then there is an environment that is smaller than $1$. Therefore the integral on the sub disc must be less than $1$, which contradicts the value in the center of the disc.

答:
If $|f|$ is bounded by $M$ in the unit disk then $g(z) = f(z)/M$ maps the unit disk into itself.

Now let $z_0$ be a zero of $f$ in the unit disk. Then $g(0) = 1/M$ and $g(z_0) = 0$ and the Schwarz-Pick theorem shows that
$$
\frac 1M = |g(0)| = \left|{\frac  {g(0)-g(z_{0})}{1-\overline {g(0)}g(z_{0})}}\right|\leq \left|{\frac  {0-z_0}{1-\overline {0}z_0}}\right| = |z_0| \, .
$$
So every zero $z_0$ of $f$ in the unit disk satisfies $|z_0| \ge 1/M$.


问:
Why does an infinite limit not exist?
I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like  $$\lim_{x\to \infty}{\cos x.}$$

答:
Generally, we get that

$\lim_{n \rightarrow \infty} x(n) = y \leftrightarrow \forall \epsilon>0: \exists N\in \mathbb{N} : \forall n > N :|x(n) - y|<\epsilon$

but, for example, for $x(n)=n$

it's not true that $\lim_{n \rightarrow \infty} n = \infty $

because it would imply that

$\forall \epsilon>0: \exists N\in \mathbb{N} : \forall n > N :|x(n) - \infty|<\epsilon $

but

$|x(n) - \infty|=\infty$

and it can never be less then $\epsilon$

so $x(n)=n$ does not converge to $\infty$

Real numbers are colloquially defined as limits of sequances, I just shown that $\infty$ is not a real number.

Title: Why does an infinite limit not exist?


答:
Generally, the notation
$$
\operatorname{Op}_i\text{el}_i
$$
means "Apply the operation Op to the operand $\text{el}_i$ for each possible $i$". We can include limitations on the $i$ if it isn't clear from context.

Thus for your case, it means "Take the max of $|x_i-y_i|$ for each possible $i$". Other common examples include

"$\sum_{n\in \Bbb N} \frac1{n^2+1}$" means "sum $\frac1{n^2+1}$ for each natural $n$"

"$\bigcup_k X_k$" means "Take the union of all the $X_k$"

Title: what does mean by $\max_i$ in metric space

Body:
what does mean by :
$$d(x,y) = \max_i|x_i - y_i|, \quad\text{where} \hspace{0.25cm}x_i,y_i \in \mathbb{R}^n $$
Can some one explain me what does it mean by "$\max_i$"?

(note: I am studying metric space in physics, maybe my question looks trivial)


问:
Question about minimal polynomials.
Let $p(x)$ be a minimal polynomial for $a$ for field $F$. This implies it is a monic polynomial of the lowest degree possible such that $p(x)=0$.

Why does $p(x)$ have to be irreducible? Why can't it be split into factors?

If $p(x)$ is reducible, which I think should be the case, then it can have some of its factors in common with other polynomials. Hence, $F[a]$ won't be a field then, in spite of $a$ being algebraic over $F$, because the polynomials which have non-zero factors in common with $p(x)$ will not have a multiplicative inverse.

答: if p(x) is reducible, suppose p(x)=f(x)g(x), then f(a)=0 or g(a)=0, but f and g both has lower degree than p, which is not possible, because p has lowest degree.</p>

ContentLicense: CC BY-SA 3.0

Title: Question about minimal polynomials.


问:
How to minimize the chance of making an error in algebraic calculations?
This is a question about solving problems under significant time constraints, for example under exam conditions. Even though this question might be seen as "subjective", it is seriously related to many people who study mathematics, so an answer will be very helpful.

The question is simple: what are some ways to reduce the amount of errors in algebraic calculations, for example integrals, derivatives, series expansions, and things like that?

This is important in pure maths, when we are, for example,  doing proofs in analysis/number theory. Needless to say, this is very very important for applied maths and physics.

Apparently, if one is really concentrated on finding the method to solve a problem, he/she becomes much more likely to make small mistakes along the way. Of course, such errors are not an issue if one has plenty of time to check, but it does become an issue when time is limited.

答:
This is arguably more a question for psychologists than for mathematicians, but here are a few techniques which could be helpful:

Check your work! If you have more than one approach which will generate a valid solution and have time to do so, test both. The odds that you'd make the same error in both cases are rather slim. Better yet, if you have a way to verify a solution which does not involve the same degree of effort as solving it in the first place, use it! (i.e. if you are integrating, take the derivative, or if you are finding eigenvalues and eigenvectors of a matrix, actually multiply each of them to be sure you have the correct result).

Verify calculations, signs, etc. at each step rather than waiting until the end to (maybe) spot an error. This may double the amount of time spent crunching numbers, but it reduces to near-zero the amount of time spent correcting past computational errors and doing work which you'll later have to throw away.

Verify whichever computations you can using an independent source. Obviously this is not suitable for an exam under most academic integrity policies, nor would it be ethical to use an external source on homework to do anything more than check for correctness (even that may be questionable depending on the context, but taking solution minus believed solution and seeing if it is or is not zero after checking the work yourself may be acceptable). However, if you are writing a paper then you could run your calculations past at least one collaborator, advisor, or peer if not a computer solver.


问:
Understanding if the following subsets of a vector space satisfy addition and scalar multiplication
Here, I'm supposed to justify if the following subsets are closed under scalar multiplication, addition, both or neither.

a) $W_1=\mathcal{S},Z_1=\{s \in\mathcal{S}|s \text{ is monotone}\}.$ Here, monotone sequences are defined as $a_{n+1} \geq a_n$ for all $n\geq 0$ or $a_{n+1} \leq a_n$ for all $n\geq 0$ for some sequence $(a_n)_n$.

b) $W_2=M_3(\mathbb{R}),Z_2=\{A \in M_3(\mathbb{R})|Tr(A)^3=Det(A) \}$.

a) I could pick $a_n=(0,1,1,...)$ and $b_n=(0,0,-1,-1,...)$ for some $a_n$, $b_n$ in $Z_1$ since this gives $(0,1,0,0,...)$ which isn't monotonic. Addition fails. I think multiplication will work though but I am not sure how to show this.

b) I'm almost certain addition fails but I am not so sure about scalar multiplication.

If someone could help me out, that would be great!

答:
For part $a)$ scalar multiplication, let $(a_n)$ be a monotone sequence and $s$ be a scalar. Then $s(a_n) = (s\cdot a_n)$ is also montone, since you can always just cancel from both sides of the $\leq$ sign.

For $b)$, consider

$$
A = \begin{bmatrix}
2&0&0\\
0&2&0\\
0&0&0
\end{bmatrix},B = \begin{bmatrix}
0&0&0\\
0&0&0\\
0&0&2
\end{bmatrix}
$$
Note that $A$ and $B$ have determinant and trace $0$ so they are in $Z_2$, but $A+B$ has trace $8$ and determinant $8$, but obviously $8^3 \neq 8$.

For scalar multiplication we don't have closure either. Consider the identity matrix $I$. It has cubed trace and determinant $1$, so it is in $Z_2$, however if for example we multiply the matrix by $2$, we get a trace of $8$ and a determinant of $8$ and as noted about $8^3 \neq 8$.


问:
Find a continuous function to satisfy differential equation
I've been working through differential equations in the textbook and have come across this example which has really stumbled me. Any advice on how to approach this would be great. Thanks

Find a continuous function $y\colon[0,\infty)\to \mathbb{R} $ with $y(0)=9$, that is differentiable on both $(0,\frac{\pi}{4})$ and $(\frac{\pi}{4},\infty)$ with derivative $y'(x)$ satisfying the differential equation:
$$y'(x)=
\begin{cases}
      2xy(x)+e^{x^2}\tan(x), & x\in(0,\frac{\pi}{4}), \\
      \sin^3(x)\cos^2(x),  & x>\frac{\pi}{4}.\\
\end{cases}
$$

答:
Roadmap:

Solve the IVP $y' = 2xy + e^{x^2} \tan x, \quad y(0)=9$. You'll get
$y(x) = -e^{x^2} (\log (\cos (x))-9)$. This is the solution on the interval $(0,\frac{\pi}{4})$.

Compute $y_*= y(\frac{\pi}{4}) = -e^{\frac{\pi ^2}{16}} \left(-9-\frac{\log (2)}{2}\right)$

Solve the IVP $y'=\sin^3 x \cos^2 x, \quad y(\frac{\pi}{4}) = y_*$. This is the solution on the interval $(\frac{\pi}{4}, +\infty)$.

Can you spot $\frac{\pi}{4}$ in the graphic?


问:
Prove that $g(x)=(x^2−1)^x$ is increasing on $(1,+\infty)$
Let $A = \mathbb{R}\setminus[−1,1]$. Let $g : A\to\mathbb{R}$ be defined by
$g(x)=(x^2-1)^x$
for all $x\in A$.

Prove that $g(x)=(x^2-1)^x$ is increasing on $(1,\infty)$

I have currently attempted to prove this by showing $g'(x)\geq 0 $ for all $ x\in A $ which gives $g'(x)= (x^2-1)^{x-1}(2x^2+(x^2-1)ln(x^2-1))$ which should be greater than or equal to $0$ however I am unsure how to show this or whether I have gone about this the right way.

Any Help will be grateful.

答:
A. For b>1 x^b is increasing.
B. For a>1 a^x is increasing.

For any x_1 < x_2 where both are bigger than root 2:

We get from A above:

[(x_1)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_1

Then we get from B above:

[(x_2)^2 - 1]^x_1 < [(x_2)^2 - 1]^x_2

putting them together the function is increasing when bigger than root 2.

Only very close to 1 do we need further analysis

d/dx((x^2 - 1)^x) = (x^2 - 1)^x ([(2 x^2)/(x^2 - 1)] + log(x^2 - 1))

So the front part of the right hand side of the above equation is always positive for x>1,so we only need to make sure that [(2 x^2)/(x^2 - 1)] + log(x^2 - 1) is always positive for x>1.

So we want to show 2x^2 + (x^2-1)log(x^2-1) >0 for x>1. Let U = x^2 -1, U>0 for x>1, so it suffices to show 2 + 2U + Ulog(U) > 0 for U>0.

d/dx(x log(x)) = log(x) + 1

Ulog(U) takes a minimum point at U= 1/e. So Ulog(U) > -1/e for U>0. That proves the inequality:

2x^2 + (x^2-1)log(x^2-1) >0 for x>1.


问:
Show that $-\int_\Omega\left(f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x)\right)\text{d}x\le2\|f\|_{L_2(\Omega)}\|\nabla f\|_{L_2(\Omega)^n}$
Let $n\in\mathbb N$. Let $f\in C_c^\infty(\mathbb R^n)$ and let $\Omega\subseteq\mathbb R^n$ be open such that it contains the support of $f$. I have encountered the following inequality

$$-\int_\Omega\left(f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x)\right)\text{d}x\le2\|f\|_{L_2(\Omega)}\|\nabla f\|_{L_2(\Omega)^n},$$

where $\nabla f$ denotes the gradient of $f$ and $\partial_n f$ denotes the first derivative of $f$ with respect to the $n$-th coordinate. I believe that this should be simple enough to derive, but I am having difficulty doing so myself. I am not sure how to go from the expression on the left - which is close to the $L_1$- scenario to the $L_2$ - scenario on the right. One thought I had was to look at
$f(x)f^*(x)=|f(x)|^2$ and to argue that both $f(x)\le|f(x)|^2$ and $f^*(x)\le|f(x)|^2$, but I realise that this is nonsensical since both $f(x)$ and $f^*(x)$ might be complex, whilst the right hand since is always purely real.

How does one derive this bound? What am I missing?

答:
The integrand is
$$
f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x) = \partial_n |f(x)|^2
$$
and thus it is a real number. Therefore we can estimate
$$
-(f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x)) \le |f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x)| \le 2 |\partial_n f(x)||f(x)|
$$
since $|f| = |f^\ast|$ and $|\partial_n f| = |\partial_n f^\ast|$. By Cauchy-Schwarz,
$$
-\int_\Omega\left(f^*(x)\partial_nf(x)+\partial_nf^*(x)f(x)\right)\text{d}x\le 2 \int_\Omega|\partial_n f(x)||f(x)| \text{d}x \le  2\|f\|_{L_2(\Omega)}\|\partial_n f\|_{L_2(\Omega)}
$$
which implies the desired estimate.


问:
Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$
I want to check if it's improper integral or not
$$ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}.$$
What I did so far is :

set $t=e^{x} \rightarrow \mathrm dt=e^x\mathrm dx \rightarrow \frac{\mathrm dt}{t}=dx
$ so the new integral is:
 $$ \int^{\infty}_0 \frac{\mathrm dt}{t(1+t^2)} =  \int^{\infty}_0 \frac{\mathrm dt}{t}-\frac{\mathrm dt}{1+t^{2}}$$
now how I calculate the improper integral, I need to right the $F(x)$ of this integral and then to check the limit?

Thanks!

答:
Hints:

First, it must be

$$\frac1{t(1+t^2)}=\frac1t-\frac t{1+t^2}\;\;,\;\;\text{then}:$$

$$\int\limits_1^\infty\left(\frac1 t-\frac t{1+t^2}\right)dt:=\lim_{b\to\infty}\left(\log\frac b{\sqrt{1+b^2}}+\log\sqrt 2\right)$$


问:
How to compare log and exponential

$x^{1/3}$

$　 (\log_{2}{x})^3$

How to compare them?

答:
As others have mentioned, there's no closed-form solution to
$$x^{1/3} = (\log_2 x)^3$$
in elementary functions, but it can be solved using the Lambert W function.

Briefly, the Lambert W function is the (multi-branched) inverse function of
$$f(x) = xe^x$$
That is,
$$x = W_k(x)e^{W_k(x)}$$
When working with real numbers, we only need the $W_{-1}$ and $W_0$ branches.

This function is generally not found in the standard math libraries of common programming languages, but it is found in most advanced mathematics packages. You can approximate it in various ways, eg, Newton's method.

Anyway, here's how to find the (real) solutions to your equation, assuming that you have some way of evaluating the Lambert W.

$$x^{1/3} = (\log_2 x)^3$$
$$x^{1/9} = \log_2 x$$
Let $x=u^9$
$$u = \log_2 u^9 = 9\log_2 u$$
$$2^u = u^9$$
$$u\log 2 = 9\log u$$
$$u = 9\log u / \log 2$$

The $\log u$ function grows slower than any power $u^n$ for $n>1$, so as Paul Rebenciuc mentioned at the end of his answer, for all sufficiently large $u$, $2^u > u^9$.

Let's pause for a moment to look at the graph of $y = 9\log x / \log 2$ and see where it crosses $y=x$.

Here are close-ups of the cross-over points.

So we have solutions around $u = 1.087$ and $u = 51.069$.

Continuing,

$$u = 9\log u / \log 2$$
Let $m = 9 / \log 2$
$$u = m \log u$$
Let $u = e^{-v}$, so $\log u = -v$
$$e^{-v} = -mv$$
$$-1/m = ve^v$$
Now we can apply the Lambert W:
$$W(-1/m) = W(ve^v) = v$$

So now we can get $u$ from $u = e^{-v}$ and $x$ from $x=u^9$. As I mentioned earlier, we need to evaluate both the $W_{-1}$ and $W_0$ branches to get both solutions.

Here's a short Sage / Python script that performs this calculation. It also verifies that $u = -mv$.

Code

```<code>m = 9 / log(2)
for v in [lambert_w(i, -1/m) for i in (-1, 0)]:
    u = exp(-v)
    print("v", v.n())
    print("u", u.n(), "-mv", (-m*v).n())
    print("u^9", (u^9).n(), "2^u", (2^u).n())
    print()
</code>```

Output

```<code>v -3.93318771959389
u 51.0695137614913 -mv 51.0695137614913
u^9 2.36295530009418e15 2^u 2.36295530009419e15

v -0.0837437490868883
u 1.08735022361793 -mv 1.08735022361793
u^9 2.12483412771106 2^u 2.12483412771106
</code>```

And here's a link containing the script which runs on the SageMath server.

As I mentioned earlier, if you don't have access to the Lambert W function, you can calculate it using Newton's method. Of course, you can also use Newton's method to solve this equation directly. Here's one approach.

Our equation is
$$u = 9\log u / \log 2$$

We transform the equation to eliminate the $\log u$ term, so we don't get errors if $u\le 0$.

$$u \log 2 / 9 = \log u$$
Let $k = \log 2 / 9$
$$e^{uk} = u$$
Let
$$y = e^{uk} - u$$
so we want to find $u$ such that $y=0$.

Differentiating,
$$\frac{dy}{du} = ke^{uk} - 1$$
Now
$$\Delta u = \Delta y / \frac{dy}{du}$$
So
$$\Delta u = \frac{u - e^{uk}}{ke^{uk} - 1}$$
From a starting approximation of $u$, $u+\Delta u$ is a better approximation.

Here's some Sage / Python code which implements that algorithm. The code is mostly pure Python, it just uses Sage to input the initial approximation.

```<code>from math import exp, log
k = log(2) / 9

@interact
def main(u=1):
    for i in range(9):
        z = exp(u * k)
        du = (u - z) / (k * z - 1)
        u += du
        print(u, du)
        if abs(du) < 1e-12:
            break
</code>```

Output

```<code>1.0873235056664654 0.08732350566646546
1.087350223615419 2.6717948953605787e-05
1.0873502236179315 2.5123282018144193e-12
1.0873502236179313 -2.4233897962905985e-16
</code>```

Here's a link containing the script

Title: How to compare log and exponential

Body:
As others have mentioned, there's no closed-form solution to
$$x^{1/3} = (\log_2 x)^3$$
in elementary functions, but it can be solved using the Lambert W function.

Briefly, the Lambert W function is the (multi-branched) inverse function of
$$f(x) = xe^x$$
That is,
$$x = W_k(x)e^{W_k(x)}$$
When working with real numbers, we only need the $W_{-1}$ and $W_0$ branches.

This function is generally not found in the standard math libraries of common programming languages, but it is found in most advanced mathematics packages. You can approximate it in various ways, eg, Newton's method.

Anyway, here's how to find the (real) solutions to your equation, assuming that you have some way of evaluating the Lambert W.

$$x^{1/3} = (\log_2 x)^3$$
$$x^{1/9} = \log_2 x$$
Let $x=u^9$
$$u = \log_2 u^9 = 9\log_2 u$$
$$2^u = u^9$$
$$u\log 2 = 9\log u$$
$$u = 9\log u / \log 2$$

The $\log u$ function grows slower than any power $u^n$ for $n>1$, so as Paul Rebenciuc mentioned at the end of his answer, for all sufficiently large $u$, $2^u > u^9$.

Let's pause for a moment to look at the graph of $y = 9\log x / \log 2$ and see where it crosses $y=x$.

Here are close-ups of the cross-over points.

So we have solutions around $u = 1.087$ and $u = 51.069$.

Continuing,

$$u = 9\log u / \log 2$$
Let $m = 9 / \log 2$
$$u = m \log u$$
Let $u = e^{-v}$, so $\log u = -v$
$$e^{-v} = -mv$$
$$-1/m = ve^v$$
Now we can apply the Lambert W:
$$W(-1/m) = W(ve^v) = v$$

So now we can get $u$ from $u = e^{-v}$ and $x$ from $x=u^9$. As I mentioned earlier, we need to evaluate both the $W_{-1}$ and $W_0$ branches to get both solutions.

Here's a short Sage / Python script that performs this calculation. It also verifies that $u = -mv$.

Code

m = 9 / log(2)
for v in [lambert_w(i, -1/m) for i in (-1, 0)]:
    u = exp(-v)
    print("v", v.n())
    print("u", u.n(), "-mv", (-m*v).n())
    print("u^9", (u^9).n(), "2^u", (2^u).n())
    print()

Output

v -3.93318771959389
u 51.0695137614913 -mv 51.0695137614913
u^9 2.36295530009418e15 2^u 2.36295530009419e15

v -0.0837437490868883
u 1.08735022361793 -mv 1.08735022361793
u^9 2.12483412771106 2^u 2.12483412771106

And here's a link containing the script which runs on the SageMath server.

As I mentioned earlier, if you don't have access to the Lambert W function, you can calculate it using Newton's method. Of course, you can also use Newton's method to solve this equation directly. Here's one approach.

Our equation is
$$u = 9\log u / \log 2$$

We transform the equation to eliminate the $\log u$ term, so we don't get errors if $u\le 0$.

$$u \log 2 / 9 = \log u$$
Let $k = \log 2 / 9$
$$e^{uk} = u$$
Let
$$y = e^{uk} - u$$
so we want to find $u$ such that $y=0$.

Differentiating,
$$\frac{dy}{du} = ke^{uk} - 1$$
Now
$$\Delta u = \Delta y / \frac{dy}{du}$$
So
$$\Delta u = \frac{u - e^{uk}}{ke^{uk} - 1}$$
From a starting approximation of $u$, $u+\Delta u$ is a better approximation.

Here's some Sage / Python code which implements that algorithm. The code is mostly pure Python, it just uses Sage to input the initial approximation.

from math import exp, log
k = log(2) / 9

@interact
def main(u=1):
    for i in range(9):
        z = exp(u * k)
        du = (u - z) / (k * z - 1)
        u += du
        print(u, du)
        if abs(du) < 1e-12:
            break

Output

1.0873235056664654 0.08732350566646546
1.087350223615419 2.6717948953605787e-05
1.0873502236179315 2.5123282018144193e-12
1.0873502236179313 -2.4233897962905985e-16

Here's a link containing the script


问:
Find the commutator subgroup of a certain group
I am given the group $H$ of matrices

$$\begin{bmatrix}
a&b\\
0&1\\
\end{bmatrix},$$

where $a \neq 0$ and $a,b  \in\mathbb{R}$. How do I find the commutator subgroup of this group?

I found by the definition of a commutator of elements from $H$ that $H' \subseteq Sl_2(\mathbb{R}) $, but I do not know how to take it from there.

答:
Take a typical commutator $ghg^{-1}h^{-1}$, where

$$g = \begin{pmatrix} a & b \\ & 1 \end{pmatrix}$$

$$h = \begin{pmatrix} a' & b' \\ & 1 \end{pmatrix}$$

and calculate $ghg^{-1}h^{-1}$.  At least get your hands dirty before asking the question.


答:
The way that it is stated, it is not true (see the answer by Duncan Ramage and the comments)

Judging by the discussion though, this nice question will lead you to the notion of connectedness. For more info on the topic, any book on topology (or metric spaces) will eventually talk about this notion.

Now back to the question, as $\mathbb{R}^n$ is connected, it cannot be written as a union of two non-empty disjoint open (or closed) sets.

There are a few ways in which you can prove the connectedness of $\mathbb{R}^n$ but at one point or another, you may want to use the definition of path connectedness (which is much more intuitive) and the implication $\text{ path connected } \implies \text{connected}$. Part of the proof of this implication resembles the proof of compactness of $[0,1]$.

As $\mathbb{R}^n$ is convex, it is path connected.

Title: Is it true that for all $U$, $W$ non-empty sets in $\mathbb{R}^{n}$ if $U \cup W=\mathbb{R}^{n}$ then $U=\mathbb{R}^n$ or $W=\mathbb{R}^{n}$

Body:
I've been thinking about this one for a while, at first I thought it only to be valid if $U,W$ were open (the union of open sets in $\mathbb{R}^{n}$ is open and $\mathbb{R}^{n}$ is an open set), then I remembered that $\mathbb{R}^{n}$ is also a closed set and that the union of closed sets is closed.

I thought about proving this by contradiction, however I am confused in how to proceed.


问:
The value of $r$ for which $\binom {20}{r} \binom{20}{0} +\binom{20}{r-1}\binom{20}{1}...\binom{20}{0} \binom{20}{r}$ is maximum is?
Most solutions write the given expression as
$$(1+x)^{20}(1+x)^{20}=(1+x)^{40}$$

But I can’t wrap my head around has to how it is even correct to do so.

When you expand the factors, what about all the other residual terms such $(\binom {20}{0})^2$ that aren’t included in the question expression?

I know this question already exists on MSE, no it does not clear the specific problem I have

答:
It’s just a matter of equating the coefficients of $x^r$ in two different ways of writing the same polynomial. For each $r$ from $0$ through $40$ the coefficient of $x^r$ on the lefthand side of the equation

$$(1+x)^{20}(1+x)^{20}=(1+x)^{40}\tag{1}$$

is

$$\begin{align*}
&\binom{20}r\binom{20}0+\binom{20}{r-1}\binom{20}1+\ldots+\binom{20}0\binom{20}r\\
&\qquad=\sum_{k=0}^r\binom{20}{r-k}\binom{20}k\,,
\end{align*}\tag{2}$$

which is the expression that you want to maximize. The coefficient of $x^r$ on the righthand side of $(1)$ is $\binom{40}r$. These coefficients must be equal, so

$$\sum_{k=0}^r\binom{20}{r-k}\binom{20}k=\binom{40}r\,,\tag{3}$$

which we know is maximized when $r=20$.

In more detail, the lefthand side of $(1)$ is

$$\left(\sum_{i=0}^{20}\binom{20}ix^i\right)\left(\sum_{j=0}^{20}\binom{20}jx^j\right)\,.\tag{4}$$

When you multiply this out, the collected $x^r$ term will be

$$\begin{align*}&\binom{20}0\binom{20}rx^{0+r}+\binom{20}1\binom{20}{r-1}x^{1+(r-1)}+\ldots\\
&\ldots+\binom{20}{r-1}\binom{20}1x^{(r-1)+1}+\binom{20}r\binom{20}0x^{r+0}\,,\end{align*}$$

whose coefficient is indeed $(2)$. The other terms that result from expanding $(4)$ contain other powers of $x$ and so don’t contribute to the $x^r$ term of the product.

Alternatively, you can interpret the sum $(2)$ combinatorially to arrive at the same result. You have a pool of $20$ men and $20$ women, and you want to form a committee of $r$ people from this pool. If you decide to have $k$ women on the committee, you must have $r-k$ men. You can choose the $r-k$ men in $\binom{20}{r-k}$ ways and the $k$ women in $\binom{20}k$ ways, so there are $\binom{20}{r-k}\binom{20}k$ ways to choose such a committee. Summing over $k$ shows that $(2)$ gives the total number of possible $r$-person committees. On the other hand, we could simply pick $r$ people from the pool of $40$ to form the committee, something that we can do in $\binom{40}r$ ways. This counts exactly the same committees and establishes the equality $(3)$ without reference to the polynomials in $(1)$. As was noted in the comments, this is an instance of the (very useful) Vandermonde identity.


问:
Area of a triangle from vector coordinates of vertices in 3D
I have three vectors $v_1, v_2$, and $v_3$, giving the vertices of a triangle. The $z$-coordinates are the same, so the $(x,y)$-coordinates alone give the vertices of an identical triangle in the $xy$-plane. Is there a general way to find the area of this triangle given the vectors?

$$
v_1 = (v_{x1}, v_{y1}, v_{z1}) \\
v_2 = (v_{x2}, v_{y2}, v_{z2}) \\
v_3 = (v_{x3}, v_{y3}, v_{z3})
$$

?

答: If $a$ and $b$ are the vectors for two sides then you can reduce the above answers to $\dfrac{1}{2} \sqrt{|a|^2|b|^2 - (a\cdot b)^2}$ with the advantage that it works in any dimension.


问:
What is the domain of $\frac{1}{\log x}$?
If you ask to me what  the domain of $\frac{1}{\log(x)}$ is, I set
$$
\log(x)\neq0 \Rightarrow x\neq1 $$ infact $\frac{1}{0}\to\infty$ but moreover (before this exercize) I would have set $x>0$ too because $\log(0)\to-\infty$. But in this case $\frac{1}{\log(0)}=\frac{1}{-\infty}\to0$ and the function is define for $x=0$ too.

Is this a rule? I have to put log arguments $\neq0$ ONLY if logs are at numerator? There are other rules like this?

[edit]

I got confused, the correct example is: $ \frac{1}{\log(|x|)}$ and so I should have put $|x|>0 => x\neq0$ in this case does something change?

答:
For the reciprocal of the logarithm of $x$ to be defined, the logarithm of $x$ must be defined in the first place. That means that we need to require $x>0$: The logarithm s neither defined for $0$ nor for negative numbers.
In addition, taking the reciprocal must be defined, which is only defined if we do not attempt to take the reciprocal of $0$. As $\log x=0\iff x=1$, we arrive at a maximal real domain for the function $x\mapsto \frac1{\log x}$ of
$$ (0,1)\cup (1,\infty).$$


问:
Prove that $f(f(x))=x$ has no roots .... $f$ having a general form
This problem gave me some headache, especially because $f$ have its own general form :

let $f(x) = ax^2 + bx + c$. Suppose that $f(x) = x$ has no real roots.

Show that equation $f(f(x))=x$ has also no real roots.

答:
This can be generalized to: If $f:\mathbb R\rightarrow \mathbb R$ is a continuous function and $f(f(x))$ has a real fixed point then $f(x)$ has a real fixed point:

If $f(f(c))=c$ and $f(c)=r$ then $f(r)=c$.

If $c\neq r$ then there is an $s$ between $r$ and $c$ such that $f(s)=s$ (In the case $c<r\Rightarrow f(r)-r=c-r>0$ and $f(c)-c=r-c<0$...).


答:
$\log x$ only tends to $-\infty$ as $x$ tends to $0^+$ i.e. it takes large negative values to the right of $0$ but it is not actually defined at $0$. So $1/\log x$ attains very small negative values on the right of $0$ but is not actually defined at $0$.

What you are talking about is the limit of the function $1/\log x$ as $x$ tends to $0$ from the right side, and it is well known that limits of functions may exist at points where the function itself doesn't exist.

Title: What is the domain of $\frac{1}{\log x}$?

Body:
If you ask to me what  the domain of $\frac{1}{\log(x)}$ is, I set
$$
\log(x)\neq0 \Rightarrow x\neq1 $$ infact $\frac{1}{0}\to\infty$ but moreover (before this exercize) I would have set $x>0$ too because $\log(0)\to-\infty$. But in this case $\frac{1}{\log(0)}=\frac{1}{-\infty}\to0$ and the function is define for $x=0$ too.

Is this a rule? I have to put log arguments $\neq0$ ONLY if logs are at numerator? There are other rules like this?

[edit]

I got confused, the correct example is: $ \frac{1}{\log(|x|)}$ and so I should have put $|x|>0 => x\neq0$ in this case does something change?


答:
The natural domain of a function (sometimes shortened as domain) is the maximum set of values for which the function is defined.

I suggest not writing any expressions like $\frac{1}{0}$ or $\frac{1}{-\infty}$ unless you are absolutely clear about the context and know consciously what you are doing about these expressions.

Your question has nothing to do with any "limiting" process.

First of all the natural domain for the function $\log x$ is the set $(0,\infty)$, or you may write $x>0$. It is meaningless to write $\log(0)$ since the logarithm is not defined at $0$.

Then you need to guarantee that $\log x\ne 0$ since it is in the denominator.

On the other hand, $\log x=0$ if and only if $x=1$. So the natural domain of the function $\frac{1}{\log x}$ is
$$
(0,\infty)\setminus\{1\}
$$
or you write it as
$$
(0,1)\cup (1,\infty)
$$


问:
Solve PDE on a subspace
Let us assume that we are given some partial differential equation. Just for example, let me consider the following.

$$ a(x,y,z) \partial_x F(x,y,z) + b(x,y,z) \partial_y F(x,y,z) + c(x,y,z) \partial_z F(x,y,z) = 0$$

I know how to solve it with the usual methods. However, what if we know that the previous partial differential equation is satisfied only on some subspace? Let us say, we know that the previous equation is true only on $F(x,y,z) = 0$. How to approach such case? And how to generalize this to an arbitrary partial differential equation?

My first idea would be to solve differential equation in general for the whole space, and then put the constraint in the end. Can someone (at least intuitively) show that in that case, we obtain the most general solution? Is it possible that previous equation is satisfied on $F(x,y,z) = 0$ but solution cannot be obtained by such method? Is there a criterion for that?

答:
I had the exact same question come up in my work recently so I will provide the answer I came to. I have no references for this though so maybe you'll come across a better answer. I will speak specifically to the following domain:

Smooth Category

Linear PDEs

Let me restate the problem in a slightly different way. Let $M$ be an $m$-dimensional smooth manifold with coordinates $x^i$ and let $N \subset M$ be an embedded submanifold with inclusion $\iota: N \to M$. Let $X\in \mathcal{T}N$ be a smooth vector field. The question is:

Find a smooth function $f:M \to \mathbb{R}$ that solves $$X\left(\iota^* f\right) = 0.$$ That is, find the functions that, on $N,$ solve the PDE induced by the smooth vector field $X.$

To me, this is the most natural interpretation of your statement. Solutions can be found by first solving the PDE on $N$ for some solution $g: N \to \mathbb{R}.$ Any smooth extension of $g$ to a function $f: M \to \mathbb{R}$ solves the PDE locally on $N$ and answers the question. This is important, since it does imply that solving the PDE globally and restricting doesn't include the entire class of solutions. Let me illustrate this by way of example.

Let $M = \mathbb{R}^3$ with coordinates $x = (x^1, x^2, x^3)$ and let $N = \{ x \in M \;\colon\; x^3 = 0 \}.$ Consider the smooth vector field $$X = -x^2 \frac{\partial}{\partial x^1}+ x^1 \frac{\partial}{\partial x^2}$$ on $N.$ Pick  $$g(x^1, x^2) = (x^1)^2 + (x^2)^2 - 1$$ which satisfies the PDE $X(g)=0$ on $N.$ Now let $\lambda: M \to \mathbb{R}$ be a smooth bump function satisfying $\left.\lambda\right|_N = 0$ but non-zero everywhere else. Extend $g$ in the natural way for simplicity and define
$$
f_1(x) = \lambda(x) x^2 + (1 - \lambda(x)) g(x).
$$
Another example is
$$
f_2(x) = g(x) + (x^1)^4 (x^3)^2.
$$
I think you can convince yourself that neither $f_1$ nor $f_2$ solve the PDE off of $N.$ I'm willing to bet that the problem doesn't disappear in the real analytic category for a similar reason (how we choose to extend the solution matters).

So I would solve the PDE in a coordinate system for the subset first then consider the family of extensions that satisfy whatever other assumptions you wish to enforce (smoothness, analyticity, etc.)

Title: Solve PDE on a subspace


问:
"False" arrow in an elementary topos
In "Topoi: The Categorial Analysis of Logic" by R. Goldblatt (page 117) the "false" arrow is defined based on the comutative diagram:
$\require{AMScd}$
\begin{CD}
0 @>0_1>> 1\\
@V ! V V @VV {\rm false} V\\
1 @>>{\rm true}> \Omega
\end{CD}
Now, the arrow into the final object is unique, so it seems that $0_1$ and the arrow denoted by $!$ must be the same. But this would make ${\rm true}={\rm false}$, which makes no sense. It looks like I am missing something very basic. Where am I mistaken?

答:
The arrow ${\rm false}$ is not defined by this square merely commuting, but by it being a pullback square.  So, you are correct that $0_1$ and $!$ are the same map, and so if you replaced ${\rm false}$ with ${\rm true}$ then the diagram would still commute.  However, to conclude that ${\rm true}={\rm false}$, you would need to know the diagram remains a pullback square, not just that it commutes.

(For some context, recall that $1\stackrel{{\rm true}}\to\Omega$ has the universal property that for any objects $A$ and $B$ and any monic morphism $f:A\to B$, there is a unique morphism $g:B\to \Omega$ such that $$\begin{CD}
A @>f>> B\\
@V ! V V @VVgV\\
1 @>>{\rm true}> \Omega
\end{CD}$$ is a pullback square.  So, this universal property is being applied to the monomorphism $0_1:0\to 1$ (which is just the unique morphism from $0$ to $1$), with the unique corresponding morphism $1\to\Omega$ being named ${\rm false}$.)

Title: "False" arrow in an elementary topos


问:
Ford-Fulkerson algorithm with a little changes

Why if we don't use residual graph  in Ford-Fulkerson to finding
maximum flow give no any approximation of optimal solution?

With such a modification our algorithm do find a augmenting path and increase flow,continue until we have no augmenting path,I read many papers but I can't prove that without residual graph  we have no any approximation of optimal solution.As the result suppose $F^*$ be optimal maximum flow,i want prove that this strategy give us $\alpha F^*$ such that $\alpha$ be a number in $0<\alpha\leq 1$.Any hint will be constructive.

答:
The issue is making the "wrong" choices, in those cases the residual graph corrects them.

To understand the issue, consider the digraph with vertices $s,t,a,b$ and arcs $\overrightarrow{sa}, \overrightarrow{sb}, \overrightarrow{ab}, \overrightarrow{ta}, \overrightarrow{tb}$ of capacity $1$ each.

This graph has an obvious flow of value $2$ namely $1$ on the route $s \to a \to t$ and $1$ on the route $s \to b \to t$.

But, and here is the issue, what if you start by setting a flow of $1$ on the route $s \to a \to b \to t$. Try to increase it, and you will see that, without using the residual graph (or undoing some of the choices) you cannot.

This can be easily extrended the following way: add vertices $c,t'$ and arcs $\overrightarrow{sc}, \overrightarrow{tc}, \overrightarrow{ct'}$ of capacity $1$ and $\overrightarrow{tt'}$ of capacity $2$. Then, this netweork has a flow of 3, but the flow $s \to a \to b \to t \to c \to t'$ of value $1$ cannot be increased. And the idea of this step can be extended inductively, to create a network which admits a flow of value $n \in \mathbb N$, but which also has a flow of value 1 which cannot be increased (unless we use the residual graph/undo some choices).


问:
$A$ is a rectangular matrix, and $(v_i)_1^k$ forms a basis of the rowspace. Is $(Av_i)_1^k$ linearly independent?

$A$ is a rectangular matrix of size $m\times n$ with real entries, and $\{v_1,v_2,\ldots,v_k\}$ forms a basis of the row space of $A$. Is the set $\{Av_1,Av_2,\ldots,Av_k\}$ linearly independent?

So we have that $\{v_1,\ldots,v_k\}$ is a basis of row space of $A$. We've to check if $\{Av_1,\ldots,Av_k\}$ is a linearly independent set. I started with
$$\sum_{i=1}^k\alpha_i Av_i =0 \stackrel{??}{\implies} \alpha_i = 0$$
but couldn't conclude anything about $\alpha_i's$. I also haven't found a counterexample to the result yet, so it may be true.

When $A$ is square and invertible, I could prove the claim.

Any hints to point me in the right direction?

答:
you're working over reals so you have a nice symmetric bilinear form given by the dot product, which is positive definite.

since  $\{\mathbf v_1, \mathbf v_2,\ldots,\mathbf v_k\}$ forms a basis, it is a linearly independent set.

Suppose for contradiction

$\{A\mathbf v_1, A\mathbf v_2,\ldots,A\mathbf v_k\}$ is linearly dependent

$\implies \mathbf 0 = \sum_{r=1}^k \alpha_r A\mathbf v_r = A\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big)$

where at least some $\alpha_j\neq 0 \implies \mathbf w :=\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big) \neq \mathbf 0$ since $\mathbf v_r$ are linearly independent.,

$\implies \mathbf w \in \ker A$  but there must exist some x such that $\mathbf x^T A=\mathbf w^T$ giving

$ 0 = \mathbf x^T\mathbf 0= \mathbf x^T \big(A\mathbf w\big) = \big(\mathbf x^T A\big)\mathbf w = \mathbf w^T \mathbf w \gt 0$

which is a contradiction.

(i.e. to further expand the logic $\mathbf w^T\in \text{row space }A \iff \mathbf w\in \text{image }A^T$ which it is since it is written as a linear combination of basis vectors for such space, which means there is at least one vector in pre-image, i.e. some $\mathbf x$ satisfying $A^T\mathbf x = \mathbf w$.)


问:
Convergence in distribution. Why continuous points are the right set to look at?
We define convergence in distribution where $$ is continuous. I understand why requiring convergence at every point is too strong. I just lack intuition why do we look at the continuous points of $F$. We could have defined the convergence on a smaller set/relaxed condition. Requiring $F_n \to F$ at all continuous points is still relatively strong so what does it buy us?

答:
Well, requiring the cdfs to converge at more places is "too strong" of a condition.

As an illustrative example, consider (constant) random variables $X_n=\frac{1}{n}$. This converges almost surely to the random variable $X=0$. But if you require the distribution functions to converge for all $x\in\mathbb R$ (in particular, at the discontinuity point $x=0$) then it would no longer be true that $X_n\stackrel{\mathcal D}{\to}X$!

So under this new definition, you can no longer say a.s. convergence $\implies$ weak convergence.


问:
Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$
How to get a closed form for the following sum:

$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$

Where $F_k$ is the $k$th Fibonacci number.

I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$

答:
First let's take a look at the power series $ \sum\limits_{n\geq 0}{F_{n}^{2}x^{n}} $, it is easy to find what its radius of convergence would be using the ratio test, but for now let's denote it $ R<1 $, and let's define a function $ f:x\mapsto\sum\limits_{n=0}^{+\infty}{F_{n}^{2}x^{n}} $.

We have for all $ x\in\left(0,R\right) $ : \begin{aligned} \frac{1}{1-x}\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{F_{k}^{2}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{F_{n}F_{n+1}x^{n}}\\&=\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}-F_{n}^{2}-\left(-1\right)^{n}\right)x^{n}}\\ \frac{f\left(x\right)}{1-x}&=\frac{f\left(x\right)}{x}-f\left(x\right)-\frac{1}{1+x}\\ \iff f\left(x\right)&=\frac{x\left(1-x\right)}{\left(1+x\right)\left(x^{2}-3x+1\right)}\end{aligned}

Now : \begin{aligned} \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\left(-1\right)^{n-k}F_{k}^{2}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}\right)\\ &=\frac{x\left(1-x\right)}{\left(1+x\right)^{2}\left(x^{2}-3x+1\right)}\\ &=\frac{1-x}{5\left(x^{2}-3x+1\right)}+\frac{1}{5\left(x+1\right)}-\frac{2}{5\left(x+1\right)^{2}} \\ &=\frac{1}{5}\sum_{n=1}^{+\infty}{F_{n}^{2}x^{n-1}}+\frac{1}{5}\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}-\frac{2}{5}\sum_{n=1}^{+\infty}{n\left(-1\right)^{n}x^{n-1}}\\ &=\frac{1}{5}\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}+\left(-1\right)^{n}\left(2n+3\right)\right)x^{n}}\end{aligned}

Thus, for all $ n\in\mathbb{N} $ : $$ \sum_{k=0}^{n}{\left(-1\right)^{k}F_{k}^{2}}=\frac{\left(-1\right)^{n}F_{n+1}^{2}+2n+3}{5} $$


问:
Identify the symmetries and sketch the curve $r=\sin (\theta/2)$
I've been at this for a while and I can't think clearly so I'm definitely doing something wrong.

The question:

Identify the symmetries of the curves in Exercises 1–12. Then sketch
  the curves.

$r = \sin (\frac{\theta}{2})$

In the book it states that first to find symmetry we have to check the following:

Symmetry about x-axis

$$( r ; - \theta )$$
$$( -r ; \pi-\theta )$$

Symmetry about y-axis

$$( -r ; -\theta )$$
$$( r ; \pi-\theta )$$

Symmetry about origin

$$( -r ; \theta )$$
$$( r ;  \pi+\theta )$$

So I checked for the first one and this is what I got:
For $-\theta$:

$r = \sin (\frac{-\theta}{2})$

$-r = \sin (\frac{\theta}{2})$

Which turns out to satisfy the y-axis.

Now here's what I don't understand. I want to check for $\pi-\theta$, So I replace that where $\theta$ would be.

$r = \sin (\frac{\pi-\theta}{2})$

$r = \sin (\frac{\pi}{2} - \frac{\theta}{2})$

Following the formula for sin(A-B), then:

$r = \sin (\frac{\pi}{2})\cos(\frac{\theta}{2}) - \sin (\frac{\theta}{2})\cos(\frac{\pi}{2})$

$\cos(\frac{\pi}{2}) = 0$ and $\sin (\frac{\pi}{2}) = 1$ then:

$r = \cos(\frac{\theta}{2})$

Which obviously doesn't make sense. So I checked online and found that it should actually be $r = \sin (\pi - \frac{\theta}{2})$ and then that it would satisfy the symmetry about x-axis. But I checked it (I'm sure by now I've done something major wrong):

$r = \sin (\pi - \frac{\theta}{2})$

$r = \sin (\pi)\cos(\frac{\theta}{2}) - \sin(\frac{\theta}{2})\cos(\pi)$

$\cos(\pi) = -1$ and $\sin (\pi) = 0$ then:

$r = \sin (\frac{\theta}{2})$

Which satisfies y-axis and not x-axis.

So my question is this, why is it that wherever I look it should be
$r = \sin (\pi - \frac{\theta}{2})$ and not $r = \sin (\frac{\pi-\theta}{2})$ and what did I do wrong in the proofs and how can I make sure not to make the same mistakes again?

Thank you.

答:
To check the symmetry about both axes and origin:

About X-axis

(r,θ): r = sin(θ/2) = ± { 1 – cos (θ)/ 2 }^1/2       (Half-angle property)

(r,-θ) : r = ±   { 1 – cos (-θ) / 2 }^1/2

r = ±  { 1 – cos (θ) / 2 }^1/2

Therefore, (r,θ) = (r,-θ)

The curve is symmetrical about X-axis

About Y-axis

(r,θ) : r = sin (θ/2)

= sin (θ/2)

= ± {1 – cos (θ)/ 2 }^1/2

(-r,-θ) : -r = ± {1 – cos (-θ)/2}^1/2

-r= ±  {1 – cos (θ)/2 }^1/2

Therefore, (r,θ) ≠ (-r,-θ)

(r , π-θ ) : r = ± {1 – cos(π-θ)/2}^1/2

r = ± {1 + cos (θ) /2}^1/2

Therefore , (r,θ) ≠ (r , π-θ )

The curve isn’t  symmetrical about Y-axis

About origin

(r,θ) ; r = ± { 1 - cos (θ) /2}^1/2

(r,π +θ) ; r = ±{1 – cos (π +θ)/2}^1/2

r = ± {1 – cos (θ)/2 }^1/2

Therefore, (r,θ) = (r, π +θ)

Thus the curve is symmetrical about origin.

hope it helps


问:
In how many ways can we distribute $26$ shirts into $5$ washing machines such that at least one washing machine contains $6$ shirts or more?
My attempt:

Choose a machine in $5$ ways. Then choose six shirts out of $26$ to put in the chosen machine which can be done in $\displaystyle{\binom{26}{6}}$ ways. Distribute the rest of the shirts into five machines in $5^{20}$ ways. In all there $\displaystyle{5\binom{26}{6}5^{20}}$ such possibilities.

Now, pick any two machines out of five in $\displaystyle{\binom52}$ ways. Then, first choose six shirts out of $26$ and for those choices pick another six out of $20$. That is, $\displaystyle{\binom{26}{6}\binom{20}{6}}$ choices. Next, distribute $14$ shirts into the five machines in $5^{14}$ ways. In all, there are $\displaystyle{\binom52\binom{26}{6}\binom{20}{6}5^{14}}$ such possibilities.

Continuing this way, our third and fourth cases should be  $\displaystyle{\binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8}$ and $\displaystyle{\binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$ possibilities, respectively.

Finally, we have

$\displaystyle{5\binom{26}{6}5^{20} + \binom52\binom{26}{6}\binom{20}{6}5^{14} + \binom53\binom{26}{6}\binom{20}{6}\binom{14}{6}5^8 + \binom54\binom{26}{6}\binom{20}{6}\binom{14}{6}\binom{8}{6}5^2}$.

Please see (and comment) if the reasoning above is correct. Thanks.

Edit:

This problem in its entirety reads like this below (if that helps). I rephrased the problem cause counting the number of favorable outcomes is the difficult part of such problems:

What is the probability that, when $26$ shirts are distributed into $5$ washing machines, that at least one washing machine contains $6$ shirts or more?

答:
As mentioned, the strong pigeonhole principle says that if you have more than $kn$ objects to distribute into $n$ locations, at least one location will receive more than $k$ objects. Here $k=n=5$ and $26>5\times 5$, allowing us to be certain that there are more than $5$ shirts in some washing machine.

However to show this "the hard way", let's try sharing the shirts among washing machines. I'll assume that the shirts are indistinguishable and the machines are labelled. Then we can split the shirts between machines via the sticks-and-stones principle in $\binom{26+4}{4}$ ways, total $$\binom{30}{4} = 27405$$

Then using inclusion-exclusion, we can see how many of these ways will break the constraints of only 5 shirts per machine. Pre-allocating $6$ shirts to each machine in turn gives $5\binom{24}{4}$, then pre-allocating $6$ shirts to pairs of machines gives $\binom{5}{2}\binom{18}{4}$, then triplets of machines $\binom{5}{3}\binom{12}{4}$ and finally four machines at once with pre-broken constraint is $\binom{5}{4}\binom{6}{4}$. We can't break the constraint on all five machines at once.

So using inclusion-exclusion to remove double-counting etc., the count of constraint-breaking options is \begin{align}
\binom{5}{1}\binom{24}{4}-\binom{5}{2}\binom{18}{4}+\binom{5}{3}\binom{12}{4}&-\binom{5}{4}\binom{6}{4} \\
&= 5\cdot 10626 -10\cdot 3060 + 10\cdot 495 - 5\cdot 15\\
&= 53130-30600+4950-75 \\
&= 27405
\end{align}

Certainty.


答:
You made a couple of mistakes. Firstly, you forgot to change the limits of the integration, so your integral is actually $\displaystyle\int_1^\infty \frac{\mathrm{d}t}{t(1+t^2)}$. Furthermore, $\frac{1}{t(1+t^2)} \neq \frac{1}{t}-\frac{1}{1+t^2}$. Rather $\frac{1}{t(1+t^2)} = \frac{1}{t}-\frac{t}{1+t^2}$.

Hence your integral becomes $\displaystyle\int_1^\infty \frac{1}{t}-\frac{t}{1+t^2}\,\mathrm{d}t = \left[\log(t)-\frac{1}{2}\log(1+t^2)\right]_1^\infty = \left[\frac{1}{2}\log\left(\frac{t^2}{1+t^2}\right)\right]_1^\infty$
$$ = \frac{1}{2}\left[\log(1)-\log\left(\frac{1}{2}\right)\right] = \frac{1}{2}\log(2).$$

Title: Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$

Body:
I want to check if it's improper integral or not
$$ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}.$$
What I did so far is :

set $t=e^{x} \rightarrow \mathrm dt=e^x\mathrm dx \rightarrow \frac{\mathrm dt}{t}=dx
$ so the new integral is:
 $$ \int^{\infty}_0 \frac{\mathrm dt}{t(1+t^2)} =  \int^{\infty}_0 \frac{\mathrm dt}{t}-\frac{\mathrm dt}{1+t^{2}}$$
now how I calculate the improper integral, I need to right the $F(x)$ of this integral and then to check the limit?

Thanks!


问:
Solving the probability difference equation $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$ yields a questionable closed-form equation?
I am trying to solve the following difference equation by writing it in closed-form, which represents the probability that a person flips an even number of heads in $n$ tries, with $p$ being the probability that a person flips a head at any given time: $p_{n} = p(1 - p_{n-1}) + (1 - p)p_{n-1}$.

To derive this equation, I did a simple case analysis. There are 2 cases that $n$ flips yields an even number of heads:

odd # of heads $n - 1$ times, then a head: $(1 - p_{n - 1})p$

even # of heads $n - 1$ times, then a tail: $p_{n-1}(1 - p)$

Adding the cases up yields the difference equation I wrote above.

My strategy was to derive the characteristic equation, as this is a non-homogeneous linear equation:

$p_{n} = p - pp_{n-1} + (1 - p)p_{n-1}$

$p_{n} = (1 - 2p)p_{n-1} + p$

$x = (1 - 2p)$

$p_{n} = C(1-2p)^{n}$

We know that $p_{1} = p$:

$C(1 - 2p) = p$

$C = \frac{p}{1 - 2p}$

And plugging back into the original equation, we get:

$p_{n} = \frac{p}{1 - 2p}(1 - 2p)^n = p(1 - 2p)^{n - 1}$.

However, this does not seem to be correct, because if $p$ is $1/2$ (if it is a $1/2$ chance of getting a head on a flip), $p_{n}$ will always be 0.

答:
Your recurrence

$$p_n=(1-2p)p_{n-1}+p$$

is correct, but you seem to have lost track of the $p$ term when you solved it. For convenience let $c=1-2p$, so that the recurrence is $p_n=cp_{n-1}+p$. Let $a_n=p_n-d$ for some constant $d$ that will be determined later; the recurrence can then be written

$$a_n+d=c(a_{n-1}+d)+p\,.$$

Rewrite this as

$$a_n=ca_{n-1}+[(c-1)d+p]$$

and choose $d$ to eliminate the expression in square brackets:

$$d=\frac{p}{1-c}=\frac{p}{1-(1-2p)}=\frac12\,.$$

Then $a_n=ca_{n-1}$, and $p_0=1$ (after $0$ tosses you’re bound to have $0$ heads, an even number), so

$$a_n=c^na_0=c^n(p_0-d)=\frac12c^n$$

for each $n\ge 0$, and

$$p_n=a_n+\frac12=\frac12(c^n+1)=\frac12\big((1-2p)^n+1\big)\,.$$

When $p=\frac12$ we have $p_0=1$ (because $0^0=1$) and $p_n=\frac12$ for all $n\ge 1$, as expected. When $p=0$ we have $p_n=1$ for all $n\ge 0$, again as expected: we’ve always tossed $0$ heads. And for $p=1$ we have $p_n=1$ when $n$ is even and $p_n=0$ when $n$ is odd, yet again as expected.


问:
Why is it a submanifold or not?
I am looking at the following set $$M:=\{(x,y) \in \Bbb R^2:x^3 =y^3\}$$
I know the equivalent definition but I can't explain why this should be a submanifold or not. Can please someone gives me some basic explanation, thank you!

答:
Hint:

The set is nothing but $\{(x,y)\in\mathbb{R}^2: y=x\}$.

Notes.

Also note that in order to talk about the notion of "submanifold", you should have some manifold in mind. Although, when the context is clear, people may not explicitly mention it.

Title: Why is it a submanifold or not?


问:
Doubt in proof while showing that injective continous map sends open sets to open sets.
I was trying to prove the following statement

If $U \subset \mathbb{R}^n $ and suppose $f : U \to \mathbb{R}^n$ be a continous map which is injective. Then $f(U)$ is open in $\mathbb{R}^n$.

I considered an closed $\epsilon$-ball around a point $u \in U$, that is $$D_n=\{x \in \mathbb{R}^n:||x-u|| \leq \epsilon \} \subset U$$. Then it is clear that boundary of $D_n$, say $B=\{x \in \mathbb{R}^n:||x-u|| = \epsilon \}$ is homeomorphic to $S^{n-1} \subset \mathbb{R}^n$.

What I cannot understand is the fact that  $f(B)$ too is isomorphic to $S^{n-1}$. Remember that $f$ is given to be injective and continous, but how does that help?

答:
$B$ is compact (bounded + closed), $f(B)$ is therefore compact.
$f|_B$ is therefore continuous, injective and surjective. It's inverse is also continuous because of compacity. Hence it is the homeomorphism that I think you were looking for.

That said, I do not know how this fact can help but a sketch of a solution which uses the same line of reasoning can be found below:

Assuming $U$ is open, one can show that locally $f$ is a homeomorphism.

We can assume wlog that $0 \in U, f(0)\in f(U)$ and we only need to be looking at a small neighbourhood of $0$. As you mentioned, $\overline B(0, \varepsilon) \subset U$ for some small $\varepsilon$.

Hence $g=f|_{\overline B(0, \varepsilon)}: \overline B(0, \varepsilon) \to f(\overline B(0, \varepsilon))$ is continuous, surjective (by definition) and injective (as per the assumption). Because $\overline B(0, \varepsilon)$ is compact, $g^{-1}$ is also continuous and $g$ is therefore a homeomorphism.

Finally consider $h=g|_{B(0, \varepsilon)}\text{ } (=f|_{B(0, \varepsilon)})$. As $g$ is a homeomorphism, so is $h$ towards its image. Finally notice that any open set in $B(0, \varepsilon)$ is also open in $\mathbb{R^n}$.


问:
Upper bound for $|\mathbb E[f(X)] - f(0)|$ when $X \sim N(0,\sigma^2)$ and $f:\mathbb R \to \mathbb R$ is piecewise-linear with $\|f'\|_\infty \le 1$.
Let $f:\mathbb R \to \mathbb R$ be a piecewise linear function such that $f(0)=0$ and $\|f'\|_\infty \le 1$ and let $X$ be a Guassian random variable with mean $0$ and variance $\sigma^2$. What is an upper-bound for $|\mathbb E[f(X)]|$ ?

Example

If $f(x) = \max(x,0)$, then $\mathbb E[f(X)]=\sigma/\sqrt{2\pi} = \mathcal O(\sigma)$.

答:
There needs to be more constraints for it to have a bound independent of $f$.

Consider: $f(x) = a - x$ for some large constant $a$. Then $\mathbb{E}[f(X)] = O(a)$ (e.g., when $a >> \sigma$).

Edit: With the new constraint, doubling the bound in the example (given in the question) gives a bound.


问:
Show that if $||Y - X_j\hat{\beta}_j||^2_2 < ||Y - X_{j^*}\hat{\beta}_{j*}||^2_2$ then $X_j^TY/||X_j||_2 \geq \pm X_{j*}^TY/||X_{j*}||_2$
Suppose $Y \in \mathbb{R}^n$, $X \in \mathbb{R}^{n \times p}$, $\beta \in \mathbb{R}^p$ and $\epsilon \in \mathbb{R}^n$ with mean 0 and covariance $\sigma^2I_n$.  Assume the following relationship between $X$ and $Y$:
$$Y = X\beta + \epsilon$$

Suppose that $p=3$, and $X_j \in \mathbb{R}^n$ is the $j$th column of the matrix $X$ for $j = 1, 2, 3$. Let $\hat{\beta}_j = \left(X_j^TX_j\right)^{-1}X_j^TY$ be the least squares estimator.

I want to verify the following statement:

"If $\hat{\beta}_3$ has the lowest residual sum of error, i.e.,
$$||Y - X_3\hat{\beta}_3||^2_2 <  ||Y - X_2\hat{\beta}_2||^2_2 \text{ and } ||Y - X_3\hat{\beta}_3||^2_2 <  ||Y - X_1\hat{\beta}_1||^2_2,$$ then $$s_3X_3^TY/||X_3||_2 \geq \pm X_1^TY/||X_1||_2 \text{ and } s_3X_3^TY/||X_3||_2 \geq \pm X_2^TY/||X_2||_2,"$$
where $s_3 = sign(X_3^TY)$.

My thought process is as follows:
Since $X_j$ is a vector, then $\hat{\beta}_j = \left(X_j^TX_j\right)^{-1}X_j^TY = X_j^TY/||X_j||^2_2$. However, I'm not sure where to go from here.

Body:
Let $Z_j:=X_j^{\top}Y/\|X_j\|_2$. Then
$$
\|Y-X_j\hat{\beta}_j\|_2^2=Y^{\top}(I-X_j(X_j^{\top}X_j)^{-1}X_j^{\top})Y=\|Y\|_2^2-Z_j^2,
$$
and
$$
\|Y-X_3\hat{\beta}_3\|_2^2\le \|Y-X_j\hat{\beta}_j\|_2^2
$$
is equivalent to
$$
Z_3^2\ge Z_j^2 \quad\Leftrightarrow\quad |Z_3|\ge |Z_j|.
$$


问:
Is the "cyclotomic diagonalization" always squarefree?
For every integer $n\ge 2$ , define $$f(n):=\Phi_n(n)$$ where $\Phi(n)$ is the $n$ th cyclotomic polynomial. This can be considered as the "cyclotomic diagonalization"

Prove or disprove the conjecture that $f(n)$ is squarefree for every integer $n\ge 2$

The partial factorizations I determined with PARI/GP (usually, the factors upto $12-13$ digits are found) gave no exponent greater than $1$ upto $n=200$. A rigorous test upto $n=46$ revealed that upto this limit there are only squarefree values.

If it cannot be proven, a good heuristic would also be welcome.

答:
Checking the pairs $n,p\le 10^6$ , I found one counterexample , namely :

$$283411^2\mid \Phi_{28341}(28341)$$


问:
Distance between two partitions with two elements in {1,...,m} satisfies triangle inequality
For $z, z' \in \{ -1,1 \}^m$, define $d(z,z') = \min_{\delta \in \{ -1,1\}}\sum_{i=1}^m \mathbb{1}_{z_i \neq \delta z'_i}$. This is kind of like a distance between two partitions of $\{1,\ldots,m\}$ of cardinal $2$.

Does $d$ satisfy the triangular inequality ?

答:
After a bit more thinking I found that the answer is yes by writing $d(z,z') = \min_{\delta \in \{ -1,1\}}\frac{1}{2}\sum_{i=1}^m |z_i - \delta z'_i|$ and then consider all the $8$ possible values the three $\delta$ appearing in the inequality can take, which we can name from left to right $ \delta_1, \delta_2, \delta_3$, and use the regular triangle inequality in $\mathbb{R}$.

For example, if $(\delta_1, \delta_2, \delta_3) = (1,1,-1)$, then :

$d(a,c) = \frac{1}{2}\sum_{i=1}^m|a_i - c_i| \leq \frac{1}{2}\sum_{i=1}^m|a_i+c_i| = \frac{1}{2}\sum_{i=1}^m|a_i - b_i + b_i + c_i| \leq \frac{1}{2}\sum_{i=1}^m|a_i - b_i| + \frac{1}{2}\sum_{i=1}^m|b_i + c_i| = d(a,b) + d(b,c)$

I wonder if there is another proof though.


问:
$\mathbb{Q} \times \mathbb{R}$ is complete metric space by using $d(a,b) = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2} $metric
Can someone solve it? Same for $\mathbb{Q} \times\mathbb{Q}$ in
$d(a,b) = \sqrt{(b_1-a_1)^2+(b_2-a_2)^2}$ metric

答:
I don't think this is true. For example consider the Cauchy sequence

$$\left\{(3,0), (3.1, 0), (3.14, 0), (3.141, 0), \dots \right\} $$

where each step adds another digit of $\pi$. This is a Cauchy sequence that will converge to $(\pi, 0)$, which is not in $\mathbb{Q} \times \mathbb{R}$.


问:
Determine whether the improper integral $\int^{\infty}_{0} x/(1+x^3)$ is convergent or divergent.
I've tried comparing this integral, but I didn't come to an answer, since the improper integral is smaller equal than the integral 1/x^2. Does anyone see how I can compare this to another integral or how I can say whether the integral is convergent or divergent?</p>

答: Since$$\lim_{x\to\infty}\frac{\frac x{x^3+1}}{\frac1{x^2+1}}=1$$and since he integral$$\int_0^\infty\frac1{x^2+1}\,\mathrm dx$$converges, then so does your integral.


问:
Ratio of segments formed by intersection of three cevians
ABC is a triangle and P some point inside it. Cevians AP, BP, CP cross the opposite sides at A',B' and C' respectively.

What is the minimum value the following expression can accomplish:

(1+2*PA'/PA)(1+2*PB'/PB)(1+2*PC'/PC) ?

The solution is 8, when each ratio is 1/2 (meaning that is a centroid) but I'm having trouble reaching that.

Now from the ratio of areas APB, BPC, CPA and the whole triangle ABC we get respectively:

Pa/P = A'P/A'A, Pb/P = B'P/B'B, Pa/P = C'P/C'C, and summing this i get

1 = PA'/(PA' + PA) + PB'/(PB' + PB) + PC'/(PC' + PC)

Does from here directly follows some relation to the wanted expression fully algebraically, and if so how? Or should I still consider them as being in the triangle?

Replacing PA'/PA=x PB'/PB=y PC'/PC=z I get

1 = 1/(1 + 1/x) + 1/(1 + 1/y) + 1/(1 + 1/z), and the wanted expression as (1+2x)(1+2y)(1+2z)

I also tried:

4 = 1 + x/(1+x) + 1 + y/(1+y) + 1 + z/(1+z)

4 = (1+2x)/(1+x) + (1+2y)/(1+y) + (1+2z)/(1+z)

答:
Use the substitution $ a = \frac{1}{ 1 + \frac{1}{x} }$, or that $ x = \frac{a}{1-a}$, and $ 1 + 2x = \frac{ 1+a}{1-a}$.

Rephrasing what you already have

WTS $ \prod \frac{ 1 + a } { 1 - a } \geq 8 $ subject to $ a + b + c = 1$.

This is much easier to work with, give it a try.

 Jensen's on the product: Show that $ \ln \frac{ 1+a}{1-a}$ is convex, so $\prod \frac{1+a}{1-a} \geq ( \frac{ 1 + 1/3 } { 1 - 1/3} ) ^3 = 8 $.

 AM-GM: substituting in $ a+b+c = 1$,  $\prod \frac{ (a+b) + (c+a) } { (b+c)}  \geq \prod \frac{ 2 \sqrt{ (a+b)(c+a) }} { (b+c) } = 8.$


问:
Computing the moments of a triangle
Assume you have a triangle in the plane defined by its three vertices at $(x_0,y_0)$, $(x_1,y_1)$, and $(x_2,y_2)$. Is there a general expression for the moments of the triangle, where the moments are defined as averages by the following expression:

$$\langle x^n y^m \rangle = \frac{1}{A}\int_\triangle x^n y^m \, dA $$

Integrating manually to compute the first few moments, one can obtain:

$$\langle x \rangle = \frac{1}{3} \left(x_0+x_1+x_2\right) $$

$$\langle x^2 \rangle = \frac{1}{6} \left(x_0^2+x_1^2+x_2^2+x_0 x_1+x_0 x_2+x_1 x_2\right) $$

$$\langle x y \rangle = \frac{1}{12}\left(2 x_0 y_0+x_1 y_0+x_2 y_0+x_0 y_1+2 x_1 y_1+x_2 y_1+x_0 y_2+x_1 y_2+2 x_2 y_2 \right)$$

Is it possible to obtain a closed form expression for general $n$ and $m$ using combinations of the coordinates?

EDIT:

Can the same be done for line segments defined by their endpoints $(x_0,y_0)$ and $(x_1,y_1)$?
Again the first few moments in this case are:

$$\langle x \rangle = \frac{1}{2} \left(x_0+x_1\right) $$
$$ \langle x^2 \rangle = \frac{1}{3} \left(x_0^2+x_0 x_1+x_1^2\right) $$
$$ \langle x^3 \rangle = \frac{1}{4} \left(x_0^3+x_0^2 x_1+x_0 x_1^2+x_1^3\right) $$
$$ \langle x y \rangle = \frac{1}{6} \left(2 x_0 y_0+x_1 y_0+x_0 y_1+2 x_1 y_1\right) $$
$$ \langle x^2 y \rangle  = \frac{1}{12} \left( 3 x_0^2 y_0+2 x_0 x_1 y_0+x_1^2 y_0+x_0^2 y_1+2 x_0 x_1 y_1+3 x_1^2 y_1\right)$$

答:
It should be possible using Green's Theorem.

Refer my earlier answer here for more details: Why doesn't a simple mean give the position of a centroid in a polygon?

ContentLicense: CC BY-SA 2.5

Title: Computing the moments of a triangle


问:
How to factor quadratic $ax^2+bx+c$?
How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$

The answer is
 $$(x+2)(x-1)$$

I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.

Please help me out; thanks!

答: $(x+a)(x+b)=x^2+(a+b)x+ab$, so if you factor the constant term you can find possibilities and look to see if the middle term works out.  In your case, $-2=-2*1=-1*2$, so you can look to see if $-2+1$ or $-1+2$ add to $+1$.  This leads you to $(x+2)(x-1)$.  If the $x^2$ term has a constant multiplying it, you have to factor it and match things up to look for possibilities.</p>

Title: How to factor quadratic $ax^2+bx+c$?


问:
Improper integrals Question
I was asked to define the next intergrals and I want to know if I did it right:

$$1) \int^\infty_a f(x)dx = \lim_{b \to \infty}\int^b_af(x)dx$$
$$2) \int^b_{-\infty} f(x)dx = \lim_{a \to -\infty}\int^b_af(x)dx$$
$$3) \int^\infty_{-\infty} f(x)dx = \lim_{b \to \infty}\int^b_0f(x)dx + \lim_{a \to -\infty}\int^0_af(x)dx$$
Thanks.

答:
The first two definitions you gave are the standard definitions, for $f$ say continuous everywhere. The third is more problematical, It is quite possible that the definition in your course is
$$\lim_{a\to-\infty, b\to \infty} \int_a^b f(x)\,dx.$$
So $a\to-\infty$, $b\to\infty$ independently.

What you wrote down would then be a fact rather than a definition.

Title: Improper integrals Question


问:
What is the domain of $\frac{1}{\log x}$?
If you ask to me what  the domain of $\frac{1}{\log(x)}$ is, I set
$$
\log(x)\neq0 \Rightarrow x\neq1 $$ infact $\frac{1}{0}\to\infty$ but moreover (before this exercize) I would have set $x>0$ too because $\log(0)\to-\infty$. But in this case $\frac{1}{\log(0)}=\frac{1}{-\infty}\to0$ and the function is define for $x=0$ too.

Is this a rule? I have to put log arguments $\neq0$ ONLY if logs are at numerator? There are other rules like this?

[edit]

I got confused, the correct example is: $ \frac{1}{\log(|x|)}$ and so I should have put $|x|>0 => x\neq0$ in this case does something change?

答:
Don't over think it.

$\log x$ can only have positive input and int can have any positive input.  So for $\log x$ to be defined $x$ must be positive and $x$ can be any positive.

For $\frac 1{M}$ the $M$ can be anything except $0$.  $M \in 0$.  so for $\frac 1{\log x}$ to be defined $\log x\ne 0$ but otherwise $\log x$ may be any value.

So $\log x \ne 0$ and $x \ne =1$.

So $x$ must be positive and can not be $1$ but $x$ can be any other positive value.

So the natural domain of $\frac 1{\log x}$ is all positive values except $1$.

We may write that as $(0,1)\cup (1, \infty)$ or $\mathbb R^+\setminus\{1\}$ or many other ways.


问:
Hints for two proofs
Please do not give me the answers. I only want hints to approach these two proof problems I'm struggling with:

There exists some differentiable function $f(x)$ such that $f'(x)\ne f(x)$ but $f''(x)=f(x)$

There exists some $x\in \mathbb R$ such that $x^2\in \mathbb \{R -\mathbb Q\}$ while $x^4\in \mathbb R$

For the first one I started out defining $f'$ as a limit but I couldn't come up with any way to utilize this for my proof. For the second one I've tried doing it by cases, considering $x<0, x=0, x>0$. But I didn't get anywhere doing this.

答:
Hint.

Solve the differential equation $y''=y$ and choose those such that $y'\ne y$.

Play with $\sqrt{2}$.

Title: Hints for two proofs


问:
Ring of endomorphism
Let $M$ be a simple $R$-module. My goal is to show that End$_R(M)$ is a field. I have proven a version of Schur's Lemma. That is, I have proven to myself that a homomorphism $\phi: M \rightarrow M'$ of simple $R$ -modules is either an isomorphism or the zero-morphism.

This version of Schur's Lemma immediately implies that End$_R(M)$ is a division ring. My problem arises with proving that End$_R(M)$ is commutative. I don't immediately see why End$_R(M)$ is commutative, and I was not given any information about $R$.

答:
With no further hypotheses this is not true; Schur's lemma tells you that $\text{End}_R(M)$ is a division ring (and slightly more precisely it's a division algebra over $Z(R)$) and that's all. Moreover every noncommutative division ring occurs, e.g. if $R$ itself is a division ring $D$ and $M = D$ is the regular representation, in which case $\text{End}_D(D) \cong D^{op}$.

It is true if $R$ is commutative; for this you can show that we have $M \cong R/m$ where $m$ is a maximal ideal, and then show that $\text{End}_R(R/m) \cong R/m$. It is worth thinking in some detail about the case that $R = k[x]$ for $k$ a field and $m = (f(x))$ is the maximal ideal generated by an irreducible polynomial.

Title: Ring of endomorphism


答:
You're exactly right that $\text{End}_R(M)$ is (in general) only a division ring. However, we can use the fact that $R$ is commutative to learn more.

In particular, if $M$ is a simple $R$ module, then $M \cong R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$.

But can you show that every endomorphism of $R / I$ looks like multiplication by $r$ for some $r \in R$? (Hint: if $\varphi$ is your endomorphism, look at $\varphi(1)$).

Then there is a surjection $R \to \text{End}_R(M)$. What is its kernel? Do you see why this shows $\text{End}_R(M)$ is a field?

I hope this helps ^_^

Title: Ring of endomorphism

Body:
Let $M$ be a simple $R$-module. My goal is to show that End$_R(M)$ is a field. I have proven a version of Schur's Lemma. That is, I have proven to myself that a homomorphism $\phi: M \rightarrow M'$ of simple $R$ -modules is either an isomorphism or the zero-morphism.

This version of Schur's Lemma immediately implies that End$_R(M)$ is a division ring. My problem arises with proving that End$_R(M)$ is commutative. I don't immediately see why End$_R(M)$ is commutative, and I was not given any information about $R$.


问:
When do we have orthogonal bases in a vector space equipped with a skew-symmetric form?
Suppose we have a vector space $V$ over some field $\mathbb F$ together with a skew-symmetric form $\langle\space ,\space \rangle$, i.e., $\langle\space ,\space \rangle$ is bilinear and $\langle v,v\rangle=0$ for all $v\in V$. Then when do we have an orthogonal bases?

Unlike the symmetric form over a field with $\mathbb F$ with $\text{char}(\mathbb F)\ne 2$ or Hermitian form over the complex field $\mathbb C$, $\langle v,v\rangle=0$ for all $v\in V$, so we can not find orthogonal bases like what we did for the above two cases.

Are there classifications for vector spaces equipped with a skew-symmetric form such that there exist orthogonal bases? Thank you!

答:
To expand on Qiaochu's comment, suppose we have a basis $(v_1, \dots, v_k)$ for $V$ such that $\langle v_i, v_j \rangle = 0$ for all $i \neq j$. Well, for $i = j$, we also have $\langle v_i, v_j \rangle = 0$ by the alternating hypothesis. Now, in this basis, the matrix representing the bilinear form is the zero matrix! Thus, we cannot have such a basis unless the bilinear form is $0$.

Edit: Qiaochu edited his comment with more details, so nevermind :)


问:
Prove that $g(x)=(x^2−1)^x$ is increasing on $(1,+\infty)$
Let $A = \mathbb{R}\setminus[−1,1]$. Let $g : A\to\mathbb{R}$ be defined by
$g(x)=(x^2-1)^x$
for all $x\in A$.

Prove that $g(x)=(x^2-1)^x$ is increasing on $(1,\infty)$

I have currently attempted to prove this by showing $g'(x)\geq 0 $ for all $ x\in A $ which gives $g'(x)= (x^2-1)^{x-1}(2x^2+(x^2-1)ln(x^2-1))$ which should be greater than or equal to $0$ however I am unsure how to show this or whether I have gone about this the right way.

Any Help will be grateful.

答:
$$g(x)=\left(x^2-1\right)^x$$
$g(x)$ is increasing if $g'(x)>0$

$$g'(x)=\left(x^2-1\right)^x \left(\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)\right)$$
$g'(x)>0$ if
$$\frac{2 x^2}{x^2-1}+\log \left(x^2-1\right)>0$$
to simplify set $x^2=z+1$
$$\frac{2z+2}{z}+\log z>0\to h(z)=2+\frac{2}{z}+\log z>0$$
This is true for any $z$, because minimum value of $h(z)$ is positive

Indeed $h'(z)=-\frac{2}{z^2}+\frac{1}{z}=0\to z=2$

$h''(z)=\frac{4}{z^3}-\frac{1}{z^2}$

$z=2$ is a minimum because $h''(2)=\frac{4}{8}-\frac{1}{4}>0$ (second derivative test)

$h(2)=3+\log 2>0$, therefore $g'(x)>0$ for any $x$.


答:
Hint

$1.$ Think about $\sin \alpha$;

$2.$ Think about $x^2=\sqrt{3}$.

Title: Hints for two proofs

Body:
Please do not give me the answers. I only want hints to approach these two proof problems I'm struggling with:

There exists some differentiable function $f(x)$ such that $f'(x)\ne f(x)$ but $f''(x)=f(x)$

There exists some $x\in \mathbb R$ such that $x^2\in \mathbb \{R -\mathbb Q\}$ while $x^4\in \mathbb R$

For the first one I started out defining $f'$ as a limit but I couldn't come up with any way to utilize this for my proof. For the second one I've tried doing it by cases, considering $x<0, x=0, x>0$. But I didn't get anywhere doing this.


问:
Finding the last two digits of $3^{400}$
I'm having some trouble with the following exercise:

Find the last two digits of the decimal representation of $3^{400}$.

This exercise Is under the Euler's theorem and Fermat's little theorem section of the worksheet so I think that we are supposed to use those to solve this. My problem is that this is the first time I'm seeing an exercise of this kind and I don't know how to proceed. Can someone give me a tip on how to do this or show me how it's done?

答:
It's a simple application of the totient theorem.

Applying it we get that

$$3^{\phi(100)}\equiv 1\pmod {100} $$ i.e.

$$3^{40}\equiv 1\pmod {100} $$

That's because $GCD(3, 100) = 1$ so we can apply the theorem.

Going on from there is pretty much easy.

Just raise the last congruence to the power of $10$ and you get

$$3^{400}\equiv 1\pmod {100} $$

So the last two digits are $0$ and $1$.


问:
For a $\sigma$-finite measure $\mu$, show that for $f$<$\infty$, non negative and measurable, $f\cdot \mu$ is $\sigma$-finite
Let ($\Omega,A,\mu$) be the $\sigma$-finite measure space in question.
If $B_1\subset B_2\subset B_3\dots$ is a nested sequence so that $\sum_{n=1}^{\infty}$$B_n$=$\Omega$ and $\mu(B_n)<\infty$.
My attempt has been the following:  $\phi_n\nearrow f$, with $\phi_n$ simple functions. Now on each set $B_n$,$\phi_n$ is bounded by some $M$>0. So $\int_{B_n}\phi_n d\mu$<$M\cdot \mu(B_n)$<$\infty$.
I was not able to prove this for measurable, non negative functions in general yet.
, because I do not know how the claim of boundedness on specific sets would translate to $f$.

答:
You mixed up indices when you wrote the inequality:
$$
\int_{B_n} \phi_n \mathrm d\mu < M \mu(B_n)
$$
You should use two different set of indices, one for the sequence $\phi_n$ and one for the sequence $B_n$. Let's use $m$ for the latter, then we have:
$$
\int_{B_m} \phi_n \mathrm d\mu < M \mu(B_m)
$$
Now, since $f$ can be unbounded on $B_m$, we must get clever. We have to slice up $B_m$ in a sequence $B_m^k$ of sets defined as
$$
B_m^k = f^{-1}([0, k)) \cap B_m = \{f < k\} \cap B_m
$$
And the bundle them up again like so
$$
B'_k = \bigcup_{\ell,m \leq k} B_m^\ell
$$
Notice that $B_k'$ still exhibits $\mu$ as a $\sigma$-finite measure but now $f$ is bounded on $B_k'$.

Now set $M_k = \sup_{B_k'} f$ and finish up by passing
$$
\int_{B_k'} \phi_n \mathrm d\mu < M_k \mu(B_k')
$$
through the monotone convergence theorem, so you can bring the limit for $n \to \infty$ (notice: $n$, not $k$) inside the integral.


问:
Why is hash function $h$ ($h(w_1 \oplus w_2) = h(w_1) \oplus h(w_2)$) not good?
Suppose $h$ is a hash function, $h \colon \{ 0, 1 \}^* \rightarrow \{ 0, 1 \} ^n$ and for all $w_1$, $w_2$ it holds: $$h(w_1 \oplus w_2) =
h(w_1) \oplus h(w_2).$$

$\oplus$ is the XOR operation.

Why isn't $h$ a cryptographically good hash function?

答:
There are some other interesting properties of such a hash function.

$h(\{0\}^t) = 0$ for any $t$; let $w_2$ be be a string of lenght $t$, then

\begin{align}
h(\{0\}^t \oplus w_2) &= h(\{0\}^t) \oplus h(w_2)\\
h(w_2) &= h(\{0\}^t) \oplus h(w_2)\\
0  &= h(\{0\}^t)
\end{align}

simple collision exists.

Generic collision resistance has $\mathcal{O}(2^{n/2})$-time with 50% probability by the birthday attack. What described in the comment and in the aswer for finding collisions is far more costly than the generic collision attack. It has $\mathcal{O}(2^n)$-time.

Direct collisions exist; let $w_1$ and $w_2$ be two random $t$ length strings then any bit flipping on both strings will have the same hash. Let represent the bit flipping at the position $a$ as $w_1^a$. Then

$$h(w_1^a) \oplus h(w_2^a) =h(w_1^a \oplus w_2^a) = h(w_1 \oplus w_2) = h(w_1) \oplus h(w_2).$$

This provides a tremendous number of collisions. For Cryptographical signatures, this is a complete failure to be a good cryptographic hash function.