问:
Can any positive real be approximated as $2^m/3^n$ with $(m,n)$ large enough?
Conjecture.

There exist positive integers $(m,n)$ large enough, such that for any positive real number $r$ and a given error $\epsilon$ :
$$
   \left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
There is numerical evidence for this conjecture. I have tried $r = \sqrt{2}$ and $\epsilon = 10^{-3}$.

Below is a little Delphi Pascal program with accompanying output.

But .. can somebody prove the conjecture?

```
program apart;<br>
procedure test(r : double; eps : double);
var
  a : double;
  m,n : integer;
begin
  a := 1;
  m := 0; n := 0;
  while true do
  begin
    if a < r then
    begin
      m := m + 1;
      a := a * 2;
    end else begin
      n := n + 1;
      a := a / 3;
    end;
    if abs(r-a) < eps then Break;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',a);
end;<br>
begin
  test(sqrt(2),1.E-3);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935359E+0000
```

UPDATE.

The answer by lhf does look like a very concise proof. But for me - as an retired physicist by education - it's a bit beyond comprehension.

Furthermore, it leaves a few issues untouched. One might ask for example whether there are estimates for $m$ and $n$ when $r$ and $\epsilon$ are given.

Note. The question can also be formulated as:
Can any positive real be approximated as $3^m/2^n$ with $(m,n)$ large enough? Which is the same as allowing negative integers with the original formulation. In this form, it shows some resemblance to the (in)famous Collatz problem.

EDIT.

As suggested by the answers, an approach with logarithms could be more effective:

```
program anders;<br>
procedure proef(r : double; eps : double);
var
  a,l2,l3,lr : double;
  m,n : integer;
begin
  l2 := ln(2); l3 := ln(3);
  lr := ln(r); a := 0;
  m := 0; n := 0;
  while true do
  begin
    a := m*l2 - n*l3 - lr;
    if abs(a) < eps then Break;
    if a < 0 then m := m + 1 else n := n + 1;
  end;
  Writeln(r,' = 2^',m,'/3^',n,' =',exp(a)*r);
end;<br>
begin
  proef(sqrt(2),1.E-3);
  proef(sqrt(2),1.E-9);
end.
```

Output:

```
 1.41421356237310E+0000 = 2^243/3^153 = 1.41493657935356E+0000
 1.41421356237310E+0000 = 2^911485507/3^575083326 = 1.41421356125035E+0000
```

The first line in the output is almost identical to the result obtained previously .

The last line in the output shows that the latter approach indeed is more effective.
The error plays the same role in both approaches. Oh well, almost.
Let's take a look at the places where the 'Break's are. First program:
$$
\left| r - \frac{2^m}{3^n} \right| < \epsilon
$$
Second program:
$$
-\epsilon < m\ln(2) - n\ln(3) - \ln(r) < +\epsilon \\
\ln(1-\epsilon) < \ln\left(\frac{2^m/3^n}{r}\right) < \ln(1+\epsilon) \\
-\epsilon < \frac{2^m/3^n}{r} - 1 < +\epsilon \\
\left| r - \frac{2^m}{3^n} \right| < \epsilon.r
$$
So $\epsilon$ in the first program is an absolute error,
while $\epsilon$ in the second program is a relative error.

Continuing story at:

Can the Stern-Brocot tree be employed for better convergence of $2^m/3^n$?

答:
Heuristics of a another proof

Lemma 1.

The fractions $2^m/3^n$ are all between $r/3$ and $2r$.

Proof.

According to the program - as displayed in the question. Any fraction smaller than $r$ is multiplied by $2$,
so $r.2$ is an upper bound for these fractions. Any fraction greater than $r$ is divided by $3$, so $r/3$
is a lower bound for these fractions. There can be no other fractions, except when the iterations start.
$$
r/3 < \frac{2^m}{3^n} < 2r
$$
Lemma 2.

In the sequence $2^m/3^n \to r$ there are no two fractions the same.

Proof.

Suppose that we have $2^{m_1}/3^{n_1} = 2^{m_2}/3^{n_2}$.

Three cases are distinguished:

$m_1 \neq m_2$ and $n_1 = n_2$. Then $2^{m_1} = 2^{m_2}$ hence $m_1 = m_2$. Contradiction.

$n_1 \neq n_2$ and $m_1 = m_2$. Then $1/3^{n_1} = 1/3^{n_2}$ hence $n_1 = n_2$. Contradiction.

$m_1 \neq m_2$ and $n_1 \neq n_2$. Then we have:
$$
\ln\left(\frac{2^{m_1}}{3^{n_1}}\right) = \ln\left(\frac{2^{m_2}}{3^{n_2}}\right) \quad \Longrightarrow \\
(m_1-m_2)\ln(2) - (n_1-n_2)\ln(3) = 0 \quad \Longrightarrow \\ \frac{m_1-m_2}{n_1-n_2} = \frac{\ln(3)}{\ln(2)}
$$
But $\,\ln(3)/\ln(2)\,$ is not a rational number. Contradiction.

So what we have is a bunch of fractions, all different, but they must fit within the interval $\,]r/3,2r[\,$.
This means that the fractions become crowded. Let's make a picture of the iteration process, logarithmic version.
The red line is given by $\,\color{red}{\ln(3)y=\ln(2)x-\ln(r)}\,$, small circles are fractions,
mapped on a grid $\,m/n \to (m,n)\,$, massively black filled dots are the fractions in the iteration process,
while increasing $m$ and $n$ with increments one at a time. The iterations domain is limited by: $\,\color{blue}{-\ln(2)\lt\ln(3)y-\ln(2)x+\ln(r)\lt\ln(3)}\,$. In our case $r = 100$. Mind the sequence at the start.

So it seems that there must be quite some fractions nearby the red line, rerpesenting the real number $r$ in question.

How can we be sure about this? Let's make picture of the crowding of the approximations $a$ in the interval $\,]r/3,2r[\,$,
logarithmic scale:
$$
a = m\ln(2)-n\ln(3)-\ln(r) \quad \mbox{with} \quad -\ln(3) < a < \ln(2)
$$
The red line is where $a = 0$, the desired value.

Further numerical/graphical experiments reveal that the distribution of the fractions seems to be uniform.
While seeking further confirmation of this we have done the following, speaking in terms of (Delphi) Pascal:

```
program opnieuw;<br>
procedure interval(var A,B : double);
var
  h : double;
begin
  A := Random; B := Random;
  if A > B then
  begin
    h := B; B := A; A := h;
  end;
end;<br>
procedure proef(r : double);
const
  veel : integer = 1000000000;
var
  x,l2,l3,lr,A,B : double;
  m,n,tel,t : integer;
begin
  l2 := ln(2); l3 := ln(3);
  lr := ln(r);
  interval(A,B);
  A := -l3 + A*(l2+l3);
  B := -l3 + B*(l2+l3);
  m := 0; n := 0;
  tel := 0; t := 0;
  while tel < veel do
  begin
    x := m*l2 - n*l3 - lr;
    if x < 0 then m := m + 1 else n := n + 1;
    if (-l3 < x) and (x < +l2) then tel := tel + 1;
    if (A < x) and (x < B) then t := t + 1;
  end;
  Writeln((B-A)/(l2+l3),' = ',t/tel);
end;<br>
begin
  Random; Random;
  proef(1000);
  proef(0.001);
  proef(sqrt(2));
  proef(1/sqrt(2));
  while true do
    proef(Random);
end.
```

Explanation. Make random intervals $\,]A,B[\,$ inside $\,]-\ln(3),+\ln(2)[\,$.
The length of the latter interval is $\,\ln(3)+\ln(2)=\ln(6)\,$, the lenghs of the former are $\,(B-A)\,$.
Count the (logarithms $x$ of the) fractions $\,(2^n/3^n)/r\,$ in both intervals.
Let $N$ be the total number (tel) of iterands and $n$ be the number (t) of iterands in $\,]A,B[\,$.
Then the distribution of the approximations $x$ is uniform if and only if:
$$
\lim_{N\to\infty}\frac{n}{N} = \frac{B-A}{\ln(6)}
$$
Let's check. Output after a billion iterations each line:

```
 6.58467502100393E-0001 =  6.58467500000000E-0001
 3.98733151378110E-0001 =  3.98733149000000E-0001
 1.56895805848762E-0001 =  1.56895804000000E-0001
 5.34354087430984E-0002 =  5.34354050000000E-0002
 4.04224734520540E-0001 =  4.04224734000000E-0001
 2.33572337077931E-0001 =  2.33572341000000E-0001
 4.06758418539539E-0001 =  4.06758418000000E-0001
 1.46495995344594E-0001 =  ....
```

But how can we prove that it is a uniform distribution?

Body:
Heuristics of a another proof

Lemma 1.

The fractions $2^m/3^n$ are all between $r/3$ and $2r$.

Proof.

According to the program - as displayed in the question. Any fraction smaller than $r$ is multiplied by $2$,
so $r.2$ is an upper bound for these fractions. Any fraction greater than $r$ is divided by $3$, so $r/3$
is a lower bound for these fractions. There can be no other fractions, except when the iterations start.
$$
r/3 < \frac{2^m}{3^n} < 2r
$$
Lemma 2.

In the sequence $2^m/3^n \to r$ there are no two fractions the same.

Proof.

Suppose that we have $2^{m_1}/3^{n_1} = 2^{m_2}/3^{n_2}$.

Three cases are distinguished:

$m_1 \neq m_2$ and $n_1 = n_2$. Then $2^{m_1} = 2^{m_2}$ hence $m_1 = m_2$. Contradiction.

$n_1 \neq n_2$ and $m_1 = m_2$. Then $1/3^{n_1} = 1/3^{n_2}$ hence $n_1 = n_2$. Contradiction.

$m_1 \neq m_2$ and $n_1 \neq n_2$. Then we have:
$$
\ln\left(\frac{2^{m_1}}{3^{n_1}}\right) = \ln\left(\frac{2^{m_2}}{3^{n_2}}\right) \quad \Longrightarrow \\
(m_1-m_2)\ln(2) - (n_1-n_2)\ln(3) = 0 \quad \Longrightarrow \\ \frac{m_1-m_2}{n_1-n_2} = \frac{\ln(3)}{\ln(2)}
$$
But $\,\ln(3)/\ln(2)\,$ is not a rational number. Contradiction.

So what we have is a bunch of fractions, all different, but they must fit within the interval $\,]r/3,2r[\,$.
This means that the fractions become crowded. Let's make a picture of the iteration process, logarithmic version.
The red line is given by $\,\color{red}{\ln(3)y=\ln(2)x-\ln(r)}\,$, small circles are fractions,
mapped on a grid $\,m/n \to (m,n)\,$, massively black filled dots are the fractions in the iteration process,
while increasing $m$ and $n$ with increments one at a time. The iterations domain is limited by: $\,\color{blue}{-\ln(2)\lt\ln(3)y-\ln(2)x+\ln(r)\lt\ln(3)}\,$. In our case $r = 100$. Mind the sequence at the start.

So it seems that there must be quite some fractions nearby the red line, rerpesenting the real number $r$ in question.

How can we be sure about this? Let's make picture of the crowding of the approximations $a$ in the interval $\,]r/3,2r[\,$,
logarithmic scale:
$$
a = m\ln(2)-n\ln(3)-\ln(r) \quad \mbox{with} \quad -\ln(3) < a < \ln(2)
$$
The red line is where $a = 0$, the desired value.

Further numerical/graphical experiments reveal that the distribution of the fractions seems to be uniform.
While seeking further confirmation of this we have done the following, speaking in terms of (Delphi) Pascal:

program opnieuw;

procedure interval(var A,B : double);
var
  h : double;
begin
  A := Random; B := Random;
  if A > B then
  begin
    h := B; B := A; A := h;
  end;
end;

procedure proef(r : double);
const
  veel : integer = 1000000000;
var
  x,l2,l3,lr,A,B : double;
  m,n,tel,t : integer;
begin
  l2 := ln(2); l3 := ln(3);
  lr := ln(r);
  interval(A,B);
  A := -l3 + A*(l2+l3);
  B := -l3 + B*(l2+l3);
  m := 0; n := 0;
  tel := 0; t := 0;
  while tel < veel do
  begin
    x := m*l2 - n*l3 - lr;
    if x < 0 then m := m + 1 else n := n + 1;
    if (-l3 < x) and (x < +l2) then tel := tel + 1;
    if (A < x) and (x < B) then t := t + 1;
  end;
  Writeln((B-A)/(l2+l3),' = ',t/tel);
end;

begin
  Random; Random;
  proef(1000);
  proef(0.001);
  proef(sqrt(2));
  proef(1/sqrt(2));
  while true do
    proef(Random);
end.

Explanation. Make random intervals $\,]A,B[\,$ inside $\,]-\ln(3),+\ln(2)[\,$.
The length of the latter interval is $\,\ln(3)+\ln(2)=\ln(6)\,$, the lenghs of the former are $\,(B-A)\,$.
Count the (logarithms $x$ of the) fractions $\,(2^n/3^n)/r\,$ in both intervals.
Let $N$ be the total number (tel) of iterands and $n$ be the number (t) of iterands in $\,]A,B[\,$.
Then the distribution of the approximations $x$ is uniform if and only if:
$$
\lim_{N\to\infty}\frac{n}{N} = \frac{B-A}{\ln(6)}
$$
Let's check. Output after a billion iterations each line:

 6.58467502100393E-0001 =  6.58467500000000E-0001
 3.98733151378110E-0001 =  3.98733149000000E-0001
 1.56895805848762E-0001 =  1.56895804000000E-0001
 5.34354087430984E-0002 =  5.34354050000000E-0002
 4.04224734520540E-0001 =  4.04224734000000E-0001
 2.33572337077931E-0001 =  2.33572341000000E-0001
 4.06758418539539E-0001 =  4.06758418000000E-0001
 1.46495995344594E-0001 =  ....

But how can we prove that it is a uniform distribution?


问:
Proving that these two definitions of this equivalence relation are the same
Consider a powerset $P(X)$ of a set $X$. On $P(X)$, we define two equivalence relations. The first is defined as "$S+T$ is a finite set". (where $S$ and $T$ are subsets of $X$, and $+$ denotes symmetric difference). The second is defined as "there exist finite sets $F$ and $G$ such that $(S \cup F) - G = T$". How does one prove that these are the same relation?

答:
HINT: If $S+T$ is finite, let $F=T\setminus S$ and $G=S\setminus T$; then

$$(S\cup F)\setminus G=(S\cup T)\setminus(S\setminus T)=T\,,$$

and you need only verify that $F$ and $G$ are finite. Conversely, if there are finite $F$ and $G$ such that $(S\cup F)\setminus G=T$, show that $S+T\subseteq F\cup G$; use the fact that if $x\in S+T$, then either $x\in S\setminus T$, or $x\in T\setminus S$.

ContentLicense: CC BY-SA 4.0


问:
How to prove the Weierstrass product inequality using recurrency?
I have to prove the following the Weierstrass Product Inequality  product inequality using recurrency .
$$ \forall n \in N \setminus\{0 , 1\}; 1 - \sum_{k=0}^n a_k < \prod (1-a_k) < \left(1 + \sum_{k=0}^n a_k\right)^{-1} $$

with $ 0 < a_k < 1 $

I tried to prove it for $n_0 = 2$ , we get :

$$ 1 - a_1 - a_2 < (1- a_1)\cdot(1-a_2)$$

I don't even find a way to prove this part .

I'm looking for some hint that would help me understand how I can prove this inequality using recurrency . are there some inequality theorem  could use ?

My intuition is telling me that the fact : $ 0 < a_k < 1 $ is key to solving this, but I still can't find the path for it .

答:
Alternative Proof (left inequality)

We prove that $\forall a_i \in [0,1], i=1, 2, \ldots, n,$
$$
\prod _ { i = 1 } ^ n (1- a _ i) + \sum _ { i = 1 } ^ n a _ i - 1 =\sum_{i=1}^{n-1}  a_i\left(1-\prod_{j>i} (1-a_j)\right)\geqslant 0.\tag1
$$

It suffices to prove just the equality above.

Denote $P_i=\prod_{j=i}^n (1-a_j)$. Note that $(1-a_i) P_{i+1}=P_i, \forall i <n$, and $1 - a_n =P_n$.

Then

$$\sum_{i=1}^{n-1} a_i\left(1-\prod_{j>i}(1- a_j)\right)=\sum_{i=1}^{n-1} a_i\left(1-P_{i+1}\right)\\
=\sum_{i=1}^{n-1} (a_i -P_{i+1}+P_i)=\sum_{i=1}^{n-1} a_i - \sum_{i=1}^{n-1} P_{i+1} + \sum_{i=1}^{n-1} P_i\\
=\sum_{i=1}^{n-1} a_i - \left(\sum_{i=2}^{n-1} P_i + P_n \right) + \left(P_1 + \sum_{i=2}^{n-1} P_i\right) = P_1 - P_n + \sum_{i=1}^{n-1} a_i \\
=\prod_{i=1}^n (1-a_i)-(1-a_n)+\sum_{i=1}^{n-1} a_i = \prod_{i=1}^n (1-a_i)+\sum_{i=1}^{n} a_i -1 .
$$

When do we have equality? WLOG assume $a_1 \leqslant a_2 \leqslant \cdots \leqslant a_n$, we must have $$a_{n-1}a_n=0 \implies a_{n-1}=0 \implies a_1=a_2 = \cdots =a_{n-1}=0.$$

On the other hand if $n-1$ of the $a_i's$ are zero, then the equality holds. Therefore $$"=" \iff (n-1) \text{ of } a_i's \text{ are equal to zero}.\tag2 $$

ContentLicense: CC BY-SA 3.0


问:
Notation on isomorphism showing that a vector space is isomorphic to its double dual.
Let $V$ be a finite dimensional vector space, and let $V^{\prime\prime}$ be the double dual space.

From S. Axler's linear algebra text, he defines a function $\Lambda : V \rightarrow V^{\prime\prime}$ by
$$(\Lambda v)(\varphi) = \varphi(v).$$

But I cannot make heads nor tails of this definition. I am confused on both sides of the equation.

For the LHS, $\Lambda v$ makes sense because that is what I would expect to appear when defining a function; for example, let $f(x) = \text{whatever}$. But then we tack on $\varphi$, which I know is in $V^{\prime}$; but if the LHS is supposed to deal with the domain $V$, then I don't understand how it is valid to juxtapose this $\varphi$.

The RHS is also concerning because $\varphi(v)$ is a scalar. Assuming that $\varphi(v)$ is supposed to be the output of $\Lambda,$ it should follow that $\varphi(v) \in V^{\prime}$.

I appreciate all clarification and insight concerning this notation and, perhaps, an explanation as to why such notation is valid (I think this last bit will follow once clarification is provided).

Thank you =)

答:
Here, $v$ is some vector and $\Lambda v$ is in $V''$, meaning that it sends some element of $V'$ (which is called $\varphi$ here) to a scalar. This is exactly what is happening on the left hand side: $\Lambda v$ sends $\varphi$ to a scalar.

On the right hand side, you already noted correctly that we see $\varphi$ sending the vector $v$ to a scalar.

Since we are talking about a fixed vector $v$, and the functional $\varphi$ is just a variable, this relation will define $\Lambda$ uniquely


问:
Why to obtain the coordinates of vectors in the basis that themselves belong?

Let $ \space T: \mathbb{R^2} \to \mathbb{R^3}$ a linear transformation defined as $ \space T(x,y)=(3x+2y,x+y,-2x-y)$, where $\beta=\{(1,-1),(0,1)\}$ is a basis of $\mathbb{R^2}$. Is not specify a basis for $\mathbb{R^3}$.

By the expression of $ \space T(x,y)$ one known that,

$$T(1,-1)=(1,0,-1)$$
$$T(0,1)=(2,1,-1)$$

The matrix will have $3$ lines by $2$ columns. Since there wasn't a specific basis for $\mathbb{R^3}$ I assumed the canonic basis. So the linear combination of each $T(1,-1)$ and $T(0,1)$ in the canonic $\mathbb{R^3}$ basis are themselves.

The matrix will be,

$\begin{bmatrix}1 & 2 \\ 0 & 1 \\ -1 & -1 \end{bmatrix}$

But,

$\begin{bmatrix}1 & 2 \\ 0 & 1 \\ -1 & -1 \end{bmatrix}\begin{bmatrix}1 \\ -1 \end{bmatrix}\neq$$\begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}$

My doubt is, why it's need to write each vector of $\beta$ as a linear combination of themselves to get their coordinates in the basis that they belong? Thanks

答:
When you consider $ \{ (1,-1),(0,1)\}$ as a base for the domain of $T$ and then you get  $$\begin{pmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{pmatrix}$$ then $(1,-1)=1(1,-1)+0(1,0)$ and so :</p>

<p>$$T(1,-1)=\begin{pmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=(1,0,-1)$$  </p>

Body:
When you consider $ \{ (1,-1),(0,1)\}$ as a base for the domain of $T$ and then you get  $$\begin{pmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{pmatrix}$$ then $(1,-1)=1(1,-1)+0(1,0)$ and so :

$$T(1,-1)=\begin{pmatrix}1 & 2\\ 0& 1\\ -1 & -1\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=(1,0,-1)$$


问:
are $f(f(x))=f(x)$ and $f(x)=x$ equivalent?
when I apply $f^{-1}$ to both sides $f(f(x))=f(x)$ $(1)$ we get $f(x)=x$ $(2)$.

are $1$ and $2$ equivalent?

答:
Let us assume we have a matrix $F$ that represents application of f.

In matrices this means $$F^2 = F$$
This is the equation that defines a projection.

Projections are diagonalizable and have eigenvalues $0$ or $1$.

We can take an example, a function $$x\to f(x) = |x|$$ on the space of integers $$x\in\{-3,-2,\cdots,2,3\}$$

Our matrix becomes $$F =  \left[\begin{array}{lllllll}&&&&&&\\&&&&&&\\&&&&&&\\&&&1&&&\\&&1&&1&&\\&1&&&&1&\\1&&&&&&1\end{array}\right]$$

It is now a simple exercise in linear algebra to find $F = SDS^{-1}$

We can verify $D_{ii} \in \{0,1\}$

For more general treatment of smooth power-series expandable functions, you might want to consider Carleman matrices.


问:
How to find path for method of steepest descent
We have integral:
$$\int_0^1\exp\left(n\left(\frac{itz}{\sqrt{a(1-a)n}}+a\ln(z)+(1-a)\ln(1-z)\right)\right)dz=\int_0^1\exp(nf(z))dz,$$
where $0<a<1$.

We want to approximate this integral when $n\rightarrow\infty$ by method of steepest descent. Why deforming $[0,1]$ to the contour through saddle point has the following expansion:
$$z=a+\frac{it\sqrt{a(1-a)}}{\sqrt{n}}+O\left(\frac{1}{n}\right)?$$
I found saddle point $z_0=a$ from $\text{Re}'(f(z))=0$. I don't know what to do next? I can’t understand how they got it? I hope for your help! Thank you!

答:
my advice would be to try rewriting your integral as
$$
I = \int_0^1 z^{an}(1-z)^{n(1-a)} \exp\left({\sqrt{n}xiz}\right) dz,
$$
where $x(t,a):=\frac{t}{\sqrt{a(1-a)}}.$

Then there are no stationary points, but you have a simple linear phase, so steepest descent can be applied easily by deforming onto the vertical contours Re(z)=0 and Re(z)=1, like so
$$
I = i\int_0^\infty (ip/x)^{an}(1-p/x)^{n(1-a)} \exp\left(-np\right) dp
-ie^{\sqrt{n}xi}\int_0^\infty (1+ip/x)^{an}(p/x)^{n(1-a)} \exp\left(-np\right) dp
$$
where the first and second integral correspond to the integral up Re(z)=0 and down Re(z)=1 respectively.

At this point, you could binomially expand the (1-p/x)^{n(1-a)} and (1+ip/x)^{an} terms, aiming for an asymptotic expansion of the integrand which holds for $p\to0$. Once you've done this, apply Watson's Lemma to both integrals.

I hope this helps :-)


问:
Can $(X, \left \| \cdot \right \|_{T})$ being a Banach space imply $T$ as a closed operator?

Let $X$ and $Y$ be Banach spaces and let $T:X \rightarrow Y$ be a
linear operator. For each element $x \in X$, define a norm (more
specifically, the graph norm) $\left \| \cdot \right \|_{T}$ on
$X$ by : $$\left \| x \right \|_{T} = \left \| x \right \| + \left \|
 Tx \right \|$$ for $x \in X$.

I managed to prove that if $T$ is a closed operator, then $(X, \left \| \cdot \right \|_{T})$ is a Banach space, using Cauchy sequences.

I was wondering if it's possible to prove the other way around, respectively if $(X, \left \| \cdot \right \|_{T})$ is a Banach space, then $T$ is a closed operator and if so, how I could achieve this.

答: Suppose $x_n \to x$ and $Tx_n \to y$. Then $x_n, Tx_n$ are Cauchy sequences in the $X$ and $Y$ norm resp., so $x_n$ is Cauchy in the $\|\cdot \|_T$ norm, so $x_n \to z$ in $\|\cdot\|_T$ norm. So then clearly $x_n \to z$ in the norm on $X$ also, so $z = x$, but also by the definition of the $T$ norm, $Tx_n \to Tz = Tx$ in the $Y$ norm, but $Tx_n \to y$ so $y = Tx$, which is precisely the definition of a closed operator.


问:
Link between thetrahedral numbers and combinatorics
The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is  $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1}{2}$ pairs. But you could also count the pairs differently (see here): You can match the first person with $n$ different persons, the second person can be paired with $n-1$ different persons, the third with $n-2$, $\ldots$. Hence $\binom{n+1}{2}=n+(n-1)+(n-2)+\ldots$.

Now the tetrahedral numbers $1, 4=1+3, 10=1+3+6,\ldots$ can be written as $\binom{n+2}{3}$. Is there also an combinatorical explaination as for triangular numbers? I just found this explanation but I can't see clearly why $\binom{n+2}{3}$ should be the same as $1+3+6+\ldots+\frac{n(n+1)}{2}$.

答:
I found an analogous explaintion using people.
Let's assume we want to calculate $\binom{5}{3}$, i.e. the number of subsets of order 3 with 5 elements / people {A,B,C,D,E}.

Subgroups of size 3 with A are all possible pairs using {B,C,D,E} and
A itself, i.e. $\binom{4}{2}$.

Subsets of size 3 with B (and without A, because we counted all
subsets with A already) are all  possible pairs using {C,D,E} and B
itself, i.e. $\binom{3}{2}$.

Subsets of size 2 with C (and without A and B, because we counted
these subsets already) are all possible pairs using {D,E} and C
itself, i.e. $\binom{2}{2}$.

Hence all possible subsets are $\binom{4}{2}+\binom{3}{2}+\binom{2}{2}=\binom{5}{3}$ and in general
$$\binom{n+2}{3}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\ldots +\binom{n+1}{2}.$$


问:
Very simple to state, but..... (Quadrilateral and its diagonal's adventure.)

Let $ABCD$ be a quadrilateral. $BA$$\cap$ $CD$=$K$, $BC$$\cap $$AD$=$M$. $O$ is the intersection of the diagonals. The segment parallel to $MK$ and goes through $O$ meet at sides $BC$ and $AD$ at $E$ and $F$, respectively. Then prove that $EO$=$OF$.

My try: Let $BF$$\cap $$MK$=$P$ and $BD$$\cap$ $MK$=$Q$. From here, if we can prove $P, D, E$ are collinear, we can say $MQ=QP$ by the Ceva theorem. Moreover it will be $EO$=$OF$. However, I can't prove it at all. Can you guys give me some hints?

I don't know any complex, vector and projective geometry, so please consider giving me an Euclidean one.

答:
Instead of $P=BF\cap MK$, let $P=BF\cap ED$.  Then use Desargues' Theorem with triangles $AFB$ and $CED$ to show $P$ is on line $MK$.

Complete the proof as you suggest, with Ceva's theorem.

Desargues' Theorem is traditionally a theorem of projective geometry, but it can be easily proved using Menalaus' Theorem, as shown on page 2 of the reference above.


问:
A proof of $n*0=0$?
The only proof I've seen for this assumes that $0$ follows all the rules of arithmetic. How can we make that assumption when dividing by $0$ is a problem? I know that some people don't agree that all of the numbers follow the rules for arithmetic; for example, people say that the proof of $.99999...=1$ is invalid because arithmetic can't deal with these "infinite numbers".</p>

答:
We take:

$$0=0$$
by zero property of addition:
$$0+0=0$$
by definition of multiplication:
$$a\cdot(0+0)=a\cdot0$$
by distributive law:
$$a\cdot0 + a\cdot0 = a\cdot0$$
by cancellation law:
$$a\cdot0=0$$

The cancellation law isn't under the field axioms and requires a proof for the above to be complete. Here's a proof:

We want to prove that if $a+c=b+c$, then $a=b$.

by the additive inverse property, we have an $c^{-1}$ such that $c^{-1}+c=0$. So by definition of addition:

$$c^{-1}+a+c=c^{-1}+b+c$$

by associativity and commutativity of addition:

$$(c^{-1}+c)+a=(c^{-1}+c)+b$$

by definition of $c^{-1}$:

$$0+a=0+b$$

by zero property of addition:

$$a=b$$

So we have proven the cancellation law.

Title: A proof of $n*0=0$?


问:
$f(c)<1/(b-a)\int^b_a f(x)dx<f(c')$
Let f a continous function.
Prove that if f is not constant ,then exists $c,c'\in (a,b)$ such that
$f(c)<1/(b-a)\int^b_a f(x)dx<f(c')$
I tried to apply the mean value theoreme for definite integrals and I get that exists $c\in(a,b)$
$f(c)=1/(b-a)\int^b_a f(x)dx$

答:
As you noted by the mean value theorem there is some $d\in (a,b)$ such that $$f(d)=\frac{1}{b-a}\int_a^bf(x)dx$$
We claim that if $f$ is not constant then $f(d)$ cannot be maximal/minimal, i.e. there are $c,c'\in (a,b)$ such that $f(c)<f(d)<f(c')$, the statement will then follow. Assume that $f(d)$ is minimal, so the function $g:(a,b)\to\Bbb R$, $g(x)=f(x)-f(d)$ is non-negative. We get: $$\int_a^bg(x)dx=\int_a^bf(x)dx-(b-a)f(d)=0$$
As $g$ is non-negative and continuous this implies that $g$ is constant, hence also $f$.

Similarly you can show that $f(d)$ is not maximal if $f$ is non-constant.

Title: $f(c)<1/(b-a)\int^b_a f(x)dx<f(c')$


问:
Does the difference between a subset and a measurable cover thereof have outer measure $0$?
Let $\Omega$ be a non-empty set, and let $\mathcal{A}$ be an algebra of subsets of $\Omega$, i.e. let $\mathcal{A}$ be a set consisting of subsets of $\Omega$ such that

$\emptyset \in \mathcal{A}$.

$A^c \in \mathcal{A}$ for every $A \in \mathcal{A}$ (where $A^c = \Omega\setminus A$).

$A \cup B \in \mathcal{A}$ for every $A, B \in \mathcal{A}$.

Let $\mu$ be a finite measure on $\mathcal{A}$, i.e. let $\mu:\mathcal{A}\rightarrow[0,\infty)$ be such that

$\mu(\emptyset) = 0$.

$\mu(\cup_{i = 1}^\infty A_i) = \sum_{i = 1}^\infty\mu(A_i)$ for every sequence $A_1, A_2, \dots \in \mathcal{A}$ of pairwise disjoint sets such that $\cup_{i = 1}^\infty A_i \in \mathcal{A}$.

Denote by $\sigma_\Omega(\mathcal{A})$ the sigma-algebra on $\Omega$ generated by $\mathcal{A}$, and denote by $\mu^*$ the outer measure on $\Omega$ induced by $\mu$.

Let $S \subseteq \Omega$.

A measurable cover of $S$ is defined to be a set $A \in \sigma_\Omega(\mathcal{A})$ satisfying:

$S \subseteq A$.

$\mu^*(S) = \mu^*(A)$.

Let $A$ be a measurable cover of $S$. Is it necessarily the case that $\mu^*(A\setminus S) = 0$?

答: Let $\Omega = \{0,1\}$ and let $\mathcal A = \{\emptyset, \Omega\}$, so $\sigma_\Omega(\mathcal A) = \mathcal A$. Then let $\mu : \mathcal{A} \to [0,\infty)$ given by $\mu(\emptyset) =0$ and $\mu(\Omega) = 1$. Then $\mu^*(\{0\}) = \mu^*(\{1\}) = 1$. But consider a measurable cover of $\{0\}$, it clear must be $\Omega$, and $\Omega \backslash \{0\} = \{1\}$ and $\mu^*(\{1\}) = 1$.


问:
How to express symmetric polynomials in terms of elementary symmetric polynomials?
Firstly, I know there are very similar questions (How to express a symmetric polynomial in terms of elementary symmetric polynomials and Expressing a symmetric polynomial in terms of elementary symmetric polynomials using computer? ) already. Unfortunately they didn't help me that much.

The task is to express some symmetric polynomials in terms of elementary symmetric polynomials. Which is always possible by a theorem which also says that if the initial polynomial is homogeneous of degree d, the resulting polynomial is isobaric of weight d.

For example
$\sum_{i=1}^n x_i=S_1 $

$\sum_{i=1}^n x_i^2=S_1^2-2S_2$ because $\left(\sum_{i=1}^n x_i\right)^2=\sum_{i=1}^n x_i^2 + 2 \sum_{1\leq k < l \leq n}x_{k}x_l=S_1^2 +2S_2$

The case $\sum_{i=1}^n x_i^3$ is already significantly more complicated:
One sees that $$S_1(S_1^2-2S_2)=\sum_{i=1}^n x_i^3+\sum_{r \neq s} x_{r} x_s^2$$ then $$S_1S_2=\sum_{i}x_i \cdot \sum_{r<s}x_r x_s=\sum_{r<s}x_r^2x_s+\sum_{r<s}x_r x_s^2+ \sum_{r<s \\ i \neq r,s}x_i x_r x_s=\sum_{r \neq s} x_{r} x_s^2+3S_3$$
Which gives us $\sum_{i=1}^n x_i^3=S_1^3-3S_1S_2+3S_3$

In the exercises it is asked to reshape other polynomials such as $\sum_{i \neq j } x_i^3 x_j$ and $\sum_{i=1}^n x_i^4$. The solutions of the exercises give the solutions after the first step, as if it was trivial. Personally I'm struggling a lot handling these expressions, am I missing something?

I would like to get some suggestions on how to learn solving these exercises. The theorem above gives some hints on how to choose the possible symmetric elementary polynomials, since they have to satisfy that relation. But how to find the coefficients in general?

Any suggestion is very much welcome.

答:
Hint:

A method to systematically transform a symmetric polynomial into a polynomial representation by elementary symmetric polynomials is presented in Paul Garrets algebra course section $15$ Symmetric polynomials.

Examples of this method are given in this answer.


问:
Statements about elements in $\mathbb{C}$
Let $\mathbb{C}=\left \{\begin{pmatrix}a & -b \\ b & a\end{pmatrix}\in M_2(\mathbb{R})\mid a, b\in \mathbb{R}\right \}$ and $u=\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \ i=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\in \mathbb{C}$.

I have shown that $\mathbb{C}$ is a subspace of $M_2(\mathbb{R})$ and that $(u,i)$ is a basis of $\mathbb{C}$.

Then :

Show that $(au+bi)(cu+di)=(ac-bd)u+(ad+bc)i$ for all $a,b,c,d\in \mathbb{R}$ and imply that $xy=yx$ for all $x,y\in \mathbb{C}$.

Let $z\in \mathbb{C}$. Show that :

(1) $z^t\in \mathbb{C}$ and $zz^t=\delta_zu$

(2) $\delta_z=0_{\mathbb{R}}\iff z=0_{\mathbb{C}}=0_{M_2(\mathbb{R})}$

(3) If $z\neq 0_{\mathbb{C}}$ then $z$ is invertible, $z^{-1}\in \mathbb{C}$ and it holds that $z^{-1}=\delta_z^{-1}z^t$

$$$$

For the first dot I have done the following:

It holds that $u^2=u$, $i^2=-u$ and $ui=i$.

So  we get \begin{equation*}(au+bi)(cu+di)=acu^2+adui+bcui+bdi^2=acu+adi+bci-bdu=(ac-bd)u+(ad+bc)i\end{equation*}
Since each element of $\mathbb{C}$ can be written as a linear combination of $u$ and $i$, we have that $x=au+bi$ and $y=cu+di$.

Then it holds that $xy=yx$ since the coefficients of $(ac-bd)u+(ad+bc)i$ are real numbers, and so the commutativity holds.

Is that correct?

Could you give me a hint for the second dot?

答:
It seems right to me. For the second dot note that if $z=au+bi$, then $z^{t}=au-bi$ and since $b\in\mathbb{R}$ we have that $-b\in\mathbb{R}$ and $z^{t}\in\mathbb{C}$.  Then use the first dot to calculate: $zz^{t}=(ac-bd)u+(ad+bc)i=(a^2+b^2)u+(-ab+ab)i=(a^2+b^2)u$ (I have used that $z^{t}=cu+di=au-bi$, so: $c=a, d=-b$). Try to do the others on your own, then don't hesitate to ask for help if you need.

P.S. If you are interested, it's not a casuality that it is named $\mathbb{C}$, in fact it is a way to represent the complex numbers. https://en.wikipedia.org/wiki/Complex_number#Matrix_representation_of_complex_numbers


问:
If $\dim V = n$, why does any set of $n$ linearly independent vectors span $V$?
I am currently learning about vector spaces, spaning and dimensions for the first time and am reading old posts on here. I came across this one: If $\dim(V) = n$, is every spaning set $\{v_1,v_2,\ldots,v_n\}$ a basis for $V$?

In it, it is mentioned that "If $\dim V = n$, then any set of $n$  linearly independent vectors will span $V$". I understand that any spanning set of $n$ vectors must consist of linearly independent vectors due to $\dim V = n$ but I don't quite get the other direction.

Could someone help me understand this better?

答: Every linear independent family can be extended to a basis. By definition of $\dim$, this basis still has $n$ members, i.e., we did not really extend our family and already had a basis.


问:
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.
Given non-negative integers such that $12^a \cdot 324^b \cdot 48^c $ is divisible by $ 6^{70} $. Find the smallest possible $ a + b + c $.

First I can make an assumption that $a+b+c$ will be the smallest if we can obtain the highest power of b, then c and later a.

Let me go in the general way, $12^a\cdot 324^b\cdot48^c$ = $ 3^a \cdot  2^{2a} \cdot 3^{4b} \cdot 2^{2b} \cdot3^c \cdot 2^{4c}$ = $3^{a+4b+c} \cdot 2^{2a+2b+4c}$

And $6^{70}$ = $2^{70}\cdot 3^{70}$

Then I need to find the solution for $a+4b+c = 70$ and $a+b+2c=35$ where a+b+c need to be the samllest.

Please anyone can show me the solution how to approach this type of questions.

答:
Another way .

We have $$ 2^{2 a+2 b+4 c}*3^{a+4 b+c} =6^{70}M$$ A minimum possible is when
$$2a+2b+4c=a+4b+c=70\Rightarrow a-2b+3c=0$$ whose integers solutions are given by $$(a,b,c)=(a,2a+3n,a+2n)$$ where $a$ is take as parameter and $n\in \mathbb Z$.

Now $$a+b+c=a+4(2a+3n)+(a+2n)=10a+14n=70$$
In the  expression $10a+14n$ we can choose searching for a minimum with non-negative integers $n=0$ and $a=0$. An easy calculation gives for $n=0$ the sum $a+b+c=28$ and for $a=0$ we have $$10a+14n=14n=70\iff n=5$$ from which$$a+b+c=0+3\cdot5+2\cdot5=25$$Thus $a+b+c=25$ is the required minimum.


答:
The dimension is defined as the cardinality of a maximal linearly independent set, or the minimal generating set. These two are shown to be the same in any textbook. We say basis to refer to these sets; being both maximal linearly independent and minimal generating sets.

So a set of $\dim V$ linearly independent vectors is a basis basically by definition

Title: If $\dim V = n$, why does any set of $n$ linearly independent vectors span $V$?

Body:
I am currently learning about vector spaces, spaning and dimensions for the first time and am reading old posts on here. I came across this one: If $\dim(V) = n$, is every spaning set $\{v_1,v_2,\ldots,v_n\}$ a basis for $V$?

In it, it is mentioned that "If $\dim V = n$, then any set of $n$  linearly independent vectors will span $V$". I understand that any spanning set of $n$ vectors must consist of linearly independent vectors due to $\dim V = n$ but I don't quite get the other direction.

Could someone help me understand this better?


问:
Showing that a transitive abelian permutation group is necessarily regular
I am trying to show that a transitive, abelian permutation group acting on a set $X$ is necessarily regular, given this hint: 'Given $g \in G$, consider the set $X^g:=\{x \in X\,|\,gx=x\}$.  Show that if $G$ is transitive and abelian, then the only possibilities for $X^g$ are $\varnothing$ or $X$.'

I know that $X^g=X$ iff $g$ is the identity of $G$.  And...that's about all I've got.  I don't even see how, once the above is shown to be true, the result follows.

I would really appreciate some hints to help me in the right direction.

This is from Isaacs' Algebra: A Graduate course.  He introduces permutation groups first, but group actions haven't entered the scene yet.  I am alright with a hint in terms of actions though.

Thanks!

答:
Let $gx=y$, $hx=y$, where $g,h$-elements of the group $G$; $x,y$-elements of the set $A$. $G$ acts on $A$
Let $t$ some element of $A$. As $G$-transitive,there exist $f$ in $G$ that
$t=fx$ As $G$ is abelian we have:
$$gt=g(fx)=(gf)x=(fg)x=f(gx)=fy=f(hx)=(fh)x=(hf)x=h(fx)=ht$$
So for any $t$ in $A$ we have $gt=ht$, so $g=t$, then $G$-regular


问:
Constant orthonormal frame
I'm reading a geometry article, and some doubts arose, and I hope someone can help me. At a certain point in the work, the author says:
Choosing a point wise constant local orthonormal frame $\{e_1, ....., e_n\}$ on the Riemannian manifold $(M, g)$ we have...

What does that mean? It would be something like $\nabla_X e_i=0$ for all index $i$ and $X$ in $M$? Given any Riemannian manifold, there is always this kind of frame?

Thanks!

答:
What is true is the following:

If $M$ is a Riemannian manifold with Levi-Civita connection $\nabla$, then for all $p\in M$ there is a local orthonormal frame $e_1,\dots,e_n$ with $\nabla_{X_p}e_i=0$ for all $X_p\in T_pM$.

Take for example an orthonormal basis $e_1(p),\dots,e_n(p)$ of $T_pM$ and parallel transport it in a normal neighborhood of $p$ along radial geodesics. The frame will stay orthonormal since $\nabla$ is a metric connection and $\nabla_{X}e_i=0$ for all $X\in T_pM$ is true by construction. The crucial point maybe is smoothness of the frame, this is explained here.

The condition that for each point, there is a local parallel frame (i.e. with vanishing derivative everywhere in this neighboorhood) is equivalent to $M$ having zero curvature, $R=0$ and is not true in general.

Title: Constant orthonormal frame


问:
Showing $\lim_{n \to \infty}\int_0^1x^nf(x)\,d \mu =0$
Let $f:[0,1] \to [0, \infty]$ be a measurable function s.t. $\int_0^1 f \,d \mu < \infty$. I wanna show that $$\lim_{n \to \infty}\int_0^1x^nf(x)\,d \mu =0$$

I thought to use the dominated convergence theorem but I don't see how to start.

EDIT:

For the next step I should calculate $\int_0^1 \lim_{n \to \infty}x^nf(x) \,d \mu$. This should be zero but how, can I see this the best way? I was thinking about to show that $\lim_{n \to \infty}x^nf(x)=0$ for $x \in [0,1)$, but what is for $x=1$?

答:
Hint:

You have the exact right idea with the dominated convergence theorem!

On $[0,1]$ we see $x^n f \to 0$ pointwise almost everywhere
(do you see why? What about if $f = \infty$ at certain points?).

Then, as mentioned in the comments, $|x^n f| \leq |f|$. Can you use this, plus the dominated convergence theorem, to prove that $\int x^n f \to \int 0 = 0$?

I hope this helps ^_^


问:
Commuting subgroup pairs in the symmetric group
We will say that two subgroups $H,K\subset S_n$ of the symmetric group are a commuting pair if $hk=kh$ for all $h\in H, k\in K.$ For example, $S_m$ and $S_{m'}$ can be made into a commuting pair in $S_{m+m'}.$ In fact it is a "maximal" commuting pair, in the sense that neither $S_m$ nor $S_{m'}$ can be enlarged.

Question: is every maximal commuting pair of this form? More precisely, can every commuting pair $(H,K)$ in $S_n$ be enlarged to a commuting pair $(S_m,S_{m'})$ in $S_n$, where $S_m$ and $S_{m'}$ are symmetric groups on disjoint subsets of $\{1,...,n\}$ of cardinality $m$ and $m'$ respectively?

EDIT: For my question above assume that $H\cap K=\{e\}$ (since otherwise, the answer is obviously No -- just take $K=H$ an abelian subgroup). More generally, can all maximal pairs $(H,K)$ be characterized (not assuming $H\cap K=\{e\}$)?

EDIT: Captain Lamma answered my question and I intend to accept to his answer, but I would like to wait a bit first and see if anyone might be able to provide something towards a potential classification of maximal commuting pairs.

答:
The answer is no, because you can have subgroups with intersect trivially and commute, but do not act on disjoint subsets.

For instance, in $S_4$, take $H=\{id, (12)(34)\}$ and $K=\{id, (13)(24)\}$.


问:
How many ways can you choose $8$ tickets if the order of selection does not matter?
A cash drawer holds $ \$1$, $ \$5$, $ \$10$, $ \$20$, and $ \$50$ bills. How many ways can you choose $8$ bills? (The order of selection does not matter.)

For me, the answer is simply $5^8 = 390625$ , but the answer is simply $495$. It seems I misunderstood "The order of selection does not matter". Can you explain where I was wrong in considering the last comment?

答: I guess that your solution is that we have $8$ choices, each from five elements, so we get $5^8$. But since problem says "the order of selection does not matter", taking two times each of $ \$1$, $\$5$, $\$10$ and $\$20$, will be the same no matter in which of 2520 different orders we want to do it.


问:
Show $Df(z)(y_0-x_0)=0$
Show that the mean value theorem for a real function on a convex domain in $\mathbb{R}^n$. Let $f$ be a smooth fucntion on a domain $D \subset\mathbb{R}^n$ and $x_0$ and $y_0$ and the line segment connecting them belong to $D$. There is a point $z$ in this line segment such that the derivative of $f$ at $z$ in direction $y_0-x_0$ is zero, i.e.,

$$Df(z)(y_0-x_0)=0.$$

答: As stated the claim is not true, I assume what you meant is $Df(z)(y_0-x_0)=y_0-x_0$. Hint in this case: apply the ordinary mean value theorem in one variable to the function $\varphi:[0,1]\to\Bbb R$ defined by $$\varphi(t)=f((1-t)x_0+ty_0)$$

Title: Show $Df(z)(y_0-x_0)=0$


答:
What they are asking is how many of each type of "ticket" (In your solution, you are assuming that for example $(1,5,50,5,10)\neq (1,5,5,10,50),$ that is you are giving the "tickets" and ordering.). So call $x_i$ the number of "tickets" of the $i-$th kind. So $x_1$ is the number of \$1, $x_2$ is the number of $5, etc.

You want to know how many of this elements you have $$\{(x_1,x_2,x_3,x_4,x_5):x_i\geq 0\text{ and }\sum _{i=1}^5x_i=8\}.$$ This is a problem called stars and bars, and the solution is given by $\binom{8+5-1}{5-1}=495$

Title: How many ways can you choose $8$ tickets if the order of selection does not matter?

Body:
A cash drawer holds $ \$1$, $ \$5$, $ \$10$, $ \$20$, and $ \$50$ bills. How many ways can you choose $8$ bills? (The order of selection does not matter.)

For me, the answer is simply $5^8 = 390625$ , but the answer is simply $495$. It seems I misunderstood "The order of selection does not matter". Can you explain where I was wrong in considering the last comment?


问:
Positive integer solution to $a^3=3b^2-2$
The equation $a^3=3b^2-2$ seem to only have one positive integer solution $(a,b)=(1,1)$, but I am unable to prove that.

What I did:

Google told me that the elliptic curve theory might help me, but I don't really know anything about it.

This thread in MO looks similar to what we have, but I was unable to apply it in my case.

答:
As you learned from searching with Google, this is an elliptic curve.  Integral points on elliptic curves are conveniently treated as a special subset of the rational points on such curves.

A lot of computation has already been done to identify the integral points and much else about elliptic curves with smallish coefficients.  There turn out to be surprising equivalences among these, starting with changes of variables that are birational, so that $a,b$ can be rewritten as rational expressions in $x,y$ in a way that also allows $x,y$ to be expressed as rational expressions in $a,b$.

Here, setting $a = x/3$ and $b = y/9$ leads to the following Weierstrass form of your equation:

$$ a^3 = 3b^2 - 2 $$
$$ \frac{x^3}{27} = 3 \cdot \frac{y^2}{81} - 2 $$
$$ y^2 = x^3 + 54 $$

Note that if $(a,b)$ is an integral point of your original equation, then $x = 3a, y = 9b$ is an integer solution of the rewritten Weierstrass equation.  In fact that equation is a special case of the form $y^2 = x^3 + k$ for integer $k$ known as a Mordell curve.  It is known that there are only finitely many integral points on such curves.

In addition to instructions for software packages to analyze this curve, the results of what is known are summarized in this page, LMFDB Elliptic curve y² = x³ + 54.  There we learn the only integral points are $(3,\pm 9)$.

It follows that the only integral points on your original curve are $(1,\pm 1)$.


答:
Since the order of selection does not matter, what matters is how many bills of each type are selected.  Thus, you want to find the number of solutions of the equation
$$x_1 + x_5 + x_{10} + x_{20} + x_{50} = 8$$
in the nonnegative integers, where $x_k$ denotes the number of bills with the denomination $\$k$ that are selected.

A particular solution of this equation corresponds to the number of ways $5 - 1 = 4$ addition signs can be placed in a row of eight ones.  For instance,
$$1 1 + 1 1 1 + 1 + 1 1 + $$
corresponds to the solution $x_1 = 2$, $x_5 = 3$, $x_{10} = 1$, $x_{20} = 2$, $x_{50} = 0$.  The number of such solutions is
$$\binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} = 495$$
since we must select which four of the  twelve positions required for eight ones and four addition signs will be filled with addition signs.


问:
Proof Check: If $a, b, c$ is a linearly independent system of vectors, is the system $a+b+c, a+b, a$ linearly dependent?
Here is my attempt at andwering this question. I would appreciate any feedback.

This system of vector is linearly independent. Assume, to the contrary, that the system is linearly dependent, i.e., $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a = 0$ implies that at least one of the coefficients is non-zero, say $\lambda_3 \neq 0.$ Note that $\lambda_3 a = (-\lambda_1)(a+b+c) + (-\lambda_2)(a+b)$. Since the system of vectors $a, b, c$ is linearly independent, the sums $a+b+c$ and $a+b$ are both non-zero. Furthermore, since $\lambda_3 \neq 0,$ it follows that at least $\lambda_1$ or $\lambda_2$ has to be non-zero, for if both of them were equal to zero, then we would have $\lambda_3 a = 0$ and so $\lambda_3 = 0,$ a contradiction; hence, assume $\lambda_1 \neq 0.$ Thus, since $(-\lambda_2)(a+b) = \lambda_3 a + \lambda_1(a+b+c) \neq 0,$ it is clearly that $\lambda_2 \neq 0.$ But this leads to absurdity since $\lambda_1,\lambda_2, \lambda_3 \neq 0$ implies that $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a \neq 0$. Therefore, we retrieve our assumption that the given system of vectors is linearly dependent, and conclude that the system must be linearly independent.

答:
It's a little easier to see what's happening if you look at things from a slightly different perspective.

$$\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a = 0 \\
\Rightarrow (\lambda_1+\lambda_2+\lambda_3)a + (\lambda_1+\lambda_2)b +\lambda_1 c=0
\\ \Rightarrow \lambda_1 =0 \Rightarrow \lambda_2=0 \Rightarrow \lambda_3 =0. $$

The second implication is true because $\{a, b, c \}$ linearly independent forces the coefficient of $c$ to be $0$.  Once you know that, $\lambda_2=0$ because the coefficient of $b$ must be $0$, and similarly $\lambda_3=0$.

A good general approach is to collect coefficients of your linearly independent set and then note that they each must be zero.


答:
If you don't really want to use Lebesgue Dominated Convergence Theorem, you can simply use Monotone Convergence Theorem.

First of all, let us see the following basic fact: For $f_{N}\geq 0$ are such that $f_{N}(x)\downarrow 0$ for every $x$, and $\int f_{1}<\infty$, then $\int f_{N}\rightarrow 0$. This is a matter of Monotone Convergence Theorem if we look at the increasing sequence $(f_{1}-f_{N})_{N=1}^{\infty}$ and obtain that $\int f_{1}-f_{N}\rightarrow\int f_{1}$.

Now let $N$ be fixed. Then
\begin{align*}
\int_{0}^{1}x^{n}f(x)dx&=\int_{0}^{1-1/N}x^{n}f(x)dx+\int_{1-1/N}^{1}x^{n}f(x)dx\\
&\leq\left(1-\dfrac{1}{N}\right)^{n}\int_{0}^{1}f(x)dx+\int_{0}^{1}1_{[1-1/N,1]}(x)f(x)dx.
\end{align*}
Let $f_{N}(x)=1_{[1-1/N,1]}(x)f(x)$, then $f_{N}(x)\downarrow 0$, apply the basic fact just before, for $\epsilon>0$, let $N$ be such that
\begin{align*}
\int_{0}^{1}1_{[1-1/N,1]}(x)f(x)dx<\epsilon.
\end{align*}
Now $1-1/N<1$ and hence for large enough $n$, we have
\begin{align*}
\left(1-\dfrac{1}{N}\right)^{n}\int_{0}^{1}f(x)dx<\epsilon,
\end{align*}
for all such $n$,
\begin{align*}
\int_{0}^{1}x^{n}f(x)dx<2\epsilon.
\end{align*}

Title: Showing $\lim_{n \to \infty}\int_0^1x^nf(x)\,d \mu =0$

Body:
Let $f:[0,1] \to [0, \infty]$ be a measurable function s.t. $\int_0^1 f \,d \mu < \infty$. I wanna show that $$\lim_{n \to \infty}\int_0^1x^nf(x)\,d \mu =0$$

I thought to use the dominated convergence theorem but I don't see how to start.

EDIT:

For the next step I should calculate $\int_0^1 \lim_{n \to \infty}x^nf(x) \,d \mu$. This should be zero but how, can I see this the best way? I was thinking about to show that $\lim_{n \to \infty}x^nf(x)=0$ for $x \in [0,1)$, but what is for $x=1$?


答: Assume that V is a vector space and that you have n linearly independent vectors from it. The dimension of the sub space spanned by the n vectors is n. However since dim V =n, this must be V itself.</p>


问:
What does it mean to find a general formula of a matrix?

Given matrix $$A = \begin{bmatrix} 2 & 4  \\-2 & -4\end{bmatrix}$$ find the general formula of $A^n$. Hint: diagonalize matrix $A$.

What does it mean to find a general formula of $A^n$? I don't understand the exercise itself. I have a guess, that seems logically correct:

$A^2=\begin{bmatrix}
2 & 4  \\
-2 & -4  \\
\end{bmatrix}\begin{bmatrix}
2 & 4  \\
-2 & -4  \\
\end{bmatrix}=\begin{bmatrix}
-4 & -8  \\
4 & 8  \\
\end{bmatrix}=(-2)^1A$

$A^3=\begin{bmatrix}
2 & 4  \\
-2 & -4  \\
\end{bmatrix}\begin{bmatrix}
2 & 4  \\
-2 & -4  \\
\end{bmatrix}\begin{bmatrix}
2 & 4  \\
-2 & -4  \\
\end{bmatrix}=\begin{bmatrix}
-8 & -16  \\
8 & 16  \\
\end{bmatrix}=(-2)^2A$

Therefore:
$A^n=(-2)^{n-1}A$

But I'm not sure whether is what's needed, and I didn't had to diagonalize the matrix using this method.

答:
Let me explain the "diagonalize" part:

If we could write $A = PDP^{-1}$ for a diagonal matrix $D$, then it would follow that $A^n = PD^n P^{-1}$. The advantage is that $D^n$ is simple to calculate: if $D = \text{diag}(a_1, \cdots, a_k)$, then $D^n = \text{diag}(a_1^n, \cdots, a^n_k)$.


问:
Surface integral first kind
I have to calculate surface integral:$$\iint_D zdD$$ where $S$ is part of cone $z=\sqrt{x^2+y^2}, 1<z<2$ cut out by $(x^2+y^2)^2=2(x^2-y^2)$. I am only not sure how to use part that $1<z<2$. I thought to plug this in the formula and get $\iint_S \sqrt{x^2+y^2}\sqrt{1+(\frac{x}{x^2+y^2})^2+(\frac{y}{x^2+y^2})^2} dS$. And then using given cylinder, and converting to polar coordinates I get $-\frac{\pi}{4}\le \phi\le \frac{\pi}{4}, 0\le r\le 2\sqrt{\cos(2\phi)}$ it shouldn't be hard to calculate this integral. Is this correct or there is something I'm missing?

答:
While most of what you have mentioned is correct, you have not considered that $z \geq 1$. The limits of $\theta$ will also be different. $-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ is valid only when $z$ starts from $0$. Also just to notice, there is no surface area of the cone cut out by the cylinder above $z = \sqrt2$.

Cone surface is given by $z = \rho$; Cylinder is given by $\rho = \sqrt{2\cos2\theta}$

Maximum radius of the cylinder's cross section is $\sqrt2 \ $ at $\theta = 0$ so for $z \gt \sqrt2$, the cylinder is completely inside the cone.

For a cone with $z=r, \ |r'_{\rho} \times r'_{\theta}| = \rho\sqrt2$.

At $z = 1, \rho = 1 = \sqrt{2 \cos2\theta} \implies -\frac{\pi}{6} \leq \theta \leq \frac{\pi}{6}$. As $z$ increases, $\rho$ increases and range of $\theta$ that cuts out the surface area of the cone reduces. So we cannot have $\theta \gt \frac{\pi}{6}$.

Also note that there is one more petal between $\frac{3\pi}{4} \leq \theta \leq \frac{5\pi}{4}$. So we need to multiply integral by $2$.

Substituting $z = \rho$ and multiplying by factor $\rho \sqrt2$, the integral becomes,

$S = 2 \sqrt2 \displaystyle \int_{-\pi/6}^{\pi/6} \int_{1}^{\sqrt{2\cos (2\theta)}} \rho^2 \  d\rho \  d\theta$

Alternatively,
$S = 2\sqrt2 \displaystyle \int_1^\sqrt2 \int_{-\frac{1}{2} \arccos\big(\frac{\rho^2}{2}\big)}^{\frac{1}{2} \arccos\big(\frac{\rho^2}{2}\big)} \rho^2 \ d\theta \  d\rho$

Title: Surface integral first kind


答:
$\lambda_1 a + \lambda_2 (a+b) + \lambda_(a+b+c) = (\lambda_1+\lambda_2+\lambda_3) a + (\lambda_2+\lambda_3) b + \lambda_3 c$

As $a,b,c$ are lineraly indepndent:

$\lambda_1+\lambda_2+\lambda_3 = 0\\
\lambda_2+\lambda_3 = 0\\
\lambda_3 = 0$

A completely different approach:

If a,b,c are linearly independent, then the matrix [a,b,c] with a,b,c as column vectors has a non-zero determinant.

$[a,b,c]\begin{bmatrix} 1&1&1\\&1&1\\&&1\end{bmatrix} = [a,a+b,a+b+c]$

$\det \begin{bmatrix} 1&1&1\\&1&1\\&&1\end{bmatrix} = 1$

$\det [a,a+b,a+b+c] = \det [a,b,c]$

$[a,a+b,a+b+c]$ is non-sigular and $a,a+b,a+b+c$ are linearly independent.

Title: Proof Check: If $a, b, c$ is a linearly independent system of vectors, is the system $a+b+c, a+b, a$ linearly dependent?

Body:
Here is my attempt at andwering this question. I would appreciate any feedback.

This system of vector is linearly independent. Assume, to the contrary, that the system is linearly dependent, i.e., $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a = 0$ implies that at least one of the coefficients is non-zero, say $\lambda_3 \neq 0.$ Note that $\lambda_3 a = (-\lambda_1)(a+b+c) + (-\lambda_2)(a+b)$. Since the system of vectors $a, b, c$ is linearly independent, the sums $a+b+c$ and $a+b$ are both non-zero. Furthermore, since $\lambda_3 \neq 0,$ it follows that at least $\lambda_1$ or $\lambda_2$ has to be non-zero, for if both of them were equal to zero, then we would have $\lambda_3 a = 0$ and so $\lambda_3 = 0,$ a contradiction; hence, assume $\lambda_1 \neq 0.$ Thus, since $(-\lambda_2)(a+b) = \lambda_3 a + \lambda_1(a+b+c) \neq 0,$ it is clearly that $\lambda_2 \neq 0.$ But this leads to absurdity since $\lambda_1,\lambda_2, \lambda_3 \neq 0$ implies that $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a \neq 0$. Therefore, we retrieve our assumption that the given system of vectors is linearly dependent, and conclude that the system must be linearly independent.


问:
If $E$ is not compact then any image of $E$ by a continuous 1-1 map is not compact.
If $E$ is not compact then any image of $E$ by a continuous 1-1 map is not compact.

First I was going to prove it by assuming otherwise, saying let there is some open cover for $E$ which doesnot have any finite subcover for $E$ then then assume $f(E)$ is compact then it has finite subcover comes from $E$ but we dont know $f$ is open or not.

Then I try to analyse the problem more but I stuck because I realy dont understand what "any image of $E$ by a continuous 1-1 map is not compact." really means?

Any hint and explanation the above sentence?

答:
It is not correct; take any infinite set $A$ and let $X$ be $A$ endowed with the discrete topology and $Y$ the space with the trivial topology. Then the map $X\to Y$, $x\mapsto x$ is continuous, 1-1, $X$ is not compact but $Y$ is.

Other examples: Take $f:[0,2\pi)\to S^1, x\mapsto e^{ix}$, or the function here.


问:
A proof of $n*0=0$?
The only proof I've seen for this assumes that $0$ follows all the rules of arithmetic. How can we make that assumption when dividing by $0$ is a problem? I know that some people don't agree that all of the numbers follow the rules for arithmetic; for example, people say that the proof of $.99999...=1$ is invalid because arithmetic can't deal with these "infinite numbers".</p>

答: I think that you're confused by what the phrase "rules of arithmetic" means. A formal statement of these rules treats $0$ differently, accounting for the fact that you cannot divide by it and so forth. $0$ does not break any rules of arithmetic at all.

Title: A proof of $n*0=0$?


问:
What motivation for assuming field K is infinite.
A theorem in Shaum's Outline of Linear Algebra says that if we suppose the field $K$ is infinite then any system of linear equations has either a unique solution, or no solution, or an infinite number of solutions.

So if we have $n$ unknowns then solutions are vectors in $K^n$ which can be real or complex.

Why do we want $K$ to be infinite? Wouldn't any subset of $K$ still ensure the trichotomy i.e. we would still have either a unique solution, or no solution, or an infinite number of solutions?

答:
As pointed out in a comment, if $K$ is a finite field, then $K^n$ has $|K|^n$ elements so there cannot be an infinite number of solutions.

The trichotomy then breaks up into $n+2$ cases. The linear system $Ax=b$ may have no solution - this occurs when $b \notin \mathrm{im}A$. On the other hand, if the solution set of $Ax=b$ is non-empty then it is a linear subspace of $K^n$. For every $d=0, 1, \dots, n$ it is possible that the solution set is $d$-dimensional and thus has $|K|^d$ elements.

Note that when $K$ is infinite, the above $n+2$ cases still appear, but the solution set is infinite for $d=1, \dots, n$.


问:
Contradiction with Nth roots on finite fields
I am having problems finding the mistake in my thought process here. One of the results from Ireland-Rosen on the topic:

If $F$ is a finite field with $q$ elements then for every $\alpha \in F^*$ , $x^n=\alpha$ has solutions iff $\alpha^{(q-1)/d} = 1$ where $d=\text{gcd}(n,q-1)$. If there are solutions, there are exactly $d$ solutions.

Now one of the suggested exercises is:

Let $F$ be a finite field with $q$ elements, $n\equiv 1 \pmod q$ and $K$ an extension of $F$ with $[K:F]=n$ then for every $\alpha \in F^*$, $x^n=\alpha$ has exactly $n$ solutions in $K$.

Don't these two results contradict each other? $K$ has $nq$ elements and therefore $x^n = \alpha$ has to have $\text{gcd}(nq-1,n)=(n(q-1)+n-1,n)=(n-1,n)=1$ or 0 solutions.

答:
You're thinking of the number of elements computed from an index as if you were talking about a group.  That's not how field index is defined.

The index, $[K:F]$ is the dimension of $K$ as a vector space over $F$.  So if the index is $2$, $|K| = |F|^2$ and if the index is $n$, $|K| = |F|^n$.


问:
Rudin's Exercise 2.27 in Principles of Mathematical Analysis: Why are countable many points of $E$ not in perfect set $P$?
Here is the theorem to be proved:

Suppose $E \subset \mathbb{R}^k$, $E$ is uncountable, and $P$ is the set of condensation points of $E$.  Prove that $P$ is perfect and that there are at most countably many points in $E$ that are not in $P$.

I tried to prove it as follows:

Let $p\in \mathbb{R}^k$ be a limit point of $P$, then for some $\delta \gt 0, N_\delta(p)$ contains $y\ne x$ such that $y\in P$. Let $r=\frac{1}{2} \min (d(p,y), \delta-d(p,y))$. We have $N_r(y)\subset N_\delta (p)\implies N_r(y) $ contains uncountable many points of $E$ (as $y$ is condensation point)$\implies N_\delta (p)$ contains uncountable many points of  $E\implies p$ is a condensation point of $E$. So $p\in P$. Hence $P$ is closed. $\tag{1}$

Let $p\in P$ be an isolated point of $P$. $\exists\delta \gt 0: N_\delta (p)\cap P=\{p\}$.  Since $p\in P$, we have that $N_\delta (p)\cap E$ is uncountable.

Suppose that $N_\delta (p)$ does not contain any other condensation point of $E$ except $p$.

Then for every $y\in N_\delta (p)-\{p\},\exists r\gt 0: N_r (y)$ is at most countable for every $y\in N_\delta (p)-\{p\}$.

We have $N_\delta (p)\subset (\cup_{y\in N_\delta (p)-\{p\}} N_r(y))\cup\{p\}$, whence it follows that $ E\cap N_\delta (p)\subset (E\cap \{p\})\cap_{y\in N_\delta (p)-\{p\}} (E\cap N_r(y)\subset E\cap N_r(y)$ for some $y\in N_\delta (p)-\{p\}$. Hence, $ N_\delta(p)$ is at most countable, which is a contradiction. Therefore, all points of $P$ are limit points of $P$. $\tag{2}$

By $(1)$ and $(2), P$ is perfect.

I got stuck at proving the second part of this question.

I tried to use hint provided in the exercise: "Let $\{V_n\}$ be a countable base of $\mathbb R^k$, let $W$ be the union of those $V_n$ for which $E\cap V_n$ is at most countable and show that $P=W^c$."

I understand that since $\mathbb R^k$ is separable, it must have a countable base. I don't know how to proceed further using the hint. I also don't understand why such $W$ exists. Can existence of such $W$ be proven?

Please help. Thanks.

答:
In your proof that $P$ is closed you need to say that for every $\delta>0$ the nbhd $N_\delta(p)$ contains a point of $P\setminus\{p\}$, not just some $\delta>0$; otherwise you cannot conclude that $p\in P$. Otherwise that proof is correct, though you could also simply let $r=\delta-d(p,y)$: that still ensures that $N_r(y)\subseteq N_\delta(p)$. However, it is simpler to prove that $\Bbb R^k\setminus P$ is open. If $x\in\Bbb R^k\setminus P$, there is an $\epsilon>0$ such that $N_\epsilon(x)\cap E$ is countable. $N_\epsilon(x)$ is an open nbhd of each of its points, so every point of $N_\epsilon(x)$ has an open nbhd that contains only countably many points of $E$, and therefore no point of $N_\epsilon(x)$ is a condensation point of $E$; in other words, $N_\epsilon(x)\cap P=\varnothing$. Since $x\in\Bbb R^k\setminus P$ was arbitrary, it follows that $\Bbb R^k\setminus P$ is open and hence that $P$ is closed.

Your argument to show that $P$ has no isolated points, however, is not correct: somehow you turned

$$E\cap\bigcup_{y\in N_\delta(p)\setminus\{p\}}N_r(y)$$

into

$$\bigcap_{y\in N_\delta(p)\setminus\{p\}}\big(E\cap N_r(y)\big)\,,$$

and the two sets are definitely not equal in general. I should also note that you cannot use the same $r$ for every $y\in N_\delta(p)\setminus\{p\}$: you need to index the $r$s by $y$, e.g., as $r(y)$ or $r_y$, since each point $y$ may require a different $r$.

The easiest way to prove that $P$ has no isolated points is to use the same idea that is used to prove that $E\setminus P$ is countable. (Note: I use countable in its more usual sense of finite or countably infinite.) Let $\mathscr{B}=\{B_n:n\in\Bbb N\}$ be a countable base for $\Bbb R^k$. Suppose that $p$ is an isolated point of $P$; then there is an $m\in\Bbb N$ such that $B_m\cap P=\{p\}$. Thus, for each $x\in B_m\setminus\{p\}$ there is a $B_{n(x)}\in\mathscr{B}$ such that $x\in B_{n(x)}\subseteq B_m$, and $B_{n(x)}\cap E$ is countable. Let $N=\big\{n(x):x\in B_m\setminus\{p\}\big\}$; then $N$ is countable, since it’s a subset of $\Bbb N$, and

$$E\cap\big(B_m\setminus\{p\}\big)=\bigcup_{x\in B_m\setminus\{p\}}(E\cap B_{n(x)})=\bigcup_{n\in N}(E\cap B_n)\,.$$

That last union is the union of countably many countable sets, so it is countable. Thus, $B_m$ is an open nbhd of $p$ that contains only countably many points of $E$, contradicting the assumption that $p\in P$. It follows that $P$ has no isolated points.

Exactly the same idea is used to show that $E\setminus P$ is countable. Let

$$N=\{n\in\Bbb N:E\cap B_n\text{ is countable}\}\,,$$

and let $W=\bigcup\limits_{n\in N}B_n$. Clearly

$$E\cap W=E\cap\bigcup_{n\in N}B_n=\bigcup_{n\in N}(E\cap B_n)$$

is the union of countably many countable sets, so $E\cap W$ is countable. It’s also clear that $W\cap P=\varnothing$: if $x\in W$, $W$ is an open nbhd of $x$ containing only countably many points of $E$, so $x$ is not a condensation point of $E$.

Finally, suppose that $x\in\Bbb R^k\setminus W$, and let $U$ be any open nbhd of $x$. $\mathscr{B}$ is a base for $\Bbb R^k$, so there is an $m\in\Bbb N$ such that $x\in B_m\subseteq U$. For each $n\in N$ we have $B_n\subseteq W$, and $x\in B_m\setminus W$, so $m\notin N$. By the definition of $N$ this means that $E\cap B_m$ is uncountable, and therefore $E\cap U$ is uncountable (since $B_m\subseteq U$). Thus, $x\in P$, and therefore $\Bbb R^k\setminus W\subseteq P$, i.e., $W\cup P=\Bbb R^k$. We already saw that $W$ and $P$ are disjoint, so they must be complementary: $P=\Bbb R^k\setminus W$, and therefore $E\setminus P=E\cap W$, which we’ve already shown to be countable.

Note, by the way, that this argument also shows directly that $P$ is closed, since it’s the complement of an open set.


问:
If $\dim V = n$, why does any set of $n$ linearly independent vectors span $V$?
I am currently learning about vector spaces, spaning and dimensions for the first time and am reading old posts on here. I came across this one: If $\dim(V) = n$, is every spaning set $\{v_1,v_2,\ldots,v_n\}$ a basis for $V$?

In it, it is mentioned that "If $\dim V = n$, then any set of $n$  linearly independent vectors will span $V$". I understand that any spanning set of $n$ vectors must consist of linearly independent vectors due to $\dim V = n$ but I don't quite get the other direction.

Could someone help me understand this better?

答:
Let $\{v_1,\ldots,v_r\}\subset V$ be a set of $r$ linearly independent vectors, and let $\{w_1,\ldots,w_n\}$ be a basis for $V$. Then there exist $\alpha_{ij}$ with $i\in\{1,\ldots,r\}$ and $j\in\{1,\ldots,n\}$ such that $v_i=\sum_{j=1}^n \alpha_{ij}w_j$. Consider the system of equations: $$\sum_{i=1}^r \alpha_{ij} x_i=0, \ j\in\{1,\ldots,n\}$$
Let $(x_1,\ldots,x_n)$ be a solution. Then $$\sum_{i=1}^r x_i v_i=\sum_{i=1}^r\sum_{j=1}^n\alpha_{ij}x_i w_j=\sum_{j=1}^n\left(\sum_{i=1}^r\alpha_{ij}x_i\right)w_j=\sum_{j=1}^n 0\cdot w_j=0$$
$\{v_1,\ldots,v_r\}$ is linearly independent, so it must be $(x_1,\ldots,x_n)=0$. Hence, the previous system of equations either has unique solution or hasn't solution at all. Anyways, this means that it must have at least as many equations as variables, i.e. $r\leq n$.

Then, any linearly independent set must have, at most, $n$ vectors. Let now $\{v_1,\ldots,v_n\}$ be a linearly independent set and let's prove it must span $V$. Assume it doesn't, and take $v\in V\setminus\operatorname{Span}(v_1,\ldots,v_n)$. Then $\{v,v_1,\ldots,v_n\}$ is a linearly independent set with $n+1$ elements; a contradiction. Hence $\{v_1,\ldots,v_n\}$ spans $V$ so it's a basis.


问:
Dot product versus matrix multiplication, is the later a special case of the first?
I seemed to have thoroughly confused myself today...

Long story short, the question is simple. Is matrix multiplication just a special case of the dot product of two sets of vectors when the sets of vectors have the same cardinality and all vectors in both sets have the same length?

I assume the answer is yes from reviewing the computation of matrix multiplication and the dot product.

答:
What @Pawel said, additionally, though, I would like to add that there is a nice duality between $1\times 2$ matrices and 2d vectors.

3Blue1Brown covers this in the 9th episode of his series 'Essence of Linear Algebra'


问:
Distance between $A= \{(x,\frac{1}{x}) \mid x\in \mathbb{R} \setminus \{0\} \}$ and $B= \mathbb{R \times \{0\}}$.

Let $A= \{(x,\frac{1}{x}) \mid x\in \mathbb{R} \setminus \{0\} \}$ and $B= \mathbb{R \times \{0\}}$. Determine the distance between $A$ and $B$ using the Euclidean metric and the discrete metric.

Letting $(x,y) \in A$ and $(x',y') \in B$ I have that $y'=0$ and $y\ne0$. However I cannot seem to be able to find any upper bound for the distance... I have that $\sqrt{(x-x')^2+(y-y')^2} = \sqrt{(x-x')^2+y^2} \geqslant \sqrt{(x-x')^2}$, but this doesn't seem to help. How should I approach this?

答:
In the Euclidean distance, for all $n \in \Bbb N^+$, $a_n:=(n,\frac1n) \in A$ and $b_n:=(n,0) \in B$ and $d(a_n, b_n)=\frac1n \to 0$ so that $d(A,B)=0$.

In the discrete metric $d(A,B)=1$ for any pair of disjoint sets.


问:
Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$
We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.

$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
 = 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$

Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.

Body:
$$1+x+x^2+\cdots+x^5=\frac{1-x^{6}}{1-x}=(1-x^{6})(1-x)^{-1}$$

so

$$(1+x+x^2+\cdots+x^5)^4=\left(\frac{1-x^{6}}{1-x}\right)^4=(1-x^{6})^4(1-x)^{-4}=(1-4x^{6}+\cdots)(1+4x+10x^2+20x^3+35x^4+56x^5)$$

There is only one $x^4$ term, and you can see immediately that its coefficient is $35$.


问:
Lattices of projections
The set of projections of a von Neumann algebra forms an orthomodular lattice.

Is there some kind of correspondence between such lattices and von Neumann algebras? More specifically, does there exist a theorem that states

"Let $\mathcal{H}$ be some Hilbert space and $\mathcal{P}$ be some family of projections on $\mathcal{H}$. Then the following are equivalent:

-The set $\mathcal{P}$ is an orthomodular lattice.

-The set $\mathcal{P}$ is the set of all projections of some von Neumann algebra $\mathcal{M}$. "?

答: No, sets of projections that are the projections of a von Neumann algebra are much more special than just being orthomodular lattices (or even more specifically, sub-orthomodular complete lattices of the orthomodular complete lattice of all projections on $\mathcal{H}$).  For a really simple example, consider $\mathcal{H}=\mathbb{C}^2$ and let $p$ and $q$ be two rank $1$ projections on $\mathcal{H}$ that are neither equal nor orthogonal complements of each other.  Then $\{0,1,p,q,1-p,1-q\}$ is a subset of the lattice of all projections on $\mathcal{H}$ that is closed under joins, meets, and complements.  However, it is not the projections of any von Neumann algebra: the von Neumann algebra it generates is the entire matrix algebra $M_2(\mathbb{C})$ which has many more projections.

Title: Lattices of projections


答: If $E$ has the discrete topology and is infinite (so non-compact), any function on $E$ is continuous, so we can map it to a compact set $f[E]$ via a continuous function that we can choose to be 1-1 if so desired. So counterexamples abound.

Body:
If $E$ is not compact then any image of $E$ by a continuous 1-1 map is not compact.

First I was going to prove it by assuming otherwise, saying let there is some open cover for $E$ which doesnot have any finite subcover for $E$ then then assume $f(E)$ is compact then it has finite subcover comes from $E$ but we dont know $f$ is open or not.

Then I try to analyse the problem more but I stuck because I realy dont understand what "any image of $E$ by a continuous 1-1 map is not compact." really means?

Any hint and explanation the above sentence?


答:
I think there's a misplaced factor of two in your post, it should be

$$
c = \frac{L}{t^-} \left[ \frac{1 + \Delta t / 2t^-}{1 + \Delta t / t^-}\right]
$$

Now assume that $\Delta t / t^- \ll 1$, you can use the binomial approximation , keeping up to first order terms you get

\begin{eqnarray}
c &=& \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 + \frac{\Delta t}{t^-}\right)^{-1}} \\
&\approx & \frac{L}{t^-} \left(1 + \frac{\Delta t}{2t^-}\right) \color{blue}{\left(1 - \frac{\Delta t}{t^-}\right)} \\
&=& \frac{L}{t^-}\left[1 - \frac{\Delta t}{t^-} + \frac{\Delta t}{2t^-} - \frac{1}{2}\left(\frac{\Delta t}{t^-}\right)^2 \right] \\
&\approx& \frac{L}{t^-}\left[1 - \frac{\Delta t}{2t^-} \right]
\end{eqnarray}

Title: Approximation and Taylor Series

Body:
Given that

$$ t^{+}=t^{-}+\Delta t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
$$ c=\frac{L}{2}\Biggl(\frac{1}{t^{+}}+\frac{1}{t^{-}}\Biggl)\,\,\,\,\,\,\,\,(2) $$
how eqns (1) and (2) can be reworked to
$$  c=\frac{L}{t^{-}}\Biggl[\frac{1+\Delta t/t^{-}}{1+2\Delta t/t^{-}}\Biggl] \,\,\,\,\,\,\,\,(3)             $$

and why (3) can be reduced to

$$  c=\frac{L}{t^{-}}[1-\Delta t/t^{-}] \,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)             $$


答:
Given
\begin{align}
 t^+ &= t^- + \Delta t,  \tag1 \\
 c &= \frac L2 \left(\frac{1}{t^+} + \frac{1}{t^-}\right),  \tag2
\end{align}

we have

\begin{align}
 c &= \frac{L}{2t^-} \left(\frac{t^-}{t^+} + 1\right) \\
 &= \frac{L}{2t^-} \left(\frac{(t^+ - \Delta t) + t^+}{t^+}\right) \\
 &= \frac{L}{2t^-} \left(2 - \frac{\Delta t}{t^+}\right) \\
 &= \frac{L}{t^-} \left(1 - \frac{\Delta t}{2t^+}\right). \\
\end{align}

If $\Delta t$ is small enough compared to $t^-$, then
$\frac{L}{t^-} \left(1 - {\Delta t}/({2t^-})\right)$
might be a reasonable approximation (plus some higher order terms), but
$\frac{L}{t^-} \left(1 - {\Delta t}/{t^-}\right)$
just seems like a mistake.

Plugging in actual numbers, if we let
$t^- = 9.75,$ $t^+ =10.25,$ and $L=1000$, then
$\Delta t = 0.5,$

\begin{align}
\frac L2 \left(\frac{1}{t^+} + \frac{1}{t^-}\right)
&= \frac{1000}{2}\left(\frac1{10.25}+\frac1{9.75}\right) \approx 100.0625,\\
\frac{L}{t^{-}}\left(\frac{1+\Delta t/t^-}{1+2\Delta t/t^-}\right)
&= \frac{1000}{9.75} \left(\frac{1+0.5/9.75}{1+2(0.5)/9.75}\right)
 \approx 97.7937, \tag A\\
\frac{L}{t^{-}}\left(1-\frac{\Delta t}{t^-}\right)
&= \frac{1000}{9.75} \left(1+\frac{0.5}{9.75}\right)
 \approx 97.3044, \tag B\\
\frac{L}{t^-} \left(1 - \frac{\Delta t}{2t^-}\right)
&= \frac{1000}{9.75} \left(1+\frac{0.5}{2(9.75)}\right)
 \approx 99.9343 \tag C\\
\frac{L}{t^{-}} &= \frac{1000}{9.75} \approx 102.5641.
\end{align}

Now let's look at the errors:

\begin{align}
 97.7937 - 100.0625 &= -2.2688, \\
 97.3044 - 100.0625 &= -2.7581, \\
 99.9343 - 100.0625 &= -0.1282, \\
102.5641 - 100.0625 &=  2.5016. \\
\end{align}

So we see that the factor of $2$ in the denominator of
$\frac{L}{t^-} \left(1 - {\Delta t}/({2t^-})\right)$
(in Equation $(C)$)
reduces the error by an entire order of magnitude,
whereas the simple approximation $L/t^-$
has about the same error as either of the approximations in
$(A)$ or $(B)$.
This is not surprising, because by subtracting
$\Delta t/t^-$ when they should be subtracting
about half that value, ${\Delta t}/({2t^+}),$
whoever came up with $(A)$ and $(B)$ overcorrected the
$L/t^-$ approximation by about the same amount as its error.

Title: Approximation and Taylor Series


问:
Why is the constant in differential equations set to c?
Might have been raised and answered a thousand times, but for example in this ODE:

$\begin{align}
&x\,y = y' \\
&\frac{1}{2}\,x^2 + C_1 = \ln(y) + C_2
\end{align}$

I agree at putting $C_1$ and $C_2$ together. But when I'd resolve it for y:

$\begin{align}
&\frac{1}{2}\,x^2 + C = \ln(y) \quad|\cdot e^{(\bullet)} \\
&e^{\frac{1}{2}\,x^2} \, e^{C} = y
\end{align}$

I'd get back $e^{C}$ (in case I'm not totally foolish). However in solutions I always see $C$.

I'd take it if $C$ would stand for $e^{C}$ but in problems with starting conditions involved this is not the case. It really just is a constant. So where am I getting wrong?

答:
From this:

$$
xy = y' \\
\frac{1}{2}x^2 + C_1 = \ln(y) + C_2$$

get this:

$$\begin{align}
\frac{1}{2}\,x^2 + K &= \ln(y) \quad \mid e^{(\bullet)} \\
e^{\frac{1}{2}\,x^2} \, e^K &= y
\end{align}$$

... calling it $K$ rather than $C$.

Then note that if $K$ is an arbitrary constant, then so is $e^K$ -- with the proviso that $e^K > 0$.

So make $C = e^K$.

So you can now say the solution is:

$$y = C e^{\frac{1}{2}\,x^2}$$

taking into account that $C > 0$.


问:
an other definition of adherent value for a sequence in a topological space
Let $(x_n)$ be a sequence in a topological space and $\ell$ an adherent value for $(x_n)$.

I need to prove the equivalence of these two definitions

$$
[\forall V\in\mathcal{V}_{\ell}, |\{n, x_n\in V\}|=+\infty]\Longleftrightarrow [\forall V\in \mathcal{V}_{\ell}, \forall n\in\mathbb{N},\exists k\in\mathbb{N}: k\geq n \Rightarrow x_n\in V ]
$$

the difficulty is to prove that $
[\forall V\in\mathcal{V}_{\ell}, |\{n, x_n\in V\}|=+\infty]\Rightarrow [\forall V\in \mathcal{V}_{\ell}, \forall n\in\mathbb{N},\exists k\in\mathbb{N}: k\geq n \Rightarrow x_n\in V ]
$

答:
It's trivial, a subset of $\Bbb N$ is infinite iff it is unbounded above (which is what the right hand condition states).

A more commmon name for $\ell$ is a cluster point (of a sequence), or accumulation point.


问:
Condition for a cubic equation to have a single root
If a cubic equation

$$ f(x) = ax^3+bx^2+cx+d$$

Is given, what is the condition for the equation to only have a single root (counting multiple roots as one)

答:
Referring to this image of a specific cubic

The geometric parameters labeled on the diagram, in terms of the coefficients, are

$$\begin{align*}
        x_N &= \dfrac{-b}{3a} \quad \text{(abscissa of inflection point)}\\
        \\
        \delta^2 &= \dfrac{b^2-3ac}{9a^2} = x_N^2 - \dfrac{c}{3a} \quad \text{(x distance squared from inflection point to turning point)}\\
        \\
        y_N &= f(x_N) = \dfrac{2b^3}{27a^2} - \dfrac{bc}{3a} + d \quad \text{(ordinate of inflection point)}\\
        \\
        h &= 2a\delta^3 \quad \text{(y distance from inflection point to turning point)} \\
\end{align*}$$

The cubic can intersect the $x$-axis only once, in the following circumstances:

$$\begin{align*} h &= 0\\
\\
h &\in i\mathbb{R}\setminus 0\\
\\
\mathrm{or}\;\left|\dfrac{-y_N}{h}\right| &> 1 \\
\end{align*}$$

For the first two conditions, a $0$ or imaginary height, $h$, means the cubic won't have the two turning points, so it could never cross the $x$-axis more than once.

For the third condition, the ordinate of the inflection point of the cubic, $y_N$, is farther from the $x$-axis than the height, $h$, so the cubic could never cross the $x$-axis more than once.


答:
This is of course :-) intended to find out the answer to:

You have a bag of $16$ (identical) frogs, and wish to distribute them to $4$ of your wizard friends. However each wizard only has enough pocket space to carry $5$ frogs. In how many ways can you distribute the frogs?

And the answer by inclusion-exclusion is

$\binom {16+4-1}{4-1} - \binom{4}{1}\binom {16-6+4-1}{4-1} + \binom{4}{2}\binom {16-12+4-1}{4-1} = \binom {19}{3} - 4\binom {13}{3} + 6\binom {7}{3} = 969 -1144+210 = 35$

Edit to add: As Doug M mentions in comments to the main question, this is also the number of ways to roll $20$ on four distinct dice, since choosing $4$ times from $\{0,1,2,3,4,5\}$ to make  a total of $16$ can be done the same number of ways as choosing $4$ times from $\{1,2,3,4,5,6\}$ to make $20$.

Title: Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$

Body:
We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.

$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
 = 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$

Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.


问:
What does degrees of freedom mean
Say I have a vector space V with dim V=n.

Explain the following to me in terms of degrees of freedom:

Any set of n linearly independent vectors from V spans V

Any set of n-1 linearly independent vectors from V doesn’t necessarily span a subspace of dimension n-1

Any set of n-2 linearly independent vectors from V doesn’t necessarily span a sub space of V of dimension n-2

E.g. In $\mathbb R^3$, take the sub space spanned by (1,0,0) and (0,1,0). (1,0,0) and (0,0,1) does not span this, even though the sub spaces are the same dimension. I would like someone to elucidate this in terms of the concept “degrees of freedom”

答:
For a set of vectors to span a subspace, the vectors need to be contained in that subspace.

Take your example: In $\mathbb{R}^3$ the subspace spanned by $(1, 0, 0)$ and $(0, 1, 0)$, say $W$ is not the same as the subspace spanned by $(1, 0, 0)$ and $(0, 0, 1)$.
For example the vector $(0, 0, 1)$ is not an element of $W$, because you can't write it as a linear combination of $a(1, 0, 0) + b(0, 1, 0)$.

The two spaces are isomorphic, though and have the same dimension.

This doesn't really relate to degrees of freedom in any way.

Title: What does degrees of freedom mean


答:
\begin{align} \left(1+x+x^2+x^3+x^4+x^5\right)^2\ \times \left(1+x+x^2+x^3+x^4+x^5\right)^2 \\
= \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{red}{5x^6}+\color{orange}{4x^7}+\color{green}{3x^8}+\color{blue}{2x^9}+\color{purple}{x^{10}}\right)\ \times \left(1+2x+3x^2+4x^3+5x^4+6x^5+\color{purple}{5x^6}+\color{blue}{4x^7}+\color{green}{3x^8}+\color{orange}{2x^9}+\color{red}{x^{10}}\right)\\
\\
\end{align}

So the coeffecient of $x^{16}$ is $\ \color{red}{5\times 1} + \color{orange}{4\times 2} + \color{green}{3\times 3} + \color{blue}{2\times 4} + \color{purple}{1\times 5} = 35.$


问:
Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?
Let $X$ be a continuous random variable with pdf,
$$f(x;\theta)=\frac {\theta^m.x^{m-1}e^{-\theta x}} {(m-1)!} ; x\geq0, \theta>0$$
Is $\frac{m-1}{x}$ an unbiased estimator of $\theta$ for given pdf?</p>

答:
it's an unbiased estimator if the expectation value of $(m-1)/x$ is equal to $\theta$. The expectation value is calculated as the integral over $x$
$$\int_0^\infty{\rm d}x \frac{m-1}{x} f(x;\theta) = \theta $$
so that it works, the answer is Yes. After a few trivial substitutions, the integral is just the Euler integral for the Gamma functions. In Mathematica, you may type

Integrate[(m - 1)/x theta^m x^(m - 1) Exp[-theta x] / Gamma[m], {x, 0, Infinity}]

and decode the result $\theta$ from the output, while accepting the sensible inequalities. Mathematica didn't exactly cancel the Gamma functions (or factorials) in an elegant way but they do cancel.


问:
What is "numerically exact approach"?
I read a paper now (field: theoretical physics, which is also my background), where they claim that some approach to solve the problem is "numerically exact".

What does this mean?

答: It means that the algorithm used to solve the problem does not involve any approximation so that the final result is exact up to machine precision ($10^{-16}$ for double precision floats for example).

Title: What is "numerically exact approach"?


问:
Difficult Laplace inverse transform
Let $$F(s)=-\frac{1}{s}+\arctan(s+2),\ \ \mathrm{with\ } \ \Re  (s)>0$$
I want to compute the inverse Laplace transform.

attempt: I need to find $f$ such that $L(f)(s)=F(s)$, i.e.,
$$\int_0^\infty e^{-ts}f(t)\ dt = -\frac{1}{s}+\arctan(s+2) $$

but how can I clear $f(t)$ from here? I know for example that if $F(s)=\frac{1}{s}$ for $s\geq 0$ then we should have $f(t)=1$, for $t\geq 0$.

答:
Let's consider the inverse Fourier transformation. By definition $L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}F(s)\exp(sx)ds$, where the integral is evaluating along the line parallel the imaginary axis $Y$ and crossing the $X$ axis at some $a$ ($a$ - real positive number). $\arctan(s+2)$ has two branch points: $s=-2+i$ and $-2-i$. We can connect these brunch points by the cut; as soon as we do not cross the cut $\arctan(s+2)$ is a single-valued function.

We have to evaluate

$L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}+\arctan(s+2)\right)\exp(sx)ds$

After integrating the second term by part we get

$L^{-1}(F)(x)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}-\frac{1}{x(1+(s+2)^2)}\right)\exp(sx)ds+\frac{1}{2\pi{i}x}\exp(sx)\arctan(s+2)|_{a-i\infty}^{a+i\infty}=$$=I_1+I_2+I_3$

To get $I_3$ we just have to take the limit:

$I_3=$$\frac{1}{2\pi{i}x}\lim_{N\to\infty}\left(\exp(ax+iNx)(-\frac{i}{2})\log\frac{1+i(a+2+iN)}{1-i(a+2+iN)}-\exp(ax-iNx)(-\frac{i}{2})\log\frac{1+i(a+2-iN)}{1-i(a+2-iN)}\right)$$=\frac{\exp(ax)}{2\pi{i}}(-\frac{i}{2})(\pi{i})\lim_{N\to\infty}\frac{\exp(iNx)-exp(-iNx)}{x}=\frac{\exp(ax)}{2}\lim_{N\to\infty}\frac{\exp(iNx)-exp(-iNx)}{2xi}=$$=\frac{\exp(ax)}{2}\lim_{N\to\infty}\frac{\sin(Nx)}{x}=\frac{\exp(ax)}{2}\pi\delta(x)=\frac{\pi}{2}\delta(x)$

To evaluate $I_1$ and $I_2$ we integrate along the line parallel the imaginary axis $Y$ (and crossing the $X$ axis at real and positive $a$), but we can close the contour along the part of big circle of radius R counter clockwise on the left side (and, partially, on the right side) of the complex plane. It can be easily evaluated that integrals along the parts of the circle (both in the left and right sides) $\to0$ as soon $R\to\infty$ (Jordan's lemma is applicable here, because integrand $\to0$ at $R\to\infty$). When evaluating $I_2$ we don't have to care about any cut - because we got rid of $\arctan$ and the integrand now is a single-valued function inside the contour.

$I_1=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\left(-\frac{1}{s}\right)\exp(sx)ds=Res_{s=0}(-\frac{1}{s}\exp(sx))=-1$

$I_2=-\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}\frac{1}{x}\frac{\exp(sx)}{1+(s+2)^2}ds=$$-$$\frac{1}{x}(Res_{s=-2+i}\frac{\exp(sx)}{1+(s+2)^2}+Res_{s=-2-i}\frac{\exp(sx)}{1+(s+2)^2})=$$=-\frac{\exp(-2x)}{x}\frac{\exp(ix)-\exp(-ix)}{2i}=-\frac{\exp(-2x)\sin(x)}{x}$

$$L^{-1}(F)(x)=-1+\frac{\pi}{2}\delta(x)-\frac{\exp(-2x)\sin(x)}{x}$$

Title: Difficult Laplace inverse transform


问:
Prove that $ f(x) = |x| $ is Riemann integrable on the interval [-1, 2] using lower and upper integrals
Prove that $ f(x) = |x| $ is Riemann integrable on the [-1, 2] using lower and upper integrals

I am confused how to partition ($P_N$) in this interval. Should we consider two different intervals [-1, 0) and [0, 2] or just one interval [-1, 2]? Then how do use the lower and upper integrals? I know that as long as limits of lower and upper integrals are equal, then the function is Riemann integrable.

答: For each $n\in\Bbb N$ let$$P_n=\left\{-1,-1+\frac1n,-1+\frac2n,\ldots,-1+\frac{3n}n(=2)\right\}.$$In other words, consider the intervals$$\left[-1,-1+\frac1n\right],\left[-1+\frac1n,-1+\frac2n\right],\ldots,\left[-1+\frac{3n-1}n,2\right].$$There are $3n$ intervals here. The function $f$ is decreasing on the first $n$ of those intervals, that is, on the intervals of the form $\left[-1+\frac{k-1}n,-1+\frac kn\right]$, with $k\in\{1,2,\ldots,n\}$. Therefore, the minimum of $f$ on each of them is attained at $-1+\frac kn$ and that minimum is $1-\frac kn$. And $f$ is increasing on the intervals of the form $\left[-1+\frac{k-1}n,-1+\frac kn\right]$, with $k\in\{n+1,n+2,\ldots,3n\}$. Therefore, the minimum of $f$ on each of them is attained at $-1+\frac{k-1}n$ and that minimum is $-1+\frac{k-1}n$. So,\begin{align}L(f,P_n)&=\sum_{k=1}^n\frac1n\left(1-\frac kn\right)+\sum_{k=n+1}^{3n}\frac1n\left(-1+\frac{k-1}n\right)\\&=\frac1n\left(n-\frac{n+1}2\right)+\frac1n\left(-2n+4n-1\right)\\&=\frac52-\frac3{2n}.\end{align}So, $\lim_{n\to\infty}L(f,P_n)=\frac52$ and a similar computation shows that $\lim_{n\to\infty}U(f,P_n)=\frac52$ too. Therefore, $f$ is Riemann-integrable and $\int_{-1}^2|x|\,\mathrm dx=\frac52$.


问:
Intuition behind optimzation problem of ridge regression?
In one of the texts that I am reading it is given that regularization parameter restricts the choice of functions in case data given is not sufficient for processing of signal.
It is given that lambda acts as relative trade-off between norm and loss function.
I have two questions:

Why we are choosing norm as the criterion in this optimization problem?

How lambda is acting as a trade-off?

My understanding says that as I increase lambda my weight has to be reduced much more as compared to when lambda is low,which happens only when we restrict lower limit of the optimization problem.So how lambda is a trade-off when both norm and loss-function reduces with increase of lambda.

Please shed some light on this.

Thank you.

Ridge Regression

答:
Why we are choosing norm as the criterion in this optimization problem?

Mainly because it has a closed form solution (unlike $l_1$ norm penalty) since it was proposed much before the intensive computing era. There is nothing special in Euclidean norm. The $l_1$ norm is much more popular these days, unlike the former, it also performs variables ("features") selection and not only shrinks the magnitude of the coefficients.

How lambda is acting as a trade-off?

For large penalty ($\lambda$) you shrink heavily the coefficients of the features, namely you don't put much faith on your raw data. For $\lambda \to \infty$, the coefficients are shrank to zero. For $\lambda = 0$, you just use the usual least squares estimates, i.e., you use only the loss function and not penalizing it or stabilizing the results.


问:
How to find attributes from a function in machine learning?
I have started learning machine learning. I have a question in mind. Is it possible to know how many attributes or features a certain model has by seeing its equation? Suppose a function given below:

$$f(x) = 2.54  x^2 + a  x + 3.10
$$
How many attributes or features does this model have in machine learning?

答:
You can look at it from two different perspectives

Non-linear model

In this case, there's only one feature $x$

Linear model

If you call $z_0 = x^0 = 1$, $z_1=x^1$, $z_2 = x^2$ then your model becomes

$$
g(z_0,z_1, z_2) = 2.54z_2 + a z_1 + 3.10 z_0
$$

which is a linear model with three features: $z_0, z_1$ and $z_2$


答:
At each point in $\Bbb{R}^3$, there are three degrees of freedom -- that is for each point of the space, we can setup up a (typically "small", because not all spaces are as nice and flat as Euclidean spaces) local coordinate system of displacements from that point which looks like a copy of $\Bbb{R}^3$.  (Note that the local coordinates of displacements do not have to run parallel to the coordinates in the space.  In fact, it is sometimes useful to have the local coordinates conform to some geometric object which we have placed in the big space and these need not be parallel to the big space's coordinates.)

A vector subspace of $\Bbb{R}^3$ is either the origin, a line through the origin, a plane through the origin, or all of $\Bbb{R}^3$ (which we won't discuss further since you seem to be interested in proper subspaces).  If we restrict to one of these (proper subspaces), we have fewer than three degrees of freedom.

For the origin, a $0$-dimenaional subspace, we need $0$ local coordinates to represent possible displacements.  That is, since any displacement violates restriction to the subspace, no displacement is possible and no local coordinates of displacement are needed.

For any point on a line through the origin, one only needs one local coordinate to represent displacements along the line, so there is only one degree of freedom.  Note that most lines through the origin are not parallel to the big space's coordinate axes, so the displacement coordinates are not necessarily the same as the big space's coordinates.

For a point on a plane through the origin, one only needs two local coordinates to represent displacements in the plane, so there are only two degrees of freedom.  Again, the local coordinates need not be parallel to the big space's coordinates.

You talk about the subspace spanned by $(1,0,0)$ and $(0,1,0)$.  This is a plane through the origin.  In particular, it is the $xy$-plane, the plane of points in $\Bbb{R}^3$ with $z$-coordinates equal to zero.  Each point on this subspace has two degrees of freedom.  We are free to choose our local coordinates to point in any direction.  We can make the parallel to and the same scale as the $\Bbb{R}^3$ coordinates, but recall that displacements are measured from the point where you start, not from the origin, so the only point on the plane that can actually have its displacement coordinates exactly agree with the $\Bbb{R}^3$ coordinates is the origin.  For other points, their local coordinates must be displaced to put the origin at their location.

You also talk about the subspace spanned by $(1,0,0)$ and $(0,0,1)$.  This is the $xz$-plane, that is the plane of all points in $\Bbb{R}^3$ with $y$-coordinate equal to zero.  Similar comments about degrees of freedom and displacement coordinates apply on this plane as for the $xy$-plane.

The intersection of these two planes is the set of all points having $z$-coordinates equal to zero and $y$-coordinates equal to zero.  So the intersection of these two planes is the $x$-axis.  Every point in both planes has a degree of freedom parallel to this intersection, so every point of both spaces can have one of its local coordinate axes run parallel to the $x$-axis.  However, the second degree of freedom direction at any such point in the $xy$-plane is not a degree of freedom direction on the $xz$-plane, so there is no choice of local coordinates for points on one plane that corresponds to a possible displacement on the other plane.  Certainly all such points can have one degree of freedom direction direction along the $x$-axis, but their second directions cannot be chosen the same.

In particular, one can take the displacement vector $(1,0,0)$ to be a degree of freedom direction for every point of both planes.  However, on the $xy$-plane, the other displacement vector at each point must be chosen from the $xy$-plane minus the $x$-axis.  ("minus the $x$-axis" because we already have a local coordinate axis pointing in that direction, and we need our coordinates to be linearly independent.)  For the $xz$-plane, for each point, the second degree of freedom direction must be chosen from the $xz$-plane subspace minus the $x$-axis.

But this means that the local coordinate directions for a point on one plane are not parallel to vectors in  $\Bbb{R}^3$ that span the other plane.

It is easy enough for two subspaces to have a nonempty intersection.  (In fact the intersection of any pair of subspaces necessarily includes the origin.)  Consider two distinct random lines in the plane that intersect at the origin.  These are one-dimensional subspaces and if we examine the restrictions to either line, we find one degree of freedom for each point in each line.  But a vector that spans one line does not even lie in the other, so cannot span the other.  A similar thing happens with the two planes you call out: they intersect, but there are vectors in one that are not in the other and vice versa, so neither can span the other.

If vector (sub-)space $V$ contains a spanning set for vector (sub-)space $W$, then necessarily $V$ contains $W$, so their intersection is all of $W$.  If two (sub-)spaces have nonempty intersection, but the intersection does not contain all of either space, then a spanning set for one cannot span the other and vice versa.

Title: What does degrees of freedom mean

Body:
Say I have a vector space V with dim V=n.

Explain the following to me in terms of degrees of freedom:

Any set of n linearly independent vectors from V spans V

Any set of n-1 linearly independent vectors from V doesn’t necessarily span a subspace of dimension n-1

Any set of n-2 linearly independent vectors from V doesn’t necessarily span a sub space of V of dimension n-2

E.g. In $\mathbb R^3$, take the sub space spanned by (1,0,0) and (0,1,0). (1,0,0) and (0,0,1) does not span this, even though the sub spaces are the same dimension. I would like someone to elucidate this in terms of the concept “degrees of freedom”


问:
What is the exact definition of conformal map regarding angle-preserving?
I've looked through many similar questions on this site and also searched Google, but I couldn't find a precise definition that works for what I'm about to ask.

I'm talking about conformal maps from $\mathbb{R}^n$ to $\mathbb{R}^n$, not in the complex sense, I already know the relation between a holomorphic map and conformality, but this is not the object of the question. So, I understand the concept of conformal map at a point. It locally preserves angles between curves and their orientation. It can be defined as a differentiable map at that point whose Jacobian is a possitive multiple of a rotation matrix. And this is the best definition for now because I couldn't find another good-enough equivalent definition.

Trying to formalise the idea of "preserves angles", it would look like this:

Let $f : D\subseteq\mathbb{R}^n \to \mathbb{R}^n$ where $D$ is open. Take $\gamma : I \to D$ to be a curve through the point ${\bf x_0}$, $\gamma(t_0)={\bf x_0}$ and $\gamma$ is differentiable at ${\bf x_0}$.
The function $f$ is conformal at ${\bf x_0}$ if there exists a rotation matrix ${\bf R}$ so that for any such $\gamma$, $f\circ\gamma$ is differentiable at $t_0$ and $(f\circ\gamma)'(t_0) = \alpha{\bf R}\gamma'(x_0),\ \ \alpha\in\mathbb{R}_{>0}$

In this form, the definition with Jacobian just implies it. It's necessary, but not sufficient. Guided by several other approaches I've seen, let's restrict the function $f$ to map continuously-differentiable curves $\gamma$ at ${\bf x_0}$ to continuously-differentiable curves at $f({\bf x_0})$. This way, I think the definitions are equivalent. However, I've never seen this condition stated anywhere.

Another thing is that in some sources, the only curves considered were smooth paths. Why would we add this restriction?

Lastly, let's suppose instead that $D$ is not open, but ${\bf x_0} \in D$ is an accumulation point of $D$. Would this work?

Could anyone provide a definition that is equivalent to $f$ is differentiable at ${\bf x_0}$ and the Jacobian is a positive scalar times a rotation matrix, but regarding curves and angles?

答:
Classically I've seen a conformal map defined as a differentiable map $\phi: M\to N$ for $M,N$ Riemannian manifolds with metrics, $g, \tilde{g}$ such that $$\tilde{g}(\phi_*(v),\phi_*(w))=\lambda^2(p)g(v,w)$$ for $v,w\in T_pM$, where $\phi_*$ is the pushforward of vectors under $\phi$, and $\lambda: N\to \mathbb{R}$ is differentiable.

In $\mathbb{R}^n$ to $\mathbb{R}^m$ Then we have $\phi: \mathbb{R}^n\to\mathbb{R}^m$ such that $J_{\phi}v\cdot J_{\phi} w=\lambda^2(x)(v\cdot w)$ for $v$ and $w$ tangent vectors to curves at $\mathbb{R}^m\ni x=\phi(y)$, $y\in\mathbb{R}^n$. We require differentiability because in general, it is difficult to talk about the orientation at which two curves intersect one another without requiring them to be differentiable. I don't think there is a more basic way to define such a map and remove the differentiability condition.


问:
Density of squared random variable
$X \sim Unif(a, b)$, $a$ and $b$ are arbitrary numbers.

I need to calculate the density of $Y = X^2$. I know about Method of transformations. But in my case the function is not invertible on $R$. However, the function is invertible on the positive and negative half-axes. Is there any way to apply this here?

I also tried to calculate directly through the distribution function. But with any $a$ and $b$ this is difficult.

答:
Consider the the equivalent events.

If $a,b>0$ or $a,b<0$, then
$f_Y(y) = \begin{cases}
\frac{1}{2|b-a|\sqrt{y}}, & a^2<y<b^2\\
0, & \text{o.w.}
\end{cases}$.

If $a<0$ and $b>0$, let $m=\min\{|a|,|b|\}$, and $M=\max\{|a|,|b|\}$, then

$f_Y(y) = \begin{cases}
\frac{1}{(M+m)\sqrt{y}}, & 0<y<m^2\\
\frac{1}{2(M+m)\sqrt{y}}, & m^2<y<M^2\\
0, & \text{o.w.}
\end{cases}$

because $y \in (0,m^2)$ occurs if $x = \pm \sqrt{y}$, but $y\in(m^2, M^2)$ only occurs if $x=\sqrt{y}$, when $M=b$, or if $x=-\sqrt{y}$, when $M=|a|$.

Title: Density of squared random variable