问:
Do singular values change when an arbitrary matrix is multiplied by a unitary matrix?
Let X be an arbitrary matrix, and let U be a unitary matrix.
Is it so that the singular values of X are the same as the singular values of UX?
Does the same apply for XU?</p>

答:
First, we can think about this geometrically. The SVD $X = \Phi\Sigma\Psi^\top$ tells us that any linear transformation ($X$) can be decomposed into a rotation ($\Psi^\top$), a dilation ($\Sigma$), and another rotation ($\Phi$). If we apply a unitary matrix $U$---which represents a rotation---to either side of $X$, we will change one of the rotations, but not the dilation. So the singular values will not be changed.

Now let's be a bit more precise. Let $X = \Phi\Sigma\Psi^\top$ be the SVD of $X$. Then $UX = (U\Phi)\Sigma\Psi^\top$. If this is the singular value decomposition for $UX$ (i.e., if $U\Phi$ is the matrix of left singular vectors for $UX$), then $X$ and $UX$ have the same singular values (i.e., they have the same $\Sigma$ in their singular value decompositions; recall that the SVD is unique up to signs). In order for that to be true, $U\Phi$ has to be unitary. We know that $U$ and $\Phi$ are individually unitary, so $U^\top U = I$ and $\Phi^\top \Phi = I$. Now observe that $(U\Phi)^\top (U\Phi) = \Phi^\top U^\top U \Phi = \Phi^\top I \Phi = \Phi^\top \Phi = I$, so $U\Phi$ is indeed unitary. Therefore, $X$ and $UX$ have the same singular values.

For the second case, we have $XU = \Phi\Sigma(\Psi^\top U)$. What can you say here?

Here's a quick numerical sanity check for the first case in Python.

```<code>>>> import numpy as np
>>> import scipy.linalg as la

# Get a random test matrix X and take its SVD.
>>> n = 10
>>> X = np.random.random((n,n))
>>> sigma = la.svdvals(X)

# Get a random unitary matrix of the same size as X.
>>> U = la.qr(np.random.random((n,n)))[0]
>>> np.allclose(sigma, la.svdvals(U @ X))
True
</code>```

You'll notice that this example uses square matrices. Does anything change if you use non-square matrices?

Body:
First, we can think about this geometrically. The SVD $X = \Phi\Sigma\Psi^\top$ tells us that any linear transformation ($X$) can be decomposed into a rotation ($\Psi^\top$), a dilation ($\Sigma$), and another rotation ($\Phi$). If we apply a unitary matrix $U$---which represents a rotation---to either side of $X$, we will change one of the rotations, but not the dilation. So the singular values will not be changed.

Now let's be a bit more precise. Let $X = \Phi\Sigma\Psi^\top$ be the SVD of $X$. Then $UX = (U\Phi)\Sigma\Psi^\top$. If this is the singular value decomposition for $UX$ (i.e., if $U\Phi$ is the matrix of left singular vectors for $UX$), then $X$ and $UX$ have the same singular values (i.e., they have the same $\Sigma$ in their singular value decompositions; recall that the SVD is unique up to signs). In order for that to be true, $U\Phi$ has to be unitary. We know that $U$ and $\Phi$ are individually unitary, so $U^\top U = I$ and $\Phi^\top \Phi = I$. Now observe that $(U\Phi)^\top (U\Phi) = \Phi^\top U^\top U \Phi = \Phi^\top I \Phi = \Phi^\top \Phi = I$, so $U\Phi$ is indeed unitary. Therefore, $X$ and $UX$ have the same singular values.

For the second case, we have $XU = \Phi\Sigma(\Psi^\top U)$. What can you say here?

Here's a quick numerical sanity check for the first case in Python.

>>> import numpy as np
>>> import scipy.linalg as la

# Get a random test matrix X and take its SVD.
>>> n = 10
>>> X = np.random.random((n,n))
>>> sigma = la.svdvals(X)

# Get a random unitary matrix of the same size as X.
>>> U = la.qr(np.random.random((n,n)))[0]
>>> np.allclose(sigma, la.svdvals(U @ X))
True

You'll notice that this example uses square matrices. Does anything change if you use non-square matrices?


问:
Complement of the product of Bernstein set and meager set
Assume $M\subset\mathbb R$ be a meager set with cardinality $\mathfrak c.$ I want to construct a Bernstein set $B$ such that $\mathbb R\setminus(B\cdot M)$ has cardinality $\mathfrak c$ , $B\cdot M=\{b\cdot m\mid b\in B,  m\in M\}$. It is an easy transfinite induction to construct a Bernstein set but I want an extra condition.

I do not know if I need a set-theoretical assumption for that or not. More precisely, It is possible or not. I hope I can find an answer or hint since I spent  plenty of time but I did not get any thing at least  possible or not.

答:
Eventually, I answered my question. Surprisingly, the answer is very simply even though  I spent a lot of time thinking to find such  function. There answer is

No Bernstein can be found with this property

Let $M$ be compact nowhere dense prefect set. For contradiction, let $B$ be a Bernstein set such that $\mathbb R\setminus(BM)$ has cardinality $mathfrak c.$ Now,  pick $x\notin BM$. Notice that $xM^{-1}$ is uncountable closed set( cantinas a perfect set) and $$xM^{-1}\cap B=\emptyset$$ which is a contradiction   since $B$ is a Bernstein set.

ContentLicense: CC BY-SA 4.0


问:
Meaning of expression similar to cross product
Given are 2 vectors $\vec{x},\vec{y}$ in $\mathbb{R}^4$. I have a function  $f(\vec{x},\vec{y})$ that looks quite symmetrically. Therefore I think that there must be a meaning or any algebraic rule. What is the meaning of $f$?

$$f(\vec{x},\vec{y})=(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2$$

The first 4 summands look like the squared norm of a cross product. Of course such a cross product has not properties as in $\mathbb{R}^3$. By cross product I just mean the application of the construction rule of a cross product in  $\mathbb{R}^3$ that is analogically extended to  $\mathbb{R}^4$.

Another interpretation would be the squared norm of a (so far unknown) vector product $\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}^6$.

Edit:

For simplicity above was given  only a part but the general form looks like
$$f(\vec{x},\vec{y},\vec{z})=\\
(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2+\\
(x_1z_2-x_2z_1)^2
+(x_2z_3-x_3z_2)^2
+(x_3z_4-x_4z_3)^2
+(x_4z_1-x_1z_4)^2
+(x_1z_3-x_3z_1)^2
+(x_2z_4-x_4z_2)^2+\\
(y_1z_2-y_2z_1)^2
+(y_2z_3-y_3z_2)^2
+(y_3z_4-y_4z_3)^2
+(y_4z_1-y_1z_4)^2
+(y_1z_3-y_3z_1)^2
+(y_2z_4-y_4z_2)^2
$$

Body:
There is a simple meaning that can be understood without knowing anything about abstract $\wedge$ notation or exterior algebra: $$f(\vec{x},\vec{y})=4 A^2,$$ where $A$ is the area of the triangle in $\mathbb{R}^4$  that is spanned by vectors $\vec{x},\vec{y}$.

Proof

The area of a triangle in any dimension is given by
$$A=\frac{1}{2}\|\vec{x}\|\|\vec{y}\|\sin\alpha$$
where $\|\vec{x}\|,\|\vec{y}\|$ are 2 side lengths that enclose angle $\alpha$.
Using dot product $\cdot$ and relations $$\sin^2\alpha+\cos^2\alpha=1$$ $$\cos\alpha=\frac{\vec{x}\cdot\vec{y}}{\|\vec{x}\|\|\vec{y}\|}$$
we get
$$4A^2=\|\vec{x}\|^2\|\vec{y}\|^2-(\vec{x}\cdot\vec{y})^2\\=(\vec{x}\cdot\vec{x})(
\vec{y}\cdot\vec{y})-(\vec{x}\cdot\vec{y})^2\\
=(x_1^2+x_2^2+x_3^2+x_4^2)(y_1^2+y_2^2+y_3^2+y_4^2)-(x_1y_1+x_2y_2+x_3y_3+x_4y_4)^2\\
=x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2\\
+x_2^2y_3^2-2x_2x_3y_2y_3+x_3^2y_2^2\\
+x_3^2y_4^2-2x_3x_4y_3y_4+x_4^2y_3^2\\
+x_4^2y_1^2-2x_1x_4y_1y_4+x_1^2y_4^2\\
+ x_1^2y_3^2-2x_1x_3y_1y_3+ x_3^2y_1^2 \\
+x_2^2y_4^2-2x_2x_4y_2y_4+x_4^2y_2^2\\
=(x_1y_2-x_2y_1)^2
+(x_2y_3-x_3y_2)^2
+(x_3y_4-x_4y_3)^2
+(x_4y_1-x_1y_4)^2
+(x_1y_3-x_3y_1)^2
+(x_2y_4-x_4y_2)^2\\
=f(\vec{x},\vec{y})$$


问:
Are these two curves birationally equivalent?
I'm trying to work out if the following two curves are birationally equivalent:

$$Y^2 = 2X^4 + 17X^2 + 12$$
$$2Y^2 = X^4 - 17$$

(I'm considering the above as the affine shorthands for the projective curves they represent)

I saw elsewhere online that I could use the following code in magma:

```<code>K<x,y>:=AffineSpace(Rationals(),2);
C1A:=Curve(K,2*x^4+17*x^2+12-y^2);
C2A:=Curve(K,x^4-17-2*y^2);
C1:=ProjectiveClosure(C1A);
C2:=ProjectiveClosure(C2A);
IsIsomorphic(C1,C2);
</code>```

However this gives me a runtime error.

Can anyone see an error with my code/approach above? Alternatively, if there's another way I can check if two curves like the above are birationally equivalent either by hand or by using Magma / Sage or something similar that would be really helpful.

答:
In PARI/GP:

```<code>? ellinit(ellfromeqn(2*x^4+17*x^2+12-y^2)).j
%1 = 384200066/111747
? ellinit(ellfromeqn(x^4-17-2*y^2)).j
%2 = 1728
</code>```

The curves are not birationally equivalent because they have different $j$-invariants.

Body:
In PARI/GP:

? ellinit(ellfromeqn(2*x^4+17*x^2+12-y^2)).j
%1 = 384200066/111747
? ellinit(ellfromeqn(x^4-17-2*y^2)).j
%2 = 1728

The curves are not birationally equivalent because they have different $j$-invariants.


问:
Convergence of improper unbounded integral

Show that $\displaystyle \int_{0}^{1}  \frac{1}{x^2 + \sqrt{x}} dx$ converges.

My method : This is a type of improper integral where the function becomes unbounded at the lowe limit of integration and both the limits of integration are finite. I took the given function as $f(x)$ being integrated from $0$  to $1$ and another function $g(x) =  \frac{1}{\sqrt{x}}$ by taking $\frac{1}{\sqrt{x}}$ common from $f(x)$ and observing that at $x=0$ the convergence and divergence of the function $f(x) $ was solely dependent by this $\frac{1}{\sqrt{x}}$

I was taught a limit test to find the convergence or the divergence. So it says that when the function is of the unbounded type with finite bounds to integrate  :

Find $\lim_{x \to 0 } \frac{f(x)}{g(x)}$ which I get as 1 in this question. Since the limit exists and is not equal to zero, whatever is the behaviour of $g(x)$, that is the behaviour of $f(x) $ also. Therefore by using the test integral I find that $g(x)$ diverges therfore $f(x) $ should also diverges. But the answer says that it converges. How is this possible ?

Test integral used :

$\displaystyle \int_{a}^{\infty}  \frac{1}{x^p} dx$ where $a>1$;
If p $\leq$ 1 then it diverges.
If p $>$ 1 than it converges.

I know that in the test integral $a>1$ is given but then my prof used it irrespective of it in other examples of the same type (that is in sums of unbounded type where the limits of integration were $0$ to $3$ and $0$ to $4\pi$ in the two examples he gave) . And I also do not understand the $\infty$ limit in the integration of test integral as in the given question I need to integrate only from $0$ to $1$

I know I have asked two questions but since they're related please don't close my answer.

答:
You use the wrong test integral. Instead, you should use
$$
\int_0^1\frac{1}{x^p}\,dx
$$
for the cases $0<p<1$ and $p>1$.

Notes.

You want to analyze the integral $\displaystyle\int_0^1\frac{1}{x^2+\sqrt{x}}dx$, which is fundamental different from integrals of the form $\displaystyle \int_1^\infty f(x)\, dx$. These are two different types of improper integrals.

In general, if $0\le f(x)\le g(x)$ and $\int_0^1g(x)dx$ is convergent, then $\int_0^1f(x)dx$ is also convergent.

In your example, $\displaystyle g(x)=\frac{1}{x^{1/2}}$.


问:
A linear functional $f$ is continous if and only if...
Let be $E$ a topological vector space and a linear functional $f$. How to show that $f$ is continous if and only if exist an oper set $U$ and a real value $t$ such that $t\notin f(U)$.

答:
Here is how it works without a norm:

If $f$ is continuous then $U:=f^{-1}(\,(0,1)\,)$ is open and clearly $f(U)$ is not all of $\Bbb R$ (I'm assuming real vector space, otherwise let $U$ be the pre-image of the complex open unit ball).

If $f$ has the property suppose that $f$ is discontinuous and get a contradiction. Let $U$ be one of the special open sets and by translation assume $0\in U$. Let $z_\alpha\to0$ a net with $f(z_\alpha)\not\to0$. Either $f(z_\alpha)$ is unbounded or admits a convergent (not to $0$) sub-net. Either way we may pass to a sub-net so that $\frac1{f(z_\alpha)}$ converges and is always defined. Now for $\lambda\in \Bbb R$ you have that
$$\frac{\lambda}{f(z_\alpha)}z_\alpha $$
still converges to $0$, hence must eventually lie entirely in $U$. However
$$f\left(\frac{\lambda}{f(z_\alpha)}z_\alpha\right)=\lambda$$
this yields the contradiction, so if $f$ has the property it cannot be discontinuous.


答: The integral of $g(x)$ is $\displaystyle \int_0^1 \frac{1}{\sqrt{x}}\,dx = 2\sqrt{x}\Big\rvert_0^1 = 2$, doesn't diverge.

Title: Convergence of improper unbounded integral

Body:
Show that $\displaystyle \int_{0}^{1}  \frac{1}{x^2 + \sqrt{x}} dx$ converges.

My method : This is a type of improper integral where the function becomes unbounded at the lowe limit of integration and both the limits of integration are finite. I took the given function as $f(x)$ being integrated from $0$  to $1$ and another function $g(x) =  \frac{1}{\sqrt{x}}$ by taking $\frac{1}{\sqrt{x}}$ common from $f(x)$ and observing that at $x=0$ the convergence and divergence of the function $f(x) $ was solely dependent by this $\frac{1}{\sqrt{x}}$

I was taught a limit test to find the convergence or the divergence. So it says that when the function is of the unbounded type with finite bounds to integrate  :

Find $\lim_{x \to 0 } \frac{f(x)}{g(x)}$ which I get as 1 in this question. Since the limit exists and is not equal to zero, whatever is the behaviour of $g(x)$, that is the behaviour of $f(x) $ also. Therefore by using the test integral I find that $g(x)$ diverges therfore $f(x) $ should also diverges. But the answer says that it converges. How is this possible ?

Test integral used :

$\displaystyle \int_{a}^{\infty}  \frac{1}{x^p} dx$ where $a>1$;
If p $\leq$ 1 then it diverges.
If p $>$ 1 than it converges.

I know that in the test integral $a>1$ is given but then my prof used it irrespective of it in other examples of the same type (that is in sums of unbounded type where the limits of integration were $0$ to $3$ and $0$ to $4\pi$ in the two examples he gave) . And I also do not understand the $\infty$ limit in the integration of test integral as in the given question I need to integrate only from $0$ to $1$

I know I have asked two questions but since they're related please don't close my answer.


问:
Are there any increasing concave functions that grow faster than $e^x$?
As titled. Intuitively I don't think there is any. But I wish to get a somehow rigorous argument, not even necessarily a proof. If anyone has any reference on this, it would also be much appreciated.</p>

答: Off the top of my head: $x^x$.

Body: Off the top of my head: $x^x$.


问:
Can a knight visit every field on a chessboard?
I was doing excercises about graphs theory and I came across a quite interesting excercise (which probably has something to do with Hamiltonian Cycle):
"Is it possible to step on every field of a 4x4 or 5x5 chessboard just once and return to the starting point using a knight?"
Does anyone have any idea how to tackle this problem? I am more interested in a outline of how to do it or just some hints.</p>

答: Hint: You must enter and exit from each square. Look at the corners. What is special about them?</p>

ContentLicense: CC BY-SA 3.0


问:
Formal definition of a complement of a language.
I am in the midst of learning decidable and undecidable language and I came across the following theorem:

A language  $A \subseteq \Sigma^*$ is decidable if and only if $A$ as well as $\Sigma^* \setminus A$ are semi-decidable.

I'm not sure what $\Sigma^* \setminus A$, the complement of the language $A$, is supposed to mean. A language is a set of strings, so would its complement then be the set of all strings that are not in $A$? Can anyone provide a more formal explanation and definition?

答:
$\Sigma$ is supposed to denote the alphabet.

A string is just an arbitrary ordering of some (or all) elements of the alphabet.

$\Sigma^n$ is the set of all possible n-sized strings.

$\Sigma^{*}$ = $\bigcup_{i}\Sigma^{i}$

You have a language A,defined over a subset of $\Sigma^{*}$.

$\Sigma^{*}\backslash A$ is simply all strings in $\Sigma^{*}$ but not in
$A$


答: $\Gamma(x)$ (the factorial for non-integer values) is such a function. As is $2^{2^x}$ and $2$ to the power of the previous function and so on. Even $3^x$ grows faster than $e^x$ because its base is larger.

Title: Are there any increasing concave functions that grow faster than $e^x$?

Body: As titled. Intuitively I don't think there is any. But I wish to get a somehow rigorous argument, not even necessarily a proof. If anyone has any reference on this, it would also be much appreciated.</p>


答:
There are plenty of examples of convex functions that grow faster than exponential functions. Some examples are:

Double exponential function https://en.wikipedia.org/wiki/Double_exponential_function

Factorials / Gamma function https://en.wikipedia.org/wiki/Factorial

Tetration https://en.wikipedia.org/wiki/Tetration


问:
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$ und show that $$\left (\prod_{k=1}^na_k\right )^{\frac{1}{n}}\leq \frac{1}{n}\sum_{k=1}^na_k$$

For the first part I applied the method of Lagrange multipliers.

We have the function $f(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2$ and the constraint $g(x,y,z)=x_1^2+ \ldots + x_n^2-1=0$.

The Lagrange function is \begin{equation*}L(x_1, \ldots , x_n  ,\lambda )=f(x_1, \ldots , x_n)+\lambda g(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2+\lambda \left (x_1^2+ \ldots + x_n^2-1\right )=\left (\prod_{j=1}^nx_j\right )^2+\lambda \left (\sum_{j=1}^nx_j^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$ :
\begin{align*}&\frac{\partial}{\partial{x_i}}(x_1, \ldots , x_n  ,\lambda )=2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i  \\ & \frac{\partial}{\partial{\lambda }}L(x_1, \ldots , x_n  ,\lambda )=\sum_{j=1}^nx_j^2-1 \end{align*} with $1\leq i\leq n$.

To get the extrema we set the partial derivatives equal to zero.

Then we get the following system:
\begin{align*}&2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i =0 \Rightarrow x_i\cdot \prod_{j=1, j\neq i}^nx_j^2 +\lambda x_i =0 \Rightarrow x_i\cdot \left (\prod_{j=1, j\neq i}^nx_j^2 +\lambda \right ) =0 \\ & \sum_{j=1}^nx_j^2-1=0  \end{align*}

How can we continue?

答:
A bit late but I thought worth mentioning it.

Set $y_k = x_k^2$ for $k=1\ldots n$. So, to maximize is

$$\prod_{k=1}^n y_k \text{  subject to  } \sum_{k=1}^ny_k =1\text{  with  } y_1,\ldots ,y_n \geq 0$$

Since the product is $0$ if any of the $y_k$ is zero, we can assume $y_1,\ldots ,y_n > 0$.

Now, Jensen (or concavity of $\ln$) gives immediately

$$\sum_{k=1}^n \ln y_k \leq n \ln \sum_{k=1}^n \frac{y_k}n = \ln \frac 1{n^n}$$

Hence,

$$\prod_{k=1}^n x_k^2 = \prod_{k=1}^n y_k \leq \frac 1{n^n}$$

and equality is reached for $y_k = \frac 1n \Leftrightarrow x_k =\pm \frac 1{\sqrt n}$.


问:
Solve $e^{2z} - 2e^z + 2= 0$
So I've started by looking at

\begin{align} e^{z} =  x:\\
x^2 - 2x + 2 = 0
\end{align}

Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}

Then I did the following:

\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2}\\
e^x= {\sqrt{2}} \quad \text{and}\quad y&=\frac{\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{\pi}{4}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{7\pi}{4})+i\sin\frac{7\pi}{4})=\frac{1}{2}-\frac{i \cdot 1}{2}\\
e^x= \sqrt{2} \quad \text{and}\quad y&=\frac{7\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{7\pi}{4}\\
\end{align}

Are the solutions the following? \begin{align} \sqrt{2} + \frac{\pi}{4} \quad \text{and} \quad \sqrt{2} + \frac{7\pi}{4} \end{align}

Forgive the shoddy formatting as this is my first ever post here. Thanks.

答:
Yes.....

Once you get $e^z = 1 +i, 1-i$ you can take the following as a formula (assuming $a,b$ are real):

$a + bi= \sqrt{a^2 + b^2}e^{\arctan \frac ab i}= e^{\ln(a^2 + b^2)}e^{\arctan \frac ab i}= e^{\ln(a^2+b^2) + \arctan \frac ab i}$

[I suppose another way putting this is the principal natural log $\operatorname{Ln}(a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi$. But as natural log is a "multivalued function" on complex numbers $\ln (a+bi) = \ln(\sqrt {a^2 + b^2}) + \arctan \frac abi + 2k\pi i$.]

so $z = \ln 2 + \arctan \pm 1 i = \ln 2 + \frac \pi 4i, \ln -\frac \pi 4 i$

.....

But as $\arg$ is congruent $\mod 2\pi i$ we can actually have an infinite number of solutions $z = \ln 2 +(\pm \frac \pi 4 + 2k\pi )i$ for any integer (positive or negative) $k$.

.....

Also you shouldn't use $x$ as a variable for two different meanings (although being in isolated steps of the proof there was no harm done).

Title: Solve $e^{2z} - 2e^z + 2= 0$


问:
Adding probability density functions.
I've just started studying probability, so I beg your pardon if my question is silly.

Consider a pdf $\large p_{f}(x)$, for a continuous random variable, that represents the distribution of images of all possible human faces. Now, this distribution is made up of many types of unique features, for instance faces with glasses, women, men, beard, no beard, and so on... For simplicity, let's say I'm only interested in "beard" with its own pdf $\large p_{b}$ and "no beard" with its own pdf $\large p_{nb}$.

Now, no human face can exist out of these two distributions, this implies the pdf $\large p_{f}(x)$ can be written only with $\large p_{b}(x)$ and $\large p_{nb}(x)$.

So how can I write $\large p_{f}(x)$ in terms of $\large p_{b}(x)$ and $\large p_{nb}(x)$ only?

If I'm talking about a discrete RV, then ofc it's simply the addition of pmf of these "beard" and "no beard" distributions. Which is not the case. So how can I add these two pdfs?

Thanks for your time.

答:
I asked the same question in multiple ways (here, here and here) so that someone would get my doubt. And finally, thanks to jlammy who provided the perfect answer to the question.

What I was looking for, was the mixture distribution, here its explained really really well.
The combination of density functions can be written as their weighted sum. Where the weights are their corresponding proportions in the population.

So in my above case, the $\large p_{f}(x)$ can simply be written as $\large p * p_{b}(x) + (1 - p) * p_{nb}$, where $\large p$ denotes the proportion of bearded humans, and $\large (1 - p)$ denotes the proportion of non bearded humans.

Title: Adding probability density functions.


问:
Branch cut of the logarithm
I have a function $F$ holomorphic on some open set, and I have $ F(0) = 1 $ and $ F $ is non-vanishing. I want to show that there is a holomorphic branch of $ \log(F(z)) $.

Now, I'm getting confused. The principal branch of logarithm removes $ (-\infty, 0] $. But if the point 0 is missing from the plane, what happens when we take $ \log{F(0)} = \log{1} + 0 = 0 $? (I'm sure we can take the principal branch, because $ \exp(z) $ satisfies the conditions in question).

Any help would be appreciated. Thanks

答: If the open set $U$ (where $F$ is defined) is not simply connected then $\log(F(z))$ may not exist (e.g. $F(z)=z$, $U=\mathbb{C}-0$). If $U$ is simply connected then $\log(F(z))$ exists by the Monodromy theorem: from a bit more general point of view, $\log$ is defined on a covering space $p:X\to\mathbb{C}-0$ ($X=\mathbb{C}$ and $p=\exp$), we have a map $F:U\to\mathbb{C}-0$, and since $U$ is $1$-connected, the map $F$ can be lifted to $X$.</p>

Title: Branch cut of the logarithm


答: You don't need to go through divisibility arguments. Since $ab^{-1}=(b^2)(b^3a^3)(b^{-2})$, $ab^{-1}$ and $b^3a^3$ are conjugate and have the same order. Using the fact about orders of inverses, the original claim follows.

Body:
Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.

attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, and
$b^3a^3 = b^{-2}(ab^{-1})b^2$.
Denote the identity element of $G$ by $e_G$.

Fact: for all $a \in G$,
$|a| = |a^{-1}|$.

Let $|ba^{-1}| = m$ and $|b^3a^3| = p$.

To show $m \mid p$:
\begin{align*}
(b^3a^3)^p &= e_G \\
(b^{-2}(ab^{-1})b^2)^p &= e_G \\
b^{-2}(ab^{-1})^pb^2 &= e_G \\
b^{-2}((ba^{-1})^{-1})^pb^2 &= e_G \\
((ba^{-1})^{-1})^p &= e_G
\end{align*}
By fact, $|(ba^{-1})^{-1}| = m$.
Hence, $m \mid p$ as desired.

To show $p \mid m$:
\begin{align*}
(ba^{-1})^m &= e_G \\
((ab^{-1})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^m)^{-1} &= e_G
\end{align*}
Here I got a little bit confuse.
Does it true that in:
\begin{equation*}
((b^2(b^3a^3)b^{-2})^m)^{-1} = e_G
\end{equation*}
implies
$((b^2(b^3a^3)b^{-2})^m) = e_G$? (This is come from fact.) If yes, then
$|b^3a^3| = m$. Hence, $p \mid m$.
And thus, $m=p$.


问:
How to glue the sheaf of differentials on a scheme
I have been trying to really understand the sheaf of differentials defined on a scheme but am having a lot of trouble. Let $X$ be a scheme separated over a field $k$. Suppose we have an open affine cover $\{ U_{i} = \operatorname{spec}A_{i} \}$ of $X$. On each of these $U_{i}$ we would like to define the quasi-coherent sheaf $\mathcal{F}_{i}$ given by $\widetilde{\Omega_{A_{i}/k}}$. The goal then is to glue all the $\mathcal{F}_{i}$.

The main goal is to define isomorphisms,
$$
\mathcal{F_{i}}|_{U_{i} \cap U_{j}} {\longrightarrow} \mathcal{F_{j}}|_{U_{i} \cap U_{j}}
$$
and this is the part I am having trouble with. Having defined each $\mathcal{F}_{i}$ to be quasi-coherent, and having judiciously chosen $X$ to be separated, we have that intersections are given by $U_{i} \cap U_{j} = \operatorname{spec}(A_{i} \otimes_{k} A_{j})$ and so

\begin{align}
\mathcal{F}_{i}|_{U_{i} \cap U_{j}} &\simeq \widetilde{\Omega_{A_{i}/k}}|_{U_{i} \cap U_{j}} \\
&\simeq \big( \Omega_{A_{i}/k} \otimes_{A_{i}} (A_{i} \otimes_{k} A_{j}) \big)^{\sim} \\
&\simeq \widetilde{\Omega_{(A_{i} \otimes_{k} A_{j})/A_{j}}}
\end{align}
Likewise we have
$$
\mathcal{F}_{j}|_{U_{i} \cap U_{j}} \simeq \widetilde{\Omega_{(A_{i} \otimes_{k} A_{j})/A_{i}}}.
$$
But now I am at a loss as to how to show there is a canonical (I want it to be canonical so it is easy to see that the cocycle condition holds) isomorphism,
$$
\Omega_{(A_{i} \otimes_{k} A_{j})/A_{j}} \longrightarrow \Omega_{(A_{i} \otimes_{k}. A_{j})/A_{i}}
$$
I tried to show this using the universal property and Yoneda's lemma, but I'm not even sure how to work with derivations from tensor products of rings. Is this actually the right approach to glue the sheaf of differentials? I am mostly learning from Ravi Vakil's notes, and he seems to glue things by showing that they restrict nicely and then using universal properties to show those restrictions are canonically isomorphic. That is what I have tried to do here, but I'm not sure if it will work.

答:
I do not think you have to use glueing for the following reason. Let $p,q:X\times_S X \rightarrow X$ be the projection morphisms and let $U:=Spec(A) \subseteq X$ be an affine open subscheme mapping to the open subscheme $V:=Spec(R)\subseteq S$. Let $\mathcal{I}\subseteq \mathcal{O}_{X\times_S X}$ be the ideal of the diagonal and define $\Omega:=\Omega^1_{X/S}:=p_*(\mathcal{I}/\mathcal{I}^2)$. It follows

$$\Omega^1(U):=(\mathcal{I}/\mathcal{I}^2)(p^{-1}(U)):=   (\mathcal{I}/\mathcal{I}^2)(U\times_S X)\cong (\mathcal{I}/\mathcal{I}^2)(U\times_V U)$$

since $\mathcal{I}/\mathcal{I}^2$ is supported on the diagonal.

It follows

$$\mathcal{I}/\mathcal{I}^2(U\times_V U):=I/I^2$$

where $I\subseteq A\otimes_R A$ is the ideal of the diagonal. Hence for any open subscheme $U$ it follows $\Omega^1(U)$ is the left $A$-module $\Omega^1_{A/R}$.

Note that for $a\in A$ and $\omega:=\sum_i u_i\otimes v_i \in \Omega^1_{A/R}$ it follows

$aw:=\sum_i (au_i)\otimes v_i =\sum_i u_i\otimes (v_ia)=\omega a$.

Hence the left and right $\mathcal{O}_X$-module structure on $\Omega^1_{X/S}$ agree.


答:
Claim 1: For any $x,y \in G$ we have $|xyx^{-1}|=|y|$

Proof: Exercise

Claim 2: For any $x \in G$ we have $|x|=|x^{-1}|$

Proof: Exercise

Now, from your last step of actual manipulation we have, $b^3a^3=b^{-2}(ab^{-1})b^2$.

So applying Claim 1 twice, we have, $|b^3a^3|=|b^{-2}(ab^{-1})b^2|=|b^{-1}(b^{-1}(ab^{-1})b)b|=|b^{-1}(ab^{-1})b|=|ab^{-1}|$

Now applying Claim 2, we get, $|b^3a^3|=|ab^{-1}|=|(ab^{-1})^{-1}|=ba^{-1}$

Title: Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$.


问:
Interesting pattern in the number of steps in the Collatz conjecture.

I was playing around with the Collatz conjecture and plotting random things when I came across this.

The x axis is the natural numbers and the y axis is the number of steps it takes for each number to reach one.

This looked interesting to me, I mean, there is definitely a pattern here, don't you agree? :)

Does this look similar to anything?

答:
You asked if your plot looks similar to anything. The simple answer is that the plot is not new as similar ones can be found on Collatz conjecture Wikipedia page.

Your other question ask if anyone agrees that there is a pattern. Yes, I do agree there is a general pattern. The graph shows that at different depths on the Collatz tree with root 1, there are clusters of numbers whose distance apart is small; hence the appearance of dashed lines. Additionally, the curved downward patterns are an indication that some small natural numbers require more steps to reach 1 than the larger natural numbers that they ascend to before the trajectories descend.


问:
Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$

If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.

If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.

Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.

Any hints? I have the exam soon =/

答:
We essentially want to show that $60 \vert (a^4-1)$. Since $\gcd(a,30) = 1$, we have
$$a \equiv \pm1, \pm 7, \pm 11, \pm 13, \pm 17, \pm 19, \pm 23, \pm 29 \pmod{60}$$
Hence,
$$a^2 \equiv \begin{cases}1 \pmod{60} & \text{if }a \equiv \pm1,\pm11,\pm19,\pm29\\  -11 \pmod{60} & \text{if }a \equiv \pm7,\pm13,\pm17,\pm23\end{cases}$$
which inturn gives us
$$a^4 \equiv 1 \pmod{60} \,\,\,\, \forall a \text{ such that }\gcd(a,30) = 1$$</p>


问:
Result relating fundamental groups and covering spaces
During my Basic Algebraic Topology class, my professor left an exercise for us to make, but I'm having some trouble getting ahold of it. Here it is:

Exercise:  Let $p:\tilde X\to X$ be a covering map of $m$ sheets, with $m$ prime, and $\tilde X$ simply connected. Prove that $\Pi_1(X)\cong\Bbb Z_m$.

There was a previous result that tells us that the fiber $p^{-1}(x)$ has the discrete topology and $\#p^{-1}(x)=\#L$, being $L$ the set of slices of our covering space. Could I use this in order to follow with anything? Or is there another way I can approach this exercise?

I still want to do the exercise by myself, so could anyone lend me some help to carry on without spoiling the whole result? Thanks in advance.

答: Choose a base point $b$ in $X$ and one from $p^{-1}(b)$, say $a$. Since $\bar X$ is path connected there exists a path from $a$ to every point in $p^{-1}(b)$. The composition of these paths with $p$ gives non homotopic loops in $X$. Also since $\bar X$ is simply connected every loop in $X$ will be homotopic to one of these. Hence there are exactly $m$ elements in $\pi_{1}(X)$ and hence it is $\Bbb Z_{m}$.


问:
How to calculate a differential of an Integral squared?
could anyone help me calculating the derivative of this function:

$G:C[0,1]\rightarrow\mathbb{R},x\mapsto \left(\int_0^1 x(s)ds\right)^2$

I tried using the Gateaux Differential - working out

$\frac{1}{s}(G(u+sh)-G(u)$ to then go $s\mapsto 0$

but this leads nowhere. I have the feeling this needs some "trick" regarding the squared integral. Could I write this differently?

Thank you in advance.

答:
Setting $G(x) = \left( \int_0^1 x(s)\,ds \right)^2$ gives
$$
G(x+h\xi) = \left( \int_0^1 (x(s) + h\xi(s))\,ds \right)^2
= \left( \int_0^1 x(s)\,ds + h \int_0^1 \xi(s)\,ds \right)^2
\\= \left( \int_0^1 x(s)\,ds \right)^2 + 2h \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) + h^2 \left( \int_0^1 \xi(s)\,ds \right)^2
$$
so
$$
\frac{G(x+h\xi) - G(x)}{h} = 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right) + h \left( \int_0^1 \xi(s)\,ds \right)^2
\\ \to 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right)
$$
as $h\to0.$

Thus,
$$
\lim_{h\to0} \frac{G(x+h\xi) - G(x)}{h}
= 2 \left( \int_0^1 x(s)\,ds \right)\left( \int_0^1 \xi(s)\,ds \right)
.
$$


问:
Open subscheme of Noetherian integral scheme is integral?
Let $(X, \mathcal O_X)$ be a Noetherian scheme. If $X$ is an integral scheme (i.e. $X$ is irreducible and reduced scheme), then is it true that for every open  subset $U$ of $X$, the open subscheme $(U, \mathcal O_U)$ is integral ? (Here $\mathcal O_U=\mathcal O_X|_U$)

I think this is true and here are my thoughts: $U$ being an open subset of $X$, is irreducible. Moreover, for every $u\in U$, we have $\mathcal O_{U,u}\cong \mathcal O_{X,u}$ is reduced , hence $U$ is reduced. Thus $U$ is integral.

Am I correct ?

答:
Definition: Let $(X, \mathcal{O}_X)$ be a scheme. We say $(X, \mathcal{O}_X)$ is "integral" iff for every open set $U\subseteq X$ it follows
$\mathcal{O}_X(U)$ is an integral domain.

Example: Consider $(V,\mathcal{O}_V)$ with $V\subseteq X$ an open subscheme. Let $U \subseteq V$ be an open set. It follows $\mathcal{O}_V(U)\cong \mathcal{O}_X(U)$ is an integral domain since $X$ is integral, hence $(V, \mathcal{O}_V)$ is an integral scheme.


问:
Accelerate convergence of sequence
Is it always possible to extract a subsequence from my generic sequence $(q_n)$, such that the convergence of the subsequence to the same limit $r$ is faster then the original?

答:
The answer is yes.

Pick any rate of convergence $r_n$ you want. Then, for each $n$ there exists some $M_n$ such that for all $m >M_n$ we have
$$|a_n-l|<r_n$$

Pick inductively
$$k_n > \max \{ k_1,k_2,..., k_{n-1}, M_n \}$$

Then $a_{k_n}$ is a subsequence and since $k_n >M_n$ we have
$$|a_{k_n} -l| <r_n$$

Title: Accelerate convergence of sequence


问:
Using jacobian to transform area elements from one system to another.
Please excuse my lack of rigor,I'm just average Physics undergraduate.

I have a transformation from $(u, v)$ to $(x, y)$. So an infinitesimal area element from $ (u, v)$ to $(x, y)$ plane  will generally map as shown.

Using Jacobian we relate these area elements as

$d A=J. d u .d v$

where
$d A \neq d x d y$.

So we can write for some $f(x, y)$ on a region

$\int f(x, y) d A=\int f(u, v) J d u d v$

But how does this follow

$\int f(x, y) d x d y=\int f(u, v) J d u d v$

Since we know that $d A \neq d x d y$ then why is $dA$ replaced by $dx dy$.
Every online source does it, replacing a parallelogram $dA$ by a square $dx.dy$.

I'd be grateful if someone helped. Thank you.

答:
Since $x = x(u,v)$ and $y=y(u,v)$ are functions of $u$ and $v$, you  can write their differentials using the standard formula:

$$ dx = x_u du + d_v dv \hspace{0.5cm} \text{ and } \hspace{0.5cm} dy = y_u du + y_v dv $$
I've used the subscript notation for partial derivatives, so $x_u$ means $\frac{\partial x}{\partial u}$, etc...

Now assuming $dA$ means $dx \, dy$ (which it usually does), you get
$$ dA = dx \, dy = x_u y_u du^2 + x_v y_v dv^2 + x_uy_v d_u d_v + x_v y_u dv du $$
The symbols "$dx$", "$dy$", "$du$", "$dv$", and "$dA$", are usually explained in low-level undergrad calculus courses as represented a "really small" length or area, as your picture indicates. But this is a convenient lie. They are really algebraic objects called "differential forms". See the Wikipedia page for more info.

The important part for this discussion is two properties of multiplication of differential forms.

If you square a differential form, you get $0$. So $du^2 = 0$ and $dv^2=0$.

It is not commutate, and $du \, dv \neq dv du$. In fact it is "anti-commutative" which means $dv du = - du dv$.

Applying these two rules to the equation above, you get

$$ dA = dx \, dy = (x_uy_v - x_vy_u) \, du \, dv = J \, du \, dv$$


答:
Let $\,\phi\,$ be Euler's Totient Function , then

$$\phi(60)=\phi(4)\phi(3)\phi(5)=2\cdot 2\cdot 4=16$$

and in fact

$$\left(\Bbb Z/60\Bbb Z\right)^*\cong C_2\times C_2\times C_4$$

Thus, the exponent of the above group is $\,4\,$ , which means

$$a\in\left(\Bbb Z/60\Bbb Z\right)^*\implies a^4=1$$

and from here the claim follows (why?  Check $(a,30)=1$))

Title: Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$

Body:
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.

If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.

Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.

Any hints? I have the exam soon =/


问:
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$
$(a+c)(b+d)=1$, $a,b,c,d \in R^{+}$. Prove that

$$\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \geq \frac{1}{3}$$

rather annoying that the powers of denominators and numerators are not the same and can't think of easy way to flatten them..

答:
The naive Titu also works

WTS

$$ \sum_{cyc} \frac{a^3}{b+c+d} = \sum_{cyc} \frac{ a^4 } { ab+ac+ad } \geq \frac{ (\sum a^2 )^2 } { 2 ( ab+ac+ad+bc+bd+da) } \geq \frac{1}{3}.$$

This follows because

 1. $ a^2 + b^2 + c^2 + d^2  \geq \frac{ 2}{3} ( ab+ac+ad+bc+bd+da)$, and

 2. $a^2 +b^2 + c^2 + d^2 \geq \frac{1}{2} [ (a+c)^2 + (b+d)^2] \geq (a+c)(b+d) = 1$


问:
Help evaluating an interesting series.
I'm trying to evaluate the sum
$$
s = 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\cdots.
$$

The hint I was given was $\frac{1}{2n+1} = \int_0^1 x^{2n} \, dx$. I'm familiar with the trick of writing the $n$th term of series with a definite integral and then interchanging the order of integral and sum, but the signs on the terms are making this difficult for me. If this were a normal alternating series, we could do something like
\begin{align*}
\sum_{n=0}^{\infty}(-1)^n\frac{1}{2n+1} &= \sum_{n=0}^\infty\int_0^1 (-1)^n x^{2n} \, dx\\
&= \int_0^1 \left(\sum_{n=0}^{\infty} (-x^2)^{n}\right)dx\\
&= \int_0^1 \frac{1}{1+x^2} \, dx = \arctan(1)-\arctan(0) = \frac{\pi}{4}.
\end{align*}

I can't however seem to proceed with the strange alternating series. I know that $(-1)^{(n^2+n+2)/2}$ gives the right pattern of $++--++--\dots$, but that doesn't combine well with the $x^{2n}$ term in the integrand after exchanging sum and integral.

Any thoughts?

Edit: If I use the imaginary unit $i$, then I can write a related sum
$$
\tilde{s} = 1+\frac{1}{3}i-\frac{1}{5}-\frac{1}{7}i+\frac{1}{9}+\frac{1}{11}\cdots.
$$
Using the above trick gives
$$
\sum_{n=0}^{\infty}(i)^{n}\frac{1}{2n+1} = \sum_{n=0}^{\infty}\int_{0}^{1}(i)^nx^{2n}dx = \int_{0}^{1} \sum_{n=0}^{\infty}(ix^2)^n =\int_{0}^{1}\frac{1}{1-ix^2}dx
$$
Which I imagine I can do with some type of partial fraction decomposition.

My hope is that from this new sum I can recover my old sum by adding real and imaginary parts. My only concern is that this would essentially be a rearrangement of the series (adding the even indexed terms together and then the odd indexed terms separately) and this series doesn't converge absolutely so I'm worried that this isn't rigorously justified.

Title: Help evaluating an interesting series.

Body:
From your answer,
$$ \tilde{s}=\int_{0}^{1}\frac{1}{1-ix^2}dx=\int_{0}^{1}\frac{1}{1+x^4}dx+i\int_{0}^{1}\frac{x^2}{1+x^4}dx $$
and then you can obtain
\begin{eqnarray}
s&=&\Re(\tilde{s})+\Im(\tilde{s})\\
&=&\int_{0}^{1}\frac{1 + x^{2}}{1 + x^{4}}\,d x\\
&=&\frac12\int_{0}^{\infty}\frac{1 + x^{2}}{1 + x^{4}}\,d x\\
&=&\frac12\int_{0}^{\infty}\frac{1 + \frac1{x^{2}}}{x^2 + \frac{1}{x^{2}}}\,d x\\
&=&\frac12\int_{0}^{\infty}\frac{1}{(x - \frac{1}{x})^2+2}\,d (x-\frac1x)\\
&=&\frac12\int_{-\infty}^\infty\frac{1}{x^2+2}dx\\
&=&\frac\pi{2\sqrt2}.
\end{eqnarray}


问:
Show that the Heaviside function is not weakly differentiable
Consider the function $$H(x):=\left.\begin{cases}0&\text{, if }x<0\\1&\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x\in\Omega:=(-1,1).$$

How can I show that there is no $g\in\mathcal L^2(\Omega)$ with $$\langle H,\varphi'\rangle_{L^2(\Omega)}=-\langle g,\varphi\rangle_{L^2(\Omega)}\tag1$$ for all $\varphi\in C_c^\infty(\Omega)$?

Clearly, if $g$ exists, then $$\varphi(0)=\langle g,\varphi\rangle\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega)\tag2.$$ In particular, $$\varphi(0)=\langle g,\varphi\rangle\;\;\;\text{for all }\varphi\in C_c^\infty(\Omega\setminus\{0\})\tag3.$$

Now we know that $C_c^\infty(\Omega\setminus\{0\})$ is a dense subset of $L^2(\Omega\setminus\{0\})$ and hence $C_c^\infty(\Omega\setminus\{0\})^\perp=\overline{C_c^\infty(\Omega\setminus\{0\})}^\perp=\{0\}$. So, $$g(x)=0\;\;\;\text{for Lebesgue almost all }x\in\Omega\setminus\{0\}\tag4.$$

But I don't see that this is a contradiction to any of the assumptions.

答:
(Supposing $g$ exists) note that since $\varphi$ is compactly supported, we actually have
$$\varphi(0) = \langle g,\varphi\rangle$$
for all $\varphi$.

Now the main idea is to pick a sequence of functions $\varphi_n$ which breaks this by concentrating around the jump of $H$ at $x=0$.

For example, we can pick smooth functions such that for each $n$,
$$\begin{gather*}
\varphi_n(0)=1 \\
\varphi_n(x) \leq 1 \\
\text{supp}(\varphi) \subseteq [-1/n,1/n]
\end{gather*}$$
You can imagine smoothing out an isosceles-triangle-shaped function with fixed height and decreasing width (in $n$), or use some standard bump function that you know. The important fact you should conclude is that
$$\langle g,\varphi_n\rangle \xrightarrow{n\to\infty} 0.$$


问:
Monotony for a not derivable function by parts
Let $f(x)=\begin{cases}
\frac{(e^{x^2}-5x^2-2)}{x^2}, \quad x\neq 0 \\
0                             , \quad x=0
\end{cases}$

This is not a derivable function in fact it is not continous since $\lim_{x\to 0}\frac{(e^{x^2}-5x^2-2)}{x^2}\neq 0$. So since I can't use the first derivative if I want to study if it is increasing (strictly) or decreasing how can I do?

答:
The fact that $f(x)$ is not differentiable at $x=0$ does not mean that it is not differentiable on its entire domain. In particular, we can differentiate $f(x)$ to get $$f'(x)=\frac{(2x^2-2)e^{x^2}+4}{x^3}$$
and we can easily see that $f'(x)<0$ for $x<0$ and $f'(x)>0$ for $x>0$, i.e. it is strictly decreasing on $(-\infty,0)$ and strictly increasing on $(0,\infty)$.

We get the same conclusion with a graph:


答:
Continue as you started: As $3 \nmid a$, we have $a \equiv \pm 1 \pmod 3$, giving
$$ a^4 - 1 \equiv 1 - 1 = 0 \pmod 3 $$
which means $3 \mid a^4 + 1$.

For $5$, we have either $a \equiv \pm 1 \pmod 5$, giving
$$ a^4 - 1 \equiv 1 - 1 \equiv 0 \pmod 5 $$
or $a \equiv \pm 2 \pmod 5$,
$$ a^4 - 1 \equiv 16 - 1 = 15 \equiv 0 \pmod 5$$
and we are done.


问:
Exponentiating an almost idempotent and almost self-adjoint element
Let $A$ be a unital $C^*$-algebra, and let $q\in A$ be such that $\|q^2-q\|<\varepsilon$, and $\|q\|<K$.

In answer to another question, s.harp showed that

$$\|e^{2\pi iq}-1_A\| < \varepsilon\ \frac{e^{2\pi K}-e^{2\pi}}{K^3-K^2}.$$

On the other hand, for any $s\in\mathbb{R}$,
$$|e^{2\pi is}-1| < 2\pi|s|.$$
Using this, a better bound should be possible in the case that $q$ is self-adjoint, or, more generally, close to self-adjoint.

Question: Suppose $q$ also satisfies $\|q-q^*\|<\delta$ in addition to the above. Then is it possible to improve the above bound for $\|e^{2\pi iq} - 1_A\|$ by taking $\delta$ into account?

答:
Here is a calculation yielding better bounds, but its probably very far away from optimal. The gain is that there is a uniform bound in $\delta$ and $\epsilon$.

Fix some constants:

$$K:=\|q\|, \quad z:= 2\pi i, \quad C:=\sup_{t\in [0,1]}\|\exp( z \,t \,(q-q^*)/2+ zq)\|$$
(one bound of $C$ is $e^{\pi\delta}e^{2\pi K}$, but thats probably a bad bound).

One bound I can get is the following:

If $\epsilon + (K+1)\frac{\delta}2<\frac{\pi^2}9+\frac{\pi}3$ then:
$$\|e^{i2\pi q}-1\|≤C\frac \delta2+\sqrt{2\left(1-\cos(2\pi R (\epsilon, \delta) )\right)}$$
With $R(\epsilon,\delta)= \sqrt{1/4 + \epsilon +(K+1)\frac\delta2}- 1/2$.

The only hard part is the bookkeeping, the path is conceptually simple start with:

$$\|\exp( z\,q)-1\| ≤ \|\exp(z\,q)-\exp(z\frac{q+q^*}2)\|+\|\exp(z\frac{q+q^*}2)-1\|\tag{$*$}$$
now since $q$ is close to $q^*$ we will see if this makes the first term small and since $\frac{q+q^*}2$ is hermitian we will apply the inequality from the hermitian case.

For the first term note:

$$\exp(z\,q)-\exp(z\frac{q+q^*}2) = \left[\exp(zt\,(q-q^*)/2+ zq)\right]_{t=0}^{t=1}=z\int_0^1dt \frac{q-q^*}2 \exp(zt\,(q-q^*)/2+ zq)$$
taking the norm and pulling it into the integral yields that the first term of $(*)$ is smaller than $\pi\,C\,\delta$.

For the second term compute:
$$\|\frac{q+q^*}2(1-\frac{q+q^*}2)\| ≤ \begin{split}\frac14\left( \|q(1-q)\|+\|q^*(1-q^*)\|+\|q^*(1-q^*)+(q-q^*)(1-q^*)\|+\\ \|q(1-q)+(q^*-q)(1-q)\|\right)\end{split}$$
which you further simplify to $\epsilon +(K+1)\frac\delta2$.

Now you may apply the hermitian case to $\frac{q+q^*}2$ using $\epsilon + (K+1)\frac{\delta}2$ as the parameter. This yields, provided $\epsilon + (K+1)\frac{\delta}2<\frac{\pi^2}9+\frac\pi3$, that
$$\|e^{i2\pi (q+q^*)/2}-1\|≤\sqrt{2\left(1-\cos(2\pi R(\epsilon, \delta))\right)}$$


问:
Product of terms involving eigenvalues and an eigenvector.
Let $A_{n\times n}$ be a matrix and $\lambda_1,\lambda_2\ldots \lambda_k$ be $k$ eigenvalues of $A$. Let $x$ be an eigenvector of $A$ corresponding to eigenvalue $\lambda_3$. Then without expanding prove that $(A-\lambda_1 I)(A-\lambda_2 I)\cdots (A-\lambda_k I)x=0.$

Please help me to prove this.

答:
Assume $n\ge 3$ so that eigenvalue $\lambda_3$ exists.

$(A-\lambda_3 I)x = 0$ by definition of eigenvalue and eigenvector.

$A$ and $I$ commute, so we can rearrange the terms.

$$\left[\prod_{j\ne 3}(A-\lambda_j I)\right](A-\lambda_3 I) x = 0.$$

Okay, after posting this, looked at the comments.  @JMoravitz already gave the answer there!


问:
Why can’t measures be defined on uncountable powersets? Example that actually applies to probability theory
What actually breaks in probability when you let $\mathcal F = 2^\Omega$ for uncountable $\Omega$?

To the best of my knowledge, this is not a duplicate question. While there are similar-sounding questions, answers to them usually cite:

Vitali sets. However, only translation-invariant measures prevent Vitali sets. Probability measures don’t need to be translation-invariant, so Vitali sets are irrelevant. You can totally have a probability measure and $\sigma$-algebra containing Vitali sets.

Lebesgue measures. Same objection as above.

Banach-Tarski. But measures must be at least rotation-invariant to prevent Banach-Tarski (or live in 2-D). Probability measures don’t need to be rotation-invariant, so Banach-Tarski is irrelevant.

AFAICT, there seems to be a logical gap in the usual explanations.

Could someone provide a concrete example of any paradoxes that emerge from uncountable sets that apply to probability theory? Any issues with letting $\mathcal F = 2^\Omega$?

Perhaps there is no such paradox. For example, $\delta_a(\cdot)$ seems like a perfectly good probability measure that works with $\mathcal F = 2^\Omega$. The answer might “There’s no paradox, but here is an extra property X we want from our probability measures, here’s why it matters, and for this restricted class of measures, they cannot be defined on $\mathcal F = 2^\Omega$.” If so, I’d love to know what this property $X$ would be.

Apologies if I’ve made an error in this post; I’m new to measure theory!

答:
I'm two years late, but you might be interested in the following construction. In a discrete, real-world probability space, we can define a notion of translation-invariance, and as a result construct non-measurable sets.

Pick a natural number $n$. Let $\Omega = \{ 1, \dots, n \}^\mathbb{N}$ for some natural number $n$. The one assumption we need for this construction is that each of the outcomes $1, \dots, n$ should be equally likely. This is hopefully fine by you because this model certainly comes up a lot—$n = 2$ is an infinite sequence of coin tosses and $n = 6$ is an infinite sequence of die rolls.

Define an equivalence relation on $\Omega$ as follows. Say $\omega_1 \sim \omega_2$ if they differ in finitely many entries. Then let $S \subseteq \Omega$ consist of one representative per equivalence class.

Note that the set of maps on $\Omega$ which change finitely many entries is countable. So call these maps $\sigma_1, \sigma_2, \dots$. Then, for any $i\neq j$, we have that $\sigma_i(S)$ and $\sigma_j(S)$ are disjoint. Moreover, $ \bigcup_{i} \sigma_i(S) = \Omega$.

By the "outcomes are equally likely" assumption, we know that the $\sigma_i(S)$ all have the same measure. Since the $\sigma_i(S)$ are disjoint, by additivity, we should have $\sum_i \mu(\sigma_i(S)) = \mu(\Omega) $. This implies $\sum_i \mu(S) = 1$, which is a contradiction.

So the physical interpretation is we can't assign probabilities to all possible subsets of sequences of (for example) infinitely many coin tosses.


问:
Composition of Linear Rotations and Reflections
Prove that if $T_{1}$ is a rotation of $R^{2}$ about O, and $T_{2}$ a reflection in a line through O, then $T_{1}$$\circ$$T_{2}$ and $T_{2}$$\circ$$T_{1}$ are both reflections in a line through O.

I'd prefer a hint than the answer, because I'm not sure how to begin. Thanks in advance for any help.

答:
There are 4 different types of (non-trivial) isometries of the plane. They are rotations, reflections, translations and glide translations.

Hint: The non-trivial isometries of the plane are classified according to the existence of fixed points, and if they preserve orientation.

Hint: Both $T_1 (T_2)$ and $T_2 (T_1)$ have/don't have fixed points, and preserve/don't preserve orientation, hence they must be $\underline{ \quad \quad \quad} $.


问:
$\lim_{ x \to 0 } \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$
I have to evaluate this limit:
$$\lim_{ x \to 0 }  \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$$

I have an indeterminated form $[\frac{0}{0}]$.
So I applied the Hopital rule
$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \sim $$
$$-e^{\frac{\log(1+\arctan x)}{x}} \cdot \frac{\frac{x}{(1+\arctan x)(1+x^2)}-\log(1+\arctan x)}{x^2} \sim$$
$$ -e \cdot \frac{1}{x^2} \cdot \left(\frac{x}{(1+x)(1+x^2)}-x\right)
\sim +e $$

According to my book and wolfram alpha the final result should be $\frac{e}{2}$ and I don't know where I made mistakes.

答:
$$(1+\arctan x)^{1/x}=\exp\left(\frac{1}{x}\log(1+\arctan x)\right)$$

And

$$\log(1+\arctan x)=\log(1+x+O(x^3))=x-\frac12x^2+O(x^3)$$

Therefore

$$\frac{e-(1+\arctan x)^{1/x}}x=\frac{e-\exp(1-\frac x2+O(x^2))}{x}=\frac{e-e(1-\frac x2+O(x^2))}{x}\\=\frac e2+O(x)\xrightarrow[x \to 0]{} \frac e2$$


问:
Properties of entire function
I met a problem, which says if $f$ and $g$ are all entire functions on $C$. And $f(g(z)) \in C/[- \infty，0]$,  for all $z \in C$. Then both $f$ and $g$ must be a constant.

I have no idea about how to reach this conclusion, is it about the CR equation or I need to expand them into some series?

答:
The correct conclusion is that at least one of $f$ and $g$ is constant.  You cannot conclude that both are constant. Write $\psi= f \circ g$.  Let $h$ be a conformal   injective map from    ${\Bbb C} \setminus (-\infty,0]$ to the unit disk. This exists by the Riemann mapping theorem; for an explicit example, fix a branch of the square root function that is well defined on   ${\Bbb C} \setminus (-\infty,0]$  and take $$h(z)=\frac{\sqrt{z}-1}{\sqrt{z}+1}\, .$$ Then $h \circ \psi$ is bounded; by Liouville [1] it is constant, whence $\psi$ is also constant. If $g$ is not constant, its image must contain a continuous curve on which $f$ is constant, so $f$ equals a constant everywhere.

[1] https://en.wikipedia.org/wiki/Liouville%27s_theorem_(complex_analysis)

Title: Properties of entire function


问:
How do you show a function is continuous at endpoints if the domain is disjoint closed intervals?
I was given the following:

The function $f(x)=−1$ for $x\in[−1,0]$ and $f(x)=1$ for $x\in(1,2]$, with domain $[−1,0]\cup(1,2]$ is continuous everywhere.

The function $f(x)=−1$ for $x\in[−1,0]$ and $f(x)=1$ for $x\in(0,2]$, with domain $ [−1,2]$ is continuous everywhere except at $x=0$.

The second example makes sense because at $x=0$, the two sided limits are not equal. But what about at $x=0$ for the first example? How do you use the epsilon-delta definition to prove $x=0$ is continuous if there are no open intervals around that point?

答:
Continuity is defined with respect to the function's domain: $f$ is continuous at $a$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x \in \text{dom}(f)$, if $|x - a| < \delta$, then $|f(x) - f(a)| < \epsilon$.

The fact that $f$ is not defined on $(0,1]$ does not interfere with showing $f$ is continuous at $0$:

Let $\epsilon > 0$ be an arbitrary value, and set $\delta = \frac 12$. If $|x - 0| < \delta$ and $x \in \text{dom}(f)$, then $x \in (-\frac 12, 0]$, so $f(x) = -1 = f(0)$. Hence $|f(x) - f(0)| = 0 < \epsilon$.

Since the choices of $\epsilon$ and $x$ are arbitrary within their given conditions, this calculation holds for all such values. Thus the definition of continuity is satisfied, and $f$ is continuous at $0$.


问:
How to check that given relation is a function?
If f is defined from set of integers to set of natural numbers,
f(x)=|x|
Then is f a function?
Obviously it's not one-one but is it even a function?
I got confused because 0 belongs to integers and f(0)=0 but 0 does not belong to natural numbers so 0 has no image under f, hence f is not a function.
Thanks in advance.</p>

答: The answer depends on whether or not $0$ is included in the natural numbers. There are two conventions, one that is all positive integers and one that is all non-negative integers (see this question for more information). If you are taking the natural numbers to not include $0$ then you are correct that $f$ is not a function. This is because a function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set. In your example, the $0$ in your domain is not associated with any element in the second set so it is not a function.


问:
Tough Number Theory mod problem
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.

I tried using mod rules to simplify but I got a very strange answer which was divisible by 7 and did not fit the answer criteria. I noted that 15 is 1 mod 7 and even 8 is 1 mod 7, but I do not know how to go ahead from there.

EDIT: I also noted that 6 is -1 mod 7 and 30^12 is 1 mod 7. But that didn't help either.

答:
Update: It's interesting to see that  $a=b$, and the number of cases are the same when there are $2$ or $3$ solutions for $z$, respectively. I have a way to show it is true but right now it's not simple enough to replace the solution below.

I'll adopt Joffan's notation and compute $a$, $b$ and $c$ separately.

First notice that $3^y$ and $5^z$ both cycle through $\{1, 2, 3, 4, 5, 6\} \pmod 7$. So if $x$ and $y$ are fixed then $5^z \equiv 2^{-x} 3^{-y} \pmod 7$ will have $3$ solutions from $1$ to $12$ if $2^x 3^y \equiv 1 \pmod 7$, and $2$ solutions otherwise.

For a given $x$, there are always at least $13 \cdot 2=26$ cases when $y$ cycle through $0$ to $12$. But there is one more solution if and only if $5^z\equiv 2^{-x} 3^{-y} \equiv 1$. Our goal is to find those cases and add to the "base cases".

Next we compute $a$. If $2^x 3^y \equiv 1$ there are two cases:

Case 1: $2^x \equiv  3^y \equiv 1$. There are $3$ cases when $3^y\equiv 1$ for each $x \in \{0,3,6,9,12\}$ so the sum of all these $x$'s is $3 \cdot (3+6+9+12)=3 \cdot 30=90$.

Case 2: $2^x \not \equiv 1$. There are $2$ cases when $3^y \equiv 2^{-x}$ for each $x\in\{1,2,4,5,7,8,10,11\}$ and the sum of all these $x$'s is $2\cdot(78-30)=96$.

Since the sum of $x$'s in  the "base" cases is $78\cdot 26=2028$, we have
$$a=2028+90+96=2214$$

Finally we compute $b$ and by symmetry $c=b$.

If $3^y 2^x \equiv 1$ we have the following cases (note that if $3^y \not\equiv 1,2,4$ then we can't have $3^y 2^x \equiv 1$):

Case 3: $3^y\equiv 2^x \equiv 1$. There are $5$ cases when $2^x\equiv 1$ for each $y\in\{0,6,12\}$ so the sum of all these $y$'s is $5\cdot(0+6+12)=90$;

Case 4: $3^y \equiv 2, 4$. There are $4$ cases when $2^x\equiv 3^{-y}$ for each $y \in \{2,4,8,10\}$ so the sum of all these $y$'s is $4\cdot (2+4+8+10)=96$;

Therefore $b=c=2028+90+96=2214, a+b+c=3\cdot 2214=6642$.

Title: Tough Number Theory mod problem


问:
DNF conversion of this statement
How to convert this to DNF?

$(x\lor{\neg{y}\lor{z}})\land{(\neg{x}\lor{\neg{z}})}$

I have tried de morgans and have got no where. I'm pretty sure its the distributive law but cant work out the steps to get in the form to do that.

答:
I never heard of DNF before, looked it up here. I take it you just want the simplest boolean out of this.

This is a case that works like distributive multiplication of two sums.

$$(x\lor{\neg{y}\lor{z}})\land{(\neg{x}\lor{\neg{z}})}\\
=x(\neg x\lor \neg z)\lor \neg y(\neg x\lor \neg z)\lor z(\neg x\lor \neg z)\\
=(x\neg x \lor x\neg z)\lor (\neg y\neg x \lor \neg y\neg z)\lor z\neg x \lor z\neg z)\\
$$
If there were no mistakes in this expansion, the Veitch diagram below helps with the solution. For shorthand, a slightly different notation is used. Note that hte conditions $x\neg x$ and $z\neg z$ cannot exist so they are to be ignored.

Title: DNF conversion of this statement


问:
How to prove geometrically that anticorrelated Gaussians are separated?
I am trying to prove the following theorem:

There exists a universal constant $K>0$ such that if $g,h$ are standard Gaussians with $\mathsf{E} gh=1-\alpha$,

$$\mathsf{P}\left(g< -1,h> 1\right)\ge K\alpha.$$

Note that without loss of generality, we can assume that $g=\langle u,X\rangle$ and $h=\langle v,X\rangle$, where $u$ and $v$ are unit vectors in $\mathbb{R}^2$ such that $\langle u,v\rangle=1-\alpha$ and $X$ is a 2D-standard Gaussian. So it is basically a geometric statement about the Gaussian area of a certain portion of the plane.

If we replace $-1$ and $1$ by $0$, then I can prove that

$$\mathsf{P}\left(g<0,h>0\right)=\frac{\arccos(1-\alpha)}{2\pi}$$

using rotational invariance of Gaussians. Is it possible to extend this reasoning to prove the above theorem?

答:
There is no positive $K$ that will make the inequality true for all $\alpha \in (0,2)$.

This proof works more generally to show that for any positive $\epsilon$, there is no positive $K$ for which the inequality
$$P[X<-\epsilon,Y>\epsilon]\ge K \alpha$$
holds for all $\alpha \in (0,2)$.

Divide both sides by $\alpha$ to re-write the inequality

$$\frac{P[X<-\epsilon,Y>\epsilon]}{\alpha} \ge K $$

The probability in the numerator can be written as

$$\int_{-\infty}^{-\epsilon} \int_{\epsilon}^{\infty}f_{X,Y,\alpha}(x,y)dx dy$$
where $f_{X,Y,\alpha}(x,y)$ is the bivariate normal density function given standard normal variables with correlation $1-\alpha$.

First, show that this probability is increasing as a function of $\alpha$ over the interval $(0,2)$.

Differentiate with respect to $\alpha$ under the integral sign and then integrate to find the derivative with respect to $\alpha$.

The derivative is

$$\frac{d}{d\alpha} \left(\int_{-\infty}^{-\epsilon} \int_{\epsilon}^{\infty}f_{X,Y,\alpha}(x,y)dx dy \right)
= \int_{-\infty}^{-\epsilon} \int_{\epsilon}^{\infty} \frac{d}{d\alpha}f_{X,Y,\alpha}(x,y)dx dy$$
$$=\frac{e^{-\frac{\epsilon^2}{\alpha}}}{2 \pi \sqrt{\alpha(2-\alpha)}}$$.

This is always positive.

Hence, the probability is strictly increasing as a function of $\alpha$.  The minimum occurs as the limit as $\alpha$ approaches 0.  In the limit, $X=Y$.  Since $X$ can't be both less than $-\epsilon$ and greater than $\epsilon$ simultaneously, the infimum is $0$.  The maximum probability occurs as the limit when $\alpha$ approaches 2.  The supremum is when $X=-Y$ and that probability is $P[X<-\epsilon]$.

Now, return to the ratio of this probability divided by $\alpha$. Find the limit as $\alpha$ goes to $0$. Use L'Hospital's rule to deduce that the limit is the same as the limit of the derivative of the numerator as $\alpha$ goes to $0$.

$$\lim_{\alpha \to 0^+} \frac{P[X<-\epsilon,Y>\epsilon]}{\alpha}=\lim_{\alpha \to 0^+} \frac{e^{-\frac{\epsilon^2}{\alpha}}}{2 \pi \sqrt{\alpha(2-\alpha)}}=0$$.

The reason there exists a positive $K$ that works in the case $\epsilon=0$ is in that case
$$\lim_{\alpha \to 0^+} \frac{P[X<-\epsilon,Y>\epsilon]}{\alpha}=\infty$$


问:
Finitely number natural $n$ such that $2^{n}=xy-1$ with $x,y$ primes number and $x\not=y$

If $x\not=y$ with $x,y$ primes number. Are there finitely many $n\in \mathbb{N}$ such that
$2^n=xy-1$?

答:
Hint: They are the sequence A092559 of integers $n$, such that $2^n+1$ is semiprime.

It starts with

$$
n=3, 5, 6, 7, 11, 12, 13, 17, 19, 20, 23, 28, 31, 32, 40, 43, 61, 64,\cdots
$$

It looks like that there are infinitely many such numbers, but I am not aware of a proof.


问:
Formal definition of a complement of a language.
I am in the midst of learning decidable and undecidable language and I came across the following theorem:

A language  $A \subseteq \Sigma^*$ is decidable if and only if $A$ as well as $\Sigma^* \setminus A$ are semi-decidable.

I'm not sure what $\Sigma^* \setminus A$, the complement of the language $A$, is supposed to mean. A language is a set of strings, so would its complement then be the set of all strings that are not in $A$? Can anyone provide a more formal explanation and definition?

答: $\Sigma^*\setminus A = \{x\mid x\in\Sigma^*\wedge x\not\in A\}$.


问:
Create a number by multi by $2$ and divide by $3$ (integer part)
How can I create a given postitive integer $N$ by multi by $2$ and divide by $3$ (integer part) ?

(Write a computer program is allowed)
For example:
$$100 = 2*2*2*2*2*2*2*2*2*2*2/3/3/3/3*2*2$$

(The numbers of $2, 3$ need to be minimum)

($/3$ mean $\text{div}\ 3$, eg: $2/3 = 0, 4/3 = 1...$)

Is solving equation

$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$
help ?

If it help then, how to solve
$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$

such that $a > c, a+b+c$ is minimum.

答:
While the 2nd answer of mine provides a (restricted) brute force method this method cannot be used when the length of the decomposition is over 30 or so: the computation time grows exponentially with the length. (If we use for the available digits for the residues modulo 3 only the residues $\{0,1\}$ we have a $2^\text{length}$ instead of a $3^\text{length}$-problem, but we see, that the method shall come to its limit soon.)

Here I've applied my "strictly decreasing" algorithm on some example numbers $a_0$ taken from Eric Roosendaal's Collatz-pages. I think that are well "complicated" numbers: record holders for glide-, delay- and path- records; and the "strictly decreasing" algo can handle numbers with dozens (or likely hundreds) of digits. The trajectory-length of the "strictly decreasing" algo is in all cases drastically shorter than the original Collatz-trajectory. On one hand there are examples where the "strictly decreasing" decompositions are not optimal, but they might be the best systematic approach available when decomposition length becomes large and memory and/or computation time becomes galactic when full combinatorical brute force is no more applicable. (This reminds me, btw., of the problem of "perfect rulers" and "optimal rulers", I think in Wikipedia is an entry)

I give the table not in mathjax because I'd like to have colors etc, and that's not my expertise... So here is a jpeg of the table, and below the raw data.

```<code>Roosendaal :
here: idx= index in E.R.'s webpage
idx   a0                  collatz  collatz   collatz    stric dec       optimal??
                           N (3^N)  S (2^S)  length(N+S)# of tokens
      page: Glide-records
32    180352746940718527    714       1189      1903        241
20            1200991791    326        547       873        113
10                626331    189        319       508         84
      page: Delay-records
100   14022512981985        590        979      1569        199
96     3743559068799        583        966      1549        179
66        1341234558        370        617       987        134
30             26623        113        194       307         77
      page: Pathrecords
93  274133054632352106267   672       1133      1805        280
88    1980976057694848447   547        928      1475        216
      page: (other)
30           903             41         75       116         54

Eric Roosendaal: http://www.ericr.nl/wondrous/
Glide-records: http://www.ericr.nl/wondrous/glidrecs.html
Delay-records: http://www.ericr.nl/wondrous/delrecs.html
Pathrecords: http://www.ericr.nl/wondrous/pathrecs.html
</code>```

The Pari/GP-code is the following (only little reworked for readable, I'll improve this later)

```<code>\\ Pari/GP
\\ from some a0 find between a1=a0*3^e3 ... a1+(3^e3 -1) a number a2 divisible by
\\    2^e2 such that a3=a2/2^e2 < a0 . If for e3=1 there is no such number,
\\ increase e3 until success or e3>maxe3.
\\ Return a short list of first solutions with increasing e3
\\
\\ user routine "mcc( [vectorv,vectorv,...,vectorv])" returns concatenation
\\  to rectangular matrix and "VE()" extracts leading rows&columns of matrix
\\
{findstepdown(a0,maxe3=4)=my(aa,ae,a1,p3,a2,m2,r1,A,res_a,res_A,res_e3,res_r,il);
      res_a=res_e3=res_A=res_r=vectorv(maxe3);il=0; \\ res_ulting value-vectors
      for(e3=1,maxe3,
          p3=3^e3;[aa=p3*a0,ae=aa+p3-1,p3-1,m2=2^ceil(ld3*e3)];
          [r1= (-aa % m2),p3-1] ; \\ >p3-1
          if(r1>=p3,next());      \\ residue must be in interval a1..a1+3^e3-1
          A=valuation(aa +r1,2);a1=(aa+r1)>>A;
          il++;
          res_e3[il]=e3;res_a[il]=a1;res_A[il]=A;res_r[il]=r1;
               \\ fill list of result vectors
       ); return(VE(mcc([res_e3,res_a,res_A,res_r]),il,4));
   }

\\ calls "findstepdown()" repeatedly until a1 approaches 1
{strictlydecreasing(a0,maxstepdown=8)=my(lambda,st,st1,st2,res,res1,e3,e2,a1,r1);
 lambda=0; st=Str("");a1=a0;print1(a0,":");
 for(k=1,40,
     res= findstepdown(a1,maxstepdown);
          res1=res[1,]; \\ use always first stepdown-result!
          e3=res1[1];a1=res1[2];e2=res1[3];r1=res1[4];lambda+=(e2+e3);
          print1([a1,lambda]);
          st1=st2="";
             for(k=1,e2,st1=Str(st1,"*2"));
             for(k=1,e3,st2=Str(st2,"/3"));
             st=Str(st1,st2,st);
     if(a1==1,break())
     );
  print("");
  print(a0,"=",Str(1,st," (#"lambda")")); }

  \\ apply this
  a0=4535
  strictlydecreasing(a0) \\ prints a decomposition of a0
</code>```

Exemplary results: decompositions of two not too large values $a_0$:

These are also a challenge for competing methods to find better/optimal decompositions.

```<code>legend: -------------------------------------------------------------
  a_0 :[consecutive a_k, cumulative partial length] [...,...] ...
626331:[528467, 8][297263, 14][3919, 28][3307, 36][2791, 44][785, 51][589, 54][221, 58][83, 62][47, 68][5, 79][1, 84]
  a_0 = decomposition-string beginning at 1 growing towards a_0:
626331=1*2*2*2*2/3*2*2*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2/3*2*2*2/3*2*2/3*2*2*2*2*2/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2*2*2/3/3/3 (#84)

26623:[14215, 19][1999, 27][1687, 35][949, 41][89, 47][67, 50][19, 57][11, 63][5, 72][1, 77]
26623=1*2*2*2*2/3*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2*2*2/3/3*2*2/3*2*2*2*2*2/3*2*2*2*2/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2*2/3/3*2*2*2*2*2*2*2*2*2*2*2*2/3/3/3/3/3/3/3 (#77)
</code>```

Body:
While the 2nd answer of mine provides a (restricted) brute force method this method cannot be used when the length of the decomposition is over 30 or so: the computation time grows exponentially with the length. (If we use for the available digits for the residues modulo 3 only the residues $\{0,1\}$ we have a $2^\text{length}$ instead of a $3^\text{length}$-problem, but we see, that the method shall come to its limit soon.)

Here I've applied my "strictly decreasing" algorithm on some example numbers $a_0$ taken from Eric Roosendaal's Collatz-pages. I think that are well "complicated" numbers: record holders for glide-, delay- and path- records; and the "strictly decreasing" algo can handle numbers with dozens (or likely hundreds) of digits. The trajectory-length of the "strictly decreasing" algo is in all cases drastically shorter than the original Collatz-trajectory. On one hand there are examples where the "strictly decreasing" decompositions are not optimal, but they might be the best systematic approach available when decomposition length becomes large and memory and/or computation time becomes galactic when full combinatorical brute force is no more applicable. (This reminds me, btw., of the problem of "perfect rulers" and "optimal rulers", I think in Wikipedia is an entry)

I give the table not in mathjax because I'd like to have colors etc, and that's not my expertise... So here is a jpeg of the table, and below the raw data.

Roosendaal :
here: idx= index in E.R.'s webpage
idx   a0                  collatz  collatz   collatz    stric dec       optimal??
                           N (3^N)  S (2^S)  length(N+S)# of tokens
      page: Glide-records
32    180352746940718527    714       1189      1903        241
20            1200991791    326        547       873        113
10                626331    189        319       508         84
      page: Delay-records
100   14022512981985        590        979      1569        199
96     3743559068799        583        966      1549        179
66        1341234558        370        617       987        134
30             26623        113        194       307         77
      page: Pathrecords
93  274133054632352106267   672       1133      1805        280
88    1980976057694848447   547        928      1475        216
      page: (other)
30           903             41         75       116         54

Eric Roosendaal: http://www.ericr.nl/wondrous/
Glide-records: http://www.ericr.nl/wondrous/glidrecs.html
Delay-records: http://www.ericr.nl/wondrous/delrecs.html
Pathrecords: http://www.ericr.nl/wondrous/pathrecs.html

The Pari/GP-code is the following (only little reworked for readable, I'll improve this later)

\\ Pari/GP
\\ from some a0 find between a1=a0*3^e3 ... a1+(3^e3 -1) a number a2 divisible by
\\    2^e2 such that a3=a2/2^e2 < a0 . If for e3=1 there is no such number,
\\ increase e3 until success or e3>maxe3.
\\ Return a short list of first solutions with increasing e3
\\
\\ user routine "mcc( [vectorv,vectorv,...,vectorv])" returns concatenation
\\  to rectangular matrix and "VE()" extracts leading rows&columns of matrix
\\
{findstepdown(a0,maxe3=4)=my(aa,ae,a1,p3,a2,m2,r1,A,res_a,res_A,res_e3,res_r,il);
      res_a=res_e3=res_A=res_r=vectorv(maxe3);il=0; \\ res_ulting value-vectors
      for(e3=1,maxe3,
          p3=3^e3;[aa=p3*a0,ae=aa+p3-1,p3-1,m2=2^ceil(ld3*e3)];
          [r1= (-aa % m2),p3-1] ; \\ >p3-1
          if(r1>=p3,next());      \\ residue must be in interval a1..a1+3^e3-1
          A=valuation(aa +r1,2);a1=(aa+r1)>>A;
          il++;
          res_e3[il]=e3;res_a[il]=a1;res_A[il]=A;res_r[il]=r1;
               \\ fill list of result vectors
       ); return(VE(mcc([res_e3,res_a,res_A,res_r]),il,4));
   }

\\ calls "findstepdown()" repeatedly until a1 approaches 1
{strictlydecreasing(a0,maxstepdown=8)=my(lambda,st,st1,st2,res,res1,e3,e2,a1,r1);
 lambda=0; st=Str("");a1=a0;print1(a0,":");
 for(k=1,40,
     res= findstepdown(a1,maxstepdown);
          res1=res[1,]; \\ use always first stepdown-result!
          e3=res1[1];a1=res1[2];e2=res1[3];r1=res1[4];lambda+=(e2+e3);
          print1([a1,lambda]);
          st1=st2="";
             for(k=1,e2,st1=Str(st1,"*2"));
             for(k=1,e3,st2=Str(st2,"/3"));
             st=Str(st1,st2,st);
     if(a1==1,break())
     );
  print("");
  print(a0,"=",Str(1,st," (#"lambda")")); }

  \\ apply this
  a0=4535
  strictlydecreasing(a0) \\ prints a decomposition of a0

Exemplary results: decompositions of two not too large values $a_0$:

These are also a challenge for competing methods to find better/optimal decompositions.

legend: -------------------------------------------------------------
  a_0 :[consecutive a_k, cumulative partial length] [...,...] ...
626331:[528467, 8][297263, 14][3919, 28][3307, 36][2791, 44][785, 51][589, 54][221, 58][83, 62][47, 68][5, 79][1, 84]
  a_0 = decomposition-string beginning at 1 growing towards a_0:
626331=1*2*2*2*2/3*2*2*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2/3*2*2*2/3*2*2/3*2*2*2*2*2/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2*2*2/3/3/3 (#84)

26623:[14215, 19][1999, 27][1687, 35][949, 41][89, 47][67, 50][19, 57][11, 63][5, 72][1, 77]
26623=1*2*2*2*2/3*2*2*2*2*2*2/3/3/3*2*2*2*2/3/3*2*2*2*2*2/3/3*2*2/3*2*2*2*2*2/3*2*2*2*2/3/3*2*2*2*2*2/3/3/3*2*2*2*2*2*2/3/3*2*2*2*2*2*2*2*2*2*2*2*2/3/3/3/3/3/3/3 (#77)


问:
If $f_n$ is the $n$th Fibonacci number then $\lim_{n\to \infty} (f_{n-2}+{1-\sqrt 5\over 2}f_{n-1})$?
If $f_n$ is the $n$th Fibonacci number then what is  the value of $$\lim_{n\to \infty}\left(f_{n-2}+{1-\sqrt 5\over 2}f_{n-1}\right)?$$
With direct calculation I saw the first four terms are approximately $(-0.62,0.39,-0.24,0.15)$. It seems that the limit approaches $0$. But how to show it analytically?

答:
We have $\psi^{n}=f_{n}\psi +f_{n-1}$, where $\psi={1-\sqrt 5\over 2}$. This is easily proved by induction since $\psi^2=\psi+1$. Therefore,
$$
f_{n-2}+{1-\sqrt 5\over 2}f_{n-1} = f_{n-2}+\psi f_{n-1} = \psi^{n-1} \to 0
$$
because $|\psi|<1$.


问:
Can we relate $H^i(Y,\mathscr{G})$ and $H^i(X,f^*\mathscr{G})$ when $f:X\to Y$ is a morphism of schemes?
Let $f:X\to Y$ be a morphism of schemes. If $\mathscr{F}$ is a sheaf over $X$ then I know that
$$H^i(X,\mathscr{F})\cong H^i(Y,f_*\mathscr{F})\qquad\text{for all}\quad i\geq 0$$
whenever $j$ is a closed immersion or $\mathscr{F}$ is quasi-coherent and $f$ is an affine morphism between noetherian separated schemes.

I wonder if there's some similar relation for the inverse image.

答:
Question: "I wonder if there's some similar relation for the inverse image."

Answer: Let $Y:=Spec(A)$ be a noetherian affine scheme and let $\mathcal{E}$ be an $\mathcal{O}_Y$-module. The isomorphism you write down above holds for any affine morphism of noetherian separated schemes and $\mathcal{F}$ quasi coherent.

Let us assume there is an isomorphism

$$ H^i(X, f^*\mathcal{E})\cong H^i(Y, \mathcal{E}).$$

The right hand since is zero for $i\geq 1$ since $Y$ is affine (Hartshorne III.3.5).

It seems the left hand side is not zero in general, but I do not  have an immediate example. If $E$ is a non-trivial finite rank projective $A$-module, it follows $f^*\mathcal{E}$ will be a non-trivial locally free sheaf on $X$ and this should have non-zero cohomology groups in general if $X$ is non-affine.

Example: If $X:=Spec(B)$ there is a canonical map

$$ \rho: H^0(Y, \mathcal{E}) \cong E \rightarrow E\otimes_A B\cong H^0(X, f^*\mathcal{E})$$

defined by

$$\rho(e):=e\otimes 1$$

and the map $\rho$ is seldom an isomorphism (even when $f$ is flat).

Example: Let $A:=k$ a field and $dim(B) \geq 1$.
It follows $B\cong \oplus_{i \in I}ke_i$ with $\# I=\infty$

$$E\otimes_A B\cong \oplus_{i\in I} Ee_i \neq E.$$

The map $f$ is flat since $B$ is a flat $A$-algebra. Hence the map $\rho$ is never an isomorphism for $X$ affine of positive dimension over a field $k$.

Example: If $B:=A/I$ we get a canonical map

$$\rho: E \rightarrow B\otimes_A E\cong E/IE=E$$

since $IE=(0)$. Hence when $f$ is a closed immersion the map $\rho$ is always an isomorphism.


答: Hint: one way would be to represent rotations and reflections by canonical matrix representatives, and then compute the matrix products.

Title: Composition of Linear Rotations and Reflections

Body:
Prove that if $T_{1}$ is a rotation of $R^{2}$ about O, and $T_{2}$ a reflection in a line through O, then $T_{1}$$\circ$$T_{2}$ and $T_{2}$$\circ$$T_{1}$ are both reflections in a line through O.

I'd prefer a hint than the answer, because I'm not sure how to begin. Thanks in advance for any help.


答:
You can avoid the exponentials by a simple trick using the standard limit $\lim_{y\to 0}\frac{e^y-1}y=1$.

\begin{eqnarray*}\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}
& = & e\cdot \frac{\left(1-e^{\frac{\ln(1+\arctan x)-x}{x}}\right)}{x}\\
& \stackrel{y=\frac{\ln(1+\arctan x)-x}{x}}{=} & e\cdot \underbrace{\frac{1-e^y}y}_{\stackrel{x\to 0}{\longrightarrow}-1}\cdot \frac{\ln(1+\arctan x)-x}{x^2}
\end{eqnarray*}
Here I use that $\lim_{x\to 0}\frac{\ln(1+\arctan x)-x}{x} =0$, which can quickly be shown by L'Hospital.

Now, we only have to deal with

\begin{eqnarray*}\frac{\ln(1+\arctan x)-x}{x^2}
& \stackrel{L'Hosp.}{\sim} & \underbrace{\frac 1{(1+x^2)(1+\arctan x)}}_{\stackrel{x\to 0}{\longrightarrow}1}\cdot \frac{1-(1+\arctan x)(1+x^2)}{2x}
\end{eqnarray*}

Finally,

\begin{eqnarray*}\frac{1-(1+\arctan x)(1+x^2)}{2x}
& \stackrel{L'Hosp.}{\sim} & \frac{-(1+\arctan x)2x-(1+x^2)\frac 1{1+x^2}}{2}\\
& \stackrel{x\to 0}{\longrightarrow} & -\frac 12\\
\end{eqnarray*}

All together

$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \stackrel{x\to 0}{\longrightarrow} e\cdot (-1) \cdot \left(-\frac 12\right) = \frac e2 $$

Body:
I have to evaluate this limit:
$$\lim_{ x \to 0 }  \frac{e-(1+\arctan x)^{\frac{1}{x}}}{x}$$

I have an indeterminated form $[\frac{0}{0}]$.
So I applied the Hopital rule
$$\frac{e-(1+\arctan x)^{\frac{1}{x}}}{x} \sim $$
$$-e^{\frac{\log(1+\arctan x)}{x}} \cdot \frac{\frac{x}{(1+\arctan x)(1+x^2)}-\log(1+\arctan x)}{x^2} \sim$$
$$ -e \cdot \frac{1}{x^2} \cdot \left(\frac{x}{(1+x)(1+x^2)}-x\right)
\sim +e $$

According to my book and wolfram alpha the final result should be $\frac{e}{2}$ and I don't know where I made mistakes.


问:
How to define morphisms using coordinates in scheme theory?
It is very usual in scheme theory to define morphisms between schemes using coordinates. For example: one could say that the morphism of rings

\begin{align*}
B^n\times B^m &\to B^{nm}\\
(x_1,\dotsc,x_n,y_1,\dotsc,y_m)&\mapsto (x_iy_j)_{ij}
\end{align*}

induces a morphism of schemes $\mathbb{A}_A^n\times_{\operatorname{Spec}A}\mathbb{A}_A^m\to \mathbb{A}_A^{nm}$. I think this comes from the functor of points somewhat, but I don't know precisely how. I would be happy if someone could explain that in detail.

答:
You find a brief description of the "functor of points" here:

https://math.stackexchange.com/posts/3977439/edit

Let $k$ be any field and let $B:=k[x_1,..,x_n]$ be the polynomial ring in $n$ variables. Let $I \subseteq B$ be an ideal with $A:=B/I$ and let $X:=Spec(A)$ and let $S:=Spec(k)$.

Let

R0. $X(k)$ be the set of $n$-tuples $(a_1,..,a_n)\in k^n$ with $f(a_1,..,a_n)=0$ for all polynomials $f \in I$.

R1. Let $Hom_{k-alg}(A,k)$  be the set of all maps of $k$-algebras $\phi: A \rightarrow k$.

Lemma. There is a one-to-one correspondence of sets $X(k) \cong Hom_{k-alg}(A,k)$.

If $a:=(a_1,..,a_n)\in X(k)$ we define a map $\phi_a: A\rightarrow k$ by letting $\phi_a(x_i):=a_i$. Why is the map $\phi_a$ well defined? To give a map of $k$-algebras $\psi: k[x_1,..,x_n]\rightarrow k$ is equivalent to specify how $\psi$ acts on the variables $x_j$. Hence we must define

$$\phi_a((f(x_1,..,x_n)):=f(a_1,..,a_n) \in k.$$
If $g(x_1,..,x_n)\in I$ it follows $\phi_a(g)=0$ by definition, hence we get a well defined map

$$\phi_a: A \rightarrow k$$

of $k$-algebras.

Conversely if $\phi: A\rightarrow k$ is a map of $k$-algebras it follows $\phi(x_i)=a_i \in k$ for all $i=1,..,n$. Since $\phi$ is well define it follows $$\phi(g(x_1,..,x_n))=g(a_1,..,a_n)=0$$

for all $g\in I$, hence it follows

$$ g(a_1,..,a_n)=0$$
for all $g\in I$ and it follows $a:=(a_1,..,a_n)\in X(k)$ by definition. You may prove that his is a one-to-one correspondence of sets. QED.

I include the following discussion in the post:

Question: "Dear @hm2020, I understand that morphisms of k-algebras $A \rightarrow k$ are determined by n-tuples $(a_1,…,a_n)\in k^n$ which satisfy $f(a_1,..,a_n)=0$ for all $f \in I$. What I don't understand is why, if $X$ and $Y$ are $A$-schemes, it suffices to define morphisms $X(B)\rightarrow Y(B)$ for all $A$-algebras $B$ in order to determine a morphism $X\rightarrow Y$. (I understand that it suffices to give $X(S)\rightarrow Y(S)$ for all $A$-schemes $S$ by the way.)"

Answer: This is "deeper" in a sense. The functor of points $h_X$ is a "sheaf in the zariski topology". If a functor $F$ is a sheaf in the zariski topology and if $F$ has an open cover of representable functors, it follows $F$ itself is representable. You need to investigate the "embedding"

$$Sch(S)⊆Funct(Sch(S),Sets)$$

and what it means for a functor $F$ to have an "open cover of representable sub-functors". This leads naturally to other grothendieck topologies (the etale topology etc), and it is a good exercise to do this construction. A scheme $X$ has an open cover of affine schemes $U_i:=Spec(A_i)$, and similarly the functor of points $h_X$ of $X$ has an open cover $h_{U_i}$ of the functor of points of the affine schemes $U_i$. To define maps of functors $h_X \rightarrow h_Y$ you can do this on an "open affine cover" and prove that this "glues" to a global map. A good exercise is to prove that the projective space bundle functor (or grassmannian functor) is representable. You do this by proving it is a sheaf in the zariski topology and by giving an open cover of representable sub-functors.  This is essentially Proposition II.7.12 in Hartshornes book. There are books where this is written out in complete details but I do not have a good reference. I believe you also find articles on this on the arXiv preprint server. It is not extremeley "deep" - it is a systematic use of the Yoneda lemma and the functor of points.


问:
Determining a confidence interval for $\sigma$ from a Rayleigh distribution
Hello stackexchangers,

Suppose we have $n$ Rayleigh distributions defined by
$$f_X(x)=\frac{x}{\sigma^2}e^{-x^2/2\sigma^2}.$$
How would you go about determining an approximative confidence interval for $\sigma$?
I got this problem from a dear friend, and he suggested that I use the method of least-squares on $\sigma$.

Guided by his wisdom, I found the least-squares prediction to be
$$\sigma^*=\bar{x}\sqrt{\frac{2}{\pi}},$$
where $\bar{x}$ is the mean value, but I am completely lost as to how to proceed.

Any hints would be greatly appreciated.

Your Mayor,

Ron Ford

答:
I will give you a rough approximation. Because we have to start somewhere, we boldly assume that $n$ is large enough for the central limit theorem to hold for $\sigma^*=\bar{X}\sqrt{\frac{2}{\pi}}$; note that your point estimator $\sigma_{\text{obs}}^*=\bar{x}\sqrt{\frac{2}{\pi}}$ of the scale parameter $\sigma$ is an observation of the random variable $\sigma^*$.

Now, since $\sigma^*$ is approximately normally distributed with expected value $\sigma$ and standard deviation $D[\sigma^*]=f(\sigma)$, an approximate confidence interval for the estimator is given by $I_\sigma=(\sigma_{\text{obs}}^*\pm\underbrace{\lambda_{0.025}}_{1.96}d)$, where we calculate $d$ as follows:

$$D[\sigma^*]=\sqrt{V[\sigma^*]}=\sqrt{V[\bar{X}\sqrt{\frac{2}{\pi}}]}=\sqrt{\frac{2}{\pi}V[\bar{X}]}=\sqrt{\frac{2}{\pi}\frac{1}{n}V[X]}=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}\sigma^2}\\\implies d=\sqrt{\frac{2}{\pi}\frac{1}{n}\frac{4-\pi}{2}{\sigma_{\text{obs}}^*}^2}$$


问:
Does dx in dA=dx*dy represent change, or is it a notation that denotes an infinitesimal arbitrary length? Change vs. an arbitrary physical length
I'm not sure how to explain this, but I have a gap in my understanding of infinitesimals/differentials. I've so far had calc 1 and 2, and have been taught that dy/dx represents a slope, which represents a change. So, my current understanding is if I had a function dy/dx(x) and plugged in x, I could deduce how much of a change in y I could expect, if I moved a unit x, at that particular x of interest.

Where my understanding falls apart is, I don't see how dx in a slope, which represents change, would also represent "change" in dA=dx*dy. I mean in the classical meaning of the word change. I know we could say "well in the same way delta X is a change in x and represents length", but I don't view it as such.

Not sure if this is related, but here's another thing bothering me, say we have P=dF/dA. Why dF and not only F? I understand dA would represent this infinitesimal area, but what does dF represent then? It's supposed to be a change in F, right? Why do we need a change?

I'd really appreciate someone clearing this up or me, because I'm expected to pass a fluid mechanics test while our uni doesn't even offer a calc 3 course. You can understand my frustration.

Thanks

答:
Well, if you take any function $f$ which admits derivative, then you may write:
$$f'(x)\qquad\text{or}\qquad\frac{df}{dx}(x)=\frac{d}{dx}f(x)$$
which are considered equivalent in notation. I guess you have been explained to take a variation $\Delta f$ and $\Delta x$, whose ratio represents a line which passes through the plot of $f$ in (at least) $2$ points. Then you think of derivation as making that line a tangent line on a point $x_{0}$. With 2 variables, say $x$ and $y$, the idea is somewhat the same. If $f(x,y)$ is now a real function in two variables, the plot will be in $\mathbb{R}^{3}$ thus it is a sort of surface. In this new setting, the idea of tangent line sounds appealing but not good enough: if you imagine a sphere and a point on its surface, then you realize there are many tangent lines through that point: but only 1 tangent plane. A variation $\Delta x\Delta y$ is now a variation in the area, so we may write $\Delta A=\Delta x\Delta y$. Passing to infinitesimal variations, we would have $d A=dxdy$. Note this little discussion lacks of mathematical rigorousness and it is meant only to help you visualize what's going on.

In terms of notation, $P=\frac{dF}{dA}$ can be written as $P=\frac{d}{dA}F$ if it pleases you more.


问:
"Generalized eigenvectors corresponding to distinct eigenvalues are linearly independent" for infinite-dimensional spaces?
This is form Axler's Linear Algebra Done Right.
Please allow me to borrow the screenshot from this Question :
8.13 Linearly independent generalized eigenvectors

From the above picture, we can see that in the proof of 8.13 the author uses the condition "V is finite-dimensional".

My question is that, if V is an infinite-dimensional space, does this conclusion still hold?

Thanks!

答:
If $v_1,\cdots,v_m$ are generalized eigenvectors for distinct eigenvalues $\lambda_1,\cdots,\lambda_m$, then the subspace
\begin{equation}
W = \text{span}\{T^i v_j: i\in{\mathbb N}, j\in\{1,\cdots,m\}\}
\end{equation}
is finite dimensional and stable by $T$. The $v_j$'s are generalized eigenvectors of the restriction of $T$ to $W$, hence the theorem applies and they are linearly independent.

That said, the spectral properties of operators in infinite dimensional spaces are not limited to just eigenvalues, eigenvectors and generalized eigenvectors.


问:
theory of equation
let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
 ( c+d)^{-1}+ (c+d)^{-1}$$

my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck

答:
Notice that

$$\frac{1}{a+b} + \frac{1}{c+d} = \frac{a+b+c+d}{(a+b)(c+d)} = \frac{3}{(a+b)(c+d)}.$$

Repeat for two more sets of two fractions. Then add the three results.

Title: theory of equation


问:
Singularity at 0 is at most a pole of order 7 if the complex function is 1/4 integrable.
So the real question is, given f analytic in the punctured disk $(D_1 \backslash \{ 0\})$ and $\int_{D_1} |f(z)|^{1/4} < \infty$, characterize the smoothness of f at 0.

It is easy to show that if $z=0$ is a pole, then it is at most of order 7.

However, I cannot show that it cannot be an essential singularity. (Or find a counter example)

Remark, if instead we had that $\int_{D_1} |f(z)|^p < \infty$, for $p >=1$, it is relatively easy to show that it cannot be an essential singularity, however it heavily depends on Holder inequality.

答:
Lemma: For an entire non constant function $g$ and $p >0$ we cannot have $\int_{2 \le |z| \le R}|g(z)|^p/|z|^2dA \le A+B \log R$ for positive constants $A,B$ and arbitrary $R >2$

Assume the Lemma and consider $z^8f(z)=\sum_{n=-\infty}^{\infty}a_nz^n=h(z)+k(z)$ the Laurent expansion of $f$ on the unit disc decomposed in the nonpositive power part $h$ and the (strictly) positive power part $k$, so $h(z)=\sum_{n=-\infty}^{0}a_nz^n, k(0)=0$ and $g(w)=h(1/w)$ is entire

Since $k$ is analytic in the unit disc, $\max_{|z| \le 1/2}|k(z)|=C$ finite, so using the inequality $|a|^p+|b|^p \ge |a+b|^p, 0 \le p \le 1$ we get $|h(z)|^{1/4} \le |z|^2|f(z)|^{1/4}+C^{1/4}, |z| \le 1/2$ or

$|h(z)|^{1/4}/|z|^2 \le |f(z)|^{1/4}+C^{1/4}/|z|^2, A_r=r \le |z| \le 1/2$ hence integrating on the annulus $r \le |z| \le 1/2$ we get (using the hypothesis for $f$):

$\int_{A_r}|h(z)|^{1/4}dA_z/|z|^2 \le C_1-C_2\log r$

Changing variables $w=1/z, dA_w=|w'(z)|^2dA_z=dA_z/|z|^4=|w^2|dA_z/|z|^2$ and $B_R$ the annulus $2 \le |w| \le R=1/r$ we get:

$\int_{B_R}|g(w)|^{1/4}dA_w/|w|^2 \le C_1+C_2\log R$, so by the Lemma we get that $g$ is constant, hence $z^8f$ analytic in the unit disc, so the conclusion that $f$ has a pole of order at most $8$, but then we get an easy contradiction with the hypothesis when the order is actually $8$

The Lemma follows from the fact that for any nonnegative number $x$ we have $p\log x \le x^p, p>0$ where $\log 0 = -\infty$ so first assuming $f(0) \ne 0$ we have $p\log |f(w)|-2\log |w| \le |f(w)|^p/|w^2|$ and now integrating on the annulus $B_R$ and using Jensen and the given inequality, it follows that $f$ has finitely many zeroes outside $|z| \ge 2$ hence in the plane and writing $f=Pe^g$ it immediately follows that $g$ must be constant, while for a polynomial $P$ of degree $n \ge 1$ it is clear that $\int_{2 \le |z| \le R}|P(z)|^p/|z|^2dA$ diverges like $R^{np}$ for large $R$ (since $|P|$ behaves like $R^n$ there) so it cannot be $O(\log R)$ and we are done for this case.

But if $f(0)=0$ writing $f(z)=z^kh(z)$ we can repeat the above for $h$ as the inequality only improves after all for large $z$ so we are done in general!


问:
Composition of Linear Rotations and Reflections
Prove that if $T_{1}$ is a rotation of $R^{2}$ about O, and $T_{2}$ a reflection in a line through O, then $T_{1}$$\circ$$T_{2}$ and $T_{2}$$\circ$$T_{1}$ are both reflections in a line through O.

I'd prefer a hint than the answer, because I'm not sure how to begin. Thanks in advance for any help.

答: You can assume without loss of generality that $O$ is the origin point. Then write the rotation matrix as $(\cos[\theta],-\sin[\theta]),(\sin[\theta],\cos[\theta])$. Try to write the reflection matrix out as well and see how the multiplication goes. If you are stuck, Artin's book may be a good reference. </p>


问:
Kodaira dimension of an elliptic curve is zero
Let $C$ be an elliptic cure, i.e. a projective curve over $\Bbb{C}$ having genus $1$.

I'm trying to understand why the Kodaira dimension of $C$ is zero (according to this article).

I know that the canonical divisor $K_C$ is trivial, so that $H^0(C,\mathcal{O}_C(nK_C))=H^0(C,\mathcal{O}_C)=\Bbb{C}$ for all $n\geq 0$.

This way, the canonical ring $\bigoplus_{n\geq 0}H^0(C,\mathcal{O}_C(nK_C))$ is just $\Bbb{C}$, so its field of fractions is also $\Bbb{C}$.

By definition of Kodaira dimension, $\kappa(C)=\text{trdeg}_{\Bbb{C}}\Bbb{C}-1=0-1=-1$.

What am I missing?

答:
The canonical ring $R=\bigoplus_{n\geq 0}H^0(C,\mathcal{O}_C(nK_C))$ is not just $\Bbb{C}$, because it has a copy of $\Bbb{C}$ at every (positive) grading, i.e. it is the graded ring $R\cong\Bbb{C}[x]$  with $deg(x)=1$.

Thus $\kappa(C)=\text{trdeg}_{\Bbb{C}}\Bbb{C}(x)-1=1-1=0$.


问:
Trajectory of stationary stochastic process not in $\mathrm{L}^2$
Let $(X(t))_{t\in T}$ with $T\subseteq \mathbb{R}^n$ be a (strictly) stationary stochastic process. In the book 'Fourier Analysis and Stochastic Processes' Pierre Brémaud states on p.157 (Chapter 3.4.1) that the trajectory of $(X(t))_{t\in T}$ is almost surely not in $\mathrm{L}^2(\mathbb R)$ (he assumes $T\subseteq \mathbb R$). Unfortunately, he does not include a proof. I've got two questions regarding his statement:

Does someone know where I can find a proof of this statement or how to proof it?

Does this statement also hold for wide-sense stationary processes, i.e. are the trajectories of wide-sense stationary processes almost surely not in $\mathrm{L}^2(\mathbb R)$?

I would be happy if someone could help! Thanks a lot!

答:
Consider the case $T=\Bbb R$. Other choices of $T$ can be handled in like fashion, the key being that the Lebesgue measure of $\Bbb R$ is infinite. Let $v:=\Bbb E[X(t)^2]$, assumed finite and (to avoid triviality) strictly positive. Because $(X(t))$ is stationary, $v$ doesn't depend on $t$. Consider now the $L^2$ norm of the sample path $t\mapsto X(t)$.
Define
$$
Z(t):=\int_t^\infty X(s)^2\,ds.
$$
By stationarity, $Z(t)$ has the same distribution as $Z(0)$ for each $t>0$. But  you also have
$$
\lim_{t\to\infty}Z(t) = 0,
$$
at each point of  the event $\{Z(0)<\infty\}$. This implies that $\Bbb P[Z(0)=0$ or $+\infty] =1$.
It follows that
$$
\|X\|_2=\sqrt{\int_{\Bbb R} X(t)^2\,dt} = 0 \hbox{ or } +\infty,\quad\hbox{a.s.}
$$
The non-triviality hypothesis $v>0$ implies that $P[\|X\|_2 =+\infty]>0$.  This probability will therefore be $1$ if $X$ is ergodic, but it's easy to make (stationary, non-ergodic) examples in which $P[\|X\|=0]>0$.


问:
Definition of ring of regular functions
In Hartshornes "Algebraic Geometry" the ring of regular functions is defined over affine and quasi-affine varieties. In the case of an affine variety I don´t understand how this ring looks like. Because a regular function is only defined on open sets, we have to add and multiply regular functions which are defined on different subsets of the affine variety. How does that work?</p>

答:
Hartshorne defines the "ring of regular functions" as follows (Chapter I.3).
Let $k$ be an algebraically closed field and let $I\subseteq A:=k[x_1,..,x_n]$ be a prime ideal. Let $Z:=Z(I)$ be the zero set of $I$. Let $U\subseteq Z$ be a zariski open set and let $f:U\rightarrow k$ be a function.

Definition (D1). We say $f$ is regular at $p\in U$ iff there is an open set $p\in U(p) \subseteq U$ and functions $u,v\in A$ with $v\neq 0$ on $U(p)$, such that $f(x)=\frac{u(x)}{v(x)}$ on $U(p)$. we say $f$ is regular at $U$ iff f is regular at all points in $U$

Let $\mathcal{O}(U)$ be the set of functions that are regular at $U$.

Lemma. $\mathcal{O}(U)$ is a $k$-algebra.

Proof. Let $f,g$ be regular at $p\in U$. There are open sets $U(p), V(p)$ containing $p$ with $f=\frac{u}{v}$ on $U(p)$ and $g=\frac{w}{z}$ on $V(p)$, with $v,z\neq 0$. it follows $f+g=\frac{uz+wv}{vz}$ on $W(p):=U(p)\cap V(p)$, and $vz\neq 0$ on $W(p)$,
hence if $f,g\in \mathcal{O}(U)$ it follows $f+g\in \mathcal{O}(U)$. Similarly $fg\in \mathcal{O}(U)$ and if $a\in k$ it follows $af\in \mathcal{O}(U)$. The Lemma follows.

For any open sets $V\subseteq U$ there are canonical restriction maps (restriction of functions)

$$\rho_{UV}: \mathcal{O}(U) \rightarrow \mathcal{O}(V)$$

and you get (...) a sheaf of $k$-algebras $\mathcal{O}_Z$ on $Z$, with the property that
$(Z, \mathcal{O}_Z)$ is a (locally) ringed space.  It is a good exercise to verify the sheaf axioms from Hartshorne, Chapter II. The definition D1 is local and you may define this for any quasi projective variety as defined in HH Chapter I.
Hence it makes sense to speak about sheaves of functions for the varieties introduced in Chapter I in Hartshornes book.

Note: Using D1 it follows the ring $\mathcal{O}_Z(U)$ is a "ring of functions" on the open set $U$, and this is a bit "more intuitive" than the construction of the structure sheaf in Chapter II in the same book. In Chapter II in the same book the sheaf of sections $\mathcal{O}_X$ of a scheme $X$ is introduced, and you should think about sections $s\in \mathcal{O}_X(U)$ as "functions" taking values in the residue field $\kappa(x)$. Given a section $s$ you get canonically an element in the local ring at $x$, $s_x\in \mathcal{O}_{X,x}$ and you may define the "value" $s(x)\in \kappa(x)$ because of the canonical map

$$ ev_x: \mathcal{O}_{X,x} \rightarrow \kappa(x)$$

and by definition $s(x):=ev_x(s_x)\in \kappa(x)$. The element $s(x)$ is the "value" of the section $s$ at the point $x$. What is "unintuitive" here is that the residue field $\kappa(x)$ may change with the point $x$.

Here you will find a discussion of the "value" of a section $s$.

About the reducedness in algebraic geometry

Example. If $X:=Spec(\mathbb{R}[x])$ and $\mathfrak{m}:=(p_z(x))$ the maximal ideal generated by a polynomial of degree two with non-real roots, it follows the residue field is the field of complex numbers. If we want to "calculate" the value of a section $s$ at the point $x$ corresponding to $\mathfrak{m}$, we have to choose an explicit isomorphism

$$\kappa(x) \cong \mathbb{R}[x]/(p_z(x)) \cong \mathbb{C}.$$

We have two choices: A complex zero $z$ of the polynomial $p_z(x)$ or the conjugate $\overline{z}$. Hence the value of $s$ at $x$ depends on this choice. What does not depend on this choice is if the section $s$ is zero or not, and this is what we are interested in when we are doing algebraic geometry: We want to study the locus of points where the section $s$ is zero.

Example. If $X:=Spec(A)$ and $s\in \Gamma(X, \mathcal{O}_X)\cong A$ is a global section, it follows the "locus of points" $Z_s$ where the global section $s$ is zero has a natural scheme structure: It is the subscheme $V((s))\subseteq X$, where $(s)\subseteq A$ is the ideal defined by the element $s$. When studying sections of sheaves in algebraic geometry, you are mostly interested in the locus where the section is zero or where it is non-zero $D(s)\cong Spec(A_s)$. At the level of topological spaces you may write $X$ as a disjoint union

$$X=D(s) \cup V(s).$$

Title: Definition of ring of regular functions