答:
It is easier to set up your integral as $dz \, dx \, dy$ given you have one of the equations in just $x, y$.

Here is how you can see it - if you take infinitely thin vertical strips from down to up between $y = 0$ to the plane, you realize you hit upon $x = 2y$ till $x = 1$ and then on it is the other plane $x + 2y + z = 2$. So you will have to break your integral into two.

But if you take horizontal strips, it is between two planes throughout your region finally both planes meeting at $y = \frac{1}{2}$.

$x = 2y, x + 2y + z = 2, y = 0, z = 0$

You rightly set the bounds for $z$.

Now set $z = 0$ to get the upper bound of $x, y$.

So $x = 2y, x + 2y = 2 - 2y$.

That gives you bounds of $2y \leq x \leq 2-2y$

You already know $x$ varies between $0$ and $\frac{1}{2}$ based on the intersection and your graph.

So your integral becomes -

$\displaystyle \int_{0}^{1/2} \int_{2y}^{2-2y} \int_{0}^{2-2y-x} \, dz \, dx \, dy$

You could have also set up as

$\displaystyle \int_{0}^{1} \int_{0}^{x/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx +
\int_{1}^{2} \int_{0}^{(2-x)/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx$

Title: Triple integral/tetrahedron

Body:
Let the Tetrahedron be bounded by the planes

$x+2y+z=2$, $x=2y$, $y=0$, $z=0$

so the limits for z are easy and would be $0 \leq z \leq  2-x-2y$
I have calculated the intersection point in the $(x,y)$ plane and I get $(1, \frac{1}{2})$.
So $y=0$ means that the lower boundary of $y$ should start with zero. I drew a sketch, but I am unsure about the x limits and y limits.

what would the functions $h_1$ and $h_2$ be?

$\int_{0}^{1} \int_{h_1(y)}^{h_2(y)} \int_{0}^{2-2y-x} dzdxdy$

on the other hand, if we had $x=0$ it would be easy to write because I could write it as
$\int_{0}^{1} \int_{x/2}^{1-x/2} \int_{0}^{2-2y-x} dzdydx$

but since we have $y=0$ I need to determine the $h_1(x)$, $h_2(x)$, where I am confused how to do it

]1


问:
Balancing chemical equations - Matrix determinant is zero for specific reaction
I am currently working on a program which takes a chemical equation as input and returns the balanced chemical equation instead. It finds the appropiate chemical coefficients with matrix operations, using the formula $x = A^{-1}b$. I am using JAMA for calculations involving matrices. According to their docs, it calculates $A$'s inverse using

Eigenvalue Decomposition of both symmetric and nonsymmetric square matrices

Let's take an example to better visualize this. Suppose the following chemical equation with denoted coefficients for every compound:

$aCa(OH)_2 + bH_3PO_4 \rightarrow cCa_3(PO_4)_2 + dH_2O$

The computer creates an equation for every element:

$P: b - 2c = 0$

$H: 2a + 3b - 2d = 0$

$Ca: a - 3c = 0$

The last equation needs to be $a = det(A)$ for three reasons. We need a square matrix (and that's why oxygen's equation is omitted) to compute it's inverse and determinant. Secondly, we need to give a numerical value to one of the variables to be able to get a numerical result, and thirdly (as far as I understood because I learned nothing about matrices in school yet) $adj(A)$ is multiplied by $1/det(A)$ to get $A^{-1}$ which causes the computed coefficients to sometimes be non-integer numbers. I am not sure of how or why setting $a$ to $det(A)$ works, neither this is the most efficient way of doing it.

$A = \left[\begin{array}{cccc}0&1&-2&0\\2&3&0&-2\\1&0&-3&0\\1&0&0&0\end{array}\right]$
$b = \left[\begin{array}{c}0\\0\\0\\det(A)\end{array}\right]$

After dividing every number to the GCD the computed vector $x$ will be:

$x = A^{-1}b = \left[\begin{array}{c}6\\4\\2\\12\end{array}\right]$ = $\left[\begin{array}{c}3\\2\\1\\6\end{array}\right]$

And of course, the balanced chemical equation is

$3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O$

This program works perfectly for even more complex equations:

$299H_2SO_4 + 10K_4Fe(CN)_6 + 122KMnO_4 \rightarrow 60CO_2 + 5Fe_2(SO_4)_3 + 188H_2O + 60HNO_3 + 162KHSO_4 + 122MnSO_4$

$9Fe_{36}Si_5 + 836H_3PO_4 + 192K_2Cr_2O_7 \rightarrow 324FePO_4 + 45SiO_2 + 128K_3PO_4 + 384CrPO_4 + 1254H_2O$

However, in the process of testing, I found a chemical equation that produces the following error: java.lang.RuntimeException: Matrix is singular.

Let's write everything down:

$aB_{10}H_{12}CNH_3 + bNiCl_2 + cNaOH \rightarrow dNa_4(B_{10}H_{10}CNH_2)_2Ni + eNaCl + fH_2O$

$B: 10a - 20d = 0$

$C: a - 2d = 0$

$Na: c - 4d - e = 0$

$H: 15a + c - 24d - 2f = 0$

$Cl: 2b - e = 0$

This time, the code omits the equation for three elements ($N, Ni, O$) because it would result in an overdetermined system and a non-square matrix.

$A = \left[\begin{array}{cccccc}10&0&0&-20&0&0\\1&0&0&-2&0&0\\0&0&1&-4&-1&0\\15&0&1&-24&0&-2\\0&2&0&0&-1&0\\1&0&0&0&0&0\end{array}\right]$

This is completely unexpected, both for me and the script. Using a matrix calculator, I found out that $A$ has no inverse: its determinant is zero. I analyzed the matrix and found that $10a - 20d = 0$ and $a - 2d = 0$ are really
the same equation. I'm not sure if this has something to do with the matrix being singular.

What should I do to make this work? Should I use a different method for finding solutions and representing my system of equations, or do I have to do something specific when $det(A) = 0$ ? I've read a little bit about Gaussian Elimination and LU Decomposition, but I don't seem to understand them very well. In case I have to use another method for solving the system, which one is the most suitable for this script? Also, I would be happy to get some details about it if possible. Keep in mind I'm a beginner in this matrix field. Any help is appreciated!

答:
Not all matrices are invertible.

If your mathematical system of equations is underdeterminated or overdeterminated, your chemical equation is chemically wrong and you have to correct it.

To check if your chemical equation is correct, you have to check the rank of your matrix.

You should use an algorithm that yields the rank of your matrix, e.g. Gaussian elimination.

You also could include electric charges in your chemical equations.

ContentLicense: CC BY-SA 4.0


问:
Given $f(x) = \frac{x^3}{3} - \sqrt{x^2+1}$, proof that $f(\alpha) = \frac{\alpha^4 - 3}{3\alpha}$
I evaluated the function on $\alpha$ and removed the radical by multiplying by $\sqrt{\alpha^2+1}$ as follows:

$$ \frac{\alpha}{3} - \frac{\alpha^2 + 1}{\sqrt{\alpha^2+1}} $$

After some algebraic manipulations, I got to this step:

$$ \frac{\alpha^4 \sqrt{1 + 1/\alpha^2}}{3\alpha\sqrt{1 + 1/\alpha^2}} - \alpha(1 + 1/\alpha^2)^{-1/2} $$

Does anyone knows if I am on the good track? or is there any hints that could help me? Thanks

Body:
$$\frac{x^3}{3}-\sqrt{x^2+1}=\frac{x^4-3}{3 x}$$
$$\frac{x^3}{3}-\frac{x^4-3}{3 x}=\sqrt{x^2+1}$$
$$\frac{1}{x^2}=x^2+1$$
Substitute $x^2=w$
$$\frac{1}{w}=w+1\to w^2+w-1=0\to w=\frac{1}{2}\left(\sqrt{5}-1\right)$$
the root $w_2=\frac{1}{2}\left(-\sqrt{5}-1\right)$ is discarded because is negative. The solution is:

$$x=\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)}$$


问:
Relationship expressed by exponentiation
I have a basic question, but, i don't really remember how to solve this kind of problem. I know the result must be $a = b$, but I don't know how to get there.

"Determine the relationship between a and b where a and b are natural
numbers that express the digit numbers of $x = 4 ^ {12} * 5 ^ {20}$ and $y =
 4 ^ {14} * 5 ^ {18}$ respectively."

Body:
Note $4=2^2 \Rightarrow 4^n=2^{2n}$ and $2^m \cdot 5^m = (2\cdot 5)^m=10^m$

So
$$x=4^{12}\cdot 5^{20}=2^{24} \cdot 5^{20}$$
$$ =2^{4+20}\cdot 5^{20}=2^{4} \cdot 2^{20}\cdot 5^{20}$$
$$ =2^{4} \cdot 10^{20}=16\cdot 10^{20}$$

Thus $x$ has $2+20=22$ digits.

Similarly show $$y=2^{10}\cdot10^{18}=1024\cdot10^{18}$$

Then $y$ has $4+18=22$ digits.


答:
Let $(x,y)$ be the tangential points for the pair of the tangent lines from the point $(h,k)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Then ,
$$y’=-\frac{b^2x}{a^2y} = \frac{y-k}{x-h}\implies
\frac{xh }{a^2}+\frac{yk }{b^2}=1$$
Substitute it into $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and obtain the products below via the Vieta’s formula
$$x_1 x_2 = \frac{a^4(b^2-k^2)}{a^2k^2+b^2h^2},\>\>\>\>\>y_1 y_2 = \frac{b^4(a^2-h^2)}{a^2k^2+b^2h^2}$$
Then, the condition ${x_1}{x_2} + {y_1}{y_2} = 0$ for the right angle subtended by the two tangential points leads to

$$\frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$$

Title: Chord of contact from point $(h, k) $ subtending right angle at centre of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Body:
Prove that the chord of contact of tangents drawn from the point (h, k) to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  subtends a right angle at the centre, if

$$\frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}+\frac{1}{b^2}$$

I want to approach this problem through my method

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ be the equation of the ellipse at $\left( {{x_1},{y_1}} \right)$.

${m_1} = \frac{{{y_1}}}{{{x_1}}} \Rightarrow m{'_1} =  - \frac{{{x_1}}}{{{y_1}}}$

${m_2} = \frac{{{y_2}}}{{{x_2}}} \Rightarrow m{'_1} =  - \frac{{{x_2}}}{{{y_2}}}$

${x_1}{x_2} + {y_1}{y_2} = 0$ is the condition

$k =  - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{x_1}^2}}{{{y_1}^2}} + {b^2}}  \Rightarrow k =  - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{y_2}^2}}{{{x_2}^2}} + {b^2}}  \Rightarrow k =  - \frac{{{x_1}}}{{{y_1}}}h + \sqrt {\frac{{{a^2}{y_2}^2 + {b^2}{x_2}^2}}{{{x_2}^2}}} $

$ \Rightarrow k =  - \frac{{{x_1}}}{{{y_1}}}h + \frac{{ab}}{{{x_2}}}\& k =  - \frac{{{x_2}}}{{{y_2}}}h + \frac{{ab}}{{{x_1}}}$

On multiplying we get $\Rightarrow {k^2} = \frac{{{x_1}}}{{{y_1}}}\frac{{{x_2}}}{{{y_2}}}{h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} + ab\left( {\frac{1}{{{x_1}}} + \frac{1}{{{x_1}}}} \right)$

$\Rightarrow {k^2} = \frac{{{x_1}}}{{{y_1}}}\frac{{{x_2}}}{{{y_2}}}{h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} - abh\left( {\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}}} \right)$

$ \Rightarrow {k^2} =  - {h^2} + \frac{{{a^2}{b^2}}}{{{x_1}{x_2}}} - abh\left( {\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}}} \right)$

How do we proceed form here


问:
Eilenberg-MacLane spaces with $G$ an abelian torsion group
let $G$ be an abelian torsion group. let $z\in H_i(K(G,1))$. throughout the coefficient group is $\mathbb Q$.

1) Why there exists a FINITE subcomplex $X$, such that $j:X\hookrightarrow K(G,1),\, j_*(x)=z$ for some
$x\in H_i(X)$ ?

2) Why the image of $\pi_1(X)$ in $\pi_1(K(G,1))=G$ is finite?

MY guess:

1) $G=\oplus_p G_p$ where $G_p$ ranges over all $p$-subgroups of $G$
so by Kunneth $$H_i(K(G,1))=\oplus_{i_1+\cdots+i_l=i}H_{i_1}(K(G_{p_1},1) \otimes \cdots\otimes H_{i_l}(K(G_{p_l},1 ) $$
and I don't know what does it mean for $z$ to be in this decomposition!!

2) if $X$ is a finite CW that implies that its $\pi_1$ is finitely generated and not finite !!! and what finite CW means here anyway: having a finite number of cells at each dimension or having no cells beyond a dimension $n$ so that $X$ equals its $n$-skeleton?

答:
Pick the usual simplicial model for $K(G,1)$. If $x\in H_i(K(G,1))$, then $x$ is the class of a cycle which involves finitely many simplices, which in turn involve finitely many group elements. Let $H$ be the subgroup generated by all these. Then $x$ is in the image of $$H_i(K(H,1))\to H_i(K(G,1)).$$ Moreover, if $X$ is the $n$-skeleton of $K(H,1)$ with $n>i+1$, then $x$ is in the image of the composition $$H_i(X)\to H_i(K(H,1))\to H_i(K(G,1)).$$

Since $H$ is finitely generated and torsion, it is finite, so $X$ is finite.

ContentLicense: CC BY-SA 3.0


问:
Having trouble with composite functions, inverse functions. Have an example question from upcoming test.
Here is a look at what I am doing. The I can evaluate the first one no problem. The second one is causing me problems and the solution given isn't helping me to get my head around it. Can someone please try and give me the steps and an explanation? Thanks

Body:
If $y=f_1(x)=3x+2$, then $y-2=3x$ and $x=\frac13(y-2)$, so $f^{-1}(y)=\frac13(y-2).$

To calculate $f_1^{-1}(f_2(1.9))$, first calculate $f_2(1.9)=4\times1.9^3=27.436$.

Now calculate $f_1^{-1}(f_2(1.9))=f_1^{-1}(27.436)=\frac13(27.436-2)\approx8.479.$


答:
For the second problem, you are given
\begin{align}
f_{1}(x) &=3x+2 \\
f_{2}(x) &=4x^3
\end{align}
and you are asked to find $$(f_{1}^{-1}\circ f_{2})(1.9).$$  To do this, you will need to first find the inverse function of $f_{1}(x)$.  To do that, we can take the equation $y = 3x+2$ and solve for $y$:
\begin{align}
y &= 3x + 2\\
y - 2 &=3x\\
\frac{1}{3}(y - 2) &= x.
\end{align}
This tells you that the inverse function is given by the equation $$f_{1}^{-1}(y) = \frac{1}{3}(y-2),$$
or to keep the input notation the same with all your functions, we can write this as $$f_{1}^{-1}(x) = \frac{1}{3}(x-2)$$ as they did in the solution.  Then, to evaluate \begin{align}
(f_{1}^{-1}\circ f_{2})(1.9) &= f_{1}^{-1}(f_{2}(1.9)).
\end{align}
This means we first need to evaluate $f_{2}(1.9)$.  This gives us $$f_{2}(1.9) = 4(1.9)^{3} = 27.436,$$ which leaves us with $f_{1}^{-1}(27.436)$.  To evaluate that, we use the inverse function we found, so $$f_{1}^{-1}(27.436) = \frac{1}{3}(27.436 - 2) = 8.478$$ to three decimal places.  So, all in all, we have that $$(f_{1}^{-1}\circ f_{2})(1.9) = 8.478$$ to three decimal places.

Title: Having trouble with composite functions, inverse functions. Have an example question from upcoming test.

Body: Here is a look at what I am doing. The I can evaluate the first one no problem. The second one is causing me problems and the solution given isn't helping me to get my head around it. Can someone please try and give me the steps and an explanation? Thanks


问:
If $X = U \begin{pmatrix} \Sigma & 0 \\0&0 \end{pmatrix} U^T$, how is $ U \begin{pmatrix} \Sigma^{-1} & 0 \\0&0 \end{pmatrix} U^T$ called?
Assume $X\succeq 0$ and is decomposed as
\begin{align}
X &= U \begin{pmatrix} \Sigma & 0 \\0&0 \end{pmatrix} U^T,
\end{align}
where $U$ is an orthonormal matrix and $\Sigma\succ0$. Is there a name for

\begin{align}
& U \begin{pmatrix} \Sigma^{-1} & 0 \\0&0 \end{pmatrix} U^T?
\end{align}

答:
Trying to invert $X$ we get
\begin{align}
X^{-1} &= U \begin{pmatrix} \Sigma & 0 \\0&0 \end{pmatrix}^{-1} U^T,
\end{align}
but the matrix in the middle is not invertible (its determinant equals zero). However if we let

\begin{align}
\begin{pmatrix} \Sigma & 0 \\0&0 \end{pmatrix}^{-1} = \begin{pmatrix} \Sigma^{-1} & 0 \\0&0 \end{pmatrix}
\end{align}
then we have performed a pseudo inverse.


问:
Prove: $\int_0^\infty \sin (x^2) \, dx$ converges.
$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.

I'm trying to understand why and would like some help in working towards a formal proof.

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

With Ramanujan-MT:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{2}}\,\dd x} =
{1 \over 4}\Gamma\pars{\color{red}{3 \over 4}}\,
{\Gamma\pars{1 - \color{red}{3/4}} \over
\Gamma\pars{2 - 2\bracks{\color{red}{3/4}}}} =
{1 \over 4\,\Gamma\pars{1/2}}\,{\pi \over \sin\pars{\pi/4}}
\\[5mm] = &\
{1 \over 4\root{\pi}}\,{\pi \over 1/\root{2}} =
\bbx{{\root{2} \over 4}\,\root{\pi}} \approx 0.6267 \\
\end{align}


问:
Rudin exercise 2.18
I am trying to solve problem 2.18 from Rudin's Priciples of Mathematical Analysis. The question asks whether there is a nonempty perfect set in R that contains no rationals. My attempt:
''Let P be such a subset of real numbers and p be a point of this set. As rational numbers are dense in real numbers there exists a rational q in R such that q is in the epsilon neighborhood of p(i.e. $d(p,q)<\epsilon$).  Also as P is perfect: the point p is a limit point of P. Hence there exists a sequence $ {p_n} $, consisting of the points from P, converging to p. So for sufficiently large n, $d(p_n,p)<\epsilon$. But from the triangle inequality for large n's again $d(p_n,q)\leq d(p_n,p)+d(p,q)<2\epsilon$. Therefore $q$ is another limit point of P implying that q is a point of P, from the definition of the perfect set. But we assumed that P is a perfect set with no rational points.'' I deduced that no such set exists but I checked the answers and read that there is such a set. That means there is a flaw in my argument but I still couldn't figured it out.

答:
I'll write the whole answer from scratch because what i said before wasnt correct. There is a flaw in your argument. What you showed is that if you pick an $\epsilon>0$ then you can find a rational number $q$ and point $p_n\in P$ such that $d(p_n,q)<\epsilon$. This is of course true since rationals are dense in $\mathbb{R}$ but this argument doesnt mean that $q$ is a limit point of $P$. To show that this particular fixed $q$ is a limit point, you need to show that for $\epsilon>0$ there is some $p\in P$ such that $d(q,p)<\epsilon$ and of course $q\neq p$. Meaning, that you have to find $p_n's$ that are $\epsilon>0$ close to $q$ independently of $q$.

Let me right down again the definition of the perfect set and give some examples (also of a perfect set that does not contains and rationals)

A set $P$ is called perfect if it is closed and if every point $p\in P$ is a limit point of $P$. This is equivalent to $P=P'$.

$1)$ Consider the set $A=\{\sqrt{2}\}\cup \{\sqrt{2}+1/n:\, n\in \mathbb{N}\}$. Then $A$ is closed and $A'=\{\sqrt{2}\}$. Meaning that $A$ is not a perfect set. Since, rationals are dense, we can find rationals arbitrary close every point of $A$, but $\mathbb{Q}\cap A=\emptyset$ and in particular, $A'\cap \mathbb{Q}=\emptyset$.

The previous example shows that you can use the argument that you tried but you cant conclude that $\mathbb{Q}\subseteq A'$.

Hint: Now, in order to find an example of a perfect set that it does not contain any rationals try to modify a well known perfect set. The example is the following:

 Consider the Cantor set $C$. We know that $C$ is perfect and it does not contains any intervals. If we pick any $\alpha \in \mathbb{R}$, then the set $\alpha+C$ is closed and perfect. In order to have $(\alpha +C)\cap \mathbb{Q}=\emptyset$ we need to pick some $\alpha \in \mathbb{R}\setminus (\mathbb{Q}-C)$. So, we need to show that $\mathbb{Q}-C\neq \mathbb{R}$. If $\mathbb{R}=\mathbb{Q}-C$ then writing $\mathbb{Q}-C=\bigcup_{n=1}^{\infty}q_n-C$ where $\{q_n:\, n\in \mathbb{N}\}$ is an enumeration of the rationals, we conclude by Baire's catergory theorem that some of the $q_n-C$ must have a non empty interior. This means that $C$ must have a non empty interior which cant be true.

Title: Rudin exercise 2.18


问:
Question about the rank-nullity theorem if the linear transformation is one to one (injective)
I have a question about my understanding of the rank-nullity theorem. Let $f: V \to W$ be a linear transformation , if I know that f is injective and that $\dim(V) = \dim (W) = 3$, I want to prove that $f$ is an isomorphism, so that it is also onto. So since $f$ is one to one, $\text{Ker}(f)= \{0\}$ so $\dim\big(\text{Ker}(f)\big)= 0$ and thus by the rank-nullity theorem, $\dim(V) = \dim\big(\text{Im}(f)\big)$. I know that $\dim\big(\text{Im}(f)\big) = 3$. At this point however, how can I connect the dimension of the Image of $f$ and that the $\dim(W) = 3$ to explain that the function onto, so that every element of $W$ is connected to a element in $V$ through $f$? I feel like they are not connected that much.

答: We know that $\text{Im}(f)$ is a subspace of $W$. And as you suggested, we also know that $\dim\big(\text{Im}(f)\big) = \dim(W) = 3$. This is only possible when $\text{Im}(f) = W$.


答: Since $\dim W=3$, $\operatorname{Im}(f)\subset W$, and $\dim\operatorname{Im}(f)=3$, $\operatorname{Im}(f)=W$: a finite-dimensional vector space and a subspace have the same dimension if and only if they are equal.

Title: Question about the rank-nullity theorem if the linear transformation is one to one (injective)

Body: I have a question about my understanding of the rank-nullity theorem. Let $f: V \to W$ be a linear transformation , if I know that f is injective and that $\dim(V) = \dim (W) = 3$, I want to prove that $f$ is an isomorphism, so that it is also onto. So since $f$ is one to one, $\text{Ker}(f)= \{0\}$ so $\dim\big(\text{Ker}(f)\big)= 0$ and thus by the rank-nullity theorem, $\dim(V) = \dim\big(\text{Im}(f)\big)$. I know that $\dim\big(\text{Im}(f)\big) = 3$. At this point however, how can I connect the dimension of the Image of $f$ and that the $\dim(W) = 3$ to explain that the function onto, so that every element of $W$ is connected to a element in $V$ through $f$? I feel like they are not connected that much.


问:
Proving a theorem using its converse
Converse of Pythagoras' theorem: If the lengths of the sides of a triangle $T$ are $a$, $b$, and $c$, and if $a^2+b^2=c^2$, then the triangle is a right triangle and the side opposite to the right angle is the one whose length is $c$.

Proof: Construct a line segment $XY$ whose length is $a$. Then construct a line segment $YZ$ whose length is $b$ which is perpendicular to $XY$. By construction, the triangle $XYZ$ is a right triangle, and therefore, by Pythagoras' theorem and because we are assuming that $a^2+b^2=c^2$, the length of $XZ$ is equal to $c$. So, the triangle $XYZ$ is similar to the original triangle $T$. Since the triangle $XYZ$ is a right triangle, then so is $T$.
What I find peculiar about this proof is the fact that it uses Pythagoras' theorem in order to prove its converse.

It is not the only situation that I am aware of in which this occurs. For instance, there is a proof of the converse of Ceva's theorem  which uses that theorem. But I am not aware of any example outside Euclidean Geometry.

Can anyone provide an example of a theorem of the type $A\implies B$ outside Geometry with a proof which uses the fact that $B\implies A$?

答:
Say natural number $n$ is good if it can be writen as a sum of two squares.

Theorem: $n$ is good iff $2n$ is good.

Proof: If $n=a^2+b^2$ then $2n = (a+b)^2+(a-b)^2$ and we are done.

Now the converse. Say $2n$ is good. Then, by already proven part, also $4n$ is good, so $$4n = x^2+y^2$$ Since $x,y$ must be both even, we can write $a=x/2$ and $b=y/2$ and we are done.

Title: Proving a theorem using its converse


问:
$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$ , $\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$
Please help me integrate

$$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$$

and

$$\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$$

I've tried the standard $u = \tan \frac{x}{2}$ substitution but it looks horrible.

Thanks in advance!

答:
For the second one, where the integral is from $0$ to $2\pi$, here is a way out.

\begin{align}
\dfrac12\int_0^{2\pi} \dfrac{dx}{1+\dfrac{\sin(x)}2} & = \dfrac12\int_0^{2\pi}\sum_{k=0}^{\infty}\left(-\dfrac12 \right)^k \sin^{k}(x) dx = \sum_{k=0}^{\infty}\dfrac1{2^{2k+1}} \int_0^{2 \pi}\sin^{2k}(x) dx\\
& = \sum_{k=0}^{\infty} \dfrac1{2^{2k+1}} \cdot \dfrac{2\pi}{2^{2k}} \dbinom{2k}k = \sum_{k=0}^{\infty} \cdot \dfrac{\pi}{16^{k}} \dbinom{2k}k= \dfrac{\pi}{\sqrt{1-4\times \left(\dfrac1{16}\right)}} = \dfrac{2\pi}{\sqrt3}
\end{align}
where we used the following facts:

$\dfrac1{1+r} = \displaystyle \sum_{k=0}^{\infty}(-r)^k$, $\displaystyle \int_0^{2\pi} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{2\pi}{2^{2k}}$ and $ \dfrac1{\sqrt{1-4x}} = \displaystyle\sum_{k=0}^{\infty} \dbinom{2k}k x^k \,\, \forall x \in \left[-\dfrac14,\dfrac14 \right)$


问:
Variance and expected value of $2^X$
I have 2 similar problems:

given $X \sim Po(3)$, find the variance of $2^X$

given $X \sim B\left(200, \frac{1}{9}\right)$, find the expected value of $2^{X+3}$

I'm not really certain what I'm expected to calculate should it be $Var(2^X)$ and $E[2^{X+3}]$ respectively or should it be $2^{Var(X)}$ and $2^{E[X+3]}$. If it is the second one then: for (1) $$X \sim Po(3)\Longrightarrow\lambda=3\Longrightarrow Var(X)=\lambda = 3 \Longrightarrow 2^{Var(X)}=2^3$$  and for (2) $$X \sim B\left(200, \frac{1}{9}\right)\Longrightarrow E[X+3]=\frac{200}{9} + 3\Longrightarrow 2^{E[X+3]}=2^{\frac{200}{9} + 3}$$

But if it's the first one I don't know how I can find the variance and expected value with the given information

after reading the answers I did for (1): $$ E[2^X]=\sum^\infty_{i=0}2^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(2\lambda)^i}{i!}=e^{-\lambda}\cdot e^{2\lambda} = e^\lambda=e^3$$
and same way
$$ E[(2^X)^2]=E[(4^X)]=\sum^\infty_{i=0}4^i\cdot e^{-\lambda}\frac{\lambda ^i}{i!} = e^{-\lambda}\sum^\infty_{i=0}\frac{(4\lambda)^i}{i!}=e^{-\lambda}\cdot e^{4\lambda} = e^{3\lambda}=e^9 $$
hence
$$ Var(X) = E[(2^X)^2] - (E[2^X])^2 = e^9 - (e^3)^2 = e^9 - e^6$$

as for (2): $$ E[2^X]=\sum^{200}_{i=0}2^i\binom{200}{i}(\frac{1}{9})^i(1-\frac{1}{9})^{200-i} = \sum^{200} _{i=0}\binom{200}{i}(\frac{2}{9})^i (\frac{8}{9})^{200-i} =(\frac{2}{9} + \frac{8}{9})^{200} = (\frac{10}{9})^{200} $$
so $$ E[2^{X+3}]=E[8\cdot 2^X]=8\cdot E[2^X]=8(\frac{10}{9})^{200} $$

答:
You need to calculate $E[2^{X+3}]$ and $\mathrm{Var}[2^X]$. That is what is asked.

To calculate $E[2^X]$, you'll need to work out $\int_x 2^x p(x) dx$, where $p(x)$ is the pdf of $B\left(200, \frac19\right)$. Similar for the variance.

Title: Variance and expected value of $2^X$


问:
Doubts about directional derivative in one variable
Sometimes I read that directional derivatives for function in one variables are right and left derivatives. But this doesn't make sense to me. The only unit vector in $\mathbb{R}$ are $\pm1 $ , so we are saying that:
$$D_1f(x)=f'_+(x)=\lim_{t\to 0^+} \frac{f(x+t)-f(x)}{t}$$
$$D_{-1}f(x)=f'_-(x)=\lim_{t\to 0^-} \frac{f(x+t)-f(x)}{t}$$
But by definition:
$$D_{\mathbf{v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$
By this logic:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}-t\mathbf{v})-f(\mathbf{x})}{t}$$
I can substitute $-t=u$ ($-t$ doesn't assume infinite times the value $0$ so I can apply composite function limit theorem), so:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{-u}=-\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{u}=-D_{\mathbf{v}} f(\mathbf{x})$$
This should mean that:
$$f'_+(x)=-f'_-(x)$$
That in general is false!
Wouldn't be more correct to say that by convention in one variable we use derivative along the unit vector $1$ and that right and left derivatives are simply the right and left directional derivatives along the unit vector $1$.
Thanks in advance.

答:
You are right. Left & right derivatives and directional derivative are different notions – they coexist in fact, also in one-dimensional spaces such as $\mathbf{R}$.

The directional derivative can be defined on a generic vector- or affine space or on a differential manifold, without the existence of a norm or scalar product. It's therefore useful not only in many dimensions, but also in one dimension when the notion of "unit vector" is undefined.

More precisely we define, at a point $\pmb{x}$, the right directional derivative $\partial_{\pmb{v}^+}f$ of a function along the vector $\pmb{v}$ as
$$\lim_{t\to 0+} \frac{f(\pmb{x}+t\pmb{v}) - f(\pmb{x})}{t}$$
or equivalently
$$\lim_{n\to +\infty} \frac{f(\pmb{x}+\pmb{v}/n) - f(\pmb{x})}{1/n}.$$
The left directional derivative is defined identically except for $t \to 0-$ or $n\to -\infty$. From these definitions we have $\partial_{\pmb{v}^+}f \equiv -\partial_{-\pmb{v}^-}f$ (so I'm not sure about your last statement "this in general is false" – it depends on how you're defining $f_{\pm}'(x)$). Note how this limit doesn't involve any scalar product or norm in the space of $\pmb{v}$.

If both left and right directional derivatives along $\pmb{v}$ exist and are the same, then we just speak of the directional derivative along $\pmb{v}$. In differential geometry it is also denoted simply as "$\pmb{v}(f)$". We can have an interesting interplay of these two notions on the boundary of a manifold with boundary, even a 1D one, where it can happen that both $\pmb{v}$ and $-\pmb{v}$ can be defined at the boundary and yet we can only speak of the left or right derivative with respect to one or the other.

Intuitively and informally speaking, the directional derivative tells us something about the rate of change of a function as we move on the vector/affine space or manifold along some direction with some velocity, represented together by the vector $\pmb{v}$. The left and right derivatives tell us something about possible discontinuities in such rate of change.

So no, it isn't true that "directional derivatives for function in one variables are right and left derivatives", but we can say that the left derivative is minus the right derivative with respect to $-\pmb{e}$, where $\pmb{e}\equiv 1$ is the canonical unit vector, and vice versa.

For derivatives on vector spaces and differential manifolds, also with boundary, see eg Choquet-Bruhat, DeWitt-Morette, Dillard-Bleick's Analysis, Manifolds and Physics. Part I: Basics (rev. ed., Elsevier 1996), and Curtis, Miller's Differential Manifolds and Theoretical Physics (Academic Press 1985). There are of course many other good books out there on these topics.


问:
Describe a simplicial complex by its subcomplexes
Let $K$ be a pure simplicial complex of dimension $d$. I would like to ask, if there is a way to describe a simplicial complex by means of certain subcomplexes rather than by simplices. Suppose I understand that $K$ is constructed from two subcomplexes $A$ and $B$ of dimension $d$ that share a common subcomplex $C$ in its boundary, say of dimension $d'<d$.

For example let $A$ be the cone over a (stellar) subdivided edge $C$, that is $A=C\ast a$, and $B$ be another cone over $C$, that is, $B=C\ast b$, so that $A$ and $B$ coincide in $C$.

I feel that this information should suffice to fully describe $K$ up to simplicial isomorphism.

My question is, how can I formalize this information/construction within the context of the usual definition of a simplicial complex? To give the full set of simplices, which one has to do be definition, seems superfluous here.

答:
You may be looking for the pushouts of simplicial complexes.  Here is a recipe to compute them in the special case of working with subcomplexes.

Recall that a simplicial complex $X$ is the data of a set $V(X)$ of vertices together with a downward-closed family $S(X)$ of subsets of $V(X)$.  Given a span $$A \hookleftarrow C \hookrightarrow B$$ of inclusions of simplicial complexes, the pushout $A \cup_C B$ has vertices $$V(A \cup_C B) = V(A) \cup_{V(C)} V(B)$$ and simplices $$S(A \cup_C B) = S(A) \cup S(B).$$

However, while the pushout still exists if $C$ is not a subcomplex of $A$ and $B$, its description is more complicated.  Moreover, it is not true that geometric realization preserves these pushouts in general, so pushouts in simplicial complexes and the corresponding pushouts in topological spaces could look different.


问:
Zariski closure of infinite subset of the circle is the circle
Let $S^1 \subseteq \mathbb{R}^2$ be the unit circle circle $\mathbb{V}(X^2+Y^2-1)$. Let $S$ be an infinite subset of $S^1$. I want to show that
$$\overline{S}^{\text{Zariski}}= \mathbb{V}(\mathbb{I}(S))= S^1$$

Is there an easy way to see this?

Clearly $S^1$ itself is Zariski closed, so it contains the Zariski-closure of $S$. However, the other inclusion seems more delicate. Any ideas how I can show this?

答:
Sabino Di Trani's answer is good. Here is a solution which is perhaps more hands-on.

Suppose $f\in \Bbb R[x,y]/(x^2+y^2-1)$ vanishes on an infinite subset of $S^1$. We'll show it's actually zero, which shows that the Zariski closure of any infinite subset of $S^1$ is all of $S^1$: $I(S)=(0)$ which implies $V(I(S))=V(0)=S^1$.

We can write $f$ as $p+yq$ for polynomials $p,q\in\Bbb R[x]$ by rewriting every instance of $y^2$ as $1-x^2$. Then $f\cdot(p-yq)=p^2-y^2q^2=p^2+(x^2-1)q^2$ is a polynomial in $x$ vanishing on an infinite set, so it's zero. As $\Bbb R[x,y]/(x^2+y^2-1)$ is a domain, we must have that either $f=p+yq$ or $p-yq$ is zero, both of which give $f=0$.


问:
Newton's method : Show inequality
I have written the formula of Newton's method to appoximate $a^{1/n}$, which is $$x_{k+1}=\left (1-\frac{1}{n}\right )x_k+\frac{a}{nx_k^{n-1}}$$

Now I want to show the inequality $x_{k+1}-a\leq \left (1-\frac{1}{n}\right )(x_k-a)$.

Since we subtract at the left side $a$ instead of $a^{1/n}$ we cannot consider it as the error, right?

I achieved to show the inequality but for $a\geq 1$:
$$x_{k+1}=\left (1-\frac{1}{n}\right )x_k+\frac{a}{nx_k^{n-1}} \leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n\left (a^{1/n}\right )^{n-1}} = \left (1-\frac{1}{n}\right )x_k+\frac{a}{na^{1-\frac{1}{n}}} = \left (1-\frac{1}{n}\right )x_k+\frac{a^{\frac{1}{n}}}{n} $$
Then subtracting at both sides "$a$" we get $$x_{k+1}-a\leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n}-a=\left (1-\frac{1}{n}\right )x_k-\left (1-\frac{1}{n}\right )a=\left (1-\frac{1}{n}\right )(x_k-a)$$
We have that $x_k\geq a^{1/n}$.

Could you give me a hint for the case $a<1$ ? Or is there an other way to show the inequality without taking cases for $a$ ?

答:
Let
$$g(x) = \left(1 - \frac{1}{n}\right)x + \frac{a}{nx^{n - 1}}.$$
Then the iteration satisfies
$$x_{k + 1} = g(x_k).$$
Let $c = a^{1/n}$. Let $e_{k} = x_{k} - c$ be the sequence of errors. Then we know $g(c) = c$ by the construction of Newton's method. Hence by the Lagrange remainder theorem,
$$x_{k + 1} = g(x_{k}) = g(c) + g'(\xi_k)(x_k - c) = c + g'(\xi_k)(x_k - c),$$
where $\xi_{k}$ is between $c$ and $x_k$.
Hence
$$e_{k + 1} = g'(\xi_k)e_{k}.$$
The power rule reveals that
$$g'(x) = \left(1 - \frac{1}{n}\right)\left(1 - \frac{a}{x^n}\right).$$
Assume that $n \in \mathbb{R} > 1$ and that $x_{k} > c$. Then we have that $e_{k} > 0$ and that
$$x_{k} > c \implies \xi_k > c \implies \xi_{k}^n > c^n = a \implies 0 < g'(\xi_k) < \left(1 - \frac{1}{n}\right) \implies 0 < e_{k + 1} < \left(1 - \frac{1}{n}\right)e_{k}$$
as desired.

Note that to solve $f(c) = 0$, Newtons method constructs $g(x) = x - \frac{f(x)}{f'(x)}$ and tries to find a fixed point of $g$ since $f(c) = 0 \iff g(c) = c$. Note that $g'(c) = 0$, and $g''(c) = \frac{f''(c)}{f'(c)}$. Thus if you expand the Taylor series one more term, you get that when it converges, Newton's method is quadratically convergent if $f'(c) \neq 0$ and $f^{(3)}$ is continuous at $c$:
$$e_{k + 1} = g'(c)e_{k} + \frac{g''(\xi_k)}{2}e_{k}^2 = \frac{g''(\xi_k)}{2}e_{k}^2 \implies \frac{e_{k + 1}}{e_{k}^2} \to \frac{g''(c)}{2} = \frac{f''(c)}{2f'(c)} \text{ as $k \to \infty$}.$$

Title: Newton's method : Show inequality


问:
How to show that an open map $f $ implies the surjectivity of $f'$ in a dense set
Let $f$ be a $C^1$ map from $U\to \mathbb{R}^m$, where $U$ is an open set in $\mathbb{R}^n$, $n\geq m$. Then we know that if $f'$ is surjective everywhere, then $f$ is open.

My question is whether the converse is valid,  if the answer is no, can we prove the following weaker conclusion:

If $f$ is open, $f'$ is surjective in a dense set of $U$

答: If f is defined on na open subset of Rm with values on Rn and is of class C1, then f open implies that f'(x) is surjective on na open and dense subset of U.</p>


问:
Balancing chemical equations - Matrix determinant is zero for specific reaction
I am currently working on a program which takes a chemical equation as input and returns the balanced chemical equation instead. It finds the appropiate chemical coefficients with matrix operations, using the formula $x = A^{-1}b$. I am using JAMA for calculations involving matrices. According to their docs, it calculates $A$'s inverse using

Eigenvalue Decomposition of both symmetric and nonsymmetric square matrices

Let's take an example to better visualize this. Suppose the following chemical equation with denoted coefficients for every compound:

$aCa(OH)_2 + bH_3PO_4 \rightarrow cCa_3(PO_4)_2 + dH_2O$

The computer creates an equation for every element:

$P: b - 2c = 0$

$H: 2a + 3b - 2d = 0$

$Ca: a - 3c = 0$

The last equation needs to be $a = det(A)$ for three reasons. We need a square matrix (and that's why oxygen's equation is omitted) to compute it's inverse and determinant. Secondly, we need to give a numerical value to one of the variables to be able to get a numerical result, and thirdly (as far as I understood because I learned nothing about matrices in school yet) $adj(A)$ is multiplied by $1/det(A)$ to get $A^{-1}$ which causes the computed coefficients to sometimes be non-integer numbers. I am not sure of how or why setting $a$ to $det(A)$ works, neither this is the most efficient way of doing it.

$A = \left[\begin{array}{cccc}0&1&-2&0\\2&3&0&-2\\1&0&-3&0\\1&0&0&0\end{array}\right]$
$b = \left[\begin{array}{c}0\\0\\0\\det(A)\end{array}\right]$

After dividing every number to the GCD the computed vector $x$ will be:

$x = A^{-1}b = \left[\begin{array}{c}6\\4\\2\\12\end{array}\right]$ = $\left[\begin{array}{c}3\\2\\1\\6\end{array}\right]$

And of course, the balanced chemical equation is

$3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O$

This program works perfectly for even more complex equations:

$299H_2SO_4 + 10K_4Fe(CN)_6 + 122KMnO_4 \rightarrow 60CO_2 + 5Fe_2(SO_4)_3 + 188H_2O + 60HNO_3 + 162KHSO_4 + 122MnSO_4$

$9Fe_{36}Si_5 + 836H_3PO_4 + 192K_2Cr_2O_7 \rightarrow 324FePO_4 + 45SiO_2 + 128K_3PO_4 + 384CrPO_4 + 1254H_2O$

However, in the process of testing, I found a chemical equation that produces the following error: java.lang.RuntimeException: Matrix is singular.

Let's write everything down:

$aB_{10}H_{12}CNH_3 + bNiCl_2 + cNaOH \rightarrow dNa_4(B_{10}H_{10}CNH_2)_2Ni + eNaCl + fH_2O$

$B: 10a - 20d = 0$

$C: a - 2d = 0$

$Na: c - 4d - e = 0$

$H: 15a + c - 24d - 2f = 0$

$Cl: 2b - e = 0$

This time, the code omits the equation for three elements ($N, Ni, O$) because it would result in an overdetermined system and a non-square matrix.

$A = \left[\begin{array}{cccccc}10&0&0&-20&0&0\\1&0&0&-2&0&0\\0&0&1&-4&-1&0\\15&0&1&-24&0&-2\\0&2&0&0&-1&0\\1&0&0&0&0&0\end{array}\right]$

This is completely unexpected, both for me and the script. Using a matrix calculator, I found out that $A$ has no inverse: its determinant is zero. I analyzed the matrix and found that $10a - 20d = 0$ and $a - 2d = 0$ are really
the same equation. I'm not sure if this has something to do with the matrix being singular.

What should I do to make this work? Should I use a different method for finding solutions and representing my system of equations, or do I have to do something specific when $det(A) = 0$ ? I've read a little bit about Gaussian Elimination and LU Decomposition, but I don't seem to understand them very well. In case I have to use another method for solving the system, which one is the most suitable for this script? Also, I would be happy to get some details about it if possible. Keep in mind I'm a beginner in this matrix field. Any help is appreciated!

答:
The determinant is zero, because the first two lines are linearly dependent,
indeed, we can consider $B_{10}C$
as one "atomic element",
or complex, if you like, which provides just one equation,

$B_{10}C:\quad a-2d=0$,

so you need to remove the first line
and replace it with the line with $Ni$ instead,

$Ni:\quad b-d=0$.

Then $\det A=4$ and the solution is

$a=4$, $b=2$, $c=12$, $d=2$, $e=4$, $f=12$,

or, equivalently,

$a=2$, $b=1$, $c=6$, $d=1$, $e=2$, $f=6$.

And indeed,

\begin{align}
2B_{10}H_{12}CNH_3 + NiCl_2 + 6NaOH
&\rightarrow
Na_4(B_{10}H_{10}CNH_2)_2Ni + 2NaCl + 6H_2O
.
\end{align}


问:
General knowledge regarding double convergence of sequences (and series).
While proving a subspace of $\textit{L}^p(\mathbb{R}^n)$ is dense I stumbled upon a family of functions that was double indexed; this doesnt fall in our usual definition of sequence, it is merely a set of functions $\{f_{i,j}\}_{i,j\in \mathbb{N}}$. The study of a collection of this type wasn't necessary for my objective, but i found a few papers online regarding double convergence and how it is defined.

As a first thing i would like to know if anybody knows about this part of analysis, becouse although it sound interesting, it's probably not all that popular.

To give some context to people who don't know what this is, ill give a first definition of convergence:

Given a measure space $(X,\textit{A},\mu)$ and given $E\in \textit{A}$ with $|E|<+\infty$ ($\star$ is this limitation necessary?$\star$), then we say that $\{f_{n,m}\}$ converges in the Pringsheim’s sense if for every $\epsilon > 0$ there exists $N∈\mathbb{N}$ such that $|f_{m,n}(x_0) −f|< \epsilon$ whenever $j$, $k ≥N$.

And all the other kind of convergences can be defined as well (a.e., in measure,...)

I now ask, what properties differenciate them from the normal sequences and convergence?

By just thinking of examples i am pretty sure a major difference is that if a double sequence converges, it still can have subsequences converging elswhere, for example:

$$f_{n,m}(x)=\frac{1}{n}+\frac{1}{m} \Longrightarrow \lim_{m\to +\infty}f_{n,m}=\frac{1}{n} \mbox{ and } \lim_{n\to +\infty}f_{n,m}=\frac{1}{m}$$
so when we fix one of the two between $n$ and $m$ we obtain a normal converging sequence:
$$\forall \epsilon > 0\mbox{, } \exists N_1\in \mathbb{N} \mbox{ such that } \forall n\ge N_1 \Rightarrow |f_{n,m}-\frac{1}{m}|\le \frac{\epsilon}{2}$$
$$\forall \epsilon > 0\mbox{, } \exists N_2\in \mathbb{N} \mbox{ such that } \forall m\ge N_2 \Rightarrow \frac{1}{m}\le \frac{\epsilon}{2}$$
so that
$$\forall \epsilon > 0\mbox{, } \exists N:=\max\{N_1,N_2\}\in \mathbb{N} \mbox{ such that } \forall n,m\ge N \Rightarrow |f_{n,m}-0|=|f_{n,m}-\frac{1}{m}+\frac{1}{m}-0| \le$$
$$\le|f_{n,m}-\frac{1}{m}|+\frac{1}{m}<\epsilon$$
We have obtained the "double convergence of the double sequence", but if i get the subsequence $\{f_{1,m}\}_{m\in \mathbb{N}}$, this converges in the normal definition to $1$; actually you could find infinite sunsequences, $\forall n\in \mathbb{N}$, $f_{n,m}\longrightarrow \frac{1}{n}$, for $m\rightarrow +\infty$.

As you might have seen, i am just looking for insight on what to think of this topic, maybe also knowing
what interesting results one can reach.

Also I'm sorry if my english isn't that good.

答:
Focusing on double sequences $(x_{mn})$ in $\mathbb{R}$ (or $\mathbb{R}^p$ or $\mathbb{C}$) many results are common to ordinary sequences, e.g., the Cauchy criterion, limits of sums and products of sequences , etc.

Perhaps the first unique and important question that is raised is the existence and equality of the double and iterated limits. That is, when do we have

$$\lim_{m,n \to \infty}x_{mn} = \lim_{m \to \infty}\lim_{n \to \infty}x_{mn} = \lim_{n \to \infty}\lim_{m \to \infty}x_{mn},$$

with the limit on the LHS defined in the Pringsheim sense. As with integrals and series, the question of when the order of limiting operations can be switched arises frequently in  applications.

With double sequences we can see a variety of possibilities:

(1) $\,\,x_{mn} = (-1)^{m+n} \left(\frac{1}{m} + \frac{1}{n} \right)$ where $\underset{m,n \to \infty} \lim x_{mn}  = 0$  but the iterated limits do not exist.

(2) $\,\,x_{mn} = \begin{cases}1, & m \neq n \\ 0, & m=n \end{cases}$ where the double limit does not exist but both iterated limits exist and are equal with $\underset{m \to \infty}\lim \underset{n \to \infty}\lim x_{mn} = \underset{n \to \infty}\lim \underset{m \to \infty}\lim x_{mn}=1.$

(3) $\,\,x_{mn} = \frac{m-n}{m+n}$ where the iterated limits exist but are unequal with  $\underset{m \to \infty}\lim \underset{n \to \infty}\lim x_{mn}  =-1$ and $\underset{n \to \infty}\lim \underset{m \to \infty}\lim x_{mn}  = 1$.

Does the double limit exist in case (3)? The answer is no and a consequence of the following "fundamental" theorem of double sequences.

If the double limit $\underset{m,n \to \infty}\lim x_{mn} =x$ exists,
and for each $m \in \mathbb{N}$ the limit $ \underset{n \to
 \infty} \lim x_{mn} = y_m$ exists, then the iterated limit exists and
$\underset{m,n \to \infty} \lim x_{mn}  = \underset{m \to \infty}\lim
 y_m = x$.

Now we can see that if both collections $\{y_m= \underset{n \to
 \infty} \lim x_{mn}\,|\, m \in \mathbb{N}\}$ and $\{z_n= \underset{m \to
 \infty} \lim x_{mn}\,|\, n \in \mathbb{N}\}$ exist along with the double limit, then we must have

$$\lim_{m,n \to \infty}x_{mn} = \lim_{m \to \infty}\lim_{n \to \infty}x_{mn} = \lim_{n \to \infty}\lim_{m \to \infty}x_{mn},$$

and if

$$\lim_{m \to \infty}\lim_{n \to \infty}x_{mn} \neq \lim_{n \to \infty}\lim_{m \to \infty}x_{mn},$$

then the double limit cannot exist.


问:
What do Sylow $p$-subgroups of a dihedral group $D_{2n}$ look like when $p$ is an odd prime?
Question Show that a Sylow $p$-subgroup of dihedral group $D_{2n}$ is cyclic and normal when $p$ is an odd prime.
$$
D_{2n}:=\left\{ \langle x,y\rangle|x^{n}=1,y^2=1,xy=yx^{-1}=yx^{n-1} \right\} .
$$

My thought

Assume that $x^i y\in S$ where $S$ is a Sylow $p$-subgroup,
$$
\left( x^iy \right) ^2=x^iy(x^iy)=x^iy(yx^{n-i})=x^i(yy)x^{n-i}=x^ix^{n-i}=1,
$$

implying $\text{order}(x^i y)=2$($i=0,1,\cdots,n-1$), so it can't be a member of $S$ as $|S|=p^r$ is odd, contradiction! Then we know that $S$ is a subgroup of $\langle x\rangle$, cyclic and normal indeed.

However, this deduction has nothing to do with the prime property of $p$, only using the fact that $p$ is odd. Therefore, I wonder if there is anything wrong with the idea or it is actually the case. Thank you in advance!

答:
Your argument is correct! Indeed, any odd order subgroup of a dihedral group is both cyclic and normal, and this follows from the argument you suggest.

(If you think about the dihedral group geometrically, then half of the group consists of rotations generating a cyclic group, and the other half consists of reflections which all have order two --- this is just another way of saying what you computed geometrically)


问:
Delta measure, calculate $\int_A f \space d \delta_p$
We are looking at the measure space $(\mathbb{R}^d, \mathcal{P}(\mathbb{R}^d),\delta_p)$, for all $A\subset \mathbb{R}^d$:

$\delta_p(A):=\begin{cases}
1, & \text{if } p \in A \\
0, & \text{otherwise}
\end{cases}$.

Let $A \subset \mathbb{R}^d$ and $f:\mathbb{R}^d \rightarrow   \overline {\mathbb {R}}$ arbitrary. Show that $f$ is measurable and calculate $\int_A f \space d \delta_p$.

This is probably not that difficult but I don't know how to do it. :(

答:
$\delta_p$ puts all the mass into the point $p$, so the integral is

$$ \int_{A}fd\delta_p= \begin{cases} f(p) & \text{if}~ p\in A\\ 0 & \text{otherwise} \end{cases}. $$


问:
Brown representability theorem
I am looking for quite a detailed proof of Brown representability theorem roughly along the lines of this sketch https://www3.nd.edu/~andyp/notes/BrownRepresentability.pdf .
Any reference suggestions?</p>

答:
Look at chapter 9 (especially Theorem 9.12) of

Switzer, Robert M. Algebraic Topology - Homotopy and Homology. Springer, 2017

or at chapter 7 (especially Theorem 7.7.14) of

Spanier, Edwin H. Algebraic topology. Springer Science & Business Media, 1989.

Title: Brown representability theorem


问:
How many different 2-regular graphs are there with 5 vertices?

How many different 2-regular (simple) graphs are there with 5 vertices?

I just asked a very similar question, and I actually already understand the answer of this question.

I think there are very much different ways to look at such a question (but maybe I'm wrong) and my head always kind of feel unsatisfied, even if I get the answer. I hope my head get a little more satisfied by actually understanding the different ways to solve this problem. I see this problem now like:

Just by trial and error I understand that the graph shoud look like:

\begin{array}{ccccc}
& & 1 & &
\\ & ╱ & & ╲ \\
5 & & & & 2 & \\ | & & & & | \\
4 & & — & &3
\end{array}

And as there are $4!$ cycles of the form $(12345)\in S_5$, there must be $4!/2$ graphs of this form as $(12345)$ and $(54321)$ are different cycles but represent the same graph.

So this is one way to look at this problem, but I'm interested in other ways to think of this question.

答:
There is indeed only one $5$-vertex $2$-regular graph, which is isomorphic to the $5$-cycle:

This graph has the dihedral group (which I will denote $D_5$ today) as its automorphism group.

Let $A$ be the set of labelled graphs isomorphic to the $5$-cycle.  The symmetric group $S_5$ acts on $A$ by permuting the vertex labels.  There is only one orbit under this group action, and this orbit is $A$.  The elements of $S_5$ that stabilise a labelled graph in $A$ are precisely the elements of $D_5$.  Hence, by the Orbit-Stabiliser Theorem $$|A|=\frac{|S_5|}{|D_5|}=\frac{5!}{10}.$$

In general, the same argument can be used to show that the number of labelled graphs isomorphic to a graph $G$ is $$\frac{|V(G)|!}{|\mathrm{Aut}(G)|}$$ where $V(G)$ is the set of vertices of $G$ and $\mathrm{Aut}(G)$ is the automorphism group of $G$.


问:
Why probability density is equal to = $\int_{a}^{b} f(x) dx$?
this is the formula used for probability density (where f(x) is the probability density function):

$$\int_{a}^{b} f(x) dx$$

but it does not make sense to me , I will explain in details:

according to Riemann integral definition , $$\int_{a}^{b} f(x) dx = \lim_{\Delta x \rightarrow
0}\sum_{k=1}^n f(x_k) \Delta x $$ , $n \rightarrow \infty \text{ as } x \rightarrow 0$

$\because$ $f(x)$ is just frequency divided by class width

$$ \therefore = \lim_{\Delta x \rightarrow 0}\sum_{k=1}^n {{f_k} \over \Delta x} \Delta x = \lim_{\Delta x \rightarrow 0}\sum_{k=1}^n {{f_k}} = \text{sum of all frequencies over the interval $[a,b[$}$$

so the integral of $f(x)$ does not give the probability of having a continuous random variable over the interval [a,b[

i am a newbie and sorry if the question sounds stupid to you and it will be very appreciated if someone could clean things up

thanks in advance guys.

答: I Think of it as a function whose area on a measurable set gives you the probability of that set. It makes it easier in many cases to calculate the probability if you know this function. Not all of the probabilities have it, but if you are dealing with one that does, you can apply all we know about integrals to it and you are done.</p>


问:
What is the notation $\textrm{LN}(X)$ in linear algebra?
What is the notation $\textrm{LN}(X)$ in linear algebra?

In the paper I am reading, for a given matrix $V$, the author decomposes matrix $A$ into $A^{\|}$ and $A^{\perp}$ such that

$AV = (A^{\perp} + A^{\|})V = A^{\perp}V + A^{\|}V = A^{\perp}V + \mathbf{0} = A^{\perp}V$,

and states that $A^{\perp} = A - \textrm{Projection}_{\textrm{LN}(V)}A$.

And I am not sure what $\textrm{LN}(V)$  exactly stands for.

Thank you,

答:
Maybe LN = left nullspace, also known as the cokernel.

Yes, the paper explicitly says:

The (left) null space $\operatorname{LN}(T)$ of $T$ describes all
vectors that are mapped to the zero vector by $T$:
$$\operatorname{LN}(T) = \{\tilde{x}^T \in \mathbb{R}^{1\times d_s}|\tilde{x}^T T = 0\}\tag4$$


问:
Understanding nonstandard Peano arithmetic
I've had the idea of nonstandard Peano arithmetic introduced to me in the comments of this question. The concept that we could write down the axioms which produce the natural numbers and also produce something else from them is totally crazy to me! I did not expect this.

I've been doing some reading here, to try and understand what these non-standard models of the Peano axioms are. It's difficult for me to follow the details of this paper, but I've understood that the most simple non-standard model of the Peano axioms is
$$
M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\}
$$

I understand this to be the natural numbers along with an additional constant $c$, is that correct?

So from this understanding, it makes sense to me to ask the questions,

What is $n+c$ for $n$ a natural number?

What is $n\cdot c$ for $n$ a natural number?

What is $c+c$?

What is $c\cdot c$?

Is $c≤n$  for $n$ a natural number?

Are these sensible questions? If not, is there some similar concrete way to understand this model?

答:
I've understood that the most simple non-standard model of the Peano axioms is
$$
M = \{\mathbb{N}, +, ×, ≤, 0, 1, c\}
$$ I understand this to be the natural numbers along with an additional constant $c$, is that correct?

It sounds like your $M$ consists of only the usual naturals and the single extra element $c$. But we know this cannot be the case: $\mathsf{PA}$ proves for example that every number has a distinct successor, so $c+1$ needs to be something other than $c$ (and clearly can't be a standard natural). The usual ways of showing that a nonstandard model exists do involve introducing a new constant $c$ and building a model around it, but that doesn't mean that that model consists only of $c$ and the standard numbers.

Indeed, there is no simplest nonstandard model of $\mathsf{PA}$ in any sense I can think of, and Tennenbaum's theorem in fact rules out nicely-describable models altogether in a fairly strong way.

I suspect the confusion arises from considering the "snappy" construction of a nonstandard model of $\mathsf{PA}$:

Let $T$ be a complete consistent theory in the language of arithmetic + a new constant symbol $c$, containing $\mathsf{PA}\cup\{c>\underline{n}: n\in\mathbb{N}\}$. By the completeness theorem, $T$ has a model $M$; but $M$, or rather the reduct to the language of arithmetic of $M$, is clearly a nonstandard model of $\mathsf{PA}$.

The issue is that this hides a hugely complicated step: the actual model consruction invoked when we apply the completeness theorem. To see what's going on here, and why it doesn't result in a simple picture at all, we have to look at how it's proved. I think the Henkinization argument is best here (Godel's original argument was different):

Henkin argued as follows. For any theory $T$ whatsoever, we can look at the set $Term_T$ of closed terms in the language of $T$ modulo $T$-provable equality, and in a natural way turn that set into a structure $\mathfrak{Term}_T$ in the language of $T$. This is the term structure of $T$.

However, in general we do not have $\mathfrak{Term}_T\models T$ - this is a fun exercise. To get around this, Henkin now turns to the task of proving the following:

Every consistent theory $T$ has some extension $S$ - possibly in a larger language - such that $\mathfrak{Term}_S\models S$.

A fortiori (the reduct to the language of $T$ of) the structure $\mathfrak{Term}_S$ will satisfy $T$, so if we can show this we'll be done. Note by the exercise above this reduct is in general not the same as $\mathfrak{Term}_T$: $S$ is playing an essential role here.

The passage from a theory $T$ to the structure $\mathfrak{Term}_T$ is totally explicit; however, the passage from a theory $T$ to an appropriate extension $S$ if necessary is not. It's this second step which is elided in the snappy construction above, and the elision of which may lead one to underestimate the complexity of the resulting situation.

(For a bit more on this, see here.)


问:
Proving a theorem using its converse
Converse of Pythagoras' theorem: If the lengths of the sides of a triangle $T$ are $a$, $b$, and $c$, and if $a^2+b^2=c^2$, then the triangle is a right triangle and the side opposite to the right angle is the one whose length is $c$.

Proof: Construct a line segment $XY$ whose length is $a$. Then construct a line segment $YZ$ whose length is $b$ which is perpendicular to $XY$. By construction, the triangle $XYZ$ is a right triangle, and therefore, by Pythagoras' theorem and because we are assuming that $a^2+b^2=c^2$, the length of $XZ$ is equal to $c$. So, the triangle $XYZ$ is similar to the original triangle $T$. Since the triangle $XYZ$ is a right triangle, then so is $T$.
What I find peculiar about this proof is the fact that it uses Pythagoras' theorem in order to prove its converse.

It is not the only situation that I am aware of in which this occurs. For instance, there is a proof of the converse of Ceva's theorem  which uses that theorem. But I am not aware of any example outside Euclidean Geometry.

Can anyone provide an example of a theorem of the type $A\implies B$ outside Geometry with a proof which uses the fact that $B\implies A$?

答:
Sorry, could not resist to write this one from geometry, due to its simplicity.

Suppose $ABCD$ is convex quadrilateral with sides $a,b,c,d$. Then $ABCD$ is tangent iff $a+c = b+d$ ($a,c$ are opposite sides).

Proof: Suppose $ABCD$ is tangent, then ... $a+c=b+d$

Now the converse. Say lines $AD$ and $BC$ meet at $E$ and draw an incircle in triangle $ABE$. Say tangent at $C$ meet $AE$ at $D'$. We have to prove $D'=D$. .
Suppose it is not. But then we have, by already proven part $a+c' = b+(d-x)$
By hypothetis we have also $a+c=b+d$ so $c=c'+x$. But this is impossible because of triangle inequality in $CDD'$. A contradiction.

Title: Proving a theorem using its converse


问:
Find all values of x for which B is not invertible.
The matrix $B$ is below:

$$ B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ -1 & x & 0 & 1 \\ 0 & 0 & x & 1 \\ x^2 & x & 7 & 0 \end{pmatrix} $$

Using Gaussian elimination I got this for the determinant.   -----

$$  \det B = 1×x×x×((x^2-x-6)/x) = x^3-x^2-6x $$

I need the values $x$ that make $B$ noninvertible.

Can you help me out from there?

答: Adding the first line to the second one and also adding the first line time $-x^2$ to the fourth on, you get$$\begin{vmatrix}1&0&1&0\\-1&x&0&1\\0&0&x&1\\x^2&x&7&0\end{vmatrix}=\begin{vmatrix}1&0&1&0\\0&x&1&1\\0&0&x&1\\0&x&-x^2+7&0\end{vmatrix}.$$Now, subtracting the second one from the fourth one, you get$$\begin{vmatrix}1&0&1&0\\0&x&1&1\\0&0&x&1\\0&x&-x^2+7&0\end{vmatrix}=\begin{vmatrix}1&0&1&0\\0&x&1&1\\0&0&x&1\\0&0&-x^2+6&-1\end{vmatrix}.$$So\begin{align}\begin{vmatrix}1&0&1&0\\-1&x&0&1\\0&0&x&1\\x^2&x&7&0\end{vmatrix}&=x\begin{vmatrix}x&1\\-x^2+6&-1\end{vmatrix}\\&=x(-x+x^2-6)\\&=x^3-x^2-6x\end{align}indeed. Therefore, $B$ is invertible if and only if $x\notin\{-2,0,3\}$.


问:
Is there a function whose second derivative is the inverse square of itself?
Namely, is there a function $f(x)$ that satisfies

$$\frac{d^2f(x)}{dx^2} = \frac{1}{(f(x))^2}$$

I've been messing with an overly simplified physical system in which a body has a gravitational attraction towards another body much more massive (allowing me to assume that the second body doesn't move), and I got the following differential equation:

$$\frac{d^2x(t)}{dt^2} = \frac{Gm}{(x(t))^2}$$

That got me wondering if such differential equation has a known solution.

答:
hint

$$y'' = \frac{1}{y^2} \implies$$

$$y'y''=\frac{y'}{y^2} \implies$$

$$\frac{y'^2}{2}= -\frac 1y +C$$

In the particular case $ C=0 $, we have

$$y'=\frac{\sqrt{2}}{\sqrt{-y}}$$
and

$$\frac 23(-y)^\frac 32=-x\sqrt{2}$$

From here, we can see that the function $$Y:x\mapsto x^{\frac 23}$$
satisfies
$$Y'=\frac 23x^{-\frac 13} \; \text { and }\;Y''=-\frac 29x^{-\frac 43}$$
$$=\frac{K}{Y^2}$$


问:
How to find Sylow 3-subgroups of $SL_n(F_3)$?
I've enumerate all 24 elements in $SL_2(F_3)$. As $24=2^3\times 3$, I need to find groups of order 3. Sylow Theorem tells me that the number of Sylow 3-subgroups $n_3 \equiv 1\ \text{mod}\ 3$ and divides $8$, so $n_3= 1 \text{ or } 4$.

I've found 2 of them: upper/lower triangular matrices with 1 on the diagonal, i.e.,
$$
S_1=\left\{ \left( \begin{matrix}
 1&  0\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  1\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  2\\
 0&  1\\
\end{matrix} \right) \right\} ,S_2=\left\{ \left( \begin{matrix}
 1&  0\\
 0&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  0\\
 1&  1\\
\end{matrix} \right) ,\left( \begin{matrix}
 1&  0\\
 2&  1\\
\end{matrix} \right) \right\} .
$$
The remaining ones are given by my friend: Groups generated by $
\left( \begin{matrix}
 0&  2\\
 1&  2\\
\end{matrix} \right) ,\left( \begin{matrix}
 2&  2\\
 1&  0\\
\end{matrix} \right)
$ respectively. However, this process is rather experimental than theoretical. I wonder if there is a logical explanation to this, or a method to figure them out more efficiently.

Any idea or help would be appreciated! Thank you!

答:
The Sylow subgroups of $\mathrm{SL}_2(\mathbf{F}_3)$ are conjugate to the group of upper triangular matrices as you have noted. Another way to think about this subgroup is as the group which fixes the line generated by the vector $\mathbf{v} = (1,0)$. Any other Sylow $P$ (which is conjugate) will also fix a line generated by a vector $\mathbf{w}$, and this line is uniquely determined by $P$, since if $P$ fixes two linearly independent vectors then it fixes everything. So now you just have to enumerate the lines and write down the determinant one elements which stabilize the lines. For example, you can take the lines generated by $(1,0)$, $(0,1)$, $(1,2)$, and $(1,1)$; in that order, you get the groups you wrote down.

Unnecessary Extras:
If you did the same for $\mathrm{SL}_3(\mathbf{F}_p)$, you want want to stabilize a line generated by $\mathbf{v} = \mathbf{v}_1$ but also a plane $\{\mathbf{v}_1,\mathbf{v}_2\}$ generated by two vectors. More generally, Sylow subgroups of $\mathrm{SL}_n(\mathbf{F}_p)$ or $\mathrm{GL}_n(\mathbf{F}_p)$ correspond to "flags" of subspaces. The number of choices of $\mathbf{v}_1$ is $(p^n - 1)/(p-1)$ for the first vector (any non-zero vector up to scalar), then $(p^{n-1} - p)/(p-1)$ for the second vector (any non-zero vector up to scalar in $V/\mathbf{v}_1$), then $(p^{n-2} - 1)/(p-1)$ choices of a vector up to scalar in $V/(\mathbf{v}_1,\mathbf{v}_2)$, and so on. Hence

$$n_p = \frac{1}{(p-1)^p} \prod (p^{n-i} - 1).$$

Comparing this to

$$|\mathrm{GL}_n(\mathbf{F}_p)| = \prod (p^{n-i} - 1) p^i, \qquad
|\mathrm{SL}_n(\mathbf{F}_p)| = \frac{1}{(p-1)} \prod (p^{n-i} - 1) p^i,$$

you deduce by Sylow III that the normalizer $N$ of $P$ has order

$$(p-1)^{n} \prod p^i$$

for $\mathrm{GL}_n(\mathbf{F}_p)$ and one fewer factor of $(p-1)$ for $\mathrm{SL}_n$. This is consistent with the fact that the normalizer of the group of upper triangular matrices with ones on the diagonal is the group of upper triangular matrices which has these orders. (In fact, this latter approach is a quicker way to compute $n_p$, but it's also nice to have the explicit description.)


问:
Determine multiplicity of root $a$ of polynomial $\frac{x-a}{2}(f'(x) + f'(a)) - f(x) + f(a)$
Let's say that this is polynomial over some field for simplicity.

To be honest, I don't really know what trick we can use here. I know that if polynomial has multiplicity $k$ of some root, then it's derivative is equal to $k - 1$. But how can we use that fact to solve this problem?

Thank you in advance!

答:
Suppose the field is $R$. Let
$$ f(x)=\sum_{i=1}^nb_i(x-a)^i+b_0. $$
Then
\begin{eqnarray}
F(x)&:=&\frac{x-a}{2}(f'(x) + f'(a)) - f(x) + f(a)\\
&=&\frac{x-a}{2}\bigg[\sum_{i=1}^nib_i(x-a)^{i-1}+b_1\bigg] - \sum_{i=1}^nb_i(x-a)^i\\
&=&\frac{1}{2}\sum_{i=3}^n\bigg[(i-2)b_i(x-a)^{i}-b_1(x-a)\bigg].
\end{eqnarray}
Suppose $n\ge3$. Then the multiplicity of $a$ of $F(x)$ is $1$ if $b_1\neq0$. If
$b_1=0$,then the multiplicity of $a$ of $F(x)$ is $k$ if
$$ f(x)=\sum_{i=k}^nb_i(x-a)^i+b_0, b_3=\cdots=b_{k-1}=0, b_k\neq0, k\ge3. $$
If $n=1,2$, $F(x)\equiv0$.


问:
sylow basis of finite solvable groups
Let $G$ be a finite solvable non-$p$-group and $A$ be a maximal subgroup of $G$.
Therefore $A$ is of primary index $p^{n} $, that is $|G : A|=p^{n}$ where $p$ is prime and $n\geq 1$.

If $|G|=p_1^{n_1}p_2^{n_2}\cdots p_l^{n_l}p^{m+n}$; $n_i\geq 1, m\geq 0$
and $P_i$ is a sylow $p_{i}$-subgroup and $P$ is a sylow $p$-subgroup of $G$,
can we obtain a sylow basis $\{P_1, P_2, \cdots,P_l, P\} ;  l\geq 1$ of $G$ such
that $P_i\subseteq A$; $1\leq i\leq l$?

Any answer will be appreciated</p>

答:
This is a special case of Hall (1937), and is mostly easily understood after you've seen Hall subgroups (subgroups whose order is coprime to their index). You can see the ideas begin in Hall (1928), but you can also see them put together in a nice way in Carter (1959). In particular Carter describes how to use the maximality condition in a non-trivial way to understand subgroups of a finite soluble group. The proof below does not use maximality and holds for any subgroup $A$ of a finite soluble group $G$ (except for the part where $\tilde P_i = P_i$).

Main construction

Set $p_0 = p$ and let $\{\tilde P_1, \tilde P_2, \cdots, \tilde P_l, \tilde P = \tilde P_0\}$ be a Sylow basis for $A$. Let $\tilde Q_i$ be the product of the $\tilde P_j$ for $j \neq i$. Since we have a Sylow basis, each $\tilde Q_i$ is a subgroup of order coprime to $p_i$ and of index (relative to $A$) a power of $p_i$ (called a Hall $p_i'$-subgroup, or a Sylow $p$-complement; the collection of $\{\tilde Q_i\}$ is called a Sylow complement system for $A$).

Let $Q_i$ be a Hall $p_i'$-subgroup of $G$ containing $\tilde Q_i$ (let me know if you need to see a proof of the existence of $Q_i$; this is in Hall's 1928 paper). Define $P_\pi = \bigcap_{p_j \notin \pi} Q_j$. Then I claim $\{ P_i \} =  \{ P_{\{p_i\}} \}$ is a Sylow basis of $G$, and that it extends the Sylow basis of $A$.

Verification

$P_\pi$ is a $\pi$-subgroup.

Proof: If $p_j \notin \pi$, then $P_\pi \leq Q_j$, and $Q_j$ is a $p_j'$-subgroup, so $P_\pi$ does not have order divisible by $p_j$.

$P_\pi$ is a Hall $\pi$-subgroup of $G$.

Proof: This is true by definition if $\pi$ is missing only zero or one primes dividing $|G|$, since then $P_\pi = G$ or $P_\pi = Q_j$. Suppose for induction's sake that $P_\pi$ is a Hall $\pi$-subgroup of $G$ for $\pi$ missing less than $k$ primes dividing the order of $G$, and let $\pi = \rho \setminus \{ p_i \}$ for $\rho \neq \pi$ be a set of primes missing exactly $k$ prime divisors of $|G|$. Then by definition $P_\pi = P_\rho \cap Q_i$. Since $P_\rho$ is a Hall $\rho$-subgroup by induction, and $Q_i$ is a $p_i'$-subgroup, we have $[G:P_\rho]$ is a $\rho'$-number and $[G:Q_i]$ is a $p_i$-number, so their indices are relatively prime and hence $[G:P_\pi] = [G:P_\rho] [G:Q_i]$ so that $P_\pi$ is a Hall $\pi$-subgroup of $G$, and the induction proceeds.

$P_\pi \cap P_\rho = P_{\pi \cap \rho}$

Proof: This is just the definition.

$P_\pi P_\rho = P_\rho P_\pi = P_{\rho \cup \pi}$

Proof: $P_\pi$ and $P_\rho$ are by definition subgroups of $P_{\rho \cup \pi}$, so that $P_\pi P_\rho$ is a subset of $P_{\rho \cup \pi}$. However $|P_\rho P_\pi| = |P_\rho| |P_\pi| / |P_\rho \cap P_\pi|$ which is exactly the size of $|P_{\rho \cup \pi}|$, so we have $P_\pi P_\rho = P_{\rho \cup \pi}$ is a group.

$A \cap P_i = \tilde P_i$. (That is the Sylow basis of $G$ that we constructed “reduces” into $A$)

Proof: $A \cap P_i = A \cap \bigcap_{j \neq i} Q_j = \bigcap{j \neq i} (A \cap Q_j) = \bigcap_{j \neq i} \tilde Q_j = \tilde P_j$. The only tricky equality is the last: by definition $\tilde P_i \leq \tilde Q_j$ so we get the $\supseteq$. The other inclusion is just an order consideration using the previous claims for $\tilde P_\pi = \bigcap_{p_j \notin \pi} \tilde Q_j$.

$P_i = \tilde P_i$ if $i \neq 0$.

Proof: This is just an order consideration. $P_i$ is a Sylow $p_i$-subgroup of $G$ containing $\tilde P_i$, a Sylow $p_i$-subgroup of $A$, but the same power of $p_i$ divides both $|G|$ and $|A|$.

Bibliography

Hall, P.
“A note on soluble groups.”
Journal of the London Mathematical Society 3, (1928) 98-105.
JFM 54.0145.01
DOI:10.1112/jlms/s1-3.2.98

Hall, P.
“On the Sylow systems of a soluble group.”
Proceedings of the London Mathematical Society, (2nd series) 43, (1937) 316-323.
JFM 63.0069.04
DOI:10.1112/plms/s2-43.4.316

Carter, R. W.
“On a class of finite soluble groups.”
Proceedings of the London Mathematical Society, (3rd series) 9 (1959) 623–640.
MR114859
DOI:10.1112/plms/s3-9.4.623

Title: sylow basis of finite solvable groups


问:
How can I project matrix $A$ onto the space of the matrix $X$?
I don't know if this question makes any sense, but how can I project matrix $A$ onto the space of the matrix $X$ (how can I compute $\textrm{Projection}_{X}A$)?

Thank you,

答:
The question makes sense: with $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$, in the vector space $\mathcal{M}_n(\mathbb{K})$, you want to find the projection on $\mbox{Span}(X)$ (assuming that $X$ is not zero, otherwise it is obvious).

A good candidate is $P(M) = \frac{\mbox{Tr}(M^TX)}{\mbox{Tr}(X^TX)}X$. Then the image of $P$ is $\mbox{Span}(X)$ and for any $\lambda \in \mathbb{K}$, you find $P(\lambda X) = \lambda X$.


问:
Showing a polynomial is of degree $N$ at most
I found the following exercise in my calculus notes:

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f\in C^\infty (\mathbb{R})$ and $f^{(k)}\equiv 0$ for all $k\geq N$ Use Taylor's formula to show that $f(x)$ is a polynomial of degree $N$ at most.

I've been trying to solve the previous exercise but I don't really know how to start. Any help is appreciated

答: The $k$-th term of the Taylor series centered at $a$, by Taylor's theorem, is $f^{(k)}(a)(x-a)^k/k!=0$, for all $k\ge N$.  The function is real analytic (equals its Taylor series), and we can actually say it is a polynomial of degree at most $N-1$.


答:
Here we avoid division by estimating the quotient and don't worry about negative residues.

We observe that

$\quad 2 \nmid 2021019 \land 2 \nmid 1431471$

and

$\quad 5 \nmid 2021019 \land 5 \nmid 1431471$

and this allows us some 'trick' flexibility in 'getting close to the modulus'.

Since $\frac{100}{14} \approx 7$ we can perform a trick and change the problem.

Problem: Find $\text{gcd}(2021019, 1431471) =$

$\quad\quad\quad \text{gcd}(5*2021019, 1431471) =$

$\quad\quad\quad \text{gcd}(10105095, 1431471)$.

Since $10105095 \approx 100 \cdot 10^5$ and $1431471 \approx 14 \cdot 10^5$ we (directly with quotient set to $7$) write

$\; 10105095 = (7)\cdot(1431471) + 84798$

New Problem: Find $\text{gcd}(1431471, 84798) =$

$\quad\quad\quad \text{gcd}(1431471, 42399)$

Since $1431471 \approx 143 \cdot 10^4$ and $42399 \approx 42 \cdot 10^3$ we estimate the quotient

(we've got an extra digit to use since $\Large \frac{10^4}{10^3} = 10$),

$\; \frac{143}{42} \approx 3.4$

and (directly with quotient set to $34$) write

$\; 1431471 = (34)\cdot(42399) - 10095$

New Problem: Find $\text{gcd}(42399, 10095) =$

$\quad\quad\quad \text{gcd}(42399, 2019)$.

When estimating the quotient it looks like $21$ works and we write

$\; 42399 = (21)\cdot(2019) + 0$

So

$\tag{ANSWER} \text{gcd}(2021019,1431471) = 2019$

Here is another method (using the tricks from the answer above) that is fast if you are an expert at subtraction and exact division with both $2$ and $10$.

$\quad\quad\quad \text{gcd}(2021019, 1431471) =$

$\quad\quad\quad \text{gcd}(2021019-1431471, 1431471) =$

$\quad\quad\quad \text{gcd}(589548, 1431471) =$

$\quad\quad\quad \text{gcd}(294774, 1431471) =$

$\quad\quad\quad \text{gcd}(147387, 1431471) =$

$\quad\quad\quad \text{gcd}(147387, 1431471-147387) =$

$\quad\quad\quad \text{gcd}(147387, 1284084) =$

$\quad\quad\quad \text{gcd}(147387, 642042) =$

$\quad\quad\quad \text{gcd}(147387, 321021) =$

$\quad\quad\quad \text{gcd}(147387, 321021-147387) =$

$\quad\quad\quad \text{gcd}(147387, 173634) =$

$\quad\quad\quad \text{gcd}(147387, 86817) =$

$\quad\quad\quad \text{gcd}(147387-86817, 86817) =$

$\quad\quad\quad \text{gcd}(60570, 86817) =$

$\quad\quad\quad \text{gcd}(6057, 86817) =$

$\quad\quad\quad \text{gcd}(6057, 86817-6057) =$

$\quad\quad\quad \text{gcd}(6057, 80760) =$

$\quad\quad\quad \text{gcd}(6057, 8076) =$

$\quad\quad\quad \text{gcd}(6057, 4038) =$

$\quad\quad\quad \text{gcd}(6057, 2019) =$

$\quad\quad\quad \text{gcd}(6057-2019, 2019) =$

$\quad\quad\quad \text{gcd}(4038, 2019) =$

$\quad\quad\quad \text{gcd}(2019, 2019)$

Title: Are there any other ways to find GCD faster?

Body:
Find $GCD$ of $2021019$ and $1431471$

My solution:

$2021019 = 1\cdot(1431471) + 589548$

$1431471 = 2\cdot(589548) + 252375$

$589548 = 2\cdot(252375) + 84798$

$252375 = 2\cdot(84798) + 82779$

$84798 = 1\cdot(82779) + 2019$

$82779 = 41\cdot(2019)$

So $GCD$ of $2021019$ and $1431471$ is $2019$

Are there any other ways to solve this faster?


问:
But what's the *actual* way to *actually* solve a second order linear PDE?
https://www.youtube.com/watch?v=5OL1WUyw2ag

Nope, that doesn't give me an explicit solution from boundary conditions, something's not right here.

This wouldn't even work on a first order problem, it would leave this weird unknown summed function left over even though the method of characteristics could easily give me an explicit solution from known boundary conditions.

I know this method in the video is related to an old and mostly useless calculus 3 trick to integrate that way if you happen to have some other piece of information about the function, but this is an ill-posed problem as far as I can tell. If solving PDE was as easy as integrating, well then the field of differential equations wouldn't even exist, you could solve $x'(t) = x(t)$ by just saying the solution is $\int x'(t)dt+C$ which obviously solves nothing.

答:
The reason no condition is given is because no condition is needed. Consider the simple ODE
$$\mathrm{D}^2_x[f](x)=0$$
Which of course has solution
$$f(x)=c_1 x+c_2$$
Since we imposed no boundary conditions on $f$, the constants are arbitrary. They can be any real or complex numbers we want.

Now consider the analogous PDE:
$$\partial^2_x [F](x,y)=0$$
The solution is exactly the same, but now instead of arbitrary constants, we have arbitrary functions -
$$F(x,y)=f_1(y)x+f_2(y)$$
Once again, since we imposed no boundary conditions, the functions are completely arbitrary. What's wrong with this approach?


问:
Proof of QR decomposition

Prove that any $n \times n$ matric over $\mathbb{C}$ can be written as $A = QR$ where $Q$ is unitariy and $R$ is upper-triangular.

I believe I understand the proof when $A$ is invertible and all of the columns are linearly independent, hence nonzero, and have non-negative lengths. I can use the Gram-Schmidt procedure to create an orthonormal basis, the columns of which are $Q$, and then the columns of the matrix $R$ contain the coefficients of the orthonormal basis relative to the old basis. I'm only having trouble generalizing. Though I can solve the $3 \times 3$ case, for example, I do not know how to prove I can do this for the $n \times n$ case. Is induction the way to go? The other problem is generalizing to the case where $A$ is not invertible.

答:
Let $a_1,a_2,...,a_n$ the columns
from $A$ and let $q_1, q_2,..,q_n$
the vectors obtained by orthonormalizing them by  Gram-Schmidt.

$$Gen(a_1,a_2,...,a_n)=Gen(q_1,q_2,...,q_n)$$

we defined

$$Q=[q_1q_2...q_n]$$
as each  $a_I$ is a linear combination
from $q_1q_2...q_n$ scalars must exist
$r_ij$ such that :

$$a_i=r_{1i}q_1+...+r_{ni}q_{n}=Q (r_{1i},..,r_{ni})^{T} \quad \text{$i=1,2,...,n$}$$

where $r_{ji=0}$ , $j=i+1,...,n$ and  $i=1,2,...,n$, by Gram-Schmidt. then,

$$A=(a_1,...,a_n)=(Q(r_{11},...,0),...,Q(r_1,...,r_{nm}))^T=QR$$

Where $R$ is the matrix whose element $(i, j)$ is $r_{ij}$. The matrices searched are the matrices $Q$ and $R$: $Q$ has its orthonormal columns and $R$ is upper triangular. Also $R$ must be invertible because otherwise $Rx=0$ will have infinite solutions and therefore also   $QRx=Ax=0$ contradicting the fact that the columns of $A$ are linearly independent.

Title: Proof of QR decomposition


问:
Finding a kernel of a linear transformation of linear transformations.
Question:
Let V,W be vector spaces over field F.
We mark L(V,W) as the vector space of linear transformations from V to W.
Let $v_0 \ne 0$. We define a transformation: $\Psi: L(V,W) \to W$ that sends a linear transformation $T \in L(V,W)$ to $T(v_0) \in W$

What is the image of $\Psi$ ? Evaluate  $dim(Ker \Psi)$

What I did:
I said that $dim(V)=m, dim(W)=n$, therefore $dim L(V,W)=mn$
I see that $\Psi$ can't be injective because the dimensions of L and W are not equal.
$Im\Psi=\Psi(T)=T(v_0)$
Can't figure how should I calculate the dim of the Kernel.
A thought I had is that since dimW < dimL then if $\Psi$ is surjective then $dimKer\Psi = mn-m$, but I don't know what if it's not surjective.

Thanks in advance

答: Well, if I'm not missing something, $\Psi$ is always surjective (for any $V, W$), at least assuming $dim(V)$ and $dim(W)$ are finite, as you implicitely seem to do when you write $dim(V)=m$ and $dim(W)=n$. To show this, complete $v_{0}$ to a base of $V$, let's say $(v_{0},\dots,v_{m-1})$ (a single non zero vector is linearly independent so you can do this). For any $w\in W$ consider the linear transformation $T\in L(V,W)$ such that $T(v_{0})=w$ and $T(v_{i})=0$ for any $i=1,\dots,m-1$. Obviously $\Psi (T)=w$ so that $Im (\Psi)=W$ and then one gets $dim(Ker(\Psi))=mn-n=dim(L(V,W))-dim(Im(\Psi))$.</p>


问:
Find which of the following groups are isomorphic
I want to check if my solution to this problem from my abstract algebra course is correct. The problem says:

Find which of the following groups are isomorphic: $$\mathbb{Z}_8 \ \
 , \ \ \mathbb{Z}_4\oplus\mathbb{Z}_2 \ \ , \ \ \mathbb{Z}_2^3 \ \ , \
 \ \mathbb{Z}_{24}^\times \ \ , \ \ \mathbb{Z}_{30}^\times \ \ , \ \ D_4.$$

What I did is calculate the order of each element from each group. Doing that I obtained:

$\mathbb{Z}_8$ : 4 elements of order $8$, 2 elements of order $4$, 1 element of order $2$.

$\mathbb{Z}_4\oplus\mathbb{Z}_2$, $\ \mathbb{Z}_{30}^\times$ : 4 elements of order $4$, 3 elements of order $2$.

$\mathbb{Z}_2^3$, $\ \mathbb{Z}_{24}^\times$ : 7 elements of order $2$.

$D_4$ : 2 elements of order $4$, 5 elements of order $2$.

I don't know if this condition is sufficient to ensure that the groups that share the "number of elements of each order" are isomorphic. Am I missing something? Any help will be appreciated, thanks in advance.

答:
Finite abelian groups are determined by the number of elements of any given order. But if you don't know the classification of abelian groups you can always be more concrete in any given case: you can map $(a,b,c) \in (\mathbb{Z}/2 \mathbb{Z})^3$ to $(-1)^a \times 5^b \times 7^c \in (\mathbb{Z}^{\times}_{24})$ and check that this is an isomorphism directly.

As noted in the comments, two groups can have the same number of elements of the same order and not be isomorphic. A good example to keep in mind is the abelian group $G = (\mathbb{Z}/3 \mathbb{Z})^3$ and the non-abelian $3$-Sylow subgroup of $\mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$:

$$\left( \begin{matrix} 1 & * & * \\ 0 & 1 & * \\ 0 & 0 & 1 \end{matrix} \right) \subset \mathrm{GL}_3(\mathbb{Z}/3\mathbb{Z})$$

Both groups have $27$ elements and every non-trivial element has order $3$. If you don't like the description of that latter group you could think about the permutation group:

$$\langle (1, 4, 7)(2, 5, 8)(3, 6, 9),(1, 3, 2)(4, 5, 6),(4, 6, 5)(7, 8, 9) \rangle \subset S_9.$$


问:
How can I take a partial derivative of an equation like $2u+v=5$?
I know that $\frac{du}{dx} = 0$ and $\frac{du}{dy} = 0$. But I don't undestand how? how can you take a derivative of this when you don't know the value of either u or v?

答:
When given an equation like this, you can apply implicit differentiation, which means that you differentiate each side of the equations as though they're actually a function that hasn't been explicitly stated. And then, since they're supposed to represent the same function, their derivatives must also be equal.

So it looks something like this:

$$\begin{eqnarray}2u + v & = & 5 \\
\frac{\partial}{\partial x} \left(2u + v \right) & = & \frac{\partial}{\partial x}(5) \\
\frac{\partial}{\partial x}(2u) + \frac{\partial}{\partial x}(v) & = & 0 \mbox{ (derivative of a constant is 0)} \\
2\frac{\partial u}{\partial x} + \frac{\partial v}{\partial x} & = & 0
\end{eqnarray}$$

and the same thing happens for $\frac{\partial}{\partial y}$.

It's equivalent to introducing the phantom function $f(x, y) = 2u + v$, finding $\frac{\partial f}{\partial x}$, then working with the equation $f(x, y) = 5$, but since we'll be discarding $f$ anyway we just never state it explicitly.


答:
I'd split this up using the minors of the second column: $$\begin{vmatrix}1&0&1&0\\-1&x&0&1\\0&0&x&1\\x^2&x&7&0\end{vmatrix}=x\begin{vmatrix}1&1&0\\0&x&1\\x^2&7&0\end{vmatrix}+x\begin{vmatrix}1&1&0\\-1&0&1\\0&x&1\end{vmatrix}$$

$$=x(0+x^2+0-0-7-0)+x(0+0+0+1-x-0)=x(x^2-x-6)=x(x+2)(x-3)$$

The matrix is non-invertible when $\det(B)=0$, and so we simply solve this, i.e. $x=0,x=-2,x=3$

Title: Find all values of x for which B is not invertible.

Body:
The matrix $B$ is below:

$$ B = \begin{pmatrix} 1 & 0 & 1 & 0 \\ -1 & x & 0 & 1 \\ 0 & 0 & x & 1 \\ x^2 & x & 7 & 0 \end{pmatrix} $$

Using Gaussian elimination I got this for the determinant.   -----

$$  \det B = 1×x×x×((x^2-x-6)/x) = x^3-x^2-6x $$

I need the values $x$ that make $B$ noninvertible.

Can you help me out from there?


问:
There exists a graph $G$ with chromatic number $\chi(G)\ge n$ but $G$ does not contain a triangle
Could you give an example or a proof for a graph that has no triangles but still has chromatic number $\chi(G) \ge n$, where $n \gt 3$ is the number of vertices? This is from coloring of a graph.

The only way I can think about it is in terms of complete graph $K_n$, but that has triangles and its chromatic number is $\chi(K_n) \ge n$. But can you give an example for a graph that is not complete and still has $\chi(G) \ge n$?

答: I assume that by $n$ you mean the number of vertices. If a graph is not complete then there is a pair of vertices with no edge between them, color them in the same color and all of the other vertices in their own unique colors to obtain an $(n-1)$-coloring. So a graph that has chromatic number $\ge n$ is complete. A complete graph can still contain no triangle, as long as it has less than 3 vertices :). If it has 3 or more then take any 3 of them to form a triangle.


问:
how to prove that a set is a partition of another set
problem 2 let f:X$\rightarrow$Y  be onto . For each b$\in$Y, let $A_b$=f^-1[{b}]. Prove that { $A_b$:b$\in$ Y} is a partition of X

I know I need to prove that for all sets in { $A_b$:b$\in$ Y} they are not empty
2 for all sets C,D$\in${ $A_b$:b$\in$ Y}, C=D or C$\cap$D=$\emptyset$
and for allx$\in$X there exists a set E$\in${ $A_b$:b$\in$ Y} such that x$\in$E
how do I prove this

答: The relation $x\sim x' \stackrel{(def.)}{\iff} f(x)=f(x')$ is an equivalence relation in $X$, and the equivalence class of $x\in X$ is given by $[x]_\sim=\{x'\in X\mid f(x')=f(x)\}= A_{f(x)}$. Since $f$ is surjective, $\forall b\in Y, \exists x\in X$ such that $b=f(x)$, and hence $\{A_b, b\in Y\}=\{A_{f(x)}, x\in X\}=\{[x]_\sim, x\in X\}=X/\sim$.


答:
An upper bound for chromatic number is $\chi(G) \le \Delta(G)+1$ (see Greedy coloring). And we know that for any simple graph $G$, we have $\Delta(G) \le n-1$ (Note that if $G$ is not simple, we can consider its condensation by removing parallel edges as they don't affect chromatic number). Thus, we have $\chi(G) \le n$. So, it is not hard to see that $\chi(G) = n$ is possible only when we have $G = K_n$. But, as you noticed, this is not a triangle-free graph for $n \ge 3$.

However, if you are not restricting yourself with number of vertices, I recommend you to see Mycielski Graph as they can generate a graph with any chromatic number you want and they are always triangle-free.

Title: There exists a graph $G$ with chromatic number $\chi(G)\ge n$ but $G$ does not contain a triangle

Body:
Could you give an example or a proof for a graph that has no triangles but still has chromatic number $\chi(G) \ge n$, where $n \gt 3$ is the number of vertices? This is from coloring of a graph.

The only way I can think about it is in terms of complete graph $K_n$, but that has triangles and its chromatic number is $\chi(K_n) \ge n$. But can you give an example for a graph that is not complete and still has $\chi(G) \ge n$?


问:
The definition of a regular operator
In this paper the notion of a regular operator is referenced. I have not heard the term before, in one sentence the author remarks:

A closed operator $T$ is regular and self-adjoint if and only if $T −\lambda$ is surjective for all $\lambda \in \Bbb C\setminus \Bbb R$, if and only if the Cayley transform of $T$ is unitary, if and only if $T$ has a functional calculus homomorphism on $C_0(\Bbb R)$.

This is confusing to me: If $T$ is self-adjoint then all these things are already true!

What am I missing, in particular: What is the correct definition of "regular" used here? Google is not helpful, unfortunately.

答:
@user854214 has already settled this question in the comments with the reference to
Pal's paper but it might perhaps be useful to see an  example illustrating  why must  the definition of a regular operator
in the context of Hilbert-modules be so much more intricate than the definition for Hilbert spaces.

Recall that if $A$ is a C*-algebra and $E$ and $F$ are Hilbert-modules over $A$, then an operator from $E$ to $F$ is said to be regular if:

$T$ is closed and densely defined,

its adjoint $T^*$ is also densely defined, and

the range of $I + T^*T$ is dense in $E$.

As noted by Pal, if $A=\mathbb C$, in which case $E$ and $F$ are nothing but Hilbert spaces, then (2) and (3) follow from
(1), so regular operators are nothing but  closed operators.

As the following example  illustrates, already $(1) \Rightarrow (2)$ may fail, even for bounded operators.

Let $A$ be any unital C*-algebra and let $I$ be a closed 2-sided ideal in $A$.  We may then view both $A$ and $I$ as (right) Hilbert
modules over $A$ with the obvious module structure and with the inner product defined by $\langle x, y\rangle  = x^*y$.

Let  $T:I\to A$ be the inclusion of $I$ into $A$.
As a bounded map, it is obvious that $T$ satisfies (1).

The domain of $T^*$ is then the set of elements $a$ in $A$ such that there
exists some (necessarily unique) $x$ in $I$ satisfying
$$
  \langle T(y), a\rangle  =   \langle y, x\rangle , \quad\forall y\in  I.
  $$
This translates to $y^*a=y^*x$, and  one then has by definition that  $T^*(a)=x$.

In this case, notice that $a-x$ belongs to
$$
  I^\perp := \{b\in A: yb = 0, \quad\forall y\in  I \},
  $$
so we have that
$$
  a = x+a-x\in  I+I^\perp,
  $$
and in fact it is easy to prove that the domain of $T^*$ is precisely $I+I^\perp$,  Moreover
$I+I^\perp$ is always closed, so (2) can only hold when $I+I^\perp=A$.  However,   more often than not, this is false.

For example, if
$A=C(X)$ and $I=C_0(U)$, for some open set $U\subseteq X$, then  $I^\perp=C_0(X\setminus \overline U)$ and
$$
  I+I^\perp=\{f\in C(X): f \text{ vanishes on } \partial U\},
  $$
which coincides with $C(X)$ iff $\partial U$ is empty.

Title: The definition of a regular operator


问:
coarser/finer comparison for relative topology and norm topology
Suppose $p\in [1, \infty]$. Show the relative topology on $l^p(\mathbb{R})$ induced from the product topology on $\mathbb{R}^{\omega}= \Pi_{l=1}^{\infty} \mathbb{R}$ is strictly coarser than the $p$-norm topology on $l^p(\mathbb{R})$, but the relative topology on $l^p(\mathbb{R})$induced by the box topology on $\mathbb{R}^{\omega}$ is strictly finer than the $p$-norm topology on $l^p(\mathbb{R})$.

Attempt:

We know that $l^p(\mathbb{R}) \subseteq \mathbb{R}^{\omega}$.

By definition the relative topology is $S = \{ V \cap l^p(
\mathbb{R}): V \in T \}$, where $T$ is the product topology on $\mathbb{R}^{\omega}$.

To show it's coarser than the $p$-norm topology, we want to show that there exists an open ball $B$ in $l^p(\mathbb{R})$ such that $B \notin S$. But I'm not sure how to find such open ball.

Now let $S' = \{ V \cap l^p(
\mathbb{R}): V \in T \}$, where $T$ is the box topology on $\mathbb{R}^{\omega}$.

I'm not sure what to do to show the $p$-norm topology is strictly coarser than $S'$.

Thanks in advance!

答:
I will show the first part and leave pointers for the second part.

It is easy to see that any basic open set of the relative product topology contains an open $p$-ball from the $p$-norm topology (justify this yourself). But, consider the unit $p$-ball in $\ell^p(\mathbb{R})$: $B_p(0, 1) = \{x \in \ell^p(\mathbb{R}) : \|x\|_p < 1\}$. This is not open in the product topology induced on $\ell^p(\mathbb{R})$. To see this, consider the sequence $(0)_{i = 1}^\infty \in B_p(0, 1)$. And let
$$U = \Big(\prod_{i = 1}^\infty (a_i, b_i)\Big) \cap \ell^p(\mathbb{R}),\ (a_j, b_j) = (-\infty, +\infty)  \text{ for all but finitely many $j \in I_{\text{finite}}$}$$ be any basic open set of the induced product topology containing $(0)_{i = 1}^\infty$. Thus, $a_i < 0 < b_i$. Then $U$ is never contained in $B_p(0, 1)$. Indeed, choose any of the infinitely many $n \in \Bbb Z_+ - I_{\text{finite}}$ such that $(a_n, b_n) = (-\infty, +\infty)$. And consider the sequence $(\delta_{n,i})_{i = 1}^\infty$ (i.e. it is $1$ at index $n$ and $0$ elsewhere). Then $\|(\delta_{n,i})_{i = 1}^\infty\|_p = \sqrt[p]{\sum_{i = 1}^\infty|\delta_{n,i}|^p} = \sqrt[p]{|\delta_{n,n}|^p} = 1$. So $(\delta_{n,i})_{i = 1}^\infty \not\in B_p(0, 1)$. But $(\delta_{n,i})_{i = 1}^\infty \in U$ easily because obviously $(\delta_{n,i})_{i = 1}^\infty$ converges in the $p$-norm and $a_i < 0 < b_i, i \neq n$ while $-\infty = a_n < 1 < b_n = +\infty$.

The situation changes with the induced box topology. Again, it is not hard to see the induced box topology contains the $p$-norm topology (justify this yourself). But, consider the following basic open set from the induced box topology:
$$
U = \Big(\prod_{i = 1}^\infty (-4^{-i/p}, 4^{-i/p})\Big) \cap \ell^p(\mathbb{R})
$$ Again, $(0)_{i = 1}^\infty \in U$ but no open $p$-ball $B_p(a, R), a \in \ell^p(\mathbb{R}), R > 0$ containing $(0)_{i = 1}^\infty$ is in $U$. To see this, first simplify: any such $B_p(a, R)$ also contains another ball centered at $(0)_{i = 1}^\infty$: $B_p(0, r), R > r > 0$ (why?). Now, the sequence $(\frac{1}{2}r\ 2^{-i/p})_{i = 1}^\infty \in B_p(0, r)$ (why does this converge in the $p$-norm? why is it in the ball?). However, $(\frac{1}{2}r\ 2^{-i/p})_{i = 1}^\infty \not\in U$ (find some $j \in \Bbb Z_+$ large enough such that $4^{-j/p} < \frac{1}{2}r\ 2^{-j/p}$; why is this enough?).


问:
Example of a space whose complex K-theory is not easily computable from singular cohomology
I am looking for a counterexample to the formula
$$ K^n(X) \cong \prod_{i\equiv n \mod 2} H^i(X) $$
where $K^*$ denotes complex topological $K$-theory, $H^*$ singular cohomology and $X$ a compact space.
I would hope for an example where $X$ is a finite $CW$-complex.

There should be such an example, as I was told that the $K$-theoretic Atiyah-Hirzebruch sprectral sequence does not collapse at $E^2$ in general.

答:
Since you mention the Atiyah-Hirzebruch spectral sequence, I want to point out how you might construct a (minimal) example if you somehow didn't know any examples of topological spaces (but do know quite a bit of algebraic topology).

As we're working with the $K$-theory of spaces, we would like to understand the structure of the connective complex $K$-theory spectrum $ku$ better and find examples of $X$ for which the AHSS does not degenerate at $E_2$.  It's well-known that the first differential in the AHSS is given by the bottom nontrivial $k$-invariant in the Postnikov tower of $ku$.

Suppose you somehow knew that this $k$-invariant is given by $$Q_1: \Sigma^{-1} H\mathbb{Z} \to \Sigma^2 H\mathbb{Z}.$$

Then, counterexamples to your question can be constructed if we can find spaces whose cohomology has a nontrivial $Q_1$ operation.  Let's work mod $2$; we have $$Q_1 = \mathrm{Sq}^1 \mathrm{Sq}^2 + \mathrm{Sq}^2 \mathrm{Sq}^1$$ in mod $2$ cohomology.  Therefore, a minimal cell complex with a nontrivial $Q_1$ has at least three cells connected by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$ in either order (i.e., a complex that looks like a question mark, or inverted by Spanier-Whitehead duality).  We can build such complexes cell-by-cell by taking iterated cofibers of maps between spheres.  In particular, since $\mathrm{Sq}^1$ detects $2$ and $\mathrm{Sq}^2$ detects $\eta$ in the homotopy groups of spheres, we can build the required three cell complex in two steps by taking cofibers of maps representing $2$ and $\eta$.  That this construction is possible is guaranteed by the relation $2\eta = 0$ in $\pi^s_*$.

So we now have a finite complex with a nontrivial $Q_1$ operation, and hence a counterexample to your formula in the question.  How does this relate to the example $\mathbb{R}P^n$ given in the other answer?  Well, if you look at the cell structure of $\mathbb{R}P^n$, you would see exactly this kind of question mark complex as a frequently occuring motif.

Of course, this answer isn't very helpful since by the time you come up with it you would have other simpler ways to see this result.  On the other hand, maybe it's a good showcase of how the tools developed in homotopy theory can answer such questions "intrinsically".


问:
Study the continuity of $f(x)=1$ if $x=1/n$ where $n\in \mathbb{Z}$ or $0$ otherwise on $[-1,1]$
$$f(x)=\begin{cases} 1 & \text{for } x=\frac{1}{n}, \; n\in\mathbb{Z}\\0 & \text{otherwise.}\end{cases}$$

I am asked to find the points of discontinuity on the interval $[-1,1]$.

I think I'm supposed to prove that the limit of the function at any given $x=\frac{n}{2}$ will be $0$, so the function is discontinuous at those points, but I don't know how to prove this.

答:
$f$ is continuous on all intervals $\left(\frac{1}{n + 1}, \frac{1}{n}\right), n \gt 0$ and $\left(\frac{1}{n}, \frac{1}{n - 1}\right), n \lt 0$.

$f$ is discontinuous at all points $\frac{1}{n}, n \in \mathbb{Z}$ because $f(\frac{1}{n}) = 1 \neq 0 = \lim_{x \to \frac{1}{n}} f(x)$. These are all removable discontinuities.

The only other point of interest is $x = 0$. $f$ is discontinuous at $0$ because $\lim_{x \to 0} f(x)$ does not exist; this is an essential discontinuity. Sequences $\left(x_n\right)$ can be chosen where $x_n \to 0$ and $f(x_n) \to 0$ or $f(x_n) \to 1$ or $f(x_n)$ flips back and forth between $0$ and $1$. For example, $(x_n) = \frac{1}{n}$ has $f(x_n) \to 1$; $(x_n) = \frac{2}{2n + 1}$ has $f(x_n) \to 0$; and $(x_n) = \frac{2}{n}$ has $f(x_n)$ alternate between $0$ and $1$.


问:
Is it possible to find a null vector of matrix $A$ of size $m\times n$ when $m >> n$?
Is it possible to find a null vector (or null space) of matrix $A$ of size $m\times n$ when $m >> n$? or is it the case that in such a case, the null space does not exist?

Thank you,

答:
It is possible for an $m \times n$ matrix to have a nontrivial nullspace even when $m \gg n$. For instance, the $m \times 2$ matrix: $$A = \begin{bmatrix}1 & 1 \\ 2 & 2 \\ 3 & 3 \\ \vdots & \vdots \\ m & m\end{bmatrix}$$ has $v = \begin{bmatrix}1 \\ -1\end{bmatrix}$ in its nullspace for any integer $m \ge 1$.

Note that the nullspace of a matrix always exists and is non-empty because the vector of all zeros lies in the nullspace. However, there are non-zero vectors in the nullspace if and only if the rank of the matrix equals the number of columns.


问:
How to find the distribution function of compound Poisson process?

Let $\{X(t),t\ge 0\}$ be a compound Poisson process with $X(t)=\sum_{i=1}^{N(t)}X_i$.

Suppose that $\lambda=1$ and $P(\{X_i=j\})=j/10$, $j=1,2,3,4$. Calculate $P\{X(n)=20\}$.

Difficulty

The summands $X_i$ take four different values, and the number of ways in which they add up to $20$ is pretty large. For example, one can calculate the probability of $N(n)=5$ and $X_1=X_2=X_3=X_4=X_5=4$, which is one of many scenarios of $X(n)=20$. But the summation appears unmanageable. Is there a better way?

答:
Here is a possible start, using the better notation $X(t) = \sum\nolimits_{i = 1}^{N(t)} {X_i }$. By the law of total probability, conditioning on $N(n)$,

$$
{\rm P}(X(n) = 20) = \sum\limits_{k = 0}^\infty  {{\rm P}(X(n) = 20|N(n) = k){\rm P}(N(n) = k)}.
$$
Hence, since the $X_i$ take values in $\{1,2,3,4\}$,
$$
{\rm P}(X(n) = 20) = \sum\limits_{k = 5}^{20} {{\rm P}(X_1  +  \cdots  + X_k  = 20)\frac{{e^{ - n} n^k }}{{k!}}}.
$$

Another possibility. Use the probability-generating function of a compound Poisson process (from which you can obtain the probability mass function of $X(n)$ upon differentiation).

EDIT: The probability-generating function of $X(n)$ is given by
$$
G_n (z):={\rm E}[z^{X(n)} ] = e^{n[P_{X_i } (z) - 1]} ,
$$
where $P_{X_i} (z) = \sum\nolimits_{j = 1}^4 {z^j \frac{j}{{10}}} $ is the probability-generating function of $X_i$ (standard exercise).
Hence, using that
$$
{\rm P}(X(n) = k) = \frac{{G_n^{(k)} (0)}}{{k!}}
$$
(where $G_n^{(k)} (0)$ is the $k$th derivative evaluated at $0$), you can in principle obtain the probability in question, that is ${\rm P}(X(n) = 20)$.


问:
Differentiation of $x^{\sqrt{x}}$, how?
The answer is (I think)
$x^{\sqrt{x}-0.5} (1+0.5\ln(x))$, but how?</p>

答:
$$y=x^{\sqrt{x}}$$
 $$\ln y=\sqrt{x}\ln x$$
now differentiate w.r.t. x using chain rule
$$\dfrac {1}{y}\cdot\dfrac {dy}{dx}=\dfrac{\sqrt{x}}{x}+\dfrac{\ln x}{2\sqrt x}$$
$$\dfrac {dy}{dx}=x^{\sqrt{x}}\cdot\left(\dfrac{2+\ln x}{2\sqrt x}\right)$$</p>

Title: Differentiation of $x^{\sqrt{x}}$, how?


问:
Does for any non-negative discrete random variable $X$ hold $P(X \neq 0) \geq \frac{(E[X])^2}{E[X^2]}$?

Suppose $X$ is a non-negative random variable on $\mathbb{Z}$ with finite second moment. Is it always true that $P(X \neq 0) \geq \frac{(E[X])^2}{E[X^2]}$?

I managed to prove only a weaker bound:

$$P(X \neq 0) = 1 - P(X = 0) \geq 1 - P(X \leq 0) = 1 - P(-X \geq 0) = 1 - P(EX -X \geq EX) = 1 - P(|EX -X| \geq EX) \geq 1 - \frac{Var[X]}{(E[X])^2} = 2 - \frac{E[X^2]}{(E[X])^2}$$

by Chebyshev inequality.

Unfortunately, this bound is always weaker than what I need due to that fact that $\inf\{t + t^{-1}| t \in \mathbb{R}_+\} = 2$.

But that stronger inequality holds for several large families of distributions:

For binomial $X$ with parameters $n$ and $p$ we have: $P(X \neq 0) = 1 - (1-p)^n \geq \frac{n p^2}{p(1-p) + np^2} = \frac{(E[X])^2}{E[X^2]}$

For negative binomial $X$ with parameters $r$ and $p$ we have: $P(X \neq 0) = 1 - (1-p)^r \geq \frac{pr}{pr+1} = \frac{(E[X])^2}{E[X^2]}$

For Poisson $X$ with parameter $\lambda$ we have: $P(X \neq 0) = 1 - e^{-\lambda} \geq \frac{\lambda}{\lambda+1} = \frac{(E[X])^2}{E[X^2]}$

However, I do not know how to prove it in general.

Body:
Yes. By C-S:

$$ E[X] = E [ X {\bf 1}_{X>0} ] \le (E [X^2])^{1/2} P(X>0)^{1/2}.$$

Therefore, assuming $E[X^2]>0$ (equivalent, under the assumption to $P(X \neq 0) > 0$),

$$ P(X>0) \ge \frac{E[X]^2}{E[X^2]}.$$