问:
Can one factor matrices?
I know that one can factor integers as a product of prime numbers. Is there an analog of it to matrices? Can we define prime matrices such that every matrix is a product of prime matrices? Is there any applications of factorization of matrices?</p>

答:
There are many ways to do so. We can view prime numbers in two different ways to consider their generalizations to other rings, including matrix rings.

The first uses the notion of ideals. Given a commutative ring $R$, an ideal is a subgroup $I$ of $R$ such that $aI\subset I$ for all $a\in R$.

A prime ideal is an ideal $\mathfrak{p}$ such that for all $a,b\in R$ we have $ab
\in R$ if and only if $a\in R$ or $b\in R$.

An irreducible element is an element $\alpha\in R$ such that for all $a,b\in R$ we have $ab =\alpha$ if and only if $a\in R^ \times$ or $b\in R^\times$.

For the integers $\mathbb{Z}$, an number $p$ is a prime number if and only if $p$ is irreducible if and only if $p\mathbb{Z}$ is a non-zero prime ideal. The fundamental theorem of arithmetic states that every non-zero ideal $I$ can be written uniquely as
$$ I=\prod_{\mathfrak{p}_i \text{ prime ideal}}\mathfrak{p_i}^{n_i}.$$

This theorem holds for any Dedekind domain, so we can view factorisation into primes in greater commutative rings as well.

In the ring $\mathbb{Z}$ an integer can also be written uniquely as a product of irreducible elements. In greater rings this need not be the case however. In the ring $R=\mathbb{Z}[\sqrt{-5}]$ (which is a Dedekind domain), we have
$$ (1-\sqrt{-5})(1+\sqrt{-5})=6=2\cdot 3$$
with all stated factors being irreducible. In fact in other rings, for example $R=\mathbb{Z}[\sqrt{-17}]$, there exist numbers that can be written as product of both 2 and 3 irreducible elements. (Theorem 5.16)

This notion of irreducible elements also applies to the matrix rings. Provided we consider matrix rings over a ring and not a field. For example, in the ring $R=M_2(\mathbb{Z})$, we have
$$\begin{pmatrix} 2 & 0\\ 0 & 2\end{pmatrix}=\begin{pmatrix} 2 & 0\\ 0 & 1\end{pmatrix}\cdot \begin{pmatrix} 1 & 0\\ 0 & 2\end{pmatrix},$$

with the two stated matrices being irreducible. Since $\mathbb{Z}$ is a principal ideal domain, in fact any other irreducible factor will be of the form $U_1AU_2$ where $A$ is any of the matrices above, and $U$ is an invertible matrix (i.e. a matrix of determinant $\pm 1$). (Theorem 5.19)
In other rings this is not necessarily the case. Let $K$ be a field and consider the ring $M_2(K[X,Y,Z])$. Then the matrix
$$A=\begin{pmatrix}X^2 & XY+Z\\
XY-Z & Y^2\end{pmatrix}$$
is irreducible of determinant $Z^2$ (Example 5.21), hence so is its adjugate matrix. Thus, we have
$$A\text{Adj(A)}=\begin{pmatrix} Z^2 & 0\\ 0 & Z^2\end{pmatrix}=\begin{pmatrix} Z & 0\\ 0 & 1\end{pmatrix}^2\cdot \begin{pmatrix} 1 & 0\\ 0 & Z\end{pmatrix}^2,$$
with the matrices on the left and right side of the equation being irreducible.

The theory of factorisations of ideals  into prime ideals in commutative rings is a well studied topic in Number Theory, and can be found in almost any book covering that topic.
Factorisation theory in non-commutative rings is not as well studied, but a summary can be found here.

Title: Can one factor matrices?


问:
General solution for a linear differential equation of second order (Weber differential equation)
$$R''(z)+\lambda^2*(\frac{-z^2}{2}+\frac{1}{2})*R(z)=0$$

The differential equation is only one part of the bigger differential equation. I have solved the other part which was only an ODE of 2 order. Is there a way to solve this?

lambda here is just a constant!
Also, according to Wolfram Alpha this is a Weber differential equation.

答:
$$R''(z)+\lambda^2(\frac{-z^2}{2}+\frac{1}{2})R(z)=0$$
$$\dfrac {R''(z)}{R(z)}=-\lambda^2(\frac{-z^2}{2}+\frac{1}{2})$$
Back to the Riccati's equation we have:
$$\left(\dfrac {R'}{R}\right)'+\left(\dfrac {R'}{R}\right)^2=K^2(z^2-1)$$
Where $ 2K^2= \lambda^2$. You can only solve this DE for some particular values of $K$.  For $K=1 \implies \lambda = \sqrt 2$ It's easy to solve. Then it's like Z Ahmed's answer.
$$\dfrac {R'}{R}=-z$$
$$R'+zR=0$$
$$(Re^{z^2/2})'=0$$
$$R(z)=Ce^{-z^2/2}$$

ContentLicense: CC BY-SA 4.0


问:
Stern Brocot tree sequence
For every positive real number, there exists an infinite sequence in the Stern Brocot tree (beggining at its root 1/1) that converges to the real number.

Can you please help me find the proof of this theorem?

My idea is to use the fact that every real number is present exactly once in the tree, and there are theoretically infinitely many ways how to "create" the branch, however I do not know how to proof the convergence.

答:
Prerequisite / recommended reading:

Stern-Brocot tree

Computing Ancestors of # for Stern-Brocot Tree

For every positive real number, there exists an infinite sequence in the Stern Brocot tree [ .. ] that converges to the real number.
And the OP asks us to prove this.
 Without loss of generality, attention shall be restricted to the (open) interval $]0,1[$.
So let $0 \lt g \lt 1$ such a real number. For numbers $g > 1$, simply take $1/g < 1$ and turn fractions upside down in the end.
At a certain level in the tree we have fractions with numerators $(m_1,m_2)$ and denominators $(m_2,n_2)$ such that:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_2}{n_2}
$$
Lemma.
$$
m_2 n_1 - m_1 n_2 = 1
$$
Proof by mathematical induction.

The tree is rooted by the fractions $0/1$ and $1/1$, hence: $1\cdot 1 - 0\cdot 1 = 1$. And it may be continued in exactly two ways, namely

1. By tightening the lower bound $m_1/n_1$:
$$
\frac{m_1+m_2}{n_1+n_2} \lt g \lt \frac{m_2}{n_2} \quad \Longrightarrow \quad m_2(n_1+n_2) - (m_1+m_2)n_2 = m_2 n_1 - m_1 n_2 = 1
$$
2. By tightening the upper bound $m_2/n_2$:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_1+m_2}{n_1+n_2} \quad \Longrightarrow \quad (m_1+m_2)n_1 - m_1(n_1+n_2) = m_2 n_1 - m_1 n_2 = 1
$$
This already proves the Lemma.

In practice, the tree search for $g$ is ended when its fractional approximation is within certain error bounds:
$$
\left|\, g - \frac{m_1+m_2}{n_1+n_2}\, \right| < \epsilon
$$
With help of the above Lemma we easily derive:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_2}{n_2} \\
\frac{m_1}{n_1} - \frac{m_1+m_2}{n_1+n_2} \lt g - \frac{m_1+m_2}{n_1+n_2} \lt \frac{m_2}{n_2} - \frac{m_1+m_2}{n_1+n_2} \\
- \frac{1}{n_1(n_1+n_2)} \lt g - \frac{m_1+m_2}{n_1+n_2} \lt \frac{1}{(n_1+n_2)n_2}
$$
It follows that convergence to $g$ is quadratically with the inverse of the denominators.
Another consequence is:
$$
\frac{m_1}{n_1} \lt \frac{m_1+m_2}{n_1+n_2} \lt \frac{m_2}{n_2}
$$
That's where the name mediant comes from.

The above may be implemented in a computer program.
The sample code below is written in my favorite programming language (Delphi) Pascal.
An approximation is sought for $g = \ln(2)/\ln(3)$, which indeed is a real number between zero and one.
```
program proef23;<br>
procedure Stern_Brocot(getal,eps : double);
{
  Walk through Stern-Brocot tree
  until |fraction - getal| < eps
}
var
  m1,m2,n1,n2,m,n : integer;
begin
{ Initialize tree }
  m1 := 0; n1 := 1;
  m2 := 1; n2 := 0;
  m := m1+m2; n := n1+n2;
  while true do
  begin
    if getal < (m1+m2)/(n1+n2) then
  { Tightening }
    begin
    { Upper Bound }
      m2 := m1+m2;
      n2 := n1+n2;
    end else begin
    { Lower Bound }
      m1 := m1+m2;
      n1 := n1+n2;
    end;
    Writeln(m1/n1:12:10,' = ',m1,'/',n1,
    ' < ',getal:12:10,' < ',m2,'/',n2,' = ',m2/n2:12:10);
    m := m1+m2; n := n1+n2;
    if abs(getal - m/n) < eps then Break;
  end;
  Writeln;
  Writeln('-1/',n1*n,' < ',getal:12:10,' - ',m,'/',n,' < 1/',n*n2);
end;<br>
begin
  Stern_Brocot(ln(2)/ln(3),1.E-6);
end.
```
Output:
```
0.0000000000 = 0/1 < 0.6309297536 < 1/1 = 1.0000000000
0.5000000000 = 1/2 < 0.6309297536 < 1/1 = 1.0000000000
0.5000000000 = 1/2 < 0.6309297536 < 2/3 = 0.6666666667
0.6000000000 = 3/5 < 0.6309297536 < 2/3 = 0.6666666667
0.6250000000 = 5/8 < 0.6309297536 < 2/3 = 0.6666666667
0.6250000000 = 5/8 < 0.6309297536 < 7/11 = 0.6363636364
0.6250000000 = 5/8 < 0.6309297536 < 12/19 = 0.6315789474
0.6296296296 = 17/27 < 0.6309297536 < 12/19 = 0.6315789474
0.6304347826 = 29/46 < 0.6309297536 < 12/19 = 0.6315789474
0.6307692308 = 41/65 < 0.6309297536 < 12/19 = 0.6315789474
0.6307692308 = 41/65 < 0.6309297536 < 53/84 = 0.6309523810
0.6308724832 = 94/149 < 0.6309297536 < 53/84 = 0.6309523810
0.6309012876 = 147/233 < 0.6309297536 < 53/84 = 0.6309523810
0.6309148265 = 200/317 < 0.6309297536 < 53/84 = 0.6309523810
0.6309226933 = 253/401 < 0.6309297536 < 53/84 = 0.6309523810
0.6309278351 = 306/485 < 0.6309297536 < 53/84 = 0.6309523810
0.6309278351 = 306/485 < 0.6309297536 < 359/569 = 0.6309314587
<p>-1/511190 < 0.6309297536 - 665/1054 < 1/599726
```

ContentLicense: CC BY-SA 3.0

Title: Stern Brocot tree sequence

Body:
Prerequisite / recommended reading:

Stern-Brocot tree

Computing Ancestors of # for Stern-Brocot Tree

For every positive real number, there exists an infinite sequence in the Stern Brocot tree [ .. ] that converges to the real number.
And the OP asks us to prove this.
 Without loss of generality, attention shall be restricted to the (open) interval $]0,1[$.
So let $0 \lt g \lt 1$ such a real number. For numbers $g > 1$, simply take $1/g < 1$ and turn fractions upside down in the end.
At a certain level in the tree we have fractions with numerators $(m_1,m_2)$ and denominators $(m_2,n_2)$ such that:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_2}{n_2}
$$
Lemma.
$$
m_2 n_1 - m_1 n_2 = 1
$$
Proof by mathematical induction.

The tree is rooted by the fractions $0/1$ and $1/1$, hence: $1\cdot 1 - 0\cdot 1 = 1$. And it may be continued in exactly two ways, namely

1. By tightening the lower bound $m_1/n_1$:
$$
\frac{m_1+m_2}{n_1+n_2} \lt g \lt \frac{m_2}{n_2} \quad \Longrightarrow \quad m_2(n_1+n_2) - (m_1+m_2)n_2 = m_2 n_1 - m_1 n_2 = 1
$$
2. By tightening the upper bound $m_2/n_2$:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_1+m_2}{n_1+n_2} \quad \Longrightarrow \quad (m_1+m_2)n_1 - m_1(n_1+n_2) = m_2 n_1 - m_1 n_2 = 1
$$
This already proves the Lemma.

In practice, the tree search for $g$ is ended when its fractional approximation is within certain error bounds:
$$
\left|\, g - \frac{m_1+m_2}{n_1+n_2}\, \right| < \epsilon
$$
With help of the above Lemma we easily derive:
$$
\frac{m_1}{n_1} \lt g \lt \frac{m_2}{n_2} \\
\frac{m_1}{n_1} - \frac{m_1+m_2}{n_1+n_2} \lt g - \frac{m_1+m_2}{n_1+n_2} \lt \frac{m_2}{n_2} - \frac{m_1+m_2}{n_1+n_2} \\
- \frac{1}{n_1(n_1+n_2)} \lt g - \frac{m_1+m_2}{n_1+n_2} \lt \frac{1}{(n_1+n_2)n_2}
$$
It follows that convergence to $g$ is quadratically with the inverse of the denominators.
Another consequence is:
$$
\frac{m_1}{n_1} \lt \frac{m_1+m_2}{n_1+n_2} \lt \frac{m_2}{n_2}
$$
That's where the name mediant comes from.

The above may be implemented in a computer program.
The sample code below is written in my favorite programming language (Delphi) Pascal.
An approximation is sought for $g = \ln(2)/\ln(3)$, which indeed is a real number between zero and one.

program proef23;

procedure Stern_Brocot(getal,eps : double);
{
  Walk through Stern-Brocot tree
  until |fraction - getal| < eps
}
var
  m1,m2,n1,n2,m,n : integer;
begin
{ Initialize tree }
  m1 := 0; n1 := 1;
  m2 := 1; n2 := 0;
  m := m1+m2; n := n1+n2;
  while true do
  begin
    if getal < (m1+m2)/(n1+n2) then
  { Tightening }
    begin
    { Upper Bound }
      m2 := m1+m2;
      n2 := n1+n2;
    end else begin
    { Lower Bound }
      m1 := m1+m2;
      n1 := n1+n2;
    end;
    Writeln(m1/n1:12:10,' = ',m1,'/',n1,
    ' < ',getal:12:10,' < ',m2,'/',n2,' = ',m2/n2:12:10);
    m := m1+m2; n := n1+n2;
    if abs(getal - m/n) < eps then Break;
  end;
  Writeln;
  Writeln('-1/',n1*n,' < ',getal:12:10,' - ',m,'/',n,' < 1/',n*n2);
end;

begin
  Stern_Brocot(ln(2)/ln(3),1.E-6);
end.

Output:

0.0000000000 = 0/1 < 0.6309297536 < 1/1 = 1.0000000000
0.5000000000 = 1/2 < 0.6309297536 < 1/1 = 1.0000000000
0.5000000000 = 1/2 < 0.6309297536 < 2/3 = 0.6666666667
0.6000000000 = 3/5 < 0.6309297536 < 2/3 = 0.6666666667
0.6250000000 = 5/8 < 0.6309297536 < 2/3 = 0.6666666667
0.6250000000 = 5/8 < 0.6309297536 < 7/11 = 0.6363636364
0.6250000000 = 5/8 < 0.6309297536 < 12/19 = 0.6315789474
0.6296296296 = 17/27 < 0.6309297536 < 12/19 = 0.6315789474
0.6304347826 = 29/46 < 0.6309297536 < 12/19 = 0.6315789474
0.6307692308 = 41/65 < 0.6309297536 < 12/19 = 0.6315789474
0.6307692308 = 41/65 < 0.6309297536 < 53/84 = 0.6309523810
0.6308724832 = 94/149 < 0.6309297536 < 53/84 = 0.6309523810
0.6309012876 = 147/233 < 0.6309297536 < 53/84 = 0.6309523810
0.6309148265 = 200/317 < 0.6309297536 < 53/84 = 0.6309523810
0.6309226933 = 253/401 < 0.6309297536 < 53/84 = 0.6309523810
0.6309278351 = 306/485 < 0.6309297536 < 53/84 = 0.6309523810
0.6309278351 = 306/485 < 0.6309297536 < 359/569 = 0.6309314587

-1/511190 < 0.6309297536 - 665/1054 < 1/599726


问:
Find the value of $\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$
The value of$\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$.

My approach :

I am able to solve this part : $(\sqrt{7})^{1/\log_47}$ by changing base :

$(\sqrt{7})^{1/\log_47}   =  (7)^{2\log_74} = 16$

But I am unable to figure how to solve this part :

$\Bigl((\log_29)^2\Bigr)^{1/\log_2(\log_29)}  $

Please help on this … thanks

答:
Your first part is incorrect. $$(\sqrt{7})^{1/\log_47}   =  (7^{1/2})^{\log_74} = 2$$
Use @00xxqhxx00's hint to solve for the 2nd part.  Let $x = \log_29$ $$\Bigl((\log_29)^2\Bigr)^{1/\log_2(\log_29)} = (x^2)^{1/\log_2 x} = (x^2)^{\log_x 2} = 4$$
Hence the required product is $2 \times 4 = 8$.


答:
There's a mistake.

$\begin{align}\sqrt7 ^ {\log_74} = 7^{\frac12\log_74} = 4^\frac12 = 2\end{align}$

Let $\log_29 = k$

So we have $(k^2)^{1/\log_2k} = 2^ 2 = 4$

Title: Find the value of $\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$

Body:
The value of$\left((\log_29)^2\right)^{1/\log_2(\log_29) }\times \left(\sqrt{7}\right)^{1/\log_47}$.

My approach :

I am able to solve this part : $(\sqrt{7})^{1/\log_47}$ by changing base :

$(\sqrt{7})^{1/\log_47}   =  (7)^{2\log_74} = 16$

But I am unable to figure how to solve this part :

$\Bigl((\log_29)^2\Bigr)^{1/\log_2(\log_29)}  $

Please help on this … thanks


问:
Can $3p^4-3p^2+1$ be square number?
I know $4p^4+1$ can't be square number, cause $4p^4<4p^4+1<4p^4+4p^2+1$ for all natural number p.
But I don't know the way to prove $3p^4-3p^2+1$ can(not) be square number. Is there a well known way to prove it?

答:
For those interested, here is a complete solution for all integers (using a bit of theory from elliptic curves - which can be found in Silverman).

Consider the curve
$$ C : y^2 = 3x^4 - 3x^2 + 1$$
which I claim is an elliptic curve over $\mathbb{Q}$. In particular we have that $C$ is birational to the (smooth) elliptic curve with minimal model
$$E : y^2 = x^3 - 15x + 22$$
over $\mathbb{Q}$.

Thus we have a birational map $\phi : C \to E$ with birational inverse $\phi^{-1} : E \to C$ (both defined over $\mathbb{Q}$), so the only rational points on $C$ are the image of $E(\mathbb{Q})$ under $\phi^{-1}$ together with the rational base points of the map $\phi$.

Notice that $E$ has a $2$-torsion point (namely $(2,0)$), so admits a descent by 2-isogeny, and we can compute the rank of $E$ to be $0$. Thus the rational points on $E$ are only the torsion points.

Computing the torsion subgroup is easy (via, e.g., Nagell-Lutz or Formal Groups) and we see that the torsion is of order $6$, generated by $P = (-1,6)$.

Using computer algebra, we compute the base points, and $\phi^{-1}(E(\mathbb{Q}))$ to see that
$$C(\mathbb{Q}) = \{(0, -1), (0, 1), (-1, -1), (-1, 1), (0, -1), (1, 1)\}$$

For those interested, below is Magma code to automate this process

```<code>A2<x,y> := AffineSpace(Rationals(), 2);
C := Curve(A2, 3*x^4 - 3*x^2 + 1 - y^2);
C := ProjectiveClosure(C);
P := C![1,1,1];
assert IsNonSingular(P);

E, phi := EllipticCurve(C, P);
_, phi_i := IsInvertible(phi);
E2, phi2 := MinimalModel(E);
_, phi2_i := IsInvertible(phi2);

assert #TorsionSubgroup(E) eq 6;

Sel2 := TwoDescent(E); //2-descent
assert #Sel2 eq 1; //shows rank is 0 since E has 2-tors pt

P := E2![-1,6];
assert Order(P) eq 6;  //generator
pts := [P, 2*P, 3*P, 4*P, 5*P, 6*P];

im_pts := {@ @};
for pt in pts do
    Include(~im_pts, phi_i(phi2_i(pt)) );
end for;

bs_pts := BasePoints(phi);
rat_pts := im_pts join bs_pts;

rat_pts;
</code>```

Title: Can $3p^4-3p^2+1$ be square number?

Body:
For those interested, here is a complete solution for all integers (using a bit of theory from elliptic curves - which can be found in Silverman).

Consider the curve
$$ C : y^2 = 3x^4 - 3x^2 + 1$$
which I claim is an elliptic curve over $\mathbb{Q}$. In particular we have that $C$ is birational to the (smooth) elliptic curve with minimal model
$$E : y^2 = x^3 - 15x + 22$$
over $\mathbb{Q}$.

Thus we have a birational map $\phi : C \to E$ with birational inverse $\phi^{-1} : E \to C$ (both defined over $\mathbb{Q}$), so the only rational points on $C$ are the image of $E(\mathbb{Q})$ under $\phi^{-1}$ together with the rational base points of the map $\phi$.

Notice that $E$ has a $2$-torsion point (namely $(2,0)$), so admits a descent by 2-isogeny, and we can compute the rank of $E$ to be $0$. Thus the rational points on $E$ are only the torsion points.

Computing the torsion subgroup is easy (via, e.g., Nagell-Lutz or Formal Groups) and we see that the torsion is of order $6$, generated by $P = (-1,6)$.

Using computer algebra, we compute the base points, and $\phi^{-1}(E(\mathbb{Q}))$ to see that
$$C(\mathbb{Q}) = \{(0, -1), (0, 1), (-1, -1), (-1, 1), (0, -1), (1, 1)\}$$

For those interested, below is Magma code to automate this process

A2<x,y> := AffineSpace(Rationals(), 2);
C := Curve(A2, 3*x^4 - 3*x^2 + 1 - y^2);
C := ProjectiveClosure(C);
P := C![1,1,1];
assert IsNonSingular(P);

E, phi := EllipticCurve(C, P);
_, phi_i := IsInvertible(phi);
E2, phi2 := MinimalModel(E);
_, phi2_i := IsInvertible(phi2);

assert #TorsionSubgroup(E) eq 6;

Sel2 := TwoDescent(E); //2-descent
assert #Sel2 eq 1; //shows rank is 0 since E has 2-tors pt

P := E2![-1,6];
assert Order(P) eq 6;  //generator
pts := [P, 2*P, 3*P, 4*P, 5*P, 6*P];

im_pts := {@ @};
for pt in pts do
    Include(~im_pts, phi_i(phi2_i(pt)) );
end for;

bs_pts := BasePoints(phi);
rat_pts := im_pts join bs_pts;

rat_pts;


问:
Linear Algebra Vector
Can you guys take a look at this?

Let $\mathcal P_3(\mathbb R)$ be  the  set  of  polynomials  of  degree smaller or equal to $3$ with real  coefficients.  Show  that

$$\{f\in \mathcal P_3(\mathbb R)\mid f(2)=0\},$$

is a vector space with the standard operations.

答:
Let $f, g, h \in P_3$. Then, $P_3$ is a vector space if the following axioms are satisfied:

$$f+(g+h) =(f+g) +h $$
$$f+g=g+f$$
$$\exists 0\in P_3 : f+0=f$$
$$\exists 1\in \Bbb R : 1f=f$$
$$\forall f \in P_3, \exists  (-f)\in P_3 : f+(-f)=0$$
$$(ab)f=a(bf)\forall a, b\in \Bbb R$$
$$a(f+g) =af+ag$$
$$(a+b) f=af+bf$$

Most of them are straightforward to show.

Title: Linear Algebra Vector


问:
How many k-dimensional $\mathbb{Z}_q$-submodule over $\mathbb{Z}_q^d$?
Denote $\mathbb{Z}_q$ be the residue ring of module $q$ with $q=p^r$ be a prime power.
Similar to the question for $k$-dimensional vector spaces over finite fields, we can also define a $k$-dimensional $\mathbb{Z}_q$-submodule in $\mathbb{Z}_q^d$ as the following:

$$ V = \mathrm{span}\{v_1,\dots,v_k\} $$
where $v_1,\dots,v_k \in \mathbb{Z}_q^d$ are linearly independent, namely, if there exist $a_1,a_2,\dots,a_k \in \mathbb{Z}_q$ such that $a_1v_1+a_2v_2+\dots+a_kv_k = 0$ then $a_1=a_2=\dots=a_k = 0.$

Question: How many $k-$ dimensional $\mathbb{Z}_q$-submodule in $\mathbb{Z}_q^d$? And, how many bases for a $k$-dimensional submodule?

The related question for the case of finite fields was done, you can follow via the link:
How to count number of bases and subspaces of a given dimension in a vector space over a finite field?

答:
Your independence condition means a rank $k$ free submodule, there are also many non-free submodules.

Let $M(d,k,p^r)$ be the set of matrices $A\in \Bbb{Z}/(p^r)^{d\times k}$ such that its reduction $A\bmod p$ (a matrix $ \in \Bbb{Z}/(p)^{d\times k}$) has its $k$ columns linearly independent in the vector space $\Bbb{Z}/(p)^d$.

$$W=\{ span(v_1,\ldots,v_k), \sum_{j=1}^k c_j v_j=0\implies c_1=\ldots =c_k=0\}= M(d,k,p^r)/M(k,k,p^r) $$
The group $M(k,k,p^r)$ acts on $M(d,k,p^r)$ on the right and $M(d,k,p^r)/M(k,k,p^r)$ is the set of orbits, the columns of $A\in M(d,k,p^r)$ represent the $v_j$ and the action of $M(k,k,p^r)$ changes the $v_j$ without changing the submodule it generates.

$$\# W= \frac{\# M(d,k,p^r)}{\# M(k,k,p^r)}=\frac{p^{kd(r-1)}\# M(d,k,p)}{p^{kk(r-1)}\# M(k,k,p)}= \frac{p^{kd(r-1)}\prod_{l=0}^{k-1}(p^d-p^l)}{p^{kk(r-1)}\prod_{l=0}^{k-1}(p^k-p^l)}$$

where $p^{kd(r-1)}$ represents the number of matrices having the same reduction modulo $p$

$\prod_{l=0}^{k-1}(p^d-p^l)$ means choosing $v_1\bmod p$ in $\Bbb{Z}/(p)^d-\{0\}$,  then choosing $v_2\bmod p$ in $\Bbb{Z}/(p)^d-span(v_1)$, choosing $v_3\bmod p$ in $\Bbb{Z}/(p)^d-span(v_1,v_2)$, and so on.


问:
Number of equivalent rectangular paths between two points
I am trying to determine the number of paths between two points.

I am representing the paths as a list of steps "ruru" = right -> up -> right -> up

For my purposes, we can assume that there will never be "backwards" steps (lrlr), and we will only be working with diagonal cases (equal number of steps in each direction, ie: ulul, NOT luu)

Example: Between points (0,0) and (2,2) there are 6 different paths (uurr, urur, urru, ruur, ruru, rruu)

How would I find this value mathematically?

If it is of any help, I found this by exhaustion: 20 paths between (0,0) and (3,3), 6 paths between (0,0) and (2,2), and 2 paths between (0,0) and (1,1).

Note: I will need to be able to do this for both 2 and 3 dimensional cases. The same assumptions are true for the 3D case, I just have a 3rd variable (f and b for forward and back) a valid 3D path might look something like: furrfu. My ideal answer would be in the form "f(dimensions,numSteps) = numPaths"

答:
To get from $\langle 0,0\rangle$ to $\langle m,n\rangle$ you must take $m$ steps to the right and $n$ steps up. These $m+n$ steps may be taken in any order. As soon as you know which $m$ of them are to the right, you know the whole path. There are $\binom{m+n}m$ ways to select $m$ of the $m+n$ steps to be the rightward steps, so there are $$\binom{m+n}m=\binom{m+n}n=\frac{(m+n)!}{m!\,n!}$$ possible paths.

The three-dimensional problem is similar: to get from the origin to $\langle k,m,n\rangle$ you must choose which $k$ of the steps are to the right and which $m$ are up; the remaining $n$ must be in the third direction. There are

$$\binom{k+m+n}k\binom{m+n}m$$

to make these choices, so that is the number of paths. This can be expressed more compactly as the multinomial coefficient

$$\binom{k+m+n}{k,m,n}=\frac{(k+m+n)!}{k!\,m!\,n!}\;.$$


问:
Partial differentiation problem (FOC)
I am having troubles to partially differentiate the following:

$ \frac{\delta (v_i-4b_i)^{1/2}*\frac{b_i-c}{a}}{\delta b_i}=0 $

Mostly I am unsure, which rules are applied here and in what order. So far, I made some attempts, but without any success. Maybe someone can help me understanding the applied procedure here and show me how to do it and what the result is.

答: $$\frac{\partial }{\partial b_i}\left( \sqrt{v_i -4b_i}\cdot \frac{b_i-c}{a}\right) \\ = \frac{\partial (\sqrt{v_i-4b_i})}{\partial b_i}\cdot \frac{b_i-c}{a} + \frac{\partial (\frac{b_i-c}{a} ) }{\partial b_i} \cdot \sqrt{v_i-4b_i} \\ = \frac{-4}{2\sqrt{v_i-4b_i}} \cdot \frac{b_i-c}{a} + \frac 1a \cdot\sqrt{v_i-4b_i}$$

Title: Partial differentiation problem (FOC)


问:
Was the Lie bracket once called the Poisson bracket? (Milnor's Morse Theory)
Multiple times in Milnor's Morse Theory, he refers to the "Poisson bracket" (or, once, an obvious typo: "poison bracket") of two vector fields as $$[X,Y](f)=X(Yf)-Y(Xf).$$ (See, e.g., the bottom of p. 4, or the line after Definition 8.5 on p. 48). Of course, this is not the Poisson bracket at all (which isn't even defined for two vector fields, as far as I'm aware), but is instead the Lie bracket.

I guess this isn't really a math question, but I was just wondering: Is there a reason why he keeps calling it the Poisson bracket? Was it historically called that, or were these just a couple typos on Milnor's part?

答:
I don't know the history, but:

The bracket in a Gerstenhaber algebra is also called an odd Poisson bracket. On the supermanifold $\Pi T^*M$ with even coordinates $x^1,\ldots,x^n$ and odd coordinates $\xi_1,\ldots,\xi_n$ the Schouten bracket (odd Poisson bracket) takes the form $$[\![F,G]\!] = \sum_{i=1}^n\Biggl((F)\frac{\overleftarrow{\partial}}{\partial \xi^i}\cdot\frac{\overrightarrow{\partial}}{\partial x^i}(G) - (F)\frac{\overleftarrow{\partial}}{\partial x^i}\cdot\frac{\overrightarrow{\partial}}{\partial \xi_i}(G)\Biggr)$$
that also justifies the terminology.

The odd coordinates satisfy $\xi_j\xi_i = -(\xi_i\xi_j)$, and for the odd derivatives one has e.g. $$\frac{\overrightarrow{\partial}}{\partial \xi_1}(\xi_1\xi_2) = \xi_2 \qquad \text{ and }\qquad  (\xi_1\xi_2)\frac{\overleftarrow{\partial}}{\partial \xi_1}=(-\xi_2\xi_1)\frac{\overleftarrow{\partial}}{\partial \xi_1} = -\xi_2.$$

Vector fields $\sum_{i=1}^n X^i \frac{\partial}{\partial x^i}$ on $M$ are identified with superfunctions $\sum_{i=1}^n X^i \xi_i$ linear in $\xi$.

Their Schouten bracket (odd Poisson bracket) $$[\![\sum_{i=1}^n X^i \xi_i, \sum_{j=1}^n Y^j \xi_j]\!] = \sum_{k=1}^n \biggl(\sum_{i=1}^n \biggl(X^i \frac{\partial}{\partial x^i}(Y^k) - Y^i\frac{\partial}{\partial x^i}(X^k)\biggr)\biggr)\xi_k$$ agrees with the Lie bracket of vector fields.

Bi-vector fields $\sum_{1\leqslant i<j \leqslant n}P^{ij}\frac{\partial}{\partial x^i}\wedge \frac{\partial}{\partial x^j}$ are identified with superfunctions $P = \frac{1}{2}\sum_{i,j=1}^n P^{ij} \xi_i\xi_j$ quadratic in $\xi$. The Jacobi identity for a Poisson bi-vector field can be expressed as $\frac{1}{2}[\![P,P]\!] = 0$.


问:
Online Math Degree Programs
Are there any real online mathematics (applied math, statistics, ...) degree programs out there?

I'm full-time employed, thus not having the flexibility of attending an on campus program. I also already have a MSc in Computer Science. My motivation for a math degree is that I like learning and am interested in the subject. I've studied through number of OCW courses on my own, but it would be nice if I could actually be able to have my studying count towards something.

I've done my share of Googling for this, but searching for online degrees seems to bring up a lot of institutions that (at least superficially) seem a bit shady (diploma mills?).

答: For future readers: If you happen to speak German, by far the best option is the "Fernuniversität in Hagen", which offers a fully accredited mathematics bachelor and masters programme, full or part time. These are both more affordable and more rigorous (at least the bachelor) than comparable programmes offered by American universities and are just as trustworthy and high quality as those offered by any other German university (even more so if you ask me because the quality of the material they provide is excellent).</p>

ContentLicense: CC BY-SA 2.5

Title: Online Math Degree Programs


问:
Given the matrix $A^3$, find A
Given the matrix $A^3= \begin{bmatrix}83&84\\42&41\end{bmatrix}$, find A

I only know how to use the basic method to find A, which is $A^3= \begin{bmatrix}a&b\\c&d\end{bmatrix}^3$, then solve a,b,c,d.
Is there any other method?

答:
Note that $B:= A^3$ has minimal polynomial $(x-125)(x+1)$ so to apply the function $x \mapsto \sqrt[3]{x}$ we have to find a polynomial $p(x) = ax+b$ such that
$$p(125) = \sqrt[3]{125} = 5, \quad p(-1) = \sqrt[3]{-1} = -1.$$
We get that $p(x) = \frac1{21}(x-20)$ so
$$\sqrt[3]{B} = p(B) = \frac1{21}(B-20I) = \begin{bmatrix}3 & 4 \\ 2 & 1  \end{bmatrix}.$$
This is certainly a cube root of $B$.

Title: Given the matrix $A^3$, find A


答:
An important relation is : $$\log_a b= \frac{1}{\log_b a}$$
And,
$$a^{\log_a b}=b$$

Your question:$$ \Bigl((\log_2 9 * \log_2 9)^{\frac{1}{\log_2 (\log_2 9)}} \Bigr)* 7^{\frac{1}{2* \log_4 7}} $$
Can be written as:
$$\Bigl((\log_2 9)^{\log_{\log_2 9}2}\Bigr)^{2} * \bigl(7^{\frac{\log_7 4}{2}}\bigr)$$
$$(2)^{2}*{2}=8$$

Hence, the answer is 8, using the above mentioned relations.


问:
$\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$
How can we mathematically precisely argue that $$\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$$ holds?

So how can we bring

$$1-\frac xn+\frac{x}{n^2} = 1- \frac{(n+1)x}{n^2} \approx 1 - \frac xn $$
and
$$\lim_{n \to \infty} \left(1-\frac xn\right)^n=e^{-x}$$
together?

Or can we use this form somehow:
$$\left(1-\frac xn+\frac{x}{n^2}\right)^n
= \left(\left( 1 - \frac{x}{\frac{n^2}{n+1}} \right)^{\frac{n^2}{n+1} }\right)^{\frac{n+1}{n}}$$

答:
Hint:

$$\lim_{n\to\infty}\left(1-\dfrac xn+\dfrac x{n^2}\right)^n =\left(\lim_{n\to\infty}\left(1+\dfrac{x(1-n)}{n^2}\right)^{\dfrac{n^2}{x(1-n)}}\right)^{\lim_{n\to\infty}\dfrac{nx(1-n)}{n^2}}$$

Now in the inner limit  set $\dfrac{x(1-n)}{n^2}=y\implies y\to0$

and $\lim_{n\to\infty}\dfrac{nx(1-n)}{n^2}=x\lim_{n\to\infty}\left(\dfrac1n-1\right)=?$


问:
Can anyone show me, how to solve these system of Equations:
$$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$

Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step by step. It is a RMO question (The Pre- Indian team selection exam for IMO)

答:
So I'll assume you're only looking for integer solutions. Call the equations (1), (2), and (3) respectively. Expand (2):

$ x^2 + y^2 + z^2 + 3x y + 3 x z + 3 y z = 1.$

From this equation, subtract the square of (1) to obtain

$ x y + x z + y z = -3.$

Subtract double this expression from the square of (1) to obtain

$ x^2 + y^2 + z^2 = 10.$

Aha! Since $x,y,z$ are integers, this means $x=\pm 3, y=\pm 1, z =0$ or some permutation thereof. (Just look at the various cases. There aren't that many.) By (1), this means $x=3,y=-1,z=0$ or some permutation thereof.


问:
Why is the left side of Euler's formula not equal to the right when I plug in numbers?
I am struggling to understand Euler’s equation and when I insert numbers, I can’t get the left side to equal the right side.  Searching Stack Exchange and other web sources, I am not finding an answer.  Here is the situation:

Given the formula

$$e^{ia} = \cos(a)+i \sin(a)$$

If I assign values:

$i = 2$ ;
$a = 0.087\, \mathrm{radians}$ (same as five degrees)

$cos(a) = 0.996$ ;
$sin(a) = 0.087$ ;
$i*sin(a) = 0.174$.

the right-hand side equals $1.171$
the left-hand side equals $1.191$

and of course, those two are not equal.

Why the two sides of this equation not equal?

答: Because $i$ is not a variable. It is the imaginary unit.


问:
Integrals involving irrational numbers

$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$

How should I approach getting an estimate on the numeric value in this case?

Title: Integrals involving irrational numbers

Body:
Use Mathematica [1]:
$$ \frac{1}{2\pi} \int_0^{2 \pi}\sin^{100}x\,  dx = \frac{12611418068195524166851562157}{158456325028528675187087900672}\approx 0.0795892$$

[1] https://www.wolframalpha.com/input/?i=Integrate%5BSin%5E100+%28x%29%2C%7Bx%2C0%2C2Pi%7D%5D%2F%282Pi%29


答:
Due to the symmetry of the graph of $\sin^{100} x$, the integral equals $$\frac{4}{2\pi} \int_0^{\pi/2} \sin^{100} x \ dx =\frac{2}{\pi} \cdot \frac{100!}{(2^{50}50!)^2}\cdot \frac{\pi}{2}= \frac{100!}{2^{100} \cdot (50!)^2}$$

using this result.

Title: Integrals involving irrational numbers

Body:
$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$

How should I approach getting an estimate on the numeric value in this case?


答:
When $L=\lim_{x\to a} f(x)^{g(x)}\to 1^{\infty}$, then
$L=\exp[\lim_{x\to a} g(x) [f(x)-1]$
$$K=\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n  $$
$$K=\exp[ \lim_{x \to \infty} n [1-x/n+x^2/n^2-1]]=\lim_{x \to \infty} -x+x/n=e^{-x}$$

Body:
How can we mathematically precisely argue that $$\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$$ holds?

So how can we bring

$$1-\frac xn+\frac{x}{n^2} = 1- \frac{(n+1)x}{n^2} \approx 1 - \frac xn $$
and
$$\lim_{n \to \infty} \left(1-\frac xn\right)^n=e^{-x}$$
together?

Or can we use this form somehow:
$$\left(1-\frac xn+\frac{x}{n^2}\right)^n
= \left(\left( 1 - \frac{x}{\frac{n^2}{n+1}} \right)^{\frac{n^2}{n+1} }\right)^{\frac{n+1}{n}}$$


Title: Integrals involving irrational numbers

Body:
Probably, you should use complex numbers.
Say,
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \sin^{100}(x) dx =
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{(e^{ix} - e^{-ix})^{100}}{2^{100}} dx =
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}}(e^{2 ix} - 1)^{100} dx.
$$
If we exand $(e^{2 ix} - 1)^{100}$ and start doing integrals, we will see that they all are zero except
$$
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{e^{-100 i x}}{2^{100}} {100\choose 50} e^{2 \times 50 \times ix} dx =
\frac{1}{2^{100}} {100\choose 50} \approx 0.0795...
$$


问:
Hyperplane arrangements including parallel hyperplanes
Suppose $A_1, \dots, A_k$ are sets of hyperplanes in $\mathbb{R}^d$ such that within any one set $A_i$, the hyperplanes are parallel. Moreover, suppose $|A_i| \leq s$ for all $i \in \{1, \dots, k\}$.

I'd like to know an upper bound on the number of connected components in $\mathbb{R}^d \setminus \left(\cup_{i = 1}^k A_i\right)$. In other words, we can think of $\cup_{i = 1}^k A_i$ as partioning $\mathbb{R}^d$ into a set of cells, and I'd like to know an upper bound on the number of cells there are in this partition.

Ignoring the fact that many of the hyperplanes are parallel, I know from this post that the number of connected components is at most $\sum_{i = 0}^d {sk \choose i}$. I wonder if there's a better bound that takes into account the fact that many of the hyperplanes are parallel.

答:
Given natural $d$, $k$, and $s$, denote the maximal number of connected component in $\mathbb{R}^d \setminus \left(\cup_{i = 1}^k A_i\right)$ by $(d,k,s)$. Clearly, $(1,k,s)=ks+1$ and $(d,1,s)=s+1$.

To see that $(2,k,s)\le \sum_{i=0}^2 {k \choose i}s^i$ by induction, remark that adding to an arrangement of $k$ families of $s$ parallel lines each, we have that each of $s$ new lines crosses at most $ks$ lines, increasing a number of connected components by at most $ks+1$. That is

$$(2,k+1,s)\le (2,k,s)+ {k \choose 2}s^2+{k\choose 1}s^1+{k\choose 0}s^0+ks^2+s^1=$$ $${k+1 \choose 2}s^2+{k+1\choose 1}s^1+{k+1\choose 0}s^0.$$

Similarly for $d\ge 2$ we obtain an inequality $(d,k+1,s)\le (d,k,s)+s(d-1,k,s)$, which implies by induction with respect to $d$ and $k$, $$(d,k,s)\le \sum_{i=0}^d {k \choose i}s^i.$$


问:
Calculate all numbers $x ∈ ℤ$ that simultaneously satisfy the following 3 congruences
Calculate all numbers $x ∈ ℤ$ that simultaneously satisfy the following 3 congruences:

$x ≡ 7 mod 11$

$x ≡ 1 mod 5$

$x ≡ 18 mod 21$

How can I solve this system for $x$? I've tried the chinese remainder theorem, but i dont get the part with the modulo inverse. I know there must be a solution like $x = n+1155k$ with $k ∈ ℤ$, how i get the $n$?  Any hints of solutions are greatly appreciated. What value of x satisfies these three equations?

答:
If $x \equiv 7\pmod {11}$, then $x = 11p + 7$ for some integer $p$. Then, we have:

$$11p + 7\equiv 1\pmod 5$$

$$p \equiv 4\pmod 5$$

Thus, $p = 5q + 4$ for some integer $q$. Then:

$$x = 11p + 7$$

$$x = 11(5q + 4) + 7$$

$$x = 55q + 51$$

Substituting into the last congruence:

$$55q + 51\equiv 18\pmod{21}$$

$$13q\equiv 9\pmod{21}$$

Note that $13^{-1}\equiv 13\pmod{21}$:

$$q\equiv 117\pmod{21}$$

$$q\equiv 12\pmod{21}$$

Then, $q = 21r + 12$ for some integer $r$:

$$x = 55(21r + 12) + 51$$

$$\boxed{x = 1155r + 711\text{ for }r\in\mathbb{Z}}$$

This strategy will work with any number of congruences. The only real work here is finding modular inverses.


答: Since the roots of the characteristic polynomial are $-1,125$, two distinct roots, in dimension two, $A^3$ is diagonalizable.  Find a basis of eigenvectors, $\{v_1,v_2\}$.  Then conjugate by the matrix $P$ whose columns are those eigenvectors, to get $A=P\begin{pmatrix}-1&0\\0&5\end{pmatrix}P^{-1}$.

Title: Given the matrix $A^3$, find A

Body:
Given the matrix $A^3= \begin{bmatrix}83&84\\42&41\end{bmatrix}$, find A

I only know how to use the basic method to find A, which is $A^3= \begin{bmatrix}a&b\\c&d\end{bmatrix}^3$, then solve a,b,c,d.
Is there any other method?


问:
How to maximize a function when you cannot solve gradient = 0
I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$
$$ a, b \in{R+*} $$
Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$

I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). But because a,b are positive real numbers, I cannot solve $$ \frac{a}{2\sqrt{x}} = 0 $$
And it's the same for y.

How do I deal with such a situation. Thank you very much !!

答:
By cauchy-schwarz-inequality, $$a\sqrt{x} + b\sqrt{y} = (\sqrt2a)(\sqrt{x}/\sqrt2) + b\sqrt{y} \le \sqrt{(2a^2+b^2)(x/2 + y)} \le \sqrt{2a^2+b^2}$$
due to the constraint $x \le 2 − 2y \iff (x+2y)/2 \le 1$, with equality holds iff $$\frac{\sqrt2a}{b} = \frac{\sqrt{x}/\sqrt2}{\sqrt{y}} \iff 2a \sqrt{y} = b \sqrt{x}$$
and $x + 2y = 2$.
\begin{align}
2a \sqrt{y} &= b \sqrt{2 - 2y} \\
2a^2 y &= b^2 (1 - y) \\
y &= \frac{b^2}{2a^2 + b^2} \in (0,1). \tag{$a,b \in \mathbb{R}_+^*$} \\
x &= 2(1-y) = \frac{4a^2}{2a^2 + b^2}
\end{align}


答:
We must have
$$
x=18+21k\qquad k\in\mathbb{Z}.
$$
Then the congruence modulo $11$ becomes
$$
x\equiv7+10k\equiv7\bmod 11
$$
so that $k=11\ell$ wth $k\in\mathbb{Z}$. Considering now the congruence modulo $5$ we get
$$
x\equiv18+21\cdot11\ell\equiv3+\ell\equiv1\bmod 5,
$$
i.e. $\ell\equiv3\bmod5$.

Thus take $\ell=3$, hence $k=33$ and finally $x=18+21\cdot33=711$.

Body:
Calculate all numbers $x ∈ ℤ$ that simultaneously satisfy the following 3 congruences:

$x ≡ 7 mod 11$

$x ≡ 1 mod 5$

$x ≡ 18 mod 21$

How can I solve this system for $x$? I've tried the chinese remainder theorem, but i dont get the part with the modulo inverse. I know there must be a solution like $x = n+1155k$ with $k ∈ ℤ$, how i get the $n$?  Any hints of solutions are greatly appreciated. What value of x satisfies these three equations?


问:
Proving that Bombieri's Theorem implies Linnik's theorem
I'm stuck on a line in the proof of Bombieri implies Linnik, where

Bombieri:
  For primitive $\chi$ mod $q$ with $q \leq T$ we define $$N(\alpha, T; \chi)=\#\{\rho=\beta+i\gamma \;:\; \Lambda(\rho,\chi)=0, \beta>\alpha, |\gamma|<T\}.$$

There exists $c>0$ such that, for all $T \geq 2$ and all $1/2 \leq \alpha <1$, $$\sum_{q \leq T}\sum_{\chi \text{ mod }q}^*N(\alpha,T;\chi)=O(T^{c(1-\alpha)})$$

where the asterisk on the sum over characters indicates that only primitive characters are to be inlcuded.

If there exists a real primitive character $\chi_1$ mod $q_1$ with $q_1<T$, such that $\Lambda(s,\chi_1)$ has zero $\beta_1>1-\frac{\delta}{\log T}$, then we have the improved estimate

$$\sum_{q \leq T}\sum_{\chi \text{ mod }q}^*N'(\alpha,T;\chi)=O(T^{c(1-\alpha)}(1-\beta_1)\log T)$$

where the ' on the $N(\alpha,T;\chi)$ indicates that the single exceptional zero is to be excluded.

Linnik: If $a$ and $q$ are coprime then there exists a prime $p \equiv a$ mod $q$ with $p<O(q^{O(1)})$.

In the exceptional zero case, assuming Bombieri, I've got to the point where

$$\sum_{p \leq x, p \equiv a \text{ mod }q }\log p \geq \frac{1}{2 \phi(q)}\left(x-\frac{x^{\beta_1}}{\beta_1}\right)+O\left(\frac{x(\log x)^2}{T}\right)$$

for $T=q^A$ with $A>3$ and $x=T^B$ with $B>12$ a large constant. I also know that $$\beta_1\leq 1-\frac{c}{\sqrt{T}(\log T)^2}.$$

I think what I want to show is that $x^{\beta_1}\leq x^{1-\epsilon}$ for some $\epsilon>0$, because since $(O(x(\log x)^2/T)=O(x^{1-\epsilon})$ for small enough $\epsilon$, Linnik's theorem will follow. But just throwing in the bound for $\beta_1$ doesn't seem to give me this.

答:
We may write
$$x - \frac{x^{\beta_1}}{\beta_1} = x\left(1 - \frac{x^{\beta_1 - 1}}{\beta_1}\right).$$

Consider two cases.  In the first case, $\beta_1 < 1 - \frac{c}{\log x}$ for a fixed constant $c$.  In this case the answer is $\gg x$, which is more than sufficient.  Otherwise,  note that for $t \in [1 - \frac{c}{\log x}, 1]$ we have
$$\frac{d}{dt} \frac{x^{t-1}}{t} = \frac{x^{t-1}\log x}{t} - \frac{x^{t-1}}{t^2} \gg \log x.$$
Therefore, by the mean value theorem,
$$ x\left(1 - \frac{x^{\beta_1 - 1}}{\beta_1}\right) \gg x(1-\beta_1) \log x.$$
Again, this bound suffices, since $q \leq T^{\frac{1}{3}}.$


问:
Positive-semidefiniteness of combination of two positive-semidefinite matrices
I am struggling for a while in proving the positive-semidefiniteness

( in the usual sense, i.e., $A$ is Hermitian and $x^*Ax\geq 0\,$ for all
$\,x \in \mathbb{C}^2\,\big)$

of the following matrix
$$A :=\,BC^\top + CB^\top+BB^\top,$$
under the hypotheses that both $B$ and $C$ have real entries and are positive-semidefinite in $\mathbb R^2 $: in the sense that
$$x^\top B x\geq 0\quad\forall\, x\in\mathbb{R}^2$$
and likewise for $C$. Furthermore, $B,C$ are not necessarily symmetric.

Numerical tests suggest that this holds true whenever the corresponding non-diagonal elements of $B$ and $C$ share the same sign, that is, if we set $B=\begin{bmatrix} b_1 & b_2\\b_3 & b_4\end{bmatrix}$ and $C=\begin{bmatrix} c_1 & c_2\\c_3 & c_4\end{bmatrix}$ then $sign(b_2)=sign(c_2)$ and $sign(b_3)=sign(c_3)$.

My most promising attempt has been effectively computing matrix $A$ and checking the principal minors. We have that
$$A=\begin{bmatrix}
 2(l_1u_1+l_2u_2)+l_1^2+l_2^2 & l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4\\
 l_1u_3+l_3u_1+l_4u_2+l_2u_4+l_1l_3+l_2l_4 & 2(l_3u_3+l_4u_4)+l_3^2+l_4^2
 \end{bmatrix},$$
and the non-negativity of the diagonal elements of $A$ comes from the shared sign of the off-diagonal elements of $B$ and $C$. It remains to ensure that $\det(A)\geq 0$. My bet is on the following computations:

\begin{equation*}
\begin{aligned}
\det(A)&=4(b_1c_1+b_2c_2(b_3c_3+b_4c_4)+2b_1^2(b_3c_3+b_4c_4)+2b_2^2(b_3c_3+b_4c_4)+2b_3^2(b_1c_1+b_2c_2)+2b_4^2(b_1c_1+b_2c_2)\\
&\quad +b_1^2b_3^2+b_1^2b_4^2+b_2^2b_3^2+b_2^2b_4^2-(b_1c_3+b_3c_1+b_4c_2+b_2c_4+b_1b_3+b_2b_4)^2\\
&=4(b_1c_1+b_2c_2)(b_3c_3+b_4c_4)+2b_1^2({b_3c_3}+b_4c_4)+2b_2^2(b_3c_3+{b_4c_4})+2b_3^2({b_1c_1}+b_2c_2)+2b_4^2(b_1c_1+{b_2c_2})\\
&\quad +{b_1^2b_3^2}+b_1^2b_4^2+b_2^2b_3^2+{b_2^2b_4^2}-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-{b_1^2b_3^2}-{b_2^2b_4^2}\\
&\quad-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-{2b_1^2c_3b_3}-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4-{2b_3^2b_1c_1}-2b_3c_1b_2b_4\\
&\quad-2b_4c_2b_2c_4-2b_4c_2b_1b_3-{2b_4^2c_2b_2}-2b_2c_4b_1b_3-{2b_2^2c_4b_2}-2b_1b_3b_2b_4\\
&=4b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+4b_2c_2b_4c_4+2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1+b_1^2b_4^2+b_2^2b_3^2\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_3c_1-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_1c_3b_2b_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad -2b_3c_1b_2b_4-2b_4c_2b_2c_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3-2b_1b_3b_2b_4\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4+\underline{2b_1^2b_4c_4+2b_2^2b_3c_3+2b_3^2b_2c_2+2b_4^2b_1c_1}+\underline{(b_1b_4-b_2b_3)^2}\\
&\quad -b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2-2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad \underline{-2b_1c_3b_2b_4-2b_3c_1b_2b_4-2b_4c_2b_1b_3-2b_2c_4b_1b_3}\\
&=2b_1c_1b_3c_3+4b_1c_1b_4c_4+4b_2c_2b_3c_3+2b_2c_2b_4c_4-b_1^2c_3^2-b_3^2c_1^2-b_4^2c_2^2-b_2^2c_4^2\\
&\quad -2b_1c_3b_4c_2-2b_1c_3b_2c_4-2b_3c_1b_4c_2-2b_3c_1b_2c_4\\
&\quad +(b_1b_4-b_2b_3)(b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3),
\end{aligned}
\end{equation*}
where I have underlined the terms involved in the final equality in order to facilitate the reading.

And now I am stuck. I don't know if there is any obscure equality to gather the remaining terms, and can not even ensure that
$$b_1b_4-b_2b_3+2b_1c_4+2c_1b_4-2b_2c_3-2c_2b_3\geq 0,$$
although intuition tells me it should be true.

If you have any ideas for the follow up from here, I would be happy to hear them.

答:
A counter-example is  proposed, in a rush after the lunch break, hence check carefully please.

Consider the real matrices
$$B\:=\:\begin{pmatrix}1& 0\\ 0& \beta\end{pmatrix} \text{ with }\:0<\beta\qquad
C\:=\:\begin{pmatrix}1& \gamma\\ \gamma& 1\end{pmatrix}
\text{ with }\:0<\gamma <1\,,$$
they are symmetric, thus Hermitian a fortiori. Within the indicated parameter ranges both matrices are positive-definite, either seen as members of
$M_2(\mathbb R)$, either assuming that $B,C\in M_2(\mathbb C)$.

We compute that
$$A\:=\:BC^\top + CB^\top+BB^\top \:=\:
\begin{pmatrix}3& (1+\beta)\gamma\\ (1+\beta)\gamma& \beta (2+\beta)\end{pmatrix}\\[5ex]
\implies \det A\:=\: 3\beta (2+\beta) - (1+\beta)^2\gamma^2$$
and conclude, if $\beta$ is sufficiently small, that $A$ has one positive and one negative eigenvalue. In other words, $A$ is (definitely!) an indefinite matrix.


问:
Evaluate $\lim_{x\to -\infty} \sqrt {4x^2-x} +2x$
Rationalising gives
$$\lim_{x\to -\infty} \frac{-x}{\sqrt{4x^2 -x} -2x}$$
$$=\lim_{x \to -\infty} \frac{-1}{\sqrt {4-\frac 1x}-2}$$
$$=-\infty$$

But the correct ans is $\frac 14$

What am I doing wrong?

答: For $x \lt 0$ you have $\sqrt{x^2} = -x$ and not $\sqrt{x^2} = x$ as the square root is by definition non negative.


问:
$1-\frac{\ln(a+b)}{3}\le \int^1_0 (a+b)^{-x^2}\,dx \le \frac{\arctan(\sqrt{\ln(a+b)})}{\sqrt{\ln(a+b))}}$
Let $a$ and $b$ real numbers such that $a >1/2 , b >1/2$. Prove that :
$1-\frac{\ln(a+b)}{3}\le \int^1_0 (a+b)^{-x^2}\,dx \le \frac{\arctan(\sqrt{\ln(a+b)})}{\sqrt{\ln(a+b)}}$
I tried to apply the Mean value theorem but I did not get to any result.

答:
$(a+b)^{-x^2} = e^{-x^2 \ln(a+b)}$, and $1+u\le e^u \le \frac{1}{1-u}$ holds for all real $u < 1$, so that
$$
 1 - x^2 \ln (a+b) \le (a+b)^{-x^2} \le \frac{1}{1+x^2 \ln(a+b)}
$$
for $0 \le x \le 1$. Now integrate over $[0, 1]$ to get the desired inequalities.


问:
Proof: Identity with integral
Consider:

$$1-e^{at} \sum_{k=0}^{n-1}\frac{(at)^k}{k!} = \int_0^t \frac{a^k}{(k-1)!}x^{k-1}e^{-ax}dx$$

How do I get to the right?

答: Hint: Take the derivative $\frac{d}{dt}$ of both sides and invoke the fundamental theorem of calculus.

Title: Proof: Identity with integral


答:
Notice that $ \left(\forall y\in\left(0,1\right)\right) $ : $$ y\leq-\ln{\left(1-y\right)}\leq\frac{y}{1-y} $$

That can be proven either using the main value theorem,, or studiing some functions.

Thus, setting $ y_{n}=\frac{x}{n}-\frac{x}{n^{2}} $, where $ x\in\mathbb{R} $, and $ n\in\mathbb{N}^{*} $ being bigger enough so that $ y_{n}\in\left(0,1\right) $. We get : $$ \ \ \ \ \ \ \ \ \ \ \ \ \frac{x}{n}\left(1-\frac{1}{n}\right)\leq -\ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)}\leq\frac{\frac{x}{n}\left(1-\frac{1}{n}\right)}{1-\frac{x}{n}+\frac{x}{n^{2}}}\\ \iff \frac{-x\left(1-\frac{1}{n}\right)}{1-\frac{x}{n}+\frac{x}{n^{2}}}\leq\ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}}\leq -x\left(1-\frac{1}{n}\right) $$

Taking $ n $ to $ +\infty $, we get that : $$ \ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}}\underset{n\to +\infty}{\longrightarrow}-x $$

Which means : $$ \left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}\underset{n\to +\infty}{\longrightarrow}\mathrm{e}^{-x} $$

Title: $\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$


问:
Help me prove $\left(\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$ isn't a perfect square
I need some help in proving

$$\left( \displaystyle\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$$

isn't a perfect square. The only way I can think is repeated integration by parts which is obviously impractical and I still likely wouldn't be able to deduce if the result was a perfect square.

Thanks very much guys! :D

答: Repeated integration by parts seems to be the way to do it. There will be an obvious pattern, so you don't need to do it 50 times. Hint: The integral has the value 50!.</p>


答:
We first look for a non-trivial solution of the system

$\begin{align*}
&x \equiv 1 \mod 11\\
&x \equiv 1 \mod 5\\
&x \equiv 1 \mod 21
\end{align*}$

We have $5 \times 21 = 105$, coprime with $11$. Hence, we can find $a_1, b_1 \in \mathbb Z$ such that $11 a_1 + 105b_1 = 1$. Using your favourite algorithm (for example, one adapted from the Euclide GCD algorithm) you find

$11\times(-19) + 105 \times 2 = 1$.

We repeat this process for the other equations:

$11 \times 21 = 231$ and

$5 \times (-46) + 231 \times 1 = 1$

And for the final equation,

$11 \times 5 = 55$ and

$21 \times 21 + 55 \times (-8) = 1$

We have computed some interesting things, but what was the point of all that ? Well, if you look carefully at what we have just done, you notice that the first Bezout relation that we computed tells us that $105 \times 2 = 210 \equiv 1 \mod 11$. Moreover, because $105 = 5 \times 21$, we automatically have that $210 \equiv 0 \mod 21$ and $210 \equiv 0 \mod 5$.

Combining those results, we have:

$7 \times 210 \equiv 7 \mod 11$ and is equal to $0 \mod 5$ and $\mod 21$

$1 \times 231 \equiv 1 \mod 5$ and is equal to $0 \mod 11$ and $\mod 21$

$18 \times 55 \times (-8) \equiv 18 \mod 21$ and is equal to $0 \mod 5$ and $\mod 11$

We compute the sum of those three number:

$x = 7 \times 210 + 1 \times 231 + 18 \times 55 \times (-8) = -6219$

We can "get this number back" in a more appropriate range:

$-6219 \equiv -6219 + (6\times 1155) \equiv 711 \mod 1155$

Everything we have done shows that $711$ is a solution to your initial system.

TL;DR: followed the method from the wikipedia page of the CRT

Title: Calculate all numbers $x ∈ ℤ$ that simultaneously satisfy the following 3 congruences


问:
Prove that the size of $G$ is at least $\frac{n² − 2n}{4}$
Let $G$ be a graph of order $n$ that has exactly two connected components, both of them
being complete graphs. Prove that the size of $G$ is at least $\frac{n² − 2n}{4}$.

答:
With two complete connected components and $n$ vertices, there are not many possibilities for $G$. In fact, $G$ is uniquely defined by the order of either component. Let $k$ be the order of either component, the other has order $n-k$.

For a fixed $k$, the size $m_k$ of $G$ is the sum of the sizes of each component which are known as each component is complete:
$$
m_k = \frac{k(k-1)}{2} + \frac{(n-k)(n-k-1)}{2}
$$

From there on, what is left to be done is showing that $\forall k \in \{0,\dots,n\}, m_k\geq \frac{n^2-2n}{4}$.


问:
prove following combinatorial identity
I have been trying to prove following thing. With no success

prove that for $n>k\geq 2$ $$\sum_{i=0}^k nP_{(k+i)} . \binom{k}{k-i} kP_{(k-i)} = (nP{k})^2$$ where $nPk = \frac{n!}{(n-k)!}$

any useful hint is helpful.

答:
$$\sum_{i=0}^k nP_{(k+i)} . \binom{k}{k-i} kP_{(k-i)} = (nP{k})^2$$

$$\sum_{i=0}^k \frac{n!}{(n-k-i)!} \frac{k!}{i!(k-i)!} \frac{k!}{i!} = (\frac{n!}{(n-k)!})^2$$

divide both sides by n! and k! and multiply both sides by (n-k)!.

$$\sum_{i=0}^k \frac{(n-k)!}{(i)!(n-k-i)!} \frac{k!}{i!(k-i)!}  = \frac{n!(n-k)!}{(n-k)!(n-k)!k!}$$

$$\sum_{i=0}^k \binom{n-k}{i} \binom{k}{k-i}  = \binom{n}{k}$$

In this form it is a special case of Vandermonde's identity.

Title: prove following combinatorial identity


答: Let the connected components $G_{1}$ and $G_{2}$ have $k$ and $n-k$ vertices respectively. Then the number of edges of $G$ is $E$ = ${k} \choose {2}$ + $n-k \choose 2$. Suppose $k$ is chosen such that $E$ is minimised, and suppose that $k>(n-k)+1$, that is $2k>n+1$. Then by moving a vertex from $G_{1}$ to $G_{2}$ and completing $G_{2}$ we lose $k-1-(n-k)=2k-n-1>(n+1)-n-1=0$ edges from $G$ which contradicts $E$ being minimum. If $k<(n-k)-1$, that is $2k<n-1$, then by moving a vertex from $G_{2}$ to $G_{1}$ we lose $(n-k-1)-(k)=n-2k-1>n-(n-1)-1>0$ edges from $G$ which again is a contradiction. Hence $k$ and $(n-k)$ differ by at most $1$. Hence $G$ has at least $\frac{\lfloor{n}\rfloor }{2} \choose 2$ + $\frac{\lceil{n}\rceil }{2} \choose 2$ edges. If $n$ is even, then this is $\frac{n}{2}.(\frac{n}{2}-1)$ as required. If $n$ is odd then this is $\frac{1}{2}(\frac{n-1}{2})(\frac{n-3}{2})+\frac{1}{2}(\frac{n+1}{2})(\frac{n-1}{2})=\frac{1}{2}(\frac{n-1}{2})(n-1) = \frac{n}{2}(\frac{n}{2}-1)-\frac{1}{4}(n-1)+\frac{n-1}{8}<\frac{n}{2}.(\frac{n}{2}-1)$

Body:
Let $G$ be a graph of order $n$ that has exactly two connected components, both of them
being complete graphs. Prove that the size of $G$ is at least $\frac{n² − 2n}{4}$.


答:
Maybe it is simpler to prove in this form:
$$\sum_{i=0}^k \binom{n}{i,i,k-i,n-k-i} = \binom{n}{k}^2$$

Title: prove following combinatorial identity

Body:
I have been trying to prove following thing. With no success

prove that for $n>k\geq 2$ $$\sum_{i=0}^k nP_{(k+i)} . \binom{k}{k-i} kP_{(k-i)} = (nP{k})^2$$ where $nPk = \frac{n!}{(n-k)!}$

any useful hint is helpful.


问:
Find a number that minimizes distance to a vector of sets of numbers

Assumptions

$V$ is a vector of sets $V_1,V_2,...,V_n$ of numbers:

$V=[V_1, V_2,..., V_n]^T, \forall_{i=1..n}V_i\subset\mathbb{R}$

$c\in\mathbb{R}$ is constant

$d(V,c)$ is an error metric: $d(V,c)=\sqrt[p]{\sum_{i=1}^{n}\min_{v\in{V_i}}(|v-c|)^p}$

$p\in\mathbb{N}\backslash\{0\}$

$c^*=\arg\min_c{d(V,c)}$

Interpretation

$V$ is a template for vector of numbers $W$, where $V_i$ sets represent acceptable values for corresponding $i$ positions in the vector $W$. E.g. for $V=[\{1,2\},\{2,3\}]^T$, possible $W$ values are $[1,2]^T,[1,3]^T,[2,2]^T,[2,3]^T$.

$c$ is constant (scalar), that substituting all elements of $W$, e.g. $W=C=[c,c]^T$ causes an error $d(V,c)$ for template $V$. Note that it is not required to $c$ be an element of set $\bigcup_i{V_i}$ (it may be any real number).

The error caused by $c$ is an minimal Minkowski distance over all possible values of $W$.

We look for $c$, such that the error caused by it is the smallest. We denote such $c$ as $c^*$.

The question

Is there a general algorithm that finds optimal $\mathbb{c}$ in polynomial time? I know that, there is an exhaustive search algorithm, but its complexity is not acceptable complexity.

Solution for $p=1$

The problem seems to be easy for $p=1$. In such case error metric transforms to $\sum_{i=1}^n{\min_{v\in{V_i}}|v-c|}$. Note that sum of linear functions is an linear function, so:

$\forall_{i=1..n}$ sort $V_i$

Let $X_i=V_i\cup{V_i'}$, where $V_i'$ is set of averages of consecutive values in $V_i$. E.g. for $V_i=\{1,2,3\}$, $V_i'=\{1.5,2.5\}$. In other words $X_i$ is a set of points, where segmentally linear error function $d(V,c)$ breaks.

$X=\bigcup_{i=1..n}{X_i}$

$c^*=\arg\min_{c\in{X}}d(V,c)$ (Since $d(V,c)$ is segmentally linear, the minimum may be located only in point, where function breaks).

So, $O(|X|)=O(\bigcup_{i=1..n}V_i)$, assuming $V_i$s are sorted, the entire algorithm works in $O(n\log{n})$, where $n=\bigcup_{i=1..n}V_i$.

Unfortunately the above algorithms does not work for Euclidean distance, or other Minkowski distances, since the breaking points of $d(V,c)$ cannot be obtained in the above way.

答:
Actually, almost the same approach can be used for $p>1$. The only exception is that the set $X$ should be treated as the set of endpoints of consecutive intervals and you'll need to find the minimum within each interval instead of knowing it has to occur at one of the endpoints (as is the case with $p=1$).

Since the $p$-th root in the definition of the metric does not play any role when it comes to optimality, it's sufficient to consider the sum itself. Within each interval, the sum simplifies to a polynomial of degree $p$. Finding the minimum value of the polynomial within an interval can be done explicitly if $p$ is small enough (trivial for $p=2$, quadratic formula for $p=3$, possibly going up to $p=5$ with Cardano and Ferrari), but for higher values of $p$, you might have to resort to approximate, numeric solutions (bisection, Newton-Raphson, ...).


问:
$E[X|W] = 0 \implies E[XW^{T}]=0$?
Let $X \in \mathbb{R}^{m}$ and $W \in \mathbb{R}^{p}$ random vectors. Does $E[X|W] = 0 \implies E[XW^{T}]=0$?

答:
Hint

$$\mathbb E[XW^T]=\mathbb E\big[\mathbb E[XW^T\mid W]\big].$$

Title: $E[X|W] = 0 \implies E[XW^{T}]=0$?


答:
With the help of some slack variables $s_k$ we transform the inequalities into equations and then with $f(x,y) = a\sqrt x+b\sqrt y$ forming the lagrangian

$$
L = f(x,y)+\lambda_1(x-s_1^2)+\lambda_2(y-s_2^2)+\lambda_3(x-2+2y+s_3^2)+\lambda_4(y-1+s_4^2)
$$

The lagrangian stationary points are the solutions for

$$
\nabla L = 0 = \left\{
\begin{array}{l}
 \frac{a}{2 \sqrt{x}}+\lambda_1+\lambda_3 \\
 \frac{b}{2 \sqrt{y}}+\lambda_2+2 \lambda_3+\lambda_4 \\
 x-s_1^2 \\
 y-s_2^2 \\
 s_3^2+x+2 y-2 \\
 s_4^2+y-1 \\
 -2 s_1 \lambda_1 \\
 -2 s_2 \lambda_2 \\
 2 s_3 \lambda_3 \\
 2 s_4 \lambda_4 \\
\end{array}
\right.
$$

giving the solution

$$
\left[
\begin{array}{ccccccc}
f(x,y)& x & y & s_1^2 & s_2^2 & s_3^2 & s_4^2\\
 \sqrt{2a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2 a^2+b^2} & \frac{4 a^2}{2 a^2+b^2} & \frac{b^2}{2
   a^2+b^2} & 0 & \frac{2 a^2}{2 a^2+b^2} \\
\end{array}
\right]
$$

NOTE

Null $s_k$'s indicates that the corresponding constraint is active.

Title: How to maximize a function when you cannot solve gradient = 0

Body:
I have to maximize this function : $$ f(x,y) = a\sqrt{x} + b\sqrt{y}$$
$$ a, b \in{R+*} $$
Knowing that $$ 0 ≤ x ≤ 2 − 2y $$ with $$ 0 ≤ y ≤ 1 $$

I said that f is a linear combination of 2 concave functions so it has a maximum (for 0<=x<=2-2y and 0<=y<=1). But because a,b are positive real numbers, I cannot solve $$ \frac{a}{2\sqrt{x}} = 0 $$
And it's the same for y.

How do I deal with such a situation. Thank you very much !!


问:
Is the ring $\frac{\Bbb{C}[X,Y,Z]}{(XY-Z)}$ a UFD?
I think the ring $\frac{\Bbb{C}[X,Y,Z]}{(XY-Z)}$ is a UFD.

If we consider the evaluation map $f: \Bbb{C}[X,Y,Z] \rightarrow \Bbb{C}[t^2,t]$ such $f(X)=f(Y)=t$ and $f(Z)=t^2$. Then clearly $(Z-XY)$ is in $\ker f$. Now suppose $p \in \ker f$. Then considering $p$ as a polynomial is $\Bbb{C}[X,Y][Z]$, we have by division algorithm $p=q(Z-XY)+r$, where $q,r \in \Bbb{C}[X,Y][Z] $. If $r \neq 0$, then $\deg r < 1$ in $\Bbb{C}[X,Y][Z]$. So $r \in \Bbb{C}[X,Y]$. Also $f(r)=0$. Now if $r= \displaystyle{\sum_{m,n}c_{(m,n)}X^mY^n}$, then $f(r)=0 \Rightarrow r(t,t)=0 \Rightarrow \displaystyle{\sum_{m,n}c_{(m,n)}t^{m+n}}=0 \Rightarrow c_{(m,n)}=0 \,\forall \, m,n$. So $r=0$. Hence we have $\frac{\Bbb{C}[X,Y,Z]}{(XY-Z)}\approx \Bbb{C}[t^2,t]$.

Now is $\Bbb{C}[t^2,t]$ a UFD?

If so then we are done !

答:
The kernel of your proposed map $\mathbb C[X,Y,Z]\to\mathbb C[t]$ with $X, Y\mapsto t$ and $Z\mapsto t^2$ is not $(Z-XY)$ but $(Z-XY,X-Y)$. However, note that dividing by $(Z-XY)$ just imposes the relation $\overline Z=\overline X\,\overline Y$ in the quotient, which effectively let's us get rid of $Z$:

Consider the ring homomorphism $g\colon \mathbb C[X,Y,Z]\to\mathbb C[X,Y]$ given by $X\mapsto X$, $Y\mapsto Y$ and $Z\mapsto XY$. This is a surjective ring homomorphism with kernel $(Z-XY)$ and hence
$$
\mathbb C[X,Y,Z]/(Z-XY) \cong \mathbb C[X,Y].
$$

For any UFD $R$, the polynomial ring $R[X]$ is a UFD as well. Since $\mathbb C$ is a UFD, so is $\mathbb C[X]$ and hence $\mathbb C[X,Y]\cong \mathbb C[X][Y]$ is a UFD as well.


问:
property of complementary cumulative distribution function
$$\Bbb{E}(X) = \int_0^\infty xf(x)dx \ge \int_0^c xf(x)dx + c \int_c^\infty f(x)dx$$

I'm having trouble understanding the above formula.

I do understand that $ \Bbb{E}(X) = \int_0^\infty xf(x)dx \ge \int_0^c xf(x)dx$ but I don't yet get where $c \int_c^\infty f(x)dx$ comes from , how it relates to it all..

Please explain in mathnoob-friendly terms.

答:
We have by aditivity of the integral in the domain, that
$$ \int_0^\infty xf(x)\, dx = \int_0^c xf(x)\, dx + \int_c^\infty xf(x)\, dx $$
Now, look at the second integral: For each $x \in [c,\infty)$, we have as $f(x) \ge 0$ ($f$ being a pdf) that $xf(x) \ge cf(x)$. Integrating this over $[c,\infty)$ gives
$$ \int_c^\infty xf(x)\, dx \ge \int_c^\infty cf(x)\, dx = c\int_c^\infty f(x)\, dx. $$
So, alltogether
$$ \int_0^\infty xf(x)\, dx \ge \int_0^c xf(x)\, dx + c\int_c^\infty f(x)\, dx $$
as wanted.</p>


问:
Quick Second Derivatives with Parametric Equations Question
So I was given the following prompt:

"A curve is defined by the parametric equations $x(t)=e^{-3t}$ and $y(t)=e^{3t}$. What is $\frac{d^2y}{dx^2}$ in terms of $t$?"

So I guess I'm a bit confused about what the answer might look like here. I understand that to find the second derivative here, I have to divide the first derivative of the derivative of the first equation by the derivative of the $x$ equation here, but I'm confused about where that would put me for an answer. Any help would be appreciated!

答:
In order to find the second derivative of a parametric equation follow this logic:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ since both $x$ and $y$ are functions of $t$ and then,
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{\frac{dx}{dt}}$$

Using the parametric equations above you should find that
$$\frac{dy}{dx} = \frac{3e^{3t}}{-3e^{-3t}}$$

and thus
\begin{align}
\frac{d^2 y}{d x^2} &= \frac{d}{dt}(-e^{6t})\frac{1}{-3 e^{-3t}} \\
&= 2e^{9t}
\end{align}


问:
Value of sum $\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ .
$\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ is equal to

$1.$ $1$.

$2.$ $x$.

$3.$ $x^2$

$4.$ $x^n$.

According to me $B_n(x)=\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^k$  form polynomials called Bernstein polynomial tends to continuous function $x$ because $f(\frac{k}{n})=\frac{k}{n}$. So limit of $B_n(x)$ will tend to $x$, but in question it is asked without limit. So what will be correct  answer and why ?.  Thank you for solution.

Body:
I'm assuming you meant $\sum_{k=0}^{n}\dbinom{n}{k}\dfrac{k}{n}x^{k}(1-x)^{n-k} $

Note that $k\dbinom{n}{k} = n\dbinom{n-1}{k-1} \implies \dfrac{k}{n}\dbinom{n}{k} = \dbinom{n-1}{k-1}$
$$ B_n(x)=\sum_{k=0}^{n}\dbinom{n}{k}\frac{k}{n}x^{k}(1-x)^{n-k} $$
$$ = x \sum_{k=1}^{n} \dbinom{n-1}{k-1} x^{k-1}(1-x)^{n-k} $$
$$ = x(x+1-x)^{n-1}$$
$$ =x $$


问:
Find function with a degree of 4 from three statements
If $f(x)$ is a function/polynomial with a degree of 4 and:
$$f(-1)=1$$
$$f(-2)=4$$
$$f(-3)=9$$
How can I find $f(x)$?

From statement 1, I know $f(x) = (x+1)(...)+1$

With statement 2, I know $f(x) = (x+1)[(x+2)(...)-3]+1$

We can use a similar way to find $f(x)$ with statement 3, but don't we end up with a cubic polynomial?

If my calculations are correct, how can we turn the cubic into a 4 degree polynomial?

答:
This is a plain application of polynomial interpolation.

Three statements make it possible to uniquely define a polynomial of degree $2$. You can still find all the polynomial of degree $n\geq 2$ that satisfy those statements (using the Lagrange basis for example).


答:
You can see this :

$$ \sum_{k=0}^n { n \choose k}\dfrac{k}{n}x^k(1-x)^{n-k}= \mathbb{E}(X/n) $$  where $X$ follows a Bernoulli of parameter $x$.

It follows from here that :

$$ \sum_{k=0}^n { n \choose k}\dfrac{k}{n}x^k(1-x)^{n-k}=nx/n=x $$

Title: Value of sum $\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ .

Body:
$\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^{n-k}$ is equal to

$1.$ $1$.

$2.$ $x$.

$3.$ $x^2$

$4.$ $x^n$.

According to me $B_n(x)=\sum_{k=0}^{n}n_{C_k}\frac{k}{n}x^k(1-x)^k$  form polynomials called Bernstein polynomial tends to continuous function $x$ because $f(\frac{k}{n})=\frac{k}{n}$. So limit of $B_n(x)$ will tend to $x$, but in question it is asked without limit. So what will be correct  answer and why ?.  Thank you for solution.


问:
Integrals involving irrational numbers

$${1 \over 2\pi} \int_{0}^{2\pi} \sin^{100}(x) dx$$

How should I approach getting an estimate on the numeric value in this case?

Title: Integrals involving irrational numbers

Body:
Integrate by parts successively to reduce the integral

$$I_{100}= \int_{0}^{2\pi} \sin^{100} x dx
=-\frac1{100}\int_{0}^{2\pi} \tan^{99} x \>d(\cos^{100}x)\\=\frac{99}{100}I_{98}  =\frac{99}{100} \frac{97}{98} I_{96}= \cdots =\frac{99}{100} \frac{97}{98} \cdots \frac12\cdot2\pi
$$


答:
Hint: Compute $\displaystyle \int_0^\infty t^n e^{-t}\,\mathrm dt$ by induction on $n$ -- you should be able to discover a nice pattern, which allows you to plug in the value $n = 50$.

After you do that, also check out the Gamma function.

I trust you can complete the proof that the resulting quantity isn't a perfect square.

Title: Help me prove $\left(\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$ isn't a perfect square

Body:
I need some help in proving

$$\left( \displaystyle\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$$

isn't a perfect square. The only way I can think is repeated integration by parts which is obviously impractical and I still likely wouldn't be able to deduce if the result was a perfect square.

Thanks very much guys! :D


问:
Applications of the concept of a net in mathematics
This Wikipedia page gives a description of the concept of a net.

Here, I am going to ask a simple question: what topics in mathematics use nets frequently, and what are some "big" theorems proven by arguments about nets?

After some search with the word "net", I have found a lot of description of its "basic" properties, but I find it difficult to discover some more "advanced" or "developed" applications of this concept.

The word "net" means a lot of different things, which makes this search extremely difficult. Could anyone provide some reference?

答:
Nets are a tool that can be handy in proving some elementary facts about some topological spaces. They “look like” sequences, and so they seem to be most commonly used in Analysis areas. You get that continuity is equivalent to preserving net convergence, compactness can be characterised by cluster points of nets (like sequential compactness in metric spaces), it gives a didactically attractive way to prove Tychonoff’s theorem etc. (If a teacher doesn’t want to go into filters, e.g.).

They’re not studied as much as a construct in their own right, just as an alternative to filters to study general convergence (filters are much more popular, BTW, especially in areas touching more on set theory; more general and an internal tool in set theory as well (forcing etc, ultraproducts in logic etc). They occur in some proofs, but can often be avoided (but some proofs look natural done in terms of nets). There are hardly any deep theorems or hard results that require them. All in all quite a marginal topic IMO.

Schuchter (handbook of foundations of analysis or some such title) has a good modern treatment, Kelley’s general topology is the classic introduction. Engelking’s general topology does a decent job too. In Munkres they’re just a “footnote” exercise.


问:
Rotate a line on a graph relative to line's center
I am trying to figure out how to rotate a line relative to it's center.

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ and the center of the line $(x_3, y_3)$, how do I rotate the the line 180 degrees?

For example if the line was $(1,1)$ and $(2,2)$, the calculated center will be $(1.5,1.5)$ and the rotated point will be $(1,2)$ and $(2,1)$.

I am not sure which formula to use to rotate the line by its center, the ones I found rotate the lines based on axes.

答:
As comments have indicated, your query details rotating a line (or line segment) $90^\circ$ rather than $180^\circ$.  This is because the slope of the original line (or line segment) is $1$, and the slope of the new line is $-1$.

In order to determine how to rotate the line around the center $(x_0, y_0)$, where $(x,y)$ is any generic point on the line to be rotated, you first have to decide what the slope of the old and new lines are.  The following approach assumes that both slopes are non-zero and finite.  Exceptional but obvious adjustments to the algorithm will be needed if either the old or new lines are either horizontal or vertical.

The point $(x,y)$ is a distance $D = +\sqrt{(x - x_0)^2 + (y - y_0)^2}$ from the point $(x_0, y_0).$  Further, assuming that the slope of the original line is $m$, you have that $\frac{y - y_0}{x - x_0} = m$.

Assume that the new line will have a slope of $n$, and will also pass through the point $(x_0, y_0)$.  Then, you want to find the point $(x_1, y_1)$ that is on the circle of radius $D$ centered at $(x_0,y_0)$ such that $\frac{y-y_0}{x-x_0} = n.$

Two ideas to consider:

being on the circle of radius $D$ merely signifies that $(x_1, y_1)$ also satisfies the constraint that $D = +\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$.

there is a hidden ambiguity within this algorithm.  That is, if the point $P_1 = (x_0 + r, y_0 + s)$ satisfies the constraints, then so does the point $P_2 = (x_0 - r, y_0 - s)$.  In such a circumstance, you need to explicitly specify whether $(x,y)$ is rotated to $P_1$ or $P_2$.

Two plausible distinct options for generically resolving the hidden ambiguity are:

$(x_1 - x_0) > 0 \iff (x - x_0) > 0.$

$(y_1 - y_0) > 0 \iff (y - y_0) > 0.$


答:
Call

$$x = e^{-3t} = z$$
$$y = e^{3t} = \frac{1}{z}$$

Thence

$$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}(1/z)}{\text{d}z} = -\frac{1}{z^2}$$

Deriving again we get the result

$$\frac{\text{d}^2y}{\text{d}x^2} = \frac{\text{d}(-1/z^2)}{\text{d}z} = \frac{2}{z^3}$$

Substituting back and you get

$$\frac{2}{(e^{-3t})^3} = 2 e^{9t}$$

Whence

$$\frac{\text{d}^2y}{\text{d}x^2} = 2e^{9t}$$

Title: Quick Second Derivatives with Parametric Equations Question

Body:
So I was given the following prompt:

"A curve is defined by the parametric equations $x(t)=e^{-3t}$ and $y(t)=e^{3t}$. What is $\frac{d^2y}{dx^2}$ in terms of $t$?"

So I guess I'm a bit confused about what the answer might look like here. I understand that to find the second derivative here, I have to divide the first derivative of the derivative of the first equation by the derivative of the $x$ equation here, but I'm confused about where that would put me for an answer. Any help would be appreciated!


问:
Gap between composite numbers that aren't divisible by 2,3 or 5
It is easy to determine if a number is divisible by 2,3 or 5, so in my experiment i am concentrating on a consecutive list of composite numbers that are not divisible by 2,3 or 5.

My question is if there is a maximum gap limit between such numbers, and if so what is it?

Please review the image as an example:

答:
I am not sure how $42$ became the answer to every question in the Universe, but it is an answer to this question.  Given any composite number $N$ that is prime to $2×3×5=30$, the next larger composite number prime to $30$ must be no greater than $N+42$.  The bound is proven to be a sharp one, but requires a large value of $N$ to saturate it.

By hypothesis, $N\in\{1,7,11,13,17,19,23,29\}\bmod 30$.  Suppose $N\equiv 1\bmod 30$.  Then we examine the next several qualifying numbers, by subtracting $1$ from each allowed residue to obtain the difference with $N$.  Thus

$N+6\equiv7\bmod30$

$N+10\equiv11\bmod30$

$N+12\equiv13\bmod30$

$N+16\equiv17\bmod30$

$N+18\equiv19\bmod30$

$N+22\equiv23\bmod30$

$N+28\equiv29\bmod30$

Now watch what happens when we take residues $\bmod 7$:

$N+6\equiv N+\color{blue}{6}\bmod7$

$N+10\equiv N+\color{blue}{3}\bmod7$

$N+12\equiv N+\color{blue}{5}\bmod7$

$N+16\equiv N+\color{blue}{2}\bmod7$

$N+18\equiv N+\color{blue}{4}\bmod7$

$N+22\equiv N+\color{blue}{1}\bmod7$

$N+28\equiv N+\color{blue}{0}\bmod7$

Note that the increments shown cover all the different residues $\bmod 7$.  Perforce if $N$ is one greater than a multiple of $30$, then some larger number prime to $30$ but less than or equal to $N+28$ must be composite by being a multiple of $7$.  Also, since the incremental residues $1,2,3,4,5,6$ all occur earlier than $28$, that gap for $N\equiv1\bmod30$ can be achieved only if $N$ is a multiple of $7$.

What if we start with $N$ having a different residue $\bmod30$, such as $7$?  If $N$ has that residue, then the increments to larger numbers prime to $30$ will be different from those above, leading to a different pattern of residues $\bmod 7$.  Thus a different bound on the gap to a guaranteed multiple of $7$.  Put in numbers:

$N+4\equiv11\bmod30$&$N+4\bmod 7$

$N+6\equiv13\bmod30$&$N+6\bmod 7$

$N+10\equiv17\bmod30$&$N+3\bmod 7$

$N+12\equiv19\bmod30$&$N+5\bmod 7$

$N+16\equiv23\bmod30$&$N+2\bmod 7$

$N+22\equiv29\bmod30$&$N+1\bmod 7$

$N+24\equiv1\bmod30$&$N+3\bmod 7$

$N+30\equiv7\bmod30$&$N+2\bmod 7$

$N+34\equiv11\bmod30$&$N+6\bmod 7$

$N+36\equiv13\bmod30$&$N+1\bmod 7$

$N+40\equiv17\bmod30$&$N+5\bmod 7$

$N+42\equiv19\bmod30$&$N+0\bmod 7$

In this case we cover all incremental residues $\bmod 7$, and thus assure a multiple of $7$, at $N+42$ making $42$ the maximum possible gap in this case.  Here again the full gap requires $N$ to be a multiple of $7$.  Given all the "extra" iterations in this case we might suppose that a different divisor, such as $11$, might produce a lower limit, but we find that covering all eleven incremental residues $\bmod 11$ also requires allowing a maximum gap of $42$.

We do a similar analysis with $N\equiv11\bmod30,N\equiv13\bmod30$, etc, and in all cases a multiple of $7$ is forced on or before $N+42$.  A gap of $42$ requires $N$ to be both $\in\{7,11\}\bmod 30$ and a multiple of $7$.  So if every question in the Universe is ultimately governed by gaps between composite numbers having no factors of $2,3,$ or $5$, then $42$ is indeed a universal answer.

Probalistic considerations would seem to suggest that the largest gaps would occur with smaller numbers, but gradually the maximum observed gap increases beyond the value of $28$ in the OP's original list.  The smallest case that exceeds a gap of $28$ is $N=1273$, which gives a gap of $36$ up to $N+36=1309$.  The numbers $1273$ and $1309$ collectively contain the prime factors $7,11,17,19$ leaving relatively few factors that might have broken up an intermediate string of four-digit primes.  In the comments a gap of $38$ is reported between $113141$ and $113179$ by Joffan.  Finally, Empy2 in another answer gives a gap of $42$, saturating the bound, with $N=1418575498571$ along with larger values.


答:
In this answer, I will determine the necessary and sufficient condition for the polynomial degree to be at least $4.$

I considered your polynomial to be a quartic of at least degree $4.$ But, this is not enough.

Method

For the quartic polynomial, at least value of $f(-4)$ must be known. Let, $f(-4)=p$, then you have

$$f(x)=a(x+1)(x+2)(x+3)+x^2$$

$$\implies a×(-6)+16=p$$

$$\implies a=\dfrac {16-p}{6}$$

$$\implies f(x)=\left(\dfrac {16-p}{6}\right)(x+1)(x+2)(x+3)+x^2$$

But, this is still not enough. Because, we have a cubic polynomial. You still need value of $f(-5).$

Let,

$$f(-5)=\left(\dfrac {16-p}{6}\right)\times (-24)+25\Longrightarrow f(-5)=4p-39$$

So, if $f(-5)=4f(-4)-39$ you still have a cubic polynomial.

Let, $f(-4)=p$ and $f(-5)≠4f(-4)-39$, then our tricky method will not work. So, you need a general method.

$$\begin{cases} a-b+c-d+e=1 \\16a-8b+4c-2d+e=4\\ 81a-27b+9c-3d+e=9 \\ 256a-64b+16c-4d+e=f(-4)\\ 625a-125b+25c-5d+e=f(-5) \end{cases}$$

You can solve this linear system with Gauss method.

A necessary and sufficient condition:

At least the values of  $f(-4)$ and $f(-5)$ must be given and the following condition must be hold:

$$f(-5)-4f(-4)+39≠0.$$

Conclusion

To obtain a polynomial of degree of at least $4$, it is necessary to know at least the values of $f(-4)$ and $f (-5)$, where $f(-5)≠4f(-4)-39$ or $f(-5)-4f(-4)+39≠0.$

To adapt to your circumstances, I wrote $f (-4)$ and $f(-5)$ instead of $f(4)$ and $f(5).$

Let's test the necessary and sufficient condition.

Let,  $$ f(-1)=1 \\ f(-2)=4 \\ f(-3)=9\\ f(-4)=16 \\ f(-5)=25 $$

Then, we have,

$$f(5)-4f(-4)+39 \Longrightarrow 25-4 \times 16+39=0$$

As it seems, the necessary and sufficient condition are not provided.  So, we can find a lower degree polynomial. And we really have such a function:

For example, the function $f(x)=x^2$ is the simplest function that provides the values we give.

Remark

Here we have only examined the requirement to give at least $f(-4)$ and $f(-5)$ values.  However, it is also possible to write it for the cases $f(4)$ and $f(5)$, or $f(7)$ and $f(9)$. However, this can never invalidate the condition $f(-5)-4f(-4)+39≠0.$

Title: Find function with a degree of 4 from three statements

Body:
If $f(x)$ is a function/polynomial with a degree of 4 and:
$$f(-1)=1$$
$$f(-2)=4$$
$$f(-3)=9$$
How can I find $f(x)$?

From statement 1, I know $f(x) = (x+1)(...)+1$

With statement 2, I know $f(x) = (x+1)[(x+2)(...)-3]+1$

We can use a similar way to find $f(x)$ with statement 3, but don't we end up with a cubic polynomial?

If my calculations are correct, how can we turn the cubic into a 4 degree polynomial?


问:
Taylor series of $e^{-{1}/{x}}$
Does the integral converge?
\begin{align*}
\int_0^\infty (1-e^{-{1}/{x}})dx
\end{align*}

Idea: First I tried to expand the $e^{-{1}/{x}}$ using the Maclaurin series and evaluated the integral but it was not a good result and after I realized the result is not correct, because we cannot expand $e^{-{1}/{x}}$ by the Maclaurin series (because all derivatives of $e^{-{1}/{x}}$ at $x=0$ are zero but $e^{-{1}/{x}}$ is not a zero function so the approximate centered at $0$ is not work)

We can find the Taylor series expansion of $e^{-{1}/{x}}$ centered at some point (like $x=1$). According to that, the integral is divergent. I just want to know

Is my understanding correct? and

Can we expand $e^{-{1}/{x}}$ by the Taylor series centered at some point (except zero) even the function is non-analytic?

and is this the right way to evaluate this kind of integral? It will be a great pleasure to me if someone can give a little explanation.

答: You integral diverges because for $x$ high enough $$1-e^{-1/x} \sim 1/x$$ .

Title: Taylor series of $e^{-{1}/{x}}$


答: As EDX hoped to write: $1-e^{-1/x} \sim 1/x$, so the integral is divergent.

Title: Taylor series of $e^{-{1}/{x}}$

Body:
Does the integral converge?
\begin{align*}
\int_0^\infty (1-e^{-{1}/{x}})dx
\end{align*}

Idea: First I tried to expand the $e^{-{1}/{x}}$ using the Maclaurin series and evaluated the integral but it was not a good result and after I realized the result is not correct, because we cannot expand $e^{-{1}/{x}}$ by the Maclaurin series (because all derivatives of $e^{-{1}/{x}}$ at $x=0$ are zero but $e^{-{1}/{x}}$ is not a zero function so the approximate centered at $0$ is not work)

We can find the Taylor series expansion of $e^{-{1}/{x}}$ centered at some point (like $x=1$). According to that, the integral is divergent. I just want to know

Is my understanding correct? and

Can we expand $e^{-{1}/{x}}$ by the Taylor series centered at some point (except zero) even the function is non-analytic?

and is this the right way to evaluate this kind of integral? It will be a great pleasure to me if someone can give a little explanation.


问:
Is this correct regarding subtraction and addition of numbers or functions
Let us say we have

$$2=2$$

Then I would do is $$2-2=0$$i.e. I shifted $2$ from RHS to LHS. Now, my textbook says that this is not true and is a wrong concept. What my textbook tells me is this:

$$ 2-2=2-2$$

This means that when two equalities are equal , we cannot shift one from RHS to LHS but can subtract the same amount on both sides.

$$f(x) = g(x) \implies f(x)-g(x) = g(x)-g(x)$$

So, I wanted to confirm if this is how we really do it or both the methods are correct?

答:
It is correct, and is more commonly known as transposition.

Well, in most textbooks, it is introduced as:

Adding, subtracting, multiplying or dividing (not by zero) on both sides doesn't affect its truth. So, $a=b$ is equivalent to
$$a-b=b-b\implies a-b=0$$

Then, the next section tells that

For the sake of convenience, we do the same in one-step by saying that we moved $b$ from RHS to LHS, and changed its sign as we moved it. This is called transposition.

So yes, maybe the book will introduce it later, or probably not. But it is correct. In higher classes, you wouldn't be spending your time and space into writing that "extra line". At this stage, if it isn't introduced to you, nonetheless, I suggest to not use it.

Hope this helps. Ask anything if not clear :)


答:
$$(x-x_{1})(y_{2}-y_{1})= (y-y_{1})(x_{2}-x_{1})$$

Is the equation for a line between $(x_{1},y_{1})$ and $(x_{2},y_{2})$

The equation:

$$(x+x_{3}-x_{1})(y_{2}-y_{1})= (y+y_{3}-y_{1})(x_{2}-x_{1})$$

Shifts the line so that $(x_{3},y_{3})$ is on the origin.

An angle which describes the line is given by the arctan(m) where m is the slope. In the case where the line is vertical choose $\frac{\pi}{2}$. Then we rotate by $\theta$ so the new slope is:

$$tan(\theta + arctan(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}))$$

So we know the new slope, and the point $(x_{3},y_{3})$ must be in the new line so:

$$(x-x_{3})tan(\theta + arctan(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}))=y-y_{3}$$

Using point slope form.

Title: Rotate a line on a graph relative to line's center

Body:
I am trying to figure out how to rotate a line relative to it's center.

Given two points $(x_1, y_1)$ and $(x_2, y_2)$ and the center of the line $(x_3, y_3)$, how do I rotate the the line 180 degrees?

For example if the line was $(1,1)$ and $(2,2)$, the calculated center will be $(1.5,1.5)$ and the rotated point will be $(1,2)$ and $(2,1)$.

I am not sure which formula to use to rotate the line by its center, the ones I found rotate the lines based on axes.


答:
We know that $f(x)-x^2$ has roots at $-1,-2,-3$. So, $$(x+1)(x+2)(x+3)\mid f(x)-x^2$$

What does this tell us? There's exactly one polynomial that has degree $<3$ that satisfies the conditions, which is $x^2$. There's exactly one cubic polynomial, up to constant multiplication, that satisfies the conditions as well, notably $c(x+1)(x+2)(x+3)+x^2$ (for any constant $c$).

All we know is that any $f$ of degree $4$ must be of the form $$c(x+a)(x+1)(x+2)(x+3)+x^2$$for some constants $a,c$.


问:
A projectile has been shot into the atmosphere. Find the polynomial that interpolates the speed and acceleration in function of time.
This is the table (tempo = time, velocidade = speed, aceleracao = acceleration)

I used Newton's interpolation to get the polynomial (5th degree).

I got
$$102 + 138(x-15) + (-124/15)(x-15)^2 + (289/225)(x-15)^2(x-30) + (-0.0689)(x-15)^2(x-30)^2 + (0.0048)(x-15)^2(x-30)^2(x-45)$$

I tested this in my calculator and the values for p(15) and p(30) are correct but p(45) is completely wrong.

Then I got the derivative to test the derivatives
$$14341.5 - 2226.13 x + 122.896 x^2 - 2.8676 x^3 + 0.024 x^4$$

p'(15) and p'(30) are correct, albeit with a slight error, but p'(45) is very off (I got 132 point something).

Is this normal? Is it possible to get a better estimate for this?

My calculations:

答:
Given the 6 degrees of freedom and the six constraints, the interpolation can be exact, if it is possible at all (which is always a possibility in practice, but unlikely for a homework problem). Any small inaccuracies you see are the result of either errors in your work or the fact that you're calculating with decimal approximations. I'm not sure where your -0.0689 and and 0.0048 terms came from. You should double-check your computations using precise fractions -- note that since all the constraints are integral, all your powers of 15 in denominators should cancel out.

If it turns out there is still an issue after checking your computations, I'll suggest an alternate cross-check method. Letting the 6 coefficients of a fifth degree polynomial be unknowns, the 6 constraints you have from a system of linear equations in those coefficients. Converting this into a matrix equation, you should be able to solve it through Gaussian elimination. You can make life easier for yourself by also having the polynomial be a function of not $t$ but $z = (t-15)/15$, so that all the constraints are at $z = 0$, $1$, and $2$.


答:
If $f$ is a polynomial of degree $4$, then

$$f(x)=b(x+1)(x+2)(x+3)(x-c)+x^2,$$

where $b,c \in \mathbb R$ are arbitrary.

But if $f$ is a polynomial of degree $2$, then

$$f(x)=x^2$$

is uniquely detemined by the above conditions.


问:
Two questions about the infinitude of prime Euclid numbers: $E_{n}=p_{1}\cdot p_{2}\cdot \ldots \cdot p_{n} + 1$
First of all I was wondering what results there are out there about the prime Euclid numbers? I would like to read up on any results that are out there.

Secondly I wanted someone to look at the following argument about the infinitude of the prime Euclid numbers.

According to Dirichlets theorem: If $a,b\in \mathbb{Z^{+}}$ such that $(a,b)=1$, then there are infinitely many prime numbers of the form $am+b$ where $m$ is a non-negative integer. Now, if we denote the $n$th primorial (i.e the product of the first $n$ primes) as $\displaystyle \prod_{i=1}^{n} p_{i}$ we define the Euclid numbers as $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1.$$ Note that $$E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}+1$$ where the first term $$2\cdot {\displaystyle \prod_{i=2}^{n} p_{i}}\equiv 2\pmod 4.$$ Thus $$E_{n}\equiv 3\pmod 4.$$ Meaning in particular that $$\boxed{E_{n}={\displaystyle \prod_{i=1}^{n} p_{i}}+1=4m+3}$$ for some non-negative integer $m$.

Since $(4,3)=1$, then by Dirichlets theorem we have that $4m+3$, where $m$ is a non-negative integer is a prime number infinitely often. And since the $RHS$ of the boxed equation above has this form, then does this imply that whatever is on the $LHS$ must also be a prime number infinitely often? In particular, does this mean that $E_{n}$ is prime infinitely often?

答:
I will try to elaborate what the comments are trying to say with a little bit more formality.

It is true that there are infinite primes of the form $4m + 3$. But you should note that you will get this infinitude of primes if you run $m$ all over the set of natural numbers. To see this let's say, take $t \in \{2n^3 - 1| n \in \mathbb{N}\} \subset \mathbb{N}$. You will  see that  you will never get a prime of the form $4t + 3$ except $7$. My point is that you can't guarantee that there will be infinite list primes of the form $4m+3$ if you run $m$ through arbitrary subset of $\mathbb{N}$.

Now, let's go back to the boxed equation and see the set through which you are running $m$:
$$E_n = 2\prod_{i = 2}^n p_i + 1 = 4m_n + 3$$
$$2\prod_{i=2}^n p_i - 2 = 4m_n$$
$$ \prod_{i=2}^n p_i = 2m_n + 1$$
This means if $m_n$ runs through $\mathbb{N},$ then $2m_n+ 1$ should run through all odd numbers greater than $2$(or $0 $ if you include $m_1 = 0$). But the product doesn't run through $\{2k + 1 | k \in \mathbb{N},\ \mu(2k+1) = 0\}$. This implies $m_n$ runs through a set that is considerably "smaller" (not smaller per se, but you get the idea) than $\mathbb{N}$. This shows the boxed argument is not sufficient to imply the infinitude of Euclid primes.


答:
For any $c>0$
$$
\int_0^\infty xf(x)\,\mathrm dx=\int_0^cxf(x)\,\mathrm dx+\int_c^\infty xf(x)\,\mathrm dx
$$
and
$$
\int_c^\infty xf(x)\,\mathrm dx\geq \int_c^\infty cf(x)\,\mathrm dx
$$
since $0\leq cf(x)\leq xf(x)$ for all $x\in [c,\infty)$.</p>

Title: property of complementary cumulative distribution function

Body:
$$\Bbb{E}(X) = \int_0^\infty xf(x)dx \ge \int_0^c xf(x)dx + c \int_c^\infty f(x)dx$$

I'm having trouble understanding the above formula.

I do understand that $ \Bbb{E}(X) = \int_0^\infty xf(x)dx \ge \int_0^c xf(x)dx$ but I don't yet get where $c \int_c^\infty f(x)dx$ comes from , how it relates to it all..

Please explain in mathnoob-friendly terms.


答:
Since we have $x\in[0,\infty)$ we can see that:
$$x\to0^+,\frac1x\to+\infty\therefore e^{-1/x}\to0$$
so for small values of $x$, $1-e^{-1/x}\approx 1$. Now if we look at large values of $x$:
$$x\to+\infty,\frac1x\to0^+\therefore e^{-1/x}\to1$$
and so for large values of $x$, $1-e^{-1/x}\to0$. However,
$$\int_1^\infty\frac{1}{x^n}dx$$
only converges for $n>1$ so for your integral to converge you need to show that your function tends to zero faster than $1/x$, which it does not so the integral will diverge

Title: Taylor series of $e^{-{1}/{x}}$


问:
complex numbers - dealing with the square root of i
I have the following quadratic equation:

$$(1-i)x^2 +ix - \frac{1}{4}i = 0$$

when I tried to solve for $x$ and get the complex solutions I got $ \sqrt{i}$ in the determinant.

I know that

$$ i^2 = -1 \implies \sqrt{i} = (-1) ^{\frac{1}{4}}$$
but don't come any further with it.

Thanks in advance for your help

答: Note that $i=e^{\pi i/2}$. Therefore, the square roots of $i$ are $\pm e^{\pi i/4}$; in other words, they are$$\pm\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right).$$You can reach the same conclusion solving the system$$\left\{\begin{array}{l}a^2-b^2=0\\2ab=1\end{array}\right.$$and taking $a+bi$.


问:
if $R^m \cong R^n$, then $m=n$?

Suppose $R$ is a commutative ring with $1$. Further we have an isomorphism $R^m \cong R^n$. Then does that imply $m=n$?

If I consider $R=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \cdots $, then I think $R^m=R^n$ as both $R^m$ and $R^n$ has countably infinite copies of $\mathbb{Z}$. Hence $R^m \cong R^n$ for any natural numbers $m \neq n$?

答:
Since you've mentioned Atiyah's "Intoduction to Commutative Algebra", chapter 2 in the comments I will assume we talk about module isomorphisms here (because that's what the chapter is about).

Note that it may happen that $R^n$ is ring isomorphic to $R^m$ for $n\neq m$. Actually the example in the question works just fine (the bijection later in this answer is a ring homomorphisms, but not a module homomorphism).

Suppose $R$ is a commutative ring with $1$. Further if we have the isomorphism given by $R^m \cong R^n$, then does that imply $m=n$?

This is answered here: R is a commutative ring with $1\ne 0$ and $R^m \cong R^n$, then $m=n$

If I consider $R=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \dots $, then I think $R^m=R^n$ as both $R^m$ and $R^n$ has countably infinite copies of $\mathbb{Z}$. Hence $R^m \cong R^n$ for any natural numbers  $m \neq n$?

I could simply answer this question with "$R^m$ is not module isomorphic to $R^n$ for $n\neq m$ because the first part says so".

But lets have a closer look at this case. Is $R^m$ really isomorphic to $R^n$ as modules when $n\neq m$? Consider the simple case $n=1$, $m=2$. So what would $f:R^2\to R$ isomorphism look like? Well, one idea is to take a pair of sequences $(r,t)\in R^2$ and combine them into one sequence by mixing $(r_0,t_0,r_1,t_1,r_2,t_2,\ldots)$. Formally with a formula like this:

$$f(r,t)_n=\begin{cases}
r_{n/2} &\text{for even } n \\
t_{(n-1)/2}&\text{otherwise}
\end{cases}$$

Clearly $f$ is a bijection. The standard one when proving that a countable set $X$ is equinumerous with $X^2$. But is it a module homomorphism? Well, remember that the underlying ring is $\mathbb{Z}\times\mathbb{Z}\times\cdots$ itself. And so given a sequence $s\in \mathbb{Z}\times\mathbb{Z}\times\cdots$ is it true that $f(sr,st)=sf(r,t)$? I think you can easily verify that it is false, for example by considering $s_n=r_n=t_n=n$.

Anyway, no matter what function you come up with, as long as it is a bijection $R^2\to R$ it cannot be a module homomorphism (due to our general result).

Title: if $R^m \cong R^n$, then $m=n$?


问:
Showing that a linear transformation is an isomorphism.
I have been given the following question:

For part (a) I simply showed the transformation satisfies the two conditions:

$T(ax) = aT(x)$

and

$T(x+y) = T(x) + T(y)$

For part (b) I evaluated T for each value of $\beta$ and wrote this as a linear combination of the values in $\gamma$, with the coefficients of the combination making up the columns of the matrix.

I am now trying to do part (c), but I have no idea how I would do this. Any help would be greatly appreciated.

答: Now you take that matrix that you got as an answer to (b) and you compute its determinant. It's not $0$, right?! So, $T$ is an isomorphism.


问:
Help me prove $\left(\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$ isn't a perfect square
I need some help in proving

$$\left( \displaystyle\int_{0}^{\infty} t^{50} e^{-t} \,\mathrm dt \right)^{1/2}$$

isn't a perfect square. The only way I can think is repeated integration by parts which is obviously impractical and I still likely wouldn't be able to deduce if the result was a perfect square.

Thanks very much guys! :D

答:
Consider $f(t) = \displaystyle-\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t}$, then $$f^\prime(t) = -\left[\sum_{i=1}^{50} \frac{50!(-t)^{n-1}}{(n-1)!}\right]e^{-t} +\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t} = t^{50}e^{-t}.$$

Thus, $\displaystyle\int_0^\infty t^{50}e^{-t}\,\mathrm{d}t = \left.-\left[\sum_{n=0}^{50} \frac{50!(-t)^n}{n!}\right] e^{-t}\right|_0^\infty = 50!$.

Since $50!$ only contains one factor of the prime $47$, it is not a perfect square. Hence $\displaystyle\left(\int_0^\infty t^{50}e^{-t}\,\mathrm{d}t\right)^\frac{1}{2}$ is not even an integer.


问:
Proportional Percentage Scaling
I've been breaking my head in trying to calculate the proportional scaling of percentages, in which the end result should always quantify to 100%.

For example:

```<code>Person A - 60%
Person B - 40%
</code>```

When adding person C with 50%, the following scaling occurs, in proportion to their original percentage of what's left:

```<code>Person A: 30%
Person B: 20%
Person C: 50%
</code>```

When amending an entry; for example, setting Person A to 10%, the following occurs:

```<code>Person A: 10%
Person B: 30%
Person C: 60%
</code>```

I was able to calculate this off the top of my head (strange - that never works), however I'm uncertain as to how to apply this in mathematical formula, and in its simplest form; in such a way that the calculation works in all scenarios, whether adding, amending or removing persons and their percentages.

At every calculation, I have existing reference to the following:

How many items (persons) there currently are

The original percentage    allocated to that person

Your help would be greatly appreciated.

Edit

I'm not sure how to explain it in a better form, but I hope the following tables may explain a bit better:

Where ? are the values to be calculated

The closest I've come so far:

```<code>var beneficiaries = [
    100
]; // Current state of beneficiaries

const calculateBeneficiaryRatio = (action, value = 0, index = 0) => {
    let results = [];
  let cur_percentage, new_percentage;

    switch(action) {
    case "add":
            for(let i in beneficiaries) {
          cur_percentage = beneficiaries[i];
          new_percentage = (100 - value) * (cur_percentage / 100);
          beneficiaries[i] = new_percentage;
        }
            beneficiaries.push(value);
        return beneficiaries;
        break;
    case "remove":
            value = beneficiaries[index];
        beneficiaries.splice(index, 1);
            for(let i in beneficiaries) {
          cur_percentage = beneficiaries[i];
          new_percentage = cur_percentage + (value / beneficiaries.length);
          beneficiaries[i] = new_percentage;
        }
        return beneficiaries;
        break;
    case "update":
            beneficiaries[index];
            for(let i in beneficiaries) {
          cur_percentage = beneficiaries[i];
          new_percentage = (100 - value) * (cur_percentage / 100) + (value / beneficiaries.length);
          beneficiaries[i] = new_percentage;
        }
        return beneficiaries;
        break;
  }
}

beneficiaries = calculateBeneficiaryRatio('add', 29);
console.log(beneficiaries);
beneficiaries = calculateBeneficiaryRatio('add', 20);
console.log(beneficiaries);
beneficiaries = calculateBeneficiaryRatio('remove', null, 1);
console.log(beneficiaries);
beneficiaries = calculateBeneficiaryRatio('update', 20, 0);
console.log(beneficiaries);
</code>```

https://jsfiddle.net/74Lf2dhk/1/

答:
Neither of your two examples matches your description: instead of $(85,5,10)$, I would have expected $(81,9,10)$, because that preserves the ratio $a:b$, i.e., $90:10=81:9$. Maybe there was a miscalculation?

If that's the case, then here's the idea: suppose you have some percentages $(a,b,c,d)$ and you're changing $d$ to $d+\Delta$. Then we need to remove a total of $\Delta$ from the other three while preserving the ratio $a:b:c$.

Looking at $a$ as a fraction of $a+b+c$, we have $\frac a{a+b+c}$. This is the fraction of $\Delta$ by which $a$ must decrease, so the new value of $a$ will be
$$a - \left(\frac a{a+b+c}\right)\Delta,$$
which after factoring out the $a$ we can rewrite as
$$a\left(1 - \frac \Delta{a+b+c}\right).$$

To put it more simply, let $T$ be the total of the first three values: $T=a+b+c$. Then the new value of $a$ will be $a\left(\frac{T-\Delta}T\right).$

The changes to $b$ and $c$ are similar; therefore, the new values of $(a,b,c,d)$ will be
$$\left(a\left(\frac{T-\Delta}T\right), b\left(\frac{T-\Delta}T\right), c\left(\frac{T-\Delta}T\right), d + \Delta\right).$$

Finally, note that the case $a=b=c=0$ will have to be given special treatment.

Title: Proportional Percentage Scaling


问:
Hamiltonian path- Lower bound of optimal solution
I am trying to find an algorithm with polynomial run time that calculates the lower bound of the optimal solution on the Hamiltonian Path problem( a lower bound to the sum weight of the hamiltonian path).

My first thought was to place the edges in ascending order and pick one edge at a time.
But this is like kruskal.
Any ideas might be helpful

答:
Yes, because a Hamiltonian path is a spanning tree, computing a minimum spanning tree (MST) provides a lower bound on every Hamiltonian path.  And you can compute a MST in polynomial time.  Here, you are enforcing connectivity but relaxing the degree-2 constraints.

Another polynomially computable lower bound arises from using a dummy node to convert to Hamiltonian tour and then solving an assignment problem to find a minimum-weight union of cycles.  Here, you are enforcing the degree-2 constraints but relaxing connectivity.


答:
Hint.

The linear map $T$ is an isomorphism if and only if any one of its matrix representations is invertible. Now look for a matrix representation of $T$.

Title: Showing that a linear transformation is an isomorphism.

Body:
I have been given the following question:

For part (a) I simply showed the transformation satisfies the two conditions:

$T(ax) = aT(x)$

and

$T(x+y) = T(x) + T(y)$

For part (b) I evaluated T for each value of $\beta$ and wrote this as a linear combination of the values in $\gamma$, with the coefficients of the combination making up the columns of the matrix.

I am now trying to do part (c), but I have no idea how I would do this. Any help would be greatly appreciated.